CONTENT
INTRODUCTION TO CHEMISTRY
0003 - 0013
Unit 2
CHEMICAL BONDING
0014 – 0156
Unit 3
SOLUTIONS & COLLIGATIVE PROPERTIES
0157 - 0240
Unit 4
REDUCTION, OXIDATION & HYDROLYSIS REACTIONS
Unit 5
CHEMICAL EQUILIBRIUM
Unit 6
IONIC EQUILIBRIUM (ELEMENTARY)
Unit 7
SURFACE CHEMISTRY
Unit 8
EQUIVALENT CONCEPTS & TITRATIONS
Unit 9
IONIC EQUILIBRIUM (ADVANCED)
0522 – 0613
Unit 10
METALLURGY
0614 – 0679
Unit 11
MOLE CONCEPTS
Unit 12
PERIODIC TABLE & PERIODICITY
0754 – 0805
Unit 13
SOLID STATE
0806 – 0870
Unit 14
ATOMIC STRUCTURE & NUCLEAR CHEMISTRY
0871 - 0974
Unit 15
THERMODYNAMICS & THERMOCHEMISTRY
0975 – 1105
Unit 16
ELECTROCHEMISTRY
1106 – 1203
Unit 17
CHEMICAL KINETICS & RADIOACTIVITY
1204 - 1298
Unit 18
ALL BASIC CONCEPTS OF ORGANIC CHEMISTRY
1299 - 1315
Unit 19
ABC-2 (Phenol & Aniline)
1316 - 1326
Unit 20
ABC-3 (Alkyl halide, Alcohol & Ether)
1327 - 1337
Unit 21
ABC-4 (Carboxylic acid & Carbonyl compounds)
1338 - 1353
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Unit 1
0241 - 0302
a
0303 – 0376
Sa
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a
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Ja
uh
0377 - 0416
0417 – 0479
0480 – 0521
0680 - 0753
Unit 22
Unit 23
Unit 24
CARBONYL COMPOUNDS(ALDEHYDES & KETONES)
1354 – 1461
AROMATIC COMPOUNDS(PHENOL, ANILINE & DIAZONIUM
COMPOUNDS)
AROMATIC COMPOUNDS(PHENOL, ANILINE & DIAZONIUM
COMPOUNDS)
1462 – 1532
1533 – 1581
GENERAL ORGANIC CHEMISTRY-II
1582 – 1652
Unit 26
IDEAL GASES
Unit 27
IONIC EQUILIBRIUM (ADVANCED)
Unit 28
ORGANIC REACTION MECHANISMS: I - IV
Unit 29
REAL GASES
Unit 30
STRUCTURAL IDENTIFICATION & POC
Unit 31
CHEMISTRY IN EVERYDAY LIFE & POC
2281 – 2335
Unit 32
BASIC INROGANIC NOMENCLATURE
2336 – 2352
Unit 33
HYDROGEN COMPOUNDS
2353 – 2378
Unit 34
d & f-BLOCK ELEMENTS & THEIR COMPOUNDS
2379 – 2445
Unit 35
p-BLOCK ELEMENTS(B & C FAMILY)
2446 – 2514
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Unit 25
1653 – 1723
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a
1724 – 1917
1816 – 2027
2028 – 2133
2134 – 2280
Unit 36
p-BLOCK ELEMENTS(HALOGEN & NOBLE GASES)
2513 – 2566
Unit 37
p-BLOCK ELEMENTS(N & O FAMILY)
2567 – 2636
Unit 38
s-BLOCK ELEMENTS
2637 – 2694
Unit 39
QUANTUM NUMBER & ELECTRONIC CONFIGURATION
2695 – 2717
Unit 40
QUALITATIVE ANALYSIS(ANION & CATIONS)
2718 – 2826
Unit 41
ENVIRONMENTAL CHEMISTRY
2827 – 2849
Unit 42
COORDINATION COMPOUNDS
2850 – 2935
INTRODUCTION TO CHEMISTRY
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Atomic hypothesis:
Keeping in view various laws of chemical combination, a theoretical proof for the validity of different
laws was given by John Dalton in the form of hypothesis called Dalton's atomic hypothesis. Postulates
of Dalton's hypothesis are as follows:
(i) Each element is composed of extremely small particles called atoms which can take part in chemical
combination.
(ii) All atoms of a given element are identical i.e., atoms of a particular element are all alike but differ
from atoms of other elements.
(iii) Atoms of different elements possess different properties (including different masses).
(iv) Atoms are indestructible i.e., atoms are neither created nor destroyed in chemical reactions.
(v) Atoms of elements combine to form molecules and compounds are formed when atoms of more
than one element combine.
(vi) In a given compound, the relative number and kind of atoms is constant.
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Modern atomic hypothesis: The main modifications made in Dalton's hypothesis as a result of new
discoveries about atoms are :
(i) Atom is no longer considered to be indivisible.
(ii) Atoms of the same element may have different atomic weights. E.g. isotopes of oxygen O16, O17 and
O18.
(iii) Atoms of different element may have same atomic weights. E.g. isobars Ca40 and Ar40.
(iv) Atom is no longer indestructible. In many nuclear reactions, a certain mass of the nucleus is
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converted into energy along with ,  and  rays.
(v) Atoms may not always combine in simple whole number ratios. E.g. in sucrose (C12H22O11), the
elements carbon, hydrogen and oxygen are present in the ratio of 12 : 22 : 11 and the ratio is not a
simple whole number ratio.
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Atomic & Molecular masses:
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Atomic mass: It is the average relative mass of atom of element as compared with
1
times the mass
12
of an atom of carbon-12 isotope.
Atomic mass =
Average mass of an atom
1/12  Mass of an atom of C12
Average atomic mass: If an element exists in two isotopes having atomic masses 'a' and 'b' in the
ratio m : n, then average atomic mass =
(m  a) + (n  b)
. Atomic mass is expressed in amu. 1 amu =
m+n
1.66 × 10–24 g. One atomic mass unit (amu) is equal to
1
th of the mass of an atom of carbon-12
12
isotope.
Gram atomic mass (GAM): Atomic mass of an element expressed in grams is called Gram atomic
mass or gram atom or mole atom.
(i) Number of gram atoms =
Mass of an element
GAM
(ii) Mass of an element in g = No. of gram atoms × GAM
3
(iii) Number of atoms in 1 GAM = 6.02 × 1023
Number of atoms in a given substance = No. of gram atoms × 6.02 × 1023 =
(iv) Number of atoms in 1 g of element =
Mass
× 6.02 × 1023
GAM
6.02  1023
GAM
(v) Mass of one atom of the element (in g) =
GAM
6.02  1023
Molecular mass: Molecular mass of a molecule, of an element or a compound may be defined as a
number which indicates how many times heavier is a molecule of that element or compound as
1
of the mass of an atom of carbon-12. Molecular mass is also expressed in amu.
12
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compared with
Mass of one molecule of the substance
1/12  Mass of one atom of C-12
Actual mass of one molecule = Mol. mass (in amu) × 1.66 × 10 –24 g
Molecular mass of a substance is the additive property and can be calculated by adding the atomic
masses of atoms present in one molecule.
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Molecular mass =
Gram molecular mass (GMM): Molecular mass of an element or compound when expressed in g is
called its gram molecular mass, gram molecule or mole molecule.
Average atomic mass and molecular mass
 A i Xi
 Xtotal
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A (Average atomic mass) =
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Mass of substance
GMM
Mass of substance in g = No. of gram molecules × GMM
Number of gram molecules =
(Average molecular mass) =
 Mi Xi
 Xtotal
Sa
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a
Where A1, A2, A3 …… are atomic mass of species 1, 2, 3,…. etc. with % as X1, X2, X3 …… etc. Similar
terms are for molecular masses.
The Mole Concept
One mole of any substance contains a fixed number (6.022  1023 ) of any type of particles (atoms or
molecules or ions) and has a mass equal to the atomic or molecular weight, in grams. Thus it is correct
to refer to a mole of helium, a mole of electrons or a mole of any ion, meaning respectively Avogadro's
number of atoms, electrons or ions.
Number of moles =
Weight (grams)
Weight
=
Weight of one mole (g/mole)
GAM or GMM
Note : 1 mole = 1 g-atom = 1 g-molecule = 1 g-ion.
Properties of Gases
The state of matter in which the molecular forces of attraction between the particles of matter are
minimum, is known as gaseous state. It is the simplest state and shows great uniformity in behaviour.
Characteristics of gases
(1) Gases or their mixtures are homogeneous in composition.
(2) Gases have very low density due to negligible intermolecular forces.
(3) Gases have infinite expansibility and high compressibility.
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(4) Gases exert pressure.
(5) Gases possess high diffusibility.
(6) Gases do not have definite shape and volume like liquids.
(7) Gaseous molecules move very rapidly in all directions in a random manner i.e., gases have highest
kinetic energy.
(8) Gaseous molecules collide with one another and also with the walls of container.
(9) Gases can be liquefied, if subjected to low temperatures & high pressures.
(10) Thermal energy of gases >> molecular attraction.
(11) Gases undergo similar change with the change of temperature and pressure. In other words,
gases obey certain laws known as gas laws.
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Measurable properties of gases
The characteristics of gases are described fully in terms of four parameters or measurable properties:
(i) The volume, V, of the gas.
(ii) Its pressure, P
(iii) Its temperature, T
(iv) The amount of the gas (i.e., mass or number of moles).
(1) Volume : (i) Since gases occupy the entire space available to them, the measurement of volume of
a gas only requires a measurement of the container confining the gas.
(ii) Volume is expressed in litres (L), millilitres (mL) or cubic centimetres (cm3), cubic metres (m3).
(iii) 1 L = 1000 mL; 1 mL = 10–3 L; 1 L = 1 dm3 = 10–3 m3
1 m3 = 103 dm3 = 106 cm3 = 106 mL = 103 L
(2) Mass : (i) The mass of a gas can be determined by weighing the container in which the gas is
enclosed and again weighing the container after removing the gas. The difference between the two
weights gives the mass of the gas.
(ii) The mass of the gas is related to the number of moles of the gas i.e.
Mass in grams m
=
Molar mass
M
(3) Temperature : (i) Gases expand on increasing the temperature. If temperature is increased twice,
the square of the velocity (v2) also increases two times.
(ii) Temperature is measured in centigrade degree (ºC) or celsius degree with the help of
thermometers. Temperature is also measured in Fahrenheit (ºF).
(iii) S.I. unit of temperature is kelvin (K) or absolute degree.
K = ºC + 273
a
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moles of gas (n) =
Sa
nk
C o F − 32
=
5
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(4) Pressure : (i) Pressure of the gas is the force exerted by the gas per unit area of the walls of the
o
(iv) Relation between ºF and ºC is
container in all directions. Thus, Pressure (P) =
Force(F) Mass(m)  Acceleration(a)
=
Area(A)
Area(A)
1
mv 2 ) of the molecules. Kinetic energy of
2
the gas molecules increases, as the temperature is increased.
(iii) Pressure of a gas is measured by manometer or barometer.
(iv) Commonly two types of manometers are used:
(a) Open end manometer; (b) Closed end manometer
(v) The S.I. unit of pressure, the pascal (Pa), is defined as 1 newton per metre square. It is very small
unit.
1Pa = 1 Nm–2 = 1 kgm–1s–2
(vi) C.G.S. unit of pressure is dynes cm–2.
(vii) M.K.S. unit of pressure is Newton m–2. The unit Newton m–2 is sometimes called pascal (Pa).
(ii) Pressure exerted by a gas is due to kinetic energy (KE =
5
(viii) Higher unit of pressure is bar, kPa or MPa.
1 bar = 105 Pa = 105 Nm–2 = 100 KNm–2 = 100 KPa
(ix) Several other units used for pressure are,
Name
Symbol
Value
bar
bar
1 bar = 105 Pa
atmosphere
atm
1 atm = 1.01325 × 105 Pa
101325
Pa = 133.322 Pa
760
Torr
1 Torr =
millimetre of mercury
mm Hg
1 mm Hg = 133.322 Pa
;
;
;
V : Volume of gas
;
R : Universal gas constant.
8.314 JK–1mol–1
;
n = Number of moles of gas
1.987 CalK–1mol–1
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Ideal Gas Equation
PV = nRT
where, P : Pressure of gas
T : Temperature of gas
Values of R : 0.082 LatmK–1mol–1
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Torr
Prefixes used in the SI System
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a
10–2
10–1
10
102
103
106
109
1012
1015
1018
1021
1024
Symbol
y
z
a
f
p
n
μ
m
Ja
Prefix
yocto
zepto
atto
femto
pico
nano
micro
milli
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Multiple
10–24
10–21
10–18
10–15
10–12
10–9
10–6
10–3
centi
deci
deca
hecto
kilo
mega
giga
tera
peta
exa
zeta
yotta
c
d
da
h
k
M
G
T
P
E
Z
Y
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 Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
How much time (in years) would it take to distribute one Avogadro number of wheat grains if 10 10 grains
are distributed each second ?
2.
The weight of one atom of Uranium is 238 amu. Its actual weight is ..... g.
3.
Calculate the weight of 12.044 × 1023 atoms of carbon.
4.
How many grams of silicon is present in 35 gram atoms of silicon (Given at. wt. of Si = 28).
5.
Find the total number of nucleons present in 12 g of 12C atoms.
6.
Find (i) the total number of neutrons, and (ii) the total mass of neutrons in 7 mg of
mass of a neutron = mass of a hydrogen atom)
7.
Calculate the number of electrons, protons and neutrons in 1 mole of
8.
How many atoms are there in 100 amu of He?
9.
The density of liquid mercury is 13.6 g/cm3. How many moles of mercury are there in 1 litre of the
metal? (Atomic mass of Hg = 200.)
10.
Calculate the atomic mass (average) of chlorine using the following data:
% Natural Abundance
Molar Mass
35Cl
75
35.0 g
37Cl
25
37.0 g
11.
Average atomic mass of Magnesium is 24.31 amu. This magnesium is composed of 79 mole % of
and remaining 21 mole % of 25Mg and 26Mg. Calculate mole % of 26Mg.
12.
The number of molecules in 16 g of methane is :
13.
Calculate the number of molecules in a drop of water weighing 0.09 g.
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1.
(Assume that the
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14C.
ions.
24Mg
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16O–2
14.
A sample of ethane has the same mass as 10.0 million molecules of methane. How many C 2H6
molecules does the sample contain ?
15.
The number of neutrons in 5 g of D2O (D is 12H ) are :
16.
Calculate the weight of 6.022 × 1023 formula units of CaCO3.
17.
From 200 mg of CO2, 1021 molecules are removed. How many moles of CO2 are left ?
18.
Find the total number of H, S and 'O' atoms in the following :
(a) 196 g H2SO4
(b) 196 amu H2SO4
(c) 5 mole H2S2O8
(d) 3 molecules H2S2O6 .
19.
If from 10 moles NH3 and 5 moles of H2SO4, all the H-atoms are removed in order to form H2 gas, then
find the number of H2 molecules formed.
20.
If from 3 moles MgSO4.7H2O, all the 'O' atoms are taken out and converted into ozone find the number
of O3 molecules formed.
21.
If the components of air are N2 - 78%; O2 - 21%; Ar - 0.9% and CO2 - 0.1% by volume (or mole), what
would be the molecular weight of air ?
7
Find the expression of Universal Gas Constant R in SI system in terms of the given properties of
oxygen gas.
Pressure = p (kPa)
Volume = V (mL)
Temperature = t (ºC)
Mass of oxygen = w (g)
23.
The volume of a gas at 0ºC and 700 mm pressure is 760 cc. The number of molecules present in this
volume is :
24.
The weight of 350 mL of a diatomic gas at 0ºC and 2 atm pressure is 1 g. The weight of one atom is :
25.
Oxygen is present in a 1-litre flask at a pressure of 7.6 × 10–10 mm of Hg at 0ºC. Calculate the number
of oxygen molecules in the flask.
26.
Fill in the blanks :
(i) 1m = ...... nm
(iv) 1dm = ...... mm
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22.
(iii) 100 Pa = ...... kPa
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(ii) 10 MJ = ...... J
(v) 10 pm = ...... cm
PART - II : OBJECTIVE QUESTIONS
Single Correct Questions (SCQ)
Which is not a basic postulate of Dalton’s atomic theory ?
(A) Atoms are neither created nor destroyed in a chemical reaction.
(B) Different elements have different types of atoms.
(C) Atoms of an element may be different due to presence of isotopes.
(D) Each element is composed of extermely small particles called atoms.
2.
The modern atomic weight scale is based on :
(A) 12C
(B) 16O
5.
(D) 18O
(C) 1 g of H2
(D) 1.66 × 10–23 kg
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(B)
1
of O–16
14
If the atomic mass of sodium is 23, the number of moles in 46 g of sodium is :
(A) 1
(B) 2
(C) 2.3
(D) 4.6
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4.
1 amu is equal to
1
(A)
of C–12
12
(C) 1H
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3.
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1.
How many grams are contained in 1 gram-atom of Na ?
(A) 13 g
(B) 23 g
(C) 1 g
(D)
1
g
23
6.
1.0 g of hydrogen contains 6 × 1023 atoms. The atomic weight of helium is 4. It follows that the number
of atoms in 1 g of He is :
1
(A)  6  1023
(B) 4 × 6 × 1023
(C) 6 × 1023
(D) 12 × 1023
4
7.
The atomic weights of two elements A and B are 40u and 80u respectively. If x g of A contains y atoms,
how many atoms are present in 2x g of B?
y
y
(A)
(B)
(C) y
(D) 2y
2
4
8.
A sample of aluminium has a mass of 54.0 g. What is the mass of the same number of magnesium
atoms? (At. wt. Al = 27, Mg = 24)
(A) 12 g
(B) 24 g
(C) 48 g
(D) 96 g.
8
9.
The number of atoms in 558.5 g of Fe (at wt.= 55.85) is :
(A) Twice that in 60 g carbon
(B) 6.022 × 1022
(C) Half in 8 g He
(D) 558.5 × 6.023 × 1023
10.
Which of the following has the Maximum mass ?
1
mole of CH4
2
(D) 3.011 × 1023 atoms of oxygen
(A) 1 g-atom of C
(B)
(C) 10 mL of water
11.
The total number of protons, electrons and neutrons in 12 g of 12
6 C is :
(B) 6.022 × 1023
(C) 6.022×1022
(D) 18
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(A) 1.084 × 1025
1 mole of element X has mass, 3/10 times the mass of 1 mole of element Y. One average atom of
element X has mass, 2 times the mass of one atom of 12C. What is the atomic weight of Y ?
(A) 80
(B) 15.77
(C) 46.67
(D) 40.0
13.
The charge on 1 gram ions of Al3+ is : (NA = Avogadro number, e = charge on one electron)
1
1
1
(A)
NAe coulomb
(B) × NAe coulomb
(C) × NAe coulomb
(D) 3 × NAe coulomb
3
9
27
14.
It is known that an atom contains protons, neutrons and electrons. If the mass of neutron is assumed to
half of its original value whereas that of proton is assumed to be twice of its original value, then the
(A) same
14
6C
will be :
(B) 114.28 % less
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atomic mass of
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12.
(C) 14.28 % more
(D) 28.56 % less
The isotopic abundance of C–12 and C–14 is 98% and 2% by mass respectively. What would be the
number of C–14 isotope in 12 g carbon sample ?
(A) 1.032×1022
(B) 3.01×1023
(C) 5.88×1023
(D) 6.02×1023
16.
In chemical scale, the relative mass of the isotopic mixture of X atoms (X 20, X21, X22) is approximately
equal to : (X20 has 99 percent abundance)
(A) 20.002
(B) 21.00
(C) 22.00
(D) 20.00
17.
Indium (atomic weight = 114.8) has two naturally occurring isotopes, the predominant one form has
isotopic weight 115 and abundance of 95.00%. Which of the following isotopic weights is the most likely
for the other isotope ?
(A) 111
(B) 112
(C) 113
(D) 114
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a
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15.
18.
The number of molecules of CO2 present in 44 g of CO2 is :
(A) 6.0  1023
19.
(B) 3  1023
(C) 12  1023
The number of mole of ammonia in 4.25 g of ammonia is :
(A) 0.425
(B) 0.25
(C) 0.236
(D) 3  1010
(D) 0.2125
20.
Which one of the following pairs of gases contains the same number of molecules :
(A) 16 g of O2 and 14 g of N2
(B) 8 g of O2 and 22 g of CO2
(C) 28 g of N2 and 22 g of CO2
(D) 32 g of O2 and 32 g of N2
21.
The weight of a molecule of the compound C60H22 is :
(A) 1.09 × 10–21 g
(B) 1.24 × 10–21 g
(C) 5.025 × 10–23 g
(D) 16.023 × 10–23 g
Number of electrons in 1.8 mL of H2O() is about :
(A) 6.02 × 1023
(B) 3.011 × 1023
(C) 0.6022 × 1021
(D) 60.22 × 1020
22.
9
23.
One mole of P4 molecules contain :
(A) 1 molecule
1
(C)
× 6.022 × 1023 atoms
4
(B) 4 molecules
(D) 24.088 × 1023 atoms
24.
A sample of ammonium phosphate (NH4)3PO4 contains 3.18 mole of H atoms. The number of mole of O
atoms in the sample is :
(A) 0.265
(B) 0.795
(C) 1.06
(D) 3.18
25.
Torr is unit of :
(A) Temperature
(B) Pressure
(C) Volume
(D) Density
The atmospheric pressure on Mars is 0.61 kPa. What is the pressure in mm Hg ?
(A) 0.63
(B) 4.6
(C) 6.3
(D) 3.2
27.
Centigrade and Fahrenheit scales are related as :
C F − 32
C F − 32
C F − 32
=
=
=
(A)
(B)
(C)
5
9
8
9
5
5
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26.
(D) None of these
At what temperature, both Celsius and Fahrenheit scale read the same value :
(A) 100º
(B) 130º
(C) 60º
(D) –40º
29.
The value of universal gas constant R depends on :
(A) temperature of gas
(B) volume of gas
(C) number of moles of gas
(D) units of volume and pressure
30.
The value of gas constant in calorie per degree temperature per mol is approximately :
(A) 1 cal
(B) 2 cal
(C) 3 cal
(D) 4 cal
31.
The value of R in SI unit is :
(A) 8.314 × 10–7 erg K–1 mol–1
(C) 0.082 litre atm K–1 mol–1
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28.
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(B) 8.314 JK–1 mol–1
(D) 2 cal K–1 mol–1
The pressure of sodium vapour in a 1.0 L container is 9.5 torr at 927ºC. How many atoms are in the
container ?
(A) 9.7 × 107
(B) 7.5 × 1019
(C) 4.2 × 1017
(D) 9.7 × 1019
33.
The pressure of a gas having 2 mole in 44.8 litre vessel at 546 K is :
(A) 1 atm
(B) 2 atm
(C) 3 atm
(D) 4 atm
According to the ideal gas laws, the molar volume of a gas is given by :
(A) 22.4 litre
(B) RT / P
(C) 8RT / PV
(D) RT / PV
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a
32.
34.
35.
Equal volumes of oxygen gas and a second gas weigh 1.00 and 19/8 grams respectively under the
same experimental conditions. Which of the following is the unknown gas?
(A) NO
(B) SO2
(C) CS2
(D) CO
36.
A high altitude balloon contains 6.0 g of helium in 10 4 L at 240 K. Assuming ideal gas behaviour, how
many grams of helium would have to be added to increase the pressure to 4.0 × 10–3 atm ?
(A) 1
(B) 1.2
(C) 1.5
(D) 2.0
37.
Four 1-1 litre flasks are separately filled with the gases H 2, He, O2 and O3 at the same temperature and
pressure. The ratio of total number of atoms of these gases present in different flask would be :
(A) 1 : 1 : 1 : 1
(B) 1 : 2 : 2 : 3
(C) 2 : 1 : 2 : 3
(D) 3 : 2 : 2 : 1
Under the same conditions, two gases have the same number of molecules. They must
(A) be noble gases
(B) have equal volumes
3
(C) have a volume of 22.4 dm each
(D) have an equal number of atoms
38.
10
39.
16 g of an ideal gas SOx occupies 5.6 L. at STP. The value of x is
(A) x = 3
(B) x = 2
(C) x = 4
(D) none of these
The ratio of the weight of one litre of a gas to the weight of 1.0 L oxygen gas both measured at S.T.P. is
2.22. The molecular weight of the gas would be :
(A) 14.002
(B) 35.52
(C) 71.04
(D) 55.56
41.
Avogadro number is :
(A) Number of atoms in one gram of the element
(B) Number of mililitre which one mole of a gaseous substance occupies at NTP (1 atm & 0ºC)
(C) Number of molecules present in one gram molecular mass of a substance.
(D) All are correct
42.
The weight of 1  1022 molecules of CuSO4.5H2O is :
(A) 41.59 g
(B) 415.9 g
(C) 4.159 g
45.
How many moles of electron weigh one kilogram :
1
6.023
(A) 6.023 × 1023
(B)
× 1031
(C)
× 1054
9.108
9.108
(D)
1
× 108
9.108  6.023
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44.
(D) None of these
Number of atoms in 560 g of Fe (atomic mass 56 gmol–1) is :
(A) Twice that in 70 g N (B) Half that in 20 g H (C) Both (A) and (B)
(D) None of these
Which has maximum number of atoms :
(A) 24 g of C (12)
(B) 56 g of Fe (56)
(D) 108 g Ag (108)
(C) 27 g of Al (27)
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43.
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40.
If we consider that 1/6, in place of 1/12 mass of carbon atom is taken to be the relative atomic mass
unit, the mass of one mole of a substance will :
(A) decrease twice
(B) increase two fold
(C) remain unchanged
(D) be a function of the molecular mass of the substance
47.
How many moles of magnesium phosphate, Mg 3(PO4)2 will contain 0.25 mole of oxygen atoms ?
(A) 0.02
(B) 3.125 × 10–2
(C) 1.25 × 10–2
(D) 2.5 × 10–2
48.
Given that the abundances of isotopes
atomic mass of Fe is :
(A) 55.85
(B) 55.95
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46.
54Fe, 56Fe
and
57Fe
are 5%, 90% and 5% respectively, the
(C) 55.75
(D) 56.05
Multiple Correct Questions (MCQ)
49.
50.
51.
52.
53.
Which property of an element may have non-integral value.
(A) Atomic weight
(B) Atomic number
(C) Atomic volume
(D) None of these
Which of the following would contain 1 mole of particles :
(A) 0.5 mole of H2
(B) 1 g of H-atoms
(C) 16 g of O-18
(D) 16 g of methane
Which of the following will have the same number of electrons :
(A) 1 g Hydrogen
(B) 2 g Oxygen
(C) 2 g Carbon
(D) 2 g Nitrogen
Which the following is equal to 10–2 atm :
(A) 0.76 cm of Hg
(B) 7.6 torr
(D) 0.0076 torr
(C) 0.076 dm of Hg
Pressure exerted by a sample of oxygen is same for the following conditions :
(A) 2 L, 27ºC
(B) 1 L, 150 K
(C) 4 L, 54ºC
(D) 10 L, 1227ºC
11
Assertion / Reasoning (A/R)
Each question has 5 choices (A), (B), (C), (D) and (E) out of which ONLY ONE is correct.
(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true and statement-2 is not correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
(E) Both statements are false.
Statement-1 : Gram molecular weight of O2 is 32 g.
Statement-2 : Relative atomic weight of oxygen is 32.
55.
Statement-1 : 1 mole of all ideal gases exert same pressure in same volume at same temperature.
Statement-2 : Behaviour of ideal gases is independent of their nature.
56.
Statement-1 : Value of the universal gas constant depends upon the choice of sytem of units.
Statement-2 : Values of universal gas constant are 8.314 J/molK, 0.0821 L.atm/molK, 2 cal/molK.
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54.
59.
(C) Both I and II
Find the average molar mass of the mixture of gases X, Y and Z.
(A) 40/7
(B) 50/7
(C) 20
Match the column:
Column-I
(Atomic mass (M))
Isotope-I Isotope-II Average
(z – 1)
(z + 3)
z
(z + 1)
(z + 3)
(z + 2)
z
3z
2z
(z – 1)
(z + 1)
z
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60.
Identify the correct statement(s) :
I. Gas Y is lighter than gas X.
II. Gas Z is lighter than gas Y
(A) I only
(B) II only
(C) 30 g
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58.
Find the molar mass of gas X.
(A) 20 g
(B) 10 g
a
57.
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Comprehension #
A vessel of 25 L contains 20 g of ideal gas X at 300K. The pressure exerted by the gas is 1 atm. 20 g of
ideal gas Y is added to the vessel keeping the same temperature. Total pressure became 3 atm. Upon
further addition of 20 g ideal gas Z the pressure became 7 atm. Answer the following questions. (Hint:
Ideal gas equation is applicable on mixture of ideal gases) [Take, R = 1/12 L.atm / mol K]
(A)
(B)
(C)
(D)
(D) 5 g
(D) None of the statements
(D) 60/7
Column-II
(% composition of heavier isotope)
(p)
(q)
(r)
(s)
25% by moles
50% by moles
% by mass dependent on z
75% by mass
12
PART – I
1.9  106 years (approx.)
2.
3.95 × 10–22
3.
24 g
4.
980 g of Si
5.
12 × 6.022 × 1023
6.
24.088 × 1020, 0.004 g.
7.
10 × 6.022 × 1023, 8 × 6.022 × 1023, 8 × 6.022 × 1023.
8.
25
9.
68 mole
10.
35.5
11.
10
12.
6.02 × 1023
13.
3.01 × 1021 molecules of H2O
14.
5.33 × 106
15.
2.5 NA
17.
0.00288
18.
(a) H = 4NA, S = 2NA, O = 8NA atoms
(c) H = 10NA, S = 10NA, O = 40 NA atoms
19.
20 NA
22.
R=
25.
2.647 × 1010
26.
(i) 1000
(ii) 107
100 g
20.
11 NA
23.
1.88 × 1022
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(b) H = 4 atoms, S = 2 atoms, O = 8 atoms.
(d) H = 6 atoms, S = 6 atoms, O = 18 atoms.
21.
28.964 u
24.
16 amu
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32pV
1000  w  (t + 273)
16.
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1.
(iii) 0.1
(iv) 100
(v) 10–9
(C)
2.
(A)
3.
(A)
4.
(B)
5.
(B)
6.
(A)
7.
(C)
8.
(C)
9.
(A)
10.
(A)
11.
(A)
12.
(A)
13.
(D)
14.
(C)
15.
(A)
16.
(A)
17.
(A)
18.
(A)
19.
(B)
20.
(A)
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a
1.
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PART – II
21.
(B)
22.
(A)
23.
(D)
24.
(C)
25.
(B)
26.
(B)
27.
(A)
28.
(D)
29.
(D)
30.
(B)
31.
(B)
32.
(B)
33.
(B)
34.
(B)
35.
(C)
36.
(D)
37.
(C)
38.
(B)
39.
(B)
40.
(C)
41.
(C)
42.
(C)
43.
(D)
44.
(C)
45.
(A)
46.
(C)
47.
(B)
48.
(B)
49.
(AC)
50.
(BD)
51.
(ABCD)
52.
(ABC)
53.
(ABD)
54.
(C)
55.
(A)
56.
(B)
57.
(A)
58.
(C)
59.
(D)
60.
(A) - (p,r) ; (B) - (q,r) ; (C) - (q,s) ; (D) - (q,r)
13
CHEMICAL BONDING
JEE(Advanced) Syllabus
Orbital overlap and covalent bond; Hybridisation involving s, p and d orbitals only; Orbital energy
diagrams for homonuclear diatomic species; Hydrogen bond; Polarity in molecules, dipole moment
(qualitative aspects only); VSEPR model and shapes of molecules (linear, angular, triangular, square
planar, pyramidal, square pyramidal, trigonal bipyramidal, tetrahedral and octahedral).
JEE(Main) Syllabus
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Chemical Bonding and Molecular Structure
Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of
covalent bond, valence bond theory, resonance, geometry of molecules, VSEPR theory, concept of
hybridization involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory
of homonuclear diatomic molecules (qualitative idea only). Hydrogen bond.
Chemical Bond
: The attractive force which holds various constituents such as atoms, ions etc.,
together in different chemical species is called a chemical bond.
Covalent Bond :
Ja
A chemical bond formed by sharing of electrons between two elements is called as covalent bond.
A⎯A (Single bond) : When 2 electrons are shared between the two combining elements.
(Double bond) : When 4 electrons are shared between the two combining elements.
(Triple bond) : When 6 electrons are shared between the two combining elements.
To explain nature of chemical bond, different theories are given
(ii) Valence bond theory
(iv) Molecular orbital theory.
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(i) Octet rule
(iii) Valence shell electron pair repulsion theory
Section (A) : Octet rule, Lewis dot structures
Octet rule : “Tendency of atoms to have eight electrons in their outermost shell is known as Lewis
Sa
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(ii)
(iii)
octet rule". To achieve inert gas configuration atoms lose, gain or share electrons.
It has been observed that atoms of noble gases have little or no tendency to combine with each other or
with atoms of other elements.
It means that these atoms must have a stable electronic configuration.
These elements (noble gases) have 8 electrons (ns 2 np6) except helium which has 2 electrons (1s2) in
their outer most shell.
Element
Ne
Ar
Kr
Xe
Rn
Outer most shell
configuration
2s22p6
3s23p6
4s24p6
5s25p6
6s26p6
2
6
It is therefore concluded that ns np configuration in the outer energy level constitues a structure of
maximum stability or minimum energy.
The Octet rule can be understood by considering the formation of the chlorine molecule, Cl 2. The Cl
atom with electronic configuration, [Ne]10 3s2 3p5, is one electron short of the argon configuration. The
formation of the Cl2 molecule can be understood in terms of the sharing of a pair of electrons between
the two chlorine atoms, each chlorine atom contributing one electron to the shared pair. In the process
both
a
(i)
(iv)
Cl
+
Cl
Cl
8e–
or Cl – Cl
Cl
8e–
Covalent bond between two Cl atoms
Figure
14
chlorine atoms attain the outer shell octet of the nearest noble gas (i.e., argon). The dots represent
electrons. Such structures are referred to as Lewis dot structures.
Lewis Dot Structures :
The Lewis dot structures can be written for other molecules also, in which the combining atoms may be
identical or different.
(2)
(3)
(4)
To decide, Central atom, following steps are followed :
In general the least electronegative atom occupies the central position in the molecule/ion. For
example in the NF3 and CO32–, nitrogen and carbon are the central atoms whereas fluorine and oxygen
occupy the terminal positions.
Generally the atom which is/are less in number acts as central atom.
Generally central atom is the atom which can form maximum number of bonds( which is generally equal
to the number of electrons present in the valence shell of the atom).
Atom of highest atomic number or largest atom generally acts as central atom.
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(1)
To write the Lewis dot structure following steps are to be followed :
The total number of electrons are obtained by adding the valence electrons of combining atoms.
For Anions, we need to add one electron for each negative chanrge.
For cations, we need to subtract one electron for each positive charge.
After then the central atom is decided.
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(i)
(ii)
(iii)
(iv)
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Hence fluorine and hydrogen can never act as central atoms.
After accounting for the shared pairs of electrons for single bonds, the remaining electron pairs are
either utilized for multiple bonding or remain as the lone pairs. The basic requirement being that each
bonded atom gets an octet of electrons.
Lewis representations of a few molecules/ions are given in the following Table
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Table
a
Each H-atom attains the configuration of helium (a duplet of electrons)
Section (B) : Formal charge & limitations of octet rule
Formal Charge :
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The formal charge of an atom in a molecule or ion is defined as the difference
between the number of valence electrons of that atom in an isolated or free state and the number of
electrons assigned to that atom in the lewis structure.
Let us consider the ozone molecule (O3). The Lewis structure of O3, may be drawn as :
The atoms have been numbered as 1, 2 and 3. The formal charge on :
1
The central O atom marked 1 = 6 – 2 –
(6) = + 1
2
1
The terminal O atom marked 2 = 6 – 4 –
(4) = 0
2
15
1
(2) = – 1
2
Hence, we represent O3 along with the formal charges as follows :
The terminal O atom marked 3 = 6 – 6 –
It should be kept in mind that formal charges do not indicate real charge separation within the molecule.
Indicating the charges on the atoms in the Lewis structure only helps in keeping track of the valence
electrons in the molecule. Formal charges help in the selection of the lowest energy structure
from a number of possible Lewis structures for a given species. Generally the lowest energy
structure is the one with the smallest formal charges on the atoms.
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Note : (i) The formal charge Is a factor based on a pure covalent view of bonding in which electron pairs are
shared equally by neighbouring atoms.
(ii) Pair of +1 and –1 formal charge on adjacent atoms is considered a coordinate bond.
– +
:C O: :C O:
(iii) Lewis dot structure with minimum formal charges is most stable.
Limitations of the Octet Rule :
uh
The octet rule, though useful, is not universal. It is quite useful for understanding the structures of most
of the organic compounds and it applies mainly to the second period elements of the periodic table.
There are three types of exceptions to the octet rule.
The incomplete octet of the central atom : In some compounds, the number of electrons surrounding
the central atom Is less than eight. This is especially the case with elements having less than four
valence electrons. Examples are LiCl, BeH2 and BCl3.
BeF2, BF3, AlCl3
2.
Odd-electron molecules : In molecules with an odd number of electrons like nitric oxide, NO and
nitrogen dioxide (NO2), the octet rule is not satisfied for all the atoms.
e.g.
NO, ClO2 , ClO3
3.
The expanded octet : Elements in and beyond the third period of the periodic table have, apart from 3s
and 3p orbitals, 3d orbitals also available for bonding. In a number of compounds of these elements
there are more than eight valence electrons around the central atom. This is termed as the expanded
octet. Obviously the octet rule does not apply in such cases.
Some of the examples of such compounds are: PF5, SF6, PCl5, HNO3, SO3, SO2, H2SO4 and a number
of coordination compounds.
Sa
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1.

4.
Interestingly, sulphur also forms many compounds in which the octet rule is obeyed. In sulphur
dichloride, the S atom has an octet of electrons around it.
Other drawbacks of the octet theory :
(i) It is clear that octet rule is based upon the chemical inertness of noble gases. However, some noble
gases (for example xenon and krypton) also combine with oxygen and fluorine to form a number of
compounds like XeF2, KrF2, XeOF2 etc.,
(ii) This theory does not account for the shape of molecules.
(iii) It does not explain the relative stability of the molecules being totally silent about the energy of a
molecule.
16
Section (C) : Resonance and bond order calculation
Resonance :
It is often observed that a single Lewis structure is inadequate for the representation of a molecule in
conformity with its experimentally determined parameters. For example, the ozone, O 3 molecule can be
equally represented by the structures I and II shown below :
lp
Ja
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Resonance in the O3 molecule
Structures I and II represent the two canonical forms.
The structure III is the resonance hybrid
In both structures we have a O–O single bond and a O=O double bond. The normal O–O and O=O
bond lengths are 148 pm and 121 pm respectively.
Experimentally determined oxygen-oxygen bond lengths in the O3 molecule are same (128 pm). Thus
the oxygen-oxygen bonds in the O3 molecule are intermediate between a double and a single bond.
Obviously, this cannot be represented by either of the two Lewis structures shown above.
According to the concept of resonance, whenever a single Lewis structure cannot describe a
molecule accurately, a number of structures with similar energy, positions of nuclei, bonding
and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which
describes the molecule accurately.
Thus for O3 the two structures shown above constitute the canonical structures or resonance structures
and their hybrid i.e., the III structure represents the structure of O 3, more accurately. This is also called
resonance hybrid. Resonance is represented by a double headed arrow.
Definition : Resonance may be defined as the phenomenon in which two or more structures involving
in identical position of atom, can be written for a particular compound
Resonance Hybrid : It is the actual structure of all different possible structures that can be written for
the molecule without violating the rules of covalence maxima for the atoms.

Resonance hybrid
a
Example :
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(i)
(ii)

(iii)

(iv)





Resonance energy = Actual bond energy – Energy of most stable resonating structure.
Stability of molecule  resonance energy.
More is the number of covalent bonds in molecule more will be its resonance energy.
Resonance energy  number of resonating structures.
17
Bond order in oxoanions and corresponding acids :
Let’s starts with example of HSO4–

Bond order =
Total No. of bonds formed between two atoms in all structures
Total No. of resonating structures
=
2+ 2+1
= 5/3
3
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Consider another example of SO4– – :
Bondorder=
2 + 2 + 1+ 2 + 1+ 1
Total No. of bonds formed between two atoms in all structures
=
=1.5
6
Total No. of resonating structures
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
Bond order =
Total No. of bonds formed between two atoms in all structures 1 + 2 + 1 + 1
=
= 5/4
4
Total No. of resonating structures
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
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Consider another example of PO4– – – :
Section (D) : VBT, overlapping of orbitals
Modern Concept of Covalent Bond (VBT) :
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a
(i) Introduced by Heitler and london.
(ii) This theory is bassed on the knowledge of Atomic orbitals electronic configuration of electrons, the
overlap criterion of atomic orbitals, the hybridization of atomic orbitals and the principles of variation and
superposition.
Consider two hydrogen atoms A and B approaching each other having nuclei H A and HB and electrons
present in them are represented by EA and EB. When the two atoms are at large distance from each
other, there is no interaction between them. As these two atoms approach each other, new attractive
and repulsive forces begin to operate.
Attractive forces arise between :
(i) nucleus of one atom and its own electron that is HA – EA and HB – EB.
(ii) nucleus of one atom and electron of other atom i.e., HA – EB. HB– EA.
Similarly repulsive forces arise between :
(i) electrons of two atoms like EA – EB , (ii) nuclei of two atoms HA – HB.
Attractive forces tend to bring the two atoms close to each other whereas repulsive forces tend to push
them apart (Fig.).
18
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
Orbital Overlap Concept
In the formation of H2 Molecule, In the minimum energy state when two H atoms are so near that their
atomic orbitals undergoes partial interpenetration. This partial merging is called overlapping.
Acc. to orbital overlap concept, The formation of a covalent bond between two atoms results by pairing
of electrons present in the valence shell having opposite spins.
The extent of overlap decides the strength of a covalent bond.
Greater the overlap, stronger is the bond formed between two atoms.
Sa
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
Figure : The potential energy curve
for the formation of H2 molecule as a
function of internuclear distance of the
H atoms. The minima in the curve
corresponds to the most stable state
of H2.
a
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Figure : Forces of attraction and repulsion during the formation of H 2 molecule.
Experimentally it has been found that the magnitude of new attractive force is more than the new
repulsive forces. As a result, two atoms approach each other and potential energy decreases.
Ultimately a stage is reached where the net force of attraction balances the force of repulsion and
system acquires minimum energy. At this stage two hydrogen atoms are said to be bonded together to
form a stable molecule having the bond length of 74 pm.
Since the energy gets released when the bond is formed between two hydrogen atoms, the hydrogen
molecule is more stable than that of isolated hydrogen atoms. The energy so released is called as bond
enthalpy, which is corresponding to minima in the curve depicted in Fig. Conversely. 435.8 kJ of energy
is required to dissociate one mole of H2 molecule.
H2(g) + 435.8 kJ mol–1 → H(g) + H(g)


Directional Properties of Bonds
The valence bond theory explains the formation and directional properties of bonds in polyatomic
molecules like CH4, NH3 and H2O, etc. in terms of overlap and hybridisation of atomic orbitals.
Overlapping of Atomic Orbitals
When two atoms come close to each other there is overlapping of atomic orbitals. This overlap may be
positive, negative or zero depending upon the properties of overlapping of atomic orbitals. The various
arrangements of s and p orbitals resulting in positive, negative and zero overlap are depicted in the
following figure.
The criterion of overlap, as the main factor for the formation of covalent bonds applies uniformly to the
homonuclear/heteronuclear diatomic molecules and polyatomic molecules.
19
Negative overlap
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Positive overlap
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Zero overlap
Sa
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Figure : Positive, negative and zero overlaps of s and p atomic orbitals
Types of Overlapping and Nature of Covalent Bonds
The covalent bond may be classified into two types depending upon the types of overlapping :
(i) sigma() bond, and (ii) pi () bond
(i)
Sigma () bond : This type of covalent bond is formed by the end to end (head-on) overlap of bonding
orbitals along the internuclear axis. This is called as head on overlap or axial overlap. This can be
formed by any one of the following types of combinations of atomic orbitals.
⚫ s-s overlapping : In this case, there is overlap of two half filled s-orbitals along the internuclear axis
as shown below :
⚫ s-p overlapping: This type of overlap occurs between half filled s-orbitals of one atom and half filled
p-orbitals of another atom.
20
+
s–orbital
p–orbital
s–p orbital
⚫ p-p overlapping : This type of overlap takes place between half filled p-orbitals of the two
approaching atoms.
+
p–orbital
p–p overlaping
pi() bond : In the formation of  bond the atomic orbitals overlap in such a way that their axes remain
parallel to each other and perpendicular to the internuclear axis. The orbitals formed due to sidewise
overlapping consists of two saucer type charged clouds above and below the plane of the participating
atoms.
+
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p–orbital
p–p overlaping
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p–orbital
Or
a
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(ii)
p–orbital
Sa
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a
Strength of Sigma and pi Bonds :
Basically the strength of a bond depends upon the extent of overlapping- In case of sigma bond, the
overlapping of orbitals takes place to a larger extent. Hence, it is stronger as compared to the pi bond
where the extent of overlapping occurs to a smaller extent. Further, it is important to note that pi bond
between two atoms is formed in addition to a sigma bond. It is always present in the molecules
containing multiple bond (double or triple bonds).
Ex-1.
What are the total number of  &  bonds in tetracyanomethane.
Sol.
From the structure it is clear that it has 8  and 8  bonds.
21
MISCELLANEOUS SOLVED PROBLEMS (MSPS)
Ex-1.
Sol.
Classify the following bonds as ionic, polar covalent or covalent and give your reasons :
(a) SiSi bond in Cl3SiSiCl3
(b) SiCl bond in Cl3SiSiCl3
(c) CaF bond in CaF2
(d) NH bond in NH3
(a) Covalent, due to identical electronegativity.
(b) Polar covalent, due to less electronegativity difference.
(c) Ionic, due to more electronegativity difference.
(d) Polar covalent, due to different electronegativity.
 Marked questions are recommended for Revision.
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Ex-2. Why is anhydrous HCl predominantly covalent in the gaseous state but is ionic in aqueous solution?
Sol.
It exists as HCl (bond formed by equal sharing of electrons) but in aqueous solution ionises as H + (or
H3O+) and Cl– due to polarity of HCl.
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PART - I : SUBJECTIVE QUESTIONS
Section (A) : Octet rule, Lewis dot structures
A-1. Draw the Lewis structures of the following molecules and ions.
PH3, H2S, BeF2, SiCl4, HCOOH, H2SO4, O22–, F2O, C34–, S32–, NOCl, SOBr2, SO2Cl2
With the help of Lewis dot structure find the number of total covalent bonds formed in the following
species.
(i) CO32–, (ii) CCl4, (iii) NF3
A-3.
Indicate what is wrong with each of the following Lewis structures? Replace each with a more
acceptable structure.
+
2–
+
(a) [ S – C = N ]–
(b) [ Cl] [ O ] [Cl ]
(c) O = N = O
(d) Cl – N = Cl
|
Cl
A-4.
In how many of the following species, the central atoms have two lone pairs of electrons ?
(i) XeF4
(ii) XeF5–
(iii) F2SeO2
(iv) XeF3+
(v) XeOF4
–
(vi) ClOF3
(vii) ICl4
(viii) SCl2
(ix) OSF4
Sa
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A-2.
Section (B) : Formal charge & limitations of octet rule
B-1.
How many compounds violate octet rule ?
(i) CO2
(ii) PCl5
(iii) SiF4
(v) IF7
(vi) PCl3
(vii) H2SO4
(iv) BrF5
(viii) BF3
B-2. Write the reason for the violation of octet rule by various molecules?
BCl3, XeF2, NO, IF7, NO2, ClF3, ClO2
Section (C) : Resonance and Bond order Calculation
C-1. Write down the resonance structure(s) for :
(i) SO42–
(ii) CH3COO–
(iii) HCO3–
(iv) NO3–
(v) PO43–
Also calculate average bond order of M–O bond in these compounds.
Where M is central atom (And M–O bonds considered are the one which involve delocalization)
C-2.
How many types of N–O bondlengths are present in (a) HNO3 (b) NO3– ?
22
C-3.
Explain the following :
C–O bond lengths in formic acid are 1.23 Å & 1.36 Å and both the C–O bond lengths in sodium formate
have same value 1.27 Å.
C-4. Compare bond length of S–O bond in SO32– and HSO3–.
Section (D) : VBT, Overlapping of orbitals
D-1.
Find number of sigma bonds and pi bonds in CH2=C=C=CH2.
(i) pz and dxy
(j) px and dxy
(k) px and dz2
(m) dx 2 − y 2 and dx 2 − y 2
(n) dxy and dxy
(o) dxy and dyz
(d) pz and pz
(h) pz and dz2
(l) px and dx 2 − y 2
a
ri
D-2. Draw the type of overlaps between
(a) s and px
(b) px and px
(c) py and py
(e) s and dz2
(f) s and dx 2 − y 2 (g) s and dyz
uh
if internuclear axis is z-axis. Identify them as , ,  bond wherever bond is formed.
PART - II : ONLY ONE OPTION CORRECT TYPE
Ja
Section (A) : Octet rule, Lewis dot structures
A-1.
Among the following which property is commonly exhibited by a covalent molecule
(A) High solubility in water
(B) High electrical conductance
(C) Low boiling point
(D) High melting point
A-2. The possible structure of monothiocarbonate ion is :
C
C
O
O
(C)
S
O
O
(D)
C
O
O
Which one is the electron deficient compound :
(A) ICl
(B) NH3
(C) BCl3
a
A-3.
(B)
S
lp
(A)
S
Sa
nk
Section (B) : Formal charge & limitations of octet rule
B-1.
The octet rule is not obeyed in :
(A) CO2
(B) BCl3
(C) PCl5
B-2. Pick out among the following species isoelectronic with CO 2.
(A) N3–
(B) (CNO)–
(C) (NCN)2–
B-3.
To which of the following species is the octet rule applicable ?
(A) BrF5
(B) SF6
(C) IF7
(D) PCl3
(D) (B) and (C) both
(D) All of these
(D) CO2
Section (C) : Resonance and Bond order Calculation
C-1. The average charge on each O atom and average bond order of I–O bond in IO65– is :
(A) –1 and 1.67
(B) – 5/6 and 1.67
(C) –5/6 and 1.33
(D) –5/6 and 1.167
C-2.
The relation between x, y and z in bicarbonate ion with respect to bond length is (A) x > y > z
(B) x > z > y
(C) z = y > x
(D) x > y = z
C-3.
Average bond order of C–C bond in C6H6 is
(A) 1
(B) 2
C-4.
(C) 1.5
Among the species, which has the weakest carbon- oxygen bond :
(A) CO2
(B) CH3COO–
(C) CO
(D) 1.33
(D) CO32–
23
Section (D) : VBT, Overlapping of orbitals
D-1. Which of the following overlaps is incorrect [assuming z-axis to be the internuclear axis] ?
(a) 2py + 2py → 2py
(b) 2pz + 2pz → 2pz
(c) 2px + 2px → 2px
(d) 1s + 2py → (1s–2py )
(A) ‘a’ & ‘b’
(B) ‘b’ & ‘d’
(C) only ‘d’
(D) None of these
D-2.
Effective overlapping will be shown by :
(A)
(B)
(C)
(D) All the above
D-5.
C34– has :
(A) two  and two -bond
(C) two  and one -bond
(B) three  and one -bond
(D) two  and three -bond
Which of the following is not correct
(A) A sigma bond is weaker than -bond
(B) A sigma bond is stronger than -bond
(C) A double bond is stronger than a single bond
(D) A double bond is shorter than a single bond
uh
D-4.
a
ri
D-3. Indicate the wrong statement according to Valence bond theory :
(A) A sigma bond is stronger then -bond
(B) p-orbitals always have only sidewise overlapping
(C) s-orbitals never form -bonds
(D) There can be only one sigma bond between two atoms
Match the column
(p)
(q)
(r)
(s)
Column-II
Number of  bonds
5
1
3
4
Sa
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a
(A)
(B)
(C)
(D)
Column-I
Molecule/ion
C2–2
C2H2
CH3OH
HNO3
lp
1.
Ja
PART - III : MATCH THE COLUMN
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
1.
In NO3– ion, The number of bond pair and lone pair of electrons present on Nitrozen atom are :
(A) 2,2
(B) 3,1
(C) 1,3
(D) 4,0
2.
How many bonded electron pairs are present in IF7 molecule?
(A) 6
(B) 7
(C) 5
(D) None of these
Which of the following is the electron deficient molecule?
(A) C2H6
(B) SiH4
(C) PH3
(D) BeCl2 (g)
Which is not an exception to the octet rule?
(A) BF3
(B) SnCl4
(D) ClO3
3.
4.
5.
(C) XeF6
Which of the following structure is the most preferred structure for SO 3 ?
(A)
(B)
(C)
(D)
24
6.
For hydrazoic acid, which of the following resonating structure will be least stable?
(A) I
(B) II
(C) III
(D) Both (I) and (III)
What is correct order of bond order of Cl–O bond.
(A) ClO4– > ClO3– > ClO2– > ClO–
(B) ClO– < ClO2– > ClO3– < ClO4–
–
–
–
–
(C) ClO3 < ClO2 < ClO4 < ClO
(D) ClO2– < ClP3– < ClO4– < ClO–
8.
Which of the following statements is not correct for sigma and pi bond formed between two carbon
atoms?
(A) Free rotation of atoms about a sigma - bond is allowed but not in case of a pi-bond
(B) Sigma -bond determines the direction between carbon atoms but a pi-bond has no primary effect in
this regard
(C) Sigma-bond is stronger than a pi-bond
(D) Bond energies of sigma- and pi-bonds are of the order of 264 kJ/mol and 347 kJ/mol. respectively.
a
ri
7.
The number of  and  bonds in dicyanogen (CN)2 are :
(A) 2 + 3
(B) 3 + 2
(C) 3 + 4
(D) 4 + 3
Ja
10.
uh
9.
Number and type of bonds between two carbon atoms in CaC2 are :
(A) one sigma () and one pi () bond
(B) one  and two  bonds
(C) one  and one and a half  bond
(D) one  bond
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
In OF2 number of bond pairs of electrons are :
2.
How many of the following molecules the central atom is surrounded by atleast 10 electrons.
(i) ClO2
(ii) NO3–
(iii) O3
(iv) PCl5
(v) SO3
(vi) SO42–
(vii) CO2
(viii) N3–
–
(ix) I3
3.
Number of molecule or ions having lone pairs  2 for central atom are :
(i) HClO4
(ii) HClO3
(iii) HClO2
(v) NH2–1
(vi) ClF3
(vii) XeF2
(ix) XeF6
(x) I3–1
(xi) N3–1
(xiii) ICl4–
(xiv) ICl2+
(xv) XeO3
Sa
nk
a
lp
1.
(iv) H2O
(viii) XeF4
(xii) O3
(xvi) XeF5–1
4.
Total no. of resonating structure in CO32– are :
5.
Compound SO3 has x bond pairs and y lone pairs. Calculate value of x + y.
6.
Find the number of molecule having two lone pairs on central atom.
(i) I3+
(ii) XeF2
(iii) XeF4
(v) NH2–
(vi) H2S
(vii) H2SO4
7.
(iv) H2O
(viii) NF3
Consider y-axis as internuclear axis, how many of following will lead to  bond formation :
(i) py – py
(ii) px – px
(iii) pz – pz
(iv) dxy – dxy
(v) dyz – dyz
(vi) px – dxy
(vii) dxy – pz
(viii) dxz – dxz
25
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
2.
Find the correct statements regarding SO4–2.
(A) Bond order of S–O bond is 1.5
(C) It violates Octet Rule.
(B) Bond order of S–O bond is 2.5
(D) All S–O bonds are equivalent.
Which of the following Lewis diagram is/are incorrect ?
••
••
••
••
+
•
(A) Na O− C l •
–
(B)
+
H


|


(C)  H − N − H 
2|

 [ S]
H


2
H H
| |
(D) H − N − N − H
••
••
Which are the exceptions of the lewis octet rule.
(A) NO3– and N2O
(B) BeH2 and NO
(C) KrF2 and ClF3
(D) All of these
4.
Which species have same bond order ?
(A) CO3–2
(B) NO3–
(D) NO
Ja
(C) NO2
lp
PART - IV : COMPREHENSION
uh
3.
a
ri
1.
Read the following passage carefully and answer the questions.
Sa
nk
a
Comprehension # 1
Definition: Resonance may be defined as the phenomenon in which two or more structures involving
in identical position of atom, can be written for a particular compound.
Resonance hybrid: It is the actual structure of all different possible structures that can be written for
the molecule without violating the rules of covalence maxima for the atoms.

Bond order =
e.g.,





Resonance hybrid
Total No. of bonds formed between two atoms in all structures
Total No. of resonating structures
O
O–
O–
C
C
C
Bond order =
2 + 1+ 1
= 1.33
3
–
–
O–
O–
O
O
O
O
Resonance energy = Actual bond energy – Energy of most stable resonating structure.
Stability of molecule  resonance energy.
More is the number of covalent bonds in molecule more will be its resonance energy.
Resonance energy  number of resonating structures.
26
1.
Which is the resonance hybride of sulphate ion :
(A)
2.
(B)
(C)
(D)
The correct order of increasing C–O bond length of CO, CO32–, CO2 is :
(A) CO32– < CO2 < CO (B) CO2 < CO32– < CO (C) CO < CO32– < CO2 (D) CO < CO2 < CO32–
4.
(D) (III) (iv) (S)
In which of the following combination octet is not violated.
(A) (IV) (i) (Q)
(B) (II) (iii) (R)
(C) (II) (i) (R)
(D) (III) (iii) (P)
In which of the following combination the sum of oxidation state of central atom and number of -bonds
is maximum.
(A) (I) (ii) (S)
(B) (II) (iii) (R)
(C) (IV) (i) (Q)
(D) (III) (iv) (R)
lp
5.
In which of the following combination are correct for hypervalent ion.
(A) (I) (ii) (R)
(B) (II) (i) (R)
(C) (IV) (i) (Q)
Ja
3.
uh
a
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Comprehension # 2
Answer Q.3, Q.4 and Q.5 by appropriately matching the information given in the three columns
of the following table.
Observe the three columns in which column-1 represents ions, column-2 represents number of
equal contributing resonating structure while column-3 represents bond order.
Column-1
Column-2
Column-3
(I)
(i)
4
(P)
1.66
SO4–2
(II)
(ii)
6
(Q)
1.25
CO3–2
–
(III)
(iii)
3
(R)
1.33
NO2
(IV)
(iv)
2
(S)
1.5
PO4–3
a
* Marked Questions may have more than one correct options.
Sa
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PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
1.
Molecular shapes of SF4, CF4 and XeF4 are respectively :
(A) the same with 2, 0 and 1 lone pair of electrons respectively.
(B) the same with 1, 1 and 1 lone pair of electrons respectively.
(C) different with 0, 1 and 2 lone pair of electrons respectively.
(D) different with 1, 0 and 2 lone pair of electrons respectively.
2.
The number of lone pair(s) of electrons in XeOF4 is :
(A) 3
(B) 2
(C) 1
[JEE–2000(S), 1/135]
[JEE–2004(S), 3/144]
(D) 4
3.
In which of the following the maximum number of lone pairs is present on the central atom ?
[JEE–2005(S), 3/144]
(A) [ClO3]–
(B) XeF4
(C) SF4
(D) I3–
4.*
The compound(s) with TWO lone pairs of electrons on the central atom is(are).
[JEE(Advanced) 2016, 4/124]
(A) BrF5
(B) ClF3
(C) XeF4
(D) SF4
5.
The sum of the number of lone pairs of electrons on each central atom in the following species is
[TeBr6]2–, [BrF2]+, SNF3, and [XeF3]–
(Atomic numbers: N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54)
[JEE Advanced 2017, 3/122]
27
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) OFFLINE PROBLEMS
The number of lone pairs on Xe in XeF2, XeF4 and XeF6 respectively are :
[AIEEE-2002, 3/225]
(1) 3, 2, 1
(2) 2, 4, 6
(3) 1, 2, 3
(4) 6, 4, 2
2.
In the anion HCOO– the two C–O bonds are found to be of equal length. What is the reason for it ?
[AIEEE-2003, 3/225]
(1) Electronic orbits of carbon atom are hybridised.
(2) The C=O bond is weaker than the C–O bond.
(3) The anion HCOO– has two resonating structures.
(4) The anion is obtained by removal of a proton from the acid molecule.
3.
Which of the following has maximum number of lone pairs associated with Xe ? [AIEEE-2011, 4/120]
(1) XeF4
(2) XeF6
(3) XeF2
(4) XeO3
4.
Which of the following exists as covalent crystals in the solid state ?
(1) Iodine
(2) Silicon
(3) Sulphur
5.
The correct statement for the molecule, CsI3, is :
[JEE(Main)-2014, 4/120]
–
+
(1) it is a covalent molecule.
(2) it contains Cs and I 3
3+
(3) it contains Cs and I¯ ions.
(4) it contains Cs+, I¯ and lattice I2 molecule.
6.
Total number of lone pair of electrons in I 3 ion is :
(2) 12
(3) 3
[JEE(Main)-2018, 4/120]
(4) 6
Ja
(1) 9
[AIEEE-2013, 4/120]
(4) Phosphorus
uh
–
a
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1.
lp
EXERCISE - 1
a
PART - I
Sa
nk
A-1.
(ii) 4
2–
A-2.
(i) 4
A-3.
(a)
(b) Covalent compound
(c)
(d)
C
C
C
2–
(iii) 3
A-4.
5 (i, ii, iv, vii, viii)
B-1.
5 (ii, iv, v, vii, viii)
28
BCl3
NO
NO2
ClO2
C-1.
(i)
Electron deficient molecule
Odd electron
Odd electron
Odd electron
;
;
;
XeF2
IF7
ClF3
Super octet molecule
Super octet molecule
Super octet molecule
Bond order =
(ii)
Bond order = 1 +
1
= 1.5
2
Bond order = 1 +
1
= 1.5
2
uh
(iii)
Bond order = 1 +
1
= 1.33
3
Bond order = 1 +
1
= 1.25
4
Ja
(iv)
C-2.
(a) 2
C-3.
Sodium formate exists as HCOO– Na+
Sa
nk
a
(b) 1
lp
(v)
C-4.
2 + 1+ 1+ 2
= 1.5
4
a
ri
B-2.
→ average B.O. = 1
1
3
→ average B.O. = 1
1
2
Hence one bond in HSO3– is longer than S–O bond in SO32–. But other two S–O bond in HSO3– are
shorter bonds.
D-1.
Number of sigma bonds is 7 & number of pi bonds is 3.
D-2.
(a)
(b)
 bond
29
 bond
(d)
(f)
(g)
(h)
uh
(e)
a
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(c)
(j)
Ja
(i)
(l)
(n)
Sa
nk
(m)
a
lp
(k)
(o)
A-1.
(C)
A-2.
(D)
PART - II
A-3.
(C)
B-1.
(D)
B-2.
(D)
B-3.
(D)
C-1.
(D)
C-2.
(D)
C-3.
(C)
C-4.
(D)
D-1.
(C)
D-2.
(C)
D-3.
(B)
D-4.
(A)
D-5.
(A)
30
PART - III
1.
(A) → q, (B) → r, (C) → p, (D) → s
EXERCISE - 2
(D)
2.
(B)
3.
PART - I
(D)
4.
(B)
5.
(A)
6.
(B)
7.
(A)
8.
(D)
9.
(B)
10.
(C)
a
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1.
PART - II
2
2.
5 (i, iv, v, vi, ix)
3.
10 (iii, iv, v, vi, vii, viii, x, xiii, xiv, xvi)
4.
3
5.
12 (x = 6, y = 6)
6.
5 (i, iii, iv, v, vi)
7.
5 (ii, iii, iv, v, vi)
PART - III
1.
(ACD)
2.
(AC)
3.
(BC)
(D)
2.
(D)
3.
(C)
lp
1.
4.
Ja
PART - IV
uh
1.
(AB)
4.
(B)
5.
(A)
4.*
(BC)
5.
6
(2)
5.
(2)
EXERCISE - 3
(D)
2.
(C)
3.
PART - I
(D)
Sa
nk
a
1.
1.
(1)
6.
(1)
2.
(3)
PART - II
JEE(MAIN) OFFLINE PROBLEMS
3.
(3)
4.
31
Section (A) : VSEPR theory
Valence shell electron pair repulsion (VSEPR) theory :
Lewis concept is unable to explain the shapes of molecules. This theory provides a simple procedure to
predict the shapes of covalent molecules. Sidgwick and Powell in 1940, proposed a simple theory
based on the repulsive interactions of the electron pairs in the valence shell of the atoms. It was further
developed and redefined by Nyholm and Gillespie (1957).
(iii)
(iv)
(v)
(vi)
a
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(ii)
The main postulates of VSEPR theory are as follows :
The shape of a molecule depends upon the number of valence shell electron pairs [bonded or
nonbonded) around the central atom.
Pairs of electrons in the valence shell repel one another since their electron clouds are negatively
charged.
These pairs of electrons tend to occupy such positions in space that minimise repulsion and thus
maximise distance between them.
The valence shell is taken as a sphere with the electron pairs localising on the spherical surface at
maximum distance from one another.
A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple
bond are treated as a single super pair.
Where two or more resonance structures can represent a molecule, the VSEPR model is applicable to
any such structure.
uh
(i)
Sa
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a
lp
Ja
The repulsive interaction of electron pairs decreases in the order :
lone pair (p) - lone pair (p) > lone pair (p) - bond pair (bp) > bond pair (bp) - bond pair (bp)
Nyholm and Gillespie (1957) refined the VSEPR model by explaining the important difference between
the lone pairs and bonding pairs of electrons. While the lone pairs are localised on the central atom,
each bonded pair is shared between two atoms. As a result, the lone pair electrons in a molecule
occupy more space as compared to the bonding pairs of electrons. This results in greater repulsion
between lone pairs of electrons as compared to the lone pair-bond pair and bond pair-bond pair
repulsions. These repulsion effects result in deviations from idealised shapes and alterations in bond
angles in molecules.
For the prediction of geometrical shapes of molecules with the help of VSEPR theory it is convenient to
divide molecules into two categories as (i) molecules in which the central atom has no lone pair and (ii)
molecules in which the central atom / ion has one or more lone pairs.
Shape (molecular geometry) of Some Simple Molecules / ions with central atom/ion having no
Lone Pairs of Electrons (E).
32
Table-1
Number of
electron
pairs
General
formula
type
2
AB2
Arrangement of
electron pairs
Molecular
Geometry
Example
B–A–B
Linear
BeCl2,HgCl2
Linear
3
AB3
BF3
Trigonal planar
CH4, NH4+
AB4
Tetrahedral
5
AB5
6
AB6
7
AB7
lp
Octahedral
a
Pentagonal bipyramidal
PCl5
Trigonal bipyramidal
Ja
Trigonal bipyramidal
Tetrahedral
uh
4
a
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Trigonal planar
SF6
Octahedral
IF7
Pentagonal bipyramidal
Sa
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Shape (molecular geometry) of Some Simple Molecules/Ions with central atom / ions having One
or More Lone Pairs of Electrons (E).
Table-2
No. of
No. of
General
Arrangement of
bonding
lone
Shape
Examples
formula type
electron pairs
pairs
pairs
AB2E
2
1
Bent
SO2,O3
AB3E
3
1
Trigonal
Pyramidal
NH3
AB2E2
2
2
Bent
H2O
AB4E
4
1
See saw
SF4
33
3
2
T–shape
CIF3
AB5E
5
1
Square
Pyramidal
XeOF4
AB4E2
4
2
Square
Planar
XeF4
AB5E2
5
2
Pentagonal
Planar
XeF2
(b)
ClO3–
Ja
(a)
lone pairs occupy the equatorial positions to have minimum
repulsion. Thus it is linear.
To minimize the repulsion between lone pair and double bond,
species acquires trigonal pyramidal.
a
Sol.
Use the VSEPR model to predict the geometry of the following :
(a) XeF2
(b) ClO3–
Species
Structure
lp
Ex-1.
XeF5–
uh
Shapes of Molecules containing Bond Pair and Lone Pair
a
ri
AB3E2
Section (B) : Hybridisation
Hybridisation :
Sa
nk
– Hypothetical concept Introduced by pauling and slater.
– Atomic orbitals of same atom combine to form new set of equivalent orbitals know as hybrid orbitals.
– This phenomenon is known hybridization.
– Process of Intermixing of the atomic orbitals of equal or slightly different energies in the formation of
new set of orbitals of equivalent energies and shape is known as hybridization.
1.
2.
3.
4.
(i)
(ii)
(iii)
(iv)
Salient features of hybridisation : The main features of hybridisation are as under :
The number of hybrid orbitals is equal to the number of the atomic orbitals that get hybridised.
The hybridised orbitals are always equivalent in energy and shape.
The hybrid orbitals are more effective in forming stable bonds than the pure atomic orbitals.
These hybrid orbitals are directed in space in some preferred direction to have minimum repulsion
between electron pairs and thus a stable arrangement is obtained. Therefore, the type of hybridisation
indicates the geometry of the molecules.
Important conditions for hybridisation :
The orbitals present in the valence shell (and sometimes penultimate shell also) of the atom are
hybridised.
The orbitals undergoing hybridisation should have almost equal energy.
Promotion of electron is not essential condition prior to hybridisation.
It is the orbital that undergo hybridization and not the electrons. For example, for orbitals of nitrogen
atom (2s2 2p1x 2p1y 2p1z ) belonging to valency shell when hybridize to form four hybrid orbitals, one of
34
which has two electrons (as before) and other three have one electron each. It is not necessary that
only half filled orbitals participate in hybridisation. In some cases, even filled orbitals of valence shell
take part in hybridisation.
Determination of hybridisation of an atom in a molecule or ion:
Steric number rule (given by Gillespie) :
Steric No. of an atom = number of atom bonded with that atom + number of lone pair(s) left on that atom.
Note : This rule is not applicable to molecules/ions which have odd e– (ClO2, NO, NO2 ), free radicals and
compounds like B2H6 which involve 3 centre 2e– bond (banana bond).
For example :
O=C=O
S.No. = 2 + 0 = 2
S.No. = 3 + 0 = 3
S.No. = 3 + 1 = 4
ns, npx / pz / py
ns, npx, pz / py, pz/px , py
ns, npx, pz , py
ns, npx, pz, py, dz2
sp3d2
Octahedral
ns, npx, pz, py,
dz2 dx2 − y2
7
sp3d3
Pentagonal bipyramidal
ns, npx, pz, py,
dz2 dx2 − y2
, dxy
lp
This type of hybridisation involves the mixing of one s and one p orbital resulting in the formation of two
equivalent sp hybrid orbitals.
Each sp hybrid orbitals has 50% s-character and 50% p-character. Such a molecule in which the
central atom is sp-hybridised and linked directly to two other central atoms possesses linear geometry.
This type of hybridisation is also known as diagonal hybridisation.
The two sp hybrids point in the opposite direction along the Z-axis with projecting bigger positive lobes
and very small negative lobes, which provides more effective overlapping resulting in the formation of
stronger bonds.
Sa
nk

Linear
Trigonal planar
Tetrahedral
Trigonal bipyramidal
a

Involving orbitals
6
sp hybridisation :

Geometry
Ja
Types of
Hybridisation
sp
sp2
sp3
sp3d
uh
Table-3
Steric
number
2
3
4
5
a
ri
S.No. = 2 + 1 = 3
Example of a molecule having sp hybridisation
BeCl2 : The ground state electronic configuration of Be is 1s 22s2. In the excited state one of the 2selectrons is promoted to vacant 2p orbital to account for its divalency. One 2s and one 2p-orbitals get
hybridised to form two sp hybridised orbitals. These two sp hybrid orbitals are oriented in opposite
direction forming an angle of 180º. Each of the sp hybridised orbital overlaps with the 2p-orbital of
chlorine axially and form two Be–Cl sigma bonds.
35
Figure : (A) Formation of sp hybrids from s and p orbitals ; (B) Formation of the linear BeCl 2 molecule.
Examples of sp hybridisation.
Important characteristic
Linear, highly posionous, weak acid
Linear,  bond planes are perpendicular
Linear, both  bond are perpendicular to each other
Non planar both hydrogen are perpendicular to each other
Iso electronic with CO2 and linear in shape. Both N–N bonds are similar
a
ri
Species
H–CN
H–CC–H
O=C=O
H2C=C=CH2
N3– (azide ion)
HgCl2
NO2+ (nitronium ion), N2O
uh
Hydrazoic acid
sp2 hybridisation :
Sa
nk
a
lp
Ja
Mixing of one s and two p orbitals to form 3 equivalent sp2 hybridized orbitals.
For example, in BCl3 molecule, the ground state electronic configuration of central boron atom is 1s 2 2s2
2p1. In the excited state, one of the 2s electrons is promoted to vacant 2p orbital as a result boron has
three unpaired electrons.
These three orbitals (one 2s and two 2p) hybridise to form three sp 2 hybrid orbitals. The three hybrid
orbitals so formed are oriented in a trigonal planar arrangement and overlap with 2p orbitals of chlorine
to form three B–Cl bonds.
Therefore, in BCl3, the geometry is trigonal planar with CI–B–Cl bond angle of 120°.
36
Figure : Formation of sp2 hybrids and the BCl3 molecule
Steric No. = 3
Geometry = Trigonal planar
Ideal bond angle = 120°
Table-4
Type
Shape
Example
AB2L
V-shape (bent)
NOCl, O3, NO2,
NO2–
B = side atom,
L = lone pair of e–
A = central atom,
Structure
Statements
Ja
Molecule
uh
a
ri
Where
AB3
Trigonal
C6H6, CO32–, HCO3–
H2CO3, graphite, BF3
B(OH)3, SO3, NO3–
C60 (Fullerene)
All three S–O bonds are equivalent. Out of 3p bond one p-p other to are
p-d.
O3
V shaped molecule. Both O–O bond length are equal.
CO3– –
All three C–O bonds are equivalent.
Bond lengths are shorter than single bond length but longer than double
bond length.
lp
SO3
NO2–
a
Bond order of N–O bond is 1.5, planar molecule.
Sa
nk
Note : In N2O5, N is sp2 hybridised but in solid, N2O5 exist in NO2+ (N is sp hybridised) and NO3– (N is sp2
hybridised) ions.
sp3 hybridisation :





mixing of one s and three p orbitals to form 4 equivalent sp3 hybridized orbitals.
4 sp3 orbitals are directed towards four corner of tetrahedron.
This type of hybridisation can be explained by taking the example of CH 4 molecule in which there is
mixing of one s-orbital and three p-orbitals of the valence shell to form four sp3 hybrid orbital of
equivalent energies and shape.
There is 25% s-character and 75% p-character in each sp3 hybrid orbital. The four sp3 hybrid orbitals so
formed are directed towards the four corners of the tetrahedron.
The angle between sp3 hybrid orbital is 109.5° as shown in figure.
37
a
ri
uh
Figure
lp
Ja
Ammonia
a
Water
Geometry = tetrahedral
Sa
nk
Steric No. = 4
Type
Shape
Example
Table-5
AB3L
pyramidal
XeO3
AB4
tetrahedral
CH4
Ideal bond angle = 109° 28
AB2L2
V-shape or bent
OBr2
ABL3
Linear
–OCl
Some other Examples of sp3 hybridisation

(a)
(b)
(c)


Cl
BeCl2(s),
Cl
Be
Cl
Cl
Be
Be
Cl
Cl
Cl
Cl
BF4– ;
Diamond, CCl4,
, all alkane ;
38
(d)

SiCl4, Silicates etc.
Structure of SiO2
(e)

a
ri
SiO2 is a covalent network solid like diamond
NH2– (amide ion)
V-shape
NH2–NH2 (hydrazine)

NH2OH (hydroxylamine)
lp

P4 (White phosphorus)
Sa
nk
(f)
lp-lp repulsion increases the N–O
a
bond length.
Ja
uh

 Structures of silicate ion : [SiO4]4–
–
–

–
sp3 = Silicon ; O = Oxygen
–
39

Structures of cyclic silicates : [Si3O9]6–

H2O2

SOCl2 (Thionyl chloride)

XeO4
a
ri
= Silicon ; O = Oxygen
Sa
nk
⚫
Trigonal pyramidal
a
lp
Ja
uh
H2O2(g)
sp3d hybridisation :
Steric number = 5
Type
Shape
Example
AB5
Trigonal bipyramidal
PCl5, PBr5, PF5 etc.
Geometry = trigonal bipyramidal
Table-6
AB4L
See-saw
SF4, XeO2F2
AB3L2
T-shape
ClF3, [XeF3]+
AB2L3
Linear
XeF2, I3–, [ICl2]–
Important points regarding sp3d :
(i)
According to VSEPR theory lone pair will occupy equatorial (E) positions but not axial (a).
(ii)
(iii)
More electronegative atoms will prefer to occupy axial positions.
Since, double bonds occupy more space. Therefore, they will also prefer equatorial positions.
40
PF5(g)
PF5(g) is trigonal bipyramidal and the electron diffraction
shows that some bond angles are 90º and others are
120º, and the axial P–F bond lengths are 1.58 Å while the
equatorial P–F lengths are 1.53 Å. But NMR studies
suggest that all five atoms are equivalent because of
pseudo rotation. PF5 remains covalent and is trigonal
bipyramidal in the solid state.
PBr5(g)
PBr5 exist as (PBr4)+Br– in solid state.
PCl3F2
PCl3F2 is non polar molecule as all three Cl atoms are at
equatorial position and both F atoms in axial position.
uh
a
ri
PCl5(g)
Ja
AB5
It is covalent in the gas but in solid state exists as ionic
solid consisting of [PCl4]+ (tetrahedral) and [PCl6]–
(octahedral). All P–Cl bonds are not of equal lengths.
Here axial bonds are longer and weaker than equatorial
bonds.
PCl2F3 is polar molecule as both Cl atoms and one F
atom are at equatorial position and both F atoms in axial
position.
SF4
SF4 molecule have T-shape geometry.
Sa
nk
a
AB4L
lp
PCl2F3
AB3L2
XeO2F2
Both F atoms will be at equatorial position.
ClF3
T-shape (It is planar molecule)
I3–
I3– has linear shape as both I atoms are at axial position.
41
XeF2 has linear shape as both F atoms are at axial
position.
XeF2
sp3d2 hybridization :
Table-7
AB6
Octahedral
SF6, PF6–, [SiF6]2–, [AlF6]3– , [XeO6]4–
Type
Shape
Example
AB5L
Square Pyramidal
BrF5, IF5, XeOF4
AB4L2
Square Planar
ICl4–, XeF4
AB6
uh
Important : Since, octahedral is a symmetrical figure hence
positions of a lone pair can be any where
but if there are two lone pairs (max.) then these must be in the trans position.
Bond angle = 90°
Due to over-crowding and maximum valency of S, SF6
is much less active (almost inert) than SF4.
SF6
Ja
(a)
(b)
Geometry = octahedral
a
ri
Steric number = 6
[XeO6]4– is perxenate ion & H4XeO6 is called
perxenic acid. But H2[XeO4] is called xenic acid.
XeOF4
Sa
nk
AB5L
a
lp
[XeO6]4–
AB4L2
Molecule has square pyramidal geometry.
XeF4
Molecule has square plannar geometry.
I2Cl6
ICl3 does not exist, but the dimer I2Cl6 is bright yellow
solid. Its structure is planar. The terminal I–Cl bonds are
normal single bonds of length 2.38 Å and 2.39 Å. The
bridging I–Cl bonds appreciably longer (2.68 Å and
2.72Å) suggesting delocalized bonding rather than
simple halogen bridges formed by coordinate bonds
from Cl2 to I.
Note : The liquid has an appreciable electrical conductance due to self ionization.
2Cl6
[Cl2]+ (bent) + [Cl4]– (square planar)
sp3d3 Hybridization :
Steric number = 7
Geometry = Pentagonal bi-pyramidal
42
Table-8
AB7
Pentagonal bi-pyramidal
IF7
Type
Shape
Example
F7
AB6L
XeF6
(g)
Bond angle = 72º & 90º
uh
a
ri
AB7
AB6L
Distorted octahedral
XeF6
Ja
Distorted octahedron with a nonbonding
electron pair either at the centre of a face or the
midpoint of an edge.
Sa
nk
XeF6(s)
a
lp
or
[XeF5]–
XeF6(s) is found to be ionic solid consisting of
[XeF5]+ and F– ions.
It is found that F– is forming a bridge between
two XeF5+ ions.
Pentagonal planar ion with two nonbonding
electron pairs above and below the plane of the
pentagon.
43
Section (C) : Bond angle, bond length comparison
(i)
(ii)
(iii)
Size of atom (see along the group)  bond length
HF < HCl < HBr < HI
F–F < Cl–Cl < Br–Br < –
CH4 < SiH4 < GeH4 < SnH4
Multiplicity of bond (nearly same period element)
single bond > double bond > triple bond
C—C > C=C > CC
F—F > O=O > NN
Electronegativity difference (See along the period)
H—C > H—N > H—O > H—F
3.
uh
Sa
nk
a
4.
Ja
2.
Hybridisaition :
sp > sp2 > sp3
> sp3d2
180° 120°
109°28’
90°
Number of lone pair : If hybridisation of the central atom is same but number of lone pair is different
then more is the number of lone pair less is the bond angle.
e.g.
CH4
NH3
H2O
Hybridisaition sp3
sp3
sp3
lone pair
.P. = 0 .P. = 1 .P. = 2
B.A.
109°28’ 107°
104°
Size or electronegativity of central atom : When hybridisation is same and no. of lone pair is same
but central atom is different then see the electronegativity of central atom. More is the electronegativity
more is the bond angle.
e.g.
NH3
PH3
AsH3
SbH3
Hybridisaition sp3
no
no
no
lone pair
.P. = 1 .P. = 1 .P. = 1 .P. = 1
B.A.
107°
93°
92°
91°
Size or electronegativity of terminal atom :
Hybridisation same, lone pair same, central atom same but terminal atom is different then greater is the
size of the terminal atom greater will be the bond angle. Only in case of flourine the electronegativity
factor is considered, due to greater electronegativity of the flourine atom the bond angle for it comes out
to be smallest( due to smaller bond bond pair repulsions)
e.g.
PF3
PCl3
PBr3
PI3
Hybridisaition sp3
sp3
sp3
sp3
lone pair
.P. = 1 .P. = 1 .P. = 1 .P. = 1
B.A.
98°
100°
101°
102°
Reason : As the E.N. of x , b.p.–b.p. repulsion will less but l.p. compression will work as usual
Drago rule : Element of 3rd and higher period (4, 5 ......) (p-Block) does not allow hybridisation in
molecule when they form bond if they have lone pair on them with less electronegative elements such
as hydrogen.
eg : PH3, (CH3)2S, AsH3, H2S not have hybridisation.
lp
1.
a
ri
HOW TO COMPARE BOND ANGLES
Bond angle depends on the following factor
I.
Hybridisaition
II.
No. of lone pair
III.
Size or electronegativity of central atom
IV.
Size or electronegativity of terminal atom
Bent rule : According to Bent’s rule, more electronegative atoms prefer hybrid orbitals having less S
character and more electropositive atoms prefer hybrid orbitals having more S character.
eg : In CH2F2, F–C–F bond angle less than 109.5° indicating less than 25% S character.
H–C–H bond angle more than 109.5° indicating more than 25% S character.
44
Ex-2.
Compare bond angle of OF2 and Cl2O.
Sol.
OF2
;
Cl2O
because of large size of Cl atom.
Ex-3.
Discuss the bond angle in carbonyl halides COF2, COCl2, COBr2, CO2.
Sol.
Discuss the bond angle hydride of nitrogen family.
B.A.
107º
92º-93º
92º
91º
NH3
PH3
AsH3 SbH3
P4O10
Sa
nk

P4O6
a

lp
Section (D) : Multicentered species
Ja
Ex-4.
Sol.
uh
a
ri
Carbonyl Hallide
B.A.   COF2 < COCl2 < COBr2 < CO2
Explanation 
(A) double bonds require more room than single bonds. Hence C=O group compresses the molecule
and bond angle  max. in COF2. Later on halogen atom becomes bigger and less (–ve) also so
 Inter atomic repulsion between the halogens causes  in B.L.
As X(halogen) becomes less and less (–ve) b.p.-b.p. repulsion also becomes imp and therefore 
P–O bond length shows that the bridging bonds on
the edges are 1.65 Å and are normal single bonds.
There is no. P–P bonds.
The P–O bond lengths shows that the bridging
bonds on the edges are 1.60 Å but the P=O bonds
on the corners are 1.43 Å and this P=O is formed
by pp–dp back bonding. A full p-orbital on the
oxygen atom overlaps sideway with an empty dorbital on the phosphorus atom. The bond angle
POP is 127º and there is no P–P bonds.
Total L.P. = 20.
Total no. of p-d bonds = 4
 Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
Section (A) : VSEPR theory
A-1.
Why NO2+ and I3– are linear species ?
A-2.
PCl5 has the shape of a trigonal bipyramidal where as IF5 has the shape of square pyramidal. Explain.
45
A-3. Write the geometry of XeF4 and OSF4 using VSEPR theory and clearly indicate the position of lone pair
of electrons.
A-4.
Explain the structure of CIF3 on the basis of VSEPR theory.
Section (B) : Hybridisation
B-1. Explain hybridisation of central atom in :
(1)
XeF2
(2)
XeF4
(5)
SF6
(6)
IF3
(9)
CH4
(10)
CCl4
(13)
H2O
(14)
NH3
(17)
NO3–
(18)
CO32–
(3)
(7)
(11)
(15)
(19)
PCl3
IF5
SiCl4
PO43–
NH4+
(4)
(8)
(12)
(16)
(20)
PCl5 (g)
IF7
SiH4
BrF5
ClO3–
B-3.
a
ri
B-2. The order of size of the hybrid orbitals is as follows sp < sp 2 < sp3. Explain.
Draw the structure of the following compounds. Clearly indicate the number of bond pairs and lone
pairs involved on central atom. Write (i) number of bond pairs and lone pairs on the central atom (ii) the
shape of the molecules (iii) hybridization of the central atom.
(a) SF4
(b) XeOF4
C-2.
uh
Section (C) : Bond angle, bond length comparison
C-1.
Draw an electron dot structure for Br3–. Deduce an approximate value of the bond angle.
Which compound has the smallest bond angle in each series ?
(a)
SbCl3
SbBr3
SbI3
(b)
PI3
AsI3
SbI3
C-4.
Ja
C-3. Compare the C–H bond strength in C2H6, C2H4 and C2H2.
The POCl3 molecule has the shape of an irregular tetrahedron with the P atom located centrally. The
Cl–P–Cl angle is found to be 103.5º. Give a qualitative explanation for the deviation of this structure
from a regular tetrahedron.
Write the Increasing order of Bond length of each :
(1) C–C, C=C, CC
(2) C–N, C–O, C–F
a
C-6.
lp
C-5. Which one has highest and least bond angle in the following ?
(1)
CH4
PH3
AsH3 SbH3
(2)
H2O
(3)
PH3
H2O
(4)
Cl2O
(5)
PF3
PH3
(6)
BF3
(7)
NH3
NF3
(8)
PF3
Sa
nk
Section (D) : Multicentered species
D-1.
Find number of p–d bonds in
(a) Disulphate
(b) triphosphate
(d) trimer of SO3
(e) P4O10
H2S
ClO2
NF3
PCl3
H2Te
CO2
(3) H–Cl, H–Br, H–I, HF
(c) trimetaphosphate
(f) P4O6
D-2. In which of the following compounds, the p–d bonding take place ?
(a) P4O10
(b) HNO3
(c) N2O5
(d) HClO3
D-3. Calculate individual and average oxidation number (if required) of the marked element and also draw
the structure of the following compounds or molecules.
(1) Na2 S2 O3
(2) Na2 S4O6
(3) H2SO5
(4) H2 S2O8
(5) H2S2O7
(6) S8
(7) HNO4
(8) C3O2
(9) OsO4
(10) PH3
(11) CrO42–
(12) Cr2O72–
(13) Cr O2Cl2 (14) CrO5
(15) Na2 H PO4 (16) FeS2
(17) C6H12O6 (18) NH4 NO3
46
PART - II : ONLY ONE OPTION CORRECT TYPE
:
Section (A) : VSEPR theory
A-1.
Which is the right structure of XeF4 ?
F
F
Xe
(A)
Xe
(B)
F
:
(C)
F
(D)
F
:
F
F
Xe
:
F
F
F
F
:
F
:
F
F
a
ri
Xe
F
F
Ja
Which of the following statement is true for IO2F2– ?
(A) The electrons are located at the corners of a trigonal bipyramidal but one of the equatorial pairs is
unshared.
(B) It has sp3d hybridisation and is T-shaped.
(C) Its structure is analogous to SF4.
(D) (A) and (C) both
lp
A-3.
uh
A-2. Identify the correct match.
(i)
XeF2
(a) Central atom has sp3 hybridisation and bent geometry.
–
(ii) N3
(b) Central atom has sp3d2 hybridisation and octahedral.
–
(iii) PCl6 (PCl5 (s) anion) (c) Central atom has sp hybridisation and linear geometry.
(iv) ICl2+ (I2Cl6 () cation) (d) Central atom has sp3d hybridisation and linear geometry.
(A) (i – a), (ii – b), (iii – c), (iv – d)
(B) (i – d), (ii – b), (iii – d), (iv – c)
(C) (i – b), (ii – c), (iii – a), (iv – d)
(D) (i – d), (ii – c), (iii – b), (iv – a)
A-4. Which reaction involves a change in the electron–pair geometry for the under lined element ?
(A) BF3 + F– ⎯→ BF4–
(B) NH3 + H+ ⎯→ NH4+
(C) 2SO2 + O2 ⎯→ 2SO3
(D) H2O + H+ ⎯→ H3O+
In which of the following molecules number of lone paris and bond pairs on central atom are not equal ?
(A) H2O
(B) I3–
(C) O2F2
(D) SCl2
A-6.
Which of the following species given below have shape similar to XeOF4 ?
(A) XeO3
(B) OF4+
(C) PCl5
(D) XeF5
Sa
nk
a
A-5.
Section (B) : Hybridisation
B-1.
4
3
2
1
The hybridization of carbon atoms in C2–C3 single bond of HC  C− CH = CH2 is :
(A) sp3–sp3
(B) sp2–sp
(C) sp–sp2
(D) sp3–sp
B-2. Specify the hybridisations of central atom in the following species respectively {N 3–, NOCl, N2O}
(A) sp, sp2, sp
(B) sp, sp, sp3
(C) sp2, sp, sp
(D) sp2, sp2, sp.
B-3.
In pent-3-en-1-yne the terminal carbon-atoms have following hybridisation
(A) sp & sp2
(B) sp2 & sp3
(C) sp2 & sp
(D) sp & sp3
B-4.
S1 : [XeF7]+ has sp3d3 hybridisation
S2 : [PCl4]+ has sp3d2 hybridisation
S3 : [SF6] has sp3d2 hybridisation
S4 : [PF4]+ has sp3 hybridisation
(A) T F F T
(B) T T F T
B-5.
(C) T F T T
BF3 + F– → BF4–
What is the hybridiation state of B in BF3 and BF4– :
(A) sp2, sp3
(B) sp3, sp3
(C) sp2, sp2
(D) F F F T
(D) sp3, sp3d
47
Section (C) : Bond angle, bond length comparison
C-1. The ONO angle is maximum in :
(A) HNO3
(B) NO2+
C-3.
Which statement is correct for
ion.
(A) It is bent molecule
(C) Central atom is sp2 hybridized
(B) Bond angle is < 120°
(D) None of these
Consider the following molecules ;
H2O
H2S
H2Se H2 Te



V
Arrange these molecules in increasing order of bond angles.
(A)       V
(B) V      
(C)     V  
C-4. In which of the following bond angle is maximum
(A) NH3
(B) NH4+
(C) PCl3
C-5.
In which of the following central atom is unhybridised?
(A) S(CH3)2
(B) SO2
(C) SiH4
D-2.
The no. of S-O-S bonds in the trimer of SO3 is
(A) 1
(B) 2
(C) 3
Which of the following species do not contain S–S linkage?
(A) H2S2O5
(B) H2S2O7
(C) H2S2O3
(D) SCl2
(D) PCl3
(D) None
(D) H2S4O6
lp
Ja
D-3. Which statement is incorrect about pyrosilicate ion.
(A) sp3 hybridisation
(B) One oxygen atom is shared between two tetrahydron
(C) there are eight Si–O bond
(D) There is one Si-Si bond
D-4. Which is correct about the cyclic silicate [Si6O18]n– :
(A) The value of n is 12
(B) each Si atom is bonded with three oxygen atoms
(C) each oxygen atom is bonded with two Si atoms
(D) all the above are correct.
(D)   V    
uh
Section (D) : Multicentered species
D-1.
(D) NO2
a
ri
C-2.
(C) HNO2
N3–
PART - III : MATCH THE COLUMN
a
Match the following :
Column–I
(A)
SF2
(p)
(B)
KrF4
(q)
(C)
NOCl
(r)
(D)
NF3
(s)
(t)
Column–II
sp3 and bent
two lone pairs on central atom
bond angle < 109º28’
sp2 and bent
sp3d2 and square planar
Sa
nk
1.
2.
Match the compounds listed in column-I with characteristic(s) listed in column-II.
Column – I
Column –II
(A) ClF2–, CIF2+
(p) Square pyramidal.
(B) IO2F2–, F2SeO (q) See–saw and pyramidal shaped respectively.
(C) IOF4–, XeOF2 (r) Linear and bent shaped respectively.
(D
BrF5, XeOF4
(s) Square pyramidal and T-shaped respectively.
(t) Both sp3d2.
3.
Match the following :
Column-I
(A)
H3P3O9
(p)
(B)
H2S2O7
(q)
(C)
H2S4O6
(r)
(D)
H4P2O5
(s)
Column-II
S–O–S bond is present
Di-basic acid
P–O–P bond is present
Central atom (S or P) in maximum oxidation state.
48
 Marked Questions may have for Revision Questions.
PART - I : ONLY ONE OPTION CORRECT TYPE
1.
The hybridisation of P in phosphate ion (PO43–) is the same as :
(A)  in Cl4–
(B) S in SO3
(C) N in NO3–
(D) S in SO32–
Choose the molecules in which hybridisation occurs in the ground state ?
(a) BCl3
(b) NH3
(c) PCl3
(d) BeF2
The correct answer is (A) a, b, d
(B) a, b, c
(C) b, c
(D) c, d
3.
The bent or V–shape of the molecule can be resulted from which of the following hybridization.
(A) sp3
(B) sp2
(C) Both (A) and (B)
(D) None of these
4.
sp3d hybridization is considered to be a combination of two hybridization. They are
(A) p3 + sd
(B) sp2 + pd
(C) spd + p2
(D) none of these
5.
If the equatorial plane is x–y plane in sp3d hybridisation then the orbital used in pd hybridisation are (A) pz and dz2
(B) px and dxy
(C) py and dyz
(D) none of these
6.
A -bonded molecule MX3 is T-shaped. The number of lone pairs of electrons can be
(A) 0
(B) 2
(C) 1
(D) none of these
7.
Which of the following should have pyramidal shape :
(A) [ClOF2]+
(B) ICl3
(C) [BrCl]–
Ja
uh
a
ri
2.
(D) All of these
Which of the following molecules has two lone pairs and bond angle (need not be all bond angles) <
109.5°?
(A) SF2
(B) KrF4
(C) Cl4–
(D) All of these
9.
The correct order of bond angle is :
(A) H2S < NH3 < BF3 < CH4
(C) H2S < NH3 < CH4 < BF3
In which of the following molecules are all the bonds not equal?
(A) NF3
(B) ClF3
(C) BF3
(D) AlF3
Which of the following is correct order of bond length ?
(A) BF4– < BF3
(B) NO2+ < NO2–
(C) CCl4 < CF4
(D) +CH3 > CH4
Sa
nk
11.
(B) NH3 < H2S < CH4 < BF3
(D) H2S < CH4 < NH3 < BF3
a
10.
lp
8.
12.
Identify the correct statement :
(A) single N–N bond is stronger than single P–P bond
(B) single N–N bond is weaker than single P–P bond
(C) N N is weaker than PP
(D) None of these
13.
In which of the following species peroxide group is not present :
(A) [B4O5(OH)4]2–
(B) [S2O8]2–
(C) CrO5
14.
Which of the following is correct ?
(A) S3O9 : contains no S–S linkage.
(C) (HPO3)3 : contains P–P linkage
(D) HNO4
(B) S2O62– : contains –O–O– linkage.
(D) S2O82– : contains S–S linkage
15.
The percentage of s–character in the orbital forming P–S bonds in P4S3 is :
(A) 25
(B) 33
(C) 75
(D) 50
16.
Indicate the incorrect statement :
(A) Number of hybrid orbitals formed is equal to no. of atomic orbitals involved.
(B) 2px and 2py - orbitals of carbon can be hybridized to yield two new more stable orbitals
49
(C) Effective hybridisation is not possible with orbitals of widely different energies
(D) The concept of hybridisation has a greater significance in the VB theory of localised orbitals than in
the MO theory.
In which of the following compounds B atoms are in sp 2 and sp3 hybridisation states ?
(A) Borax
(B) Diborane
(C) Borazole
(D) All
18.
S1 : Oxidation number of N in N2O5 is 5
S2 : The anhydride of Hypochlorous acid is Cl2O
S3 : As the electronegativity of central atom in a molecule having same hybridisation state and same
terminal atoms increases, bond angle increases.
S4 : For heteronuclear diatomic species A–B, the bond length decreases as the difference in
electronegativity values increases (considering A and B of similar size) in all compound.
(A) T T T F
(B) F T T T
(C) F F T F
(D) T T F T
a
ri
17.
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
(c) CO32–
(g) PCl3
(d) SO3
(h) XeF4
uh
2.
Find the number of planar molecule.
(a) BF3
(b) BCl3
(e) NH3
(f) NCl3
Find the number of species having bond angle less than 109°28'.
(a) H2S
(b) SO4– –
(c) CCl4
(e) PH3
(f) SiH4
(g) NH4+
–
(i) NH2
(j) SO3
(k) H2O
(d) NH3
(h) PF3
Ja
1.
Find out total number of  bond in following xenon oxyflourides.
XeOF2, XeO2F4, XeO3, XeO4, XeO3F2, XeOF4, XeO2F2
4.
P4O10 has two different types of P–O bonds. Find the no. of P–O bonds with shorter bond length.
5.
Difference in the oxidation number of sulphur atom is in Na2S4O6 is x, that of H2S2O5 is y. Find value of
x × y is :
6.
In a P4O6 molecule, the total number of P–O–P bonds is :
lp
3.
Which statement is correct about hybridization ?
(A) In hybridisation orbitals take part.
(B) In hybridisation electrons take part.
(C) In hybridisation fully filled, half filled or empty orbitals can take part.
(D) Hybridised orbitals only contains bond pair electron.
Sa
nk
1.
a
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
2.
Which of the following represent the given mode of hybridisation sp2–sp2–sp–sp from left to right
(A) H2C=C=C=CH2
(B) HCC–CCH
(C) H2C=CH–CN
(D) H2C=CH–CCH
3.
Which is/are in linear shape ?
(A) NO2+
(B) XeF2
(C) 3–
(D) 3+
4.
Which is true about NH2–, NH3, NH4+ ?
(A) Hybridization of N is same.
(B) No. of lone pair of electron on N are same.
(C) Molecular geometry (i.e. shape) is different. (D) Bond angle is same.
5.
Which of the following molecule (s) has/have bond angle close to 90º ?
(A) NH3
(B) H2S
(C) PH3
6.
(D) ICl3
Which of the following statements are true about borax :
(A) Boron atoms are present in 2 different oxidation state and it different by 1
(B) the average oxidation state of boron is same that in B2H6.
(C) Boron atoms are present in different hybridization
50
(D) 2 boron atoms are connected with 4 oxygen each and 2 boron atoms are connected with 3 oxygen
atom each.
Identify the correct statement
(A) H2S2O7 has peroxy linkage
(B) H2S2O6 has S–S linkage
(C) H2S2O8 has peroxy linkage
(D) H2SO3(Sulphurous acid) has S in +4 oxidation state
8.
In which of the following compound(s) oxidation number of one central atom is/are  6 ?
(A) N2O6
(B) CrO5
(C) H3PO5
(D) H2S2O8
a
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7.
PART - IV : COMPREHENSION
Read the following passage carefully and answer the questions.
Comprehension # 1
VSEPR THEORY
Ja
uh
The trigonal bipyramid is not a regular shape since the bond angles are not all the same. It therefore
follows that the corners are not equivalent in ClF3 molecule. Lone pairs occupy two of the corners, and
F atoms occupy the other three corners. These different arrangements are theoretically possible, as
shown in figure.
Sa
nk
a
lp
(i) The most stable structure will be the one of lowest energy, that is the one with the minimum
repulsion between the five orbitals. The greatest repulsion occurs between two lone pairs. Lone pair
bond pair repulsions are next strongest, and bond pair-bond pair repulsions the weakest.
1.
2.
A rule of thumb can be theorised, that the position having maximum repulsion amongst them are
occupied at equatorial points. Therefore (3) structure is right.
(ii) Since double bond occupies more space compared to single bond therefore it will prefer equatorial
position.
(iii) More electronegative element will occupy axial position in case of trigonal bipyramidal geometry
(iv) In case of sp3d2 hybridisation lone pairs should be placed opposite to each other because all the
corners are identical.
Geometry (i.e. arrangement of electron pairs around central atom) of ClOF3 is similar to the :
(A) XeF4
(B) SOCl2
(C) 3¯
(D) ClO4¯
The shape of SF5– can be :
–
F
F
F
S
F
S
F
••

F
–
F
F
S
F
F

–
F
••
F
••
F
–
F
S
F
F

F
F
F
F
V
51
(A)  only
3.
4.*
(B)  and  only
(C) V only
(D) , , & 
Actual shape of the molecule BrF5 is similar to the molecule :
(A) PCl5
(B) XeF4
(C) PCl4+
Which of the following do not exist ?
(A) SH6
(B) HFO4
(C) SI6
(D) None of these
(D) HClO3
a
ri
Comprehension # 2
Answer Q.5, Q.6 and Q.7 by appropriately matching the information given in the three columns
of the following table.
Observe the three columns in which column-1 : compound, column-2 : shape while in column-3 :
Hybridisation are given.
Column-1 (Compound)
Column-2 (Shape)
Column-3 (Hybridisation)
(I)
XeF4
(i)
Tetrahedral
(P)
sp3
(II)
ClF3
(ii)
Square planar
(Q)
sp2
(III)
SiF4
(iii) Bent
(R)
sp3d
(IV)
CH3OCH3
(iv) T-shape
(S)
sp3d2
Which of the following combination is true for compound which have 2 lone pair of electrons on central
atom?
(A) (I), (ii), (S)
(B) (I), (ii), (R)
(C) (I), (i), (P)
(D) (II), (iv), (Q)
6.
Which combination is correct for the compound having bond angle > 109º28' ?
(A) (III), (i), (P)
(B) (IV), (iii), (P)
(C) (IV), (ii), (P)
(D) (I), (ii), (S)
7.
Which of the following is true for a planar compound ?
(A) (III), (i), (P)
(B) (I), (iv), (P)
(C) (II), (iv), (R)
(D) (II), (i), (P)
lp
Ja
uh
5.
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
The hybridization of atomic orbitals of nitrogen in NO2+, NO3– and NH4+ are :
[JEE–2000(S), 1/135]
(A) sp, sp3 and sp2 respectively
(B) sp, sp2 and sp3 respectively
(C) sp2, sp and sp3 respectively
(D) sp2, sp3 and sp respectively
Sa
nk
1.
a
* Marked Questions may have more than one correct options.
2.
The number of P—O—P bonds in tricyclic metaphosphoric acid is :
(A) zero
(B) two
(C) three
[JEE–2000(S), 1/135]
(D) four
3.
Draw the molecular structures of XeF2, XeF4 and XeO2F2, indicating the location of lone pair of
electrons.
[JEE–2000(M), 3/135]
4.
The correct order of hybridisation of the central atom in the following species; NH 3, PCl5 and BCl3 is :
[JEE–2001(S), 1/135]
(A) dsp2, sp2, sp3
(B) sp3, dsp3, sp2
(C) dsp2, sp3, dsp3
(D) dsp2, sp2, dsp3
5.
The number of S–S bonds, in sulphur trioxide trimer (S3O9) is :
(A) three
(B) two
(C) one
6.
Which of the following are isoelectronic and isostructural ?
NO3–, CO32–, ClO3–, SO3
–
(A) NO3 , CO32–
(B) SO3, NO3–
(C) ClO3–, CO32–
[JEE–2001(S), 1/135]
(D) Zero
[JEE–2003(S), 3/144]
(D) CO32–, SO3 .
52
7.
Which of the following represent the given mode of hybridisation sp2–sp2–sp–sp from left to right.
[JEE–2003(S), 3/144]
(A) H2C=CH–CN
(B) HCC–CCH
(C) H2C=C=C=CH2
(D)
8.
Using VSEPR theory, draw the shape of PCl5 and BrF5.
9.
Amongst the following the acid having –O–O– bond is :
(A) H2 S2 O3
(B) H2 S2 O5
(C) H2 S2 O6
10.
Use VSEPR model to draw the structures of OSF4 and XeF4 (indicate the lone pair(s) on central atom)
and specify their geometry.
[JEE–2004(M), 2/144]
11.
Write the structure of P4O10 .
12.
The percentage of p-character in the orbitals forming P–P bonds in P4 is :
[JEE–2007, 3/162]
(A) 25
(B) 33
(C) 50
(D) 75
13.*
The nitrogen oxide(s) that contain(s) N—N bond(s) is(are) :
(A) N2O
(B) N2O3
(C) N2O4
14.
Based on VSEPR theory, the number of 90 degree F–Br–F angles in BrF5 is :
[JEE–2010, 3/163]
15.
The shape of XeO2F2 molecule is
(A) trigonal bipyramidal
(C) tetrahedral
[JEE–2012, 3/136]
[JEE–2004(S), 3/144]
(D) H2 S2 O8
a
ri
[JEE–2005(M), 1/144]
[JEE–2009, 4/160]
(D) N2O5
uh
16.
[JEE–2003(M), 2/144]
(B) square plannar
(D) see-saw
The total number of lone pairs of electrons in N2O3 is :
[JEE(Advanced) 2015, 4/168]
17.
Among the triatomic molecules/ions, BeCl2,
N2O,
O3, SCl2, ICl2–, I3– and XeF2, the total
number of linear molecules(s)/ion(s) where the hybridization of the central atom does not have
contribution from the d-orbital(s) is
[JEE(Advanced) 2015, 4/168]
[Atomic number : S = 16, Cl = 17,  = 53 and Xe = 54]
18.*
The crystalline form of borax has
(A) tetranuclear [B4O5(OH)4]2– unit
(B) all boron atoms in the same plane
(C) equal number of sp2 and sp3 hybridized boron atoms
(D) one terminal hydroxide per boron atom
19.
The total number of compounds having at least one bridging oxo group among the molecules given
below is _______ .
[JEE Advanced 2018, 3/120]
N2O3, N2O5, P4O6, P4O7, H4P2O5, H5P3O10, H2S2O3, H2S2O5
NO2+,
[JEE Advanced 2016, 4/124]
a
lp
Ja
N3–,
Sa
nk
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) OFFLINE PROBLEMS
1.
The hybridisation of the underline atom changes in :
(1) AlH3 changes to AlH4–
(2) H2O changes to H3O+
(3) NH3 changes to NH4+
(4) in all cases
[AIEEE-2002, 3/225]
2.
Bond angle of 109º 28' is found in :
(1) NH3
(2) H2O
[AIEEE-2002, 3/225]
(4) NH4+
(3) CH3+
3.
Which of the following compounds has the smallest bond angle in its molecule ? [AIEEE-2003, 3/225]
(1) SO2
(2) H2O
(3) H2S
(4) NH3
4.
The pair of species having identical shapes for molecules of both species is :
[AIEEE-2003, 3/225]
(1) CF4, SF4
(2) XeF2, CO2
(3) BF3, PCl3
(4) PF5, IF5.
5.
The maximum number of 90º angles between bond pair–bond pair of electrons is observed in :
[AIEEE-2004, 3/225]
(1) dsp3
(2) sp3d
(3) dsp2
(4) sp3d2
6.
The correct order of bond angles (smallest first) in H2S, NH3, BF3 and SiH4 is :
(1) H2S < SiH4 < NH3 < BF3
(2) NH3 < H2S < SiH4 < BF3
[AIEEE-2004, 3/225]
53
(3) H2S < NH3 < SiH4 < BF3
(4) H2S < NH3 < BF3 < SiH4
7.
Which one of the following does not have sp2 hybridized carbon?
(1) Acetone
(2) Acetic acid
(3) Acetonitrile
8.
The molecular shapes of SF4, CF4 and XeF4 are :
[AIEEE-2005, 3/225]
(1) the same with 2, 0 and 1 lone pairs of electrons on the central atom, respectively.
(2) the same with 1, 1 and 1 lone pair of electrons on the central atom, respectively.
(3) different with 0, 1 and 2 lone pairs of electrons on the central atom, respectively.
(4) different with 1, 0 and 2 lone pairs of electrons on the central atom, respectively.
9.
The number and type of bonds between two carbon atoms in calcium carbide are :
[AIEEE-2005 (3/225), 2011 (4/120)]
(1) one sigma, one pi (2) one sigma, two pi
(3) two sigma, one pi
(4) two sigma, two pi
10.
The hybridisation of orbitals of N atom in NO3–, NO2+ and NH4+ are respectively : [AIEEE-2011, 4/120]
(1) sp, sp2, sp3
(2) sp2, sp, sp3
(3) sp, sp3, sp2
(4) sp2, sp3, sp
11.
The structure of IF7 is :
(1) square pyramid
(2) trigonal bipyramid
(3) octahedral
The molecule having smallest bond angle is :
(1) NCl3
(2) AsCl3
(3) SbCl3
a
ri
[AIEEE-2011, 4/120]
(4) pentagonal bipyramid
[AIEEE-2012, 4/120]
(4) PCl3
uh
12.
[AIEEE-2004, 3/225]
(4) Acetamide
In which of the following pairs the two species are not isostructural ?
(1) CO32– and NO3–
(2) PCl4+ and SiCl4
(3) PF5 and BrF5
[AIEEE-2012, 4/120]
(4) AlF63– and SF6
14.
The species in which the N atom is in a state of sp hybridization is :
(1) NO2–
(2) NO3–
(3) NO2
[JEE(Main)-2016, 4/120]
(4) NO2+
Ja
13.
lp
JEE(MAIN) ONLINE PROBLEMS
The number and type of bonds in C22– ion in CaC2 are : [JEE(Main) 2014 Online (09-04-14), 4/120]
(1) One -bond and one -bond
(2) One -bond and two -bonds
(3) Two -bonds and two -bonds
(4) Two -bonds and one -bonds
2.
Which one of the following does not have a pyramidal shape?
[JEE(Main) 2014 Online (11-04-14), 4/120]
(1) (CH3)3N
(2) (SiH3)3N
(3) P(CH3)3
(4) P(SiH3)3
Sa
nk
a
1.
3.
The geometry of XeOF4 by by VSEPR theory is :
[JEE(Main) 2015 Online (10-04-15), 4/120]
(1) pentagonal planar (2) octahedral
(3) square pyramidal
(4) trigonal bipyramidal
4.
Which of the following compound has a P–P bond ?
[JEE(Main) 2015 Online (11-04-15), 4/120]
(1) H4P2O5
(2) (HPO3)3
(3) H4P2O6
(4) H4P2O7
5.
Choose the incorrect formula out of the four compounds for an element X below :
(1) X2O3
(2) X2Cl3
(3) X2(SO4)3
(4) XPO4
6.
The group of molecules having identical shape is :
[JEE(Main) 2016 Online (09-04-16), 4/120]
(1) PCl5, IF5, XeO2F2
(2) BF3, PCl3, XeO3
(3) ClF3, XeOF2, XeF3+ (4) SF4, XeF4, CCl4
7.
Assertion: Among the carbon allotropes, diamond is an insulator, Whereas, graphite is a good
conductor of electricity.
Reason: Hybridization of carbon in diamond and graphite are sp3 and sp2, respectively.
[JEE(Main) 2016 Online (10-04-16), 4/120]
(1) Both assertion and reason are correct, but the reason is not the correct explanation for the
assertion.
(2) Assertion is incorrect statement, but the reason is correct.
(3) Both assertion and reason are correct, and the reason is the correct explanation for the assertion.
(4) Both assertion and reason are incorrect.
54
8.
The bond angle H–X–H is the greatest in the compound : [JEE(Main) 2016 Online (10-04-16), 4/120]
(1) NH3
(2) PH3
(3) CH4
(4) H2O
9.
Identify the pair in which the geometry of the species is T-shape and square-pyramidal, respectively :
[JEE(Main) 2018 Online (15-04-18), 4/120]
–
–
–
(1) ICl2 and ICl5
(2) IO3 and IO2F2
(3) ClF3 and IO4–
(4) XeOF2 and XeOF4
10.
H N N N
In hydrogen azide (above) the bond orders of bonds (I) and (II) are :
[JEE(Main) 2018 Online (15-04-18), 4/120]
(I)
(II)
(I)
(II)
(1)
<2
>2
(2)
>2
>2
(3)
>2
<2
(4)
<2
<2
11.
In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridization are
respectively :
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 33 and 25
(2) 67 and 75
(3) 50 and 75
(4) 33 and 75
12.
The decreasing order of bond angles in BF3, NH3, PF3 and I3− is :
uh
a
ri
(II)
(1) I3− > BF3 > NH3 > PF3
(2) BF3 > I3− > PF3 > NH3
(3) BF3 > NH3 > PF3 > I3−
(4) I3− > NH3 > PF3 > BF3
The number of P–O bonds in P4O6 is :
(1) 6
(2) 9
(3) 12
[JEE(Main) 2018 Online (15-04-18), 4/120]
(4) 18
Ja
13.
(I)
In XeO3F2, the number of bond pair(s), -bond(s) and lone pair(s) on Xe atom respectively are :
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 5, 2, 0
(2) 4, 2, 2
(3) 5, 3, 0
(4) 4, 4, 0
15.
Among the oxides of nitrogen :
[JEE(Main) 2018 Online (16-04-18), 4/120]
N2O3, N2O4 and N2O5 ; the molecule(s) having nitrogen-nitrogen bond is/are :
(1) N2O3 and N2O4
(2) N2O4 and N2O5
(3) N2O3 and N2O5
(4) Only N2O5
16.
Which of the following conversions involves change in both shape and hybridisation ?
[JEE(Main) 2018 Online (16-04-18), 4/120]
(1) H2O → H3O+
(2) BF3 → BF4–
(3) CH4 → C2H6
(4) NH3 → NH+4
17.
The incorrect geometry is represented by :
[JEE(Main) 2018 Online (16-04-18), 4/120]
(1) NF3 – trigonal planar
(2) BF3 – trigonal planar
(3) AsF5 – trigonal bipyramidal
(4) H2O – bent
Sa
nk
a
lp
14.
18.
The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF 4, respectively, are :
[JEE(Main) 2019 Online (10-01-19), 4/120]
(1) sp3d2 and 1 (2) sp3d2 and 2
(3) sp3d and 1
(4) sp3d and 2
19.
Two pi and half sigma bonds are present in :
(1) N2+
(2) O2
(3) O2+
[JEE(Main) 2019 Online (10-01-19), 4/120]
(4) N2
20.
The pair that contains two P–H bonds in each of the oxoacid is:
[JEE(Main) 2019 Online (10-01-19), 4/120]
(1) H4P2O5 and H3PO3 (2) H4P2O5 and H4P2O6 (3) H3PO2 and H4P2O5 (4) H3PO3 and H3PO2
21.
The element that shows greater ability to form p-p multiple bonds, is:
[JEE(Main) 2019 Online (12-01-19), 4/120]
(1) Ge
(2) Sn
(3) C
(4) Si
55
EXERCISE - 1
PART - I
A-2.
+
In NO2+ the N has sp hybridisation; so it is linear O = N = O
In I3– there are 5 electron pairs around central iodine atom (3 lone pairs
and 2 bond pairs). The hybridisation of iodine is thus sp 3d. To have
minimum repulsions between lp-lp and lp-bp it acquires linear shape as
shown below.
a
ri
A-1.
In PCl5 there are 5 electron pairs around central phosphorus atom and all are bond
pairs. The hybridisation of phosphorus is thus sp3d. To have minimum repulsions
between bp-bp it acquires trigonal bipyramidal shape as shown below.
Cl
Cl
Cl
P
Cl
..
F
F
Ja
F
uh
Cl
In IF5 there are 6 electron pairs around central iodine atom. The hybridisation of iodine is thus sp 3d2.
6 electron pairs contain 5 bond pairs and one lone pair so it will be square pyramidal to have minimum
repulsions between lp-bp and bp-bp.
F
A-3.
Square planar
Trigonal bipyramidal
..
Sa
nk
F F
a
lp
F
A-4.
87.5º
..
Cl
87.5º
F nearly 'T' shaped.
F F
B-1.
1.
3.
⎯⎯
→ sp3
⎯⎯
→ sp3d2
2.
4.
⎯⎯
→
sp3d
56
9.
CH4 : –
⎯⎯
→ sp3d2
⎯⎯
→ sp3
⎯⎯
→ sp3
Sa
nk
15.
⎯⎯
→ sp3
a
H2O : –
8.
F7 : –
⎯⎯
→ sp3d3
10.
CCl4 : –
12.
lp
11. SiCl4 : –
13.
⎯⎯
→ sp3d
a
ri
F5 : –
F3 : –
Ja
7.
6.
14.
16.
⎯⎯
→ sp3
uh
⎯⎯
→ sp3d2
5.
⎯⎯
→ sp3
SiH4 : –
..
⎯⎯
→
N H3 :–
BrF5
sp3
⎯⎯
→ sp3d2
17.
⎯⎯
→ sp2
18.
⎯⎯
→ sp2
19.
⎯⎯
→ sp3
20.
⎯⎯
→ sp3
B-2.
True, As the s character in hybrid orbital decrease, size of hybrid orbital increases.
B-3.
(a) SF4
4 bond pair & 1 lone pair, Hybridization = sp3d Shape : see saw
(b) XeOF4
5 bond pair & 1 lone pair, Hybridization = sp3d2 Shape : Square pyramidal
57
C-1.
C-2.
In Br3– there are 5 electron pairs around central bromine atom (3
lone pairs and 2 bond pairs). The hybridisation of bromine is thus
sp3d. To have minimum repulsions between lp-lp and lp-bp it
acquires linear shape as shown below.
linear
(a) Cl, the most electronegative of the halogens in this
series, pulls shared electrons the most strongly away from
Sb, reducing electron density near Sb. The consequence
is that the lone pair exerts the strongest influence on
shape in SbCl3.
Order of C–H bond strength is C2H2 > C2H4 > C2H6 as %s character decreases in the same order.
Ja
C-3.
uh
a
ri
97.1º
98.2º
99º
(b) Phosphorus is the most electronegative of the central
atoms. Consequently, it exerts the strongest pull on
shared electrons, concentrating these electrons near P
and increasing bonding pair-bonding pair repulsions–
hence, the largest angle in PI3. Sb, the least
electronegative central atoms, has the opposite effect:
Shared electrons are attracted away from Sb, reducing
102º
100º
99º
repulsions between the Sb–I bonds.
The consequence is that the effect of the lone pair is greatest in SbI 3, which has the smallest angle.
Atomic size arguments can also be used for these species. Larger outer atoms result in larger angles;
larger central atoms result in smallest angles.
C-4.
, double bond occupies large area and has large electron density. So there is intrinsic
repulsion between P=O and P–Cl bond pairs. To minimize this repulsion bond angle decrease from
109.5º to 103.5º.
Lowest bond angle
SbH3
H2Te
PH3
Cl2O
PH3 (no hybridisation)
NF3 (sp3 hybridisation)
NF3
PF3
lp
Highest bond angle
(1)
CH4
(2)
CO2
(3)
H2O
(4)
ClO2
(5)
PF3 (sp3 hybridisation)
(6)
BF3 (sp2 hybridisation)
(7)
NH3
(8)
PCl3
Sa
nk
a
C-5.
C-6.
(1) CC < C=C < C–C
D-1.
(a) 4
D-2.
(a) P4O10,
(b) 3
(c) 3
(d) 6
(2) C–F < C–O < C–N
(e) 4
(3) HF < H–Cl < H–Br < H–I
(f) 0
:
(b) and (c) nitrogen does not have empty d-orbital.
Cl
(d) HClO3,
p-d
O
O
OH
58
D-3.
(1) +2 (6, –2) (2) +5/2(5, 5, 0, 0)
(3) +6
(4) +6 (+6, +6) (5) +6 (+6, +6)
(6) 0
(7) +5
(8) 4/3 (+ 2, +2, 0)
(9) +8
(10) –3
(11) +6
(12) +6 (+6, +6)
(13) +6
(14) +6
(15) +5
(16) +2
(17) 0
(18) –3, +5
Note: Inside the bracket ( ) answer for individual oxidation number is given.
PART - II
(C)
A-2.
(D)
A-3.
(D)
A-4.
(A)
A-5.
(B)
A-6.
(D)
B-1.
(B)
B-2.
(A)
B-3.
(D)
B-4.
(C)
B-5.
(A)
C-1.
(B)
C-2.
(D)
C-3.
(B)
C-4.
(B)
C-5.
(A)
D-1.
(C)
D-2.
(B)
D-3.
(D)
D-4.
(A)
a
ri
A-1.
PART - III
(A – p, q, r) ; (B – q, r, t) ; (C – s) ; (D – r)
3.
(A – r,s); (B – p,q,s) ; (C – q); (D – q,r)
2.
(A – r) ; (B – q) ; (C – s) ; (D – p, t)
uh
1.
EXERCISE - 2
PART - I
(D)
2.
(C)
3.
6.
(B)
7.
(A)
8.
11.
(B)
12.
(B)
16.
(B)
17.
(A)
(C)
4.
Ja
1.
(B)
5.
(A)
9.
(C)
10.
(B)
13.
(A)
14.
(A)
15.
(A)
18.
(D)
6 (a, d, e, h, i, k)
3.
16
10
6.
6
4.
(AC)
5.
(BCD)
4.*
(ABC)
5.
(A)
lp
(D)
PART - II
5 (a, b, c, d, h)
2.
4.
4
5.
Sa
nk
a
1.
PART - III
1.
(AC)
2.
(CD)
3.
(ABC)
6.
(BCD)
7.
(BCD)
8.
(BD)
1.
(C)
2.
(D)
6.
(B)
7.
(C)
PART - IV
3.
(D)
EXERCISE - 3
PART - I
1.
(B)
2.
(C)
3.
According to VSEPR theory
Number of electron pairs = 5,
Number of bond pairs = 2,
So, Number of lone pairs = 3.
XeF2
59
Thus XeF2 is linear with 3 lone pairs occupying 3 equatorial positions of
trigonal bipyramidal so as to minimize the repulsions.
(linear)
Number of electron pairs = 6,
Number of bond pairs = 4,
So, Number of lone pairs = 2.
Thus XeF4 is linear with 2 lone pairs occupying 2 axial positions of
octahedral pyramidal so as to minimize the repulsions.
(D)
6.
(A)
7.
XeO2F2
(see-saw)
4.
(B)
(A)
8.
There are 5 electron pairs and all are bonds pairs in PCl 5. So to have the minimum repulsions between
bond pairs it acquires trigonal bipyramidal shape. In BrF 5, there are 6 electrons pairs out of which one
lone pair and rest all are bond pairs. So to have the minimum repulsions between bond pairs and lone
pairs it acquires square pyramidal shape.
,
PCl5 (trigonal bipyramidal),
9.
(D)
BrF5 (square pyramidal)
lp
10.
Ja
uh
5.
(square planar)
a
ri
Number of electron pairs (including super electron pairs) = 5,
Number of bond pairs = 4,
So, Number of lone pairs = 1.
Thus XeO2F2 is see-saw with 1 lone pairs occupying one equatorial
position and two double bonds occupying other two equatorial positions
of trigonal bipyramidal so as to minimize the repulsions.
XeF4
a
According to VSEPR theory two lone pairs out of six electron pairs are
trans to each other to have minimum repulsion. The shape of XeF 4 is
square planar and geometry is octahedral with sp3d2 hybridisation. The
molecule looks like :
Sa
nk
In OSF4, there are five electron pairs and all are bond pairs. So geometry
is trigonal bipyramidal. As double bond creates more repulsion than
singles bond, the double bond acquires one of equatorial position of
trigonal bipyramidal to have minimum repulsions.
The structure looks like:
11.
Structure of P4O10.
12.
(D)
13.*
(ABC)
14.
0 or 8
15.
(D)
16.
8
60
17.
4
18.*
(ACD)
19.
6
PART - II
JEE(MAIN) OFFLINE PROBLEMS
1.
(1)
2.
(4)
3.
(3)
4.
(2)
5.
(4)
6.
(3)
7.
(3)
8.
(4)
9.
(2)
10.
(2)
11.
(4)
12.
(3)
13.
(3)
14.
(4)
(3)
5.
(2)
(2)
2.
(2)
3.
(3)
4.
6.
(3)
7.
(1)
8.
(3)
9.
11.
(2)
12.
(1)
13.
(3)
14.
16.
(2)
17.
(1)
18.
(1)
19.
21.
(3)
(4)
10.
(1)
(3)
15.
(1)
(1)
20.
(3)
Sa
nk
a
lp
Ja
uh
1.
a
ri
JEE(MAIN) ONLINE PROBLEMS
61
Section (A) : Back bonding
Back bonding generally takes place when out of two bonded atoms one of the atom has vacant orbitals
(generally this atom is from second or third period) and the other bonded atom is having some nonbonded electron pair(generally this atom is from the second period). Back bonding increases the bond
uh

a
ri
strength and decreases the bond length. For example, in BF3 the boron atom completes its octet by
accepting two 2p-electrons of fluorine into 2p empty orbital.
Decrease in B–F bond length is due to delocalised p–p bonding between filled p-orbital of F atom
and vacant p-orbital of B atom.
Ja
The extent of back bonding is much larger if the orbitals involved in the back bonding are of same size,
for example the extent of back bonding in boron trihalides is as follows : BF3 > BCl3 > BBr3
There is p-p back bonding in boron trihalide. The extent of back bonding decreases from BF3 to BI3
because of increasing size of p-orbitals participating in back bonding that is from 2p(in F) to 4p(in Br).

The extent of back bonding decreases if the atom having vacant orbitals is also having some nonbonded electron pairs on it. So among the atoms of third period the extent of back bonding follows the
lp
order :
Si > P > S > Cl
1
Lewis acid character
The extent of p-p overlapping 
Ex-1.
Compare B-F bond length in BF3 and [BF4]–.
Sol.
BF3 + KF ⎯⎯→ K+ [BF4]–
Sa
nk
a

[No back bonding]
B–F Bond length (order) [BF4]– > BF3
Ex-2.
Sol.
Explain all boron trihalides are lewis acids also explain their order.
Boron trihalides are electron deficient molecules therefore act as a Lewis acids
Cl
But in BCl3 
P-P back bonding 2p of B & 3p of Cl less effective
B
Cl
Cl
tendency to accept L.P in BCl3 > BF3. Lewis acid strength BF3 < BCl3 < BBr3.
62
Ex-3.
Trisilylamine is a planar molecular and does not act as a lewis base while trimethyl is a pyramidal and
act as lewis base. Explain it :
SiH3
:
N
Sol.
N
|
CH3 CH CH3
3
S.N = 3
SiH3
:
SiH3
Planar
N (CH3)3 pyramidal
Localised .P. S.No. = 4
delocalised .P.
S.N. = 3 sp2
sp3
Ex-4.
Silyl isocyanate (SiH3NCO) is linear but methyl isocyanate (CH 3NCO) is bent explain !
H
H
H Si N C O
H C N C O
H
H
Vacant
No vacant
d-orbitals
orbitals
.P. of N can be delocalised
(Back bonding)
S.N of N = 2 (sp)
S.N = 3 (sp2)
p – d back bonding.
Ja
uh
Sol.
a
ri
* But in a similar compound N(PH3)3 the shape is found to be pyramidal, so N atom must be sp3
hybridised due to much less extent of back bonding into the vacant orbitals of P.
Section (B) : Electron deficient bonding
•
Definition of Lewis acid and base :
A lewis acid is an electron pair acceptor and lewis base is electron pair donor.
Ex.
NH3 + H+ = [NH3 ⎯⎯
→ H]+
(Base)
lp
••
( Acid)
a
Proton is a lewis acid and Ammonia is lewis base since the lone pair of electron of the nitrogen atom
can be donated to a proton.
Sa
nk
Coordinate Bond :
A bond in which sharing of a pair of electron between two atoms, where both electrons originate from
one atom and none from other is called coordinate bond.
H
H
H
H N + [H]+ = H N
H
H
F
H
F
H
H N +B F =H N
H
F
H
B F
F
Electron deficient bonding :
There are many compounds in which some electron deficient bonds are present apart from normal
covalent bonds or coordinate bonds which are 2c-2e bonds (two centre two electron bonds). These
electron deficient bonds have less number of electrons than the expected such as three centre-two
electron bonds(3c-2e) present in diborane B2H6, Al2(CH3)6, BeH2(s), bridging metal carbonyls.
B2H6 (diborane)  It is having (3centre – 2electron) bond / (banana bond) / (electron deficient bond)
63
Compound
Bridge
Hybridisation
Planar/Non
Planar
3c – 2e–
sp3–s–sp3
Non planar
Structure
H
H
H
B
B
H
H
H
3C – 2e– bond
B2H6
B2H6 have 4 2c-2e bonds and 2 3c-2e bonds. Bridging
bonds have larger bond length than terminal bonds.
Angle between terminal bonds is more than angle
between bridging bonds if all 4 terminal bonds are in
one plane then bridging bonds are in perpendicular
plane.
2
H
y
H x
H
1
B
B
Ht
H
Hb
Al2H6
3c – 2e–
sp3–s–sp3
a
ri
3C – 2e– bond
H
Non planar
H
H
Al
Al
H
H
H
H
3c – 2e–
3c – 2e–
sp2–s–sp2
sp3–s–sp3
H–Be
planar
Be–H
H
uh
Be2H4
H
H
Be
Non planar
Be
Be
H
Ja
(BeH2)n
H
All bonds except the extreme ends are 3C – 2e– bonds.
H H H
C
Al
Al
C
sp3–sp3–sp3
a
3c – 2e–
Sa
nk
Al2(CH3)6
lp
H3C
Al2(C6H5)6
3c – 2e–
sp3–sp2–sp3
H3C
CH3
H HH
H H H
Non planar
H3C
CH3
C
Al
Al
H3C
H5C6
CH3
C
H
H
H
C6H5
C6H5
Al
H5C6
Non planar
CH3
Al
C6H5
C6H5
H
C
HC
CH
CH
HC
C
Al
Al
64
Compound
Bridge
Hybridisation
Planar/
Non Planar
Structure
Cl
Cl
Al
Al2Cl6
3c – 4e–
sp3–sp3–sp3
Non Planar
Cl
Al
Cl
Cl (Driving force is to
complete the octet)
Cl
Br
3c – 4e–
sp3–sp3–sp3
Non Planar
Br
Br
Br
Al
Al
Br
Br
uh
Al2Br6
a
ri
Only covalent bond, no electron deficient
bond is present.
3c – 4e–
sp2–sp3–sp2
sp3–sp3–sp3
Cl
Be–Cl
Cl
Dimeric form
Cl–Be
Planar
Cl
Be
Be
Non Planar
Cl
Be
Cl
Cl
Polymeric form
Sa
nk
a
(BeCl2)n
(solid)
3c – 4e–
lp
Be2Cl4
Ja
Dimeric form
l2CI6
SnCl2(Solid)
3c –
4e–
3c – 4e–
sp3d2–sp3–
Cl
sp3d2
sp3–sp3–sp3
Planar
Cl
Cl
l
Cl
I
Cl
Cl
Dimeric form
Sn
Non Planar
Cl
Sn
Sn
(Driving force is to
complete the octet)
Cl Cl Cl
Cl Cl
Polymeric form
65
Section (C) : H-bonding & intermolecular force of attraction.
Hydrogen Bond :
Discovered by Huggius, Latimer and Rodbush.
When hydrogen is bonded to strongly electronegative element 'X'. The electron pair shared between
the two atoms moves far away from hydrogen atom. As a result the hydrogen atom becomes highly
electropositive with respect to the other atom 'X'. Since there is displacement of electrons towards X,
the hydrogen acquires fractional positive charge (+) while X' attain fractional negative charge (–). This
results in the formation of a polar molecule having electrostatic force of attraction which can be
representes as :
H+ – X– – – – H+ – X– – – – H+ – X –
(iv)
a
ri
(iii)
Main condition for Hydrogen bonding :
Hydrogen should be covalently bonded with high electronegative element like F, O, N.
Atomic size of electronegative element should be small
Decreasing order of atomic size is :
N > O > F
Decreasing order of electronegatively is :
F (4.0) > O (3.5) > N (3.0)
1
Strength of Hydrogen bond  Electronegativity of element 
atomic size of element
Hydrogen bonding occurs is HCN, due to (–CN) triple bond (sp hybridisation), electronegativities of
carbon and nitrogen increases.
H–CN– ....... +H–CN– ....... +H–CN
uh
(i)
(ii)
TYPES OF H-BONDS :
Intermolecular
Hetero Intermolecular
Intermolecular Hydrogen bond :
Hydrogen bond formation between two or more molecules of either the same or different compounds
known as Inter molecular Hydrogen bonding.
(i)
Homo intermolecular : Hydrogen bond between molecules of same compounds.
F–
+
+
+
H
H
H
+
+
+
+
H
H
H
H
–
–
–
Eg.
O
O
O
–
–
+
+
+
+
F
F
H
H
H
H
Sa
nk
a
(A)
Intramolecular
lp
Homo Intermolecular
Ja
Types of Hydrogen Bonding
Eg.
Boric acid
66
In the solid state, the B(OH)3 units are hydrogen bonded together in to two dimensional sheets
with almost hexagonal symmetry. The layered are quite a large distance apart (3.18 Å).
Hetro intermolecular : Hydrogen bond between molecules of different compounds.
Eg. alcohol, water
O – H+  –O – H+  –O – H+  –O – H+ 
(ii)
R
alcohol
R
alcohol
H
water
Intra molecular Hydrogen bond : It takes place within the molecule.
(i) Hydrogen bonded with electronegative elements of a functional group, form Hydrogen bond with
another electronegative element present on nearest position on the same molecule.
(ii) This type of Hydrogen bond is mostly occured in organic compounds.
(iii) It result in ring formation (Chelation).
a
ri
(B)
H
water
uh
Necessary conditions for the formation of intramolecular hydrogen-bonding :
(a) the ring formed as a result of hydrogen bonding should be planar.
(b) a 5- or 6-membered ring should be formed.
(c) interacting atoms should be placed in such a way that there is minimum strain during the ring
closure.
Nickel dimethyl glyoximate (a chelate)
Ja
Ex-1.
Extra stability of the complex is because of intramolecular hydrogen bonding in addition to the chelating
effect.
Ex-2.
O
S
O

lp
O
Persulphate ion (HSO5–)
H
a
O
O
K1 of peroxomono sulphuric acid (i.e., caros acid) is greater than K2. After the loss of one hydrogen, the
persulphate ion gets stabilised due to intramolecular hydrogen bonding and thus the removal of second
hydrogen becomes difficult.
Sa
nk

Ex-3.
Aceto acetic ester (enolic form)
The intramolecular hydrogen bonding attributes the stability of enolic form of aceto acetic ester.
H
O
H
O
O
Ex-4.
N
H+
O–
O
o-nitrophenol
C
O
O+
H–
Salicylaldehyde
C
O
H+
F–
o-fluorophenol
O
O
H
2,6-dihydroxyl benzoate
67
Strength of H-bond :
The strength of H-bond is usually very low (5-10 kJ/mol) but in some cases this value may be as high
as 50 kJ/mol. The strongest H-bonds are formed by F atoms. Deuterium is more electropositive than H,
therefore it also form stronger bonds. The strength of the H-bond can be compared by the relative bond
energies and the geometry of the various compounds as given below.
F– + HF ⎯⎯
→ [FHF]– ; H = – 161 ± 8 kJ mol–1
(CH3)2CO + HF ⎯⎯
→ (CH3)2CO ...... HF ; H = – 46 kJ mol–1
H2O + HOH ⎯⎯
→ H2O ...... HOH (ice) ; H = – 25 kJ mol–1

a
ri
HCN + HCN ⎯⎯
→ HCN ...... HCN ; H = – 12 kJ mol–1
The magnitude of H-bonding depends on the physical state of the compounds. H-bonding is maximum
in the solid state and minimum in the gaseous state. Thus hydrogen bonds have strong influence on the
structure and properties of the compounds.
Order of H-bond strength
>
O
H - - - - - - :O
>
N
H - - - - - - :N
>
N
H - - - - - - :O
uh
Hydrogen bond is weaker than covalent bond.
The hydrogen bonds in HF link the F atom of one molecule with the H-atom of another molecule, thus
forming a zig-zag chain (HF)n in both the solid and also in the liquid.
lp
Ja
Some hydrogen bonding also occurs in the gas, which consists of a mixture of cyclic (HF)6 polymers,
dimeric (HF)2, and monomeric HF.
Very strong hydrogen bonding occurs in the alkali metal hydrogen fluorides of formula M[HF 2];
in KHF2, for example, an X-ray diffraction study together with a neutrons diffraction study shows that
there is a liner symmetrical anion having an over all, F–H–F distance of 2.26 Å, which may be
compared with the H–F bond length of 0.92Å in hydrogen fluoride monomer.
F– – –x – H+ ——
F–
y
Bond length x = y
a
It is found urea and phosphoric Acid form H bond with each other
Sa
nk
NH2CONH2 + H3PO4 →
2 types of H–bond
B.L.OH  x = 122 pm ;
(i)
B.L. N–H  y = 100 pm
Effect of Hydrogen bond on physical properties :
Solubility :
(A) Inter molecular Hydrogen bonding
(a) Few organic compounds (Non-polar) are soluble in water (Polar solvent) due to Hydrogen bonding.
e.g alcohol, acetic acid etc. are soluble in water.
–
+
–
+
 O – H  O – H 
R
H
Other examples- Glucose, Fructure etc, dissolve in water.
(b) Ketone, ether, alkane etc. are insoluble (no Hydrogen bond)
But C2H2 is highly soluble in acetone due to H bonding
68
a
ri
None of the above 2 has H-bonding individually.
But C2H2 is not soluble in water because water molecules already so much associated through H-bond
that it is almost impossible for C2H2 molecules to break that association.
So, it is not soluble in H2O (l)
(c) Solubility order- CH3OCH3 < CH3OH
Primary amine > secondary amine > tertiary amine
(B) Intra molecular Hydrogen bonding :
(a) It decreases solubility as it form chelate by Hydrogen bonding, so hydrogen is not free for other
molecule.
(b) It can’t form H-bond with water molecule so can’t dissolves.
O–
H
H+
C
O
Salicylaldehyde
C
C
O
+
–
O–H ···· O
H
Ja
H
+
O–H ······
uh
(C) Inter molecular Hydrogen bond
+
–
O–H ······ O=C–H
p-hydroxy benzaldehyde
It can form hydrogen bond with water molecule so it can dissolved
Viscosity:
Hydrogen bond associates molecules together, so viscosity increases
lp
(ii)
<
a
CH3OH
Sa
nk
H2O
water
(iii)
<
CH2–OH
CH3–OH
CH2–OH
<
CH3–OH <
alcohol
CH–OH
CH2–OH
CH3–O–CH3
ether
Melting point and boiling point
(a)
Due to intermolecular hydrogen bond melting point & boiling poin of compounds increases.
H2O > CH3OH > CH3–O–CH3
(b)
Trihydric alcohol > dihydric alcohol > monohydic alcohol
Monocarboxylic acid form stronger hydrogen bond than alcohol of comparable molecular
weight. Therefore boiling point of carboxylic acid is higher than alcohol.
(c)
Decreasing order of melting point & boiling point isomer amines1°-amine > 2°-amine > 3°amine
R
R – NH2 > R – NH – R > R – N – R (no hydrogen with nitrogen atom)
69
(d)
uh
a
ri
(iv)
Boiling points of VA, VIA, VIIA hydrides decreases on decreasing molecular weights.
VA
VIA
VIIA
NH3
H2O
HF
boiling point
HF > HI > HBr > HCl
PH3
H2S
HCl
H2O > TeH2 > SeH2 > H2S
AsH3
SeH2
HBr
NH3 > SbH3 > AsH3 > PH3
SbH3
TeH2
HI
But sudden increase in boiling point of NH3, H2O and HF is due to hydrogen bonding
H2O > HF > NH3
(e)
Intramolecular hydrogen bonding gives rise to ring formation, so the force of attraction among
these molecules are vander waal's force. So, melting point and boiling point are low.
Molecular weight: Molecular weight of CH3COOH is double of its molecular formula, due to dimer
formation occur by hydrogen bonding.
O–......+H–O
C–R
R–C
O–H+......–O
The crystal structures of NaHCO3 and KHCO3 both show hydrogen bonding, but are different.
(a) In NaHCO3, the HCO3– ions are linked into an infinite chain.
(b) in KHCO3, RbHCO3, CsHCO3, HCO3– forms a dimeric anion.
Solubility in water NaHCO3 < KHCO3 < RbHCO3 < CsHCO3
(b)
H−O− − −H−O
|
|
H
H
Sa
nk
º
EH
=0
+
/H
lp
Ja
Physical state: H2O is liquid while H2S is gas.
(a) Water and Ice : Both have hydrogen bonding even then density of ice is less than water. Volume of
ice is more because of open cage like crystal structure, from by association of water molecules with the
help of hydrogen bond.
H2O becomes solid (Ice) due to four hydrogen bond among water molecule are formed in tetrahedral
manner.
H
H 3 H
H
1
O 2
O
O
O
4
H
H
H
H
a
(v)
2
(c)
D−O− − −D−O
|
|
D
D
º
ED
= −0.01V
+
/D
2
D is more electro positive than H
So, it is capable of forming slightly more H bond than H that why D 2O molecule are more
associated and its boiling point is more than the boiling point of H 2O liquid.
Ice floats on water
H
H
O
–
O
H
H +
+
H
– O H
+
–
+
H
O
–
O H
H
H
70
a
ri
One H2O is tetrahedrally bonded with 4H2O molecules.
Intermolecular specing increases
So, density of ice decreases
uh
No. of water molecule attached to 1 H2O molecule = 4
No. of H bonds in a molecule = 2
lp
Ja
(d) D2O(s)
Density of D2O(s) > density of H2O(l)
So, it sinks in H2O(l) but floats on D2O(l)
Due to open structure, ice is capable of forming Clatherates. (cage like compounds)
a
Xe will be captured by ice.
Xe(g) because of its bigger atomic size is trapped in the cages formed by H 2O molecules in the
structures of ice. He cannot be trapped due to smaller size. Such compounds are called clatherates
compounds.
(e) Density of water max at 4ºC
slightly
H2O(s) ⎯⎯⎯⎯→ H2O(l)
Sa
nk
warm
ice
T < 4ºC
density decreases
>0ºC
In this temperature region some of the ice melts and hence some H 2O molecule go into the cages of
remaining ice structure.
So, volume decreases density increases becoming max at 4ºC but beyond this temp thermal effects
become dominating volume increases then density decreases.
(f) An interesting hydrate of Hydronium ion with the formula H3O+(H2O)20 in which H2O molecules are
hydrogen bonded. Some lower species like H5 O2+ have also been observed where the 2 water
molecules are linked three H bonds.
Similarly H3 O2− ions are also observed which is actually hydrate OH– ion.
71
Base strength
CH3NH2, (CH3)2NH, (CH3)3N, form hydrogen bond with water. So, less hydrolysis i.e. it gives OH – ions.
While (CH3)4N+ OH– (ammonium compound) will give OH– ion in large amount due to no hydrogen
bonding.
H
H
CH3–N+ – H + OH–
CH3–N.......H–O
H
+
H
CH3–N–CH3
H
Ex.
H
a
ri
H
(Ammonium compound)
No Hydrogen atom bonded directly with Nitrogen
atom so no hydrogen bonding occurs.
Which is a stronger base why?
(a)
Trimethyl ammonium hydroxide
(b)

+
uh
(vi)
−
Tetramethyl ammonium hydroxide N(CH3 )4 OH
Ja
b is more basic
(CH3)3N: - - - - H – O
Due to nitramolecular H bonding, release of OH – ion become difficult  it is a weaker base.
lp
Effect of intramolecular H-bonding
(i)
Strength of acid :
(a) The formation of intramolecular H-bonding in the conjugate bas of an acid gives extra stability to
conjugate base and hence acid strength increases eg. Salicyclic acid is stronger than benzoic acid and
2, 6-dihydroxy benzoic acid > salicyclic acid.
O
O–
O
O
H
C
C
H
Sa
nk
a
O
O
+ H+
Conjugate base
–1/2
H
O–1/2
C
H
O
O
O
+
+ H
2,6-dihydroxyl benzoate
ion
+
(b) C2H5SH is more acidic than C2H
5OH. In C2H5OH, hydrogen bonds forms, so H is not free
(c) HF is weaker acid than HI, due to hydrogen bond in H–F, H+ is not free.
(ii)
Stability of chloral hydrate :
If two or more OH group on the same atom are present it will be unstable, but chloral hydrate is stable
(due to Hydrogen bonding)
H
O
H
Cl
F
H
F
F
O
O
C F
F
C C
Cl C C
H
H
F C C
F
O
F
O
O
H
Cl
H
H
Error! Not a valid link. F
Chloral hydrate
72
O
H
Ph
Ph
O
O
H
C
O
C
O
C
O
H
H
O
Maleic acid (cis) is stronger acid than fumaric acid (trans).
O
O
H
H
C
C
C
OH
C
O–
C
OH
C
O
C
H+
H
H+
a
ri
H
C
O
O
Stable conjugate base of
maleic acid
H
COOH
C
(Maleic acid)
uh
(iii)
O
C
HOOC
N
H
C=O
H O
Ja
Fumaric (No-intramolecular
hydrogen bonding)
Note : The relative strength of various bonds is as follows
Ionic bond > covalent bond > Metallic bond > Hydrogen bond > Vander waal's bond.
Sol.
C2H2 is not soluble in H2O but it is highly soluble in acetone.
lp
Ex-5.
a
In hybridisation as %s character increase electronegativity increase hence C2H2 forms H–bonds with
O–atom of acetone and get dissolved. But H 2O molecules are so much associated that it is not possible
for C2H2 molecules to break that association, hence C 2 H2 is not soluble in H2O.
Why crystalline sodium peroxide is highly hygroscopic in nature.
Na2O2 forms stable hydrates on account of H-bonding.
− − − O22− − − − (H2O)8 − − − O22− − − − (H2O)8 − − −
Ex-7.
Explain that tetramethyl ammonium hydroxide is a stronger base than that of trimethyl ammonium
hydroxide.
Sa
nk
Ex-6.
Sol.
Sol..
+
CH3


CH3
|
 –

|
CH3 − N → CH3  O H
CH3 − N → H    O − H


|
|


CH
CH3
3


In the trimethyl compound the O–H group is hydrogen bonded to Me3NH group and this makes it more
difficult for the OH group to ionize and hence it is a weak base.
In the tetramethyl compound, hydrogen bonding cannot occur, so the OH – group ionizes easily and thus
it is a much stronger base.
73
Intermolecular forces (Van der Waal’s Forces) :
(a)
Intermolecular attractions hold two or more molecules together. These are weakest chemical forces and
can be of following types.
(a) Ion-dipole attraction
(b) Dipole-dipole attraction (Keesom forces)
(c) Ion-induced dipole attraction (Debye force) (d) Dipole-induced dipole attraction
(e) Instantaneous dipole- Instantaneous induced dipole attraction (Dispersion force or London forces)
Strength of van der Waal’s forces a > b > c > d > e
Ion-dipole attraction : Exists between an ion and a polar molecule. Its strength depends on (i) size of
ion (ii) charge on the ion (iii) dipole moment of the polar molecule. It is thought to be directional. Iondipole forces are important in solutions of ionic compounds in polar solvents where solvated species
such as Na(OH2)x+ and F(H2O)y– (for solution of NaF in H2O) are found. Hence this force is responsible
for hydration.
+
H

+
Na
O
+
H
Dipole-dipole attraction : This is electrostatic attractions between the oppositively charged ends of
permanent dipoles. Exists between polar molecules and due to this force gas can be liquefied.



H
Cl
H
Cl–
–
+
Ion-induced dipole attraction : Exists between ion and non-polar molecules (e.g., an atom of a noble
gas such as Xenon).


Na+ Cl
Cl
Dipole-induced dipole attraction : Exists between polar and non-polar molecules.
–
(d)
uh
(c)
+
+
Ja
(b)
Head to tail arrangement of dipoles
+
H
–

Cl
+

–
Cl
Cl
(non-polar)
(polar)
Antiparallel arrangement of dipoles
lp
Instantaneous dipole- Instantaneous induced dipole attraction :
Exists among the non-polar molecules like H2, O2, Cl2 etc. in solid or liquid states. Even in atoms in
molecules which have no permanent dipole, instantaneous dipoles will arise as a result of momentary
unbalances in electron distribution.
Sa
nk
a
(e)
a
ri
–
London forces are extremely short range in action and the weakest of all attractive forces. The London
forces increase rapidly with molecular weight, or more properly, with the molecular volume and the
number of polarizable electrons.
e.g.   
 ,
Cl
Cl Cl
Cl
–
+
–
+
74
Note : Fluoro carbon have usually low boiling points because tightly held electrons in the fluorine atoms have
a small polarizability.

Strength of vander waal force  molecular mass.

van der Waal’s force  boiling point.
Ex-9
Sol.
Arrange the inert gases, according to their increasing order of boiling points
He < Ne < Ar < Kr < Xe
(boiling point)
Because strength of van der Waal’s force increases down the group with increase in molecular mass.
a
ri
Sol.
Give the order of boiling point of following
Cl2, HCl
Cl2–Cl2
<
HCl–HCl
(boiling point)
dispersion force
dipole-dipole attraction
As dipole-dipole attraction is stronger than dispersion force.
Ja
 Marked questions are recommended for Revision.
uh
Ex-8.
PART - I : SUBJECTIVE QUESTIONS
Section (A) : Back bonding
The B–F bond length in Me3N.BF3 is 1.35 Å, much longer than 1.30 Å in BF3. Explain.
lp
A-1.
A-2. Explain why SiH3NCO is linear (except H atoms) but CH3 NCO is non linear.
Draw the structure : Identify the type (p–p, p–d) of bonds and number of these bonds in the
following molecule :
(i) SO3
(ii) H3PO4
(iii) N2
(iv) HClO4
a
A-3.
Section (B) : Electron deficient bonding
Sa
nk
B-1. Explain why
(i) NH3 is better e– pair donor than PH3 .
(ii) NH3 is a better base than CH3CN?
B-2.
Why BCl3 and SiF4 act as Lewis acids ? Explain.
B-3. BF3 exists but BH3 does not. explain Why.
B-4.
Which orbitals are involved in banana bonding in Al2(CH3)6.
Section (C) : H-bonding & intermolecular force of attraction.
C-1. In which of the following the hydrogen bonding is strongest. Explain briefly ?
(a) O – H – – – S ()
(b) S – H – – – O ()
(c) F – H – – – F– (s)
(d) F – H – – – O ()
C-2.
Why D2O has higher viscosity than H2O ?
C-3.
Why glucose, fructose, sucrose etc. are soluble in water though they are covalent compounds ?
C-4. Ethanol has higher boiling point than diethyl ether. Why ?
75
PART - II : ONLY ONE OPTION CORRECT TYPE
Section (A) : Back bonding
For BF3 molecule which of the following is true ?
(A) B-atom is sp2 hybridised.
(B) There is a p–p back bonding in this molecule.
(C) Observed B–F bond length is found to be less than the expected bond length.
(D) All of these
A-2.
Which of the following statements regarding the structure of SOCl2 is not correct ?
(A) The sulphur is sp3 hybridised and it has a tetrahedral shape.
(B) The sulphur is sp3 hybridised and it has a trigonal pyramid shape.
(C) The oxygen-sulphur bond is p-d bond.
(D) It contain one lone pair of electrons in the sp3 hybrid orbital of sulphur.
a
ri
A-1.
A-3. Respective order of strength of back-bonding and Lewis acidic strength in boron trihalides is :
(A) BF3 < BCl3 < BBr3 and BF3 < BCl3 < BBr3
(B) BF3 > BCl3 > BBr3 and BF3 > BCl3 > BBr3
(C) BF3 > BCl3 > BBr3 and BF3 < BCl3 < BBr3
(D) BF3 < BCl3 < BBr3 and BF3 > BCl3 > BBr3
Which of the following contains a coordinate covalent bond
(A) HNO3
(B) BaCl2
(C) HCl
B-2. Bonds present in CuSO4. 5H2O(s) is
(A) Electrovalent and covalent
(C) Electrovalent, covalent and coordinate
B-3.
Electron deficient molecule among the following is :
(A) I2Cl6
(B) B2H6
(C) Al2Cl6
(D) All of these
For B2H6
S1 : Each boron is sp3 hybridised
S2 : four terminal 'H' & two 'B' atom are in same plane but two bridge hydrogen in different plane.
S3 : It has 4  bond & 2 bridge bond
S4 : 8  bonds are present in it
(A) T T F F
(B) T T T F
(C) F F T F
(D) F T F T
lp
B-4.
(D) H2O
(B) Electrovalent and coordinate
(D) Covalent and coordinate
Ja
B-1.
uh
Section (B) : Electron deficient bonding
(B) Boron atom is in ground state
(D) There are two, three centre two electron bonds
Sa
nk
a
B-5. Which is not true about B2H6
(A) Both ‘B’ atoms are sp3 hybridised
(C) Two hydrogens occupy special positions
Section (C) : H-bonding & intermolecular force of attraction.
C-1.
Which of the following is not correctly matched with respect to the intermolecular forces existing
amongst the molecules (Hydrogen bonding is not taken as dipole-dipole attraction) ?
(A) Benzene – London dispresion forces
(B) Orthophosphoric acid – London dispresion force, hydrogen bonding.
(C) Hydrochloric acid – London dispresion force, dipole-dipole attraction
(D) Iodine monochloride – London dispersion force
C-2. Which of the following factor is responsible for van der Waals forces ?
(A) Instantaneous dipole-induced dipole interaction.
(B) Dipole-induced dipole interaction and ion-induced dipole interaction.
(C) Dipole-dipole interaction and ion-induced dipole interaction.
(D) All of these.
C-3.
Which of the following bonds/forces is weakest ?
(A) Covalent bond
(B) Ionic bond
(C) Hydrogen bond
(D) London force
C-4. In which of the following compound, intra-molecular H-bonding is not observed :
(A) o-hydroxy benzyaldehyde
(B) o-nitrophenol
(C) Chloral hydrate
(D) Boric acid
76
C-5.
Consider the following sets of H-bonds
P:
–O–H-----:
Q:
|
– O – H - - - - - : O—
..
R:
–
S:
|
H - - - - -: O —
H-----:
The correct order of H-bond strengths is :
(A) Q > P > S > R
(C) R > S > P > Q
(B) R > Q > S > P
(D) P > Q > R > S
C-7. The correct order of boiling point is :
(A) H2O < H2S < H2Se < H2Te
(C) H2O > H2S > H2Se > H2Te
a
ri
C-6. Amongst NH3, PH3, AsH3 and SbH3 the one with highest boiling point is :
(A) NH3 because of lower molecular weight
(B) SbH3 because of higher molecular weight
(C) PH3 because of H-bonding
(D) AsH3 because of lower molecular weight
(B) H2O > H2Se > H2Te > H2S
(D) H2O > H2Te > H2Se > H2S
(D) HNO3
uh
C-8. Which of the following compounds has the highest boiling point
(A) HCl
(B) HBr
(C) H2SO4
PART - III : MATCH THE COLUMN
(p)
(q)
(r)
(s)
Ja
Column-II
Strength of hydrogen bonding
Dipole moment
Boiling point
Bond energy
Match the column:
Column-I
(A) Liquid bromine
(B) Solid hydrogen fluoride
(C) Solution of sodium fluoride in water
(D) Liquid methylamine
(E) Noble gas clathrate in ice.
Sa
nk
3.
Match the following :
Column-I
(A)
HCl < HF
(B)
PH3 < NH3
(C)
H2O < D2O
(D)
F2 < Cl2
Column-II
sp3 hybridization
p–p back bond
p–d back bond
3c–2e bond
lp
2.
Match the following :
Column-I
(A)
BF3
(p)
(B)
(SiH3)3N
(q)
(C)
B2H6
(r)
(D)
SiO2
(s)
a
1.
Column-II
(p) Hydrogen bond
(q) Ion-dipole force
(r) Dispersion force.
(s) Dipole induced dipole interaction.
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
1.
2.
In which of the following compounds B–F bond length is shortest ?
(A) BF4–
(B) BF3  NH3
(C) BF3
(D) BF3  N(CH3)3
Which of the following statement is false for trisilylamine ?
(A) Three sp2 orbitals are used for  bonding, giving a plane triangular structure.
(B) The lone pair of electrons occupy a p-orbital at right angles to the plane triangle and this overlaps
with empty p–orbitals on each of the three silicon atoms resulting in  bonding.
(C) The N–Si bond length is shorter than the expected N–Si bond length.
(D) It is a weaker Lewis base than trimethyl amine.
77
3.
In which of the following molecules/species all following characteristics are found ?
(a) Tetrahedral hybridisation
(b) Hybridisation can be considered to have taken place with the help of empty orbital(s).
(c) All bond lengths are identical i.e. all A–B bond lengths are identical.
(A) B2H6
(B) Al2Cl6
(C) BeCl2 (g)
(D) BF4–
4.
H-bonding is maximum in
(A) C6H5OH
(B) C6H5COOH
5.
(C) CH3CH2OH
(D) CH3COCH3
Which one of the following does not have intermolecular H-bonding ?
(A) H2O
(B) o-nitro phenol
(C) HF
(D) CH3COOH
Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is
false.
S1 : HF boils at a higher temperature than HCl.
S2 : HBr boils at lower temperature than HI.
S3 : Bond length of N2 is less than N2+.
S4 : F2 has higher boiling point than Cl2.
(A) T F T T
(B) T T F F
(C) T T T F
(D) T T T T
7.
Select the correct statement for the sulphuric acid.
(I) It has high boiling point and viscosity.
(II) There are two types of bond lengths in its bivalent anion.
(III) p-d bonding between sulphur and oxygen is observed.
(IV) Sulphur has the same hybridisation that is of boron in diborane.
(A) II and III only
(B) II, III and IV only
(C) I, III and IV only
uh
Which of the following is least volatile ?
(A) HF
(B) HCl
Ja
8.
a
ri
6.
(C) HBr
(D) III and IV only
(D) HI
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
How many of the following has hydrogen bonding
(a) NH3
(b) CH4
(c) H2O
(e) HF
(f) HCOOH
(g) B(OH)3
(i) HCO3– ion
lp
1.
(d) HI
(h) CH3COOH
1.
Which compounds are lewis acids ?
(A) AlCl3
(B) BCl3
3.
(C) H2O
(D) NH3
Which of the following is/are electron deficient compounds ?
(A) NaBH4
(B) B2H6
(C) AlCl3
(D) LiAlH4
Which of the following have coordinate bonds ?
(A) NH4Cl
(B) NaCl
(C) O3
(D) Cl2
Sa
nk
2.
a
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
4.
Which of the following is/are correct.
(A) Boiling point of alcohol is higher than than of diethyl ether
(B) Density of water is higher than ice.
(C) Glycerol is more viscous than ethanol
(D) Ammonia is more easily liquified than HCl due to H-bonding in NH3
5.
Which of the following statements is correct regarding phosphoric acid ?
(A) p–d back bonding exist between O and P.
(B) The anion is resonance stablized.
(C) It is a dibasic acid.
(D) Inter molecular H bonding between molecules make it a syrupy (viscous) liquid.
6.
Which of the following is correct order of strength of hydrogen bonding?
(A) N—H- - -N > N—H- - -O
(B) F—H- - -N > O—H- - -N
(C) N—H- - -Cl > N—H- - -N
(D) O—H- - -F > O—H- - -O
78
PART - IV : COMPREHENSION
Read the following passage carefully and answer the questions.
Comprehension # 1
Bridge bonding is a specific kind of bonding in pages of chemistry. In general -bond pair delocalisation
is very difficult. But electron deficiency of the central atom forces to delocalised and forms this kind of
bond.
1.
The state of hybridisation of central atom in dimer form of both BH3 and BeH2 is
(A) sp2, sp
(B) sp3, sp2
(C) sp3, sp3
(D) sp2, sp3
2.
Which of the following molecule has complete octet
(A) B2H6
(B) Al2Cl6
(C) Be2Cl4
a
ri
(D) BeH2
The B2H6 molecule is dissolved in tetrahydrofuran. Which atom(s) is/are having changes of
hybridisation with respect to reactant and final product of the process given.
(A) B only
(B) B and O
(C) B, O and C
(D) None of these
4.
In which of the dimerisation process, the achievement of the octet is not the driving force.
(A) 2AlCl3 ⎯→ Al2Cl6
(B) BeCl2 ⎯→ BeCl2 (solid)
(C) 2lCl3 ⎯→ I2Cl6
(D) 2NO2 ⎯→ N2O4
5.
The molecule is not having 3c – 2e bond.
(A) BeH2 (dimer)
(B) BeH2 (solid)
uh
3.
(C) C2H6
(D) B2H6
(C) (I) (iii) (q)
(D) (IV) (iii) (p)
Correct combination is/are :
(A) (I) (iii) (q)
(B) (II) (iii) (s)
(C) (III) (iv) (p)
(D) (IV) (i) (s)
Incorrect combination is/are :
(A) (I) (iii) (p)
(B) (II) (ii) (q)
(C) (III) (i) (r)
(D) (IV) (ii) (p)
Sa
nk
7.
Correct combination is/are :
(A) (I) (iii) (q)
(B) (II) (i) (p)
a
6.
lp
Ja
Comprehension # 2
Answer Q.6, Q.7 and Q.8 by appropriately matching the information given in the three columns
of the following table.
Observe the three columns in which column-1 represents species, column-2
represents hybridization and shape while column-3 represents properties.
Column-1
Column-2
Column-3
Species
Hybridization & shape
Properties
(I)
O2
(i)
sp3
(P)
Peramegnetic
(II) XeF2
(ii)
sp3d
(Q)
Diamegnetic
(III) H2O
(iii)
Linear
(R)
H-bond formation
(IV) ICl2+
(iv)
Angular (V-shape)
(S)
Polar nature
8.
* Marked Questions may have more than one correct option.
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
1.
Amongst H2O, H2S, H2Se and H2Te the one with highest boiling point is :
[JEE–2000, 1/135]
(A) H2O because of H-bonding.
(B) H2Te because of higher molecular weight.
(C) H2S because of H-bonding.
(D) H2Se because of lower molecular weight.
2.
Identify the correct order of boiling points of the following compounds :
CH3 CH2CH2CH2OH
CH3CH2CH2CHO
CH3CH2CH2COOH
1
(A) 1 > 2 > 3
2
(B) 3 > 1 > 2
3
(C) 1 > 3 > 2
[JEE–2002, 3/150]
(D) 3 > 2 > 1
79
Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1 complex of BCl 3
& NH3.
[JEE–2002(S), 3/150]
(A) N : tetrahedral sp3, B : tetrahedral sp3
(B) N : pyramidal sp3, B : pyarmidal sp3
(C) N : pyramidal sp3, B : planar sp2
(D) N : pyramidal sp3, B : tetrahedral sp3
4.
Which one is more soluble in diethyl ether anhydrous AlCl 3 or hydrous AlCl3 ? Explain in terms of
bonding.
[JEE–2003(M), 2/144]
5.
AlF3 is insoluble in anhydrous HF but when little KF is added to the compound it becomes soluble. On
addition of BF3, AlF3 is precipitated. Write the balanced chemical equations.
[JEE–2004(M), 2/144]
6.
Predict whether the following molecules are iso-structural or not. Justify your answer.
[JEE–2005(M), 2/144]
(i) NMe3
(ii) N(SiMe3)3
7.
The number of water molecule (s) directly bonded to the metal centre in CuSO 4. 5H2O is
[JEE–2009, 4/160]
uh
a
ri
3.
Ja
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) OFFLNE PROBLEMS
.
An ether is more volatile than an alcohol having the same molecular formula. This is due to:
[AIEEE-2003, 3/225]
(1) dipolar character of ethers
(2) alcohols having resonance structures.
(3) inter-molecular hydrogen bonding in ethers
(4) inter-molecular hydrogen bonding in alcohols.
lp
1.
The states of hybridization of boron and oxygen atoms in boric acid (H 3BO3) are respectively :
[AIEEE-2004, 3/225]
(1) sp2 and sp2
(2) sp2 and sp3
(3) sp3 and sp2
(4) sp3 and sp3
3.
The structure of diborane (B2H6) contains :
(1) four 2c–2e bonds and four 3c–2e bonds
(2) two 2c–2e bonds and two 3c–3e bonds
(3) two 2c–2e bonds and four 3c–2e bonds
(4) four 2c–2e bonds and two 3c–2e bonds
[AIEEE-2005, 4½/225]
Sa
nk
a
2.
4.
A metal, M forms chlorides in +2 and +4 oxidation states. Which of the following statements about these
chlorides is correct ?
[AIEEE-2006, 3/165]
(1) MCl2 is more volatile than MCl4
(2) MCl2 is more soluble in anhydrous ethanol than MCl4
(3) MCl2 is more ionic than MCl4
(4) MCl2 is more easily hydrolysed than MCl4
5.
Among the following mixtures, dipole-dipole as the major interaction, is present in ? [AIEEE 2006, 3/165]
(1) Benzene and carbon tetrachloride
(2) Benzene and ethanol
(3) Acetonitrile and acetone
(4) KCl and water
6.
Which of the following hydrogen bonds is the strongest ?
(1) O – H ... O
(2) O – H ... F
(3) F – H ... H
[AIEEE-2007, 3/120]
(4) F – H ... F
80
The bond dissociation energy of B–F in BF3 is 646 kJ mol–1 whereas that of C–F in CF4 is 515 kJ mol–1.
The correct reason for higher B–F bond dissociation energy as compared to that of C–F is :
[AIEEE-2009, 4/144]
(1) stronger  bond between B and F in BF3 as compared to that between C and F in CF4.
(2) significant p–p interaction between B and F in BF3 whereas there is no possibility of such
interaction between C and F in CF4.
(3) lower degree of p-p interaction between B and F in BF3 than that between C and F in CF4.
(4) smaller size of B-atom as compared to that of C-atom.
8.
What is the best description of the change that occurs when Na2O(s) is dissolved in water ?
(1) Oxide ion accepts sharing in a pair of electrons
[AIEEE-2011, 4/120]
(2) Oxide ion donates a pair of electrons
(3) Oxidation number of oxygen increases
(4) Oxidation number of sodium decreases
9.
Which one has the highest boiling point ?
(1) He
(2) Ne
a
ri
7.
[JEE(Main)-2015, 4/120]
(4) Xe
(3) Kr
The intermolecular interaction that is dependent on the inverse cube of distance between the molecules
is:
(1) ion-ion interaction
(2) ion-dipole interaction
[JEE(Main)-2015, 4/120]
(3) London force
(4) hydrogen bond
11.
Which one of the following statements about water is FALSE ?
[JEE(Main)-2016, 4/120]
(1) Water can act both as an acid and as a base.
(2) There is extensive intramolecular hydrogen bonding in the condensed phase.
(3) Ice formed by heavy water sinks in normal water.
(4) Water is oxidized to oxygen during photosynthesis.
12.*
Which of the following are Lewis acids ?
(1) PH3 and SiCl4
(2) BCl3 and AlCl3
Ja
uh
10.
[JEE(Main)-2018, 4/120]
(4) AlCl3 and SiCl4
lp
(3) PH3 and BCl3
JEE(MAIN) ONLINE PROBLEMS
a
The number of 2-centre-2-electron and 3-centre-2-electron bonds in B2H6, respectively, are:
[JEE(Main) 2019 Online (10-01-19), 4/120]
(1) 2 and 4
(2) 2 and 2
(3*) 4 and 2
(4) 2 and 1
Sa
nk
1.
2.
The hydride that is NOT electron different is :
(1) GaH3
(2*) SiH4
(3) AlH3
[JEE(Main) 2019 Online (11-01-19), 4/120]
(4) B2H6
81
EXERCISE - 1
PART – I
F
A-1.
In Me3 N ⎯⎯→
B
F , the electron deficiency of boron is compensated by the lone pair of electron
A-2.
a
ri
F
donated by nitrogen atom. Where as in BF3 it is compensated by back bonding with F atom; back
bonding is delocalised thus B–F bond has partial double bond character.
(Shows p–d bonding)
(i)
two p–d bond and one p–p bond.
(ii)
one p–d bond
Ja
A-3.
uh
(Cannot show p–d bonding)
(iii) NN two p–p bond.
(iv)
three p–d bond.
(i) In NH3 molecule N atom has lone pair in sp3 hybrid orbital while in PH3 as suggested by
its
bond angle (92º) the lone pair must be present in ‘S’ orbital which is much more contracted than sp 3.
Hence PH3 becomes a poor donor than NH3.
(ii) CH3CN has lone pair on sp hybridized nitroge n atom while NH3 has lone pair on sp3 hybridized
nitrogen atom.
B-2.
In BCl3, octet of Boron is incomplete. In SiF4, silicon has vacant d-orbitals, by which it can accept
electron pair.
B-3.
BF3 molecule being electron deficient gets stabilised through p–p back bonding. where as BH3
removes its electron deficiency through dimerisation and thus exists as B 2H6.
B-4.
sp3 hybridised orbital of both aluminium and sp3 hybridised orbital of carbon.
C-1.
Very strong hydrogen bonding occurs in the alkali metal hydrogen fluorides of formula M[HF 2]; there is
a linear symmetrical anion having an over all, F–H–F distance of 2.26 Å.
Sa
nk
a
lp
B-1.
[F – H – – – F]– ⎯⎯
→ [F – – – H – F]–
F– + HF ⎯⎯
→ [FHF]– ; H = – 161 ± 8 kJ mol–1
C-2.
Deuterium is more electropositive than hydrogen. Therefore, stronger H-bonding is found in D2O than in
H2O. D2O is also denser than H2O.
C-3.
These compounds contain polar–OH groups which can form H-bonds with water.
C-4.
In ethanol, there is H-bonding but in diethyl ether, there is no H-bonding (because O-atom is attached
to C-atom) but there exists weak dipole-dipole attraction in diethyl ether.
82
PART – II
A-1.
(D)
A-2.
(A)
A-3.
(C)
B-1.
(A)
B-2.
(C)
B-3.
(B)
B-4.
(B)
B-5.
(B)
C-1.
(D)
C-2.
(D)
C-3.
(D)
C-4.
(D)
C-5.
(D)
C-6.
(B)
C-7.
(D)
C-8.
(C)
1.
(A - q) ; (B – r) ; (C – s,p) ; (D – p, r)
2.
(A – p, q, r, s) ; (B – p, q, r, s) ; (C – p, q, r, s) ; (D – r, s)
3.
(A – r) ; (B – p, r) ; (C – p, q, r) ; (D – p, r) ; (E – p, r, s).
a
ri
PART – III
PART – I
(C)
2.
(B)
3.
(D)
6.
(C)
7.
(C)
8.
(A)
4.
(B)
5.
(B)
4.
(ABCD)
5.
(ABD)
4.
(C)
5.
(C)
Ja
1.
uh
EXERCISE - 2
PART – II
1.
7 (Except (b, d)
(AB)
6.
(AB)
(B)
2.
(BC)
(B)
Sa
nk
1.
2.
6.
(C)
7.
2.
3.
a
1.
lp
PART – III
(D)
(B)
(AC)
PART – IV
3.
(D)
8.
(D)
EXERCISE – 3
PART – I
(A)
3.
(A)
4.
In diethyl ether (C2H5— —C2H5) oxygen atom has two lone pairs of electrons, thus acts as lewis base
while in anhydrous AICI3 aluminium has vacant 3p-orbital of valence shell and thus acts as Lewis acid.
AlCl3 accepts a lone pair of electrons from diethyl ether to complete its octet forming a complex
C2H5
AlCl3 . Hence, anhydrous AICI3 is more soluble in diethyl ether by means of solvolysis
O:
C2H5
in comparison to hydrous AICI3 (i.e., AICI3.6H2O). Hydrous AICI3 is a polar compound, while ether is
non-polar, so on basis of Thumb`s rule, like dissolve in like solvents. Hence hydrous AICI 3 is least
soluble in ether.
:
1.
83
There is inter molecular hydrogen bonding in HF and because of this it is weakly dissociated. So AlF 3 is
not soluble in anhydrous HF. On the other hand KF is ionic compound and thus it is highly dissociated
giving a high concentration of F– ion which leads to the formation of a colourless soluble complex,
AlF3 + KF ⎯→ K3[AlF6].
BF3 is more acidic than AlF3 because of the small size of B than that of Al. Thus BF 3 pulls out F– from
[AlF6]3– forming [BF4]– and AlF3. Hence AlF3 is precipitated on adding BF3 to [AlF6]3–.
K3[AlF6] + 3BF3 ⎯→ 3K[BF4] + AlF3  .
6.
(i) N(SiMe3)3 is trigonal planar because in it silicon uses its vacant d-orbital for p-d back bonding with
lone pair of electrons of central N-atom and the p-d bonding is delocalised as given in the structure.
So, N(SiMe3)3 with steric number three is trigonal planar.
(ii) In N(Me3), there is no such p-d delocalisation of lone pair of electrons on N atom as carbon does
not have vacant d-orbital. So N(Me)3 with steric number four is trigonal pyramidal with a lone pair at the
apex.
(ii)
Hence both are not isostructural.
7.
4
PART – II
uh
(i)
a
ri
5.
(4)
2.
(1)
3.
6.
(4)
7.
(2)
8.
11.
(2)
12.*
(2,4)
(4)
4.
(3)
5.
(3)
(2)
9.
(4)
10.
(2)
lp
1.
Ja
JEE(MAIN) ONLNE PROBLEMS
JEE(MAIN) OFFLINE PROBLEMS
(3)
2.
(2)
Sa
nk
a
1.
Section (A) : Molecular Orbital Theory (MOT)
(i)
(ii)
(iii)
(iv)
(v)
The molecular orbital theory was developed by F. Hund and R.S. Mulliken in 1932. The salient features
are:
Just as electrons of any atom are present in various atomic orbitals, electrons of the molecule are
present in various molecular orbitals.
Molecular orbitals are formed by the combination of atomic orbitals of comparable energies and proper
symmetry.
An electron in an atomic orbital is influenced by one nucleus, while in a molecular orbital it is influenced
by two or more nuclei depending upon the number of the atoms in the molecule. Thus an atomic
orbital is monocentric while a molecular orbital is polycentric.
The number of molecular orbitals formed is equal to the number of combining atomic orbitals. When two
atomic orbitals combine, two molecular orbitals called bonding molecular orbital and anti-bonding
molecular orbital are formed.
The bonding molecular orbital has lower energy and hence greater stability than the corresponding
antibonding molecular orbital.
84
(vi)
(vii)
Just as the electron probability distribution around a nucleus in an atom is given by an atomic orbital,
the electron probability distribution around a group of nuclei in a molecule is given by molecular orbital.
The molecular orbitals like the atomic orbitals are filled in accordance with the Aufbau principle
obeying the Pauli Exclusion principle and the Hund’s Rule of Maximum Multiplicity. But the filling
order of these molecular orbitals is always experimentally decided, there is no rule like (n + l) rule in
case of atomic orbitals.
Figure: The relative energy levels of
molecular orbitals and their constituent
atomic orbitals for H2.
Sa
nk
a
lp
Ja
uh
The molecular wave functions m and *m are bonding and
antibonding molecular orbitals; orbitals of this type, in which
the line joining the two nuclei is a symmetry axis for the
electron distribution, are known as  orbitals if bonding and
* orbitals if antibonding, so we may alternatively denote
them 1s and 1s orbitals. When the two 1s wave functions
are added, they reinforce one another everywhere, and
especially in the region between the two nuclei; the build-up
of electron density there diminishes the internuclear
repulsion and a strong bond results. When one of the two 1s
wave functions is subtracted from the other, they exactly
cancel in a plane midway between the nuclei, and the
molecular wave function changes sign at this nodal plane.
This lack of electron density raises the internuclear repulsion,
the total energy becomes higher, the two nuclei are not
bonded together, and the orbital is described as antibonding.
a
ri
Formation of Molecular Orbitals : Linear Combination of Atomic Orbitals(LCAO)

Consider the example of H2 molecule and label both nucleis as HA and HB.

Both nucleus will have a wave function A & B associated to it.

Bonding molecular orbital will be formed when both wave functions will combine in same phase.
m = [A + B]

Antibonding molecular orbital will be formed when both wave functions will combine in opposite phase.
*m = [A – B]
Note: The above equation should be regarded as the summation of the wave functions not as the
mathematical addition or subtraction of wave function.
Figure : Another representation of the formation of molecular orbitals for H 2. Since the  and *
orbitals are respectively centrosymmetric and non-centrosymmetric these orbitals may also be denoted
by the symbols g and *u.
Molecular orbital wave functions are designated as g and u. g and u refer to the symmetry of the
orbital about its centre. If the wave function is centrosymmetric, i.e. has the same sign at the same
distance in opposite directions from the centre of symmetry. The orbital is said to gerade (German,
even); if it changes sign on inversion about the centre it is said to ungerade (German, uneven).
85
Alternative method for determining the symmetry of the molecular orbital is to rotate the orbital about
the line joining the two nuclei and then about a line perpendicular to this. If the sign of the lobes remain
the same, the orbital is gerade, and if the sign changes, the orbital is ungerade.
Thus  and * molecular orbitals are gerade and * and  molecular orbitals are ungerade.
2.
3.
a
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1.
CONDITIONS FOR THE COMBINATION OF ATOMIC ORBITALS :
The linear combination of atomic orbitals to form molecular orbitals takes place only if the following
conditions are satisfied :
The combining atomic orbitals must have the same or nearly the same energy. This means that
1s orbital can combine with another 1s orbital but not with 2s orbital because the energy of 2s orbital is
appreciably higher than that of 1s orbital. It, therefore, means that only a limited number of
combinations of atomic orbitals are possible.
The combining atomic orbitals must have the same symmetry about the molecular axis. By
convention z-axis is taken as the molecular axis. It is important to note that atomic orbitals having same
or nearly the same energy will not combine if they do not have the same symmetry. For example, 2p z
orbital of one atom can combine with 2pz orbital of the other atom but not with the 2px or 2py orbitals
because of their different symmetries.
The combining atomic orbitals must overlap to the maximum extent. Greater the extent of overlap,
the greater will be the electron-density between the nuclei of a molecular orbital.
(2)
Sa
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(3)
Difference between  molecular orbitals and the  orbitals.
For  overlap the lobes of the atomic orbitals are perpendicular to the line joining the nuclei, while for 
overlap the lobes point along the line joining the two nuclei.
For  molecular orbitals,  is zero along the internuclear line and consequently the electron density 2
is also zero. This is in contrast to  orbitals.
The symmetry of  molecular orbitals is different from that shown by  orbitals. If the bonding  MO is
rotated about the inter nuclear line a change in the sign of lobe occurs. The  bonding orbitals are
therefore ungerade, where as all  bonding MO’s are gerade. Conversely the antibonding  MO’s are
gerade while all  antibonding MO’s are ungerade.
a
(1)
lp
Ja
uh
TYPES OF MOLECULAR ORBITALS :
Molecular orbitals of diatomic molecules are designated as (sigma), (pie), (delta) etc.
In this nomenclature, the sigma () molecular orbitals are symmetrical around the bond-axis while
pi () molecular orbitals are not symmetrical. For example, the linear combination of 1s orbitals
centered on two nuclei produces two molecular orbitals which are symmetrical around the bond-axis.
Such molecular orbital’s are of the  type and are designated as  1s and *1s [Fig.(a)]. If internuclear
axis is taken to be in the z-direction, it can be seen that a linear combination of 2pz-orbitals of two
atoms also produces two sigma molecular orbitals designated as 2pz and * 2pz. [Fig. (b)] Molecular
orbitals obtained from 2px and 2py orbitals are not symmetrical around the bond axis because of the
presence of positive lobes plane. Such molecular orbitals, are labelled as  and * [Fig. (c)]. A
 bonding MO has large electron density above and below the inter nuclear axis. The * antibonding
MO has a node between the nuclei.
-type of molecular orbitals are obtained by involvement of d-orbitals into bonding.
ENERGY LEVEL DIAGRAM FOR MOLECULAR ORBITALS :
The energy levels of molecular orbitals have been determined experimentally from spectroscopic data
for homonuclear diatomic molecules of second row elements of the periodic table. The increasing order
of energies of various molecular orbitals for O2 and F2 is given below :
1s < *1s < 2s < *2s < 2pz < (2px = 2py) < (*2px = *2pz) < *2pz.
The increasing order of energies of various molecular orbitals for Be 2, B2, C2, N2 etc., is :
 1s < * 1s < 2s < *2s < (2px = 2py) < 2pz < (*2px = *2py) < *2pz
The important characteristic feature of this order is that the energy of 2pz molecular orbital is higher
than that of 2px and 2py molecular orbitals.
86
Antibonding sigma
molecular orbital
2pz
2Pz
2pz
2pz
– + – +
*1s
2pz
+ + –
– + –
2pz
2pz
Sa
nk
a
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Ja
2pz
→
– +
(b)
– + –
Bonding sigma
molecular orbital
→
→
*2pz
– +
uh
Atomic
orbital
→
Energy
Atomic
orbital
a
ri
Molecular
orbitals
Figure : Bonding and antibonding molecular orbitals formed through combinations of (a) 1s atomic orbitals;
(b) 2pz atomic orbitals and (c) 2px atomic orbitals.
Section (B) : Application of MOT


ELECTRONIC CONFIGURATION AND MOLECULAR BEHAVIOUR
The distribution of electrons among various molecular orbitals is called the electronic configuration of
the molecule. From the electronic configuration of the molecule, it is possible to get important
information about the molecule as discussed below.
(i) The molecule is stable if Nb is greater than Na, and (ii) The molecule is unstable if Nb is less than Na
In (i) more bonding orbitals are occupied and so the bonding influence is stronger and a stable
molecule results. In (ii) the antibonding influence is stronger and therefore the molecule is unstable.
Nb is number of electrons in bonding molecular orbitals and N a is number of electrons in antibonding
molecular orbitals.
BOND ORDER
Bond order (B.O.) = ½(Nb – Na)
A positive bond order (i.e., Nb > Na) means a stable molecule while a negative (i.e., Nb < Na) or zero
(i.e., Nb = Na) bond order means an unstable molecule.
BOND-LENGTH
The bond order between two atoms in a molecule may be taken as an approximate measure of the
bond length. The bond length decreases as bond order increases.
87
2.
Sa
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a
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Ja
uh
3.
a
ri
1.
MAGNETIC NATURE
If all the molecular orbitals in a molecule are doubly occupied, the substance is diamagnetic (repelled
by magnetic field) e.g., N2 molecule. However if one or more molecular orbitals are singly occupied it is
paramagnetic (attracted by magnetic field), e.g., O2 molecule.
BONDING IN SOME HOMONUCLEAR DIATOMIC MOLECULES
Hydrogen molecule (H2) : (1s)2
N − Na 2 − 0
=
=1
Bond order = b
2
2
This means that the two hydrogen atoms are bonded together by a single covalent bond. The bond
dissociation energy of hydrogen molecule has been found to be 438 kJ mol –1 and bond length equal to
74 pm. Since no unpaired electron is present in hydrogen molecule, therefore, it is diamagnetic.
Helium molecule (He2) : (1s)2 (*1s)2
Bond order of He2 is ½(2 – 2) = 0
The molecular orbital description of He2 predicts two electrons in a bonding orbital and two electrons in
an antibonding orbital, with a bond order of zero - in other words, no bond. This is what is observed
experimentally. The noble gas He has not significant tendency to form diatomic molecules and, like the
other noble gases, exists in the form of free atoms. He2 has a very low binding energy, approximately
0.01J/mol ; for comparison, H2 has a bond energy of 436 kJ/mol.
Lithium molecule (Li2) : (1s)2 (*1s)2 (2s)2
Its bond order, therefore, is 1/2(4 – 2) = 1. It means that Li2 molecule is stable and since it has no
unpaired electrons it should be diamagnetic. Indeed diamagnetic Li 2, the molecules are known to exist
in the vapour phase. The MO model predicts a single Li-Li bond in Li2, in agreement with gas phase
observations of the molecule.
Beryllium (Be2) : (1s)2 (*1s)2 (2s)2 (*2s)2
Be2 has the same number of antibonding and bonding electrons and consequently a bond order of
zero. Hence, like He2, Be2 is not a stable chemical species.
Boron (B2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p1x = 2p1y) (pz)0
Here is an example in which the Molecular orbital model has distinct advantage over the Lewis dot
picture, B2 is found only in the gas phase; solid boron is found in several very hard forms with complex
bonding, primarily involving B12 icosahedra. B2 is paramagnetic. This behaviour can be explained if its
two highest energy electrons occupy separate p-orbitals. The Lewis dot model cannot account for the
paramagnetic behaviour of this molecule.
B2 is also a good example of the energy level shift caused by the mixing of s and p orbitals. In the
absence of mixing, the g (2p) orbital is expected to be lower in energy than the u(2p) orbitals and the
resulting molecule would be diamagnetic. However, mixing of the g(2s) orbital with the g(2p) orbital
lowers the energy of the g(2s) orbital and increases the energy of the g(2p) orbital to a higher level
than the  orbitals, giving the order of energies shown above. As a result, the last two electrons are
unpaired in the degenerate (having the same energy)  orbitals, and the molecule is paramagnetic.
Overall, the bond order is 1, even though the two p electrons are in different orbitals. The bond order of
B2 is 1/2(6 – 4) = 1.
Carbon molecule (C2): (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y) or KK (2s)2 (*2s)2 (2p2x = 2p2y)
The simple MO picture of C2 predicts a doubly bonded molecule with all electrons paired, but with both
highest occupied molecular orbitals (HOMOs) having  symmetry. It is unusual because it has two 
bonds and no  bond. The bond dissociation energies of B2, C2 and N2 increase steadily, indicating
single, double, and triple bonds with increasing atomic number. Although C 2 is not a commonly
encountered chemical species (carbon is more stable as diamond, graphite, and the fullerenes), the
acetylide ion, C22-, is well known, particularly in compounds with alkali metals, alkaline earths, and
lanthanides. According to the molecular orbital model, C22– should have a bond order of 3 (configuration
u2u2g2). This is supported by the similar C–C distances in acetylene and calcium carbide (acetylide) :
Table-1
C – C Distance (pm)
C = C (gas phase)
132
H–CC–H
120.5
CaC2
119.1
4.
The bond order of C2 is 1/2 (8 – 4) = 2 and C2 should be diamagnetic. Diamagnetic C2 molecules have
indeed been detected in vapour phase. It is important to note that double bond in C 2 consists of both pi
bonds because of the presence of four electrons in two pi molecular orbitals. In most of the other
molecules a double bond is made up of a sigma bond and a pi bond.
88
Nitrogen molecule (N2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) (2pz)2
N2 has a triple bond according to both the Lewis and the molecular orbital models. This is in agreement
with its very short N–N distance (109.8 pm) and extremely high bond dissociation energy (942kJ/mol.).
Atomic orbitals decrease in energy with increasing nuclear charge Z ; as the effective nuclear charge
increases, all orbitals are pulled to lower energies. The shielding effect and electron-electron
interactions cause an increase in the difference between the 2s and 2p orbital energies as Z increases,
from 5.7 eV for boron to 8.8 eV for carbon and 12.4 eV for nitrogen. As a result, the g(2s) and g(2p)
levels of N2 interact (mix) less than the B2 and C2 levels, and the g (2p) and u(2p) are very close in
energy.
The bond order of N2 is 1/2(10 – 4) = 3. It contains one sigma and two  bonds.
6.
Anionic nitrogen species (N2–) : Though 15 electrons but derived from N2, hence electronic
configuration will be according to N2
Electronic configuration : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) (2pz)2 , (*2px)1
The bond order of N2 is 1/2(10 – 5) = 2.5. It is paramagnetic species.
7.
N2+ : Bond order = 2.5, out of N2+ and N2¯, N2¯ is less stable though both have equal bond order but
N2– has greater number of antibonding electrons.
8.
Oxygen molecule (O2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px1 = *2p1y)
O2 is paramagnetic. This property, as for B2, cannot be explained by the traditional Lewis dot structure
, but is evident from the molecular orbital picture, which assigns two electrons to the degenerate
g* orbitals. The paramagnetism can be demonstrated by pouring liquid O2 between the poles of a
strong magnet; some of the O2 will be held between the pole faces until it evaporates. The bond order
1
1
of O2 is
Nb − Na  = [10 − 6] = 2 . So in oxygen molecule, atoms are held by a double bond.
2
2
Moreover, it may be noted that it contains two unpaired electrons in *2px and *2py molecular orbitals,
therefore, O2 molecule should be paramagnetic, a prediction that corresponds to experimental
observation. Several ionic forms of diatomic oxygen are known, including O 2+, and O22–. The
internuclear O–O distance can be conveniently correlated with the bond order predicated by the
molecular orbital model, as shown in the following table.
Table-2
Internuclear
Number of unpaired(s)
Bond Order
Distance (pm)
Electrons
O2+ (dioxygenyl)
2.5
112.3
1
O2 (dioxygen)
2.0
120.07
2
O2– (superoxide)
1.5
128
1
O22– (peroxide)
1.0
149
0
Note : Oxygen-oxygen distances in O2– and O22– are influenced by the cation. This influence is especially
strong in the case of O22– and is one factor in its unsually long bond distance.
Note : Oxygen-oxygen distances in O2– and O22– are influenced by the cation. This influence is
especially strong in the case of O22– and is one factor in its unsually long bond distance.
The extent of mixing is not sufficient in O 2 to push the g(2p) orbital to higher energy than the g(2p)
orbitals. The order of molecular orbitals shown is consistent with the photoelectron spectrum.
Fluorine molecule (F2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p2y)
The molecular orbital picture of F2 shows a diamagnetic molecule having a single fluorine-fluorine bond,
in agreement with experimental data on this very reactive molecule.
The net bond order in N2, O2, and F2 is the same whether or not mixing is taken into account, but the
order of the filled orbitals is different. The switching of the order of the g(2p) and u(2p) orbitals can
occur because these orbitals are so close in energy ; minor changes in either orbital can switch their
order. The energy difference between the 2s and 2p orbitals of the atoms increases with increasing
nuclear charge, from 5.7 eV in boron to 27.7 eV in fluorine. Because the difference becomes greater,
the s-p interaction decreases and the "normal" order of molecular returns in O2 and F2. The higher g
orbital is seen again in CO.
Neon molecule (Ne2) : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px2 = *2p2y) (*2pz)2
All the molecular orbitals are filled, there are equal numbers of bonding and antibonding electrons and
the bond order is therefore zero. The Ne2 molecule is a transient species, if it exists at all.
Note : HOMO : Highest Occupied Molecular Orbital., LUMO : Lowest Unoccupied Molecular Orbital
Bond lengths in homonuclear diatomic molecules
Sa
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a
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Ja
uh
a
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5.
89
uh
a
ri
Figure shows the variation
of bond distance with the
number
of
valence
electrons in second-period
p
block
homonuclear
diatomic molecules. As the
number
of
electrons
increases, the number in
bonding
orbitals
also
increases,
the
bond
strength becomes greater,
and the bond length
becomes shorter. This
continues up to 10 valence
electrons in N2 and then the
trend reverses because the
additional electrons occupy
antibonding orbitals. The
ions N2+, O2+, O22+ are also
shown in the figure and
follow a similar trend.
Figure
Sol.
Ex-3.
Ja
Sa
nk
Sol.
lp
Ex-2.
Though O2 molecule is paramagnetic yet it is a colourless gas. Why ?
It is because the energy gap between HOMO and LUMO levels in O 2 molecule is so large that
radiations of visible light cannot excite a e– from HOMO to LUMO. In fact O2 gas shows absorption in
UV zone. So it is colourless.
Correct order of bond energy is:
(A) N2 > N2+ >N2– >N22– (B) N2+ >N2– >N22– >N2 (C) N2 >N2¯ = N2+ >N22– (D) N2– >N2 = N2+ >N22–
(A) Bond order is directly proportional to the bond energy.
Bond order of N2 = 3, N2+, N2– = 2.5, N22– = 2
But N2– has more electrons in antibonding MO’s and thus N 2+ is more stable than N2–. So correct order
of bond energy will be N2 > N2+ >N2– >N22–
Which of the following species have a bond order of 3 ?
(A) CO
(B) CN–
(C) NO+
(D) O2+
–
+
(A,B,C) Species CO, CN , NO are isoelectronic with 14 electrons to N2 which has bond order of 3 (i.e.
10 − 4
= 3), so their bond order will be equal to three.
3
Which of the following are diamagnetic ?
(A) C2
(B) O22–
(C) Li2
(D) N2+
(A,B,C) Species C2, O22–, Li2 have all the electrons paired but N2+ has one unpaired electron in bonding
molecular orbital so it is paramagnetic.
a
Ex-1.
Sol.
Ex-4.
Sol.
Section (C) : Metallic bonding
Most metals crystallise in close-packed structures. The ability of metals to conduct electricity and heat
must result from strong electrons interactions among 8 to 12 nearest neighbours (which is also called
coordination number). Bonding in metals is called metallic bonding. It results from the electrical
attractions among positively charged metal ions and mobile, delocalised electrons belonging to the
crystal as a whole.
Two models are considered to explain metallic bonding: (A) Band model (B) Electron-sea model
90



a
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Sa
nk
a
lp

Figure : Band of orbital in crystal of sodium and overlapping of a half–filled “3s” band with an empty
“3p” band of NaN crystal
According to band theory, the highest-energy electrons of metallic crystals occupy either a partially filled
band or a filled band that overlaps an empty band.
A band within which (or into which ) electrons must move to allow electrical conduction is called a
conduction band.
The electrical conductivity of a metal decreases as temperature increases. The increase in temperature causes
thermal agitation of the metal ions. This impedes the flow of electrons when an electric field is applied.
Crystalline non-metals, such as diamond and phosphorus, are insulators, they do not conduct
electricity. It is due to the fact that their highest-energy electrons occupy filled bands of molecular
orbitals that are separated from the lowest empty bond (conduction band) by an energy difference
called the band gap.
In an insulator, this band gap is an energy difference that is too large for electrons to jump to get to the
conduction band.
uh

Band Model :
Ja
(A)


(B)



Figure : Distinction among metals, insulators and semiconductors. In each case an unshaded area
represents a conduction band.
Elements that are semiconductors have filled bands that are only slightly below, but do not overlap with
empty bands.
They do not conduct electricity at low temperatures, but a small increase in temperature is sufficient to
excite some of the highest-energy electrons into the empty conduction band.
Electron-Sea Model
In lithium the ions would be Li+ and one electron per atom would be contributed to the sea. These free
electrons account for the characteristic metallic properties.
If the ends of a bar of metal are connected to a source of electric current, electrons from the external
source enter the bar at one end. Free electrons pass through the metal and leave the other end
at the same rate.
In thermal conductivity no electrons leave or enter the metal but those in the region being heated gain
kinetic energy and transfer this to other electrons.
91
According to the electron-sea model, the case of deformation of metals can be thought of in this way: If
one layer of metal ions is forced across another, perhaps by hammering, the internal structure
remains unchanged as the sea of electrons rapidly adjusts to the new situation.
uh
a
ri

Why transition element have high heat of atomization.
Transition metals may use inner -d-electrons along with the outer s-electrons for bonding as (n–1) d
and ns have nearly same energy. So in them number of metallic bonds per atoms is quite large (more
than two always). Hence element have high heat of atomization.
lp
Ex-5.
Sol.
Ja
Figure : Effect of Distortion (by hammering) on Metal Sheet (assuming Electron Sea Model)
MISCELLANEOUS SOLVED PROBLEMS (MSPS)
Super oxides are coloured and paramagnetic why ?
Super oxides contain one unpaired electron in anti bonding molecular orbital and are coloured due to
transition of HOMO orbital electron within visible region.
Ex-2.
Sol.
Of the species O2+ ,O2− , O2 and O22− which would have the maximum bond strength ?
O2+ has higher bond order i.e. 2.5 than O2(2) and O2– (1.5) and bond strength is directly proportional to
bond order.
Sa
nk
a
Ex-1.
Sol.
 Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
Section (A) : MOT
A-1. Find out the bond order of :
(a) H2
(b) H2+
A-2.
(c) He2
(d) Li2
(e) Be2
(f) B2
Identify the molecules or atoms or ions from the following molecular orbital energy level formulations.
The species should be selected from (B2, C2, O22+, O2, F2, N2)
(a) KK  (2s)2  * (2s)2 (2p x )1 (2p y )1
(b) KK  (2s)2  * (2s)2 (2p x )2 (2p y )2
92
(c) KK  (2s)2  * (2s)2 (2p z )2 (2p x )2 (2p y )2
(d) KK  (2s)2  * (2s) (2p z )2 (2p x )2 (2p y )2 (2p x )1  * (2p y )1
(e)
(f) KK σ (2s)2 σ *(2s)2 π( 2p y )2 (2p x )2 σ( 2p z )2
A-3. What is the bond order of underlined species in NO [BF4]?
Section (B) : Applilcation of MOT
B-1.
How would you explain that B2 molecule is not diamagnetic?
B-3.
Which of the following are gerade molecular orbitals?
(i) *2s
(ii) 2pz
(iii) 2py
(iv) *2px
a
ri
B-2. Explain why NO+ is more stable towards dissociation into its atoms than NO ?
B-4. Arrange following compounds in the order of increasing order of O–O bond length.
(i) O2
(ii) O2[BF4]
(iii) KO2
Section (C) : Metallic bonding
C-2.
uh
C-1. Zinc has lowest melting point in 3d-series elements. Why ?
Among Be and Li, which should have higher melting point and why ?
PART - II : ONLY ONE OPTION CORRECT TYPE
Ja
Section (A) : MOT
A-1. During the formation of a molecular orbital from atomic orbitals of the same atom, probability of electron
density is :
(A) none zero in the nodal plane
(B) maximum in the nodal plane
(C) zero in the nodal plane
(D) zero on the surface of the lobe
If Z-axis is the molecular axis, then -molecular orbitals are formed by the overlap of
(A) s + pz
(B) px + py
(C) pz + pz
(D) px + px
lp
A-2.
Which of the following pairs have identical values of bond order ?
(A) N2+ and O2+
(B) F2 and Ne2
(C) O2 and B2
A-5. Which of the following molecules/ions exhibit sp mixing?
(A) B2
(B) C22–
(C) O2+
Sa
nk
A-4.
a
A-3. Bond order is a concept in the molecular orbital theory. It depends on the number of electrons in the
bonding and antibonding orbitals. Which of the following statements is true about it ? The bond order
(A) Can have a negative quantity
(B) Has always an integral value
(C) Can assume any positive or integral or fractional value including zero
(D) Is a non zero quantity
(D) C2 and N2
(D) Both (A) and (B)
A-6. The common features of the species N22– , O2 and NO– are :
(A) bond order three and isoelectronic.
(B) bond order two and isoelectronic.
(C) bond order three but not isoelectronic.
(D) bond order two but not isoelectronic.
A-7. Which of the following molecular orbitals has two nodal planes.
(A) 2s
(B) 2py
(C) *2py
(D) *2px
Section (B) : Applilcation of MOT
B-1. Among the following species, which has the minimum bond length ?
(A) B2
(B) C2
(C) F2
B-2.
Which of the following species is paramagnetic ?
(A) NO–
(B) O22–
(C) CN–
(D) O2–
(D) CO
B-3. The following molecules / species have been arranged in the order of their increasing bond orders,
Identify the correct order.
(I) O2
(II) O2–
(III) O22–
(IV) O2+
93
(A) I I I < I I < I < IV
B-4.
(B) IV < I I I < I I < I
Which one is paramagnetic from the following
(A) O2–
(B) NO
(C) I I I < I I < IV < I
(D) I I < I I I < I < IV
(C) Both (A) and (B)
(D) CN–
B-5. Which of the following orders is correct in respect of bond dissociation energy ?
(A) N2+ > N2–
(B) O2+ > O3
(C) NO+ > NO
(D) All of these
Section (C) : Metallic bonding
C-1. Iron is harder than sodium because :
(A) iron atoms are smaller.
(C) metallic bonds are stronger in sodium.
(B) iron atoms are more closely packed.
(D) metallic bonds are stronger in iron.
The enhanced force of cohesion in metals is due to :
(A) The covalent linkages between atoms
(B) The electrovalent linkages between atoms
(C) The lack of exchange of valency electrons
(D) The delocalization of valence electron between metallic kernels.
uh
C-2.
(D) FFTT
a
ri
B-6. S1  The HOMO in F2– is *2px = *2py molecular orbitals.
S2  Bond order of O2– is more then O2+.
S3  NO+ is more stable than N2+.
S4  C2 is more stable than C2+.
State, in order, whether S1, S2, S3, S4 are true or false
(A) FFFT
(B) FTTT
(C) FTFT
Ja
C-3. In the following metals which one has lowest probable interatomic forces
(A) Copper
(B) Silver
(C) Zinc
(D) Mercury
Match the following :
Column – I
O2 and NO–
(B)
O2+ and NO
(C)
CN–
CO and
C2 and CN+
(p)
Same magnetic property and bond order as that in N2+
a
(A)
Column – II
(q)
Same bond order but not same magnetic property as that in O2
(r)
Same magnetic property and bond order as that N22 –
Same magnetic property and bond order as that in NO+
Sa
nk
1.
lp
PART - III : MATCH THE COLUMN
(D)
(s)
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
1.
2.
Number of antibonding electrons in N2 is :
(A) 4
(B) 10
(C) 12
(D) 14
Following is the molecular orbital configuration of a diatomic molecule
 2p2y
 1s2 * 1s2  2s2 * 2s2  2p2x 
2
  2pz
Its bond order is :
(A) 3
(B) 2.5
(C) 2
(D) 1
94
3.
4.
The bond order of He 2+ molecule ion is :
(A) 1
(B) 2
(C) 1/2
(D) 1/4
Which species can exist among the following :
(A) B2
(B) Be2
(C) Ne2
(D) He2
Among the following which one will have the largest O – O bond length ?
(A) KO2
(B) O2
(C) O2+ [AsF6] –
(D) K2O2
6.
The correct order in which the O–O bond length increases in the following is :
(A) H2O2 < O2 < O3
(B) O2 < H2O2 < O3
(C) O2 < O3 < H2O2
(D) O3 < H2O2 < O2
7.
Which of the following is a wrong order with respect to the property mentioned against each ?
(A) O22– > O2 > O2+ [Paramagnetic moment]
(B) (NO)¯ > (NO) > (NO)+ [bond length]
+
+
(C) H2 > H2 > He2 [bond energy]
(D) NO2+ > NO2 > NO2¯ [bond angle]
8.
Which of the following option with respect to increasing bond dissociation energies is correct ?
(A) NO < C2 < O2– < He2+
(B) C2 < NO < He2+ < O2–
+
–
(C) He2 < O2 < NO < C2
(D) He2+ < O2– < C2 < NO
9.
Pick out the incorrect statement.
(A) N2 has greater dissociation energy than N2+ (B) O2 has lower dissociation energy than O2+
(C) Bond length in N2+ is less than N2
(D) Bond length in NO+ is less than in NO.
10.
The species which are diamagnetic :
(A) O2–
(B) NO2
uh
(C) ClO2
(D) N2O4
Ja
11.
a
ri
5.
Which of the following is observed in metallic bonds ?
(A) Mobile valence electrons
(B) Localised electrons
(C) Highly directed bond
(D) None of these
PART - II : SINGLE OR DOUBLE INTEGER TYPE
Find out the no. of correct statements :
(a) Bond length N2+ > Bond length N2
(c) Bond length CN– < Bond length of CN
(e) Bond length O2 > Bond length of O2+
Sa
nk
a
2.
Find the no. of species having fractional bond order ?
(a) N2+
(b) N2–
(c) O2
(e) F2
(f) B2
(g) C2+
lp
1.
3.
4.
(d) O2+
(h) CN–
(i) NO+
(b) Bond length NO+ < Bond length of NO
(d) Bond length O2– < Bond length of O2–2
(f) Bond length B2 > Bond length of B2–
In how many conversions, the bond length increases ?
(i) NO ⎯→ NO+ (ii) N2+ N2–
(iii) O2 ⎯→ O2+
(v) NH3 ⎯→ NH4+
(vi) NH3⎯→NH2–
(iv) H2 ⎯→ H2+
(vii) BF3⎯→BF4–
Which of the following have bond order less than two ?
(a) NO3–
(b) CO32 –
(c) F2
(e) Br2
(f) O22–
(g) O2–
2+
+
(i) O2
(j) Li2
(k) He2+
(d) Cl2
(h) N2–
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
1.
2.
3.
Which of the following have bond order three ?
(A) O22+
(B) NO+
(C) CN–
(D) CN+
The species which are paramagnetic is/are :
(A) NO
(B) NO2
(D) N2O4
(C) ClO2
Which of the statement(s) are correct ?
(A) There is a single bond in FO+
(B) The F and O are further apart in FO– than in FO+.
(C) There is a double bond in FO–.
(D) It would take more energy to break F–O bond in FO+ than in FO–.
95
4.
Among the following, the species with one unpaired electron are :
(A) O2+
(B) NO
(C) O2–
(D) B2
Identify correct statements
(A) Down the group strength of metallic bond increases nearly all in transition elements.
(B) Down the group strength of metallic bond increases in alkali metals.
(C) Down the group strength of metallic bond decreases in alkali metals.
(D) Down the group strength of metallic bond decreases in transition metals.
6.
Which of the following statements are correct for band theory of metallic bond.
(A) Valence band is empty or half filled in metal.
(B) Conduction band is empty in metal
(C) Energy gap between conduction and valence band is very large in non-conductors.
(D) Overlapping of conduction & valence band occurs in semi-conductors
7.
The force that binds a metal atom to a number of electrons with in its sphere of influence is known as a
metallic bond. Now, which of these is /are true for this found.
(A) Metallic bond is non-directional in nature.
(B) Metallic bonds are weaker than covalent bond.
(C) Energy required to vapourise a mole of metal (say, copper) to the vapour state is larger than the
energy required to vapourise a mole of a covalent substance (say, graphite)
(D) The valency electrons in a metallic bond are mobile.
uh
a
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5.
PART - IV : COMPREHENSION
Ja
Read the following passage carefully and answer the questions.
Comprehension # 1
Sa
nk
a
lp
The distribution of electrons among various molecular orbitals is called the electronic configuration of
the molecule which provides us the following very important informations about the molecule .
(A) Stability of molecule : The molecule is stable if number of bonding molecular orbital electrons
(Nb) is greater than the number of antibonding molecular orbital electrons (N a) and vice- versa.
1
(B) Bond order : Bond order =
(Nb – Na)
2
A positive bond order means a stable molecule while a negative or zero bond order means an unstable
molecule.
(C) Nature of the bond : Bond order 1, 2,or 3 corresponds to single, double or triple bonds
respectively.
(D) Bond length : Bond length decreases as bond order increases.
(E) Magnetic nature : Molecular orbitals in a molecule are doubly occupied, the substance is
diamagnetic and if one or more molecular orbitals are singly occupied, it is paramagnetic.
1.
Which of the following statements is incorrect ?
(A) Among O2+, O2 and O2– the stability decreases as O2+ > O2 > O2–
(B) He2 molecule does not exit as the effect of bonding and anti-bonding molecular orbitals cancel each
other
(C) C2, O22– and Li2 are diamagnetic
(D) In F2 molecule, the energy of 2 Pz is more than  2px and  2 Py
2.
The bromine (Br2) is coloured because:
(A) the difference in energy (E) between HOMO and LUMO is large and the electronic excitation take
place by absorption of light which falls in ultra violet region.
(B) the difference in energy (E) between HOMO and LUMO is small and the electronic excitation take
place by absorption of light which falls in infrared region.
(C) the bromine molecule is paramagnetic and the difference in energy (E) is such that the electronic
excitation take place in visible light.
(D) the difference in energy (E) between HOMO and LUMO is such that the electronic excitation take
place by absorption of light which falls in visible region and bromine molecule is diamagnetic.
96
3.
N2 has greater bond dissociation energy than N2+ , where as O2 has a lower bond dissociation energy
than O2+ because:
(A) Bond order is reduced when O2 is ionized to O2+ and bond order is increased when N2 is ionized to N2+
(B) Bond order is increased when O2 is ionized to O2+ and bond order is decreased when N2 is ionized to N2+
(C) Bond order is deceased when O2 is ionized to O2+ and bond order is decreased when N2– is ionized to N2+
(D) None of these.
4.
5.
Order of boiling point of K, Ca, Sc is
(A) K > Ca > Sc
(B) Ca > K > Sc
a
ri
Comprehension # 2
In a molten metal, the metallic bond is still present, although the order structure has been broken down.
The metallic bond isn’t fully broken until the metal boils. That means boiling point is actually a better
guide to the strength of the metallic bond then melting point is. On melting the bond is loosened, not
broken.
(C) Sc > Ca > K (D) K > Sc > Ca
Order of boiling point & melting point of Zn, Cd, Hg, respectively is :
(A) Zn > Cd > Hg & Zn > Cd > Hg
(B) Hg > Cd > Zn & Zn > Cd > Hg
(C) Hg > Cd > Zn & Hg > Cd > Zn
(D) Zn > Cd > Hg & Hg > Cd > Zn
Sa
nk
a
lp
Ja
uh
Comprehension # 3
Two models are considered to explain metallic bonding :
(A) Band model
(B) Electron-sea model
(A)
Band Model :
The interaction of two atomic orbitals, say the
3s-orbitals of two sodium atoms, produces two
molecular orbitals, one bonding orbital and one
antibonding orbital. If N atomic orbitals interact,
N molecular orbitals are formed. Atoms interact
more strongly with nearby atoms than with
those farther away. The energy that separates
bonding and antibonding molecular orbitals
decreases as the interaction (overlap) between
the atomic orbitals decreases. When we
consider all the possible interactions among
Figure-1. The band of orbitals resulting from
one mole of Na atoms, there is formation of
interaction of the 3s-orbitals in a crystal of sodium
series of very closely spaced molecular orbitals
(3s and
3*s). This consists of a nearly continuous band of orbitals belonging to the crystal as a whole. One
mole of Na atoms contributes one mole (6.02 × 1023) of valence electrons thus, 6.02 × 1023 orbitals in
the band are half-filled.
The empty 3 p atomic orbitals of Na atoms also
interact to form a wide band of 3 × 6.07 × 1023
orbitals.
The 3s and 3p atomic orbitals are quite close in
energy, so that these bands of molecular orbitals
overlap. The two overlapping bands contain 4 × 6.02
×1023 orbitals. Because each orbital can hold two
electrons, the resulting combination of bands is only
one-eighth full.
According to band theory, the highest-energy
electrons of metallic crystals occupy either a partially
filled band or a filled band that overlaps an empty
band. A band within which (or into which) electrons
must move to allow electrical conduction is called a
conduction band. The electrical conductivity of a
Figure-2. Overlapping of a half–filled “3s” band
metal decreases as temperature increases. The
with an empty “3p” band of NaN crystal
increase in temperature causes thermal agitation of
the metal ions. This impedes the flow of electrons
when an electric field is applied.
97
uh
a
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(B)
Crystalline non-metals, such as diamond and phosphorus, are insulators, they do not conduct
electricity. It is due to the fact that their highest-energy electrons occupy filled bands of molecular
orbitals that are separated from the lowest empty bond (conduction band) by an energy difference
called the band gap. In an insulator, this band gap is an energy difference that is too large for electrons
to jump to get to the conduction band.
Elements that are semiconductors have filled bands that are only slightly below, but do not overlap with
empty bands. They do not conduct electricity at low temperatures, but a small increase in temperature
is sufficient to excite some of the highest-energy electrons into the empty conduction band.
Electron-Sea Model :
Metals have ability to conduct electricity, ability to conduct heat, ease of deformation [that is, the ability
to be flattened into sheets (malleability) and to be drawn into wires (ductility)] and lustrous appearance.
One over simplified model that can account for some of these properties is the electron-sea model. The
metal is pictured as a network of positive ions immersed in a “sea of electrons”. In lithium the ions
would be Li+ and one electron per atom would be contributed to the sea. These free electrons account
for the characteristic metallic properties. If the ends of a bar of metal are connected to a source of
electric current, electrons from the external source enter the bar at one end. Free electrons pass
through the metal and leave the other end at the same rate.
In thermal conductivity no electrons leave or enter the metal but those in the region being heated gain
kinetic energy and transfer this to other electrons.
According to the electron-sea model, the case of deformation of metals can be thought of in this way : If
one layer of metal ions is forced across another, perhaps by hammering, the internal structure remains
unchanged as the sea of electrons rapidly adjusts to the new situation.
Considering band model, select the incorrect statement :
(A) Li metal should have partilly filled valence band and empty conduction band.
(B) Mg metal should have fully filled valence band and overlapping conduction band.
(C) Electrical conductivity of a metal decreases as temperature increases.
(D) The energy spread of each atomic energy level of an element behaving like a semiconductor is
infinitisimally small.
7.
All metal written below have usually low melting points except :
(A) Caesium
(B) Gallium
(C) Gold
(D) Mercury
lp
Which of the following physical properties can be explained by electron sea model :
(A) Electrical conduction
(B) Thermal conduction
(C) Malleability
(D) All of these
a
8.
Ja
6.
Sa
nk
Comprehension # 4
Answer Q.9, Q.10 and Q.11 by appropriately matching the information given in the three
columns of the following table.
Observe the three columns in which column-1 represents molecule, column-2 represents bond orders
while column-3 represents molecule properties. Properties of molecule explained by molecular orbital
theory. Like megnatic nature, orbital mixing etc.
Column 1
Column 2
Column 3
(I) B2
(i) Bond Order = 2
(P) Diamagnetic in Nature
(II) O2+
(ii) Bond Order = 2.5
(Q) SP Mixing Occure
(III) F2
(iii) Bond Order = 1
(R) Paramagnetic in Nature
(S) Highest Occupied Molecular orbital (HOMO) is Bonding
(IV) C2
(iv) Bond Order = 3
Molecular orbital (BMO)
9.
10.
11.
Which is incorrect combination?
(A) (I) (iii) (Q)
(B) (II) (ii) (P)
(C) (III) (iii) (P)
(D) (IV) (i) (P)
Which is correct combination for Diamagnetic species ?
(A) (III) (i) (P)
(B) (IV) (ii) (R)
(C) (IV) (i) (S)
(D) (III) (iii) (Q)
Which is correct combination?
(A) (III) (ii) (S)
(B) (III) (iii) (Q)
(D) (I) (iii) (R)
(C) (II) (ii) (P)
98
* Marked Questions may have more than one correct option.
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
1.
Write the Molecular orbital electron distribution of O 2. Specify its bond order and magnetic property.
[JEE–2000(M), 3/135]
2.
Which of the following molecular species has unpaired electron(s) ?
(A) N2
(B) F2
(C) O2–
[JEE–2002(S), 3/150]
(D) O22–
According to molecular orbital theory, which one of the following statements about the molecular
species O2+ is correct ?
[JEE–2004(S), 3/144]
(A) It is paramagnetic and has less bond order than O 2
(B) It is paramagnetic and more bond order than O2
(C) It is diamagnetic and has less bond order than O2
(D) It is diamagnetic and has more bond order than O2
4.
Arrange the following three compounds in terms of increasing O—O bond length :
O2, O2 [AsF6], K [O2]
Justify your answer based on the ground state electronic configuration of the dioxygen species in these
three compounds.
[JEE–2004(M), 2/144]
5.
The species having bond order different from that in CO is :
(A) NO–
(B) NO+
(C) CN–
uh
Ja
6.
a
ri
3.
Among the following, the paramagnetic compound is :
(A) Na2O2
(B) O3
(C) N2O
[JEE–2007, 3/162]
(D) N2
[JEE–2007, 3/162]
(D) KO2
Statement-1 : Band gap in germanium is small, because
Statement-2 : The energy spread of each germanium atomic energy level is infinitesimally small.
[JEE–2007, 3/162]
(A) Statement-1 is true, statement-2 is true; statement-2 is a correct explanation for statement-1.
(B) Statement-1 is true, statement-2 is true; statement-2 is NOT a correct explanation for statement-1.
(C) Statement-1 is true, statement-2 is false.
(D) Statement-1 is false, statement-2 is true.
8.
Statement-1 : Boron always forms covalent bond, because
Statement-2 : The small size of B3+ favours formation of covalent bond.
[JEE–2007, 3/162]
(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
(B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1.
(C) Statement-1 is True, Statement-2 is False.
(D) Statement-1 is False, Statement-2 is True.
Sa
nk
a
lp
7.
9.
Match each of the diatomic molecules in Column I with its property/properties in Column II.
Column I
Column II
[JEE–2009, 8/160]
(A)
B2
(p)
Paramagnetic
(B)
N2
(q)
Undergoes oxidation
(C)
O2–
(r)
Undergoes reduction
(D)
O2
(s)
Bond order  2
(t)
Mixing of 's' and 'p' orbitals
10.
Assuming that Hund's rule is violated, the bond order and magnetic nature of the diatomic molecule B2
is :
(A) 1 and diamagnetic
(B) 0 and diamagnetic
[JEE–2010, 5/163]
(C) 1 and paramagnetic
(D) 0 and paramagnetic
11.
Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is :
[JEE(Advanced) 2014, 3/120]
(A) Be2
(B) B2
(C) C2
(D) N2
99
Match the orbital overlap figures shown in List-I with the description given in List-II and select the
correct answer using the code given below the lists.
[JEE(Advanced) 2014, 3/120]
List-I
List-II
P.
1.
p–d  antibonding
Q.
2.
d–d  bonding
R.
3.
p–d  bonding
S.
4.
d–d  antibonding
Code :
(A)
(C)
Q
1
3
R
3
1
S
4
4
(B)
(D)
P
4
4
Q
3
1
According to Molecular orbital Theory,
(A) C22– is expected to be diamagnetic
(B) O22+ is expected to have a longer bond length than O2
(C) N2+ and N2– have the same bond order
(D) He2+ has the same energy as two isolated He atoms
R
1
3
S
2
2
[JEE(Advanced) 2016, 4/124]
Ja
uh
13.*
P
2
2
a
ri
12.
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) ONLINE PROBLEMS
1.
Increasing order of bond strength of O2, O2–, O22– and O2+ is :
(1) O2+ < O2 < O2– < O22–
(2) O2 < O2+ < O2– < O22–
–
2–
+
(3) O2 < O2 < O2 < O2
(4) O22– < O2– < O2 < O2+
2.
The bond order in NO is 2.5 while that in NO + is 3. Which of the following statements is true for these
two species?
[AIEEE-2004, 3/225]
(1) Bond length in NO+ is greater than in NO
(2) Bond length in NO is greater than in NO+
(3) Bond length in NO+ is equal to that in NO
(4) Bond length is unpredictable
3.
Which one of the following species is diamagnetic in nature ?
[AIEEE-2005, 1½/225]
(1) He2+
(2) H2
(3) H2+
(4) H2–.
Which of the following molecules/ions does not contain unpaired electrons?
[AIEEE-2006, 3/165]
(1) O22–
(2) B2
(3) N2+
(4) O2
lp
a
Sa
nk
4.
5.
6.
[AIEEE-2002, 3/225]
Which of the following species exhibits the diamagnetic behaviour?
(1) O22–
(2) O2+
(3) O2
[AIEEE-2007, 3/120]
(4) NO
In which of the following ionization processes, the bond order has increased and the magnetic
behaviour has changed ?
[AIEEE-2007, 3/120]
(1) O2 ⎯⎯
→ O2+
(2) N2 ⎯⎯
→ N2+
(3) C2 ⎯⎯
→ C2+
(4) NO ⎯⎯
→ NO+
7.
Which one of the following pairs of species has the same bond order?
(1) CN– and CN+
(2) O2– and CN–
(3) NO+ and CN+
[AIEEE-2008, 3/105]
(4) CN– and NO+
8.
Using MO theory predict which of the following species has the shortest bond length ?
[AIEEE-2009, 4/144]
(1) O2+
(2) O2–
(3) O22–
(4) O22+
9.*
Which one of the following molecules is expected to exhibit diamagnetic behaviour ?
[JEE(Main)-2013, 4/120]
(1) C2
(2) N2
(3) O2
(4) S2
100
In which of the following pairs of molecules/ions, both the species are not likely to exist ?
[JEE(Main)-2013, 4/120]
(1) H2+, He22–
(2) H2–, He22–
(3) H22+, He2
(4) H2–, He22+
11.
Stability of the species Li2, Li2– and Li2+ increases in the order of :
(1) Li2 < Li2+ < Li2–
(2) Li2– < Li2+ < Li2
(3) Li2 < Li2– < Li2+
[JEE(Main)-2013, 4/120]
(4) Li2– < Li2 < Li2+
12.
Which of the following species is not paramagnetic?
(1) CO
(2) O2
(3) B2
[JEE(Main)-2017, 4/120]
(4) NO
13.
According to molecular orbital theory, which of the following will not be a viable molecule ?
[JEE(Main)-2018, 4/120]
–
2–
2+
(1) H2
(2) H2
(3) He2
(4) He2+
JEE(MAIN) ONLINE PROBLEMS
a
ri
10.
Which of the following has unpaired electron(s) ?
[JEE(Main) 2014 Online (09-04-14), 4/120]
(1) N2
(2) O2–
(3) N22+
(4) O22–
2.
The correct order of bond dissociation energy among N2, O2, O2– is shown in which of the following
arrangements?
[JEE(Main) 2014 Online (11-04-14), 4/120]
(1) N2 > O2– > O2
(2) O2– > O2 > N2
(3) N2 > O2 > O2–
(4) O2 > O2– > N2
3.
Which one of the following molecules is paramagnetic ? [JEE(Main) 2014 Online (19-04-14), 4/120]
(1) N2
(2) NO
(3) CO
(4) O3
4.
After understanding the assertion and reason, choose the correct option.
Assertion : In the bonding molecular orbital (MO) of H2, electron density is increased between the
nuclei.
Reason : The bonding MO is A + B, which shows destructive interference of the combining electron
waves.
[JEE(Main) 2015 Online (10-04-15), 4/120]
(1) Assertion is correct, reason is incorrect.
(2) Assertion is incorrect, reason is correct.
(3) Assertion and reason are correct, but reason is not the correct explanation for the assertion.
(4) Assertion and reason are correct and reason is the correct and reason is the correct explanation for
the assertion.
5.
In the molecular orbital diagram for the molecular ion, N2+ , the number of electrons in the 2p molecular
orbital is :
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 0
(2) 2
(3) 3
(4) 1
Sa
nk
a
lp
Ja
uh
1.
6.
Which of the following best describes the diagram below of a molecular orbital ?
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) A non-bonding orbital
(3) A bonding  orbital
7.
+
–
–
+
(2) An antibonding  orbital
(4) An antibonding  orbital
According to molecular orbital theory, which of the following is true with respect to Li 2+ and Li2– ?
[JEE(Main) 2019 Online (09-01-19), 4/120]
(1) Li2+ is unstable and Li2– is stable
(2) Li2+ is stable and Li2– is unstable
(3) Both are stable
(4) Both are unstable
101
8.
In which of the following processes, the bond order has increased and paramagnetic character has
changed to diamagnetic?
[JEE(Main) 2019 Online (09-01-19), 4/120]
(1) NO ⎯→ NO+
(2) O2 ⎯→ O22–
(3) O2 ⎯→ O2+
(4) N2 ⎯→ N2+
a
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EXERCISE - 1
PART – I
(a) 1
(b) 1/2 (c) 0
(d) 1
(e) 0
(f) 1
A-2.
(a) B2 (b) C2 (c) O22+ (d) O2 , (e) F2 (f) N2
A-3.
3
B-1.
Boron (B2) : B2 is a good example of the energy level shift caused by the mixing of s and p orbitals. In
uh
A-1.
the absence of mixing, the g (2p) orbital is expected to be lower in energy than the u(2p) orbitals and
Ja
the resulting molecule would be diamagnetic. However, mixing of the g(2s) orbital with the g(2p)
orbital lowers the energy of the g(2s) orbital and increases the energy of the g(2p) orbital to a higher
level than the  orbitals, giving the order of energies shown below. As a result, the last two electrons
are unpaired in the degenerate (having the same energy)  orbitals, and the molecule is paramagnetic.
B-2.
lp
(1s)2 (*1s)2 (2s)2 (*2s)2 (2p1x = 2p1y) (pz)0.
NO+ and NO are derivative of N2; so NO+ bond order = 3 and NO bond order = 2.5; B.O.  bond
strength.
(ii) & (iv)
B-4.
O–O bond length order is ii < i < iii
Sa
nk
a
B-3.
C-1.
Weakest metallic bonding amongst the 3d-series elements → no unpaired electrons available for
metallic bonding in case of zinc.
C-2.
Be should have higher melting point as it contain 2 electrons for metallic bonding where as Li contain
only one. Further more, size of Be is smaller than that of Li.
PART – II
A-1.
(C)
A-2.
(D)
A-3.
(C)
A-4.
(A)
A-5.
(D)
A-6.
(B)
A-7.
(C)
B-1.
(B)
B-2.
(A)
B-3.
(A)
B-4.
(C)
B-5.
(D)
B-6.
(D)
C-1.
(D)
C-2.
(D)
C-3.
(D)
102
PART – III
(A – r) ; (B – p) ; (C – s) ; (D – q)
1.
EXERCISE – 2
PART – I
(A)
2.
(A)
3.
(C)
4.
6.
(C)
7.
(A)
8.
(D)
9.
11.
(A)
PART – II
(A)
4 (a, b, d, g)
2.
6 (a, b, c, d, e, f)
3.
4 (ii, iv, vi, vii)
4.
9 (a, b, c, d, e, f, g, j, k)
(ABC)
2.
(ABC)
6.
(BC)
7.
(ABD)
3.
Ja
1.
(D)
(C)
10.
(D)
uh
1.
PART – III
5.
a
ri
1.
(BD)
4.
(ABC)
5.
(AC)
lp
PART – IV
(D)
2.
(D)
3.
(B)
4.
(C)
5.
(A)
6.
(D)
7.
(C)
8.
(D)
9.
(B)
10.
(C)
11.
(D)
Sa
nk
a
1.
1.
EXERCISE – 3
PART – I
Molecular orbital electronic configuration of O2 is as follows (Z is taken as molecular axis).
1s2  *1s2  2s2  *2s2  2 p 2  2 p 2 =  2 p
z
Bond order =
x
2
y
 *2 p 1 =  *2 p 1 
x
y
10 – 6
= 2.
2
As it contains two unpaired electrons in bonding  molecular orbitals O2 is paramagnetic.
So, Magnetic moment =
2.
(C)
3.
n (n + 2) =
2 (2 + 2) = 2.83 B.M.
(B)
103
4.
The electronic configuration of O2 will be:
O2 = 1s2 *1s2 2s2 *2s2 2p2z 2p2x = 2p2y *2p1x = *2p1y
Now
bond order =
Nb − Na
2
Where, Nb = Number of electrons in bonding orbitals
Na = Number of electrons in antibonding orbitals
bond order =
10 − 6
=2
2
1s2 *1s2 2s2 *2s2 2p2z 2p2x = 2p2y *2p2x = *2p1y
Bond order =
10 − 7
3
=
= 1.5
2
2
In O2 [AsF4]–, O2 is O2+ .
The electronic configuration of O2+ will be
1s2 *1s2 2s2 *2s2 2p2z 2p2x, = 2p2y *2p1x
10 − 5
= 2.5
2
uh
bond order =
(D)
5.
(A)
9.
(A) - p, q, r, t ; (B) - q, r, s, t ; (C) - p, q, r ; (D) - p, q, r, s
10.
(A)
(C)
12.
(C)
(C)
1
.
Bond length
8.
(A)
13.*
(AC)
lp
11.
7.
Ja
Hence bond length order will be O+2 < O2 < O–2 because Bond order 
6.
a
ri
Similarly electronic configuration of O2– (in KO2) will be
PART – II
a
JEE(MAIN) OFFLINE PROBLEMS
(4)
2.
6.
(4)
7.
11.
(2)
12.
(2)
3.
(2)
4.
(1)
5.
(1)
(4)
8.
(4)
9.*
(1, 2)
10.
(3)
(1)
13.
(2)
4.
(1)
5.
(4)
Sa
nk
1.
JEE(MAIN) ONLINE PROBLEMS
1.
(2)
2.
(3)
3.
(2)
6.
(4)
7.
(3)
8.
(1)
104
Ionic or Electrovalent Bond :
It is cleared from the Kossel and Lewis approach that the formation of an ionic compound would
primarily depends upon :
* The ease of formation of the positive and negative ions from the respective neutral atoms.
* The arrangement of the positive and negative ions in the solid, that is the lattice of the crystalline
compound.
Conditions for the formation of ionic compounds :
Electronegativity difference between two combining elements must be larger.
Ionization enthalpy (M(g) → M+(g) + e–) of electropositive element must be low.
Negative value of electron gain enthalpy (X (g) + e– → X–(g)) of electronegative element should be high.
Lattice enthalpy (M+(g) + X– (g) → MX (s)) of an ionic solid must be high.
a
ri
(i)
(ii)
(iii)
(iv)
Lattice Enthalpy :
General properties of ionic compounds :
(a)
Physical state : At room temperature ionic compounds exist either in solid state or in solution phase
but not in gaseous state.
Isomorphism : Simple ionic compounds do not show isomerism but isomorphism is their important
characteristic. Crystals of different ionic compounds having similar crystal structures are known to be
isomorphs to each other and the phenomenon is known as isomorphism.
e.g., FeSO4 .7H2O
|
MgSO4 . 7H2O
Ja
(b)
uh
The lattice enthalpy of an ionic solid is defined as the energy required to completely separate one mole
of a solid ionic compound into gaseous constituent ions.
Sa
nk
a
lp
Conditions for isomorphism
(i)
The two compounds must have the same formula type e.g., MgSO4 & ZnSO4 ; BaSO4 & KMnO4
are isomorphous because they have same formula type. All alums are isomorphous because
they have same general formula :
M2SO4.M2 (SO4)3.24H2O
M = monovalent ; M = trivalent
(ii)
The respective structural units, atoms or ions need not necessarily be of same size in the two
compounds but their relative size should be little different.
(iii)
The cations of both compound should be of similar shape or structure (isostructural). Similarly
anions of both compounds should be isostructural.
(a)
SO42– and MnO4– have same shape i.e. tetrahedral, so isomorphous.
(b)
NaNO3 & NaClO3 they have same formula type yet they are not isomorphous because NO 3– is
trigonal planar but ClO3– pyramidal.
NO3– (sp2)
(c)
(d)
ClO3– (sp3)
(iv)
The respective structural units should have same polarisation property.
Electrical conductivity : Ionic solids are almost non-conductors. However they conduct a very little
amount of current due to crystal defects. All ionic solids are good conductors in molten state as well as
in their aqueous solutions because their ions are free to move.
Solubility of ionic compounds : Soluble in polar solvents like water which have high dielectric
constant.
105
Ex-1.
Sol.
Arrange in order of increasing ionic radii in water and their mobility : Be 2+ , Mg 2+, Ca2+, Sr2+.
(i) Size  1/degree of hydration (i.e. with increase in size, number of water molecules around central
metal ions decrease). So order of increasing radii is Sr 2+ < Ca2+ < Mg2+ < Be2+.
(ii) Heavily hydrated ions move slowly so the order of increasing mobility is
Be2+ < Mg2+ < Ca2+ < Sr2+.
Section (A) : Fajan’s Rule and its applications
Covalent character in ionic compounds (Fajan’s rule) :
a
ri
When anion and cation approach each other, the valence shell of anion is pulled towards cation
nucleus and thus shape of anion is deformed. This phenomenon of deformation of anion by a cation is
known as polarisation and the ability of cation to polarize a near by anion is called as polarizing power
of cation.
uh
Size of cation increases
Polarisation decreases
Covalent character decreases
Size of anion : Larger is the anion, greater is its polarisability and, therefore, more will be the
polarisation. Thus more will be covalent character in compound.
Size of anion  polarisation
e.g.,
LiF
LiCl
LiBr
Lil
lp
(ii)
Ja
(i)
Fajan’s pointed out that greater is the polarization of anion in a molecule, more is covalent character in
it.
More distortion of anion, more will be polarisation then covalent character increases.
Fajan’s gives some rules which govern the covalent character in the ionic compounds, which are as
follows:
Size of cation : Smaller is the cation more is its polarizing power and thus more will be the polarisation
of anion. Hence more will be covalent character in compound.
Size of cation  1 / polarisation.
e.g.
BeCl2
MgCl2
CaCl2
SrCl2
BaCl2
a
Sa
nk
(iii)
– Size of anion increases
– Polarisation increases
– Covalent character increases
Charge on cation : Higher is the oxidation state of cation, more will be the deformation of anion and
thus, more will be covalent character in compound.
Charge on cation  polarisation.
e.g.,
NaCl MgCl2 AlCl3
Na+
Mg2+ Al3+
– Charge of cation increases
– Polarisation increases
– Covalent character increases
(iv)
Charge on anion : Higher is the charge on anion more will be the polarisation of anion and thus more
will be covalent character in the compound.
Charge on anion  polarisation.
e.g.,
AlF3
Al2O3 AlN
F– ,
O2– , N3–
– Charge on anion increases
– Polarisation increases
– Covalent character increases
(v)
Pseudo inert gas configuration of cation : Cation having pseudo inert gas configuration has more
polarizing power than the cation that has inert gas configuration. Thus NaCl having inert gas
106
configuration will be more ionic whereas CuCl having pseudo inert gas configuration will be more
covalent in nature.
Cu+ =
[Ne] 3s2 p6 d10
Na+ = 1s2 2s2 p6
18e–
8e–
Pseudo inert gas configuration
Inert gas configuration
(poor shielding of d-electrons) (more shielding of s and p electrons)
(i)
(ii)
Figure
Variation of melting point [melting point of covalent compound < melting point of ionic
compound] :
BeCl2 , MgCl2 , CaCl2, SrCl2, BaCl2
––––––––––––––>–––––––––––––––––––––>–
Ionic character increases, melting point increases ; since size of cation increases & size of anions
is constant.
CaF2, CaCl2, CaBr2, Ca2
––––––––––––––>––––––––––––––––––>–
Covalent character increase, melting point decrease ; since size of anions increase & size of cations is
constant.
a
AgCl is colourless whereas AgI is yellow, because of :
(A) Ag+ have 18 electron shell to screen the nuclear charge.
(B) Ag+ shows pseudo inert gas configuration.
(C) distortion of I– is more pronounced than Cl– ion.
(D) existence of d–d transition.
Sa
nk
Ex-2.
lp
Ja
(iv)
uh
a
ri
(iii)
Application & Exceptions of Fajan’s Rules :
Ag2S is less soluble than Ag2O in H2O because Ag2 S is more covalent due to bigger S2– ion.
Fe(OH)3 is less soluble than Fe(OH)2 in water because Fe3+ is smaller than Fe2+ and thus charge is
more.
Therefore, Fe(OH)3 is more covalent than Fe(OH)2 .
The colour of some compounds can be explained on the basis of polarisation of their bigger negative
ions.
For example :
AgCl is white AgBr, Ag, Ag2CO3 are yellow. Similarly, SnCl2 is white but Sn2 is black. PbCl2 is white
but Pb2 is yellow.
The bigger anions are more polarised and hence their electrons get excited by partial absorption of
visible light.
Sol.
(C), the bigger anions are more polarised and hence their electrons get excited by partial absorption of
visible light.
Section (B) : Dipole moment.
Polarity of bonds :
In reality no bond or a compound is either completely covalent or ionic. Even in case of covalent bond
between two hydrogen atoms, there is some ionic character.
When a covalent bond is formed between two similar atoms, for example in H 2, O2, Cl2, N2 or F2 the
shared pair of electrons is equally attracted by the atoms. As a result electron pair is situated exactly
between the two identical nuclei. The bond so formed is called nonpolar covalent bond. Contrary to this
in case of a heteronuclear molecule like HF, the shared electron pair between the two atoms gets
displaced more towards fluorine since the electronegativity of fluorine is far greater than that of
hydrogen. The resultant covalent bond is a polar covalent bond.
107
As a result of polarisation, the molecule possesses the dipole moment which can be defined as the
product of magnitude of the partial charge (+ or –) developed on any of the covalently bonded atoms
and the distance between two atoms.
Dipole moment (µ) = Magnitude of charge (q) × distance of separation (d)
Dipole moment is usually expressed in Debye units (D). The conversion factors are

1 D = 3.33564 × 10–30 Cm, where C is coulomb and m is meter.

1 Debye = 10–18 e.s.u. cm.
Further dipole moment is a vector quantity and is depicted by a small arrow with tail on the positive
centre and head pointing towards the negative centre. For example the dipole moment of HF may be
represented as
uh
a
ri
The shift in electron density is represented by crossed arrow (
) above the Lewis structure to
indicate the direction of the shift.
In case of polyatomic molecules the dipole moment not only depend upon the individual dipole
moments of bonds known as bond dipoles but also on the spatial arrangement of various bonds in the
molecule. In such case, the dipole moment of a molecule is the vector sum of the dipole moments of
various bonds. i.e. a molecule will have a dipole moment if the summation of all of the individual
moment vector is non-zero.
2
2
R = P + Q + 2PQcos  , where R is resultant dipole moment.
Ja
For example of H2O molecule, which has a bent structure, the two O—H bonds are oriented at an angle
of 104.5º. Net dipole moment of 6.17 × 10–30 Cm (1D = 3.33564 × 10–30 Cm) is the resultant of the
dipole moments of two O—H bonds.
lp
Net Dipole moment,  = 1.85 D = 1.85 × 3.33564 × 10–30 Cm = 6.17 × 10–30 Cm.
Following compounds have zero dipole moment :
BF3, CO2, SO3, CF4, PF5(g), SF6, XeF2, CS2, CCl4, PCl5(g), XeF4

Some important orders
HF
>
HCl
>
HBr
>
1.92 D
1.08
0.78
CH3Cl >
CH2Cl2 >
CHCl3 >
1.86
1.6
1.0
Usually for disubstituted Benzene order is o > m > p
But it all depends on the substituents
Sa
nk
a


H
0.38
CCl4
0
Note : For geometrical isomers usually the dipole moment of cis is more than trans but again there
can be exceptions.
For example :
cis - form (low dipole moment)

trans - form (high dipole moment)
The presence of a centre of symmetry, i, requires that the dipole moment be zero, since any charge on
one side of the molecule is canceled by an equal charge on the other side of the molecule.
108
Thus [CoF6]3–, trans - N2F2 and the staggered conformer of ferrocene do not have dipole moments.
Some important points about dipole moment :

Ex-3.
Sol.
a
ri
uh

Ja

A polyatomic molecule having polar covalent bonds but zero dipole moment indicates the symmetrical
structure of the molecule. e.g. B–F bonds are polar in BF3 but BF3 has =0 due to its symmetrical
geometry.
If molecule have  = 0, then it should be linear or having symmetrical geometry.
e.g. linear – CO2, CS2, BeCl2(g) ; symmetrical geometry – BF3, CH4, PCl5, SF6, IF7, XeF4.
If molecule has   0 then it should be angular or having unsymmetrical geometry.
SnCl2 , PbCl2 , SO2 , angular molecular geometry.
NH3, H2O, NF3, SF4, H2S, unsymmetrical molecular geometry.
(Observed)
µExperimental
 100 =
 100
% Ionic character =
µTheoretical
(100% Ionic compound)
The resultant dipole moment of water is 1.85 D ignoring the effects of lone pair. Calculate, the dipole
moment of each OH bond (given that bond angle in H2O = 104°, cos 104° = – 0.25).
R2 = P2 + Q2 + 2PQ cos 
 1
(1.85)2 = x2 + x2 + 2x2  − 
 4
(1.85)2 = 2x2 –
3x 2
2

x = 1.51 D
For HCl molecule observed dipole moment is 1.03 D and bond length is 1.275 Å. Calculate % ionic
character.
Dipole moment = 4.8 × 10–18 × 1.275 × 10–8 = 4.8 × 1.275
1.03  100
% ionic character =
 17%
1.275  4.8
Sa
nk
Sol.

a
Ex-4.
x2
2
lp

Ex-5.
Sol.
Why NH3 is having more dipole moment than NF 3.
n
molecule Nitrogen is more electronegative than
Hydrogen. So the net dipole moment is towards.
Nitrogen atom but in NF3 molecule.
Fluorine is more electronegative than. Nitrogen so the
net dipole moment is towards fluorine atoms. In NH3
the bond pair moments and lone pair moments are in
the same direction while in NF3 the lone pair moment
and bond pair moments are in opposite direction.
Ex-6.
The geometry of SO3 and its dipole moment are :
(A) pyramidal and non-zero.
(B) trigonal planar and non-zero.
(C) trigonal planar and zero.
(D) T-shaped and zero.
(C)
Ans.
109
The steric number of sulphur = 3; so hybridisation is sp2. There is no lone pair on sulphur atom,
Sol.
therefore, according to VSEPR theory, the repulsions between the bond pairs of electrons will be
similar. Hence the molecule will be symmetrical (trigonal planar) with zero dipole moment.
Section (C) : Acidic and basic character
Types of Oxides and acidic / basic nature of oxyacids and hydra acid :
Ex.
Acidic oxides
(a) Solution in water will be acidic in nature
H2O
SO2 ⎯⎯⎯
→ H2SO3 Sulphurous acid

(b) will react with base but not with an acid
H O
2
SO3 ⎯⎯⎯
→ H2SO4 sulphuric acid
H O
Ex
(d)
(e)
Ex.
Amphiprotic oxide : which can accept and release H+ ions
 H2O
Periodicity in nature of oxides
L → R, metallic character  and non-metallic character . So, basic character of oxides  and acidic
character 
SO3
Na2O
Cl2O7
SiO2
Al2O3
P2O5
MgO
strongly
strongly
strongly
weakly
Amph
basic
acidic
basic
acidic
acidic
acidic
T → B, metallic character so, basic character of oxide will 
LiOH
LI2O
,
NaOH
Na2O,
KOH
K2O
RbOH
Rb2O,
CsOH
Cs2O
Basic character increases down the group
If the same element is forming oxides in diff oxidation states, then greater the oxidation no greater will
be acidic nature
Sa
nk
a
(a)
uh
(c)
H O
H O
2
2
Na2O ⎯⎯⎯
CaO ⎯⎯⎯
→ 2NaOH
→ Ca(OH)2
Generally, metallic oxides are basic oxides
Amphoteric oxides
Can react with an acid as well as with a base.
Generally, metalloids or elements close to metalloids can form amphoteric oxides.
BeO

BeO + 2HCl ⎯→ BeCl2 + H2O, BeO + 2NaOH⎯→NaAlO2 (Sodium meta aluminate)
ZnO

ZnO + HCl ⎯→ ZnCl2,
ZnO + NaOH ⎯→Na2ZnO2 (sodium zincate)
SnO

(stannous oxide)
SnO + HCl⎯→ SuCl2,
SnO + NaOH ⎯→ Na2SnO2 (sodium stanite)
SnO2 
(stannic oxide)
SnO2 + 4HCl ⎯→ SnCl4
SnO2 + NaOH ⎯→ Na2SnO3 (sodium stanate)
Neutral oxides : will not react with acids or bases
CO, N2O, NO, OF2
Ja
Ex.
lp
(b)
2
CO2 ⎯⎯⎯
→ H2CO3 Carbonic acid
Generally non-metallic oxides are acidic oxides
Basic oxides
(a) Solution in water will be basic in nature
(b) will react with an acid but not with a base
a
ri
(a)
(b)
(c)
+2
+7
+6
+3
+4
Mn2O7
MnO
MnO3
Mn2O3 MnO2
<
<
< more
more <
neutral acidic
basic
acidic
basic
110
oxides are anhydrides of oxyacid or hydroxides
+4
−H O
+5
+5
+4
+6
H3PO 4 ⎯→ P2O5 (P4 O10 ) ,
HNO3 ⎯→ N2O5,
HClO2 ⎯→ Cl2O3,
−H O
2
H2CO3 ⎯⎯⎯→
CO2
H3PO3 ⎯→ P2O3 (P4O6)
HNO2 ⎯→ N2O3 ,
HClO ⎯→ Cl2O,
+4
+5
−H O
2
H2SO 4 ⎯⎯⎯→
SO3 ,
2
H2SO3 ⎯⎯⎯→
SO2 ,
HO
+5
HClO4 ⎯→ Cl2O7,
NaOH ⎯→ Na2O,
HClO3 ⎯→ Cl2O5
Ca(OH)2 ⎯→ CaO
+3
uh
a
ri
2
Mixed anhydrides. ClO2 ⎯⎯⎯
→ HClO3 + HClO2
For oxy acids
(a)
On moving L → R non-metallic character  acidic strength
H3BO3 < H2CO3 < HNO3
H2SiO3 < H3PO4 < H2SO4 < HClO4
(b)
On moving T → B, non-metallic character  acidic strength 
HNO3 > H3PO4 > H3AsO4
HClO4 > HBrO4 > HO4
(c)
If same element is forming oxycids in different oxidation states then, Greater the oxidation no. of the
element greater will be the acidic strength
HNO3 > HNO2 > H2N2O2
HClO4 > HClO3 > HClO2 > HClO
Ex-2.
(a) Which one has highest and lowest melting point and why?
NaCl
KCl
RbCl
CsCl
(b) Why melting points of cesium halide (CsX) decrease in the order given below ?
CsF > CsCl > CsBr > CsI.
(a) NaCl will have highest lattice energy on account of the smaller Na + while CsCl has lowest lattice
energy on account of the larger Cs+. Hence NaCl has highest melting point and CsCl has lowest
melting point .
(b) As size of anions increase their polarisability increases thus their covalent character increases and
melting point decrease.
Sa
nk
Sol.
lp
Sol.
Classify the following bonds as ionic, polar covalent or covalent and give your reasons :
(a) SiSi bond in Cl3SiSiCl3
(b) SiCl bond in Cl3SiSiCl3
(c) CaF bond in CaF2
(d) NH bond in NH3
(a) Covalent, due to identical electronegativity.
(b) Polar covalent, due to less electronegativity difference.
(c) Ionic, due to more electronegativity difference.
(d) Polar covalent, due to different electronegativity.
a
Ex-1.
Ja
MISCELLANEOUS SOLVED PROBLEMS (MSPS)
Ex-3.
Sol.
Which of the compounds MgCO3 and ZnCO3 is thermally more stable ? Explain.
Mg+2 has less polarising power due to inert gas configuration while Zn +2 has higher polarising power
due to pseudo inert gas configuration. A cation i.e. Zn2+ with a greater, polarising power exercise a
strong pull on the electron cloud of the neighbouring O-atom of the CO32– ion and as such the metal
carbonate (ZnCO3) gets readily decomposed into CO 2 and the oxide of the metal, ZnO. Thus ZnCO 3 is
less stable than MgCO3.
Ex-4.
Sol.
Why is anhydrous HCl predominantly covalent in the gaseous state but is ionic in aqueous solution?
It exists as HCl (bond formed by equal sharing of electrons) but in aqueous solution ionises as H + (or
H3O+) and Cl– due to polarity of HCl.
Ex-5.
Sol.
Why BeF2 has zero dipole moment whereas H2O has some dipole moment ?
BeF2 has linear molecule and H2O has bent molecule.
=0
111
0
Ans.
Sol.
Ex-8.
Sol.
Ex-9.
The dipole moment of KCl is 3.336 × 10–29 Coulomb meter. The interionic distance in KCl is 2.6Å. Find
the % ionic character in KCl.
The theoretical dipole moment in KCl = e × d = 1.602 × 10–19 × 2.6 × 10–10 = 4.1652 × 10–29 C meter
exp erimental dipole moment
% ionic character =
× 100 = (3.336 × 10–29/4.1652 × 10–29) × 100
theoretical dipole moment
The gaseous potassium chloride molecule has a measured dipole moment of 10.0 D, which indicates
that it is a very polar molecule. The separation between the nuclei in this molecule is 2.67 × 10 –8 cm.
Calculate the percentage ionic character in KCl molecule.
Dipole moment of compound would have been completely ionic
= (4.8 × 10–10 esu) (2.67 × 10–8 cm) = 12.8 D
10.0
so % ionic character =
× 100% = 78.125 %  78% Ans.
12.8
Sa
nk
a
lp
Sol.
A diatomic molecule has a dipole moment of 1.2 D. If its bond distance is equal to 1.0Å then the fraction
of an electronic charge on each atom is :
(A) 25%
(B) 37%
(C) 52%
(D) 42%
(A)
Assuming complete charge transfer then dipole moment = (4.8 × 10–10 esu) (10–8 cm) = 4.8 D
1.2
so % ionic character =
× 100 % = 25%
4.8
a
ri
Ex-7.
uh
Sol.
Why crystals of hydrated calcium sulphate are soft and easily cleaved where as anhydrous calcium
sulphate are very hard and very difficult to cleave ?
With in the Ca2+ / SO42– layer the ions are held together by strong electrovalent bonds but these
separated Ca2+ / SO42– layers are linked by relative weak H–bond. The weak H-bonds link SO42– ion in
the intermediate region.
Ja
Ex-6.
 Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
Section (A) : Fajan’s Rule and its applications
A-1. Arrange the following in the increasing order of their covalent character.
(a) NaF, Na3N and Na2O
(b) NaCl, MgCl2, AlCl3, SiCl4 and PCl5
A-2.
SnCl4 has melting point - 15ºC where as SnCl2 has melting point 535ºC. Why?
A-3.
SnCl2 is white but SnI2 is red. Why ?
Section (B) : Dipole moment
B-1. Inorganic benzene is more reactive than organic benzene. Why?
B-2.
trans-1, 2 dichloro ethene have zero dipole moment while its cis- form has some dipole moment.
Explain.
B-3.
Why CCl4 have zero dipole moment but CHCl3 have some dipole moment ?
B-4. Arrange in increasing order of dipole moment ; H2O, H2S, BF3.
112
B-5. Dipole moment of LiF was experimentally determined and was found to be 6.32 D. Calculate
percentage ionic character in LiF molecule Li—F bond length is 156 pm.
Section (C) : Acidic & basic character
C-1._
(a) Write the formula of anhydride of the following acids :
(i) Nitrous acid (HNO2) (ii) Sulphuric acid (H2SO4)
(b) What are amphoteric oxides ? Give 2 examples.
C-2.
Arrange in the increasing order of acidic character : CO 2, N2O5, SiO2, SO3.
C-3. Arrange in the increasing order of basic character : MgO, SrO, K2O, NiO, Cs2O
PART - II : ONLY ONE OPTION CORRECT TYPE
A-1.
a
ri
Section (A) : Fajan’s Rule and its applications
Which of the following is in order of increasing covalent character ?
(A) CCl4 < BeCl2 < BCl3 < LiCl
(B) LiCl < CCl4 < BeCl2 < BCl3
(C) LiCl < BeCl2 < BCl3 < CCl4
(D) LiCl < BeCl2 < CCl4 < BCl3
A-3.
uh
A-2. Which of the following combination of ion will have highest polarisation ?
(A) Fe2+, Br–
(B) Ni4+, Br–
(C) Ni2+, Br–
(D) Fe, Br–
The correct order of decreasing polarizability of ion is :
(A) Cl–, Br–, I–, F–
(B) F–, I–, Br–, Cl–
(C) I–, Br–, Cl–, F–
(D) F–, Cl–, Br–, I–
Section (B) : Dipole moment
Which has maximum dipole moment ?
(A)
(B)
(C)
lp
B-1.
Ja
A-4. SnCl4 is a covalent liquid because :
(A) electron clouds of the Cl– ions are weakly polarized to envelop the cation.
(B) electron clouds of the Cl– ions are strongly polarized to envelop the cation.
(C) its molecules are attracted to one another by strong van der Waals forces.
(D) Sn shows inert pair effect.
(D)
Which of the following has the least dipole moment ?
(A) NF3
(B) CO2
(C) SO2
(D) NH3
Sa
nk
B-3.
a
B-2. Of the following molecules, the one, which has permanent dipole moment, is :
(A) SiF4
(B) BF3
(C) PF3
(D) PF5
B-4.
Which of the following compounds possesses zero dipole moment?
(A) Benzene (C6H6)
(B) Carbon tetrachloride
(C) Boron trifluoride
(D) All of these
B-5.
The geometry of H2S and its dipole moment are :
(A) angular and non zero
(B) angular and zero
(C) linear and non zero
(D) linear and zero
B-6. Which of the following has been arranged in order of decreasing dipole moment ?
(A) CH3Cl  CH3F  CH3Br  CH3I
(B) CH3F  CH3Cl  CH3Br  CH3I
(C) CH3Cl  CH3Br  CH3I  CH3F
(D) CH3F  CH3Cl  CH3I  CH3Br
B-7. The dipole moment of chlorobenzene is 1.73 D. The dipole moment of p-dichlorobenzene is expected
to be:
(A) 3.46 D
(B) 0.00 D
(C) 1.73 D
(D) 1.00 D
113
Section (C) : Acidic & basic character
C-1._
C-2._
C-3.
Which of the following is the strongest oxy-acid among the following :
(A) H2SO4
(B) H3PO4
(C) HClO4
(D) H2SiO3
Which of the following is the anhydride of Nitric acid (HNO 3) :
(A) NO2
(B) N2O3
(C) N2O5
(D) N2O
The order of basic character of given oxides is :
(A) Na2O > MgO > CuO > SiO2
(B) MgO > SiO2 > CuO > Na2O
(C) SiO2 > MgO > CuO > Na2O
(D) CuO > Na2O > MgO > SiO2
C-4. Amphoteric behaviour is shown by the oxides of :
(A) Al and Ca
(B) Pb and Ba
(C) Cr and Mg
Which one of the following oxides is neutral ?
(A) CO
(B) SnO2
(C) ZnO
a
ri
C-5.
(D) Sn and Zn
(D) SiO2
(p)
(q)
(r)
(s)
(t)
Column-II
Increasing order of covalent character
Decreasing order of thermal stability
Decreasing order of lattice energy
Increasing order of thermal stability
Increasing order of ionic character
Ja
Match the column:
Column-I
(A)
CsCl, CsBr, CsI
(B)
LiOH, NaOH, KOH
(C)
LiH, NaH, KH
(D)
Mg3N2, Ca3N2, Sr3N2
lp
1.
uh
PART - III : MATCH THE COLUMN
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
Among the following compounds the one that is polar and has central atom with sp 3 hybridisation is :
(A) H2CO3
(B) SiF4
(C) BF3
(D) HClO2
2.
Which of the following are polar.
(A) XeF4
(B) SO3
Sa
nk
a
1.
(C) XeOF4
(D) ICl4–
3.
Which of the following statemets is true?
(A) The dipole moment of NF3 is zero
(B) The dipole moment of NF3 is less than NH3
(C) The dipole moment of NF3 is more than NH3 (D) The dipole moment of NH3 is zero
4.
Which of the following would be expected to have a dipole moment of zero on the basis of symmetry?
(A) SOCl2
(B) OF2
(C) SeF6
(D) ClF5
5.
If molecule MX3 has Zero dipole moment, the hybrid orbitals used by M (Atomic No. < 21) are
(A) Pure p
(B) sp hybrid
(C) sp2 hybrid
(D) sp3 hybrid
6.
Which of the following are incorrect for dipole moment ?
(A) Lone pair of elements present on central atom can give rise to dipole moment
(B) Dipole moment is vector quantity
(C) PF5(g) molecule has non zero dipole moment
(D) Difference in electronegativities of combining atom can lead to dipole moment
7.
Which of the following has minimum melting point
(A) CsF
(B) HCl
(C) HF
(D) LiF
114
Which of the following is false ?
(A) Van der Waals forces are responsible for the formation of molecular crystals.
(B) Branching lowers the boiling points of isomeric organic compounds due to reduction in the van der
Waals force of attraction.
(C) In graphite, van der Waals forces act between the carbon layers.
(D) Boiling point of NH3 is greater than SbH3.
9.
Which of the following contains both electrovalent and covalent bonds ?
(A) CH4
(B) H2O2
(C) NH4Cl
(D) none
10.
The correct order of the increasing ionic character is :
(A) BeCl2 < MgCl2 < CaCl2 < BaCl2
(B) BeCl2 < MgCl2 < BaCl2 < CaCl2
(C) BeCl2 < BaCl2 < MgCl2 < CaCl2
(D) BaCl2 < MgCl2 < CaCl2 < BeCl2
11.
Least melting point is shown by the compound :
(A) PbCl2
(B) SnCl4
(C) NaCl
(D) AlCl3
CuI2 is unstable even at ordinary temperature because :
(A) the Cu2+ ion with a comparatively small radius has a strong polarising power.
(B) the Cu2+ ion with a 17 electron outer shell has weak polarising power.
(C) the I– ion with a larger radius has a high polarisability.
(D) both (A) and (C)
uh
12.
a
ri
8.
1.
Find total no. of polar molecules.
(a) PF3Cl2
(b) SF4
(e) SF6
(f) XeF2
Ja
PART - II : SINGLE OR DOUBLE INTEGER TYPE
(c) PCl5
(g) NO2+
(d) PCl3F2
(h) BF2Cl
How many of the following compounds are planar as well as non polar compound :
(a) C3O2
(b) CH2=C=CH2
(c) BF3
(d) CCl4
(e) SF6
(f) XeF4
(g) IF5
(h) IF7
(i) SF4
(j) ClF3
3.
How many of the following are correct orders of property indicated against it ?
(i) I– > Br– > Cl– > F–
(order of polarizability)
(ii) Li+ > Na+ > K+ > Cs+
(order of polarising power)
(iii) Li+ > Mg2+ > Al3+
(order of polarising power)
(iv) LiI > NaI > KI
(order of ionic character)
(v) AgI > AgBr > AgCl
(order of solubility in water)
(vi) (Si–Si bond) Si > SiO2 (Si–O bond) (order of % covalent character of bond.
Sa
nk
a
lp
2.
(i) BF3
4.
How many of the following oxides are acidic ?
(a) Mn2O7
(b) SO2
(e) NO2
(f) PbO2
(i) ZnO
(j) Fe2O3
(c) CO
(g) Na2O
(k) SiO2
(d) N2O
(h) CrO3
(l) SnO
PART - III : ONE OR MORE THAN ONE OPTION CORRECT TYPE
1.
The halogen form compounds among themselves with formula XX, XX3, XX5 and XX7 where X is the
heavier halogen. Which of the following pairs representing their structures and being polar and nonpolar are correct?
(A) XX – Linear – polar
(B) XX3 – T-shaped – polar
(C) XX5 – square pyramidal – polar
(D) XX7 – Pentagonal bipyramidal – non-polar
2.
Which of the following is/are correct statement(s) for dipole moment ?
(A) Lone pair of electrons present on central atom can give rise to dipole moment.
(B) Dipole moment is vector quantity.
(C) CO2 molecule has dipole moment.
115
(D) Difference in electronegativities of combining atoms can lead to dipole moment.
3.
Which of the following are polar ?
(A) XeF4
(B) XeF6
(D) XeF5–
(C) XeOF4
Which of the following compounds contain(s) both ionic and covalent bonds?
(A) NH4Cl
(B) KCN
(C) CuSO4·5H2O
(D) NaOH
5.
Which of the following factors do not favour electrovalency ?
(A) Low charge on ions
(B) High charge on ions
(C) Large cation and small anion
(D) Small cation and large anion
6.
Which statement(s) is/are correct ?
(A) Polarising power refers to cation.
(B) Polarisability refers to anion.
(C) Small cation is more efficient to polarise anion.
(D) Molecules in which cation having pseudo inert gas configuration are more covalent.
a
ri
4.
PART - IV : COMPREHENSION
uh
Read the following passage carefully and answer the questions.
lp
Ja
Comprehension # 1
The degree of polarity of a covalent compound is measured by the dipole moment (bond) of the bond
defined as:
bond = Charge on one of the poles  bond length
bond is a vector quantity. The dipole moment of a molecule is the vector addition of all the bond dipole
moments present in it. For a triatomic molecule, containing two bond's like H 2O, molecule is given by
2molecule = 2bond + 2bond + 2bond.bond cos 
 = bond angle
The % ionic character of a bond is calculated using the equations

% ionic character = obs  100
ionic
ionic = dipole moment when the molecule is assumed to be completely ionic.
The dipole moment of
Sa
nk
2.
Which of the follwing molecule has non-zero dipole moment (A) XeF2
(B) ClF3
(C) XeO2F4
a
1.
(A) 0 D
3.
is 1.5 D. The dipole moment of
(B) 1.5 D
(D) XeF4
will be -
(C) 2.86 D
(D) 2.25 D
Which of the following compound has Zero dipole moment (A) PCl3
(B) PCl2F3
(C) PCl3F2
(D) PClF4
Comprehension # 2
Molecular geometry is the general shape of a molecule as determined by the relative positions of the
atomic nuclei. VSEPR model predicts the shape of the molecules & ions in which valence shell electron
pairs are arranged about the atom as far away from one another as possible, thus minimizing pair
repulsion information about the geometry of a molecule can sometimes be obtained from an
experimental quantity called dipole moment.
4.
The dipole moment of a triatomic molecule AX 2 was found to be equal to the bond moment of A–X
bond. Which of the following information regarding geometry of the molecule can be drawn from the
above observation.
(A) Molecule is linear
(B) Molecule is V shaped with X – A – X = 90°
(C) Molecule is V shaped with X–A–X = 120°
(D) Molecular geometry can not be predicted with the given information
116
5.
Which of the following inter-halogen compounds is non-polar in nature:
(A) ClF3
(B) BrF5
(C) IF7
(D) BrCl
7.
Which of the following will be most covalent ?
(A) NaCl
(B) Na2S
(C) MgCl2
Which of the following is least ionic ?
(A) Be2
(B) BeCl2
(C) BeBr2
(D) MgS
uh
6.
a
ri
Comprehension # 3
A covalent bond in which electrons are shared unequally and the bonded atoms acquire a partial
positive and negative charge, is called a polar covalent bond. Bond polarity is described in terms of
ionic character.
Similarly in ionic bond, some covalent character is introduced because of the tendency of the cation to
polarise the anion. The magnitude of covalent character in the ionic bond depends upon the extent of
polarization caused by cations.
In general :
(i) Smaller the size of cation, larger is its polarizing power.
(ii) Larger the anion, more will be its polarisability.
(iii) Among two cations of similar size, the polarizing power of cations with pseudo - inert gas
configuration (ns2np6nd10) is larger than cation with noble gas configuration (ns 2np6) e.g. polarizing
power of Ag+ is more than K+.
(D) BeF2
Arrange the following compounds in increasing order of their ionic character :
SnCl2, SnCl4, SiCl4, SnF4, SnF2
(A) SnF2 < SnCl2 < SnF4 < SnCl4 < SiCl4
(B) SnF2 < SnCl2 < SnF4 < SiCl4 < SnCl4
(C) SiCl4 < SnCl4 < SnF4 < SnCl2 < SnF2
(D) SnCl4 < SnF4 < SnCl2 < SnF2 < SiCl4
9.
Which is the correct order of covalent character
(A) BeF2 < BeCl2 < BeBr2 < Bel2
(B) BeCl2 < BeF2 < Bel2 < BeBr2
(C) Bel2 < BeBr2 < BeCl2 < BeF2
(D) Bel2 < BeCl2 < BeBr2 < BeF2
10.
Which of the following combination of cation and anion has maximum covalent character.
(A) K+ , Cl–
(B) Na+, Cl–
(C) Cs+ , Cl–
(D) Mg+2, Cl–
lp
Ja
8.
Sa
nk
a
Comprehension # 4
Answer 11, 12 and 13 by appropriately matching the information given in the three columns of
the following table.
According to Fajan covalency is favoured by :
(i) Small size of cation
(ii) Large size of anion
(iii) High charge on cation, anion or both
(iv) Cation with non-noble gas configuration
Column-1
(I)
NaF, NaCl, NaBr, NaI
(i)
Size of cation increases
(P)
(II)
NaCl, MgCl2, AlCl3
MgCO3, CaCO3, SrCO3,
BaCO3
(ii)
Size of anion increases
(Q)
Column-3
Covalent
character
increases
Ionic character increases
(iii)
Charge on anion increases
(R)
Melting point increases
LiOH, NaOH, KOH, RbOH
(iv)
Charge
density
(Magnitude)
(S)
Solubility increases
(III)
(IV)
11.
12.
Column-2
decreases
Which of the following combination is incorrect ?
(A) I, ii, P
(B) I, iv, P
(C) I, ii, R
(D) IV, i, S
Which of the following is correct combination ?
(A) II, i, P
(B) II, iv, Q
(D) III, i, Q
(C) III, iv, D
117
13.
The incorrect combination is
(A) III, iv, Q
(B) IV, iv, S
(C) IV, iv, Q
(D) III, iv, P
* Marked Questions may have more than one correct option.
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
Arrange the following compounds in order of increasing dipole moment toluene (I), m-dichlorobenzene
(II), o-dichlorobenzene (III), p-dichlorobenzene (IV) :
[IIT JEE 1996]
(A) I < IV < II < III
(B) IV < I < II < III
(C) IV < I < III < II
(D) IV < II < I < III
2.
Which contains both polar and non-polar bonds ?
(A) NH4Cl
(B) HCN
(C) H2O2
a
ri
1.
[IIT JEE 1997]
(D) CH4
The geometry of H2S and its dipole moment are :
(A) angular and non-zero
(B) angular and zero
(C) linear and non-zero
(D) linear and zero
[IIT JEE 1999]
4.*
The molecules that will have dipole moment are
[IIT JEE 1992]
(A) 2, 2-dimethyl propane
(B) trans-2-pentene
(C) cis-3-hexene
(D) 2, 2, 3, 3-tetramethyl butane
5.
The correct order of acidic strength is :
(A) Cl2O7 > SO3 > P4O10
(C) Na2O > MgO > Al2O3
Ja
uh
3.
[JEE-2000, 1/135]
lp
(B) CO2 > N2O5 > SO3
(D) K2O > CaO > MgO
The set with correct order of acidity is :
(A) HClO < HClO2 < HClO3 < HClO4
(C) HClO < HClO4 < HClO3 < HClO2
7.
Identify the correct order of acidic strengths of CO2, CuO, CaO, H2O.
(A) CaO < CuO < H2O < CO2
(B) H2O < CuO < CaO < CO2
(C) CaO < H2O < CuO < CO2
(D) H2O < CO2 < CaO < CuO
Sa
nk
8.*
[JEE-2001, 1/135]
(B) HClO4 < HClO3 < HClO2 < HClO
(D) HClO4 < HClO2 < HClO3 < HClO
a
6.
[JEE-2002, 3/150]
The correct statement(s) regarding, (i) HClO, (ii) HClO 2, (iii) HClO3 and (iv) HClO4 is(are)
(A) The number of Cl=O bonds in (ii) and (iii) together is two
[JEE(Advanced) 2015, 4/168]
(B) The number of lone pairs of electrons on Cl in (ii) and (iii) together is three
(C) The hybridization of Cl in (iv) is sp3
(D) Amongst (i) to (iv), the strongest acid is (i)
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) OFFLINE PROBLEMS
1.
Which one of the following is an amphoteric oxide?
(1) ZnO
(2) Na2O
(3) SO2
[AIEEE-2003, 3/225]
(4) B2O3.
2.
Which of the following pair of molecules will have permanent dipole moments for both members?
[AIEEE-2003, 3/225]
(1) SiF4 and NO2
(2) NO2 and CO2
(3) NO2 and O3
(4) SiF4 and CO2
3.
Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid strength is :
[AIEEE-2004, 3/225]
118
(1) Al2O3 < SiO2 < SO2 < P2O3
(3) SO2 < P2O3 < SiO2 < Al2O3
(2) SiO2 < SO2 < Al2O3 < P2O3
(4) Al2O3 < SiO2 < P2O3 < SO2
The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences
represents the increasing order of the polarizing power of these cationic species, K +, Ca2+, Mg2+, Be2+?
[AIEEE-2007, 3/120]
(1) K+ < Ca2+ < Mg2+ < Be2+
(2) Ca2 < Mg2+ < Be2+ < K+
(3) Mg2+ < Be2+ < K+ < Ca2+
(4) Be2+ < K+ < Ca2+ < Mg2+
5.
Among the following the maximum covalent character is shown by the compound :
[AIEEE-2011, 4/120]
(1) FeCl2
(2) SnCl2
(3) AlCl3
(4) MgCl2
6.
For which of the following molecule significant   0 ?
CN
Cl
OH
(ii)
(iii)
Cl
(1) Only (i)
CN
(2) (i) and (ii)
OH
(3) Only (iii)
[JEE(Main)-2014, 4/120]
SH
(iv)
uh
(i)
a
ri
4.
SH
(4) (iii) and (iv)
In the following reactions, ZnO is respectively acting as a/an :
[JEE(Main)-2017, 4/120]
(a) ZnO + Na2O ⎯→ Na2ZnO2
(b) ZnO + CO2 ⎯→ ZnCO3
(1) base and base
(2) acid and acid
(3) acid and base
(4) base and acid
8.
Which of the following compounds contain(s) no covalent bond(s) ?
KCl, PH3, O2, B2H6, H2SO4
(1) KCl
(2) KCl, B2H6
(3) KCl, B2H6, PH3
[JEE(Main)-2018, 4/120]
(4) KCl, H2SO4
lp
Ja
7.
JEE(MAIN) ONLINE PROBLEMS
Amongst LiCl, RbCl, BeCl2 and MgCl2 the compounds with the greatest and the least ionic character
respectively are :
[JEE(Main) 2014 Online (19-04-14), 4/120]
(1) LiCl and RbCl
(2) RbCl and BeCl2
(3) MgCl2 and BeCl2
(4) RbCl and MgCl2
2.
Which of the alkaline earth metal halides given below is essentially covalent in nature ?
[JEE(Main) 2015 Online (11-04-15), 4/120]
(1) SrCl2
(2) CaCl2
(3) BaCl2
(4) MgCl2
3.
Molecular AB has a bond length of 1.61 Å and a dipole moment of 0.38 D. The fractional charge on each atom
(absolute magnitude) is : (e0 = 4.802 × 10–10 esu)
[JEE(Main) 2015 Online (11-04-15), 4/120]
(1) 0.5
(2) 0.05
(3) 0
(4) 1.0
4.
Which intermolecular force is most responsible in allowing xenon gas to liquefy?
[JEE(Main) 2016 Online (09-04-16), 4/120]
(1) Instantaneous dipole-induced dipole
(2) Ionic
(3) Ion-dipole
(4) Dipole-dipole
5.
Which of the following is a Lewis acid ?
(1) PH3
(2) NF3
Sa
nk
a
1.
[JEE(Main) 2018 Online (15-04-18), 4/120]
(3) NaH
(4) B(CH3)3
119
EXERCISE - 1
PART - I
(a) NaF < Na2O < Na3N
(b) NaCl < MgCl2 < AlCl3 < SiCl4 < PCl5
A-2.
According to Fajan's rule, as charge on cation increases its polarising power increases resulting in to
the greater polarisation of anion. Thus covalent character increases and melting point decreases.
A-3.
Bigger anion has higher polarisability; more polarisation greater is the intensity of colour (valence shell
electrons are loosely bound with the nucleus).
B-1.
Inorganic benzene (N3B3H6) contains polar covalent B –N bonds while benzene (C6H6) contains nonpolar covalent C–C bonds.
uh
a
ri
A-1
Ja
B-2.
CCl4 is a symmetrical and non polar molecule while CHCl3 is an unsymmetrical and polar molecule.
B-4.
BF3 < H2S < H2O.
B-5.
84.5%
C-1.
(a) (i) N2O3
C-2.
As En (difference in electronegativities between element and oxygen) decreases, the acidic character
increases. So, SiO2 < CO2 < N2O5 < SO3 .
lp
B-3.
Sa
nk
a
(ii) SO3
C-3.
Higher the metallic character, greater will be the basic character of its oxide as En (difference in
electronegativities between element and oxygen) increases.
So, NiO < MgO < SrO < K2O < Cs2O
A-1.
(C)
A-2.
(B)
A-3.
(C)
A-4.
(B)
B-1.
(A)
B-2.
(C)
B-3.
(B)
B-4.
(D)
B-5.
(A)
B-6.
(A)
B-7.
(B)
C-1.
(C)
C-2.
(C)
C-3.
(A)
C-4.
(D)
C-5.
(A)
1.
(A – p,q,r) ; (B – s,t) ; (C –q,r,t) ; (D – s,t)
PART - II
PART - III
120
EXERCISE - 2
PART - I
1.
(D)
2.
(C)
3.
(B)
4.
(C)
5.
(C)
6.
(C)
7.
(B)
8.
(D)
9.
(C)
10.
(A)
11.
(B)
12.
(D)
PART - II
3 (a, b, h)
2.
3 (a, c, f)
3.
3 (i, ii, vi)
4.
1.
(ABCD)
6.
(ABCD)
2.
(ABD)
3.
(BC)
PART - IV
(B)
2.
(B)
3.
(C)
6.
(D)
7.
(A)
8.
(C)
11.
(C)
12.
(D)
13.
4.
5.
(BD)
4.
(C)
5.
(C)
9.
(A)
10.
(D)
(BC)
5.
(A)
4.
(1)
5.
(3)
4.
(1)
5.
(4)
Ja
1.
(ABCD)
uh
PART - III
5 (a, b, e, h, k)
a
ri
1.
(D)
EXERCISE - 3
lp
PART - I
(B)
2.
(C)
3.
(A)
6.
(A)
7.
(A)
8.*
(BC)
Sa
nk
a
1.
4.*
PART - II
JEE(MAIN) OFFLINE PROBLEMS
1.
(1)
2.
(3)
3.
(4)
6.
(4)
7.
(3)
8.
(1)
JEE(MAIN) ONLINE PROBLEMS
1.
(2)
2.
(3)
3.
(2)
121
 Marked questions are recommended for Revision.
This Section is not meant for classroom discussion. It is being given to promote self-study
and self testing amongst the Resonance students.
PART - I : PRACTICE TEST-1 (IIT-JEE (MAIN Pattern))
Max. Time : 1 Hr.
Max. Marks : 120
Important Instructions
a
ri
5.
The test is of 1 hour duration.
The Test Booklet consists of 30 questions. The maximum marks are 120.
Each question is allotted 4 (four) marks for correct response.
Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each
question. ¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No
deduction from the total score will be made if no response is indicated for an item in the answer sheet.
There is only one correct response for each question. Filling up more than one response in any question
will be treated as wrong response and marks for wrong response will be deducted accordingly as per
instructions 4 above.
uh
1.
2.
3.
4.
If the electronic configuration of an element is 1s 2 2s2 2p6 3s2 3p6 3d2 4s2, then the electrons involved in
bonding will be.
(1) 3p6
(2) 3p6 4s2
(3) 3p6 3d2
(4) 3d2 4s2
2.
Two ice cubes are pressed together until they form one block. Which of the following force is primarily
responsible for holding the cubes together?
(1) Dipole–dipole interaction
(2) Van der Waals forces
(3) Ionic interaction
(4) Hydrogen bonding
3.
A sigma bond may be formed by the overlap of 2 atomic orbitals of atoms A and B. If the bond is
formed along the x-axis, which of the following overlaps is acceptable ?
(1) s orbital of A and pz orbital of B
(2) px orbital of A and py orbital of B
(3) pz orbital of A and px orbital of B
(4) px orbital of A and s orbital of B
4.
NH3 and BF3 combine readily because of the formation of :
(1) -back bond
(2) a hydrogen bond
(3) a coordinate bond
(4) an ionic bond
5.
Maximum bond energy is in :
(1) F2
(2) N2
Sa
nk
a
lp
Ja
1.
6.
7.
8.
9.
(3) O2
(4) equal in all
The structure of XeF2 involves hybridization of the type :
(1) sp3
(2) sp3d3
(3) sp3d
(4) sp3d2
The bond angle and hybridization in ether (CH3OCH3) is :
(1)106° 51', sp3
(2) 104°31', sp3
(3) 109°28' sp3
(4) >109°28' sp3
Which of the following have maximum boilling point
(1) CH4
(2) CF4
(3) CCl4
(4) CD4
In which of the following molecule / ion all the bonds are not equal ?
(1) XeF4
(2) BeF4–
(3) C2H4
(4) SiF4
10.
The correct order of increasing X–O–X bond angle is (X = H, F or Cl) :
(1) H2O > Cl2O > F2O
(2) Cl2O > H2O > F2O
(3) F2O > Cl2O > H2O
(4) F2O > H2O > Cl2O
11.
Which of the following is paramagnetic ?
(1) O2–
(2) CN–
(3) CO
(4) NO+
Which of the following is non-polar :
(1) NF3
(2) BF3
(3) PF3
(4) SF4
12.
122
13.
Strongest hydrogen bond present in :
(1) O–H ------ S
(2) S–H ------ O
(3) F–H ------ F
(4) F–H ------ O
Resonance structure of a molecule should not have
(1) Identical bond position
(2) Identical arrangements of atoms
(3) Nearly the same energy content
(4) The same number of paired electrons
15.
The paramagnetic property of oxygen is well explained by :
(1) Molecular orbital theory
(2) Resonance theory
(3) Valence bond theory
(4) VSEPR theory
16.
Which of the following statement is correct regarding molecular orbital theory (MOT) :
(1) Energy of bonding orbital is less than anti-bonding orbital.
(2) Energy of bonding orbital is more than anti-bonding orbital.
(3) Bonding orbitals are monocentric.
(4) Bonding orbital follow n +  rule
17.
PF2Cl3 is non polar because :
(1) P–Cl bond is non-polar
(3) P–Cl bond is polar
a
ri
14.
(2) Its dipole moment is zero
(4) P & Cl have equal electronegativity
The hybrid states of central atom in diborane, diamond and graphite are respectively :
(1) sp2, sp3, sp2
(2) sp3, sp3, sp2
(3) sp3, sp3, sp3
(4) sp, sp2, sp3
19.
Which of the set of species have same hybridization state but different shapes:(1) NO2+, NO2, NO2–
(2) ClO4–, SF4, XeF4
+
+
(3) NH4 , H3O , OF2
(4) SO4–2, PO4–3, ClO4–
20.
The bonds present in N2O5 are :
(1) Only ionic
(3) Only covalent
Ja
uh
18.
(2) Covalent & coordinate
(4) Covalent & ionic
The correct statement for the reaction:
NH3 + H+ ⎯→ NH4+
(1) Hybridisation state is changed
(2) Bond angle increases
(3) NH3 act as a Lewis acid
(4) Regular geometry becomes irregular
22.
The correct order of decreasing polarisabality of ions is :
(1) Cl– > Br– > I– > F–
(2) F– > I– > Br– > Cl–
–
–
–
–
(3) F > Cl > Br > I
(4) I– > Br– > Cl– > F–
23.
CCl4 is more covalent than LiCl because :
(1) There is more polarization of Cl in CCl4
(3) CCl4 has more weight
(2) There is more polarization of Cl in LiCl
(4) All of the above
Sa
nk
a
lp
21.
24.
An ionic compound A+ B– is most likely to be formed when (1) Ionization energy of A is low
(2) Electron affinity of B is high
(3) Electron affinity of B is low
(4) Both (1) and (2)
25.
Which of the following statements regarding HClO3 is/are correct.
(1) oxidation state of chlorine is +5
(2) it has two p-d bonds
(3) it has two type of Cl–O bond
(4) all of these
26.
The correct sequence of increasing covalent character is represented by (1) BeCl2 < NaCl < LiCl (2) NaCl < LiCl < BeCl2 (3) BeCl2 < LiCl < NaCl (4) LiCl < NaCl < BeCl2
27.
Correct order of bond length is :
(1) N–H > P–H > Sb–H (2) N–H < P–H < Sb–H (3) P–H > N–H > Sb–H (4) Sb–H > N–H > P–H
28.
Which of the follwing species contains three bond pairs and one lone pair around the central atom ?
(1) NH2–
(2) PCl3
(3) H2O
(4) BF3
29.
During change of O2 to O2– ion, the electron adds in which one of the following orbitals?
(1) * 2pz orbital
(2)  2pz orbital
(3) * 2px / * 2pyorbital
(4)  2px /  2py orbital
30.
The number of S–S bonds in sulphur trioxide trimer (S3O9) is :
(1) 3
(2) 0
(3) 1
(4) 2
123
Practice Test-1 (IIT-JEE (Main Pattern))
OBJECTIVE RESPONSE SHEET (ORS)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
Que.
Que.
a
ri
Ans.
Ans.
PART - II : NATIONAL STANDARD EXAMINATION IN CHEMISTRY (NSEC) STAGE-I
2.
Which of the following has maximum bond energy ?
(A) O2–
(B) O2+
(C) O2
The hybrid orbital of the central atom in AIF 4– is :
(A) sp
(B) sp2
(C) sp3
[NSEC-2000]
(D) O22–
uh
1.
[NSEC-2000]
(D) sp3d
The paramagnetism of oxygen is best explained by
(A) MO theory
(B) Valence Bond Theory
(C) VSEPR theory
(D) Lewis dot structure
4.
Which of the following compounds has both ionic and covalent bonding ?
[NSEC-2000]
(A) NaBr
(B) Ba(CN)2
(C) PCI3
(D) CH3CH2OH
5.
Which of the following molecule is non-polar ?
(A) CO
(B) SO2
In ammonia H-N-H bond angle is
(A) 106.7oC
(B) 104.5oC
(C) 109.5oC
[NSEC-2000]
[NSEC-2000]
(D) H2O
[NSEC-2001]
(D) 120oC
The cyanide ion, CN– and the nitrogen molecule, N2 are isoelectronic. However, in contrast to CN–, N2
is chemically inert due to :
[NSEC-2001]
(A) unsymmetrical electron distribution
(B) low bond energy
(C) absence of bond polarity
(D) presence of a greater number of elelctrons in bonding orbital
Sa
nk
a
7.
(C) SO3
lp
6.
Ja
3.
8.
Valence bond theory was proposed by :
(A) Heitler and London (B) Slater and Mullican (C) Pauling
[NSEC-2001]
(D) Rutherford
9.
The maximum possible number of hydrogen bonds a water molecule can form is :
(A) 2
(B) 3
(C) 4
(D) 1
10.
Hydrogen fluoride is liquid at room temperature because of
[NSEC-2002]
(A) covalent nature
(B) hydrogen bonding (C) van der Waals force (D) strong ionic bond.
11.
Overlapping of the atomic orbitals results in the formation of a covalent bond of the type [NSEC-2002]
(A) sigma ()
(B) pi ()
(C) sigma or pi
(D) sigma and pi.
12.
Which of the following pairs has the strongest hydrogen bonding between themselves ?
(A) SiH4 and SiF4
(B) CH4 and CH3OH
(C) CH3COCH3 and CHCI3
(D) HCO2H and CH3CO2H
[NSEC-2002]
13.
SF6 molecule has ..........geometry.
(A) pyramidal
(B) bi-pyramidal
[NSEC-2002]
(C) tetrahedral
[NSEC-2001]
(D) octahedral
124
Isosteres are compounds having similar geometry and isoelectronic species are species having the
same number of electrons. The pair of species CO2 and NO2+ is
[NSEC-2003]
(A) isosteric and isoelectronic
(B) isosteric, but not isoelectronic
(C) isoelectronic, but not isosteric
(D) neither isosteric nor isoelectronic.
15.
Considering z-axis to be the internuclear axis, the combination of orbitals on Li and CI atoms
respectively, that can lead to a stable sigma bond
[NSEC-2003]
(A) 2s and 3py
(B) 1s and 3py
(C) 1s and 3pz
(D) 2s and 3pz.
16.
The number of hydrogen bonds formed by each H2O molecule in an ice crystal is :
(A) 6
(B) 4
(C) 2
(D) 3
[NSEC-2003]
17.
The structures of AICI3 and PCI3 can be described as
(A) both planar
(B) both pyramidal
(C) planar and pyramidal respectively
(D) pyramidal and planar respectively.
[NSEC-2004]
18.
Bond orders of NO and NO+ are respectively
(A) 2.5 and 3
(B) 2 and 4
[NSEC-2004]
(C) 3.5 and 2.5
The molecule with non -zero dipole moment is
(A) BF3
(B) PCI3
(C) SiCI4
20.
21.
(D) 4 and 2
[NSEC-2004]
The molecular orbital with highest energy in a nitrogen molecule is
(A) 2p
(B) 2p
(C) *2p
Which of the following has zero dipole moment ?
(A) NH3
(B) NF3
(C) BF3
(D) XeF4
uh
19.
a
ri
14.
[NSEC-2004]
(D) *2p
[NSEC-2005]
(D) CHCl3.
According to VSEPR theory the shape of SF4 molecule is (atomic number of sulphur=16) [NSEC-2005]
(A) squashed tetrahedral
(B) square pyramidal
(C) tetrahedral
(D) square planar.
23.
In which of the following cases, does breaking of the covalent bond take place?
[NSEC-2005]
(A) boiling of water
(B) melting of KCN
(C) boiling of CF4
(D) melting of SiO2.
24.
The hybrid orbitals used by chlorine in CIF3 molecule are of the type
(A) sp3
(B) sp2
(C) sp2d
lp
Ja
22.
[NSEC-2005]
(D) sp3d.
In solid CuSO4. 5H2O, copper is coordinated to
(A) one water molecule
(B) three water molecules
(C) five water molecules
(D) four water molecules.
26.
Which of the following contain maximum number of electrons in the antibonding molecular orbital?
[NSEC-2005]
(A) O22(B) O2
(C) O2–
(D) O2+.
Sa
nk
27.
28.
[NSEC-2005]
a
25.
Identify the molecule that has more covalent bond character.
(A) Na2S
(B) MgCI2
(C) NaH
(D) SnCI4
[NSEC-2005]
The non-linear molecule is
(A) SO2
(B) CO2
(D) C2H2.
[NSEC-2006]
(C) HCN
29.
In the ammonium ion,
(A) the four hydrogens are at the corners of a square
(B) all bonds are ionic
(C) the nitrogen atom carries a formal charge
(D) all bonds are co-ordinate ones.
[NSEC-2006]
30.*
Lewis dot structures of compounds of representative elements normally follow the octet rule.
Which of the following does not obey the octet rule ?
[NSEC-2006]
(A) CO32–
(B) O3
(C) SO2
(D) 3–
31.
Trisilylamine N (Si(CH3)3)3 is
(A) acidic
(B) basic
. .
32.
[NSEC-2006]
(C) neutral
(D) amphoteric.
Which of the following compounds of phosphorus does not have resonating structures ? [NSEC-2006]
(A) H3PO2
(B) H3PO3
(C) H3PO4
(D) (CH3)2HPO3.
125
33.
In thiosulphuric acid H2S2O3, the oxidation states of sulphur atoms are
(A) +  ,+ 
(B) 0, + V
(C) +  , + 
[NSEC-2006]
(D) – , + V.
34.
Which type of bond exists between the two boron atoms in a diborane molecule ?
(A) 2-Center-2electron
(B) 3-Center-2-electron
(C) 3-Center-3-electron
(D) 4-Center-4-electron
35.
According to Molecular Orbital Theory, the oxygen molecule is(A) Diamagnetic
(B) Paramagnetic
(C) Ferromagnetic
[NSEC-2007]
(D) non magnetic
36.
Which of the following has zero dipole moment?
(A) NH3
(B) NF3
(C) BF3
(D) CHCl3
Which of the following pairs is a Lewis acid & a Lewis base ?
(A) CI & Ag+
(B) NH3 & BF3
(C) SO42– & HSO4–
(D) H+ & OH–
[NSEC-2007]
[NSEC-2007]
a
ri
37.
[NSEC-2007]
38.
Assuming a Lewis structure for SO2 in which all the atoms obey the octet rule, the formal charge on S
is:
[NSEC-2007]
(A) +1
(B) 0
(C) +2
(D) 2
39.
In which of the following pairs are both molecules polar ?
(A) O2 and H2O
(B) BF3 and PCl3
(C) SO2 and SCl2
40.
The maximum possible number of hydrogen bonds a water molecule can form is
(A) 2
(B) 4
(C) 3
(D) 1
[NSEC-2008]
41.
Which of the following has the highest bond order ?
(A) O2
(B) O2+
(C) O2
[NSEC-2008]
uh
[NSEC-2008]
(D) CS2 and NO2
(D) O22–
The hybridization of the atomic orbitals of sulphur in SO 3, SO42– and SF4 are respectively: [NSEC-2009]
(A) sp, sp3, sp2
(B) sp, sp2, sp3d
(C) sp2, sp, sp3
(D) sp2, sp3, sp3d
43.
The molecule which does not have a net dipole moment is –
(A) H2O
(B) NH3
(C) BF3
(D) BrF5
The molecular geometry for ammonia is –
(A) saw horse
(B) trigonal planar
(D) pyramidal
(C) tetrahedral
lp
44.
Ja
42.
[NSEC-2009]
[NSEC-2009]
The sequence of molecular orbitals for the carbide ion (C 22–) is –
(A)
1s2
*1s2 2s2
*2s2 2p4
2
2
2
(B)
1s
*1s
2s
*2s2 2p4
2p2
2
2
2
2
4
(C)
1s
*1s
2s
*2s
2p
2p2
*2p2
2
2
2
2
4
2
(D)
1s
*1s
2s
*2s
2p
2p
*2p4
[NSEC-2009]
46.
The structure of SF4 is
(A) Trigonal bipyramidal
(C) Squashed Tetrahedral
[NSEC-2010]
Sa
nk
a
45.
47.
48.
49.
(B) Square planar
(D) Octahedral
The pair which contains both molecules polar is
(A) O2 & H2O
(B) CO2 & PCI3
(C) SO2 & SC!2
[NSEC-2010]
(D) CS2 & NO2
The species that contains maximum number of electrons in the antibonding molecular orbitals is :
[NSEC-2010]
(A) O22–
(B) O2
(C) O2–
(D) O2+
The compound that has the highest ionic character associated with the X-Cl bond is
[NSEC-2010]
(A) PCl5
(B) BCI3
(C) CCl4
(D) SiCl4
50.
According to VSEPR theory the shape of IF5 molecule will be
(A) tetrahedral
(B) trigonal bipyramid (C) square pyramid
[NSEC-2011]
(D) trigonal planar
51.
The formal changes on the atoms underlined are
C6H5 – CN–O
(A) C = 0, N = –1, O = +1
(B) C = –1, N = +1, O = –1
(C) C = 0, N = +1, O = –1
(D) C = +1, N = 0, O = –1
[NSEC-2011]
52.
The compound that does not have a  bond is :
(A) SO2
(B) SF6
(C) O2
[NSEC-2011]
(D) SO3
126
53.
The C–O–C bond angle in dimethylether is
(A) 109°28'
(B) 110°
[NSEC-2011]
(C) 120°
(D) 180°
54.
The relative basics strengths of NH3, CH3NH2 and NF3 are in the order :
(A) CH3NH2 > NH3 > NF3
(B) NH3 > CH3NH2 > NF3
(C) NF3 > CH3NH2 > NH3
(D) CH3NH2 > NF3 > NH3
[NSEC-2012]
55.
The molecule that has maximum covalent character :
(A) NaH
(B) Na2S
(C) CaCl2
[NSEC-2012]
(D) SnCl4
Which of the following compounds has zero dipole moment ?
(A) NH3
(B) NF3
(C) BF3
(D) CHCl3
57.
[NSEC-2013]
Which one of the following is not a valid structure ..
for dinitrogen oxide ?
N
N
..
(I)
(II)
(IV)
(III)
(IV)
59.
60.
61.*
The bond order of
(A) 1
(C) III
ion is :
(B) 2
(C) 2.5
(D) IV
[NSEC-2013]
(D) 3
Which of the following molecular structures is NOT possible ?
(A) OF2
(B) SF2
(C) OF4
(D) SF4
The molecule having the highest dipole moment is
(A) CO2
(B) CH4
(C) NH3
(D) NF3
Ja
58.
NO+
(B) II
uh
(A) I
[NSEC-2013]
..
O
..
a
ri
56.
The species which has triangular planar geometry is
(A) NF3
(B) NO3–
(C) AlCl3
[NSEC-2013]
[NSEC-2014]
[NSEC-2014]
(D) SbH3
The correct order of stability for the following species is
(A) Li2 < He2+ < O2+ < C2
(B) C2 < O2+ < Li2 < He2+
+
+
(C) He2 < Li2 < C2 < O2
(D) O2+< C2 < Li2 < He2+
[NSEC-2014]
63.
The hybridization of boron in the stable borane having the lowest molecular weight is –
(A) sp2
(B) sp3
(C) sp
(D) sp3d
[NSEC-2014]
64.
For SF4, the molecular geometry and hybridization of the central atom respectively are:
(A) Square planar, dsp2
(B) Tetrahedral, sp3
3
(C) Seesaw, sp d
(D) Square pyramid, sp3d
[NSEC-2014]
Sa
nk
a
lp
62.
65.
The species that cannot exist is
(A) SiF62–
(B) BF63–
(C) SF6
(D) AlF63–
The species having highest bond energy is
(A) O2
(B) O2+
(C) O2–
(D) O22–
67.
The structure of a molecule of N(SiMe3)3 is
(A) Pyramidal with angle close to 110°
(C) Bent T-shaped with angle close to 89°
[NSEC-2015]
(B) T-shaped with angle 90°
(D) Trigonal planar with bond angle close to 120°
68.
The order of p-d interaction in the compounds containing bond between Si/P/S/Cl and oxygen is in
the order
[NSEC-2015]
(A) P > Si > Cl > S
(B) Si < P < S < Cl
(C) S < Cl < P < Si
(D) Si > P > S > Cl
69.
[NSEC-2016]
66.
[NSEC-2014]
[NSEC-2015]
Which one of the following information about the compounds is correct ?
Oxidation
No. of P–OH
No. of P–H
Compounds
state
bonds
bonds
of P
[I] H3PO2 Hypophosphorous acid
1+
2
1
[II] H4P2O5 pyropophosphorous acid
3+
2
2
[III] H4P2O6 Hypophosphoric acid
4+
2
2
[IV] H4P2O7 pyrophosphoric acid
5+
3
1
No. of P=O
bonds
0
2
2
4
127
(A) I
(B) III
(C) IV
(D) II
70.
The pair that is isostructural (i.e. having the same shape and hybridization) is
[NSEC-2016]
(A) NF3 and BF3
(B) BF4– and NH4+
(C) BCl3 and BrCl3
(D) NH3 and NO3–
71.
Number of P–S single bonds and P–S double bonds (P=S) in P4S10 are respectively
(A) 10, 6
(B) 16, 0
(C) 14, 2
(D) 12, 4
72.
Which of the following compounds contain 3-centered 2-electron bonding ?
[NSEC-2016]
(i) [BeF2]n
(ii) [Be(CH3)3]n
(iii) [BeCl2]n
(iv) [BeH2]n
(A) (i) and (ii)
(B) (ii) and (iii)
(C) (ii) and (iv)
(D) (iii) and (iv)
73.
In ammonia the bond angle is 107º 48' while in SbH 3 the bond angle is about 91º18'. The correct
explanation among the following is/are
[NSEC-2016]
(A) The orbitals of Sb used for the formation of Sb-H bond are almost pure p-orbitals.
(B) Sb has larger size compared to N.
(C) Sb has more metallic character than N.
(D) All the statements are correct.
74.
Assuming that Hund’s rule is violated by the diatomic molecule B 2, its bond order and magnetic nature
will be respectively
[NSEC-2016]
(A) 1, diamagnetic
(B) 1, paramagnetic
(C) 2, diamagnetic
(D) 2, paramagnetic
75.
In the Lewis structure of ozone (O3), the formal charge on the central oxygen atom is
(A) +1
(B) –1
(C) 0
(D) –2
76.
Which of the following represents the correct order of dipole moment?
(A) NH3 > NF3 > H3O
(B) NH3 > H2O > NF3
(C) H2O > NH3 > NF3
77.
Which of the following has the shortest bond length ?
(A) O2
(B) O2−
(C) O2+
(D) O2−2
Which of the following is not paramagnetic?
(A) S2–
(B) N2–
(D) NO
79.
a
ri
uh
Ja
F4+
II.
(A) I, II, IV
BF4−
(B) II, III, IV
[NSEC-2018]
[NSEC-2018]
III. SF4
IV. TeCl4
(C) I, III, IV
(D) I, II, III
a
Which among the following is nonlinear ?
–
(A) N3
82.
[NSEC-2017]
Among the following pairs, the one in which both the compounds as pure liquids can show significant
auto ionization is
[NSEC-2018]
(A) H2O and H2S
(B) BrF3 and Cl3
(C) PF5 and PCl5
(D) HF and HCl
Sa
nk
81.
[NSEC-2017]
[NSEC-2017]
(D) H2O > NF3 > NH3
In which of the following, all the bond lengths are not the same ?
I.
80.
(C) O2–
lp
78.*
[NSEC-2016]
–
[NSEC-2018]
–
(B) ClF2
(C) Br3
+
(D) BrCl2
The most stable Lewis structure of N2O is
(A)
• •• •• •••
• O=N=N•
(B)
••
•
•
• N=O=N•
[NSEC-2018]
(C)
• ••
•
• N–NO •
••
(D)
• ••
•
• O–NN •
••
128
PART - III : HIGH LEVEL PROBLEMS (HLP)
SUBJECTIVE QUESTIONS
Draw the structure of the following compound and identify the hybridisation of the central atom, also
count the Sigma and -bond.
(i) XeO2F2
(ii) PF3Cl2
(iii) NH2OH
(iv) Anion of PCl5(s)
(v) P4
2.
One of the first drugs to be prepared for use in treatment of acquired immuno deficiency syndrome
(AIDS) was azidothymidine (AZT).
Ja
uh
a
ri
1.
(a) How many carbon atoms have sp3 hybridisation ?
(b) How many carbon atoms have sp2 hybridisation ?
(c) How many nitrogen atoms (central atom not terminal) have sp hybridisation ?
(d) How many  bonds are in the molecule ?
Draw structures for the polymeric (BeH2)n and (BeCl2)n. Explain in brief why the hydride bridge in
(BeH2)n is considered to be electron deficient but not the halide bridge in (BeCl 2)n?
4.
Explain the structure, hybridisation and oxidation state of S in sulphuric acid, Marshall’s acid, Caro’s
acid and oleum.
5.
Give the number of characteristic bond(s) found in the various oxy-acids of phosphorus as given below.
(P) Number of P-O-P bond(s) in cyclotrimetaphosphoric acid.
(Q) Number of P-P bond(s) in hypophosphoric acid.
(R) Number of P-H bond(s) in hypophosphorus acid.
(S) Number of P-OH bond(s) in pyrophosphoric acid.
Sa
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a
lp
3.
6.
Draw the molecular orbital (both bonding and antibonding) and identify the number of nodal planes in
the following combination of atomic orbitals with z as internuclear axis : dyz and dyz
7.
Arrange the following cations in the order of increasing polarising power.
(i) V3+, Sc3+, Ti3+, Cr3+
(ii) Zn2+, Cd2+, Hg2+
8.
Why lithium salts are most hydrated amongst alkali metals salts ?
ONLY ONE OPTION CORRECT TYPE
9.
10.
Example of super octet molecule is :
(A) SF6
(B) PCl5
(C) IF7
Correct order of bond energy of C–O bond is :
(A) CO32– > CO2 > CO
(C) CO > CO2 > CO32–
(B) CO2 > CO > CO32–
(D) None of these.
(D) All of these
129
11.
Arrange the following ions in the order of decreasing X–O bond length where X is the central atom
(A) CIO4–, SO42–, PO43–, SiO44–
(B) SiO44– , PO43–, SO42–, CIO4–
4–
3–
–
2–
(C) SiO4 , PO4 , CIO4 , CIO4
(D) SiO42–, SO42–, PO43–, CIO4–
12.
Which one (s) of the following structures cannot represent resonance forms for N 2O (diamagnetic)?
(a)
(b)
(c)
(d)
(A) a and c
(e)
(B) c, d and e
(C) d and e
(D) c and d
Which of the following overlaps gives a  bond with x as internuclear axis?
(A) pz and pz
(B) s and pz
(C) s and px
(D) d x 2 − y 2 and d x 2 − y 2
14.
The structure of F2SeO is analogous to :
(A) SO3
(B) CIO–3
15.
(C) XeO3
In which of the following N is in the sp2 hybridisation state
(A) (CH3)3N
(B) CH2 = NH
(C) CH3CN
a
ri
13.
(D) (B) and (C) both
(D) NO2+
In H2SO4 molecule
(A) S-atom is sp3d2 hybridised and there are 4 lone pair electrons on ‘O’ atoms.
(B) S-atom is sp3 hybridised and there are no lone pair of electrons in the molecule.
(C) S-atom is sp3 hybridised and there are 8 lone pair of electrons on the ‘O’ atoms.
(D) S-atom is sp2 hybridised and these are 8 lone pair of electrons ‘O’ atoms
17.
In which of the following pairs hybridisation of the central atoms are different ?
(A) ClF3 , ClF3O
(B) ClF3O, ClF3O2
(C) [ClF2O]+, [ClF4O]– (D) [ClF4O]–, [XeOF4]
18.
All the following species have all their bond lengths identical except :
(A) AsF3
(B) AsF4–
(C) AsF4+
(D) AsF6–
Which one has highest bond angle.
(A) NH3
(B) PH3
(D) CH4
Ja
19.
uh
16.
(C) H2O
21.
Which of the following overlappings will result into strongest bond formation. (Consider X to be
internuclear axis)
(A) 1s–1s overlap
(B) 2Py–2Py overlap
(C) 2Px–2Pz overlap
(D) 2Px–2Px overlap
:
Sa
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a
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lp
In the thiocyanate ion, SCN– three resonating structure are possible with the electron-dot method as
shown in figure :
1–
:
S=C=N
(x)
The decreasing order of % contribution in resonance hybrid is :
(A) y > x > z
(B) y > z > x
(C) z > x > y
(D) cannot predicted.
:
20.
22.
Phosphorous penta-chloride in gaseous phase exists as a monomer. In solid state, it exists as PCl4+
and PCl6– ions. The hybrid state of P-atom in PCl5 is sp3d. The hybrid states of P–atoms in PCl4+ and
PCl6– will be :
(A) sp3d, sp3d2
(B) sp3,sp3d2
(C) sp3d2, sp3d
(D) sp3,sp3d
23.
In which of the following cyclic compound the nitrogen atom is sp3 hybridised
()
()
()
(V)
NH2
(A)  & 
24.
(B) , , 
NO2
(C)  & V
(D) ,  & V
Which of the statements is correct about SO2 ?
(A) two , two  and no lone pair of electrons
(B) two  and one 
(C) two , two  and one lone pair
(D) none of these
130
Arrange the following in the increasing order of deviation from normal tetrahedral angle :
(A) P4 < PH3 < H2O
(B) PH3 < H2O < P4
(C) P4 <H2O < PH3
(D) H2O < PH3 < P4
26.
In XeF2 molecule the angle between two lone pair orbitals is , the angle between lone pair orbital and
bond pair orbital is  and the angle between bond pair orbitals is  :
(A)  =  = 
(B)  >  > 
(C)  >  > 
(D)  >  > 
27.
In O2F2, which of the following statement is incorrect.
(A) O–F bond length in O2F2 is longer than O–F bond length in OF2.
(B) The O.N. of oxygen in O2F2 is +1.
(C) The O–O bond length in O2F2 is shorter than O–O bond length in H2O2.
(D) None of these
28.
Identify the species containing Banana bonds
(A) (BeH2)n
(B) BF3
(C) (AlCl3)2
a
ri
25.
(D) (BeCl2)n
In the coordinate valency
(A) Electrons are equally shared by the atoms
(B) Electrons of one atom are shared between two atoms
(C) Hydrogen bond is formed
(D) Electrons are completely donated to other atom with no sharing
30.
What is the nature of the bond between B and O in (C 2H5)2OBH3.
(A) Covalent
(B) Co-ordinate covalent
(C) Ionic bond
(D) Banana shaped bond
Which of the following species does not contain N–N covalent bond ?
(A) N2O3
(B) N2O22–
(C) N2O5
(D) N2O4
33.
Compounds in which oxidation state of at least one oxygen is (–1) :
(1) H2O2
(2) O2F2
(3) H2SO5
(4) CrO5
(5) H2S2O8
(6) [Na2B2H4O8]
(A) 1,2,3,5
(B) 1,3,4,5,6
(C) 1,3,4,5
(D) 1,2,3,4,5,6
Which of the following statements is incorrect ?
(A) Among O2+, O2 and O2- the stability decreases as O2+ > O2 > O2(B) He2 molecule does not exist as the effect of bonding and anti-bonding orbitals cancel each other.
(C) C2 ,O22 - and Li2 are diamagnetic
(D) In F2 molecule, the energy of  2pz is more than  2px and  2 py
Which of the following statement is incorrect ?
(A) During N+2 formation, one electron is removed from the bonding molecular orbital of N2.
(B) During O+2 formation, one electron is removed from the antibonding molecular orbital of O 2.
(C) During O–2 formation, one electron is added to the bonding molecular orbital of O 2.
(D) During CN– formation, one electron is added to the bonding molecular orbital of CN.
Sa
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34.
Ja
32.
lp
31.
uh
29.
35.
Which the following molecules / species have identical bond order and same magnetic properties ?
(I) O2+ ;
(II) NO ;
(III) N2+
(A) (I) , (II) only
(B) (I) and III only
(C) (I), (II) and (III)
(D) (II) and (III) only
36.
Which of the following has the minimum heat of dissociation of N → B bond ?
(A) [(CH3)3N → BF3]
(B) [(CH3)3N → B(CH3)F2]
(C) [(CH3)3N → B(CH3)2F]
(D) [(CH3)3N → B(CH3)3]
37.
Which of the following does not have coordinate bond ? (Consider that every atom should follow octet
rule and H-atom should follow duet rule).
(A) SO2
(B) HNO2
(C) H2SO3
(D) HNO3
38.
Gaseous SO3 molecule
(A) is planar triangular in shape with three -bonds from sp2–p overlap and three -bonds formed by one
p–p overlap and two p–d overlap.
(B) is a pyramidal molecule with one double bond and two single bonds
(C) planar triangular in shape with two double bonds between S and O and one single bond
(D) is planar triangular in shape with three  bonds from sp2–p overlap and three -bonds formed by two
p–p overlap and one p–d overlap.
131
Consider the following statements.
S1 : Fluorine does not form any polyhalide because it does not have d-orbitals in valence shell.
S2 : In ClF3 , the three lone pairs of electrons occupy the equatorial position.
S3 : In B2 and N2 molecules mixing of s- and p- atomic orbitals takes place.
Of these statements :
(A) S1, S2 and S3 are correct
(B) S1 and S2 are correct
(C) S1 and S3 are correct
(D) S2 and S3 are correct
40.
When N2 goes to N2+, the N–N bond distance ... and when O2 goes to O2+, the O–O bond distance....
(A) Decrease, Increases
(B) Increases, Decrease
(C) Increases, Increases
(D) None of these
41.
According to Molecular orbital theory which of the following is correct ?
(A) LUMO level for C2 molecule is  2px orbital (B) In C2 molecules both the bonds are  bonds
(C) In C22– ion there is one  and two  bonds
a
ri
39.
(D) All the above are correct
Which of the following is incorrect ?
(A) The O–O bond length in H2O2 is larger than that in O2F2.
(B) The O–O bond length in H2O2 is very slightly smaller than in O22– ion.
(C) O2– and O2+ species are paramagnetic and have same number of unpaired electrons.
(D) None
43.
A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as :
(A) 2
(B) 1.5
(C) 1.0
(D) 0.5
44.
Which one of the following oxides is expected to exhibit paramagnetic behaviour
(A) CO2
(B) SO2
(C) ClO2
(D) SiO2
45.
Which of the following is correct :
(A) N–O bond length in NO gaseous molecule will be greater than in NOCl gaseous molecule.
(B) Carbon-carbon bond length in CaC2 will be more than in C2H4
(C) O–O bond length in KO2 will be more than in Na2 O2 .
(D) All the four hydrogen atoms in CH4 are not coplanar
46.
Consider the following statements ;
(I) The hybridisation found in cation of solid PCl5 is sp3.
(II) In AB2L2 type the BAB bond angle is always greater than the normal tetrahedral bond angle.
(III) In ClO3–, NH3 and XeO3, the hybridisation and the number of lone pairs on the central atoms are
same.
(IV) In P4 molecule, there are six P–P bonds and four lone pairs of electrons.
of these statements :
(A) I, II and III are correct only
(B) I, III and IV are correct only
(C) III and IV are correct only
(D) All are correct
Sa
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42.
47.
(a) There are only 12 bonding electrons available in one molecule of diborane.
(b) B3N3H6 is an electron deficient compound.
(c) Al2Cl6 sublimes on heating and give AlCl3 vapours at high temperature.
(d) In Si2O76– anion, one oxygen of a SiO44– tetrahedron is shared with another SiO44– tetrahedron.
(A) T F T T
(B) F T F F
(C) T F T F
(D) F T F T
48.
The point of dissimilarity between (SO3)3 and (HPO3) (cyclic trimers) is.
(A) having six membered ring.
(B) having central atom in same hybridization
(C) having planar ring.
(D) being isoelectronic.
49.
Which of the following statements is/are true about hypophosphoric acid.
(A) Oxidation state of phosphorus atoms present in hypophosphoric acid is +3
(B) Phosphorous atom is present in 2 different oxidation state which are +3, +5
(C) Oxidation state of phosphorous in hypophosphoric acid is +4
(D) Oxidation state of phosphorus in hypophosphoric acid is + 5.
132
The gaseous HX molecule has a measured dipole moment of 4.0 D, which indicates that it is a very
polar molecule. The separation between the nuclei in this molecule is 2.67 × 10–8 cm then the
percentage ionic character in HX molecule is :
(A) 78%
(B) 31.25%
(C) 50.25%
(D) None of these
51.
Which of the following models best describes the bonding within a layer of the graphite structure ?
(A) metallic bonding
(B) ionic bonding
(C) non-metallic covalent bonding
(D) van der Waals forces
52.
Which of the following compounds would have significant intermolecular hydrogen bonding ?
(A) HCl
(B) H2S
(C) chloral hydrate
(D) CH3OH
53.
Which of the following statement is not true ?
(A) CCl4 has higher boling point that CHCl3.
(B) The HF2 – ion exists in the solid state and in liquid HF solution , but not in dilute aqueous solutions.
(C) Hydrogen bonding maintains the planar H3BO3 units in layers in solid state.
(D) None of these.
54.
S1 : In the solid B(OH)3 units are hydrogen bonded together into two-dimensional sheets with almost
hexagonal symmetry.
S2 : Na2CO3 is isomorphous with Na2SO3 as both have similar formula type.
S3 : XeO3F2 has one lone pair of electron on central xenon atom.
S4 : D2O has higher boiling point than H2O
(A) T F T F
(B) T F F T
(C) T T F F
(D) T T T T
55.
Which is correct about D2O
(A) Its boiling point is higher than that of H2O() (B) O–D----O bond is stronger than O–H----O bond.
(C) D2O(s) sinks in H2O ().
(D) all the above are correct.
56.
Among the following compounds, the correct order of the polarity of the bonds is :
SbH3, AsH3, PH3, NH3.
(A) SbH3 < AsH3 < PH3 < NH3
(B) AsH3 < SbH3 = PH3 < NH3
(C) PH3 < AsH3 < SbH3 < NH3
(D) AsH3 < PH3 < SbH3 < NH3
57.
Given the species N2, CO, CN– and NO+.Which of the following statement is incorrect .
(A) All the species are diamagnetic
(B) All the species are isoelectronic
(C) All the species have dipole moment
(D) All the species are linear
58.
CH3Cl has more dipole moment than CH3F because :
(A) electron affinity of chlorine is greater than that of fluorine.
(B) the charge separation is larger in CH3Cl compared to CH3F.
(C) the repulsion between the bond pairs and non-bonded pairs of electrons is greater in CH3Cl than CH3F.
(D) chlorine has higher electronegativity than fluorine.
Sa
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50.
59.
Select the correct statement.
(A) The order of Xe–F bond length in various fluorides of Xenon is XeF2 < XeF4 < XeF6
(B) PH5 can undergo sp3d hybridisation to have octahedral geometry.
(C) Dipole moment of CH3F is greater than that of CH3Cl
(D) Increasing strength of hydrogen bonding is Cl–H----Cl < N–H----N < O–H----O < F–H----F
60.
The pairs of bases in DNA are held together by
(A) Hydrogen bonds
(B) Ionic bonds
(C) Phosphate groups
(D) Deoxyribose groups
61.
The bond that determines the secondary structure of a protein is
(A) Coordinate bond
(B) Covalent bond
(C) Hydrogen bond
(D) Ionic bond
62.
Which of the following has highest melting point according to Fajan's rule :
(A) NaCl
(B) MgCl2
(C) AlCl3
(D) LiCl
63.
An element X occurs in short period having configuration ns 2np1. The formula and nature of its oxide is :
(A) XO3, basic
(B) XO3, acidic
(C) X2O3, amphoteric (D) X2O3, basic
133
Which of the following compounds of elements in group IV is expected to be most ionic ?
(A) PbCl2
(B) PbCl4
(C) CCl4
(D) SiCl4
65.
Which of the following cannot be explained on the basis of Fajan’s Rules ?
(A) Ag2S is much less soluble than Ag2O
(B) Fe(OH)3 is much less soluble than Fe(OH)2
(C) BaCO3 is much less soluble than MgCO3
(D) Melting point of AlCl3 is much less than that of NaCl
66.
S1 : AgI is less soluble in water than AgF due to more polarisation of I- in comparison to F– ion.
S2 : Melting point of BaCl2 is higher than the melting point of BeCl2 due to greater ionic nature of BaCl2.
S3 : Order of hydrated radii is : Al3+ (aq) > Mg2+ (aq) > Na+ (aq)
(A) T T T
(B) T T F
(C) T F T
(D) F T T
67.
Correct order of strength of metallic bond in Li, K, Fe, W
(A) W > Fe > Li > K
(B) Fe > W > K > Li
(C) Li > K > W > Fe
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64.
(D) K > Li > Fe > W
The ground state electronic configurations of some elements, P, Q, R, S, and T (these symbols
represent the some of the known elements given in the periodic table) are as follows.
P
:
1s2 2s2 2p6 3s2 3p2
Q
:
1s2 2s2 2p6 3s2 3p6 4s1
R
:
1s2 2s2 2p6 3s2 3p1
S
:
1s2 2s2 2p6 3s2 3p6 3d5 4s1
T
:
1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6.
Match the electronic configurations of the elements with the properties given below and select the
correct sequence by choosing the correct codes given.
(i) Element forms a cation which is isoelectronic with P3–.
(ii) Element which in its compounds can show a maximum oxidation state of +6 and that is coloured too.
(iii) Element has largest atomic radius and highest first ionisation energy in the respective period.
(iv) Element which has intermediate value of electronegativity and its oxide forms salts with strong acids
and bases.
(A) Q R T P
(B) Q S T R
(C) Q R S T
(D) P Q R S
69.
Increasing order of extent of hydrolysis CCl4, MgCl2, AlCl3, PCl5, SiCl4
(A) MgCl2 < AlCl3 < CCl4 < AlCl4 < PCl5
(B) MgCl2 > AlCl3 > CCl4 > AlCl4 > PCl5
(C) CCl4 < MgCl2 < AlCl3 < SiCl4 < PCl5
(D) SiCl4 < PCl5 < CCl4 < MgCl2 < AlCl3
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68.
MATCH THE COLUMN
Match the species given in column-I with the type of hybridisation given in column-II.
Column-I
Column-II
(A)
(p)
sp3d
3–
(B)
XeO3
(q)
sp3
(C)
ClOF3
(r)
sp2
+
(D)
XeF5
(s)
sp3d2
71.
Match the column :
Column – I
(A) SO3 (gas)
(B) OSF4
(C) SO3F –
(D) ClOF3
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72.
(p)
(q)
(r)
(s)
Match the column :
Column–I
(A) IF2–
(p)
(B) ClF3
(q)
(C) XeO3F2
(r)
(D) SF4
(s)
Column – II
Polar with p–d bonds and identical S–O bond, lengths.
One lone pair and p–d bond.
Non–polar with p–p and p–d bonds. Identical S–O bond lengths.
Polar with p–d bond.
Column–II
sp3d
polar
one of the bond angles is 180º
one lone pair
134
SINGLE AND DOUBLE VALUE INTEGER TYPE
Number of hypervalent among following are :
(i) H2CO3, (ii) HNO3, (iii) HNO2, (iv) H3PO4, (v) H3PO3, (vi) H3PO2, (vii) HClO4, (viii) HClO3, (ix) HClO2,
(x) HClO, (xi) H2SO3, (xii) H2SO4, (xiii) H2N2O2, (xiv) H2SO5, (xv) H2S2O8, (xvi) SO3, (xvii) SO2,
(xviii) N2O5, (xix) P4O10
74.
Compound
No. of peroxide linkage(s)
–
1.
HXeO4
–
x
2.
K3CrO8
–
y
3.
H2TiO4
–
z
4.
Na2B4O7
–
w
The value of x + y + z + w is/are .............
75.
How many compound which show  zero dipole moment.
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73.
,
A bond X–Y has a dipole moment of 1.8 × 10–29 Cm and a bond length 150 pm. What will be the
percentage of ionic character is given bond.
77.
Numbe of non-polar molecule among the following is x and number of planar molecule is y. Give x × y.
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76.
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BF3, CO2, SO2, PCl5, ClF3, NH3, CH4,
78.
Among the following, give the number of correct order of boiling point or melting point.
(1) Zn > Cd > Hg
Boiling point
(2) Zn > Cd > Hg
Melting point
(3) K > Ca > Sc
Boiling point
(4) Na > Mg > Al
Boiling point
(5) Sc > Mn > Zn
Melting point
(6) Pt > Pd > Ni
Melting point
(7) Cr > Mn < Fe
Melting point
(8) Ba > Li > Na
Melting point
79.
Calculate total number of coordinate bonds in the following molecules.
(a) PCl6–
(b) NH3.BF3
(c) HNO3
80.
(d) CO
Determine number of bonding electrons in H3SiNCO.
ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
81.
82.
Hypervalent compound is(are) :
(A) SO32−
(B) PO43−
(C) SO42−
(D) CIO4−
Which of the following overlaps is/are incorrect [assuming X-axis to be the internuclear axis] :
(a) 2py + 2py → 
(b) 2pz + 2pz → 
(c) 2px + 2px → 
(d) 1s + 2py → 
(e) 2py + 2pz → 
(f) 1s + 2s → 
(A) ‘a’ & ‘b’
(B) ‘b’ & ‘d’
(C) ‘d’ & ‘f’
(D) ‘c’ & ‘e’
135
83.
In which of following, vacant orbital take part in hybridisation :
(A) BF3
(B) PCl6–
(C) BF4–
(D) NH3
Which is not true about VSEPR theory
(A) Lone pair-lone pair repulsion is maximum.
(B) Lone pair and double bond occupy axial position in trigonal bipyramidal structure.
(C) More electronegative atoms occupies axial position in trigonal bipyramidal structure.
(D) Bigger atoms occupy axial positions in trigonal bipyramidal structure.
85.
Select the correct statement.
(A) Perxenate ion is [XeO6]4– with octahedral geometry.
(B) XeF2 is linear molecule with 3 lone pairs (l.p)
(C) XeOF4, XeF4, XeO2F2 all contains one lone pair only
(D) None of these
86.
Identify the correct option(s)
(A) NH4+ > NH3 > NH2– order of bond angle
(B) (CH3)3 B is a trigonal planar molecule (not considering the H-atoms on ‘C’)
(C) In NH4Cl ‘N’ atom is in sp3d hybridisation
(D) In S8 molecule a total of 16 electrons are left on all the ‘S’ atoms after bonding.
87.
Which of the following statements is/are correct for CIO4 – oxoanion ?
(A) It does not have any tendency of polymerisation.
(B) It has strong p–p bonding between chlorine and oxygen.
(C) All Cl–O bonds are identical and chlorine atom is sp3 hybridised.
(D) The chemical bonding takes place in ground state and charge dispersion is more than CIO3– oxo anion.
88.
Among the following back bonding is exhibited by
(A) BF3
(B) (CH3)3N
(C) CH3NCO
91.
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90.
(D) SiH3NCO
Which of the following reactions is/are likely to be impossible.
(A) (CH3)2O + BF3 →
(B) (SiH3)2O + BF3 → (C) (SiH3)3N + BF3 → (D) All the above
Which of the following statements is/are correct ?
(A) Out of trimethylamine and trimethylphosphine, trimethylamine has higher dipole moment.
(B) Out of (SiH3)2O and (CH3)2O, (SiH3)2O is more basic.
(C) C–C bond length (in pm) in C2 molecule is greater than O–O bond length in O2 molecule.
(D) N(SiMe3)3 and BF3 molecules are isostructural about central atom.
In N2O3, the oxidation state of nitrogen atom is
(A) same for both the nitrogen
(C) one’s is +2 and other is +4
a
89.
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84.
(B) different for both the nitrogen and they differ by 2.
(D) No, nitrogen atom is having +3 oxidation state.
Which of the following statements is incorrect about P 4O10 molecule ?
(A) Each ‘P’ atom can be considered to be sp3 hybridised
(B) There are four POP bonds in the molecule
(C) There are two types of P–O bond lengths
(D) POP angle is 180°.
93.
In the following, identify the incorrect statements.
(A) N2F3+ is a planar at each nitrogen atom
(B) In F2 molecule, the energy of  2pz is more than  2px and  2py.
(C) The O–O bond length in H2O2 is smaller than in O2F2.
(D) B2, O2 and F2 are paramagnetic molecules.
94.
In which of the following compound(s), is/are central atom have two different individual oxidation
state(s) ?
(A) C3O2
(B) Br3O8
(C) S4O62–
(D) P2O74–
95.
Which of the following statements are correct
(A) o-hydroxy benzaldehyde is a liquid at room temperature due to intra molecular H-bonding, where as
p-hydroxybenzaldehyde is a solid at room temperature.
(B) Order of boiling point H2O > H2Se > H2S
(C) Order of boiling point HF > HI > HBr > HCl
(D) Order of boiling point SbH3 > NH3 > AsH3 > PH3
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92.
136
Which of the following molecules have intermolecular hydrogen bonds ?
(A) KH2PO4
(B) H3BO3
(C) C6H5CO2H
(D) CH3OH
97.
Which is correct statement :
(A) Borazine has higher intermolecular force of attraction as compared to benzene.
(B) The O–O bond length in O2[AsF4] is shorter then KO2.
(C) The bond angle order in halogen –S– halogen is OSF2 < OSCl2 < OSBr2
(D) None of these
98.
The critical temperature of water is higher than that of O2 because the H2O molecule has :
(A) fewer electrons than O2
(B) two covalent bonds
(C) V-shape
(D) H-bonding
99.
Which of the following is/are correct statement(s).
(A) Increasing covalent character : NaCl < MgCl2 < AlCl3
(B) Increasing covalent character : LiF < LiCl < LiBr < LiI.
(C) Increasing polarizability : F– < Cl– < Br – < I–
(D) Decreasing ionic nature : MCl3 > MCl2 > MCl
100.
Select the correct statement(s) :
(A) Among d-block elements, group 12 elements have the least enthalpy of atomisation.
(B) Cohesive forces in metals are non-directional bonds.
(C) Group 1 elements have greater strength of metallic bonding than corresponding group 2 elements
due to increasing electronegativity on moving left to right in periodic table.
(D) Hardness of copper being more as compared to potassium can be explained by metallic bond
strength.
101.
Select the correct statement(s) :
(A) In s-block elements, melting point generally decreases on moving down the group.
(B) Band model proposed to explain metallic bonding is based on molecular orbital theory.
(C) d-block elements get a dip at/near middle of series in enthalpy of atomisation values.
(D) Electricity conduction in a single layer of graphite can be considered similar to electricity conduction
in metals according to electron sea model.
102.
Which of the following statements is / are true for the metallic bond ?
(A) It is an electrical attraction between delocalised electrons and the positive part of the atom.
(B) Transition metals may use inner d– electrons along with the outer s–electrons for metallic bonding.
(C) Strength of metallic bond does not depend on the type of hybrid orbitals participating in metallic
bonding.
(D) Strength of metallic bond is inversely proportional to the radius of metallic atom in s-block.
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96.
COMPREHENSION
Read the following passage carefully and answer the questions.
Comprehension # 1
The electronic configuration of three elements, A, B and C are given below. Answer the questions 01 to
04 on the basis of these configurations.
A
1s2
2s2
2p6
B
1s2
2s2
2p6
3s2
3p3
2
2
6
2
C
1s
2s
2p
3s
3p5
103.
Stable form of A may be represented by the formula :
(A) A
(B) A2
(C) A3
(D) A4
137
104.
105.
106.
Stable form of B may be represented by the formula :
(A) B
(B) B2
(C) B3
(D) B4
The molecular formula of the compound formed from B and C will be
(A) BC3
(B) B2C
(C) BC5
(D) A and C both
The bond between B and C will be
(A) Ionic
(B) Covalent
(D) Coordinate
(C) Hydrogen
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Comprehension # 2
Na(BH4) is ionic compiound containing BH4- tetrahydrido borate ion and in solid state NaBH4 has
sodium chloride structure. Not all tetrahydrido borates are ionic the beryllium, aluminium and transition
metal borohydrides become increasingly covalent and volatile. In these type of tertahydrido borate the
BH4- ion form covalent bond with metal ion. One or more H atoms in BH 4– act as a bridging and bond to
metal forming a 3 center bond with 2 electron shared by 3 atoms. The BH 4– is usually in that it may form
1, 2 or 3 such 3 center bonds to the metal ion when form covalent bond.
Li[AlH4] is used as a reducing agent in many reactions and it is prepared by excess LiH and AlCl 3.
select incorrect staements about Li[AlH4]
(A) hybridisation of Al is same as B in Na(BH4).
(B) Geometry around Al is same as in AlCl4–
(C) AlH4–, BH4–, AlCl4–, are iso structural
(D) AlH4–, BH4–, AlCl4–, are iso electronic
108.
Select correct satements about Al(BH4)3
(A) All three tetrahydrido borate form two hydrogen bridges
(B) two BH4– form two hydrogen bridges and one form one hydrogen bridge
(C) one BH4– form two hydrogen bridges and two form one hydrogen bridge
(D) B form only 2c–2e bond
109.
Total number of 2c–2e bond and 3c–2e bond in Al(BH4)3 are respectively
(A) 6, 12
(B) 6, 6
(C) 12, 12
(D) 12, 6
110.
Total number of 2c – 2e bond and 3c – 2e bond in Be(BH4)2 are respectively
(A) 8, 4
(B) 4, 8
(C) 4, 4
(D) 8, 8
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107.
PART - IV : PRACTICE TEST-2 (IIT-JEE (ADVANCED Pattern))
Max. Marks : 66
a
Max. Time : 1 Hr.
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Important Instructions
A. General %
1.
2.
B.
3.
4.
5.
6.
7.
8.
C.
9.
The test is of 1 hour duration.
The Test Booklet consists of 22 questions. The maximum marks are 66.
Question Paper Format
Each part consists of five sections.
Section-1 contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE is correct.
Section-2 contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE OR MORE THAN ONE are correct.
Section-3 contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from
0 to 9 (both inclusive).
Section-4 contains 1 paragraphs each describing theory, experiment and data etc. 3 questions relate to
paragraph. Each question pertaining to a partcular passage should have only one correct answer among
the four given choices (A), (B), (C) and (D).
Section-5 contains 1 multiple choice questions. Question has two lists (list-1 : P, Q, R and S; List-2 : 1, 2,
3 and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY ONE
is correct.
Marking Scheme
For each question in Section 1, 4 and 5 you will be awarded 3 marks if you darken the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one
(– 1) mark will be awarded.
138
10.
11.
For each question in Section 2, you will be awarded 3 marks. If you darken all the bubble(s)
corresponding to the correct answer(s) and zero mark. If no bubbles are darkened. No negative marks will
be answered for incorrect answer in this section.
For each question in Section 3, you will be awarded 3 marks if you darken only the bubble corresponding
to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for
incorrect answer in this section.
1.
Select the correct order of following property
(A) % s-character : sp3 > sp2 > sp
(B) O–N–O bond angle : NO3– > NO2+ > NO2
(C) All angles in CH2F2 are not identical
(D) C–F bond length : CF4 > CH3F > CH2F2 > CF3H
2.
F–As–F bond angle in AsF3Cl2 can be nearly :
(A) 90° & 180° only
(B) 120° only
(D) 90° only
4. PCl2F3
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4.
The correct sequence of polarity of the following molecule :
1. Benzene
2. Inorganic Benzene 3. PCl3F2
(P stands for polar and NP stands for non-polar)
1
2
2
4
1
(A)
P
NP
NP
P
(B)
NP
(C)
NP
P
NP
P
(D)
NP
2
NP
P
Which of the following molecule(s) is/are having p – p back bonding
(A) BF3
(B) BeF2
(C) BCl3
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3.
(C) 90° & 120° only
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SECTION-1 : (Only One option correct Type)
This section contains 7 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which Only ONE option is correct.
2
NP
P
4
P
NP
(D) All of these
In which of following option the (X–A–X) adjacent angle difference in cationic part and anionic part is
maximum in the solid state.
(A) PCl5
(B) Cl2O6
(C) N2O5
(D) BeH2
6.
The boiling point of a compound is raised by (A) intermolecular hydrogen bonding
(C) Intramolecular hydrogen bonding
(B) High volatility
(D) Non-polarity
In which of the following compound peroxide linkage is absent ?
(A) Per oxy mono sulphuric acid
(B) Per titanic acid
(C) Per chloric acid
(D) Per nitric acid
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7.
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Section-2 : (One or More than one options correct Type)
This section contains 5 multipole choice questions. Each questions has four choices (A), (B),
(C) and (D) out of which ONE or MORE THAN ONE are correct.
8.
9.
The molecule is/are having N–N bond.
(A) N2O
(B) N2O3
(C) N2O5
London force works in
(A) Polar molecule
(C) All polar and non-polar molecule
(B) Non-polar molecule
(D) Only in polar molecule
(D) N2O4
10.
There is change in the type of hybridisation when :
(A) NH3 combines with H+
(B) AlH3 combines with H–
–
(C) NH3 forms NH2
(D) SiF4 forms SiF62–
11.
Select the correct order (A) Bond length = O22– > O2– > O2 > O2+
(B) Bond strength = O2+ > O2 > O2– > O22–
(C) Unpaired electron(s) = O2 > O2+ > O2– > O22–
(D) No. of antibonding electron(s) = O22– > O2– > O2 > O2+
12.
Choose the option(s) regarding correct order of acidic nature :
(A) MgO < ZnO < P2O5 < SO3
(B) MgO < ZnO < SO3 < P2O5
139
(C) Li2O < NO < CO2 < SO2
(D) Li2O < BeO < CO2 < NO
Section-3 : (One Integer Value Correct Type.)
This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive)
X is the number of maximum atom(s) is/are present in same plane of B3N3H6. Find value of X / 2
14.
Find the number of molecules which are not having 3C–2e bond from the following :
(a) Al2(CH3)6
(b) Si2H6
(c) B2H6
(d) C2H6
(e) Si2Cl6
(f) Al2Cl6
(g) B3N3H6
15.
Which of the following species having axial bond length greater than equatorial bond length :
PCl5, IF7, SF6, CCl4
16.
The total number of orbitals involved in the overlapping in XeO 2F4? Give answer after dividing by 4.
17.
How many different type of bonds can be formed if any two d orbitals of two different atoms are
overlapping with each other ?
18.
In which of the following all bond length are not equal ?
(a) PCl5
(b) SF4
(c) ClF3
(e) [SF5]+
(f) [ClF4]+
(g) [XeF3]+
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13.
(d) XeF2
(h) O3
(i) P4 (white)
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SECTION-4 : Comprehension Type (Only One options correct)
This section contains 1 paragraphs, each describing theory, experiments, data etc. 3 questions
relate to the paragraph. Each question has only one correct answer among the four given
options (A), (B), (C) and (D)
Paragraph for Questions 19 to 21
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Hybridisation is the mixing of atomic orbital of comparable energy and the number of hybrid orbitals
formed is equal to the number of pure atomic orbitals mixed up and hybrid orbitals are occupied by bond pair and lone pair.
Which of the following geometry is most unlikely to form from sp3d hybridisation of the central atom
(A) Linear
(B) Regular Tetrahedral
(C) T-shaped
(D) See-Saw
20.
The orbital, which does not participate in sp3d2 hybridisation.
(A) px
(B) dxy
(C) dx2 − y2
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19.
21.
(D) pz
''The hybird orbitals are at angle of X° to one another'' this statement is not valid for which of the
following hybridisation. (X  180º)
(A) sp3
(B) sp2
(C) sp3d2
(D) sp
SECTION-5 : Matching List Type (Only One options correct)
This section contains 1 questions, each having two matching lists. Choices for the correct
combination of elements from List-I and List-II are given as options (A), (B), (C) and (D) out of
which one is correct
22.
Match each List-I with List-II and select the correct answer using the code given below the lists.
List-I
List-II
(Pair of species)
(Idenitcal Property in pairs of species)
(P)
PCl3F2.PCl2F3
(1)
Hybridisation of central atom
(Q)
BF3 & BCl3
(2)
Shape of molecule / ion
(3)
 (dipole moment)
(4)
Total number of electrons
CN2–2
(R)
CO2 &
(S)
C6H6 & B3N3H6
140
Code :
(A)
(B)
(C)
(D)
P
1,2
1,2,3,4
1,2,3
1,2,3,4
Q
1,2,3
1,2,3
1,2
1,2,3
R
1,2,3,4
1,2
1,2,3,4
1,2,3,4
S
1,2,3,4
1,2,3,4
1,2,3,4
1,2
Practice Test-2 ( (IIT-JEE (ADVANCED Pattern))
OBJECTIVE RESPONSE SHEET (ORS)
Que.
1
2
3
4
5
6
7
11
12
13
14
15
16
17
21
22
8
9
10
Que.
Ans.
Que.
18
19
20
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Ans.
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Ans.
PART - I
(4)
2.
(4)
3.
6.
(3)
7.
(4)
8.
11.
(1)
12.
(2)
16.
(1)
17.
(2)
21.
(2)
22.
(4)
26.
(2)
27.
(4)
4.
(3)
5.
(2)
(3)
9.
(3)
10.
(2)
a
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1.
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(2)
13.
(3)
14.
(1)
15.
(1)
18.
(2)
19.
(3)
20.
(2)
23.
(1)
24.
(4)
25.
(4)
28.
(2)
29.
(3)
30.
(2)
PART - II
1.
(B)
2.
(C)
3.
(A)
4.
(B)
5.
(C)
6.
(A)
7.
(C)
8.
(A)
9.
(C)
10.
(B)
11.
(D)
12.
(D)
13.
(D)
14.
(A)
15.
(D)
16.
(B)
17.
(C)
18.
(A)
19.
(B)
20.
(C)
21.
(C)
22.
(A)
23.
(D)
24.
(D)
25.
(D)
26.
(A)
27.
(D)
28.
(A)
29.
(C)
30.*
(CD)
31.
(C)
32.
(A)
33.
(D)
34.
(B)
35.
(B)
36.
(C)
37.
(D)
38.
(A)
39.
(C)
40.
(B)
41.
(B)
42.
(D)
43.
(C)
44.
(D)
45.
(B)
46.
(C)
47.
(C)
48.
(A)
49.
(D)
50.
(C)
51.
(C)
52.
(B)
53.
(B)
54.
(A)
55.
(D)
56.
(C)
57.
(A)
58.
(D)
59.
(C)
60.
(C)
61.*
(BC)
62.
(C)
63.
(B)
64.
(C)
65.
(B)
141
66.
(B)
67.
(D)
68.
(B)
69.
(D)
70.
(B)
71.
(D)
72.
(C)
73.
(A)
74.
(A)
75.
(A)
76.
(C)
77.
(D)
78.*
(A or C)
79.
(C)
80.
(B)
81.
(D)
82.
(D)
PART - III
•• F
Xe
(i)
sp3d ; 4, 2
Cl
O
O
P
(ii)
Cl
F
F
..
N
(v)
P
P
60º
sp3 ; 6
(in molecule)
P
6 4 1 5
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2.
sp3d2 ; 6
Ja
P
(iv)
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..
sp3 (N-atom) ; 4 (in molecule)
O
.. – H
(iii) H
H
sp3d ; 5
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1.
F
F
3.
a
(BeH2)n contains 2e–3c bonds whereas
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(BeCl2)n contain the usual 2e – 2c bonds.
4.
Sulphuric acid (H2SO4)
O
||
H − O − S − O − H S.No. = 4, Hybridisation = sp3, Oxidation state of Sulphur = + 6
||
O
Marshall's acid (H2S2O8)
Oxidation state of both sulphur atoms = + 6
Caro's acid (H2SO5)
Oxidation state of sulphur = +6
142
Oleum (H2S2O7)
Oxidation state of both sulphur atoms = +6
⎯⎯
→
6.
Bonding (3 nodal planes)
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⎯⎯
→
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5.
Anti bonding (4 nodal planes)
(i) Sc2+ < Ti3+ < V3+ < Cr3+
(ii) Cd2+ < Zn2+ < Hg2+
8.
It has highest polarising power due to smallest ionic radius amongst alkali metal, therefore, greater
degree of hydration is observed in Li+ salts.
9.
(D)
10.
(C)
11.
(B)
12.
(C)
13.
(C)
14.
(D)
15.
(B)
16.
(C)
17.
(C)
18.
(B)
19.
(D)
20.
(A)
21.
(A)
22.
(B)
23.
(A)
24.
(C)
25.
(D)
26.
(D)
27.
(D)
28.
(A)
29.
(B)
30.
(B)
31.
(C)
32.
(B)
33.
(D)
34.
(C)
35.
(C)
36.
(D)
37.
(B)
38.
(A)
39.
(C)
40.
(B)
41.
(D)
42.
(D)
43.
(B)
44.
(C)
45.
(D)
46.
(B)
47.
(A)
48.
(C)
49.
(C)
50.
(B)
51.
(C)
52.
(D)
53.
(D)
54.
(B)
55.
(D)
56.
(C)
57.
(C)
58.
(B)
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7.
59.
(D)
60.
(A)
61.
(C)
62.
(A)
63.
(C)
64.
(A)
65.
(C)
66.
(A)
67.
(A)
68.
(B)
69.
(C)
70.
(A - p) ; (B - q) ; (C - p) ; (D - s)
72.
(A – p , r) ; (B – p , q) ; (C – p , r) ; (D – p , q , s)
73.
13 (Except, i, iii, iii, x, xiii, xviii)
74.
6
75._
4
76.
77.
25
78.
6
79.
4
80.
20
81. (ABCD)
82.
(BD)
83.
(BC)
84.
(BD)
85.
(AB)
86.
(AB)
87.
(AC)
88.
(AD)
89.
(BC)
90.
(ACD)
91.
(BCD)
92.
(BD)
93.
(BCD)
94.
(ABC)
95.
(ABCD)
96. (ABCD)
97.
(ABC)
98.
(D)
99.
(ABC)
100.
(ABD)
101. (ABCD)
102.
(ABD)
103.
(A)
104.
(D)
105.
(D)
106.
107.
(D)
108.
(A)
109.
(B)
110.
(C)
71. (A – r) ; (B – s) ; (C – p, s) ; (D – q, s)
75
(B)
143
PART - IV
1.
(C)
2.
(A)
3.
(B)
4.
(D)
5.
(C)
6.
(A)
7.
(C)
8.
(ABD)
9.
(ABC)
10.
(BD)
11.
(ABD)
12.
(AC)
13.
6
14.
5 (b, d, e, f, g) 15.
1
16.
4
17.
3
18.
6 (a, b, c, e, f, g) 19.
(B)
(B)
21.
(C)
22.
(A)
a
ri
20.
PART - I
Element belongs to d-block in d-block elements (n – 1) d and ns electron take part in the bonding.
3.
Proper ovaerlapping occurs between px and s orbital.
4.
H3N : + BF3 ⎯⎯
→ H3N → BF3
lewis lewis
base acid
5.
Bond energy  Bond order.
Ja
uh
1.
Steric number = 5 ; Hybridisation = sp3d.
7.
Hybridisation = sp3 ; Bond angle = 110º
Boiling point of covalent molecules is decided by van der Waal force of attraciton and vander Waal
force of attraction depends on molecular weight.
a
8.
lp
6.
Sa
nk
9.
10.
Cl–O–Cl bond angle is more due to large size of Cl and F–O–F bond angle is least due to bent rule.
11.
O2– has one unpaired electron in *2py while other do not have unpaired electrons. (explained by MOT)
12.
BF3 has triangular planar with no lone pair.
( = 0)
13.
Strongest H-bonding is in F–H ------ O as F is most electronegative element hence creating good partial
postive charge on H. While O is good donor of electron as compare to F due to use electronegativity.
14.
It is fact.
15.
Oxygen has unpaired in *2px and *2py which explain its paramagnetic behaviour. (Explained by MOT)
144
17.
18.
Diborane : (B2H6) :
Diamond has sp3 hybrid carbon and graphite has sp2 hybrid carbon.
NH4+, H3O+ and OF2 all have sp3 hybridisation and their shapes are tetrahedral, trigonal pyramidal and
bent respectively.
a
ri
19.
20.
..
+
NH3 + H+ ⎯⎯
→ NH4
(sp3 )
(sp3 )
Trigonal pyramidal
Tetrahedral
(Bond angle  107º)
(Bond angle  109.5º)
Ja
21.
uh
It have both covalent as well as coordinate bond.
Polarisability of anion increase on increasing its size.
23.
In CCl4, C has more charge therefore Cl is more polarized in CCl 4.
24.
For the formation of ionic compound atom should have low ionization energy to form cation while other
should have high electron affinity to form anion.
26.
Covalent character ionic compound is explained by Fajan rule. Covalent character increase on
decreasing the size of cation and increasing charge on cation.
27.
More is the electronegativity difference stronger is the bond.
lp
22.
O2 → [KK] 2s2 *2s2 2pz2 2px2 2py2 *2px1 *2py1
O2– → [KK] 2s2 *2s2 2pz2 2px2 2py2 *2px2 *2py1
As in O2 HOMO is *2px and *2py and they have one electron each so next electron can be added to
these orbitals.
Sa
nk
29.
a
28.
30.
5.
(P)
(Q)
O O
|| ||
HO — P — P — OH
|
|
OH OH
PART - III
Three P–O–P bonds.
One P–P bond.
145
(R)
Two P–H bond.
(R)
Four P–OH bond.
In SF6, PCl5 and IF7 the valence shell has 12, 10 and 14 electrons. As all contain more than 8 electrons
in their valence shell they are example of super octet molecules.
10.
Bond energy  Bond order
Species
Bond order
CO
3
CO2
2
CO32–
1.33
11.
Across a period, effective nuclear charge increases. As a result the size of atom, and, therefore, the
size of d-orbital deceases leading to progressively stronger p–d bonds.
12.
(d) Position of atoms are different.
(e) Has unpaired electrons and this is not possible since the molecule is diamagnetic.
+
13.
Ja
+ +
x
(A)
pz
 bond
lp
zero overlap
y
+
,
Sa
nk
 bond
x
+
s
a
px
14.
15.
(D)
(A)
(A) both are sp3d
,
overlap is not sufficient so, such  bond do not exist.
pyramidal : (A)
16.
17.
x
–
pz pz
–
+
(B)
– –
(C)
uh
a
ri
9.
(B)
(C)
(B)
(C)
(D)
Steric No = 4
hybridization = sp3
(B) both are sp3d
(C) [ClF2O]+ is sp3 but [ClF4O]– is sp3d2
146
(D) both are sp3d2
The shape of AsF4– is see-saw.
19.
PH3 – No Hybridisation
NH3 and H2O bond angle is less than 109º28' due to LP–BP and LP–LP repulsion.
CH4 bond angle is 109º28'
20.
(i) Negative charge is more stable on ‘S’ than on N (size factor is dominant over EN factor).
(ii) Less formal charge provides more stability.
21.
S–S over lap will result into  bond.
a
ri
18.
23.
uh
22.
I and III are sp3 hybridised while II and IV are sp3 hybridised.
P4 bond angle = 60°
PH3 bond angle  90° (Drago's rule)
H2O bond angle = 104.5°
lp
25.
a
26.
(A) Bent rule.
(C) H2O2 has more lone pair-lone pair repulsion.
Sa
nk
27.
Ja
24.
30.
31.
(A) N2O3
Dinitrogen trioxide
(B) N2O22–
Hyponitrite ion
(C) N2O5
Dinitrogen pentoxide
147
(D) N2O4
Dinitrogen tetroxide
(A) Stability
O2+ > O2 > O2–
Bond order
2.5 2
1.5
(C) In all these molecules all electrons are paired in molecular orbitals.
34.
(C) O2 : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p1y)
O2– : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px2 = *2p1y)
35.
(I) O2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p0y)
Bond order = 1/2(10 – 5) = 2.5.
(II) NO is derivative of O2 and isoelectronic with O2+ :
so (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y ) (*2px1 = *2p0y)
Bond order = 1/2(10 – 5) = 2.5.
(III) N2+ : (1s)2 (*1s)2 (2s)2 (*2s)2 (2p2x = 2p2y ) (2pz)1
Bond order = 1/2(9 – 4) = 2.5.
O2+ , NO and N2+ have same bond order i.e. 2.5 and have same magnetic property having one unpaired
electron.
36.
Boron trimethyl is a weaker Lewis acid than the boron trihalides or monoborane. The electron donating
effect of the methyl groups hinders the complex formation with trimethyl amine. Hence the bond N → B
is weakest in [(CH3)3 N → B(CH3)3]. Me3N as donor (capacity). BBr3 > BCl3 > BF3 ~ BH3 > BMe3.
37.
OS=O
39.
S1 : as it does not have d–orbitals.
uh
a
ri
33.
lp
S2 : ClF3
Ja
H–O–N = O
Sa
nk
a
S3 : In B2 mixing of the g(2s) orbital with the g(2p) orbital lowers the energy of the g(2s) orbital and
increases the energy of the g(2p) orbital to a higher level than the  orbitals. As a result, the last two
electrons are unpaired in the degenerate (having the same energy)  orbitals, and the molecule is
paramagnetic.
In N2 the g(2s) and g(2p) levels of N2 interact (mix) less than the B2 and C2 levels, and the g (2p) and
u(2p) are very close in energy.
40.
Bond order of N2 = 3 bond order of N2+ = 2.5 ;
41.
M.O for C2 = 1s2 < 1*s2 < 2s2 < 2*s2 <  2p2 y =  2p2z   2p x
B.O. of O2 = 2;
B.O. of O2+ = 2.5
LUMO
HOMO
It is important to note that double bond in C2 consists of both pi bonds because of the presence of four
electrons in two pi molecular orbitals C22–
42.
.
(A) H2O2 = 1.48 Å due to repulsions between non-bonded pairs of electron on O-atoms and O2F2 =
1.217 Å.
(B) In O22− very slightly increases due to charge (–ve) on two O atoms.
(C)
148
43.
OF is derivative of O2 and isoelectronic with O2–.
So (1s)2 (*1s)2 (2s)2 (*2s)2 (2pz)2 (2p2x = 2p2y) (*2px2 = *2p1y)
The bond order of OF 1/2(10 – 7) = 1.5.
44.
ClO2 has are unpaired electron.
45.
(A)
NO B.O = 3 triple bond.
O = N – Cl there is a double bond between N and ‘O’ atom
CaC2  Ca2+ + C22– (–C  C–)
C2H4  H
H
C=C
H
H
KO2  O2–
Na2  O22–
H
(D)
H
B.O = 1.5
B.O = 1
H
Cpy =pyCpz =pzC
H
(S + Py + Px)
(xy – plane)
(S + Pz + Px) (Spx)
xz - plane
46.
a
ri
(C)
uh
(B)
(I) [PCl4]+ → sp3
has  102° bond angle due to lp – lp repulsion, as bond pairs are closer to F-atoms.
Ja
(II)
(III) All have sp3 hybridisation and one lone pair.
S
S
O
O
48.
O
O
O
3
SP
O
Sa
nk
S
O
O
O
a
O
lp
(IV)
HO
O
O
P
P
O
O
3
SP
OH
P
O
OH
50.
Dipole moment of compound if it would have been completely ionic
= (4.8 × 10–10 esu) (2.67 × 10–8 cm) = 12.8 D
4.0
so % ionic character =
× 100% = 31.25 %
12.8
51.
Graphite has layered structure. Layers are held by van der Waal’s forces and distance between two
layers is 340 pm. Each layer is composed of planar hexagonal rings of carbon atoms. C–C bond length
within the layer is 141.5 pm Each carbon atom in hexagonal ring undergoes sp 2 hybridisation and make
three sigma bonds with three neighbouring carbon atoms. Fourth electron forms a  bond. The
electrons are delocalised over the whole sheet.
53.
(A) A fact
(B) In the solid state and in liquid HF, the HF 2– ion is held together by hydrogen bonding. In aqueous
solutions there is hydrogen bonding but each HF molecule forms hydrogen bonds with the much more
prevalent H2O present, instead of with other HF molecules and H3O+ and F – are much more likely to
be formed.
(C) H3BO3 (solid) has inter molecular hydrogen bonding.
149
H
B
H
H
H
H
H
B
H
H
H
B
H
H
B
H
H
H
B
B
H
H
H
H
54.
(A) intermolecular hydrogen bonded. (true)
(false)
Ja
SO32– – trigonal pyramidal (sp3)
uh
(B) structure of anions are different CO32– – trigonal planar (sp2)
a
ri
H
no lone pair of electrons (false)
(D) True.
D2O = 374.4 K ; H2O = 373.0 K
lp
(C)
D is less electronegative than H-atom so the results
56.
This is based on the decreasing electronegativity difference between H(2.1) and Sb(1.9), As(2.0),
P(2.1) and N(3.0).
58.
Fluorine is more electronegative but dipole moment is the product of charge and distance between
combining atoms (covalently bonded) ; due to more charge separation in CH 3Cl, it has higher dipole
moment.
Sa
nk
a
55.
59.
(A) As + charge on central atom increases, the attraction between + and – increases and thus Xe–F
bond length decreases. The correct order is XeF2 > XeF4 > XeF6
(B) PH5 can not undergo sp3d hybridisation as there is much large difference in size of s, p and d
orbitals. PH5 does not exist as no partial positive charge develops on P atom.
(C) Dipole moment of CH3Cl is greater than CH3F due to greater charge sepration on carbon and
chlorine atoms in CH3Cl.
(D) it is a correct order.
The strength of hydrogen bond depends upon :
(i) size (ii) electronegativity and (iii) ease of donation of electron pair by electronegative element.
Higher the value of electronegativity and smaller the size of the covalently bonded atom to H atom
stronger is the hydrogen bonding.
60.
Fact based
61.
Fact based
62.
NaCl is an ionic compound.
150
The element present in short period will be of 13th group (group no = 10+ 3 = 13 ) i.e. aluminium and its
oxide will be Al2O3 which is amphoteric in nature; as react with acids as well as bases forming salts and
water.
64.
As Pb2+ has low polarising power. So PbCl2 is ionic.
65.
Solubility BaCO3 and MgCO3 can be explained on the basis of their hydration and lattice energies.
66.
AgI is less soluble than AgF due to covalent nature. Li I is more soluble then LiF due to high size
difference of ions more ionic nature, more melting point. Higher the charge density, larger will be the
hydrated radii.
67.
Strength of metallic bond  number of unpaired electrons.
68.
(i) Alkali metal, potassium forms K+ (number of electrons = 18) which is isoelectronic with P3– (number
of electrons = 18) → Q.
(ii) Transition element, chromium show maximum oxidation state of +6 in Cr 2O72– and CrO42– and they
are coloured → S.
(iii) Noble gas, Kr has largest atomic radius because radius is expressed as vander Waal's radius and
there is interelectronic repulsions due to completely filled outer most shell. Inert gas has highest first
ionisation energy in the respective period because of stable valence shell electron configuration → T.
(iv) Element (Z = 13), aluminium has intermediate value of electronegativity and therefore its oxide is
amphoteric in nature → R.
70.
Steric no. of 3– = 5, sp3d;
Steric no. of ClOF3 = 4 + 1 = 5, sp3d ;
71.
(A)
Ja
a
lp
polar, p-d bonds present.
No lone pair.
(C)
Steric no. of XeO3 = 4, sp3;
Steric no. of XeF5+ = 5 + 1 = 6, sp3d2
non-polar, p-d bonds present.
All S–O bonds are identical.
No lone pair.
(B)
uh
a
ri
63.
→ polar
Sa
nk
→ p-d bonds present.
→Due to resonance identical
S-O bond length
(D)
→ polar
→ p-d bonds present.
→ lone pair present.
72.
(A)
sp3d;  F – I – F = 1800 ;  = 0
(B)
sp3d; polar;  F – Cl – F  180º
151
(C)
sp3d, non–polar,  F – Xe – F = 1800
(D)
sp3d, polar;  F – S – F  1800 ; one lone pair.
73.
H3PO4, H3PO3, H3PO2, HClO4, HClO3, HClO2, H2SO3, H2SO4, H2SO5, H2S2O8, SO3, SO2, P4O10
74.
1.

x=1
2.
3.

z=1
4.
cal = q × d = 1.6 × 10–19 C × 150 × 10–12 m = 2.4 × 10–19 Cm
cal
 100 = 75%
Percentange ionic character =
obs
77.
X = BF3, CO2, PCl5, CH4,
78.
(1, 2, 5, 6, 7, 8)
a
ri
w=O
obs = 1.8 × 10–29 Cm
;
Y = BF3, CO2, SO2, ClF3,
a
Sa
nk
(a) PCl6–
(c) HNO3

lp
76.
79.
y=4
uh
and NH3 will have non-zero dipole moment.
Na2B4O7
Ja
75._

(Coordinate Bond = 1)
(Coordinate Bond = 1)
(b) NH3.BF3
(D) CO
(Coordinate Bond = 1)
(Coordinate Bond = 1)
80.
LP of N can be delocalised back bonding (p – d). SN of N = 2 sp
81.
The species in which the central atom has expanded octet, are called hypervalent compounds.
82.
(b) If X-axis is internuclear axis, then it will result in  bond.
(c) It will result in  bond.
(d) & (e) It shows zero overlap with no bond formation.
83.
In PCl6– and BF4–, vacant orbital take part in hybridization
84.
Bigger atoms, LP and double bond occupy equatorial positions in trigonal bipyramidal structure.
152
F
O
O¯
85.
O¯
(A)
Xe
(B)
Xe
O¯
O¯
F
XeF2 (Linear)
O
Perxenate ion [XeO6]4-
F
O
F
F
(D)
Xe
O
F
XeO2F2(one l.p.)
F
F
XeOF4 (one l.p.)
(A) Due to the presence of lone pair bond angle decreases.
(B)
CH3
|
B
H3C
uh
86.
Xe
O
a
ri
(C)
CH3
sp hyb.
Ja
2
(C) NH4Cl is an ionic compound and ‘N’ is in sp3 hybridisation.
(D) S8 molecule has 16 electron parirs left behind after the bonding.
lp
87.
Sa
nk
a
Valence shell electron configuration of Cl, 3s2 3p5
To obtain effective p–d overlap the size of the d orbital must be similar to size of the p orbital. Hence
in chlorine, p–d bonding is so strong that no polymerization of oxoanions occurs.
. .
89.
(CH3 )2 O : can act as lewis base but (SiH3)2O & (SiH3)3N can not, as one pairs on O & N are
deloecalised in to empty orbitals of Si.
90.
(A) Nitrogen is more electronegative than phosphorus.
So, dipole moment of trimethylamine is greater than trimethy phosphine.
(B)
In trisilyl ether the lone pair of electron on oxygen atom is less easily available for donation because of
p-d delocalisation due to presence of the vacant d-orbital with Si. This however is not possible with
carbon in CH3–O–CH3 due to the absence of d-orbital making it more basic.
153
(C) Bond order of C2 and O2 are same i.e., 2. In C2 molecules both bonds are -bonds whereas, there is
one  and one -bond in O2 molecule
C2 = 131 pm ; O2 = 121 pm.
(D)
uh
The P–O bond lengths shows that the bridging bonds on the edges are 1.60 Å but the P=O bonds on
the corners are 1.43 Å and this P=O is formed by p–d back bonding. A full p-orbital on the oxygen
atom overlaps sideway with an empty d-orbital on the phosphorus atom. The bond angle POP is 127º
and there is no P–P bonds.
(A) Structure is similar to that of ethane. Each N atom is tetrahedrally surrounded by one N, two H and
a lone pair. The two halves of the molecules are rotated 95º about N–N bond and occupy a gauche
(non-eclipsed) conformation. The bond length is 1.45 Å.
lp
93.
P4O10
Ja
92.
a
ri
91.
a
(B) Has partial double bond character due to p-d delocalisation.
(C) OF2=103º (approximate) and OCl2 = 112º (approximate).
Sa
nk
(D) Exist in polymeric structure as
+2
94.
C3O2
0
+2
O=C=C=C=O
S4O62–
in solid state.
Br3O8
P2O74–
96.
All the gives specices contain H-bonded to O, and so are capable of H-bonding.
98.
The critical temperature depends on the magnitude (of strength) of intermolecular force of attraction
between the molecules. If a molecule has dipole moment it means there is dipole dipole attraction
between the molecules and thus it will affect the critical temperature.
99.
According to Fajan's rule.
100.
Group1 elements have lesser strength of metallic bonding than corresponding group 2 elements due to
only one electron per atom participating in metallic bonding in group 1. In group 2 elements, two
electrons per metal atom participate in metallic bonding.
154
(A) due to decreasing strength of metallic bond.
(B) Fact
(C) Elements having half-filled stable configurations have lesser strength of metallic bonding.
(D) Electron delocalisation is the main reason for electricity conduction is both.
102.
(A) It results from the electrical attractions among positively charged metal ions and mobile, delocalised
electrons belonging to the crystal as a whole.
(B) In general, greater the number of (n–1)d and ns electrons, stronger is the resultant bonding.
(C) Strength of metallic bond depend on the type of hybrid orbitals participating in metallic bonding.
(D) As the size of atom increases the attraction between the positive part of the atom and delocalised
electrons decreases and thus the strength of the metallic bond decreases.
103.
A = Ne
104.
B4 = P4
105.
BC3 = PCl3
BC5 = PCl5
uh
a
ri
101.
Ja
108.
lp
110.
PART - IV
In CH2F2 bond angle are not identical due to bent rule.
a
1.
F–As–F angles are 90º and 180º.
Sa
nk
2.
3.
Benzene is non polar.
Inorganic benzene is non polar.
PCl3F2 
Non polar ;
PCl3F3 
4.
In BF3, BeF2, BCl3 p-p back bonding takes place.
In BF3 and BeF2 2p–2p back bond and BCl3 2p–3p back bond is formed.
5.
PCl5 (s)  [PCl4]+ (109.5º), [PCl6]– (90º)
Cl2O6 (s)  [ClO2+] (120º), [ClO4–] (109.5º)
N2O5  [NO2+] (180º), [NO3–] (120º)
Polar
155
7.
8.
N2O

N2O5

+
N  Ν– O –
;
;
N2O3

N2O4

9.
It is fact.
10.
NH3 (sp3) + H+ ⎯→ NH4+ (sp3) ;
AlH3 (sp2) + H– ⎯→ AlH4– (sp3) ;
11.
Using MOT, Bond order of O22– = 1, O2– = 1.5, O2 = 2, O2+ = 2.5.
15.
PCl5
16.
Total no. of  bonds = 6
Total no. of  bonds = 2
total no. of bonds= 6+2 = 8
No. of orbitals involved= 2*8=16
Ja
uh
NH3 (sp3) ⎯→ NH2– (sp3) + H+
SiF4 (sp3) ⎯→ SiF62– (sp3d2)
Sa
nk
a
lp
17.
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(Perchloric acid)
(No-peroxide linkage)
18.
(Source : Schaum series page no. 129)
PCl5, SF4, ClF3, [SF5]+, [ClF4]+, [XeF3]+.
PCl5 → sp3d
SF4 →
ClF3 → sp3d
[XeF2] → sp3d
+
3
[SF5] → sp d
[ClF4]+ → sp3d
[XeF3]+ → sp3d
156
SOLUTIONS & COLLIGATIVE PROPERTIES
JEE(Advanced) Syllabus
Concentration in terms of mole fraction, molarity, molality and normality.
Raoult's law; Molecular weight determination from lowering of vapour pressure, elevation of
boiling point and depression of freezing point.
a
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JEE(Main) Syllabus
Ja
uh
Different methods for expressing concentration of solution- molality, molarity, mole fraction,
percentage (by volume and mass both), vapour pressure of solutions and Raoult's Law-Ideal and
non-ideal solutions, vapour pressure-composition, plots for ideal and non-ideal solutions;
Colligative properties of dilute solutions-relative lowering of vapour pressure, depression of
freezing point, elevation of boiling point and osmotic pressure; Determination of molecular mass
using colligative properties; Abnormal value of molar mass, van.t Hoff factor and its
significance.
Section (A) : General Introduction & types of solution
1.
Introduction :
lp
A solution is a homogeneous mixture of two or more substances which are chemically non-reacting. We
come across many types of solutions in our daily life. e.g., solid-liquid, liquid-liquid, gas-gas. In this
chapter we will learn several properties of solutions and their applications.
Solution : A homogeneous mixture of two or more substances is known as solution.
D2
Solute : The substance present in smaller amount in a solution is called solute.
D3
Solvent : The substance present in larger amount in a solution is called solvent.
D4
Binary solutions : Those solutions which contain two components are called binary solutions, e.g., salt
solution, benzene and toluene.
D5
Ternary solutions : Those solutions which contain three components are called ternary solutions, e.g.,
ethanol + water + acetic acid.
D6
Aqueous solution : When solute is dissolved in water, it is called aqueous solution, e.g., sugar
solution, ethanol in water.
D7
Non-aqueous solution : When solute is dissolved in solvent other than water, it is called non-aqueous
solution, e.g., iodine dissolved in alcohol (Tincture of iodine).
D8
Miscible liquids : Those liquids which mix with each other and form homogeneous mixture are called
miscible liquids.
D9
Immiscible liquids : Those liquids which do not mix with each other are called immiscible liquids.
D10
Alloys : Solid solutions of the two or more metals are called alloys. One of them can be a non-metal
also.
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D1
157
2.
Types of Solutions :
S.No.
Solute
Solvent
Types of
Solutions
Examples
Solid Solutions
1
Solid
Solid
Solid in solid
All alloys like brass, bronze, an alloy of copper and gold, etc.
2
Liquid
Solid
Liquid in solid
Amalgam of mercury with Na, CuSO4.5H2O. FeSO4.7H2O
3
Gas
Solid
Gas in solid
Solution of H2 in Pd, dissolved gases in minerals.
Liquid Solutions
Solid
Liquid
Solid in liquid
Sugar solution, salt solution, I2 in CCl4
5
Liquid
Liquid
Liquid in liquid
Benzene in toluene, alcohol in water.
6
Gas
Liquid
Gas in liquid
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4
CO2 in water, NH3 in water etc.
Gaseous Solutions
Solid
Gas
Solid in gas
Iodine vapours in air, camphor vapours in N 2.
8
Liquid
Gas
Liquid in gas
Water vapours in air, CHCl3 vapours in N2.
9
Gas
Gas
Gas in gas
Air (O2 + N2)
uh
7
Ja
The concentration of a solution can be expressed by different concentration terms which are described
as follows.
Section (B) : Concentration terms (Revision of mole)
Concentration Terms :
3.1
% Concentration
Mass percentage : It is the amount of solute in grams dissolved per 100 g of solution. e.g., 10%
solution of sodium chloride means 10 g of solid sodium chloride present in 100 g of solution.
weight of solute(g)
% w/w =
× 100
weight of solution(g)
Ex.
10% w/w urea solution = 10 g of urea is present in 100 g of solution.
= 10 g of urea is present in 90 g of water.
F1
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D11
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3.
Mass by volume percentage (% w/v) : It is defined as mass of solute dissolved per 100 ml of solution.
It is commonly used in medicine and pharmacy.
F2
% wt/vol. (w/v)
% w/v = wt. of solute/100 mL of solution
gram of solutes
% w/v =
× 100
volume of solution in mL
Ex.
10% (w/v) urea solution. = 10 g of urea is present in 100 mL of solution.
But not 10 g of urea present in 90 ml of water
for dilute solution : volume solution = volume solvent.
D13
Volume percentage (% v/v) : It is defined as volume of a solute dissolved per 100 ml of solution.
volume of solute
% v/v =
× 100
volume of solution
Strength of solution in g/L
Weight of solute (in gram) per litre (1000 mL) of solution.
Sa
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D12
3.2
Ex.
10% (w/v) sucrose solution, then specify its concentration in g/L
100 mL .......... 10 g
10
1000 mL ........
× 1000 = 100 g/L
100
158
Example-1 :
Solution :

Der.1
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F3
Molarity (M) : It is expressed as the number of moles of solute per litre of solution.
Molarity = No. of moles of solute per litre of solution.
Let
n = No. of moles of solute ; N = No. of moles of solvent ; V = volume of solution
1000
n
 W
M=
=   x
 M
V (in L)
V (in mL)
no. of moles of solute = molarity x volume ( in L)
no. of m. moles of solute = molarity x volume ( in mL)
uh
D14
3.3
If we have 6% w/w urea solution with density 1.060 g/mL, then calculate its strength in g/L ?
6 g urea is present in 100 g solution.
100
6 g in
mL
1.060
100
mL ⎯⎯
→ 6 g.
1.060
6
1000 mL =
× 1.060 × 1000 = 10.6 × 6 = 63.6
100
If V1 mL of C1 molarity solution is mixed with V2 mL of C2 molarity solution (same substance or solute)

Cf (V1+V2) = C1V1 + C2V2
F4
Molality (m) : It is defined as number of moles of solute per 1000 g or 1 kg of solvent.
Molality = No. of moles of solute per kg(1000 g) of solvent.
Let w gram of solute (Molar mass = Mg/mole) is dissolved in 'W' gram of solvent.
1000
 w
molality =   ×
 M
W(g)
lp
D15
3.4
Ja
 C V + C2 V2 
Total moles
Cf =  1 1
where Cf = molarity of final solution
 =
Total
volume
+
V
V
1
2


If 20 ml of 0.5 M Na2SO4 is mixed with 50 ml of 0.2 M H2SO4 & 30 ml of 0.4 M Al2(SO4)3
solution. Calculate [Na+], [H+], [SO42–], [Al3+]. [Assuming 100% dissociation]
moles
Molarity =

10 m. moles of Na2SO4 
20 m. moles of Na+
volume
20
(i)
[Na+] =
= 0.2 M
100
(ii)
[H+] = ?
10 m. moles H2SO4
20 m. moles H+
20
[H+] =
= 0.2 M
100
56
10 + 10 + 36
(iii)
[SO42–] =
=
= 0.56 M
100
100
24
(iv)
[Al3+] =
= 0.24 M
100
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Example-2 :
a
moles  1000
W(g) o f solvent
Molality not depends on temperature.
molality =
Solution :
159
Example-3 :
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Solution :
(a) Derive a relationship between molality & molarity of a solution in which w g of solute of
molar mass M g/mol is dissolved in W g solvent & density of resulting solution = 'd' g/ml.
(b) Calculate molality of 1.2 M H2SO4 solution ? If its  = 1.4 g/mL
(a) Say 1 L solution taken,
mass of 1 lit solution = (1000 d) g
moles of solute = (molarity)
mass of solute = (molarity) x m
mass of solvent = W = 1000 d – (molarity) × m
(molarity)  1000
molality =
[Where no.of moles of solute = molarity]
1000 d − molarity  M.Wt
1.2  1000
(b) Molality =
= 0.936
1000  1.4 − 1.2  98
3.6
Ja
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3.5 D16 Normality : It is defined as number of gram equivalents of solute dissolved per litre of solution.
no. of equivalents of solute
F5
No.of equivalents per litre of solution =
volume of solution (in L)
= n-factor  molarity No. of equivalents = normality × volume (in L)
Molar mass
F6
Equivalent mass =
n − factor
Mass of the species
Mass of the species
F7
No. of equivalent =
=
Molar mass
equivalent mass
n − factor
'n' - factor
lp
(i) For oxidizing/reducing agents : no. of e– involved in oxidation/reduction half reaction per mole of
oxidising agent /reducing agent.
e.g. : 5e– + 8H+ + MnO4– → Mn2+ + H2O
n- factor = 5
(ii) For acid/ base reactions : no. of H+ ions displaced/ OH– ions displaced per mole of acid/ base.
e.g. : NaOH n - factor = 1
H2SO4 n - factor = 2
n - factor = charge on the cation = 2 x 3 = 6
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e.g. : Al2(SO4)3
a
(iii) For salt :
n = Total charge on cations.
or
total charge on anions
3.7 D17 Mole-fraction (x) : It is the ratio of number of moles of a particular component to the total number of
nA
moles of all the components. e.g., mole-fraction of component A, xA =
, where nA is the number
nA + nB
of moles of component 'A' and nB is the number of moles of component 'B'.
F8
For binary mixture.
n
moles of solute
Xsolute =
=
total moles in solutions n + N
moles of solvent
N
=
Total moles in solutions n + N
+ XSolvent = 1
F9
XSolvent =
F10
Xsolute
3.8 D18 Parts per million (ppm) : The number of parts of solute present in 1 million parts of solution are called
its ppm. When a solute is present in small quantities (very minute amounts), it is easier to express the
concentration in parts per million.
wt. of solute (in g)
F11
(a)
ppm (w/w) =
× 106
wt. of solution (in g)
160
F12
(b)
F13
(c)
wt. of solute (in g)
× 106
vol. of solution (in mL)
moles of solute
ppm (moles/moles) =
× 106
moles of solution
ppm (w/v) =
Table : 1
Molarity (M)
Units
mol solute
L solution
None
Mass %
%
Molality (m)
mol solute
kg solvent
Disadvantages
Useful in stoichiometry; measure
by volume
Temperature-dependent; must
know density to find solvent mass
Temperature-independent; useful
in special applications
Temperature-independent; useful
for small amounts
Measure by mass ; must know
density to convert to molarity
Measure by mass ; must know
density to convert to molarity
Temperature-independent useful
in special applications
Measure by mass ; must know
density to convert to molarity
uh
Mole fraction
(x)
Advantage
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Name
Ja
Note : All volume related concentration terms are temperature dependent.
If we have 10 molal urea solution, Calculate mole fraction of urea in this solution & also
calculate % w/w of urea (MW = 60).
Solution :
10 moles urea in 1000 g of water
10
10
Xurea =
=
= 0.1526
1000
65.55
10 +
18
10  60
% w/w weight of urea =
x 100 = 37.5%
10  60 + 1000
* Note : For dil. aq. solution molality molarity, as d  1 g/mL
molarity 1000
molality =
1000  d − molarity  m
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Example-4 :
Example-5 :
Calculate molarity of CaCO3(aq.)solution which has concentration of CaCO 3=200ppm.
Solution :
200 g of CaCO3 in 106 g of water.
200
= 2 moles of CaCO3 in 103 liters of water. (density =1g/mL)
100
2
So molarity =
= 2 × 10–3 M.
103
161
Section (C) : Vapour Pressure
Vapour Pressure of a pure liquid or pure solid :
4.1
The origin of saturated vapour pressure: The evaporation of a liquid
Some of the more energetic particles on the surface of the liquid move fast enough to escape from the
attractive forces holding the liquid together. They evaporate. The diagram shows a small region of a
liquid near its surface.
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4.
lp
Ja
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Note: evaporation only takes place on the surface of the liquid. That's quite different from boiling which
happens when there is enough energy to disrupt the attractive forces throughout the liquid. That's why,
if you look at boiling water, you see bubbles of gas being formed all the way through the liquid.
If you look at water which is just evaporating in the sun, you don't see any bubbles. Water molecules
are simply breaking away from the surface layer. Eventually, the water will all evaporate in this way.
The evaporation of a liquid in a closed container: Evaporation and Condensation
Now imagine what happens if the liquid is in a closed container. Common sense tells that water in a
sealed bottle doesn't disappear over time. But, there is still constant evaporation from the surface.
Particles continue to break away from the surface of the liquid - but this time they are trapped in the
space above the liquid.
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As the gaseous particles bounce around, some of them will hit the surface of the liquid again, and be
trapped there. This is called as condensation. There will rapidly be an equilibrium set up in which the
number of particles leaving the surface is exactly balanced by the number rejoining it. At equilibrium,
rate of evaporation becomes equal to the rate of condensation.
In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid.
When these particles hit the walls of the container, they exert a pressure. This pressure is called the
saturated vapour pressure (or just vapour pressure) of the liquid.
Suppose,we take 10 gram liquid in a chamber. Initially, there are no molecules in the gaseous phase.
Thus, no pressure is exerted on the mercury and the level of mercury in both columns is equal.
Suppose, after vapourization of 5 gram of the liquid vapour-liquid equillibrium is established. Now,
molecules in the gaseous phase exert pressure on the mercury and level of mercury in the column
attached to the chamber decreases.
Note: From the figure, we can see that, Vapour Pressure of the liquid = h mm Hg.
162
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Note: Volatile solids (like Iodine) also evaporate and have a vapour pressure at any given temperature,
just like liquids.
Important Points related to vapour pressure :
D19
Vapour pressure of a pure liquid : The pressure exerted by the vapours over the liquid surface at
equilibrium is called vapour pressure. It increases with the increase in temperature.
D20
Vapour pressure of solution : The pressure exerted by the vapours of solvent 'A' and solute 'B' in
equilibrium with the liquid phase is called the vapour pressure of solution.
D21
Partial vapour pressure : The pressure of vapours of a component ‘A’ over a solution of ‘A’ and ‘B’ is
called partial vapour pressure of component A. It is denoted by pA.
(1)
Definition of vapour pressure : The partial pressure of vapours in equilibrium with pure solid or pure
liquid at a given temperature.
At eq. the rate of evaporation = rate of condensation
e.g.
H2O(l)
H2O(g)
Kp = pHeqO,g
Ja
uh
4.2
2
(2)
Hence V.P is equilibrium constant (KP) of the reaction, liquid
vapours.
Since vapour pressure is an equilibrium constant. so it's value is dependent only on temperature.
Nature of liquid : The value of a liquid's vapour pressure depends on the magnitude of the
intermolecular forces in the liquid. The smaller the intermolecular forces, higher the vapour pressure
becuause loosely held molecules escape more easily into vapour phase.
(b)
Temperature of the given liquid : At higher temperature, more molecules from the liquid have enough
KE to escape from the surface of the liquid. That will increase the saturated vapour pressure.
a
lp
(a)
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Vapourization (liquid to vapour) is always endothermic. It needs heat to convert the liquid into the
vapour.
According to Le Chatelier, increasing the temperature of a system in a dynamic equilibrium favours the
endothermic change. That means that increasing the temperature increases the amount of vapour
present, and so increases the saturated vapour pressure.
F14
The dependence of vapour pressure with temperature is given by Clausius-Clapeyron equation
p 
Hv  1
1
ln  2  = −
−  , where H = molar enthalpy of vapourisation of the given liquid,

p
R
T
T
1
 1
 2
p2 = vapour pressure of the liquid at T2 and p1 = vapour pressure of the liquid at T1.
(3)
Vapour pressure of a liquid does not depend on :
(i) the amount of liquid taken
(ii) surface area of the liquid
(iii) volume or shape of the container
(4)
Partial pressure of vapours: If vapours of a liquid are present in a gaseous mixture then,
Pr essure of vapours
Partial pressure of vapours of the liquid =
Total pressure
Saturation: A gas or gaseous mixture is said to be saturated with the vapours of a liquid if the partial
pressure of the liquid vapours is equal to its (saturated) vapour pressure.
163
(5)
F15
(6)
(a) Saturated (Equillibrium) Vapour pressure of water, at a given temprature, is called aqueous tension.
The value of aqueous tension is different at different temprature.
Partial pressure of water vapour at given temperature
(b) Relative Humidity (R.H.) =
× 100%
Vapour pressure of water at the same temperature
Saturated vapour pressure and boiling point:
A liquid boils when its saturated vapour pressure becomes equal to the external pressure on the liquid.
When that happens, it enables bubbles of vapour to form throughout the liquid. If the external pressure
is higher than the saturated vapour pressure, these bubbles are prevented from forming, and we just
get evaporation at the surface of the liquid.
* Pext > Psat  evaporation
* Pext = Psat  Boiling
Saturated vapour pressure and solids: Sublimation
Solids can also loose particles from their surface to form a vapour, except that in this case we call the
effect sublimation rather than evaporation. Sublimation is the direct change from solid to vapour without
going through the liquid stage.
The forces of attraction in many solids are too high to allow much loss of particles from the surface.
However, there are some solids which easily form vapours.
Naphthalene (used in "moth balls") has quite a strong smell. Molecules must be breaking away from the
surface as a vapour, because otherwise we wouldn't be able to smell it.
Ja
(7)
uh
a
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If the liquid is in an open container and exposed to normal atmospheric pressure, the liquid boils when
its saturated vapour pressure becomes equal to 1 atmosphere. This happens with water when the
temperature reaches 100°C. But at different pressures, water will boil at different temperatures. For
example, at the top of Mount Everest the pressure is so low that water will boil at about 70°C.
(a) Boiling point : The boiling point is the temperature at which the vapour pressure of a liquid is equal
to the external pressure.
(b) At boiling temperature, vapour pressure of the pure liquid i.e. Pº = Pext
(c) At normal boiling point, vapour pressure of the pure liquid i.e. Pº = 1atm
Sa
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(10)
lp
(9)
a
(8)
Solid carbon dioxide never forms a liquid at atmospheric pressure and always converts directly from
solid to vapour. That's why it is known as dry ice.
If partial pressure of vapours of a liquid is increased beyond the saturated (equillibrium) vapour
pressure value (Pº), its vapours begin to condense till their partial pressure becomes equal to the
saturated vapur pressure.
If the partial pressure of vapour is less than v.p. of liquid, the liquid (if present) will vaporize till
(a) its v.p. is attained or (b) the liquid completely gets vaporized
If a volatile solid/ liquid is brought in contact with a gas (or vaccuum), vapours of that solid/ liquid
occupy the gas phase till the gas phase becomes saturated with that solid /liquid vapours.
Example-6 :
Ans.
Solution :
The vapor pressure of water at 80º C is 355 torr. A 100 ml vessel contained water−saturated
oxygen at 80º C, the total gas pressure being 760 torr. The contents of the vessel were pumped
into a 50.0 ml, vessel at the same temperature. What were the partial pressures of oxygen and
of water vapor, what was the total pressure in the final equilibrated state ? Neglect the volume
of any water which might condense.
PO2 = 810 mm Hg, PH2O = 355 mm Hg , Ptotal = 1165 mm Hg
In 100 ml vessel which contained water - saturated oxygen, the pressure of O2 gas = 760 – 355
= 405 torr when the contents of this vessel were pumped into 50 ml vessel, at the same
temperature, the pressure of oxygen gets doubled i.e. PO2 = 810 torr.
But pressure of water vapour will remain constant, as some vapour in this 50 ml vessel, gets
condensed.
So, PH2O = 355 torr & Total pressure = 810 + 355 = 1165 torr.
164
Section (D) : Solutions of Solid and Gases in Liquids
5. D22 Saturated solution : A solution in which no more solute can be dissolved at the same temperature is
called saturated solution.
Unsaturated solution : It is a solution in which more amount of solute can be dissolved at the same
temperature.
D24
Supersaturated solution : It is a solution which contains more mass of the dissolved solute than the
saturated solution at the same temperature and pressure. It should be prepared in a dust-free vessel
and at a higher temperature. It is metastable. Mechanical stress, such as shaking or addition of solute,
causes deposition of solute.
D25
Solubility : Solubility of a substance is its maximum amount that can be dissolved in a specified
amount of solvent (generally 100 g of solvent) at a specified temperature to form a saturated solution.
Solubility of one substance into another depends on
(i) nature of the solute and solvent.
(ii) Temprature
(iii) Pressure
5.1
Solubility of a solid in a liquid :
Like dissolves like: Polar solutes dissolve in polar solvents and non polar solutes in non-polar
solvents.
Sodium chloride and sugar dissolve readily in water, naphthalene and anthracene do not. On the other
hand, naphthalene and anthracene dissolve readily in benzene but sodium chloride and sugar do not.
Dissolution : When a solid solute is added to the solvent, some solute dissolves and its concentration
increases in solution. This process is known as dissolution.
Crystallisation : Some solute particles in solution collide with the solid solute particles and get
separated out of solution. This process is known as crystallisation.
At equilibrium, rate of dissolution is equal to rate of crystallisation. At this stage the concentration of
solute in solution will remain constant under the given conditions, i.e., temperature and pressure. Such
a solution is said to be saturated with the given solute.
Solute + Solvent
Solution
5.2
Effect of temperature on solubility of a solid in a liquid
Consider the equilibrium: Solute + Solvent
Solution. By Le Chatelier's Principle:
If above process is exothermic i.e. H < 0, then as T increases, solubility decreases.
If above process is endothermic i.e. H > 0, then as T increases, solubility increases.
5.3
Effect of pressure on solubility of a solid in a liquid (no effect)
Pressure does not have any significant effect on solubility of solids in liquids. It is so because solids and
liquids are highly incompressible and practically remain unaffected by changes in pressure.
6.
Solubility of gases in liquids :
6.1
Factors Affecting Solubility of Gas In Liquid :
(i) Nature of gas
(ii) Nature of liquid
Sa
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Ja
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a
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D23
6.2
(iv) Pressure
Henry's Law (effect of pressure on solubility of gases in liquids) :
Statement : The solubility of a gas in a liquid at a given temperature is directly proportional to its partial
pressure at which it is dissolved.
Der.2 Let
x = Mole fraction of unreacted gas in solution at a given temperature as a measure of its
solubility.
p = Partial pressure of gas in equilibrium with the solution.
By Henry's law:
That is;
6.3
(iii) Temperature
F16
xp
or
px
p = K Hx
or
x=
p
,
KH
where KH = Henry's law constant.
Characteristics of Henry's Law constant (KH).
(i) Unit same as those of pressure: torr or bar.
(ii) Different gases have different value of KH for the same solvent.
(iii) The KH value of a gas is different in different solvents and it increase with the increase in
temperature.
p
(iv) Higher the value of KH of a gas, lower will be its solubility. Since, x =
.
KH
165
Plot of p Vs x is a straight line passing through the origin with slope equal to KH.
Plot of p Vs x for solution of HCl in cyclohexane.
Solution :
Henry's law constant for O2 dissolved in water is 4.34 × 104 atm at 25°C. If the partial pressure
of oxygen in air is 0.4 atm. Calculate the concentration (M) of dissolved oxygen in water in
equilibrium with air at 25°C :
Given KH = 4.34 × 104 atm
pO2 = 0.4 atm
lp
Ja
acc. to Henry's Law
p = KHX
0.4
= 9.2 × 10–6
XO2 =
4.34  104
1000
Moles of water ( nH2O ) =
= 55.8 mol
18
nO2
XO2 =
nO2 + nH2O
uh
Example-7 :
a
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Note : If a mixture of gases is brought in contact with solvent each constituent gas dissolves in proportion to its
partial pressure. It means Henry's law applies to each gas independent of the pressure of other gas.
Since nO2 is very small in comparison to nH2O
nO2
nH2O
a
XO2 =
XO2 × nH2O = nO2
So
nO2 = 9.2 × 10–6 × 55.55 = 5.11 × 10–4 mol
So
M = 5.11 × 10–4
Sa
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or
Example-8 :
A gas 'X' is present with saturated water vapour over water liquid at
total pressure of 1.5 atm. Vapour pressure of H2O at same
temperature is 0.5 atm. What is the solubility of gas 'X' in terms of
moles in 10 moles H2O().
Solution :
PT = Px + PH2O
Px = 1.5 – 0.5 = 1 atm
slope of graph = KH = 5 × 103
nx
Px = KH
nH2O
1 = 5 × 103 ×
nx =
nx
10
1
= 2 × 10–3
500
166
6.4
Effect of temperature : Solubility of gases in liquids decreases with rise in temperature.
Explanation: When dissolved, the gas molecules are present in liquid phase and the process of
dissolution can be considered similar to condensation and heat is evolved in this process. We have
learnt that dissolution process involves dynamic equilibrium and thus must follow Le Chatelier's
principle. As dissolution of gases in liquids is an exothermic process, the solubility should decrease with
increase of temperature.
Note : KH values for both N2 and O2 increase with increase of temperature indicating that the solubility of
gases increases with decrease of temperature. It is due to this reason that aquatic spcies are more
comfortable in cold water rather than warm water.
Applications of Henry's law :
It has several applications in biological and industrial phenomena.
(i) To increase the solubility of CO 2 in soft drinks and soda water the bottle is sealed under high
pressure.
(ii) Scuba divers must cope with high concentrations of dissolved gases while breathing air at high
pressure underwater. Increased pressure increases the solubility of atmosphere gases in blood. When
the divers come towards surface, the pressure is gradually decreased. This releases the dissolved
gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates
a medical condition known as bends, which are painful and dangerous to life. To avoid bends, as well
as, the toxic effects of high concentrations of nitrogen in the blood, the tanks used by scuba divers are
filled with air diluted with helium (11.7% helium, 56.2% nitrogen and 32.1% oxygen).
(iii) At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to
low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. Low
blood oxygen causes climbers to become weak and unable to think clearly, symptoms of a condition
known as anoxia.
6.6
Limitation of Henry's law :
Henry’s law is valid only under following condition.
(i) The pressure of gas is not too high. (ii) The temperature is not too low.
(iii) The gas should not undergo any chemical reaction with the solvent.
(iv) The gas should not undergo dissociation in solution.
lp
Ja
uh
a
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6.5
Section (E) : Immiscible Liquids
7.
Vapour Pressure of liquid-liquid solution: volatile solute + volatile solvent
Completely Immiscible Liquids and Steam Distillation
For mixture of two completely immiscible liquids, each liquid exerts its own vapor pressure, independent
of the other, and the total vapor pressure is the sum of the separate vapour pressures of the two
components in the pure state at the given temperature.
Thus, pA = PA0 &
pB = PB0
0
n
P
PT = PA0 + PB0 ; A0 = A
nB
PB
Sa
nk
a
7.1
F17
where nA and nB are the numbers of moles of each component present in the vapor phase, which on
condensing form distillate.
The ratio of A to B in the vapor in terms of the actual weights w A and wB is
F18
(i) A system of two immiscible liquids will boil when the total vapor pressure P is equal to the
atmospheric pressure or the given external pressure. The boiling point of the mixture is thus lower than
that of either constituent.
(ii) Further, since the total vapor pressure is independent of the relative amounts of the two liquids, the
boiling point, and hence the composition of the vapor and distillate, will remain constant as long as the
two layers are present. e.g. Water and Benzene has boiling point = 68.9ºC.
But boiling point = H2O (pure) C6H6 (pure)
100ºC
80.2ºC
167
Steam Distillation: The boiling point of a solution of two immiscible liquids is less than the individual boiling
points of both the liquids. This concept is used in steam distillation.
A liquid (generally organic) that is immiscible with water, and that has a higher boiling point than water
can be boiled (distilled) at a much lower temperature by passing steam through it. In this way, the
organic liquid can be purified from impurities using steam distillation.
Section (F) : Completely miscible liquids : Raoult’s law
Statement of Raoult's law (for volatile liq. mixture) : In solution of volatile liquids, the partial vapour
pressure of each component is directly proportional to its mole fraction.
pA  xA
F19
pA = xAPAº
pA = Partial vapour pressure of component A
xA = Mole fraction of component ‘A’ in solution.
PAº = Vapour pressure of pure component ‘A’ at given temperature
Der.3
Derivation of total pressure over solution using Raoult’s law and Dalton’s law:
Let A, B be to two volatite liquids in a closed container as shown.
A
x
x
x
x
x
x
uh
x
x
a
ri
7.2
B
Ja
A+B
XA
XB
lp
F20
pA = xAPAº
Similarly, for liquid B we have,
pB = xBPBº
Total pressure over the solution PT , according to Dalton's law is
PT = pA + pB = xAPA0 + xBPB0
Sa
nk
a
Determining composition of vapour phase:
Let, yA = mole fraction of A in vapour phase above the solution and
yB = mole fraction of B in vapour phase above the solution
Now, we have, pA = yA PT .....Dalton's law of partial pressure for a gaseous mixture
...........Raoult's law
pA = xAPAº
Thus, pA = yA PT = xA PAº
Also, pB = yBPT = xBPBº
y P
y P
xA + xB = 1 = A 0 T + b 0 T
PA
PB
F21
Thus,
1
y
y
= A + B
0
PT
PA PB0
Graphical Representation of Raoult's Law:
pA= xAPAº
&
pB = xB PBº
PT = xAPAº + xB PBº
PT = (PAº – PBº) xA + PB0
PT = (PBº – PAº) xB + PA0
This represents equation of straight line. PT v.s x
168
Application of Raoult’s Law:
Phase Diagrams of Two-Component Ideal Solutions: Bubble and Dew Points
The compositions of the liquid and vapour that are in mutual equilibrium are not necessarily the same.
Common sense suggests that the vapour should be richer in the more volatile component. This can be
easily derived from Raouls’s Law.
1
y
y
1− yA
y
= A0 + B0 = A0 +
PT
PA
PA PB
PB0
PT =
uh
7.3
(1)
If PAº > PBº, A is more volatile than B. B.P. of A < B.P. of B.
a
ri
Note:
PA0 PB0
Ja
PA0 + (PB0 − PA0 )y A
From the above equation, the plot of PT with respect to y is a curve instead of a straight line. We can
superimpose this curve on the diagram for graphical representation of Raoult’s Law to get following
phase diagrams.
Sa
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a
lp
(A) Pressure versus Composition Phase Diagram (At constant T): P vs. x and y
In this kind of phase diagram the temperature has a fixed value. The mole fraction of one component is
plotted on the horizontal axis and the pressure is plotted on the vertical axis. If A is more volatile than B,
then we get following diagram.
Explanation :
(i) On horizontal-axis we have plotted both x and y that is both liquid composition and vapour
composition.
(ii) The lower curve (Dew Point Curve) represents the total pressure as a function of the composition
i.e. mole fraction in the vapor phase at equilibrium with the liquid phase. It is plotted using P T vs. yA
equation:
PT =
PA0 PB0
PA0 + (PB0 − PA0 )y A
(iii) The upper curve i.e. Bubble Point Curve (a straight line in the case of an ideal solution) represents
the total pressure as a function of composition i.e. mole fraction in the liquid. It is plotted using P T vs. xA
equation:
PT = (PAº – PBº) xA + PB0
169
(iv) The area between these two curves is vapour-liquid equilibrium region. Vapours cannot exist above
the bubble point curve and liquid can not exist below the dew point curve.
Ans.
Ja
Solution :
Benzene and toluene form nearly ideal solutions. At 300 K, Pº toluene = 30 mm Hg and Pºbenzene =
100 mm Hg.
A liquid mixture is composed of 3 mol of toluene and 2 mol of benzene.
(a) If the pressure over the mixture at 300 K is reduced. At what pressure does the first vapour
form?
(b) What is the composition of the first trace of vapour formed ?
(c) If the pressure is reduced further, at what pressure does the last trace of liquid disappear?
(d) What is the composition of the last trace of liquid?
(a) 58 mm Hg (b) yb = 20/29 (c) 250/6 mm Hg (d) xb = 1/6
3
2
(a) P = XAP0A + XBP0B =
× 30 +
× 100 = 58 mm Hg
5
5
(b) Composition of the first trace of vapour formed
100 20
2
20
9
X P0b
Ybenzene = b
=
×
=
&
Ytoluene = 1 –
=
58
5
29
29
29
PTotal
uh
Example-9 :
a
ri
(v) Suppose, initially the pressure over the solution is very high so that no vapour exist above the liquid.
As we gradually decrease the pressure, a point (Bubble point) comes when we cross the Bubble-Point
curve and first bubble of vapour starts forming. Hence, the name bubble point curve. Now we have
entered the vapour-liquid equilibrium region. On further decreasing the pressure, a point (Dew Point)
comes when we cross the Dew-Point curve when almost all the liquid has evaporated into vapour i.e.
only the last drop of liquid (Dew) remains. Beyond this point no liquid exists in the system.
(c) The last trace of liquid disappear at
YA = 3 / 5
YB = 2 / 5
Sa
nk
a
lp
Y
Y
1
= A + 0B
p P0 A
PB
250
2
3
1
=
+
or
p=
mm Hg
5  100
6
5  30
p
(d) The composition of the last trace of liquid
2 250

YbenzenePTotal
6 =1
X’benzene =
= 5
Pb º
6
100
so
X’toluene =
5
.
6
170
(B) Temperature versus Composition Phase Diagrams i.e. T vs x and y diagram
uh
a
ri
(i) In this type of phase diagram the pressure is held fixed. The mole fraction of one component is
plotted on the horizontal axis and the boiling temperature is plotted on the vertical axis.
Ja
Sa
nk
a
lp
(2)
(ii) The upper curve (Dew-Point curve) gives the boiling temperature at the given pressure as a function
of the mole fraction in the vapor phase, and the lower curve (Bubble-Point curve) gives the boiling
tempera ture at the given pressure as a function of the mole fraction in the liquid phase.
Note: If a liquid has a high vapour pressure at a particular temperature, it means that its molecules can
escape easily from the surface. The liquid with the higher vapour pressure at a particular temperature is
the one with the lower boiling point. If A is more volatile than B, then we get following diagram.
Distillation:
(i) The compositions of the liquid and vapour that are in mutual equilibrium are not necessarily the
same. Common sense suggests that the vapour should be richer in the more volatile component.
(ii) In a simple distillation, the vapour over a boiling mixture is withdrawn and condensed in a separate
container. The liquid collected in the new container is called condensate or distillate and the liquid
remaining in the original container is called residue. The condensate has higher mole fraction of the
more volatile component than the original mixture. The residue has higher mole fraction of the less
volatile component than the original mixture. This is the principle of distillation and is used to separate a
more volatile liquid from a less volatile liquid.
(iii) In fractional distillation, the boiling and condensation cycle is repeated successively to get a
distillate which gets richer in the more volatile component after each cycle.
Example-10 :
An equimolar mixture of benzene & toluene is prepared. The total vapour pressure of this
mixture as a function of mole fraction of benzene is found to be PT = 200 + 400 Xben.
(a) Calculate composition of vapours of this mixture [Assume that the number of moles going
into vapour phase is negligible in comparsion to number of moles present in liquid phase].
(b) f the vapour above liquid in part (a) are collected and condensed into a new liqiuid,
calculate composition of vapours of this new liquid.
Solution :
(a)
P0Beazene= 600 mm of Hg ;
P0Toloune = 200 mm of Hg
1
1
PT =
× 600 + × 200 = 400 mm of Hg
2
2
Pbenz = xben ' Pºben
= ybenPT.
1/ 2  600
3
1
ybenzene =
=
= 75% ;
yToloune = = 25%
4
4
400
171
(b)
3
1
× 600 +
× 200 = 500
4
4
yToluene = 0.1 = 10%
PT =

yben =
3 / 4  600
= 0.9 = 90%
500
Limitations of Raoult’s Law: Raoult's Law only works for ideal solutions. Very dilute solutions obey
Raoult's Law to a reasonable approximation.
(1)
Ideal Solutions : Those solutions which obey Raoult's law over the entire range of conc. are called
ideal solutions. When the forces of attraction between A—A, B—B is similar to A—B, then A and B will
form ideal solution.
lp
Ja
uh
a
ri
7.4
a
Properties of ideal solution :
(i) Raoult's law is obeyed
(ii) Hmix = 0, i.e., there should not be enthalpy change when components of ideal solutions are mixed.
(iii) Vmix = 0, (1L + 1L = 2L) i.e., there should not be change in volume on mixing. e.g.; n-hexane and
n-heptane; ethyl bromide and ethyl iodide; benzene and toluene; chlorobenzene and bromobenzene
form ideal solutions.
Sa
nk
Section (G) : Non-ideal Solutions
(2)
(A)
Non_Ideal Solutions :
Those solutions which do not obey Raoult's over the entire range of concentration are called non-ideal
solutions.
When the forces of attraction between A—A, B—B is different from A—B then 'A' and 'B' form non-ideal
solutions. For these solutions :
(i) Raoult's law is not obeyed. (ii) Hmix  0 ; (iii) Vmix  0.
Types of Non-Ideal Solutions:
Non-Ideal Solutions Showing Positive Deviation From Raoult’s Law :
(i) In this case, partial pressure of each component A and B is higher than that calculated from Raoult's
law, and hence total pressure over the solution is also higher than if the solution were ideal, as shown
in figure.
(ii) Boiling point of such a solution is relatively lower than the boiling points of both A and B.
172
P0A > P0B
a
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Note: Dashed lines represent vapour pressures and total pressure corresponding to ideal solution
uh
Ja
(B)
Examples: water and ethanol, chloroform and water, ethanol and CCI 4, methanol and chloroform,
benzene and methanol, acetic acid and toluene, acetone and ethanol, methanol and H 2O, C2H5OH +
cyclohexane.
Explanation: The fact that the vapour pressure is higher than ideal in these mixtures means that
molecules are breaking away more easily than they do in the pure liquids. That is because the
intermolecular forces between molecules of A and B are less than they are in the pure liquids.
You can see this when you mix the liquids. Less heat is evolved when the new attractions are set up
than was absorbed to break the original ones. Heat will therefore be absorbed when the liquids mix.
The enthalpy change of mixing is endothermic.
Non-Ideal Solutions Showing Negative Deviation From Raoult’s Law :
(i) In this case, partial pressure of each component A and B is lower than that calculated from Raoult's
law, and hence total pressure over the solution is also lower than if the solution were ideal, as shown in
figure.
(ii) Boiling point of such a solution is relatively higher than the boiling points of both A and B.
a
lp
P0A > P0B
Sa
nk
Note: Dashed lines represent vapour pressures and total pressure corresponding to ideal solution.
Examples: chloroform and acetone, chloroform and methyl acetate, H2O and HCI, H2O and HNO3,
acetic acid and pyridine, Phenol & Aniline.
Explanation: These are cases where the molecules break away from the mixture less easily than they
do from the pure liquids. New stronger forces must exist in the mixture than in the original liquids.
You can recognise this happening because heat is evolved when you mix the liquids-more heat is given
out when the new stronger bonds are made than was used in breaking the original weaker ones.
173
a
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uh
D26
Azeotropic Mixtures : Very large deviations from ideality lead to a special class of mixtures known as
azeotropes, azeotropic mixtures, or constant-boiling mixtures.
Azeotropes : Liquid mixtures which distill over without changes in composition are called constant
boiling mixtures or Azeotropes or Azeotropic mixtures.
A boiling liquid mixture at the azeotropic composition produces a vapour of exactly the same
composition, and the liquid does not change its composition as it evaporates. Two types of azeotropes
are known.
Ja
7.5
lp
(1) D27 Minimum Boiling Azeotropes : Non-ideal solutions showing large positive deviation from Raoult's law
form minimum boiling azeotropes which boil at temperature lower than boiling point of its components
'A' and 'B', e.g., water and benzene, chloroform and methanol.
a
The figures below show the Temprature vs. composition (at constant pressure) phase diagram on the
left side and Pressure vs. composition (at constant temprature) phase diagram for a minimum-boiling
azeotropic mixture chloroform and methanol.
Sa
nk
Note: The detailed values in the diagram are not important. Only the qualitative shape of the diagram is
important.
174
a
ri
uh
Ja
At the lowest point in the temperature vs. composition phase diagram, the concentration in the vapour
phase is the same as the concentration in the liquid phase (y = x). This concentration is known as the
azeotropic composition. At this point, the mixture boils at a constant temperature and without change in
composition.
lp
(2) D28 Maximum Boiling Azeotropes : Non-ideal solutions showing large negative deviation from Raoult's
law form maximum boiling azeotropes which boil at temperature higher than the boiling point of its
components A and B respectively, e.g., a mixture of HCl and H2O containing 20.2% HCl by weight boils
at 108.5ºC higher than either pure HCI (– 85°C) or water (100°C).
The figures below show the Temprature vs. composition (at constant pressure) phase diagram on the
left side and Pressure vs. composition (at constant temprature) phase diagram for a maximum-boiling
azeotropic mixture water and formic acid.
Sa
nk
a
Note: The detailed values in the diagram are not important. Only the qualitative shape of the diagram is
important.
175
(3)
Separation of Azeotropic Mixtures
When the azeotropic composition has been reached, the condensate has the same composition as the
azeotropic liquid. The vapour phase is not richer in any component than the liquid phase. Thus,
azeotropic mixtures can’t be separated beyond the azeotropic composition using distillation.
Section (H) : Degree of Ionisation/Dissociation for Weak Electrolytes
8. Colligative properties & constitutional properties :
⚫ Constitutional Properties :
Properties which are dependent on nature of particles are constitutional properties like electrical
conductance.
D30
⚫ Colligative properties :
The properties of the solution which are dependent only on the total no. of particles relative to
solvent/solution or total concentration of particles in the solution and are not dependent on the nature of
particle i.e. shape, size, neutral /charge etc. of the particles.
3. Depression in freezing pt. (Tf)
8.1
2. Elevation in boiling point (Tb)
uh
There are 4 colligative properties of solution.
 P 
1. Relative lowering in vapour pressere 

 P 
a
ri
D29
4. Osmotic pressure ()
Abnormal Colligative Properties :
Ja
Vant–Hoff correction :
⚫ For electrolytic solutes the number of particles would be different from the number of particles
actually added, due to dissociation or association of solute.
⚫ The actual extent of dissociation/association can be expressed with a correction factor known as
vant Haff factor (i).
moles of particles in solution after dissociation / association
moles of solute dissolved
Vant–Hoff factor : i =
D31
⚫ If solute gets associated or dissociated in solution then experimental / observed / actual value of
colligative property will be different from theoretically predicted value so it is also known as abnormal
colligative property.
lp
F22
⚫ This abnormality can be calculated in terms of Vant-Hoff factor.
i = exp/ observ ed/ actual / abnormal v alue of colligative property
=
a
Theoretical v alue of colligative property
exp. / observed no. of particles or concentration
Theoretical no. of particles or concentration
=
(Theoretical molar mass of substance)
(E xperimental molar mass of the subs tance)
Sa
nk
i > 1 dissociation
i < 1 association
Der.4
Case - I : Electrolyte dissociates
Relation between i &  (degree of dissociation) :
Let the electrolyte be AxBy
t=0
teq
F23
AxBy (aq.) ⎯⎯
→ xAy+ + yBx–
C
0
0
C(1 – )
xC yC
Net concentration = C – C + xC  + yC = C [1 + ( x+y –1) ] = C [1 + ( n –1) ].
n=x+y
= no. of particles in which 1 molecule of electrolyte dissociates
C [1 + (n − 1)]
i=
C
i = 1 + ( n – 1) 
e.g.
NaCl (100% ionised), i = 2. ; BaCl2 (100% ionised), i = 3. ; K4[Fe(CN)6] (75% ionised), i = 4.
176
Der.4
Case - II : Electrolyte associates
Relation between degree of association  & i.
t=0
teq
nA ⎯⎯
→
C
C
C ( 1– )
n
An.
0
C
n
1 
= C [ 1 +  − 1  ]
n 
Net concentration = C – C +
a
ri
1 
i = 1 +  − 1 
n 
if dimerise n = 2 ; trimerise n = 3 ; tetramerise n = 4.
1
1
e.g. CH3COOH 100% dimerise in benzene, i = ; C6H5COOH 100% dimerise in benzene, i =
2
2
Section (I) : Relative lowering of vapour pressure
8.2
Relative lowering in vapour pressure (RLVP) :
uh
F24
a
lp
P0
Ja
Vapour Pressure of a solution of non-volatile solute in a volatile solvent
Vapour Pressure of a solution of a non volatile solute (solid solute) is always found to be less than the
vapour pressure of pure solvent.
Reason : Some of the solute molecules will occupy some surface area of the solutions so tendency of
the solvent particles to go into the vapour phase is slightly decreased hence Pº > P S, where Pº is
vapour pressure of pure solvent and PS is vapour pressure of the solution.
Pure solvent
= P
Sa
nk
Lowering in VP = Pº – PS
and Relative lowering in Vapour Pressure =
P
P0
Raoult's law (For non–volatile solutes): The vapour pressure of a solution of a non-volatile solute is
equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.
OR Relative Lowering in Vapour Pressure = mole fraction of the non volatile solute in solution.
Der.6
F25
PS = xsolventP0 = (1 − xsolute )P0
P0 -Ps
n
n+N
P
where n = number of moles of non-volatile solute and N = number of moles of solvent in the solution.
RLVP =
P0
P0 -Ps
=
0
= xSolute =
N
n+N
=1+
n
n
0
N P -Ps
=
n
P0
–1=
Ps
0
P -Ps
177
P0 -Ps
n
=
N
Ps
P0 -Ps
w
w
M
M
=
×
=
×
Ps
W
W
m
m
1000
1000
w
=
×
W
1000
m
×
M
1000
M
P0 -Ps
= (molality) ×
1000
Ps
where w and W = mass of non-volatile solute and volatile solvent respectively
m and M = molar mass of non-volatile solute and volatile solvent respectively
P0 -Ps
M
P0 -Ps
i.n
If solute gets associated or dissociated ; P
=
that is
= i × (molality) ×
s
N
1000
P
s
Example-11 :
Calculate weight of urea which must be dissolved in 400 g of water so final solutions has
vapour pressure 2% less than vapour pressure of pure water :
Let vapour pressure be Pº of water
P0 – Ps = .02 V
Ps = 0.98 V
18
0.02
w

=
×
; where w = weight of urea.
400
0.98
60
2  60  400
w=
g = 27.2 g.
18  98
Ja
uh
Solution :
a
ri
F26
×
10 g of a solute is dissolved in 80 g of acetone. Vapour pressure of this sol = 271 mm of Hg. If
vapour pressure of pure acetone is 283 mm of H
58
P0 -Ps
283-271 10
w
M
=
×

=
×
W
m
271
m
80
Ps
m = 163 g/mol.
Example-13 :
Vapour pressure of solute containing 6 g of non volatile solute in 180 g of water is 20 Torr. If 1
mole of water is further added into it, vapour pressure increases by 0.02 torr. Calculate vapour
pressure of pure water and molecular weight of non-volatile solute.
Pº-20
18
P0 -Ps
w
M
6
=
×

=
×
W
20
m
m 180
Ps
a
Solution :
lp
Example-12 :

Pº = 20.22 Torr.
Sa
nk
Pº-20.02 6
18
=
×
198
20.02
m
m = 54 g/mol.
Example-14 :
What is the final volume of both container.
Solution :
i1C1 = i2C2
0.1 2
0.1 1
=
.
100 + x 100 – x
200 – 2x = 100 + x.
x = 33.3 ml.
So, final volume of container containing NaCl = 133.3 ml.
178
Example-15 :
Solution :
If 0.1 M solutions of K4 [ Fe ( CN ) 6 ] is prepared at 300 K then its density = 1.2 g/mL. If solute
is 50% dissociated calculate P of solution if P of pure water = 25 mm of Hg. (K = 39, Fe = 56)
1
i = 1 + (5 – 1) ×
= 3.
2
0.1 1000
m=
1000  1.2 – 1.2  368
im  M
3  0.1 1000
P0 -Ps

 18
=
= 

1000
Ps
 1000  1.2 – 1.2  368  1000
a
ri
P0
= 1 + 7.12 × 10–3
Ps
Ps = 24.82 mm of Hg
p = 25 – 24.82 = 0.18 mm of Hg
Section (J) : Elevation of Boiling Point & Depression of Freezing Point
Elevation in Boiling point of a solution of non-volatile solute in volatile solvent (Tb)
D32
Boiling point: The temperature at which vapour pressure of a liquid becomes equal to the external
pressure present at the surface of the liquid is called b.p of liquid at that pressure.
Normal Boiling Point: The boiling temperature when Pext = 1 atm = 760 mm of Hg is called normal
boiling point of the liquid (T b).
lp
Ja
D33
uh
8.4
a
The vapour pressure curve for solution lies below the curve for pure water. The diagram shows that TB
denotes the elevation of boiling point of a solution in comparison to solvent.
Sa
nk
Elevation of Boiling point of any solution :
Since vapour pressure of solution is smaller than vapour pressure of pure solvent at any temperature
hence to make it equal to Pext. we have to increase the temperature of solution by greater amount in
comparison to pure solvent.
Der.7
Tb = Tb – Tbº
Tb  m
m = Molality
F27
Tb = Kbm
Note:
(i) If solute gets associated/dissociated then Tb = i × Kb × molality
ΔTb
K
(ii) Units of Kb:
=

Thus units of Kb = K kg mol–1
molality
mol/kg
(iii) Kb is dependent on property of solvent and known as ebullioscopic constant of solvent
D34
It is equal to elevation in boiling point of 1molal solution. It is also called molal elevation constant. The
units of Kb, is K/m or °C/m or K kg mol–1.
179
2
F28
F29
Kb =
2
RTb M
RTb
=
1000  Hvap 1000  L vap
where, Hvap is molar enthalpy of vaporisation (cal/mol or J/mol)
LVap is Latent Heat of Vapourisation in cal/g or J/g
M is molar mass of the solvent in gram
R = 2 cal mol–1 K–1 or 8.314 J/mol-K
Tb = Boiling point of pure liquid solvent (in kelvin)
 Hvap 
Lvap = 

 M 
(iv) Elevation in boiling point is proportional to the lowering of vapoure pressure i.e. ΔTb  P
a
ri
For water Lvap = 540 Cal/g, Tb = 100ºC
2  373  373

Kb =
= K kg mol–1 = 0.52 K k/g
1000  540
Ja
Hvapourisation
Tb,solvent
lp
Svapourisation =
uh
Extra Information (Not for Boards)
As only solvent particles are going into vapours,
We have Hvapourisation of solvent = Hvapourisation of solution
S means "entropy".
Svapourisation of pure solvent =
Tb,solvent
H
Tb, solution
a
Svapourisation of solution =
H
Sa
nk
Since, Svapourisation of pure solvent > Svapourisation of solution ;
So, Tb, solvent  Tb, solution.
Due to presence of solute, it is difficult to vapourise the solution, i.e it is difficult to boil the solution. So,
there is elevation in boiling point with respect to pure solvent.
Pressure (atm)
1 atm
Liquid
Psolv
Solid
Psoln
Gas
fpsolv
fpsoln
Tf
bpsolv
bpsoln
Tb
Temperature (ºC)
Phase diagram for a pure solvent and a solution of a nonvolatile solute. Because the vapour pressure
of the solution is lower than that of the pure solvent at a given temperature, the temperature at which
the vapour pressure reaches atmospheric pressure is higher for the solution than for the solvent. Thus,
the boiling point of the solution is higher by an amount Tb.
180
Example-16 :
Solution :
A solution of 122 g of benzoic acid in 1000 g of benzene shows a boiling point elevation of 1.4º.
Assuming that solute is dimerized to the extent of 80 percent, calculate normal boiling point of
benzene. Given molar enthalpy of vapourization of benzene = 7.8 Kcal/mole.
H
  1 
122  1000
Tb – Ti =
× Kb  1 +  − 1  0.8
122  1000
  2 
H
H
Tb – Ti = Kb × 0.4
2
Kb =
RTb
1000  L vap.
H
2
2  Tb  78
1000  7.8  1000
0.4  2 2
1.4 =
Tb
105
Tb = 418.33 K
H
H
Wwater =
1000
g.
3
lp
Ja
Solution :
1 Lit. of aq. solution of urea having density = 1.060 g/mL is found to have Tb = 0.5ºC.
If temperature of this solution increase to 101.5ºC, then calculate amount of water which must
have gone in vapour state upto this pt. given Kb = 0.5 K kg mol–1 for water
mass of solution = 1.060 × 103 = 1060 g
0.5 = 0.5 m

m = 1 ; if moles of urea = x
x
1=
;x=1
1060 − 60x
mass of water = 1060 – 60 = 1000 g
1.5 = Tb = (molality)f × Kb
1000
(Molality)f = 3 
3=1×
Wwater
uh
Example-17 :
a
ri
Kb =
1000
2000
=
g = 666.67 g
3
3
a
mass of water vaporised = 1000 –
Depression in freezing point of a solution of non-volatile solute in volatile solvent (Tb)
D35
Freezing point : Temperature at which vapour pressure of solid becomes equal to vapour pressure of
liquid is called freezing point of liquid or melting point of solid.
Sa
nk
8.5
Hsub > Hvap.
Hsub = Hfusion + Hvap.
H2O(s)
H2O (g)
Kp = /eq. PH2O (g)
= vapour pressure of solid
Reason for Depression In Freezing Point :
At the freezing point, the vapour pressure of solid and liquid is equal. When non-volatile solute is
dissolved in the solvent, the vapour pressure of solvent in the solution decreases. It means vapour
pressure of solid and liquid solvent will become equal at lower temperature, i.e., freezing point of
solvent in solution is lower than that of pure solvent.
181
Depression In Freezing Point :
The difference between freezing point of pure solvent T fº and freezing point of solution T is called
depression in freezing point (Tf) as shown in figure.
a
ri
D36
Der.8
Tf = Tfº – Tf
Tf  m
m = Molality
F30
Tf = Kfm
uh
Diagram showing Tf depression of the freezing point of a solvent in a solution.
2
F31
Cryoscopic constant Kf = molal depression constant =
2
RTf M
RTf
=
1000  Hfusion
1000  L fusion
F32
Ja
where, Hfusion is molar enthalpy of fusion (cal/mol ; J/mol)
Lfusion is Latent Heat of fusion in cal/g or J/g
M is molar mass of the solvent in gram
Tf = freezing point of solvent
 Hfusion 
Lfusion = 

 M 
Kf, Freezing point Depression Constant (Molal Depression Constant) :
It is equal to depression in freezing point of 1 molal solution. It is also called cryoscopic constant. The
units of Kf is K/m or °C/m or K kg mol–1.
2  273  273
For water Tf = 273 K & LFusion = 80 cal/g. Thus, Kf =
= 1.86 K kg mol–1
1000  80
Note:
(i) Depression in freezing point is proportional to the lowering of vapoure pressure i.e. Tf  P
(ii) If solute gets associated/dissociated then Tf = i × Kf × molality
ΔTf
K
=
(iii) Units of Kf :
.
Thus units of Kf = K kg mol–1
molality mol/kg
(iv) At freezing point or below it, only solvent molecules will freeze not solute molecules (solid will be of
pure solvent)
(v) Kf = depression in freezing point of 1 molal solution.
Hfusion
Hfusion
(vi)
SFusion
=
or
Tf.p. =
S fusion
Tf.p.
Sa
nk
a
lp
D37
HFusion for solvent and solution is same.
But for entropy, we can see following diagram.
 SSolution > Ssolvent
So, freezing point of solution < freezing point of solvent
182
Example-18 :
Solution :
Van’t Hoff factors of aqueous solutions of X, Y, Z are 1.8, 0.8 and 2.5. Hence, their (assume
equal concentrations in all three cases)
(A) b.p. : X < Y < Z
(B) f. p. Z < X < Y
(C) osmotic pressure : X = Y = Z
(D) v. p. : Y < X < Z
As van’t Hoff factor increases RLVP increases i.e.,
V.P. decreases y > x > z
Elevation in b.p. increases i.e., b.p. increases y < x < z
Depresion in f.p increases i.e., f.p decreases
y>x>z
Osmotic pressure increases
so
y < x < z. Ans. (B)
1000 g H2O have 0.1 mole urea and its freezing point is – 0.2ºC
and now it is freezed upto – 2ºC then how much amount of ice will
form.
Solution :
It is assumed that solute do not freeze and do not vapourise
0.1
TF = 0.2 = Kf
× 1000
.......(i)
1000
wt. of solvent 0.2
0.1
TF = 2 = Kf
× 1000 .......(ii) on dividing =
.
wt. of solvent
2
1000
Weight of remaining H2O is 100 g and weight of ice is 900 g.
Example-20 :
If boiling point of an aqueous solution is 100.1ºC. What is its freezing point? Given latent heat
of fusion and vaporization of water are 80 cal g–1 and 540 cal g–1 respectively.
For a given aqueous solution
Tb = Kb × molality
Tf = Kf × molality
Tb
1000 lf
K'
RTb2
= b =
×
.
'
Tf
1000 lv
RTf2
Kf
Sa
nk

a
Solution :
lp
Ja
uh
a
ri
Example-19 :



Tb
T2  l
= b2 f
Tf
Tf  lv
Tb = 100 + 273 = 373 K.
Tf = 0 + 273 = 273 K.
lf = 80 cal g–1.
lv = 540 cal g–1.
373  373  80
0.1
=
.
Tf
273  273  540
Tf = 0.362.
Tf = 0.0 – 0.362 = – 0.362ºC.
183
Example-21 :
Solution :
A 0.001 molal solution of a complex reprsented as Pt(NH 3)4Cl4 in water had a freezing point
depression of 0.0054ºC. Given Kƒ for H2O = 1.86 molality–1. Assuming 100% ionisation of the
complex, write the ionisation nature and formula of complex.
Let n atoms of Cl be the acting as ligand. Then formula of complex and its ionisation is :
[Pt(NH3)4Cln]Cl(4 – n) ⎯⎯
→ [Pt(NH3)4Cln]+(4 – n) + (4 – n) Cl–
1
0
0
0
1
(4 – n)
a
ri
Thus particles after dissocation = 4 – n + 1 = 5 – n
and
therefore, van't Hoff factor (i) = 5 – n
Now
Tƒ = K'ƒ × molality × van't Hoff factor
0.0054 = 1.86 × 0.001 × (5 – n)

n = 2.1  2 (integer value)
Thus complex and its ionisation is :
[Pt(NH3)4Cl2]Cl2 ⎯⎯
→ [Pt(NH3)4Cl2]2+ + 2Cl–
Example-22 :
Ja
uh
Solution :
Depression of freezing point of 0.01 molal aq. CH 3COOH solution is 0.02046°. 1 molal urea
solution freezes at – 1.86°C. Assuming molality equal to molarity, pH of CH3COOH solution is :
(A) 2
(B) 3
(C) 3.2
(D) 4.2
For urea
T
1.86
Tf = kf × m
or
kf = f =
= 1.86
m
1
Now for CH3COOH
Tf = i kf m
0.02046
so
i=
= 1.1
1.86  0.01
Now
i=1+
so
 = 1.1 – 1 = 0.1
CH3COOH
CH3COO–
C
0
C – C
C
[H+] = C = 0.01 × 0.1 = 0.001
so
pH = 3.
+
lp
Now
H+
0
C
a
Ans. (B)
Section (K) : Osmotic Pressure
Osmosis & Osmotic pressure :
Diffusion : Spontaneous flow of particles from high concentration region to lower concentration region
is known as diffusion.
Sa
nk
8.6
D38
H2O
Ex.
Cu2+
CuSO4
Blue
Saturated
(Solution)blue
Uniform
Figure
D39
Osmosis :
The spontaneous flow of solvent particles from solvent side to solution side or from solution of low
concentration side to solution of high concentration side through a semipermeable membrane (SPM) is
known as osmosis.
D40
Semipermeable Membrane (SPM): A membrane which allows only solvent particles to move across it.
(a)
Natural : Semi permeable membrane
Animal/plant cell membrane formed just below the outer skins.
(b)
Artificial membranes also : A copper ferrocyanide.
Cu2[Fe(CN)6] & Silicate of Ni, Fe, Co can act as SPM.
184
Ex.
Figure
In which soluton side complex [Cu(NH3)4]2+ will form and deep blue colour will obtain.
In neither of side colour complex will form. No solute particle passes through SPM.
Sa
nk
a
Osmotic Pressure :
lp
Solution :
Ja
uh
Example-23 :
a
ri
Conclusion : After some time in (A) grape or egg will shrink and in (B) grape or egg will swell.
e.g.
(i) A raw mango placed in concentrated salt solution loses water & shrivel into pickle.
(ii) People taking lot of salt, experience water retention in tissue cells. This results in puffiness
or swelling called edema.
h
--------------------------Solvent
Solution
SPM
Figure
The equilibrium hydrostatic pressure developed by solution column when it is seperated from solvent by
semipermeable membrane is called osmotic pressure of the solution.
 = gh
;
 = density of solution
g = acceleration due to gravity ;
h = eq. height
1 atm = 1.013 x 105 N/m2
185
Reverse Osmosis :
If the pressure applied on the solution side is more than osmotic pressure of the solution then the
solvent particles will move from solution to solvent side. This process is known as reverse osmosis.
Berkely : Hartely device/method uses the above pressure to measure osmotic pressure.
e.g. used in desalination of sea-water.
uh
D42
Figure
Osmotic Pressure : The external pressure which must be applied on solution side to stop the process
of osmosis is called osmotic pressure of the solution.
If two solutions of concentration C1 and C2 are kept separated by SPM, and C1 > C2 then particle
movement take place from lower to higher concentration. So, extra pressure is applied on higher
concentration side to stop osmosis.
And
Pext. = (1 – 2)
a
ri
D41
Vant – Hoff Formula (For calculation of osmotic pressure)
Ja
  concentration (molarity)
T
C – mol/lit.
 = CST
R – 0.082 lit.atm. mol–1 K–1
S = ideal solution constant
 = atm.
T – kelvin
= 8.314 J mol–1 K–1 (exp value)
= R (ideal gas) constant
n
 = CRT =
RT (just like ideal gas equation)
V
⚫ In ideal solution solute particles can be assumed to be moving randomly without any interactions.
C = total concentration of all types of particles.
(n1 + n2 + n3 + .........)
V
a
F33
lp
⚫
Sa
nk
= C1 + C2 + C3 + s................. =
Example-24 :
Solution :
If V1 mL of C1 solution + V2 mL of C2 solution are mixed together then calculate final
concentration of solution and final osmotic pressure. If initial osmotic pressure of two solutions
are 1 and 2 respectively ?
C V + C2 V2
Cf = 1 1
V1 + V2
  
 
1 = C1RT, C1 =  1  ; 2 = C2RT, C2 =  2 
 RT 
 RT 
 C V + C2 V2 
=  1 1
 RT
 V1 + V2 
  V + 2 V2 
=  1 1

 V1 + V2 
186
Type of solutions :
D44
(a) Isotonic solution : Two solutions having same osmotic pressure are consider as isotonic solution.
1 = 2 (at same temperature)
(b) Hypotonic & Hypertonic solutions : If two solutions 1 and 2 are such that 2 > 1 , then 2 is called
hypertonic solution and 1 is called hypotonic solution.
Pressure
is applied
Solution 2
Solution 1
Hypertonic
Hypotonic
2
1
Figure
a
ri
D43
uh
Conclusion :
Pressure is applied on the hypertonic solution to stop the flow of solvent partices, this pressure become
equal to (2 – 1) and if hypotonic solution is replaced by pure solvent then pressure becomes equal to
2.
Note : Osmotic pressure of very dilute solutions is also quite significant. So, its measurement in lab is
very easy.
Plasmolysis : When the cell is placed in solution having osmotic pressure greater than that of the cell
sap, water passes out of the cell due to osmosis. Consequently, cell material shrinks gradually. The
gradual shrinking of the cell material is called plasmolysis.
Ja
D45
Calculate osmotic pressure of 0.1 M urea aqueous solution at 300 K ,
R = 0.082 lit atm K–1
 = CRT
 = 0.1 x 0.082 x 300
 = 2.46 atm.
Example-26 :
If 10 g of an unknown substance (non-electrolytic) is dissolved to make 500 mL of
solution,then osmotic pressure at 300 K is observed to be 1.23 atm find molecular
weight?
10  1000
1.23 =
x 0.082 x 300
M  500
20 0.082
M=
x
x 300
 400 g/mol
1.23 100
a
Sa
nk
Solution :
lp
Example-25 :
Solution :
Example-27 :
Solution :
Example-28 :
Solution :
If 6 g of urea, 18 g glucose & 34.2 g sucrose is dissolved to make 500 mL of a solution at 300 K
calculate osmotic pressure ?
molecular weight of urea = 60 g, Glucose = 180 g , Sucrose = 342 g
 = C x 0.082 x 300
0.3  1000  0.082  300
 14.76 atm
=
500
If 200 mL of 0.1 M urea solution is mixed with 300 mL of 0.2 M glucose solution
Calculate osmotic pressure
0.02 moles urea
0.08
 0.082  300 = 3.94 atm.
0.06 moles glucose   =
0.5
at 300 K.
187
Example-29 :
Solution :
If urea (aq) solution at 500K has O.P. = 2.05 atm. & glucose solution at 300 K has OP = 1.23
atm. If 200 ml of Ist solution & 400 ml of 2nd solution are mixed at 400 K then calculate O.P. of
resulting solution at 400 K (assume molarity is not dependent on temp.)
2.05
Curea =
= 0.05 
Vurea = 200 mL
R  500
1.23
Cglucose =
= 0.05 
Vglucose = 400 mL
R  300
C V + C2 V2
0.05  200 + 0.05  400
Ctotal = 1 1
=
= 0.05
V1 + V2
600
Example-31 :
Soluiton :
uh
Solution :
0.1 M urea 0.1 M NaCl 0.1 M BaCl2
,
,
(A)
(C)
(B)
Order of 
C > B > A.
Order of R.L.V.P
C > B > A.
Order of V.P
A > B > C.
Order of TB
C > B > A.
Order of TB of solution
C > B > A.
Order of TF
C > B > A.
Order of TF of solution
A > B > C.
(Calculating osmotic pressure when reaction is not taking place)
Calculate osmotic pressure of a solutions having 0.1 M NaCl & 0.2 M Na 2SO4 and 0.5 MHA.
(Given : Weak acid is 20% dissociated at 300 K).
 = NaCl + Na2SO4 + HA
Ja
Example-30
a
ri
 = CRT = 0.05 × 0.082 × 400
 = 1.64 atm
= 0.1 RT × 2 + 0.2 RT × 3 + 0.5 RT × 1.2
= 0.0821 × 300 (0.2 + 0.6 + 0.6) = 34.482 atm.
If 6 g of CH3 COOH is dissolved in benzene to make 1 litre at 300 K. Osmotic pressure of
solution is found to be 1.64 atm. If it is known that CH 3COOH in benzene forms a dimer.
Calculate degree of association of acetic acid in benzene ?
Sa
nk
Example-33 :
lp
Soluiton :
If 0.04 M Na2SO4 solutions at 300 K is found to be isotonic with 0.05 M NaCl (100 %
disscociation) solutions. Calculate degree of disscociation of sodium sulphate ?
i1 C1 RT = i2 C2 RT
i1 C1 = i2 C2
0.04 (1 + 2) = 0.05 × 2
 = 0.75 = 75%.
a
Example-32 :
Soluiton :
;
1 
i = 1 +  − 1 .
n 

1 
1.64 = 0.0821 × 300 × [ 1 +  − 1  ] × 0.1
n 

1.64 = 0.0821 × 300 [1 – ] 0.1
2
1.64
2-
=
0.0821 30
2
1.64 2-
=
4 = 6 – 3
2
2.46
3 = 2
 = 2/3
(Calculating osmotic pressure when reaction is taking place)
188
Solution :
If 200 ml of 0.2 M HgCl2 solution is added to 800 ml of 0.5 M KI (100% dissociated) solution.
Assuming that the following complex formation taken place to 100% extent.
Hg2++ 4–
[HgI4]2–
0.04 0.4
Calculate osmotic pressure of resulting initially solution at 300K ?
HgCl2 + 4KI
K2[HgI4] + 2KCl.
40
400
0
0
0
400 –160
40
80
240
80
40
1000
1000
1000
 = (i1C1 + i2C2 + i3C3) RT.
uh
Example-36 :
Ja
Solution :
If 200 ml of 0.2 M BaCl2 solution is mixed with 500 ml of 0.1 M Na2SO4 solution. Calculate
osmotic
pressure of resulting solutions ?
BaCl2 + Na2SO4 → BaSO4 + 2 NaCl
0.04 moles
0.05
0
0
0
0.01
No effect
0.08
0.08
0.01
No effect
0.7
0.7
 = (i1C1 + i2C2) RT.
0.08
0.01
= (3 ×
+2×
) 0.082 × 300. = 6.685 atm.
0.7
0.7
a
ri
Example-35 :
= (0.24 × 2 + 3 × 0.04 + 0.08 × 2) 0.082 × 300. = 18.69 atm.
(Note: Attempt this problem after you have studied co-ordination compounds)
Ba2+ ions, CN– & Co2+ ions form a water soluble complex with Ba2+ ions as free cations. For a
0.01 M solution of this complex, osmotic pressure = 0.984 atm & degree of dissociation = 75%.
Then find coordination number of Co2+ ion in this complex (T=300 K, R=0.082 L atm. mol–1 k–1)
Solution :
Say
lp
Example-37 :
Sa
nk
a
C.N. = x
0.984 = i CRT
0.984 = i 0.01 × 0.082 × 300 = i × 0.246
i = 4 = 1 + (n –1) 

n=5
Charge on co-ordination sphere = charge on Cobalt ion - charge on x cyanide ions = – (x – 2)
i.e. co-ordination sphere is [Co(CN)x ]–(x – 2)
Charge on Barium ion is +2
Thus, formula of the complex will be Ba(x – 2) [Co(CN)x]2 by charge balance.
x–2+2=5
x=5

CN = 5

Formula is Ba3[Co(CN)5)2..
189
lp
a
Sa
nk
a
ri
uh
Ja
Summary
190
MISCELLANEOUS SOLVED PROBLEMS (MSPs)
A 6.90 M solution of KOH in water has 30% by weight of KOH. Calculate density of solution.
Let V = 1 lt , then moles of solute = 6.9
wt of solute = 6.9 × 56 gm
M
6.9  56
 100 = 30 ;
% = solute × 100
So, =
d = 1.288 gm/litre
Msolution
d  1000
2.
10 ml of sulphuric acid solution (sp. gr. = 1.84) contains 98% by weight of pure acid. Calculate the
volume of 2.5 M NaOH solution required to just neutralise the acid.
98
Wt of solute = 10 × 1.84 ×
g
100
98
18.4
So moles of solute =
×
= 0.184
100
98
nH+ = 2 × 0.184
Sol.
2 × 0.184 =
4.
Ans.
Sol.
uh
A sample of H2SO4 (density 1.8 g mL–1) is labelled as 74.66% by weight. What is molarity of acid ?
(Give answer in rounded digits)
Let V = 1lt = 1000 ml
So mass of solution = 1800 gm
74.66
So mass of solute = 1800 ×
g
100
18  74.66
So molarity = moles of solute in 1lt =
= 13.71 M Ans.
98
Ja
Sol.
V = 147.2
The density of 3M solution of Na2S2O3 is 1.25 g mL–1. Calculate.
(a) the % by weight of Na2S2O3
(b) mole fraction of Na2S2O3
(a) 37.92 (b) 0.065
(a) msolution = 1000 × 1.25 = 1250 gm
Mass of Na2S2O3 = 3 × 158 = 474
lp
3.
2.5  V
1000
a
ri
1.
Sol.
474
× 100 = 37.92
1250
3
(b) XNa2S2O3 =
= 0.065.
3 + 43.11
Sa
nk
a
% (w/w) =
5.
Calculate Molality of aqueous urea solution which has Xurea = 0.2
Sol.
Molality =
6.
If 200 mL of 0.1 M urea solution is mixed with 300 mL of 0.2 M Glucose solutions at 300 K calculate
osmotic pressure
3.94 atm
C1V1 + C2 V2
Cnet =
V1 + V2
Ans.
Sol.
Now
7.
Ans.
Sol.
1000X1
1000  0.2
=
= 13.88.
(1– X1 )m2
0.8  18
 = Cnet RT
p = (0.04 + 0.12) × 0.0821 × 300 = 3.94 atm.
A 500 gm liquid consist of 15 gm ethane at any temp. T, at a pressure = 2 atm. Find Pressure of gas
required to dissolve 30 gm gas in 300 gm liquid.
P2 = 6.66 atm.
15 / 500
3
2
2
=

=
 P = 6.66 atm.
P
30 / 300
10 P
191
Check List
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% w/w
% wt/vol. (w/v)
molarity
molality
No. of equivalents per litre of solution
Equivalent mass
No. of equivalent
Xsolute
XSolvent
Xsolute + XSolvent = 1
ppm (w/w)
ppm (w/v)
ppm (moles/moles)
Clausius-Clapeyron equation
Relative Humidity (R.H.)
Henry's Law
Ja
F1
F2
F3
F4
F5
F6
F7
F8
F9
F10
F11
F12
F13
F14
F15
F16
Sa
nk
a
lp
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D39
D40
D41
D42
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D44
D45
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a
ri
Definitions (D)
Solution
Solute
Solvent
Binary solutions
Ternary solutions
Aqueous solution
Non-aqueous solution
Miscible liquids
Immiscible liquids
Alloys
Mass percentage
Mass by volume percentage
Volume percentage
Molarity (M)
Molality (m)
Normality
Mole-fraction (x)
Parts per million (ppm)
Vapour pressure of a pure liquid
Vapour pressure of solution
Partial vapour pressure
Saturated solution
Unsaturated solution
Supersaturated solution
Solubility
Azeotropes
Minimum Boiling Azeotropes
Maximum Boiling Azeotropes
Constitutional Properties
Colligative properties
Abnormal colligative property
Boiling Point
Normal Boiling Point
Elevation Of Boiling Point
Freezing Point
Depression In Freezing Point
Kf, Freezing point Depression Constant
(Molal Depression Constant)
Diffusion
Osmosis
Semi-permeable Membrane
Osmotic Pressure
Reverse Osmosis
Isotonic solution
Hypotonic & Hypertonic solutions
Plasmolysis
D1
D2
D3
D4
D5
D6
D7
D8
D9
D10
D11
D12
D13
D14
D15
D16
D17
D18
D19
D20
D21
D22
D23
D24
D25
D26
D27
D28
D29
D30
D31
D32
D33
D34
D35
D36
D37
Formule (F)
192
Distillation of Immiscible liquids

Composition of distillate obtained from distillation of Immiscible liquids

Raoult's Law
Total pressure of vapour in ideal solutions of two miscible liquids

relation between total pressure and vapour composition over an ideal solution of two miscible liquids
Vant–Hoff factor

Relation of i with 

Relation of i with 

RLVP

Relation of P with molality

Tb

Kb

Lvap

Formula for Depression of freezing point

Cryoscopic constant Kf

Lfusion

Osmotic Pressure

Der1
Der2
Der3
Der4
Der5
Der6
Der7
Der8
Molarity of mixture of two solutions of known molarity and volume of same solute
Henry's Law

Total pressure over solution using Raoult’s law and Dalton’s law
Relation of i with 

Relation of i with 

RLVP

Tb

Tf



Sa
nk
a
lp
Ja
Derivations (Der.)
uh
a
ri
F17
F18
F19
F20
F21
F22
F23
F24
F25
F26
F27
F28
F29
F30
F31
F32
F33
193
 Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
Section (A) : General Introduction & types of solution
A-1. What are the characteristics of the supersaturated solution.
A-2. What do you mean by Hygroscopic compound ?
Why are some solution processes exothermic whereas others are endothermic ?
Section (B) : Concentration terms (Revision of mole)
Commit to memory :
Molality =
no. of moles of solute
;
volume of solution (in L)
no. of moles of solute
× 1000
W(g) of solvent
molarity  1000
(where d is density of solution in g/ml)
1000d − molarity  m.wt.of solute
w msolute (g)
=
 100 ;
w msolution (g)
xsolute + xsolvent = 1
%
xsolute =
n
(where, n is moles of solute, N is moles of solvent)
n+N
Ja
B-1.
Molality =
uh
Molarity =
a
ri
A-3.
Calculate the concentration of NaOH solution in g/ml which has the same molarity as that of a solution
of HCl of concentration 0.0365 g/ml.
lp
B-2. The density of 3M solution of sodium thiosulphate (Na2S2O3) is 1.58 g/ml. Calculate
(i) amount of Na2S2O3 in % w/w
(ii) mole fraction of Na2S2O3
(iii) molality of Na+ and S2O32– ions.
a
B-3. Calculate the molality and molarity of a solution made by mixing equal volumes of 30% by weight of
H2SO4 (density = 1.20 g/mL) and 70% by weight of H2SO4 (density = 1.60 g/mL).
Section (C) : Vapour Pressure
Sa
nk
Commit to memory :
*
*
*
The partial pressure of vapours of X in equilibrium with X at a given temperature is called as its vapour
pressure. (X is given pure solid or pure liquid)
A gas is said to be saturated with vapours of a liquid if the partial pressure of liquid vapours is equal to
its (saturated) vapour pressure.
Partial pressure of vapours of X (pure solid or liquid) will remain equal to its vapour pressure till the
vapours of X are in equilibrium with X.
C-1. The vapour pressure of water at 80ºC is 355 torr. A 100 ml vessel contained water−saturated oxygen at
80º C, the total gas pressure being 760 torr. The contents of the vessel were pumped into a 50.0 ml,
vessel at the same temperature. What were the partial pressures of oxygen and of water vapour and
the total pressure in the final equilibrium state? Neglect the volume of any water which might condense.
C-2.
A vessel has nitrogen gas and water vapours in equilibrium with liquid water at a total pressure of 1
atm. The partial pressure of water vapours is 0.3 atm. The volume of this vessel is reduced to one third
of the original volume, at the same temperature, then find total pressure of the system. (Neglect volume
occupied by liquid water)
194
Section (D) : Solutions of Solid and Gases in Liquids
Commit to memory :
*
Henry’s law : The solubility of gas in a liquid at a given temperature is directly proportional to its partial
pressure above liquid in which it is dissolved.
P = KHx (where x is mole fraction of unreacted, dissolved gas and P is its partial pressure above liquid.)
D-1.
The partial pressure of ethane over a solution containing 6.56 × 10 –3 g of ethane is 1 bar. If the solution
contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas ?
D-2. If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 300 mole
of water, if N2 exerts a partial pressure of 1 bar. Given that Henry's law constant for N 2 at 293 K is 75.00
kbar.
a
ri
Section (E) : Immiscible Liquids
Commit to memory :
If A and B are volatile immiscible liquids, then above their mixture.
PT = PºA + PºB (PT = Total pressure above mixture of A and B, PºA = vapour pressure of A,
PºB = vapour pressure of B)
nA
PAº
=
(nA and nB are moles of A and B in distillate)
nB
PBº
uh
*
w A PA MA
=
(wA and wB are masses of A and B in distillate, MA and MB are molar masses of A and B)
w B PB MB
Boiling point of a mixture of water and nitrobenzene is 99°C, the vapour pressure of water is 733 mm of
Hg and the atmospheric pressure is 760 mm of Hg. The molecular weight of nitrobenzene is 123. Find
the ratio of weights of the components of the distillate.
lp
E-2.
Ja
E-1. A mixture of an organic liquid A and water distilled under one atmospheric pressure at 99.2ºC. How
many grams of steam will be condensed to obtain 1.0 g of liquid A in the distillate ? (Vapour pressure of
water at 99.2ºC is 739 mm Hg. Molecular weight of A = 123)
Section (F) : Completely miscible liquids : Raoult’s law
Commit to memory :
Sa
nk
a
Statement of Raoult's law (for volatile liq. mixture) : In solution of volatile liquids, the partial vapour
pressure of each component is directly proportional to its mole fraction in solution.
pA  xA
pA = xAPAº
....... (1)
where pA = Partial vapour pressure of component A, xA = Mole fraction of component ‘A’ in solution,
PAº = Vapour pressure of pure component ‘A’ at given temperature
Hence PT = xAPAº + xBPBº
....... (2)
(PT = Total pressure of vapour above solution, PBº = vapour pressure of pure component 'B' at given
temperature.)
1
y
y
Thus,
= A
+ Bo (for complete derivation, refer page no. 12 of sheet)
o
PT
PA PB
(yA = mole fraction of A in vapour phase above the solution and y B = mole fraction of B in vapour phase
above the solution)
F-1. Two liquids A and B form an ideal solution. At 300 K, the vapour pressure of a solution containing 1
mole of A and 3 moles of B is 550 mm of Hg. At the same temperature, if one mole of B is added to this
solution, the vapour pressure of the solution increases by 10 mm of Hg. Determine the vapour pressure
of A and B in their pure states.
F-2. Two liquids, A and B, form an ideal solution. At the specified temperature, the vapour pressure of pure
A is 200 mm Hg while that of pure B is 75 mm Hg. If the vapour over the mixture consists of 50 mol
percent A, what is the mole percent A in the liquid ?
F-3.
Two solutions of A and B are available. The first is known to contain 1 mole of A and 3 moles of B and
its total vapour pressure is 1.0 atm. The second is known to contain 2 moles of A and 2 moles of B; its
vapour pressure is greater than 1 atm, but it is found that this total vapour pressure may be reduced to
195
1 atm by the addition of 6 moles of C. The vapour pressure of pure C is 0.80 atm. Assuming ideal
solutions and that all these data refer to 25°C, calculate the vapour pressure of pure A and of pure B.
F-4.
At 80oC, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg.
Calculate the composition of a liquid in mole per cent which at 80 oC is in equilibrium with the vapour
containing 30 mole per cent of benzene.
F-5. Vapour pressure of C6H6 and C7H8 mixture at 50ºC is given P (mm Hg) = 180X B + 90, where XB is the
mole fraction of C6H6. A solution is prepared by mixing 12 mol benzene and 8 mol toluene and if
vapours over this solution are removed and condensed into liquid and again brought to the temperature
50ºC, what would be mole fraction of C6H6 in the vapour state. (At. wt. of C = 12, H = 1)
Section (G) : Non-ideal Solutions
a
ri
Commit to memory :
+ve deviation
–ve deviation
PT.exp > (xAPºA + xBPºB)
PT.exp < (xAPºA + xBPºB)
(where PT.exp is experimental total pressure above mixture of volatile liquids A and B).
Hmix = +ve
Hmix = –ve
Vmix = +ve
Vmix = –ve
Smix = +ve
Smix = +ve
Gmix = –ve
Gmix = –ve
G-1.
A non ideal solution was prepared by mixing 30 ml chloroform and 50 ml acetone. Comment on volume
of mixture.
uh
*
Ja
G-2. Total vapour pressure of mixture of 1 mole of volatile component A (Pº A = 100 mm Hg) and 3 mole of
volatile component B (PºB = 80 mm Hg) is 90 mm Hg. Find out nature of solution and sign of entropy of
solution.
Section (H) : Degree of Ionisation/Dissociation for Weak Electrolytes
Commit to memory :
For dissociation/association
i = 1 + (n – 1)
(where n is total number of particles produced per solute particle after association/dissociation, i = vant
Hoff factor,  = degree of dissociation / association.
Moles of solute particles after association / dissociation of X
Observed value of colligative property
i=
= Theoretical value of that colligative property
Moles of X without association / dissociation
lp
*
Theoretical molar mass of solute
Observed molar mass of solute
a
i =
Sa
nk
H-1. Complete the following table.
Dissociation / association
Solute
reaction
KCl
H2SO4
CH3COOH (in water)
CH3COOH (in benzene)
Urea
NaBr
A
3A → A3
Degree of dissociation
/ association
1
1
0.2
0.5
n
i
0.8
1
H-2. Calculate the percentage degree of dissociation of an electrolyte XY2 (Normal molar mass = 164) in
water if the observed molar mass by measuring elevation in boiling point is 65.6.
196
Section (I) : Relative lowering of vapour pressure
Commit to memory :
Pº − Ps
Pº − PS
n
= xsolute;
=
Ps
Pº
N
(where Pº = vapour pressure of pure solvent, Ps = partial pressure of vapour above solution, n =
dissolved moles of solute, N = moles of solvent.)
Pº − Ps msolute(g)
Msolvent
=
×
(where m = given mass in solution, M = molar mass.)
Ps
Msolute
msolvent(g)
RLVP =
M
Pº − Ps
= ( molality ) × solvent
Ps
1000
If solution is of single solute and it gets associated or dissociated;
M
Pº − Ps i.n
Pº − Ps
=
; that is
= i × (molality) × solvent
Ps
Ps
N
1000
a
ri
*
Twenty grams of a solute are added to 100 g of water at 25ºC. The vapour pressure of pure water is
23.76 mmHg; the vapour pressure of the solution is 22.41 Torr.
(a) Calculate the molar mass of the solute.
(b) What mass of this solute is required in 100 g of water to reduce the vapour pressure to one-half the
value for pure water ?
l-2.
The degree of dissociation of Ca(NO3)2 in a dilute aqueous solution containing 7 g salt per 100 g of
water at 100ºC is 70%. If the vapour pressure of water at 100ºC is 760 mm of Hg, calculate the vapour
pressure of the solution.
Ja
uh
I-1.
Section (J) : Elevation of Boiling Point & Depression of Freezing Point
Commit to memory :
Tb = i × Kb × m (where Tb = elevation in boiling point of solution, Kb = ebullioscopic constant,
m = molality of single solute, Tb = boiling point of solvent (in K).
2
Kb =
2
RTb M
RTb
=
1000  Hvap 1000  L vap
lp
*
(M = Molar mass of solvent)
Tf = i × Kf × m (where Tf = depression in freezing point of solution, Kf = cryoscopic constant,
Tf = freezing point of solvent (in K).
2
2
RTf M
RTf
=
1000  Hfusion
1000  L fusion
a
Kf =
Sa
nk
J-1. (a) A solution containing 0.5 g of naphthalene in 50 g CCl 4 yield a boiling point elevation of 0.4 K, while
a solution of 0.6 g of an unknown solute in the same mass of the solvent gives a boiling point elevation
of 0.65 K. Find the molar mass of the unknown solute.
(b) The boiling point of a solution of 0.1 g of a substance in 16 g of ether was found to be 0.100ºC
higher that of pure ether. What is the molecular mass of the substance. Kb(ether) = 2.16 K kg mol–1.
J-2. The amount of benzene that will separate out (in grams) if a solution containing 7.32 g of
triphenylmethane in 1000 g of benzene is cooled to a temperature which is 0.2°C below the freezing
point of benzene ? (Kf = 5.12 K-Kg/mol)
J-3. The boiling point of a solution of 5 g of sulphur in 100 g of carbon disulphide is 0.474ºC above that of
pure solvent. Determine the molecular formula of sulphur in this solvent. The boiling point of pure
carbon disulphide is 47ºC and its heat of vaporisation is 84 calories per gram.
[Hint : Kb =
RTb2
2  (320)2
=
= 2.438]
1000L v
1000  84
J-4. A 0.01 molal solution of ammonia freezes at –0.02ºC. Calculate the van’t Hoff factor, i and the
percentage dissociation of ammonia in water. (K f(H2O) ) = 1.86 deg molal–1.
197
Section (K) : Osmotic Pressure
Commit to memory :
*
 = CRT =
n
RT (n = Total moles of solute particles in solution, V = Total volume of solution in L)
V
K-1. (a) Predict the osmotic pressure order for the following (assume salts are 100% dissociated).
I
0.1 M urea
II
0.1 M NaCl
III
0.1 M Na2SO4
IV
0.1 M Na3PO4
(b) If equal volumes of all these solutions are mixed then calculate the osmotic pressure of the net
resultant solution obtained at 300K.
a
ri
K-2. A solution containing 3.00 g of calcium nitrate in 100 c.c. of solution had an osmotic pressure of 11.2
atmosphere at 12°C. Calculate the degree of ionisation of calcium nitrate at this dilution and
temperature.
K-3. 17.4% (w/v) K2SO4 solution at 27ºC is isotonic with 5.85% (w/v) NaCl solution at 27ºC. If NaCl is 100%
ionised, what is % ionisation of K2SO4 in aq. solution ?
uh
K-4. At 2ºC the osmotic pressure of a urea solution is found to be 500 mm of Hg. The solution is diluted and
the temperature is raised to 27ºC, when the osmotic pressure is found to be 109.09 mm of Hg.
Determine the extent of dilution.
PART - II : ONLY ONE OPTION CORRECT TYPE
Section (A) : General Introduction & types of solution
Which statement best explains the meaning of the phrase “like dissolves like “ ?
(A) A Solute will easily dissolve a solute of similar mass
(B) A solvent and solute with similar intermolecular forces will readily form a solution
(C) The only true solutions are formed when water dissolves a non-polar solute
(D) The only true solutions are formed when water dissolves a polar solute
Ja
A-1.
lp
A-2. An ionic compound that attracts atmospheric water so strongly that a hydrate is formed is said to be :
(A) Dilute
(B) Hygroscopic
(C) Immiscible
(D) Miscible
Section (B) : Concentration terms (Revision of mole)
a
B-1. Persons are medically considered to have lead poisoning if they have a concentration greater than 10
micrograms of lead per decilitre of blood. Concentration in parts per billion is :
(A) 1000
(B) 100
(C) 10
(D) 1
Sa
nk
Section (C) : Vapour Pressure
C-1. A liquid is kept in a closed vessel. If a glass plate (negligible mass) with a small hole is kept on top of
the liquid surface, then the vapour pressure of the liquid in the vessel is :
(A) More than what would be if the glass plate were removed
(B) Same as what would be if the glass plate were removed
(C) Less than what would be if the glass plate were removed
(D) Cannot be predicted
C-2. The vapour pressure of water depends upon :
(A) Surface area of container
(C) Temperature
C-3.
(B) Volume of container
(D) All
Among the following substances, the lowest vapour pressure is exerted by :
(A) Water
(B) Mercury
(C) Acetone
(D) Ethanol
C-4. At higher altitudes, water boils at temperature < 100ºC because
(A) temperature of higher altitudes is low
(B) atmospheric pressure is low
(C) the proportion of heavy water increases
(D) atmospheric pressure becomes more.
198
Section (D) : Solutions of Solid and Gases in Liquids
D-1. The solubility of gases in liquids :
(A) increases with increase in pressure and temperature
(B) decreases with increase in pressure and temperature
(C) Increases with increase in pressure and decrease in temperature
(D) decreases with increase in pressure and increase in temperature
D-2. Which of the following curves represents theHenry’slaw?
(B)
(C)
(D)
a
ri
(A)
D-3. According to Henry’s law, the solubility of a gas in a given volume of liquid increases with increase in :
(A) Temperature
(B) Pressure
(C) Both (A) and (B)
(D) None of these
uh
D-4. Some of the following gases are soluble in water due to formation of their ions :
I : CO2 ;
II : NH3 ;
III : HCl ;
IV : CH4 ;
V : H2
Water insoluble gases can be :
(A) I, IV , V
(B) I, V
(C) I, II, III
(D) IV, V
Section (E) : Immiscible Liquids
Ja
D-5. The solubility of N2(g) in water exposed to the atmosphere, when its partial pressure is 593 mm is
5.3 × 10–4 M. Its solubility at 760 mm and at the same temperature is :
(A) 4.1 × 10–4 M
(B) 6.8 × 10–4 M
(C) 1500 M
(D) 2400 M
lp
E-1. When a liquid that is immiscible with water was steam distilled at 95.2ºC at a total pressure of 748 torr,
the distillate contained 1.25 g of the liquid per gram of water. The vapour pressure of water is 648 torr
at 95.2ºC, what is the molar mass of liquid?
(A) 7.975 g/mol
(B) 166 g/mol
(C) 145.8 g/mol
(D) None of these
Section (F) : Completely miscible liquids : Raoult’s law
For a binary ideal liquid solution, the total pressure of the solution is given as :
(A) Ptotal = PºA + (PºA – PºB) XB
(B) Ptotal = PºB + (PºA – PºB) XA
(C) Ptotal = PºB + (PºB – PºA) XA
(D) Ptotal = PºB + (PºB – PºA) XB
a
F-1.
Sa
nk
F-2. An ideal solution contains two volatile liquids A (pº = 100 torr) and B (pº = 200 torr). If mixture contain 1
mole of A and 4 mole of B then total vapour pressure of the distillate is:
(A) 150
(B) 180
(C) 188.88
(D) 198.88
F-3.
At 323 K, the vapour pressure in millimeters of mercury of a methanol-ethanol solution is represented
by the equation p = 120 XA + 140, where XA is the mole fraction of methanol.Then the value of
p
lim A is
X
x A →1 A
(A) 250 mm
(B) 140 mm
(C) 260 mm
(D) 20 mm
F-4. Given at 350 K pA° = 300 torr and pB° = 800 torr, the composition of the mixture having a normal boiling
point of 350 K is :
(A) XA = 0.08
(B) XA = 0.06
(C) XA = 0.04
(D) XA = 0.02
F-5.
Two liquids A and B have PºA and PºB in the ratio of 1 : 3 and the ratio of number of moles of A and B in
liquid phase are 1 : 3 then mole fraction of ‘A’ in vapour phase in equilibrium with the solution is equal
to:
(A) 0.1
(B) 0.2
(C) 0.5
(D) 1.0
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Section (G) : Non-ideal Solutions
G-1. The vapour pressure of the solution of two liquids A(pº = 80 mm) and B(pº = 120 mm) is found to be
100 mm when xA = 0.4. The result shows that
(A) solution exhibits ideal behaviour
(B) solution shows positive deviations
(C) solution shows negative deviations
(D) solution will show positive deviations for lower concentration and negative deviations for higher
concentrations.
G-3. A solution of sulphuric acid in water exhibits :
(A) Negative deviations from Raoult’s law
(C) Ideal properties
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G-2. Consider a binary mixture of volatile liquids. If at XA = 0.4 the vapour pressure of solution is 580 torr
then the mixture could be (pAº = 300 torr, pB° = 800 torr) :
(A) CHCl3 – CH3COCH3
(B) C6H5Cl – C6H5Br
(C) C6H6 – C6H5CH3
(D) nC6H14 – nC7H16
(B) Positive deviations from Raoult’s law
(D) The applicability of Henry’s law
When KCl dissolves in water (assume endothermic dissolution), then :
(A) H = + ve, S = + ve, G = + ve
(B) H = + ve, S = – ve, G = – ve
(C) H = + ve, S = + ve, G = – ve
(D) H = – ve, S = – ve, G = + ve
G-5.
The dissolving process is exothermic when :
(A) The energy released in solvation exceeds the energy used in breaking up solute-solute and solventsolvent interactions.
(B) The energy used in solvation exceeds the energy released in breaking up solute-solute and solventsolvent interactions.
(C) The energy released in solvation is about the same as the energy used in breaking up solute-solute
and solvent-solvent interactions.
(D) The energy used in solvation is about the same as the energy used in breaking up solute-solute and
solvent-solvent interactions.
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G-6. Which of the following is less than zero for ideal solutions ?
(A) Hmix
(B) Vmix
(C) Gmix
(D) Smix
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Section (H) : Degree of Ionisation/Dissociation for Weak Electrolytes
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H-1. One mole of a solute A is dissolved in a given volume of solvent. The association of the solute take
place as follows :
nA
An
If  is the degree of association of A, the van’t Hoff factor i is expressed as :

1–  +

n
(A) i = 1– 
(B) i = 1 +
(C) i =
(D) i = 1
n
1
H-2.
The degree of dissociation of an electrolyte is  and its van’t Hoff factor is i. The number of ions
obtained by complete dissociation of 1 molecule of the electrolyte is :
i+  –1
i + 1+ 
i–1
(A)
(B) i –  – 1
(C)
(D)

1– 

H-3. If Mnormal is the normal molecular mass and  is the degree of ionization of K3[Fe(CN)6], then the
abnormal molecular mass of the complex in the solution will be :
(A) Mnormal (1 + 2)–1
(B) Mnormal (1 + 3)–1
(C) Mnormal (1 + )–1
(D) equal to Mnormal
H-4. A complex containing K+, Pt(IV) and Cl– is 100% ionised giving i = 3. Thus, complex is :
(A) K2[PtCl4]
(B) K2[PtCl6]
(C) K3[PtCl5]
(D) K[PtCl3]
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Section (I) : Relative lowering of vapour pressure
If P0 and P are the vapour pressures of a solvent and its solution respectively and N 1 and N2 are the
mole fractions of the solvent and non-volatile solute respectively, then correct relation is :
(A) P = P0N2
(B) P = P0N1
(C) P0 = PN1
(D) P = P0(N1/N2)
I-2.
Relative decrease in vapour pressure of an aqueous NaCl is 0.167. Number of moles of NaCl present
in 180g of H2O is :
(A) 2 mol
(B) 1 mol
(C) 3 mol
(D) 4 mol
I-3.
The vapour pressure of pure benzene, C6H6 at 50°C is 268 Torr. How many moles of non-volatile solute
per mol of benzene is required to prepare a solution of benzene having a vapour pressure of 167 Torr
at 50°C?
(A) 0.377
(B) 0.605
(C) 0.623
(D) 0.395
I-4.
If relative decrease in vapour pressure is 0.4 for a solution containing 1 mol NaCl in 3 mol H 2O, NaCl is
.... % ionised.
(A) 60%
(B) 50%
(C) 100%
(D) 40%
I-5.
The vapour pressure of a solution of a non-volatile solute B in a solvent A is 95% of the vapour
pressure of the solvent at the same temperature. If the molecular weight of the solvent is 0.3 times the
molecular weight of the solute, what is the ratio of weight of solvent to solute.
(A) 0.15
(B) 5.7
(C) 0.2
(D) none of these
I-6.
Relative decrease in vapour pressure of an aqueous solution containing 2 moles [Cu(NH3)3Cl]Cl in 3
moles H2O is 0.50. On reaction with AgNO3, this solution will form(assuming no change in degree of
ionisation of substance on adding AgNO3)
(A) 1 mol AgCl
(B) 0.25 mol AgCl
(C) 0.5 mol AgCl
(D) 0.40 mol AgCl
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I-1.
Section (J) : Elevation of Boiling Point & Depression of Freezing Point
J-1. An aqueous solution containing 1g of urea boils at 100.25°C. The aqueous solution containing 3 g of
(C) 100°C
(D) 100.25°C
Elevation in boiling point was 0.52 °C when 6 g of a compound x was dissolved in 100 g of water.
Molecular weight of x is : (K = 0.52 kg mol–1)
(A) 120
(B) 60
(C) 100
(D) 342
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glucose in the same volume will boil at (A) 100.75 °C
(B) 100.5 °C
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J-3. A solute’S’ undergoes a reversible trimerization when dissolved in a certain solvent. The boiling point
elevation of its 0.1 molal solution was found to be identical to the boiling point elevation in case of a
0.08 molal solution of a solute which neither undergoes association nor dissociation. To what percent
had the solute ‘S’ undergone trimerization?
(A) 30%
(B) 40%
(C) 50%
(D) 60%
J-4. A complex of iron and cyanide ions is 100% ionised at 1m (molal). If its elevation in b.p. is 2.08. Then
the complex is (Kb = 0.52° mol–1 kg) :
(A) K3[Fe(CN)6]
(B) Fe(CN)2
(C) K4[Fe(CN)6]
(D) Fe(CN)4
J-5. PtCl4.6H2O can exist as a hydrated complex 1 molal aq. solution has depression in freezing point of
3.72°. Assume 100% ionisation and Kf(H2O) = 1.86° mol–1 kg, then complex is (A) [Pt(H2O)6]Cl4
(B) [Pt(H2O)4Cl2]Cl2 . 2H2O
(C) [Pt(H2O)3Cl3]Cl . 3H2O
(D) [Pt(H2O)2Cl4] . 4H2O
J-6.
How many moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a
difference of 104°C between boiling point and freezing point.
(Kf = 1.86 K Kg mol–1, Kb = 0.52 K Kg mol–1)
(A) 1.68
(B) 3.36
(C) 8.40
(D) 0.840
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J-7. Which of the following has been arranged in order of decreasing freezing point?
(A) 0.05 M KNO3 > 0.04 M CaCl2 > 0.140 M sugar > 0.075 M CuSO4
(B) 0.04 M BaCl2 > 0.140 M sucrose > 0.075 M CuSO4 > 0.05 M KNO3
(C) 0.075 M CuSO4 > 0.140 M sucrose > 0.04 M BaCl2 > 0.05 M KNO3
(D) 0.075 M CuSO4 > 0.05 M NaNO3 > 0.140 M sucrose > 0.04 M BaCl2
J-8.
A solution of x moles of sucrose in 100 grams of water freezes at −0.2ºC. As ice separates the freezing
point goes down to – 0.25ºC. How many grams of ice would have separated?
(A) 18 grams
(B) 20 grams
(C) 25 grams
(D) 23 grams
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J-9. Sea water is found to contain 5.85 % NaCl and 9.50% MgCl 2 by weight of solution. Calculate its normal
boiling point assuming 80% ionisation for NaCl and 50% ionisation of MgCl 2 (Kb(H2O) = 0.51 kgmol–1
K).
(A) Tb = 101.9°C
(B) Tb = 102.3°C
(C) Tb = 108.5°C
(D) Tb = 110.3°C
Section (K) : Osmotic Pressure
The relationship between osmotic pressure at 273 K when 10 g glucose (P 1), 10 g urea (P2) and 10 g
sucrose (P3) are dissolved in 250 mL of water is :
(A) P1 > P2 > P3
(B) P3 > P1 > P2
(C) P2 > P1 > P3
(D) P2 > P3 > P1
K-2.
Osmotic pressure of blood is 7.40 atm at 27°C. Number of moles of glucose to be used per litre for an
intravenous injection that is to have the same osmotic pressure as blood is :
(A) 0.3
(B) 0.2
(C) 0.1
(D) 0.4
K-3.
The total concentration of dissolved particles inside red blood cells is approximately 0.30 M and the
membrane surrounding the cells is semipermeable. What would the osmotic pressure (in atmosphere)
inside the cells become if the cells were removed from the blood plasma and placed in pure water at
298 K ?
(A) 7.34 atm
(B) 1.78 atm
(C) 2.34 atm
(D) 0.74 atm
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K-1.
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K-4. Consider following cases :
I : 2M CH3COOH solution in benzene at 27°C where there is dimer formation to the extent of 100%
II : 0.5 M KCl aq. solution at 27°C, which ionises 100%
Which is/are true statements(s) :
(A) both are isotonic
(B) I is hypertonic
(C) II is hypotonic
(D) none is correct
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PART - III : MATCH THE COLUMN
1.
If at a particular temperature, the density of 18 M H2SO4 is 1.8 g cm–3. Then :
Column – I
Column – II
(A)
Molality
(p)
0.1
(B)
% concentration by wt. of solute
(q)
0.9
(C)
mole fraction of H2SO4
(r)
500
(D)
mole fraction of H2O
(s)
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2.
Match the following columnColumn – I
(A)
Acetone + CHCl3
(B)
Ethanol + Water
(C)
C2H5Br + C2H5I
(D)
Acetone + Benzene
(p)
(q)
(r)
(s)
(t)
Column – II
Smix. > 0
Vmix. > 0
Hmix. < 0
Maximum boiling azeotropes
Minimum boiling azeotropes
202
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
All of the water in a 0.20 M solution of NaCl was evaporated and a 0.150 mol of NaCl was obtained.
What was the original volume of the sample ?
(A) 30 mL
(B) 333 mL
(C) 750 mL
(D) 1000 mL
2.
A 20.0 mL sample of CuSO4 solution was evaporated to dryness, leaving 0.967 g of residue. What was
the molarity of the original solution ? (Cu = 63.5)
(A) 48.4 M
(B) 0.0207 M
(C) 0.0484 M
(D) 0.303 M
3.
The vapour pressure of water at 20°C is 17.54 mmHg. What will be the
vapour pressure of the water in the apparatus shown after the piston is
lowered, decreasing the volume of the gas above the liquid to one half of its
initial volume (assume temperature constant).
(A) 8.77 mmHg
(B) 17.54 mmHg
(C) 35.08 mmHg
(D) between 8.77 and 17.54 mmHg
4.
A sample of air is saturated with benzene (vapor pressure = 100 mm Hg at 298 K) at 298K, 750mm Hg
pressure. If it is isothermally compressed to one third of its initial volume, the final pressure of the
system is
(A) 2250 torr
(B) 2150 torr
(C) 2050 torr
(D) 1950 torr
5.
Water and chlorobenzene are immiscible liquids. Their mixture boils at 89ºC under a reduced pressure
of 7.7 × 104 Pa. The vapour pressure of pure water at 89ºC is 7 × 10 4 Pa. Weight percent of
chlorobenzene in the distillate is:
(A) 50
(B) 60
(C) 78.3
(D) 38.46
6.
If two liquids A (PºA =100 torr) and B (PºB = 200 torr) are completely immiscible with each other, each
one will behave independently of the other, are present in a closed vessel. The total vapour pressure of
the system will be:
(A) less than 100 torr (B) less than 200 torr (C) between 100 to 200 torr (D) 300 torr
7.
Given P-x curve for a non-ideal liquid mixture (Fig.). Identify the correct T-x curve for the same mixture.
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1.
(A)
(B)
(C)
(D)
8.
If vapour pressures of pure liquids ‘A’ & ‘B’ are 300 and 800 torr respectively at 25°C. When these two
liquids are mixed at this temperature to form a solution in which mole percentage of ‘B’ is 92, then the
total vapour pressure is observed to be 0.95 atm. Which of the following is true for this solution.
(A) Vmix > 0
(B) Hmix < 0
(C) Vmix = 0
(D) Hmix = 0
9.
Barium ions, CN¯ and Co2+ form an ionic complex. If that complex is supposed to be 75% ionised in
water with vant Hoff factor ‘i’ equal to four, then the coordination number of Co 2+ in the complex can be:
(A) Six
(B) Five
(C) Four
(D) Six and Four both
203
In the following aqueous solutions (a) 1 m sucrose, (b) 1 m potassium ferricyanide and (c) 1 m
potassium sulphate
Maximum value of vapour pressure of solution is that of :
(A) a
(B) b
(C) c
(D) equal
11.
When only a little quantity of HgCl2(s) is added to excess KI(aq) to obtain a clear solution, which of the
following is true for this solution? (no volume change on mixing).
The reaction is 4KI(aq.) + HgCl2(s) —→ K2[HgI4] (aq.) + 2KCl (aq.)
(A) Its boiling and freezing points remain same (B) Its boiling point is lowered
(C) Its vapour pressure become lower
(D) Its boiling point is raised
(E) Its freezing point is lowered.
12.
x mole of KCI and y mole of BaCl2 are both dissolved in 1 kg of water. Given that x + y = 0.1 and K f for
water is 1.85 K/molal, what is the observed range of Tf, if the ratio of x to y is varied ?
(A) 0.37º to 0.555º
(B) 0.185º to 0.93º
(C) 0.56º to 0.93º
(D) 0.37º to 0.93º
13.
FeCl3 on reaction with K4[Fe(CN)6] in aq. solution gives blue colour. These
are separated by a semipermeable membrane PQ as shown. Due to osmosis
there is(A) blue colour formation in side X
(B) blue colour formation in side Y
(C) blue colour formation in both of the sides X and Y
(D) no blue colour formation
14.
Two beakers, one containing 20 ml of a 0.05 M aqueous solution of a non volatile, non electrolyte and
the other, the same volume of 0.03 M aqueous solution of NaCl, are placed side by side in a closed
enclosure. What are the volumes in the two beakers when equilibrium is attained ? Volume of the
solution in the first and second beaker are respectively.
(A) 21.8 ml and 18.2 mL
(B) 18.2 mL and 21.8 mL
(C) 20 mL and 20 mL
(D) 17.1 mL and 22.9 mL
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10.
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
What volume of 98% sulphuric acid (in ml) should be mixed with water to obtain 200 mL of 15% solution
of sulphuric acid by weight ? Given density of H 2O = 1.00 g cm–3, sulphuric acid (98%) = 1.88 g cm–3
and sulphuric acid (15%) = 1.12 g cm–3.
2.
At 300 K, 40 mL of O3 (g) dissolves in 100 g of water at 1.0 atm. What mass of ozone (in gram)
dissolved in 1600 g of water at a pressure of 4.0 atm at 300 K?
3.
An ideal aqueous solution containing liquid A(M.Wt. = 128) 64% by weight has a vapour pressure of
145 mm Hg. If the vapour pressure of A is x mm of Hg and that of water is 155 mm Hg at the same
temperature. Then find x/5. The solutions is ideal.
4.
A and B form ideal solutions; at 50ºC, PAº is half PBº. A solution containing 0.2 mole of A and 0.8 mole
of B has a normal bolling point of 50ºC. Find 18 × PBº. (PBº is in atm)
5.
The vapour pressure of pure liquid A at 300 K is 577 Torr and that of pure liquid B is 390 Torr. These
two compounds form ideal liquid and gaseous mixtures. Consider the equilibrium composition of a
mixture in which the mole fraction of A in the vapour is 0.35. Find the mole % of A in liquid.
6.
How many facts related to CHCl3 + ethylmethylketone solution are correct ?
(a) It shows negative derivation.
(b) It forms maximum boiling azeotropic mixture
(c) S > 0
(d) G < 0
(e) Components can be separated by fractional distillation.
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1.
204
How many of the following solutions show negative deviation from Raoult's Law ?
Liquid A
+
Liquid B
(i)
(CH3)2CO
+
CS2
(ii)
CCl4
+
C6H6
(iii)
CCl4
+
CHCl3
(iv)
H2O
+
C2H5OH
(v)
(C2H5)2O
+
(CH3)2CO
(vi)
CH3COOH
+
C5H5N (pyridine)
(vii)
C6H5NH2
+
(CH3)2CO
(viii)
C6H5Cl
+
C6H5Br
(ix)
Cyclohexane +
Ethanol
8.
0.1 mole XY2 is dissolved in 2L water, where it ionizes to give X 2+ and Y22–. Observed osmotic pressure
is 3 atm. Molar mass of X is 24 and Y is 32. Find M observed + 2i (where Mobserved is observed molar mass
of XY2). (Use R = 1/12 L-atm/mol.K and temperature is 87°C)
9.
How many grams of sucrose (C12H22O11) must be dissolved in 90 g of water to produce a solution over
which the relative humidity is 80%? Assume the solution is ideal. Give your answer after dividing by 10.
10.
1.22 g of a monobasic acid is dissolved in 100 g of benzene. Boiling point of solution increases by
0.13ºC with respect to pure benzene. Find the molecular mass of acid in benzene solvent (in u). Report
your answer after dividing it by 100 and Round it off to nearest integer.
(Kb of benzene = 2.6 K kg mol–1).
11.
1 g of a monobasic acid dissolved in 200 g of water lowers the freezing point by 0.186ºC. On the other
hand when 1 g of the same acid is dissolved in water so as to make the solution 200 mL, this solution
requires 125 mL of 0.1 M NaOH for complete neutralization. Calculate % dissociation of acid ? (Kf
K − kg
=1.86
)
mol
12.
At 27ºC, a 1.2% solution (wt./vol.) of glucose is isotonic with 4.0 g/litre of solution of solute X. Find the
molar mass of X, if the molar mass of glucose is 180. (R = 0.082 L atm mol –1 K–1, Molar mass of
glucose = 180 g/mole)
13.
10 g of solute A and 20 g of solute B both are dissolved in 500 ml. of water. The solution has the same
osmotic pressure as 6.67 g of A and 30 g of B are dissolved in the same volume of water at the same
temperature. If the ratio of molar masses of A and B is x/y, find x + y.
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7.
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PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
1.
We have 100 mL of 0.1 M KCl solution. To make it 0.2 M,
(A) evaporate 50 mL water
(B) evaporate 50 mL solution
(C) add 0.1 mol KCl
(D) add 0.01 mol KCl
2.
Which of the following concentration factors can be calculated if the mole fraction and density of an
aqueous solution of HCl are known ?
(A) Molality
(B) Molarity
(C) Percent by mass
(D) Normality
3.
The vapour pressure of a dilute solution of a solute is influenced by :
(A) Temperature of solution
(B) Mole fraction of solute
(C) M.pt. of solute
(D) Degree of dissociation of solute
4.
According to Henry’s law, the partial pressure of gas (Pg) is directly proportional to mole fraction of gas
in dissolved state , i.e., Pgas = KH . Xgas where KH is Henry’s constant. Which are correct ?
(A) KH is characteristic constant for a given gas–solvent system
(B) Higher is the value of KH , lower is solubility of gas for a given partial pressure of gas
(C) KH has temperature dependence
(D) KH increases with temperature
205
5.
Select correct statements :
(A) Gases which have high value of van der Waals constant ‘a’ are easily liquefied
(B) Easily liquefiable gases are water soluble
(C) Gases which forms ions in a solvent are soluble in that solvent
(D) Under same conditions, NH3 has low solubility in water than that of CO2.
6.
Two liquids X and Y are perfectly immiscible. If X and Y have molecular masses in ratio 1 : 2, the total
vapour pressure of a mixture of X and Y prepared in weight ratio 2 : 3 should be (P x0 = 400 torr,
Py0 = 200 torr)
(A) 600 torr
(B) 400 torr
(C) 800 torr
(D) 1000 torr
7.
Which is/are true about ideal solutions ?
(A) The volume change on mixing is zero
(C) The entropy of mixing is zero
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(B) The enthalpy of mixing is zero
(D) The enthalpy of mixing is negative
At 40ºC, vapour pressure in Torr of methanol and ethanol solution is P = 119x + 135 where x is the
mole fraction of methanol. Hence
(A) vapour pressure of pure methanol is 119 Torr
(B) vapour pressure of pure ethanol is 135 Torr
(C) vapour pressure of equimolar mixture of each is 127 Torr
(D) mixture is completely immiscible
9.
Which is/are correct statement(s) ?
(A) When mixture is more volatile than both pure components, there is positive deviation from Raoult’s
law.
(B) When mixture is less volatile than both pure components, there is negative deviation from Raoult’s
law.
(C) Ethanol and water form ideal solution
(D) CHCl3 and water form ideal solution
10.
At 35°C, the vapour pressure of CS2 is 512 mm Hg, and of acetone is 344 mm Hg. A solution of CS2
and acetone in which the mole fraction of CS2 is 0.25, has a total vapour pressure of 600 mm Hg.
Which of the following statements is/are correct ?
(A) A mixture of 100 mL of acetone and 100 mL of CS2 has a volume of 200 mL
(B) When acetone and CS2 are mixed at 35°C, heat must be absorbed in order to produce a solution at
35°C.
(C) Process of mixing is exothermic
(D) Entropy of mixing is zero
11.
For the given electrolyte AxBy, the degree of dissociation '' can be given as
1− i
i −1
(A)  =
(B) i = (1 – ) + x + y
(C)  =
(D) None
1− x − y
x + y −1
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8.
12.
In which of the following pairs of solutions will the values of the vant Hoff factor be the same?
(A) 0.05 M K4 [Fe(CN)6] and 0.10 M FeSO4
(B) 0.10 M K4[Fe(CN)6] and 0.05 M FeSO4 (NH4)2SO4. 6H2O
(C) 0.20 M NaCl and 0.10 M BaCl2
(D) 0.05 M FeSO4 (NH4)2SO4.6H2O and 0.02 M KCl.MgCl2.6H2O
13.
2 g of non-volatile hydrocarbon solute dissolved in 100 g of hypothetical organic solvent (molar mass =
50) was found to lower vapour pressure from 75.50 to 75 mm of Hg at 20°C. Given that hydrocarbon
contains 96% of C. Then which of the following are true ?
(A) molecular wt of solute = 150
(B) molecular formula = C12H6
(C) molecular wt. of solute = 132
(D) none of these
14.
In the depression of freezing point experiment, it is found that the :
(A) Vapour pressure of the solution is less than that of pure solvent
(B) Vapour pressure of the solution is more than that of pure solvent
(C) Only solute molecules solidify at the freezing point
(D) Only solvent molecules solidify at the freezing point
206
Consider following solutions :
I : 1 M aqueous glucose solution
II : 1M aqueous sodium chloride solution
III : 1M aqueous ammonium phosphate solution
IV : 1M benzoic acid in benzene
Select correct statements for the above solutions :
(A) All are isotonic solutions
(B) III is hypertonic of I, II and IV
(C) IV is hypotonic of I, II and III
(D) II is hypotonic of III but hypertonic of I and IV
16.
Which facts are true when we use van’t Hoff equation PV = nST for osmotic pressure P of dilute
solutions?
(A) The equation is identical to that of ideal gas equation
(B) The solute particles in solution are analogous to the gas molecules and the solvent is analagous to
the empty space between the gas molecules
(C) Solute molecules are dispersed in the solvent the way the gas molecules are dispersed in empty
space
(D) The equation is not identical to that of ideal gas equation
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15.
PART - IV : COMPREHENSIONS
Comprehension # 1
PA = x APA
PB = xBPB
&
Ja
Read the following passage carefully and answer the questions.
(Note : PA > PB ; A is more volatile than B)
lp
PT = x APA + xBPB
What is vapour pressure of pure liquids ?
(A) PB = 92 mm, PT = 179 mm
(B) PB = 271 mm, PT = 92 mm
(C) PB = 180 mm, PT = 91 mm
(D) none of these
Sa
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1.
a
Vapour pressure of mixtures of Benzene (C6H6 ) & toluene (C7H8) at 50ºC are given by PM = 179 XB +
92, where XB is mole fraction of C6H6.
2.
Vapour pressure of liquid mixture obtained by mixing 936 g C6H6 & 736 g tolene is :
(A) 300 mm Hg
(B) 250 mm Hg
(C) 199.4 mm Hg
(D) 180.6 mm Hg
Comprehension # 2
Answer the questions (given below) which are based on the
following diagram. Consider some facts about the above phase
diagram: Vapour pressure diagram for real solutions of two
liquids A and B that exhibit a positive deviation from Raoult’s
law. The vapour pressure of both A and B are greater than
predicted by Raoult’s law. The dashed lines represented the
plots for ideal solutions.
207
A : This is observed when A...B attractions are greater than average of A...A and B...B attraction:
B : Hmix = +ve, Vmix = +ve
C : Boiling point is smaller than expected such that vaporisation is increased
D : Mixture can form azeotropic mixture
Select correct facts
(A) A, B, C
(B) B, C, D
(C) A, C, D
(D) A, B, C, D
4.
Total vapour pressure of mixture of 1 mol of volatile component A (p°A = 100 mm Hg) and 3 mol of
volatile component B (p°B = 60 mm Hg) is 75 mm. For such case :
(A) There is positive deviation from Raoult’s law
(B) Boiling point has been lowered
(C) Force of attraction between A and B is smaller than that between A and A or between B and B.
(D) All the above statements are correct.
a
ri
3.
Comprehension # 3
Addition of non-volatile solute to a solvent always increases the colligative properties such as osmotic
pressure, P, Tb and Tf. All these colligative properties are directly proportional to molality if solutions
are dilute. The increases in colligative properties on addition of non-volatile solute is due to increase in
number of solute particles.
For different aqueous solutions of 0.1 M NaCI, 0.1 M urea, 0.1 M Na 2SO4 and 0.1 M Na3PO4 solution at
27ºC, the correct statements are : (Consider the solutions to be dilute)
1. The order of osmotic pressure is, NaCl > Na2SO4 > Na3PO4 > urea
T
2.  = b × ST for urea solution
Kb
3. Addition of salt on ice increases its melting point
4. Addition of salt on ice brings in melting of ice earlier
(A) 2, 3, 4
(B) 2, 4
(C) 1, 2, 3
(D) 3, 4
6.
1 g mixture of glucose and urea present in 250 mL aqueous solution shows the osmotic pressure of
0.74 atm at 27ºC. Assuming solution to be dilute, which are correct ?
1. Percentage of urea in mixture is 17.6.
2. Relative lowering in vapour pressure of this solution is 5.41 × 10–4.
3. The solution will boil at 100.015ºC, if Kb of water is 0.5 K molality–1.
4. If glucose is replaced by same amount of sucrose, the solution will show higher osmotic pressure at
27ºC.
5. If glucose is replaced by same amount of NaCl, the solution will show lower osmotic pressure at
27ºC.
(A) 1, 2, 3
(B) 1, 2, 3, 5
(C) 2, 4, 5
(D) 1, 4, 5
a
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Ja
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5.
Sa
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Comprehension # 4
Let us consider a binary solution of two volatile liquids 'A' and 'B', when taken in a closed container.
Both the components would evaporate and an equilibrium would be established between vapour phase
and liquid phase. Let the total vapour pressure at this stage be ptotal and pA and pB are partial pressures
of A and B. Mole fractions of these components in liquid solution are x A and xB, that of vapour phase
are yA and yB respectively p ºA & pBº are vapour pressure of pure A & pure B.
208
Column-II
Column-III
(P)
O
(iii) Non ideal solution
(negative deviation)
(c) Form azeotropic mixture
(iv) Immiscible liquids
(d) solution having vapours of
fixed composition
 A—B
interactions
9.
vapour
Mole fraction of A
1
(R) Ssurrounding = +ve
(S)
Ja
Mole fraction of A
(D) (i) (a) (P)
For water + ethanol → solution correct set is :
(A) (ii) (b) (Q)
(B) (ii) (c) (S)
(C) (iv) (d) (S)
(D) (iv) (c) (R)
For water + H2SO4 → solution correct set is
(A) (i) (a) (P)
(B) (ii) (b) (Q)
(C) (iii) (b) (R)
(D) (iv) (c) (P)
For a mixture of water and chlorobenzene correct set is
(A) (i) (a) (P)
(B) (ii) (b) (Q)
(C) (iii) (b) (R)
(D) (iv) (d) (Q)
º
PA
1
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a
10.
y
For Hexane + Heptane → solution correct set is :
(A) (i) (a) (R)
(B) (ii) (b) (P)
(C) (ii) (d) (R)
lp
8.
Liquid + vapour
V.P
(Q) PT = P°A + P°B
O
7.
Liquid
uh
A—A
(b) interactions
B—B
interactions
Vapour Pressure of A
(ii) Non ideal solution
(Positive deviation)
x
a
ri
(a) pT = pºA x A + pBº xB
(i) Ideal solution
Vapour Pressure
Column-I
* Marked Questions may have more than one correct option.
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
1.
2.
3.
In the depression of freezing point experiment, it is found that :
[JEE 1999, 3/80]
I. The vapour pressure of the solution is less than that of pure solvent.
II. The vapour pressure of the solution is more than that of pure solvent.
III. Only solute molecules solidify at the freezing point.
IV. Only solvent molecules solidify at the freezing point.
(A) I, II
(B) II, III
(C) I, IV
(D) I, II, III.
The van’t Hoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation is :
[JEE 1999, 3/80]
(A) 91.3%
(B) 87%
(C) 100%
(D) 74%
To 500 cm3 of water, 3.0 × 10–3 kg of acetic acid is added. If 23% of acetic acid is dissociated, what will
be the depression in freezing point? kf and density of water are 1.86 K kgmole–1 and 0.997 g cm–3
respectively
[JEE 2000, 3/35]
(A) 0.186 K
(B) 0.228 K
(C) 0.372 K
(D) 0.556 K
209
During depression of freezing point in a solution, the following are in equilibrium : [JEE 2003, 3/84]
(A) Liquid solvent-solid solvent
(B) Liquid solvent-solid solute
(B) Liquid solute-solid solute
(D) Liquid solute-solid solvent
5.
A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution of glucose at same temperature. The
apparent percentage dissociation of Na2SO4 is :
[JEE 2004, 3/84]
(A) 25%
(B) 50%
(C) 75%
(D) 85%
6.
1.22 g of benzoic acid is dissolved in 100 g of acetone and 100 g of benzene separately. Boiling point
of the solution in acetone increases by 0.17ºC, while that of, in the benzene increases by 0.13ºC ; K b for
acetone and benzene is 1.7 K kg mol–1 and 2.6 K kg mol–1. Find molecular weight of benzoic acid in two
cases and justify your answer.
[JEE 2004, 4/60]
7.
The elevation in boiling point of a solution of 13.44 g of CuCl 2 in 1kg of water using the following
information, will be (Molecular weight of CuCl2 = 134.4 and Kb = 0.52 K molal–1) : [JEE 2005, 3/84]
(A) 0.16
(B) 0.05
(C) 0.1
(D) 0.2
8.
When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol–1), a
freezing point depression of 2 K is observed. The van’t Hoff factor (i) is :
[JEE-2007, 3/162]
(A) 0.5
(B) 1
(C) 2
(D) 3
a
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4.
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Comprehension #
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when
solute molecules are added to get homogeneous solution. These are called colligative properties.
Application of colligative properties are very useful in day-to-day life. One of its example is the use of
ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles
A solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is 0.9
Given : Freezing point depression constant of water (Kfwater) = 1.86 K kg mol–1
Freezing point depression constant of ethanol (Kfethanol) = 2.0 K kg mol–1
Boiling point elevation constant of water (Kbwater) = 0.52 K kg mol–1
Boiling point elevation constant of ethanol (Kbethanol) = 1.2 K kg mol–1
Standard freezing point of water = 273 K
Standard freezing point of ethanol = 155.7 K
Standard boiling point of water = 373 K
Standard boiling point of ethanol = 351.5 K
Vapour pressure of pure water = 32.8 mm Hg
Vapour pressure of pure ethanol = 40 mm Hg
Molecular weight of water = 18 g mol–1
Molecular weight of ethanol = 46 g mol–1
In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be
non-volatile and non-dissociative.
9.
10.
The freezing point of the solution M is
(A) 268.7 K
(B) 268.5 K
(C) 234.2 K
[JEE 2008, 3/163]
(D) 150.9 K
The vapour pressure of the solution M is
(A) 39.3 mm Hg
(B) 36.0 mm Hg
(C) 29.5 mm Hg
[JEE 2008, 3/163]
(D) 28.8 mm Hg
11.
Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9. The
boiling point of this solution is
[JEE 2008, 3/163]
(A) 380.4 K
(B) 376.2 K
(C) 375.5 K
(D) 354.7 K
12.
The Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole
fraction of N2 in air is 0.8. The number of moles of N 2 from air dissolved in 10 moles of water of 298 K
and 5 atm pressure is :
[JEE 2009, 3/160]
(A) 4 × 10–4
(B) 4.0 × 10–5
(C) 5.0 × 10–4
(D) 4.0 × 10–6
13.
The freezing point (in ºC) of a solution containing 0.1 g of K 3[Fe(CN)6] (Mol. Wt. 329) in 100 g of water
(Kf = 1.86 K kg mol–1) is :
[JEE 2011, 3/180]
(A) – 2.3 × 10–2
(B) – 5.7 × 10–2
(C) – 5.7 × 10–3
(D) – 1.2 × 10–2
210
For a dilute solution containing 2.5 g of a non- volatile non- electrolyte solute in 100 g of water, the
elevation in boiling point at 1 atm pressure is 2ºC. Assuming concentration of solute is much lower than
the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take K b = 0.76 K kg
mol–1)
[IIT 2012, 3/136]
(A) 724
(B) 740
(C) 736
(D)718
15.*
Benzene and naphthalene form an ideal solution at room temperature. For this process, the true
statement(s) is (are) :
[JEE(Advanced) 2013, 4/120]
(A) G is positive
(B) Ssystem is positive
(C) Ssurroundings = 0
(D) H = 0
16.
MX2 dissociates into M2+ and X– ions in an aqueous solution, with a degree of dissociation () of 0.5.
The ratio of the observed depression of freezing point of the aqueous solution to the value of the
depression of freezing point in the absence of ionic dissociation is
[JEE(Advanced) 2014, 3/120]
17.
If the freezing point of a 0.01 molal aqueous solution of a cobalt(III) chloride-ammonia complex (which
behaves as a strong electrolyte) is –0.0558ºC, the number of chloride(s) in the coordination sphere of
the complex is [Kf of water = 1.86 K kg mol–1]
[JEE(Advanced) 2015 4/168]
18. *
Mixture(s) showing positive deviation from Raoult's law at 35ºC is(are) [JEE(Advanced) 2016, 4/124]
(A) carbon tetrachloride + methanol
(B) carbon disulphide + acetone
(C) benzene + toluene
(D) phenol + aniline
19.*
For a solution formed by mixing liquids L and M, the vapour pressure
of L plotted against the mole fraction of M in solution is shown in the
following figure. Here xL and xM represent mole fractions of L and M,
Z
respectively, in the solution. The correct statement(s) applicable to this
system is (are)
[JEE(Advanced) 2017, 4/122]
pL
(A) The point Z represents vapour pressure of pure liquid M and Raoult's
law is obeyed from xL = 0 to xL = 1.
(B) Attractive intermolecular interactions between L-L in pure liquid L
and M-M in pure liquid M are stronger than those between L-M when
mixed in solution
0
1
XM
(C) The point Z represents vapour pressure of pure liquid M and
Raoult's law is obeyed when xL → 0
(D) The point Z represents vapour pressure of pure liquid L and Raoult's law is obeyed when xL → 1
20.
Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the
freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol –1. The
figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [Molecular
weight of ethanol is 46 g mol–1]
[JEE(Advanced) 2017, 3/122]
Among the following, the option representing change in the freezing point is
Sa
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a
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Ja
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a
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14.
Water
Ice
Water + Ethanol
V.P./bar
270 273
(C)
(B)
1
Ice
Water + Ethanol
271 273
(D)
Water + Ethanol
270 273
1
T/K
Water
Ice
V.P./bar
1
T/K
V.P./bar
(A)
V.P./bar
Water
1
T/K
Water
Ice
Water + Ethanol
271 273
T/K
211
21.
Liquids A and B form ideal solution over the entire range of composition. At temperature T, equimolar
binary solution of liquids A and B has vapour pressure 45 Torr. At the same temperature, a new
solution of A and B having mole fractions xA and xB, respectively, has vapour pressure of 22.5 Torr. The
value of xA / xB in the new solution is ____.
[JEE(Advanced) 2018, 3/120]
(given that the vapour pressure of pure liquid  is 20 Torr at temperature T)
22.
The plot given below shows P–T
curves (where P is the pressure
and T is the temperature) for two
solvents X and Y and isomolal 760
solutions of NaCl in these
solvents.
NaCl
completely
dissociates in both the solvents.
On addition of equal number of
moles of a non-volatile solute S in
equal amount (in kg) of these
solvents, the elevation of boiling
point of solvent X is three times
that of solvent Y. Solute S is
known to undergo dimerization in
these solvents. If the degree of
dimerization is 0.7 in solvent Y,
the degree of dimerization in
solvent X is ____.
[JEE(Advanced) 2018, 3/120]
1
2
3
4
Pressure(mmHg)
1. Solvent X
a
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2. Solution of NaCl in solvent X
3. Solvent Y
uh
367
368
362
360
4. Solution of NaCl in solvent Y
Ja
Temperature (K)
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
lp
JEE(MAIN) OFFLINE PROBLEMS
6.02 × 1020 molecules of urea are present in 100 ml of its solution. The concentration of urea solution is:
[AIEEE-2004, 3/225]
(1) 0.001 M
(2) 0.01 M
(3) 0.02 M
(4) 0.1 M.
2.
Which one of the following aqueous solutions will exhibit highest boiling point ? [AIEEE-2004, 3/225]
(1) 0.01 M Na2SO4
(2) 0.01 M KNO3
(3) 0.015 M urea
(4) 0.015 M glucose
3.
If  is the degree of dissociation of Na2SO4, the vant Hoff’s factor (i) used for calculating the molecular
mass is :
[AIEEE-2005, 1½/225]
(1) 1 + 
(2) 1 – 
(3) 1 + 2
(4) 1 – 2.
Sa
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a
1.
4.
Equimolar solutions in the same solvent have :
(1) same boiling point but different freezing point
(2) same freezing point but different boiling point
(3) same boiling and same freezing points
(4) differnet boiling and freezing points
[AIEEE-2005, 3/225]
5.
Two solutions of a substance (non electrolyte) are mixed in the following manner. 480 ml of 1.5 M first
solution + 520 mL of 1.2 M second solution. What is the molarity of the final mixture?
[AIEEE-2005, 3/225]
(1) 1.20 M
(2) 1.50 M
(3) 1.344 M
(4) 2.70 M
6.
Benzene and toluene form nearly ideal solutions. At 20ºC, the vapour pressure of benzene is 75 torr
and that of toluene is 22 torr. The partial vapour pressure of benzene at 20ºC for a solution containing
78 g of benzene and 46 g of toluene in torr is :
[AIEEE-2005, 3/225]
(1) 50
(2) 25
(3) 37.5
(4) 53.5
7.
Density of a 2.05 M solution of acetic acid in water is 1.02 g/mL. The molality of the solution is
[AIEEE-2005, 3/165]
(1) 3.28 mol Kg–1
(2) 2.28 mol Kg–1
(3) 0.44 mol Kg–1
(4) 1.14 mol Kg–1
212
8.
A mixture of ethyl alcohol and propyl alcohol has a vapour pressure of 290 mm at 300 K. The vapour
pressure of propyl alcohol is 200 mm. If the mole fraction of ethyl alcohol is 0.6, its vapour pressure (in
mm) at the same temperature will be :
[AIEEE-2007, 3/120]
(1) 700
(2) 360
(3) 350
(4) 300
9.
A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60g mol –1) in the
same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm–3, molar mass
of the substance will be
[AIEEE-2007, 3/120]
(1) 105.0 g mol–1
(2) 210.0 g mol–1
(3) 90.0 g mol–1
(4) 15.0 g mol–1
The vapour pressure of water at 20ºC is 17.5 mm Hg. If 18 g of glucose (C 6H12O6) is added to 178.2 g
of water at 20°C, the vapour pressure of the resulting solution will be :
[AIEEE-2008, 3/105]
(1) 15.750 mm Hg
(2) 16.500 mm Hg
(3) 17.325 mm Hg
(4) 17.675 mm Hg
10.
At 80ºC, the vapour pressure of pure liquid 'A' is 520 mm Hg and that of pure liquid 'B' is 1000 mm Hg.
If a mixture solution of 'A' and 'B' boils at 80º C and 1 atm pressure, the amount of 'A' in the mixture is
(1 atm = 760 mm Hg)
[AIEEE-2008, 3/105]
(1) 34 mol percent
(2) 48 mol percent
(3) 50 mol percent
(4) 52 mol percent
12.
A binary liquid solution is prepared by mixing n-heptane and ethanol. Which one of the following
statement is correct regarding the behaviour of the solution ?
[AIEEE-2009, 4/144]
(1) The solution is non-ideal, showing +ve deviation from Raoult’s Law.
(2) The solution in non-ideal, showing –ve deviation from Raoult’s Law.
(3) n-heptane shows +ve deviation while ethanol shows –ve deviation from Raoult’s Law.
(4) The solution formed is an ideal solution.
13.
Two liquids X and Y form an ideal solution. At 300K, vapour pressure of the solution containing 1 mol of
X and 3 mol of Y is 550 mmHg. At the same temperature, if 1 mol of Y is further added to this solution,
vapour pressure of the solution increases by 10 mmHg. Vapour pressure (in mmHg) of X and Y in their
pure states will be, respectively :
[AIEEE-2009, 8/144]
(1) 300 and 400
(2) 400 and 600
(3) 500 and 600
(4) 200 and 300
14.
If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous
solution, the change in freezing point of water (Tf), when 0.01 mole of sodium sulphate is dissolved in
1 kg of water, is (Kf = 1.86 K kg mol–1)
[AIEEE-2010, 4/144]
(1) 0.0372 K
(2) 0.0558 K
(3) 0.0744 K
(4) 0.0186 K
15.
On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid
components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of the
solution obtained by mixing 25.0 g of heptane and 35 g of octane will be (molar mass of heptane = 100
g mol–1 and of octane = 114 g mol–1)
[AIEEE-2010, 4/144]
(1) 72.0 kPa
(2) 36.1 kPa
(3) 96.2 kPa
(4) 144.5 kPa
a
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a
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11.
Kf for water is 1.86 K kg mol–1. If your automobile radiator holds 1.0 kg of water, how may grams of
ethylene glycol (C2H6O2) must you add to get the freezing point of the solution lowered to –2.8ºC ?
[AIEEE 2012, 4/120]
(1) 72 g
(2) 93 g
(3) 39 g
(4) 27 g
17.
Consider separate solution of 0.500 M C2H5OH(aq), 0.100 M Mg3(PO4)2(aq), 0.250 M KBr(aq) and
0.125 M Na3PO4(aq) at 25C. Which statement is true about these solution, assuming all salts to be
strong electrolytes ?
[AIEEE 2014, 4/120]
(1) They all have the same osmotic pressure.
(2) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure.
(3) 0.125 M Na3PO4(aq) has the highest osmotic pressure.
(4) 0.500 M C2H5OH(aq) has the highest osmotic pressure.
18.
The vapour pressure of acetone at 20ºC is 185 torr. When 1.2 g of a non-volatile substance was
dissolved in 100 g of acetone at 20ºC, its vapour pressure was 183 torr. The molar mass (g mol –1) of
the substance is:
[JEE(Main) 2015, 4/120]
(1) 32
(2) 64
(3) 128
(4) 488
19.
18 g glucosse (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this
aqueous solution is.
[JEE(Main) 2016, 4/120]
(1) 76.0
(2) 752.4
(3) 759.0
(4) 7.6
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16.
213
20.
The freezing point of benzene decreases by 0.45ºC when 0.2 g of acetic acid is added to 20 g of
benzene. If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in
benzene will be : (Kf for benzene = 5.12 K kg mol–1)
[JEE(Main) 2017, 4/120]
(1) 80.4%
(2) 74.6%
(3) 94.6%
(4) 64.6%
21.
For 1 molal aqueous solution of the following compounds, which one will show the highest freezing
point ?
[JEE(Main) 2018, 4/120]
(1) [Co(H2O)4Cl2]Cl.2H2O
(2) [Co(H2O)3Cl3].3H2O
(3) [Co(H2O)6]Cl3
(4) [Co(H2O)5Cl]Cl2.H2O
a
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JEE(MAIN) ONLINE PROBLEMS
Choose the correct statement with respect to the vapour pressure of a liquid among the following :
[JEE(Main) 2014 Online (19-04-14), 4/120]
(1) Increases linearly with increasing temperature
(2) Increase non-linearly with increasing temperature
(3) Decreases linearly with increasing
(4) Decreases non-linearly with increasing temperature
2.
The observed osmotic pressure for a 0.10 M solution of Fe(NH 4)2(SO4)2 at 25ºC is 10.8 atm
experimental (observed) and theoritical values of Van't Hoff factor (i) will be respectively : (R = 0.082 L
atm k–1 mol–1)
[JEE(Main) 2014 Online (19-04-14), 4/120]
(1) 5 and 4.42
(2) 4 and 4.00
(3) 5 and 3.42
(4) 3 and 5.42
3.
For an ideal solution of two components A and B, which of the following is true ?
[JEE(Main) 2014 Online (19-04-14), 4/120]
(1) Hmixing < 0 (zero)
(2) Hmixing > 0 (zero)
(3) A–B interaction is stronger than A–A and B–B interactions
(4) A–A, B–B and A–B interactions are identical
4.
A solution at 20ºC is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of
pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the
total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be,
respectively :
[JEE(Main) 2015 Online (10-04-15), 4/120]
(1) 35.0 torr and 0.480
(2) 30.5 torr and 0.389
(3) 38.0 torr and 0.589
(4) 35.8 torr and 0.280
5.
Determination of the molar mass of acetic acid in benzene using freezing point depression is affected
by :
[JEE(Main) 2015 Online (10-04-15), 4/120]
(1) partial ionization
(2) dissociation
(3) complex formation
(4) association
Sa
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Ja
uh
1.
6.
The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L –1. The solubility (in g L–1)
at 750 torr partial pressure is :
[JEE(Main) 2016 Online (09-04-16), 4/120]
(1) 0.02
(2) 0.015
(3) 0.0075
(4) 0.005
7.
An aqueous solution of a salt MX2 at certain temperature has a van’t Hoff factor of 2. The degree of
dissociation for this solution of the salt is :
[JEE(Main) 2016 Online (10-04-16), 4/120]
(1) 0.67
(2) 0.33
(3) 0.80
(4) 0.50
8.
5 g of Na2SO4 are was dissolved in x g of H2O. The change in freezing point was found to be 3.82ºC. If
Na2SO4 is 81.5% ionised, the value of x (Kf for water = 1.86ºC kg mol–1) is approximately :
(molar mass of S = 32 g mol–1 and that of Na = 23 g mol–1) [JEE(Main) 2017 Online (08-04-17), 4/120]
(1) 45 g
(2) 65 g
(3) 25 g
(4) 15 g
9.
A solution is prepared by mixing 8.5 g of CH2Cl2 and 11.95 g of CHCl3. If vapour pressure of CH2Cl2
and CHCl3 at 298 K are 415 and 200 mmHg respectively, the mole fraction of CHCl3 in vapour form is :
(Molar mass of Cl = 35.5 g mol–1)
[JEE(Main) 2017 Online 09-04-17), 4/120]
(1) 0.675
(2) 0.162
(3) 0.486
(4) 0.325
214
Two 5 molal solutions are prepared by dissolving a non-electrolyte non-volotatile solute separately in
the solvents X and Y. The molecular weights of the solvents are M X and MY, respectively where MX =
3
MY. The relative lowering of vapour pressure of the solution in X is "m" times that of the solution in Y.
4
Given that the number of moles of solute is very small in comparison to that of solvent, the value of "m"
is :
[JEE(Main) 2018 Online 15-04-18), 4/120]
4
3
1
1
(1)
(2)
(3)
(4)
3
2
4
4
11.
The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol–1) needed to be dissolved in
114 g octane to reduce its vapour pressure to 75 %, is :
[JEE(Main) 2018 Online 16-04-18), 4/120]
(1) 37.5 g
(2) 75 g
(3) 150 g
(4) 50 g
12.
Which one of the following statements regarding Henry's law is not correct ?
[JEE(Main) 2019 Online 09-01-19), 4/120]
(1) Different gases have different KH (Henry' law constant) values at the same temperature.
(2) The value of KH increases with increase of temperature and KH is function of the nature of the gas
(3) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the
solution.
(4) Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids.
13.
A solution containing 62 g ethylene glycol in 250 g water is cooled to –10ºC. If Kf for water is 1.86 K kg
mol–1, the amount of water (in g) separated as ice is:
[JEE(Main) 2019 Online 09-01-19), 4/120]
(1) 16
(2) 32
(3) 48
(4) 64
14.
Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapor pressures of
pure A and pure B are 7 × 103 Pa and 12 × 103 Pa, respectively. The composition of the vapor in
equilibrium with a solution containing 40 mole percent of A at this temperature is :
[JEE(Main) 2019 Online 10-01-19), 4/120]
(1) xA = 0.28 ; xB = 0.72
(2) xA = 0.76 ; xB = 0.24
(3) xA = 0.37 ; xB = 0.63
(4) xA = 0.4 ; xB = 0.6
15.
Elevation in the boiling point for 1 molal solution of glucose is 2 k. The depression in the freezing point
for 2 molal solution of glucose in the same solvent is 2 k. the relation between k b and Kf is :
[JEE(Main) 2019 Online 10-01-19), 4/120]
(1) Kb= 0.5Kf
(2) Kb =Kf
(3) Kb = 1.5 Kf
(4) kb = 2 kf
16.
The freezing point of a diluted milk sample is found to be –0.2°C, while it should have been
–0.5°C for pure milk. How much water has been added to pure milk to make the diluted sample ?
[JEE(Main) 2019 Online 11-01-19), 4/120]
(1) 1 cup of water to 3 cups of pure milk
(2) 1 cup of water of 2 cups of pure milk
(3) 2 cups of water to 3 cups of pure milk
(4) 3 cups of water to 2 cups of pure milk
Sa
nk
a
lp
Ja
uh
a
ri
10.
17.
K2HgI4 is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is :
[JEE(Main) 2019 Online 11-01-19), 4/120]
(1) 1.8
(2) 1.6
(3) 2.0
(4) 2.2
18.
Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If
molecular weight of X is A, then molecular weight of Y is: [JEE(Main) 2019 Online 12-01-19), 4/120]
(1) 4A
(2) 2A
(3) 3A
(4) A
19.
Molecules of benzoic acid (C6H5COOH) dimerise in benzene. ‘w’ g of the acid dissolved in 30 g of
benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to
form dimer in the solution is 80, then w is :
[JEE(Main) 2019 Online 12-01-19), 4/120]
(Given that Kf = 5 kg mol–1, Molar mass of benzoic acid = 122 g mol–1)
(1) 2.4 g
(2) 1.8 g
(3) 1.0 g
(4) 1.5 g
215
EXERCISE - 1
PART - I
Some of the characteristics of supersaturated solution are given below
(i) If a crystal of solute is added to supersaturated solution, crystallisation occurs rapidly.
(ii) Supersaturated solutions contain more solute than they should have at a particular temperature.
A-2.
Certain compounds combine with the moisture of atmosphere and are converted into hydroxides or
hydrates. Such substances are called hygroscopic. e.g., anhydrous CuSO 4, quick lime (CaO),
anhydrous Na2CO3 etc.
A-3.
The overall energy change associated with dissolution depends on the relative magnitude of the solute–
solute, solvent–solvent and solute–solvent interactions. The process is exothermic if the new interaction
release more energy than disrupting the old interactions requires, it is endothermic if opposite is true.
B-1.
0.04 g/mL
B-3.
Molality = 11.44 m, Molarity = 7.55 M
C-1.
PO2 = 810 mm Hg, PH2O = 355 mm Hg , Ptotal = 1165 mm Hg
D-1.
7.62 bar
F-1.
pA0 = 400 mm of Hg, pB0 = 600 mm of Hg
F-3.
PB0 = 0.7 atm & PA0 = 1.9 atm
F-4.
14.16 mole percent benzene
F-5.
Y’B = 0.932.
G-2.
There is positive deviation from Raoult’s law, S > 0.
G-1.
4 mmol.
V < 80ml.
H-1.
5.15 g
2.4 atm
E-2.
WW
= 3.973
WN
F-2.
27.3 mole %
Degree of dissociation /association
n
i
1
1
0.2
0.5
No association or dissociation
0.8
1
2
3
2
0.5
Not defined
2
0.33
2
3
1.2
0.75
1
1.8
0.33
lp
KCl
H2SO4
CH3COOH (in water)
CH3COOH (in benzene)
Urea
NaBr
A
Dissociation/association
reaction
KCl → K+ + Cl–
H2SO4 → 2H+ + SO42–
CH3COOH → H+ + CH3COO–
2CH3COOH → (CH3COOH)2
No association or dissociation
NaBr → Na+ + Br–
3A → A3
a
Solute
E-1.
C-2.
uh
D-2.
(i) 30%, (ii) 0.046 (iii) Na+ = 5.42 m, S2O32– = 2.71 m
Ja
B-2.
a
ri
A-1.
75%
I-1.
(a) 60 g/mol, (b) 333.3 g
J-1.
(a) M = 94.52, (b) m = 135
J-4.
1.075, 7.5.
K-1.
(a) Urea < NaCl < Na2SO4 < Na3PO4
K-2.
 = 0.81
K-3.
50% K2SO4
K-4.
A-1.
(B)
A-2.
(B)
B-1.
(B)
C-1.
(B)
C-2.
(C)
C-3.
(B)
C-4.
(B)
D-1.
(C)
D-2.
(A)
D-3.
(B)
D-4.
(D)
D-5.
(B)
E-1.
(C)
F-1.
(B)
F-2.
(C)
F-3.
(C)
F-4.
(A)
F-5.
(A)
G-1.
(C)
G-2.
(A)
G-3.
(A)
G-4.
(C)
G-5.
(A)
G-6.
(C)
H-1.
(C)
H-2.
(A)
H-3.
(B)
H-4.
(B)
I-1.
(B)
I-2.
(B)
I-3.
(B)
I-4.
(C)
I-5.
(B)
I-6.
(A)
J-1.
(D)
J-2.
(B)
J-3.
(A)
J-4.
(A)
J-5.
(C)
J-6.
(D)
J-7.
(A)
J-8.
(B)
J-9.
(B)
K-1.
(C)
K-2.
(A)
K-3.
(A)
K-4.
(A)
Sa
nk
H-2.
J-2.
232
l-2.
746.2 mm of Hg
J-3.
S8
(b) 6.15 atm
Volume must have been made 5 times
PART - II
216
PART - III
1.
(A) – (r); (B) – (s); (C) – (q); (D) – (p)
2.
(A) – (p,s,r); (B) – (p,q,t); (C) – (p); (D) – (p,q,t)
EXERCISE - 2
PART - I
1.
(C)
2.
(D)
3.
(B)
4.
(C)
5.
(D)
6.
(D)
7.
(B)
8.
(B)
9.
(B)
10.
(A)
11.
(B)
12.
(A)
13.
(D)
14.
(B)
1.
18
2.
5
3.
21
4.
20
5.
27
6.
4 (A,B,C,D)
7.
2
8.
48
9.
43
10.
2
11.
60
12.
60
13.
4
1.
(ABD)
2.
(ABCD)
3.
(ABD)
4.
(ABCD)
5.
(AC)
6.
(A)
7.
(AB)
8.
(B)
9.
(AB)
10.
(B)
11.
(ABC)
12.
(BD)
13.
(AB)
16.
(ABC)
1.
(B)
2.
(C)
3.
6.
(A)
7.
(D)
8.
PART - IV
14.
(AD)
15.
(BCD)
4.
(D)
5.
(B)
9.
(C)
10.
(D)
Ja
(B)
uh
PART - III
a
ri
PART - II
(B)
EXERCISE - 3
PART - I
2.
(B)
6.
122, 224
7.
(A)
11.
(B)
12.
(A)
16.
2
17.
(1)
21.
19
22.
3.
(B)
4.
(A)
5.
(C)
8.
(A)
9.
(D)
10.
(B)
13.
(A)
14.
(A)
15.
(BCD)
18.
(AB)
19.
(BD)
20.
(A)
(3)
5.
(3)
lp
(C)
a
1.
Sa
nk
0.05
PART - II
JEE(MAIN) OFFLINE PROBLEMS
3.
(3)
4.
1.
(2)
2.
(1)
6.
(1)
7.
(2)
8.
(3)
9.
(2)
10.
(3)
11.
(3)
12.
(1)
13.
(2)
14.
(2)
15.
(1)
16.
(2)
17.
(1)
18.
(2)
19.
(2)
20.
(3)
21.
(2)
JEE(MAIN) ONLINE PROBLEMS
1.
(2)
2.
(1)
3.
(4)
4.
(3)
5.
(4)
6.
(2)
7.
(4)
8.
(1)
9.
(4)
10.
(2)
11.
(Bonus)
12.
(4)
13.
(4)
14.
(1)
15.
(4)
16.
(4)
17.
(1)
18.
(3)
19.
(1)
217

Marked questions are recommended for Revision.
This Section is not meant for classroom discussion. It is being given to promote selfstudy and self testing amongst the students.
PART - I : PRACTICE TEST-1 (IIT-JEE (MAIN Pattern))
Max. Marks : 120
Max. Time : 1 Hr.
5.
The test is of 1 hour duration.
The Test Booklet consists of 30 questions. The maximum marks are 120.
Each question is allotted 4 (four) marks for correct response.
Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each
question.
¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction
from the total score will be made if no response is indicated for an item in the answer sheet.
There is only one correct response for each question. Filling up more than one response in any question
will be treated as wrong response and marks for wrong response will be deducted accordingly as per
instructions 4 above.
uh
1.
2.
3.
4.
a
ri
Important Instructions %
Mole fraction of C3H5(OH)3 in a solution of 36 g of water and 46 g of glycerine is :
(1) 0.46
(2) 0.36
(3) 0.20
(4) 0.40
2.
Colligative properties have many practical uses, some of them may be :
I : Melting of snow by salt
II : Desalination of sea water
III : Determination of molar mass
IV : Determination of melting point and boiling point of solvent
Actual practical uses are :
(1) I, II
(2) III, IV
(3) I, II, III
(4) II, III, IV
3.
Select correct statement(s) :
(1) When solid CaCl2 is added to liquid water, the boiling temperature rises
(2) When solid CaCl2 is added to ice at 0°C, the freezing temperature falls
(3) Both (1) and (2)
(4) None of the above
a
lp
Ja
1.
If pKa = – log Ka = 4, and Ka = Cx2 then Van’t Hoff factor for weak monobasic acid when C = 0.01 M is :
(1) 1.01
(2) 1.02
(3) 1.10
(4) 1.20
5.
Consider following terms (m = molality) :
Tb
I : mKb ;
II : mKbi
III :
IV : Kb
i
Terms which can be expressed in degree (temperature) are
(1) III, IV
(2) I, II
(3) I, II, IIII
Sa
nk
4.
(4) I, III
6.
Elevation in b.p. of an aqueous urea solution is 0.52°. (Kb = 0.52° mol–1 kg) Hence, mole-fraction of
urea in this solution is :
(1) 0.982
(2) 0.567
(3) 0.943
(4) 0.018
7.
Insulin (C2H10O5)n is dissolved in a suitable solvent and the osmotic pressure  of the solution of various
concentration (in kg/m3) is measured at 20ºC. The slope of a plot of  against c is found to be 8.134 ×
10–3 (SI units) The molecular weight of the insulin (in kg/mol) is :
(1) 4.8 × 105
(2) 9 × 105
(3) 293 × 103
(4) 8.314 × 105
218
8.
What is the normal boiling point of the solution represented by the phase diagram ?
(1) A
(2) B
(3) C
(4) D
An aqueous solution of a solute AB has b.p. of 101.08°C (AB is 100% ionised at boiling point of the
solution) and freezes at – 1.80°C. Hence, AB (Kb / Kf = 0.3)
(1) is 100% ionised at the f.p. of the solution
(2) behaves as non-electrolyte at the f.p. of the solution
(3) forms dimer
(4) none of the above
10.
Density of 1M solution of a non-electrolyte C6H12O6 is 1.18 g/mL. If Kf (H2O) is 1.86° mol–1 kg, solution
freezes at :
(1) – 1.58°C
(2) – 1.86°C
(3) – 3.16°C
(4) 1.86°C
11.
Mole fraction of a non-electrolyte in aqueous solution is 0.07. If Kf is 1.86° mol–1 kg, depression in f.p.,
Tf, is:
(1) 0.26°
(2) 1.86°
(3) 0.13°
(4) 7.78°
12.
What is the normal freezing point of the solution represented by the phase diagram ?
(3) T3
(4) T0
Total vapour pressure of mixture of 1 mol of volatile component A (p A° = 100 mmHg) and 3 mol of
volatile component B (pB° = 60 mmHg) is 75 mm. For such case :
(1) there is positive devitation from Raoult’s low
(2) boiling point has been lowered
(3) force of attraction between A and B is smaller than that between A and A or between B and B
(4) all the above statements are correct
Sa
nk
13.
(2) T2
a
(1) T1
lp
Ja
uh
a
ri
9.
14.
A colligative property of a solution depends on the :
(1) arrangement of atoms in solute molecule
(2) total number of molecules of solute and solvent
(3) number of molecules of solute in solution
(4) mass of the solute molecules
15.
Which has maximum freezing point ?
(1) 6g urea solution in 100 g H2O
(3) 6g sodium chloride in 100 g H2O
16.
(2) 6g acetic acid solution in 100g H2O
(4) All have equal freezing point
Select correct statements :
(1) The fundamental cause of all colligative properties is the higher entropy of the solution relative to
that of the pure solvent
(2) The freezing point of hydrofluoride solution is larger than that of equimolal hydrogen chloride
solution
(3) 1M glucose solution and 0.5 M NaCl solution are isotonic at a given temperature
(4) All are correct statements
219
The vapour pressure of a pure liquid A is 40 mmHg at 310 K. The vapour pressure of this liquid in a
solution with liquid B is 32 mmHg. Mole fraction of A in the solution, if it obeys Raoult’s law is :
(1) 0.8
(2) 0.5
(3) 0.2
(4) 0.4
18.
Mole fraction of the toluene in the vapour phase which is in equilibrium with a solution of benzene
(p° = 120 Torr) and toluene (p° = 80 Torr) having 2.0 mol of each is
(1) 0.50
(2) 0.25
(3) 0.60
(4) 0.40
19.
An azeotropic solution of two liquids has a boiling point lower than either of them when it :
(1) shows negative deviation from Raoult’s law (2) shows positive deviation from Raoult’s low
(3) shows ideal behaviour
(4) is saturated
20.
Depression of freezing point of 0.01 molal aq. CH3COOH solution is 0.02046°. 1 molal urea solution
freezes at – 1.86°C. Assuming molality equal to molarity, pH of CH3COOH solution is :
(1) 2
(2) 3
(3) 3.2
(4) 4.2
21.
Which of the following azeotropic solutions has the boiling point more than boiling point of the
constituents A and B ?
(1) CH3CH2OH and CH3COCH3
(2) CS2 and CH3COCH3
(3) CHCl3 and CH3COCH3
(4) CH3CHO and CS2
22.
A 0.50 molal solution of ethylene glycol in water is used as coolant in a car. If the freezing point
constant of water is 1.86° per molal, at which temperature will the mixture freeze ?
(1) 1.56°C
(2) – 0.93°C
(3) – 1.86°C
(4) 0.93°C
23.
The depression of freezing points of 0.05 molal aqueous solution of the following compounds are
measured.
1. NaCl
2. K2SO4
3. C6H12O6
4. Al2(SO4)3
Which one of the above compounds will exhibit the maximum depression of freezing point ?
(1) 3
(2) 2
(3) 4
(4) 1
24.
On mixing 10 mL of acetone with 40 mL of chloroform , the total volume of the solution is :
(1) < 50 mL
(2) > 50 mL
(3) = 50 mL
(4) Cannot be predicted
25.
Select correct statement ?
(1) Heats of vaporisation for a pure solvent and for a solution are similar because similar intermolecular
forces between solvent molecules must be overcome in both cases
(2) Entropy change between solution and vapour is smaller than the entropy change between pure
solvent and vapour
(3) Boiling point of the solution is larger than that of the pure solvent
(4) All are correct statements
26.
What will be the molecular weight of NaCl determined experimentally from elevation in the boiling point
or depression in freezing point method ?
(1) < 58.5
(2) > 58.5
(3) = 58.5
(4) None of these
Sa
nk
a
lp
Ja
uh
a
ri
17.
27.
Which characterises the weak intermolecular forces of attraction in a liquid ?
(1) High boiling point
(2) High vapour pressure
(3) High critical temperature
(4) High heat of vaporization
28.
On the basis of intermolecular forces predict the correct order of decreasing boiling points of the
compounds:
(1) CH3OH > H2 > CH4 (2) CH3OH > CH4 > H2 (3) CH4 > CH3OH > H2 (4) H2 > CH4 > CH3OH
29.
During depression of freezing point in a solution the following are in equilibrium :
(1) Liquid solvent, solid solvent
(2) Liquid solvent, solid solute
(3) Liquid solute, solid solute
(4) Liquid solute, solid solvent
30.
Which of the following liquid pairs shows a positive deviation from Raoult’s law ?
(1) Acetone – chloroform
(2) Benzene – methanol
(3) Water – nitric acid
(4) Water – hydrochloric acid
220
Practice Test-1 (IIT-JEE (Main Pattern))
OBJECTIVE RESPONSE SHEET (ORS)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
Que.
Que.
Ans.
a
ri
Ans.
PART - II : NATIONAL STANDARD EXAMINATION IN CHEMISTRY (NSEC) STAGE-I
Solutions having the same osmotic pressure are called
(A) isotonic solutions
(B) molar solutions
(C) hypotonic solutions
(D) ideal solutions
2.
A colligative property of a solution depends on the
[NSEC-2000]
(A) arrangement of atoms in solute molecule.
(B) total number of molecules of solute and solvent
(C) number of molecules of solute in solution.
(D) mass of the solute molecules.
3.
Consider 1 M solutions of the following salts. State which solution will have the lowest freezing point.
[NSEC-2000]
(A) Na2SO4
(B) BaCI2
(C) NaCI
(D) AI(NO3)3
4.
When 0.6 g of urea dissolved in 100 g of water, the water will boil at (K b for water = 0.52 kJ. mol–1 and
normal boiling point of water = 100ºC) :
[NSEC-2001]
(A) 373.052 K
(B) 273.52 K
(C) 372.48 K
(D) 273.052 K
5.
The osmotic pressure of a solution is given by the equation :

CR
CT
(A) p =
(B)
= RT
(C) p =
R
T
C
Ja
lp
[NSEC-2001]
(D) p =
C.
R
T
a
The relative lowering of vapour pressure is equal to the mole fraction of the solute. This is the statement
of:
[NSEC-2001]
(A) Raoult’s law
(B) Boyle’s law
(C) Osmotic pressure law
(D) Graham’s law
Sa
nk
6.
[NSEC-2000]
uh
1.
7.
Azeotropes are :
(A) liquid mixtures which distil unchanged in composition
(B) liquids which can mix with each other in all proportions
(C) solids which form solid solutions of definite compositions
(D) gases which can be separated.
[NSEC-2002]
8.
Swimming for a long time in salt water makes the skin of one’s finger tips wrinkled. Which one of the
following properties is responsible for this observation ?
[NSEC-2002]
(A) osmosis
(B) dialysis
(C) electrodialysis
(D) coagulation.
9.
Which gas is mixed with oxygen by sea-divers at the high underwater pressure ?
(A) Nitrogen
(B) Neon
(C) Helium
(D) Argon.
10.
A 2% solution of glucose has the same elevation in the boiling point as that of a 5% solution of a nonvolatile solute. The molar mass of the solute is :
[NSEC-2003]
(A) 180
(B) 450
(C) 72
(D) 18
11.
For water, the maximum number of phases that can be in equilibrium is
(A) 1
(B) 2
(C) 3
(D) 4.
[NSEC-2002]
[NSEC-2004]
221
12.
The boiling point of a glucose solution is higher than that of water because
(A) glucose forms extensive hydrogen bonding with water
(B) glucose does not dissociate in water
(C) its vapour pressure is higher than that of water at a given temperature
(D) its vapour pressure is lower than that of water at a given temperature.
[NSEC-2004]
13.
A 1.0 molal solution with the lowest freezing point is that of :
(A) FeCI3
(B) HCI
(C) KCI
[NSEC-2004]
(D) MgCI2.
14.
From among the following, the aqueous solution which has the highest freezing point depression is :
[NSEC-2005]
(A) 0.1 M Sr(NO3)2
(B) 0.1 M KCI
(C) 0.1 M HNO3
(D) 0.1 M glucose.
15.
In chemical industries, the preferred method of purification of liquids is :
(A) differential extraction
(B) fractional distillation
(C) chromatography
(D) leaching.
16.
The solubility of a gas in a liquid is driectly proportional to the partial pressure of the gas over the
solution. This statement is known as:
[NSEC-2007]
(A) Raoult’s law
(B) Henry’s law
(C) Boyle’s law
(D) Charles’ and Gay Lussac’s Law
17.
Which of the following is not a colligative property ?
(A) solubility.
(B) vapor pressure lowering.
(C) boiling point elevation.
(D) osmotic pressure.
[NSEC-2007]
18.
Which of the following has the lowest freezing point and the highest boiling point?
(A) 1.5 m magnesium phosphate
(B) 1.0 m sodium chloride
(C) 1.5 m aluminum nitrate
(D) 1.5 m calcium chloride
[NSEC-2007]
19.
A solution of urea was found to be isotonic with a solution of salt XY of molecular weight 74.6. If 0.15
moles of urea are dissolved in a certain volume V mL of the isotonic solution, the amount of salt in the
solution will be :
[NSEC-2008]
(A) 22.4g
(B) 5.6 g
(C) 11.2 g
(D) 7.46 g
20.
The desalination of sea water involves the phenomenon of :
(A) Sedimentation
(B) Distillation
(C) Precipitation
21.
According to this phase diagram, which phases can exist at pressures lower than the triple point
pressure ?
[NSEC-2008]
(D) Reverse osmosis
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[NSEC-2006]
(A) gas only
(C) liquid only
(B) solid and gas only
(D) solid and liquid only
22.
For a dilute solution, Raoult's law states that :
[NSEC-2009]
(A) the lowering of vapour pressure is equal to the mole fraction of the solute.
(B) the relative lowering of vapour pressure is equal to the mole fraction of the solute
(C) the vapour pressure of solution is equal to the mole fraction of solution.
(D) the relative lowering of vapour pressure is proportional to the amount of solute in the solution.
23.
20 g of Compound X are dissolved in 500 mL of water and the osmotic pressure of the resulting
solution is 500mm of Hg of 10°C. The average molecular mass of X is
[NSEC-2010]
(A) 1115.42
(B) 1150.70
(C) 1412.84
(D) 1163.88
24.
A mixture of two liquids which boils without change in composition is called
(A) Stable mix ture
(B) Binary liquid mixture
(C) Azeotropic mixture
(D) Zerotropic mixture
[NSEC-2010]
222
25.
The aqueous solution having osmotic pressure nearest to that of an equimolar solution of K4[Fe(CN)6] is
[NSEC-2010]
(A) K2SO4
(B) Na3PO4
(C) Al2(SO4)3
(D) C6H12O6
26.
The elevation in boiling point of a solution containing 13.44 g of CuCl2 in 1 kg of water is
(Kb = 0.52 K kg mol–1)
[NSEC-2011]
(A) 0.05
(B) 0.10
(C) 0.16
(D) 0.21
27.
The freezing point of a solution containing 8.1 g of HBr in 100 g of water, assuming the acid to be 90%
ionized is [H = 1, Br = 80, Kf for water = 1.86 K kg mol–1]
[NSEC-2011]
(A) 0.85ºC
(B) –3.53ºC
(C) 0ºC
(D) –0.35ºC
28.
The inorganic precipitate which acts as a semipermeable membrane is
(A) Calcium phosphate (B) Nickel phosphate
(C) Plaster of paris
29.
Which of the following observation indicates colligative properties ?
I. A 0.5 M NaBr solution has a higher vapour pressure than 0.5 M BaCl2.
II. A 0.5 M NaOH solution freezes at a lower temperature than pure water.
III. Pure water freezes at a higher temperature than pure ethanol.
(A) Only I
(B) Only II
(C) Only III
(D) I and II
30.
Osmotic pressure of a 2 % w/v solution of glucose is same as 5% w/v solution of a nonvolatile nonelectrolyte solute. The molar mass of the solute is :
[NSEC-2014]
(A) 180
(B) 450
(C) 72
(D) 45
31.
The colligative property used in the determination of molar mass of a polymer is :
(A) lowering of the vapour pressure
(B) elevation in the boiling point
(C) depression in the freezing point
(D) osmotic pressure.
32.
The vapor pressure of benzene is 53.3 kPa at 60.3ºC, but it fall to 51.5 kPa when 19 g of a nonvolatile
organic compound is dissolved in 500g benzene. The molar mass of the nonvolatile compound is
[NSEC-2015]
(A) 82
(B) 85
(C) 88
(D) 92
33.
The vapor pressure of two pure isomeric liquids X and Y are 200 torr and 100 torr respectively at a
given temperature. Assuming a solution of these components to obey Raoult’s law, the mole fraction of
component X in vapor phase in equilibrium with the solution containing equal amounts of X and Y, at
the same temperature, is
[NSEC-2015]
(A) 0.33
(B) 0.50
(C) 0.66
(D) 0.80
34.
In cold climate, the water in a radiator of car gets frozen causing damage to the radiator. Ethylene
glycol is used as an antifreezing agent. The amount of ethylene glycol that should be added to 5 kg of
water to prevent it from freezing at –7ºC is :
[NSEC-2016]
(Given : Kf for water = 1.86 K mol–1 kg ; Molar mass of ethylene gycol = 62 g mol–1)
(A) 1165 g
(B) 46.7 g
(C) 116.7 g
(D) 93.4 g
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[NSEC-2012]
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35.
[NSEC-2012]
(D) Copper ferrocyanide
Which of the following is correct ?
A liquid with
(A) low vapour pressure will have a low surface tension and high boiling point
(B) high vapour pressure will have high intermolecular forces and high boiling point
(C) low vapour pressure will have high surface tension and high boiling point
(D) low vapour pressure will have low surface tension and low boiling point
[NSEC-2017]
223
PART - III : HIGH LEVEL PROBLEM (HLP)
THEORY
Ostwald–Walker Method :
Experimental or lab determination of relative lowering of vapoure pressure i.e.
dry air
Ps
ΔP
P
0
or
P
.
Ps
P
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anhydrous CaCl2
(dehydrating agent)
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solution
solvent
(i) Initially note down the weights of the solution set, solvent set containers & of dehydrating agent
before start of experiment.
(ii) Note down the same weights after the experiment is complete.
Loss of weight of solution containers  Ps.
Loss of in weight of solvent containers  (Pº – Ps)
gain in weight of dehydriating agent  Pº.
Pº – Ps
loss in weight of solvent
=
Pº
loss in weight of solution
M
Pº – Ps
M
w
loss is weight of solvent
=
=
×
= (molality) ×
W
gain is weight of dehydrating agent
m
1000
Ps
lp
Suppose, dry air was passed through a solution of 5 g of a solute in 80 g of water & then it is passed
through pure water. Loss in weight of solution was 2.50 g & loss in weight of pure water was 0.04 g.
Then, to find molecular weight of the solute, we have
Pº – Ps
loss in wt. of solvent
=
loss in wt. of solution
Ps
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a
0.04 w
5 18
Pº – Ps
M

=
=
×
=
. Thus. m = molecular weight of the solute = 70 g/mol.
W
m 80
2.50 m
Ps
Example-1 :
If same volume solution of different solute is used then what is order of (a) vapour pressure (b)
moles of solute (c) molar mass of solute.
Solution :
PA  1 g.
PC – PB  1 g
PC > PA > PB
;
;
;
PA – PB  0.5 g, PB  0.5 g.
PC  1.5 g.
nC < nA < nB . ;
MC > MA > MB.
224
SUBJECTIVE QUESTIONS
1.
Dry air was passed through bulbs containing a solution of 40 grams of non-volatile electrolytic solute in
360 grams of water, then through bulbs containing pure water at the same temperature and finally
through a tube in which pumice moistened with strong H 2SO4 was kept. The water bulbs lost 0.0870
grams and the sulphuric acid tube gained 2.175 grams. Calculate the molecular weight of solute.
2.
Calculate the freezing point of a solution of a non-volatile solute in a unknown solvent of molar mass 30
g/mole having mole fraction of solvent equal to 0.8. Given that latent heat of fusion of solid solvent = 2.7
kcal mol–1, freezing point of solvent = 27°C and R = 2 cal mol–1 k–1.
ONLY ONE OPTION CORRECT TYPE
The vapour pressure of two pure liquids A and B, that form an ideal solution are 100 and 900 torr
respectively at temperature T. This liquid solution of A and B is composed of 1 mole of A and 1 mole of
B. What will be the pressure, when 1 mole of mixture has been vapourized ?
(A) 800 torr
(B) 500 torr
(C) 300 torr
(D) None of these
4.
A maxima or minima obtained in the temperature composition curve of a mixture of two liquids indicates
(A) an azeotropic mixture
(B) an eutectic formation
(C) that the liquids are immiscible with one another
(D) that the liquids are partially miscible at the maximum or minimum
5.
At a constant temperature, S will be maximum for which of the following processes :
(A) Vaporisation of a pure solvent
(B) Vaporisation of solvent from a solution containing nonvolatile and nonelectrolytic solute in it
(C) Vaporisation of solvent from a solution containing nonvolatile but electrolytic solute in it
(D) Entropy change will be same in all the above cases
6.
The freezing point of aqueous solution that contains 3% urea, 7.45% KCl and 9% of glucose is (given K f
of water = 1.86 and asume molarity = molality).
(A) 290 K
(B) 285.5 K
(C) 267.42 K
(D) 250 K
7.
Select correct statement :
(A) Osmosis produced by semipermeable membrane.
(B) Desalination of sea-water is done by reverse osmosis
(C) Both are correct statements
(D) None is correct statement
8.
Osmotic pressure of a solution of glucose is 1.20 atm and that of a solution of cane sugar is 2.5 atm.
The osmotic pressure of the mixture containing equal volumes of the two solutions will be
(A) 2.5 atm
(B) 3.7 atm
(C) 1.85 atm
(D) 1.3 atm.
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3.
9.
Pressure cooker reduces cooking time because
(A) the heat is more evenly distributed inside the cooker.
(B) a large flame is used.
(C) boiling point of water is elevated.
(D) whole matter is converted into steam.
10.
Consider two liquids A & B having pure vapour pressures PA & PB forming an ideal solution. The plot of
1
1
v/s
(where XA and YA are the mole fraction of liquid A in liquid and vapour phase respectively)
YA
XA
is linear with slope and Y intercepts respectively :
PºA
PºA
(PAº − PBº )
(PBº − PAº )
(A) º and
(B)
and
PBº
PBº
PB
PºB
(C)
PºB
PºA
and
(PAº − PBº )
PBº
(D)
PºB
PºA
and
(PBº − PAº )
PBº
225
11.
Assuming the formation of an ideal solution, determine the boiling point of a mixture containing 1560 g
benzene (molar mass = 78) and 1125 g chlorobenzene (molar mass = 112.5) using the following
against an external pressure of 1000 Torr.
benzene
2200
1800
Vapour
1350
Pressure
1000
540
400
300
200
chlorobenzene
(A) 90ºC
(C) 110º
(D) 120ºC
Dry air is slowly passed through three solutions of different concentrations, c 1, c2 and c3 ; each
containing (non volatile) NaCl as solute and water as solvent, as shown in the Fig. If the vessel 2 gains
weight and the vessel 3 loses weight, then
(A) c2 > c3
(B) c1 < c2
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12.
(B) 100ºC
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90 100 110 120
t(cº)
(C) c1 < c3
(D) Both (A) and (B)
For a solution of 0.849 g of mercurous chloride in 50 g of HgCl2() the freezing point depression is
1.24ºC. Kf for HgCl2 is 34.3. What is the state of mercurous chloride in HgCl2 ? (Hg – 200, Cl – 35.5)
(A) as Hg2Cl2 molecules
(B) as HgCl molecules
(C) as Hg+ and Cl– ions
(D) as Hg22+ and Cl– ions
14.
The vapor pressures of chlorobenzene and water at different temperatures are
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13.
At what temperature will Cl steam-distillation under a total pressure of 800 mmHg?
(A)  95ºC
(B)  92ºC
(C)  94ºC
(D)  89ºC
15.
MATCH THE COLUMN
Assuming all the solutes are non volatile and all solutions are
and anion.
Column – I
10 ml 0.1 M NaOH aqueous solution is added
(A)
(p)
to 10 ml 0.1 M HCl aqueous solution
10 ml 0.1 M NaOH aqueous solution is added
(B)
(q)
to 10 ml 0.1 M CH3COOH aqueous solution
10 ml 0.1 M HCl aqueous solution is added to
(C)
(r)
10 ml 0.1 M NH3 aqueous solution
10 ml 0.1 M HCl aqueous solution is added to
(D)
(s)
10 ml 0.1 M KOH aqueous solution
ideal and neglect the hydrolysis of cation
Column – II
Osmotic pressure of solution
increases
Vapour pressure of solution increases
Boiling point of solution increases
Freezing point of solution increases
226
SINGLE AND DOUBLE VALUE INTEGER TYPE
The vapour pressure of fluorobenzene at t°C is given by the equation
1250
log p (mm Hg) = 7.0 –
t + 220
Calculate the boiling point of the liquid in ºC if the external (applied) pressure is 5.26% more than
required for normal boiling point. (log 2 = 0.3)
17.
Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO 2 of 4 atm
over the liquid at 25ºC. The Henry’s law constant for CO2 in water at 250C is 3.1 × 10–2 mol/litre–atm.
Write answer after multiplying by 100.
18.
Two liquids A and B are miscible over the whole range of composition and may be treated as ideal
(obeying Raoult’s law.) At 350 K the vapour pressure of pure A is 24.0 kPa and of pure B is 12.0 kPa. A
mixture of 60% A and 40% B is distilled at this temperature. A small amount of the distillate is collected
and and redistilled at 350 K; what is the mole percent of B in the second distillate ?
19.
Determine i (vant-Hoff factor) for a tribasic acid H3A. Assuming first dissociation to be 100%, second
dissociation 50%, third dissociation 20%. (Round off your answer to nearest integer).
20.
Dry air was successively passed through a solution of 5g solute in 80 g water and then through pure
water. The loss in weight of solution was 2.5 g and that of pure water was 0.04 g. What is mol. wt of
solute ?
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16.
ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
Which of the following is/are correct for an ideal binary solution of two volatile liquids (eg. benzene &
toluene) ?
(A) Its vapor is always richer in the more volatile component (compared to the liquid).
(B) The liquid will gradually become richer in the less volatile component if such a mixture is boiled
(distilled).
(C) The PT (ie. the total pressure) above the solution will be the sum of the vapor pressures of the two
pure components.
(D) The boiling point of the solution will be less than the boiling points of the two components.
22.
The diagram given below represents boiling point composition diagram of solution of components A and
B, which is/are incorrect among the following?
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21.
(A) The solution shows negative deviation
(C) The solution is ideal solution
23.
(B) A-B-interactions are stronger than A-A and B-B
(D) The solution shows positive deviation.
For a dilute solution having molality m of a given solute in a solvent of mol.wt. M, b.pt. T b and heat of
  Tb 
vaporisation per mole H; 
is equal to :

  m  m →0
(A) Molal elevation constant of solvent
RTb2 M
(B)
; where M in kg vap H and R in J mol–1
 vap H
(C)
RTb2 M
; where M in kg ; vapS and R in J mol–1
 vap S
227
(D)
RTb2 M
; where M in g ; R and vapH expressed in same unit of heat.
1000  vap H
PART - IV : PRACTICE TEST-2 (IIT-JEE (ADVANCED Pattern))
Max. Time : 1 Hr.
Max. Marks : 66
Important Instructions
A. General %
The test is of 1 hour duration.
The Test Booklet consists of 22 questions. The maximum marks are 66.
Question Paper Format :
Each part consists of five sections.
Section 1 contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE is correct.
5.
Section 2 contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE OR MORE THAN ONE are correct.
6.
Section 3 contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from
0 to 9 (both inclusive).
7.
Section 4 contains 1 paragraphs each describing theory, experiment and data etc. 3 questions relate to
paragraph. Each question pertaining to a partcular passage should have only one correct answer among
the four given choices (A), (B), (C) and (D).
8.
Section 5 contains 1 multiple choice questions. Question has two lists (list-1 : P, Q, R and S; List-2 : 1, 2,
3 and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY ONE
is correct.
C. Marking Scheme :
9.
For each question in Section 1, 4 and 5 you will be awarded 3 marks if you darken the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one
(– 1) mark will be awarded.
10. For each question in Section 2, you will be awarded 3 marks. If you darken all the bubble(s)
corresponding to the correct answer(s) and zero mark. If no bubbles are darkened. No negative marks will
be answered for incorrect answer in this section.
11. For each question in Section 3, you will be awarded 3 marks if you darken only the bubble corresponding
to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for
incorrect answer in this section.
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1.
2.
B.
3.
4.
SECTION-1 : (Only One option correct Type)
This section contains 7 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which Only ONE option is correct.
1.
Assuming each salt to be 90% dissociated which of the following will have highest osmotic pressure ?
(A) Decimolar Al2(SO4)3
(B) Decimolar BaCl2
(C) Decimolar Na2SO4
(D) A solution obtained by mixing equal volumes of (A), (B) and (C) and filtering.
2.
The melting points of most of the solid substances increases with an increase of pressure acting on
them. However, ice melts at a temperature lower than its usual melting point , when the pressure
increases. This is because :
(A) Ice is less dense than water
(B) Pressure generates heat
(C) The bonds break under pressure
(D) Ice is not a true solid
228
4.
The phase diagrams for the pure solvent (solid lines) and the solution (non-volatile solute, dashed line)
are recorded below : The quantity indicated by L in the figure is :
(A) p
(B) Tf
(C) Kbm
(D) Kfm
Available are 1L of 0.1 M NaCl and 2L of 0.2 M CaCl2 solutions. Using only these two solutions what
maximum volume of a solution can be prepared having [Cl–] = 0.34 M exactly. Both electrolytes are
strong
(A) 2.5 L
(B) 2.4 L
(C) 2.3 L
(D) None of these
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3.
A teacher one day pointed out to his students the peculiar fact that water is unique liquid which freezes
exactly at 00 C and boils exactly at 1000 C. He asked the students to find the correct statement based
on this fact :
(A) Water dissolves anything however sparingly the dissolution may be
(B) Water is a polar molecule
(C) Boiling and freezing temperatures of water were used to define a temperature scale
(D) Liquid water is denser than ice
6.
Three different ideal solutions (I, II, III) each containing total 10
moles of A & B in different composition are taken as shown in
figure and pressure over the solutions is gradually reduced.
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5.
Initially external pressure is same for all three solutions. At a
particular external pressure (P1), II solution is found to have
XA = 0.4
(liq composition)
&
YA = 0.8
(Vap. composition)
I II III
P1
Pext
lp
Then select correct statement :
For an ideal binary solution with Pº A / PºB which relation between XA (mole fraction of A in liquid phase)
and YA (mole fraction of A in vapour phase) is correct , X B and YB are mole fraction of B in liquid and
vapour phase respectively : (Given : PºA > PºB)
(A) XA = YA
(B) XA > YA
YA
XA
(C)
<
(D) XA, YA, XB and YB cannot be correlated
YB
XB
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7.
a
Pure A
Pure B
(A) At the same external pressure, for solution I, XA < 0.4, YA < 0.8.
(B) At the same external pressure, for solution III, XA > 0.4, YA > 0.8.
(C) For all three solutions at same external pressure (P1) liquid & vapour composition will be same
(D) None of these.
Section-2 : (One or More than one options correct Type)
This section contains 5 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.
8.
For KH, Henry’s constant, which are correct ?
(A) KH is characteristic constant for a given gas–solvent system.
(B) Higher is the value of KH, lower is solubility of gas for a given partial pressure of gas.
(C) KH has temperature dependence.
(D) KH increases with temperature.
9.
The vapour pressure of ideal solution of benzene and toluene
is 550 torr at 80ºC then what would be correct statement about
same solution at 100ºC.
229
(A) Vapour pressure of solution = 725 torr.
(B) at 725 torr pressure at 90ºC, no vapour form
9
2
(C) Composition of vapour is
and
11
11
(D) Composition of liquid remain same at equilibrium condition at any temp.
Select incorrect statement :
(A) Na+ and K+ ions are responsible for maintaining isotonic property inside and outside of the cell of
organism.
(B) Aquatic species are more comfortable in lakes present at sea level in comparison to lakes present
at high altitude.
(C) Solubility of N2 decreases, in presence of He when oxygen cylinder is utilised by Scuba divers.
(D) The KH value of CO2 is higher than KH of N2.
11.
Following is false when in a volatile solvent A and a non volatile solute B is mixed (where symbols have
their usual meaning) :
P0 − PB0 nB
P0 − PB0
nB
P 0 − PA nB
PA
=
=
=
(A) A
(B) A
(C) PA0 =
(D) A
PA
nA
PA
nA + nB
PA
nA
1 − XB
12.
0.2 moles of A and 0.3 moles of B are taken in separate beakers and enclosed in chamber I. Another
0.2 moles of A and 0.3 moles of B are mixed in a beaker and enclosed in chamber II. At equilibrium.
Which of the following are not true. (A and B are volatile liquids and they form ideal solution on mixing)
(A) The vapour pressure in chamber I is greater than vapour pressure in chamber II.
(B) The vapour pressure in chamber I is less than vapour pressure in chamber II.
(C) The vapour pressure in both chambers are equal.
(D) The vapour pressure in chamber II can not be determined.
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10.
A very small amount of a non-volatile solute (non-associative, non-dissociative) is dissolved in 100 cm3
of a solvent. At room temperature, vapour pressure of this solution is 98.7 mm of Hg while that of pure
solvent is 100 mm of Hg. If the freezing temperature of this solution is 0.72 K lower than that of pure
solvent, what is the value of cryoscopic constant of solvent (in K Kg/mol) ? Round off your answer to
the nearest whole number. Report your answer as 0 (zero) if you find data insufficient.
Given : Molar mass of solvent = 78 g/mol.
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a
13.
lp
Section-3 : (One Integer Value Correct Type.)
This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive)
14.
At 10ºC the osmotic pressure of urea solution is 500 mm. The solution is diluted and the temperature is
raised to 25ºC, when the osmotic pressure is found to be 105.3 mm. Determine extent of dilution.
15.
In aqueous solution of 1 × 10 –3 molal Kx[Fe(CN)6] depression in freezing point is 7.2 × 10 –3 K.
Determine sum of primary and secondary valency of complex (K f of H2O = 1.8 K Kg/mole). (Assume
that % ionisation of complex is 100%)
16.
Calculate molarity of final solution obtained by mixing I and II HNO3 solution.
I
II
w
%
12.6
6.3
w
d (in g/mL)
1.5
1
Volume (in L) 5
5
17.
8 × 10–x moles of gas A is dissolved in 36 mL of water when pressure of gaseous mixture above water
is 4 atm. Mole percentage of gas A in mixture is 25. Henry constant for gas A in water is 2.5 × 103 atm.
Find x.
230
18.
A solution containing 0.1 g of a non volatile organic substance P(Molecular mass 100) in 100 g of
benzene raises the boiling point of benzene by 0.2ºC while a solution containing 0.1 g of another non
volatile substance Q in same amount of benzene raises the boiling point of benzene by 0.4ºC. If ratio of
x
molecular masses of P and Q is
then. Find minimum value of x + y.
y
Paragraph for Questions 19 to 21
a
ri
SECTION-4 : Comprehension Type (Only One options correct)
This section contains 1 paragraphs, each describing theory, experiments, data etc. 3 questions
relate to the paragraph. Each question has only one correct answer among the four given
options (A), (B), (C) and (D)
uh
A solution is made by mixing 1 mole benzene ( PB = 100 mm Hg) & 1 mole toluene ( PT = 40 mm
Hg).Suppose, initially the pressure over the solution is very high so that no vapour exist above the
liquid. As we gradually decrease the pressure, a point (bubble point) comes when we aross the bubble
point curve & first bubble of vapour starts forming (hence called bubble point curve). Now we have
entered the vapour-liquid equilibrium region. On further decreasing the pressure, a point (dew point)
comes when we cross the dew point curve then almost all the liquid has evaporated into vapour i.e.
only the last drop of liquid (dew) remains. Beyond this point no liquid exist in the system. Then answer
the following questions :
If the pressure over the mixture at 300 K is reduced, at what pressure does the first bubble form :
(A) 140 mm Hg
(B) 90 mm Hg
(C) 65 mm Hg
(D) 70 mm Hg
20.
What is the composition of first bubble formed ?
(A) YA = 2/7, YB = 5/7
(B) YA = 3/7, YB = 4/7
(C) YA = 1/7, YB = 6/7
(D) none of these
21.
What will be the pressure when 1 mole of mixture has been vapourised ?
(A) 70 mm Hg
(B) 63.25 mm Hg
(C) 100 mm Hg
(D) 40 mm Hg
lp
Ja
19.
Sa
nk
a
SECTION-5 : Matching List Type (Only One options correct)
This section contains 1 questions, each having two matching lists. Choices for the correct
combination of elements from List-I and List-II are given as options (A), (B), (C) and (D) out of
which one is correct
22.
Match each List-I with an appropriate pair of characteristics from List-II and select the correct answer
using the code given below the lists.
0.1 mol of each solute in the list-I are dissolved in 10 mole water separately.
List-I
List-II
P.
AlCl3 if  = 0.8
1.
i = 3.4
Q.
BaCl2 if  = 0.9
2.
has minimum osmotic pressure among the given solutions.
R.
Na3PO4 if  = 0.9
3.
has minimum freezing point among the given solutions.
S.
K4[Fe(CN)6] if  = 0.7
4.
has RLVP = 37/1037.
Code :
(A)
(C)
P
4
2
Q
2
1
R
3
3
S
1
4
(B)
(D)
P
1
1
Q
2
2
R
4
3
S
3
4
231
Practice Test-2 ((IIT-JEE (ADVANCED Pattern))
OBJECTIVE RESPONSE SHEET (ORS)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Ans.
Que.
Ans.
a
ri
Que.
PART - I
(3)
2.
(3)
3.
(3)
4.
(3)
5.
(3)
6.
(4)
7.
(3)
8.
(4)
9.
(2)
10.
(2)
11.
(4)
12.
(1)
13.
(4)
14.
(3)
15.
(1)
16.
(4)
17.
(1)
18.
(4)
19.
(2)
20.
(2)
21.
(3)
22.
(2)
23.
(3)
24.
(1)
25.
(4)
26.
(1)
27.
(2)
28.
(2)
29.
(1)
30.
(2)
lp
Ja
1.
uh
Ans.
PART - II
(A)
2.
(C)
3.
(D)
4.
(A)
5.
(B)
6.
(A)
7.
(A)
8.
(A)
9.
(C)
10.
(B)
11.
(C)
12.
(A)
13.
(A)
14.
(A)
15.
(B)
16.
(B)
17.
(A)
18.
(A)
19.
(B)
20.
(D)
21.
(B)
22.
(B)
23.
(C)
24.
(C)
25.
(C)
26.
(C)
27.
(B)
28.
(D)
29.
(D)
30.
(B)
31.
(D)
32.
(B)
33.
(C)
34.
(A)
35.
(C)
Sa
nk
a
1.
PART - III
1.
M = 48
2.
10.33°C
3.
(C)
4.
(A)
5.
(A)
6.
(C)
7.
(C)
8.
(C)
9.
(C)
10.
(B)
11.
(B)
12.
(D)
13.
(A)
14.
(B)
15.
(A) – q, s; (B) – q, s; (C) – q, s; (D) – q, s
16.
85
17.
12 mol/litre.
18.
14
20.
70
21.
(AB)
22.
(ABC)
23.
(ABD)
19.
3
232
PART - IV
(A)
2.
(A)
3.
(C)
4.
(A)
5.
(C)
6.
(C)
7.
(C)
8.
(ABCD)
9.
(AB)
10.
(CD)
11.
(AB)
12.
(ABD)
13.
4
14.
5
15.
9
16.
2
17.
4
18.
2
x
=
Hence x + y = 3
1
y
19.
(D)
20.
(A)
21.
(B)
22.
(B)
PART - I
Mole of H2O =
36
=2 ;
18
Mole of glycerine =
Mole fractions of glycerine =
0.5
n1
=
2.5
n1 + n2
X0 = 0.2 Ans.
Ja
total mole = 2 + 0.5 = 2.5 ;
46
= 0.5
92
uh
1.
a
ri
1.
I.
II.
III.
3.
When non volatile solute added to solvent. Due to elevation in boiling point  boiling pointand due to
dispression in freezing point, freezing temperature 

6.
As

7.
i = 1 + x = 1.1
Tb = i Kbm
iKb m can be expressed in degree (Unit of temperature)
Kb m can be expressed in degree (Unit of temperature)
Tb
and
can be expressed in degree (Unit of temperature)
i
But unit of Kb is mol–1 kg K
As
so
and
Sa
nk
5.
HA
H+ + A¯
i = [1 + (y – 1) x] = 1 + x
pKa = 4 = – log Ka
Ka = 10–4 = Cx2
1 × 10–4 = 0.01 × x2  x = 0.1 
a
4.
Melting of snow by salt : Depression in freezing point
Desalination of sea water : Reverse osmosis
Osmosis is used to determine the molar mass.
lp
2.
Tb = molality × Kb
0.52 = m × 0.52
molality = 1 mol kg–1
urea = 1 mol
1000
moles of water =
18
 = CRT
c
 = RT
M
 RT
=
c
M
RT
M=
/c
= 55.55 ;
mole fraction of urea =
1
= 0.018
1 + 55.55
C = moles/liter, c = kg/m3
[/c = 8.314 × 10–3 ]
233
[T = 293 k]
8.314  293
M=
= 293 × 103
8.314  10 −3
8.
11.
a
ri
1M C6H12O6 (molar mass = 180 g mol–1)
1000 mL solution has = 180 g solute
1180 g solution has = 180 g solute
1000 g solvent has = 180 g solute
Thus, molality = 1 molal

Tf = Kf molality = 1.86 × 1 = 1.86º

uh
10.
F.P. = – 1.86ºC
Firstly we have to convert mole fraction into molality.
0.07  1000
x solute
Molality =
=
= 4.18
0.93  18
x solvent Msolvent /1000
Now,
Ja
9.
Normal boiling point of the solution is that temperature at which vapour pressure of solution equals to
1 atm.
Given Tb = 1.08ºC, i = 2 at boiling pt. of solution.
k
ik m
Tb
and
Tf = 1.80ºC, and b = 0.3
so
= b b
so
if = 1
if k f m
Tf
kf
i.e., AB behaves as non–electrolyte at the f.p of the solution.
Tf = kf m = 1.86 × 4.18 = 7.78º.
From given graph, we can say T1 is that temp at which solid state and liquid (solution) are in
equilibrium.
13.
P = PA°XA + PB°XB
100 60  3
+
= 70 mm < 75 mm (experimental)
4
4
Thus, there is positive deviation (1) is true, mixture is more volatile due to decrease in b.p. Thus, (2) is
true also force of attraction is decreased thus (3) is true.
14.
Colligative property of a solution depends on no. of particles of solute in solution.
Sa
nk
a
lp
12.
15.
Value of van’t Hoff factor is least for urea solution, so there will be least depression in freezing point
i.e., maximum freezing point.
16.
In HF hydrogen bonding is present so there is association of molecules due to this van’t hoff factor is
less, so depression in f.p decreases therefore f.p. value is larger than HCl. Similarly value of i = 2 for
NaCl and i = 1 for Glucose.
32
pA = XA pºA ;
32 = XA 40

XA =
= 0.8.
40
17.
18.
PT = XA pºA + XB pºB
2
2
=   × 80 +   × 120 = 100 Torr
4
4
Now mole fraction in vapour phase =
19.
40
X A PA0
=
= 0.4.
100
PT
Boiling point get lowered when vapour pr. increases and it happens when there is a positive deviation
from Raoult’s law.
234
20.
Tf = kf × m
For urea,
or
kf =
Now for CH3COOH
Tf = i kf m
so
i = = 1.1
Now
i=1+
Now
CH3COOH
CH3COO–
C
0
C – C
C
[H+] = C = 0.01 × 0.1 = 0.001
Tf
1.86
=
= 1.86
m
1
so
+
so
 = 1.1 – 1 = 0.1
H+
0
C
pH = 3.
Mixtures of CHCl3 and CH3COCH3 shows negative deviation from Raoult’s law, so vapour pressure
decreases and boiling point increases.
22.
Tf = kf m. ;
23.
More the value of van’t hoff factor, more will be the depression in freezing point.
24.
Acetone and chloroform forms hydrogen bonding so volume decreases.
25.
All are facts.
We should remember that, Entropy of solution is more than entropy of pure solvent. So the difference in
entropy change will be less in case of solution.
26.
Mobserved =
27.
Due to weak force of attraction more vapour will be forrmed so vapour pressure will be high.
28.
The order of force attraction and boiling point is CH 3OH > CH4 > H2 .
29.
At freezing point liquid solvent and solid solvent are in equilibrium.
30.
There is very weak attraction between benzene and methanol as compare to attraction between
molecules of methanol.
58.5
;
i
so Tf = – 0.930C.
uh
Tf = 1.86 × 0.5 = 0.93.
a
ri
21.
lp
Ja
i > 1.
PART - III
360
2.175
M
= 1+
×
18
0.087
40
a

 25 = 1 +
M
2
Ans.
M = 48
MRTf2
2  300  300  30
=
= 2.00 K kg mol–1
1000 Hf
1000 2700
Sa
nk
1.
40
P − PS 0.087
M
=
=
2.175 40 360
P
+
M
18
2.
Kf =
or

or

mole fraction of water = 0.8

mole fraction of solute = 0.2
n
N
0.2 =
and
0.8 =
n+N
n+N
0.2 1
n
wM
=
=
=
N
mW
0.8 4
1
1
w  30
w
=

=
4
mW
mW
4  30
1000  2.0
1000  w  K f
Tf =
=
= 16.67

F.P. of solution. = 10.33°C.
4  30
m W
235
Let nB mole of B present in 1 mole of mixture that has been vaporized. Thus, y B =
Mole fraction of B in the remaining liquid phase will be x B =
xB =
P − PTº
1 − nB
1
........ (1)
PBº − PTº
[
PB
Pº x
 B B
........ (2)
P
P
After substitution of values of xB and yB in (1) and (2)
P − Pº
we get 1 – nB = º Tº
........ (3)
PB − PT
yB =
and
nB =
or
so
(1 − nB )PBº
P
PBº
nB =
P + PB
1–
........ (4)
PBº
P − Pº
= º Tº
P + PB PB − PT

PBº . PTº = 100  900
P=
P = PºT + (PºB – PºT) xB]

uh
and
nB
1
a
ri
3.
300 torr
(a) An azeotropic mixture boil at perticular temperture without changing its composition.
5.
Entropy of solvent is less than that of solution.
6.
Tf = i.m. Kf
Tf = i1m1Kf + i2 m2 Kf + i3 m3 Kf = (m1 + 2m2 + m3) Kf
3 7.45  2
9
+
+
74.5
180 × 1000 × 1.86
Tf = 60
100
Tf = 3 × 1.86
= 5.58
Tf of solution = 273 – 5.58 = 267.42 K Ans.
7.
Desalination is an application of reverse osmosis.
8.
fVf = 1V1 + 2V2 ;
9.
(C) B.P. of water is elevated.
lp
Ja
4.
1.2V + 2.5V
3.7V
=
;
2V
2V
Sa
nk
a
f =
10.
P’A = PAºXA and P’B = PºBXB
P’A = PM . YA and P’B = PM .YB
PA
P

= B
YA
YB
or
PA X A
P X
= B B
YA
YB
or
PB
P
= A + (PºB – PºA)
XA
YA
or
11.
f = 1.85 atm.
y = mx + C
=

PB (1 − XA )
(1 − YA )
Slope = m =
20 mole C6H6, 10 mole C6H5Cl 
at t = 100°C  ps = 300 x
XB =
PAº
PBº
or
PA X A
(1 – YA) = PBº – PBº XA
YA
or
(Pº − Pº )
1 PAº
1
=
. º + B º A
YA PB
XA
PB
and intercept C =
(PBº − PAº )
PBº
.
2
1
, XC =
3
3
1
2
+ 1350 x
= 100 + 900 (=1000).
3
3
236
12.
13.
Wt gain means
V.Pincoming > V.Poutgoing
weight loss means
V.Pincoming < V.Poutgoing
So p1 > p2 < p3 and c1 < c2 > c3
 0.849 / M 
1.24 = 34.3 
  M = 469.68  as Hg2Cl2 molecules.
 0.05 
At 90ºC,
Total PT = 730 mm Hg
At 100ºC total PT = 289 + 760 = 1049 mm Hg
so for 800 mm Hg, temperature will lie in between 90º –100ºC
(800 − 730)
Using extrapolation method, Temperature = 90 +
× (100 – 90) = 92.19ºC.
(1049 − 730)
15.
No of particles , so vapour pressure , i 
Tf = mKf i ;
Tf  , freezing point .
16.
1.0526 × 760 = pext
According to Henry’s law,
a
= KH
P

so
so
Ans. 85
a = 3.1 × 10–2 × 4 = 0.12 mol/litre.
192
16  12
0.4
6
× 24 +
× 12 =
=
= 19.2 kPa
10
10
10
10
144
6
PA = XA PA =
× 24 =
= 14.4 = XA × 19.2
10
10
144
XA =
= 75%
XB = 25%
192
3
3
1
PT =
× 24 +
× 12 = 18 + 3 = 21 k Pa
&
PA =
× 24 = 18 = XA (21)
4
4
4
18
6
XA =
=
= 85.7% of A  XB = 14.3%
21
7
19.
H3A ⎯→ H+ + H2A–
1
0
0
0
1
1–0.5
H2A–
1–0.5
H+ +
1+0.5
HA2–
0.5
0.5–0.1
HA2–
0.5–0.1
H+ +
1+0.5
+0.1
A3–
0.1
Sa
nk
Initial mole
Ja
PT =
so
t = 84.878 °C
lp
18.

a
17.
pext = 799.976 ~ 800 Torr
1250
2.9 = 7 –

t + 220
uh

;
1250
log 800 = 7 –
t + 220
a
ri
14.
(
+
–
2–
3–
observed mole of all the species moles of H + H2 A + HA + A
i=
=
Theoretical moles
1
)
=
1.6 + 0.5 + 0.4 + 0.1
=2.6 3
1
20.
Loss in weight of solution  PS ;
Loss in weight of solution  P0 – PS
P0 − Ps
0.04 5  18
w M

=
Also,
By. eq (1) and (2) we get
 m = 70.31
2.5 80  m
Ps
m W
21.
PT = PA + PB  PA0 + PB0
boiling point of the solution will be in between the boiling point of two liquids.
22.
For – ve deviation
A––B>A––A
A––B>B––B
– ve deviation solution are non ideal solution.
237
23.
Tb = mKb i;
RT 2
Kb = 1000 Hvapour
PART - IV
 = MRT i :
2.
This is due to cage like structure ice.
3.
L indicates elevation in boiling pt. i.e., kbm.
0.1x + 0.4y
Let volumes taken by ‘x’ & ‘y’ litres, so
= 0.34 & Vg = (x + y) (to be maximised) so y = 4x so
x+y
4.
y , i ,  
for maximum volume
y = 2L & x =
1
L
2
5.
Freezing point and boiling point are used in temperature scale.
6.
On the same tie line liquid & vapour composition will be same.
7.
Mole fraction of more volatile substance is greater in vapour phase.
9.
P = PA XA +
a
ri
1.
lp
Ja
uh
500
XB
(XB = mole fraction of benzene)
760
550 = 400 × (1 × XB) + 600(XB)
150 = (600 – 400) XB
150
3
=
=
XB
200 4
1
XA =
4
At 100ºC mole fraction will be same initially but get change at equilibrium.
3
1
P=
× 500 +
× 800 = 125 + 600 = 725 torr
4
4
XA PA = YAP
Sa
nk
a
1
× 500 = YA = 725
4
125
= YA
725
24
5
So YA =
YB =
29
29
At 90ºC mole fraction will be same initially but get change at equilibrium.
550 < P < 725
So at P = 725 only liquid state exist.
11.
12.
13.
PA0
PA0 – PA
=
nA
P0 − PA
n
+ 1 or A
= B
PA
nB
nA
In both chamber vapour compositions are same so vapour pressure are equal.
Po – Ps
Po

Now,

= Xsolute
100 – 98.7
= Xsolute = 0.013
100
Tf = Kf × m
0.013  1000
0.72 = Kf ×
0.978  78
Kf  4.2 K Kg/mol
238

14.
Reported answer = 4
Initially P =
500
atm, T = 283 K
760

pV = nRT
Let V = V1
 500 
Or 
...(1)
 × V1 = R × 5 × 283
 760 
Let on dilution the volume becomes V2 and temp is raised to 25ºC i.e. 298 K
 105.3 
 =
 atm
 760 
 105.3 
 760  × V2 = n × R × 298


a
ri
(1)/(2)
283 105.3
V1
=
×
298
500
V2
...(2)
Or V2 = 5V1
Tf = iKfm
7.2 × 10–3 = i × 1 × 10–3 × 1.8
7.2
i=
=4
1.8
now , i = 1 + (x + 1 – 1)  = 1 + x
4–1=x
x=3
So oxidation number of Fe
3 + y – 6 = 0, y = +3
Sum of primary and sec valency = 3 + 6 = 9
16.
Molarity of I solution =
17.
PA = KHXA
25
4×
= 2.5 × 103 XA 
100
a
lp
12.6  1.5  10
=3
63
6.3  10  10
Molarity of II solution =
= 10
63
3  5 + 1 5 4
M V + M2 V2
= =2
M= 1 1
=
5+5
2
V1 + V2
Ja
15.
uh

Sa
nk
2
2
;
Number of moles of water = = 2
1
5000
2
Number of moles of gas A dissolved 
× 2 = 0.8 × 10–3 = 8 × 10–4
5000
XA =
19.
P = XAPA0 + XBPB0 = 0.5 x 40 + 0.5 x 100 = 70
20.
YA =
0.5
x 40
2
5
=
; YB =
70
7
7
21.
1– x
B=x
x
T=1–x
P = 40 (1 – x) + 100 x
x
1− x
Y
Y
1
1
100x + (1 − x)40
= A0 + A 
=
+
=
P
P
40
100
40 x 100
PA
PB
So
p2 = 40 x 100
p = 20 10 = 63.25
239
P.
AlCl3 → i = 1 + (4 – 1) × 0.8 = 1 + 2.4 = 3.4
Q.
BaCl2 → i = 1 + (3 – 1) × 0.9 = 1 + 1.8 = 2.8
R.
Na3PO4 → i = 1 + (4 – 1) × 0.9 = 1 + 2.7 = 3.7
S.
K4[Fe(CN)6] → i = 1 + (5 – 1) × 0.7 = 1 + 2.8 = 3.8
so, K4[Fe(CN)6] has highest colligative property and hence minimum freezing point and BaCl2 has
lowest colligative property, so lowest osmotic pressure.
0.1 3.7
0.1 i
3.7
37
RLVP =
=
=
=
0.1 i + 10 0.1 3.7 + 10 10.37
1037
Sa
nk
a
lp
Ja
uh
a
ri
22.
240
REDUCTION, OXIDATION & HYDROLYSIS
REACTIONS
JEE(Advanced) Syllabus
Reduction of Alkenes, Alkynes, Alcohols, Aldehydes, Ketones, Acids, Acid halides, Esters, Amides
Cyanide & Anhydrides.
a
ri
Oxidation of Alkenes, Alkynes, Alcohols, Aldehydes, Ketones, Acids, Acid halides, Esters, Amides &
Anhydrides.
Hydrolysis reaction of Acid halides, Esters, Amides, Anhydrides, Cyanides, Ethers etc.
JEE(Main) Syllabus
uh
Reduction and oxidation reactions of Alkenes, Alkynes, Alcohols, Aldehydes, Ketones Cyanides, Acids
& Acid derivatives.
Section (A) : Reduction-1
Introduction :
1.1
lp
Ja
Reduction covers both the addition of hydrogen (or deuterium) to a double bond and the replacement of
an atom or group by hydrogen (or deuterium). In other words, reduction means hydrogenation or
hydrogenolysis.
Reduction can be carried out in following ways:
(a)
Catalytic hydrogenation
(b)
Metal/proton (acid) reduction
(c)
Metal hydrides reduction
(d)
Miscellaneous Reductions
Catalytic hydrogenation :
Sa
nk
a
Hydrogenation using H2(g) on metal surface or other suitable catalyst is called catalytic hydrogenation.
A catalyst provides a new pathway for the reaction that involves lower free energy of activation.
Heterogeneous hydrogenation catalysts typically involve finely divided platinum.
The catalysts used can be divided into two broad classes, (a) Heterogeneous catalysts (b)
Homogeneous catalysts. Both of which mainly consist of transition metals and their compounds :
(i)
Heterogeneous catalysts : (catalysts insoluble in the reaction medium)
In heterogeneous catalytic hydrogenation catalysts are used in powdered form. Raney nickel (Ni),
Palladium on charcol (Pd/C), Platinum metal or its oxide are common heterogeneous catalysts.
241
Facts of catalysis :
(i) Substrate molecules are assumed to undergo homolysis into atoms at the surface of the catalyst.
(ii) The substrate is chemisorbed on the surface of the catalyst and hydrogenation takes place.
(iii) The process is exothermic.
Note : (If H2 and D2 mixture is used with a Pt catalyst, the two isotopes quickly scramble to produce a random
mixture of HD, H2 and D2. (No scrambling occurs in the absence of the catalyst.) The product has both
D and H atoms).
Catalytic Reduction of Functional Groups Using H2/Pd(C) or H2/Pt or H2/Ni
Ni or Pd
a
ri
R–CH=CH–R + H2 ⎯⎯⎯⎯⎯→ R–CH2–CH2–R + heat
Substrate
Product
RCOCl
RCH2OH
RNO2
RNH2
RCH2CH2R
RCCR
RCHO
RCH2OH
RCH=CHR
RCH2CH2R
RCOR
RCHOHR
RCN
RCH2NH2
(ii)
uh
Note : (i) Generally RCOOH, RCOOR, RCONH2 groups are not reduced by catalytic hydrogenation.
(ii) Stereochemistry of catalytic hydrogenation: The above reaction is syn addition (addition takes
place from the same side of alkene or alkyne).
Homogeneous catalysis : (catalysts soluble in the reaction medium). It uses reactants and
(iii)
Ja
catalyst in the same phase. Both hydrogen atoms usually add from the same side of the molecule.
Common example : Use of Wilkinsion catalyst Rh [(C6H5)3P]3Cl.
Lindlar’s catalyst : [H2/Pd, CaCO3, quinoline]
lp
It is a poisoned palladium catalyst. It is composed of powdered calcium carbonate coated with
palladium and poisoned with quinoline. It is used to carry out partial reduction of alkyne to alkene and
acid chloride to aldehyde.
Stereochemistry : Syn addition.
H Lindlar ' s catalyst
2
→
CH3–CH2–CC–CH3 ⎯⎯⎯⎯⎯⎯⎯⎯
(syn addition)
a
or Ni2B
Note : Nickel boride Ni2B (P-2 catalyst) (made from Nickel acetate and sodium borohydride) is an excellent
alternative catalyst for the conversion of alkyne into alkene. Stereochemistry : Syn addition.
NaBH
Sa
nk
4
→ Ni2B.
Ni(OCOCH3)2 ⎯⎯⎯⎯
C2H5OH
(iv)
Rosenmund catalyst : [H2/Pd, BaSO4, quinoline] Hydrogenation in presence of H2/Pd/BaSO4
is called Rosenmund Reduction. It reduces alkyne to alkene and acid halide to aldehyde. It is
poisoned palladium catalyst, composed of powdered barium sulphate coated with palladium, poisoned
with quinoline or sulphur.
Stereochemistry : Syn addition.
(a)
CH3–CC–CH3
(b)
1.2
CH3–CHO
Metal/proton (acid) reduction :
Reduction by dissolving metals is based on the fact that the metal acts as a source of electrons.
Step-1 : Metals give electrons to the electrophilic species and form anion
Step-2 : Proton is abstracted from the acidic source.
242
–
–
–
H+
e
→ A• + B–H
A–B ⎯⎯→ A• + B : or A : + B• ⎯⎯⎯
–


e
A=B ⎯⎯→ : A– B• or A• – B : A•–BH
(i)
Birch reduction [Na or Li/NH3(liq.) or (ethyl alcohol)] :
Alkyne and Aromatic Compounds are reduced by Na or Li/NH3.
Alkynes are reduced to trans alkene. Stereochemistry : anti addition.
Na / NH3
R–CC–R ⎯⎯⎯⎯
→
(anti addition)
a
ri
Benzene ring is reduced at 1, 4-position.
Typical example of reduction for aromatic system :
Presence of alkyl, alkoxy, amines reduces the benzene ring at ortho position.
Na / NH
3
⎯⎯⎯⎯
→
Na / NH
(ii)
Ja
3
⎯⎯⎯⎯
→
uh
Presence of nitro, cyano, carboxylic, keto or aldehyded group reduces the benzene ring at ipso
position.
Bouvealt-Blanc reduction [Na/C2H5OH] :
CH3CHO
Acetaldehyde
CH3CH2OH
Ethanol
Stephen’s Reductions : [SnCl2/HCl]
Sa
nk
(iii)
a
lp
Reduction of aldehydes, ketones, acidhalides, esters or cyanide by means of excess of Na/C 2H5OH is
called Bouvealt-Blanc reduction.
Na + C2H5OH is mild reducing agent.
Reagent
Na/C2H5OH
Na/C2H5OH
Na/C2H5OH
Na/C2H5OH
Na/C2H5OH
Reactant
Aldehyde
Ketone
Cyanide
Ester
acid halide
Product
1° alcohol
2° alcohol
1° amine
1° alchol
1° alchol
When reduction of cyanide is carried out with acidified stannous chloride (SnCl 2/HCl) at room
temperature, imine hydrochloride is obtained. Which on subsequent hydrolysis with boiling water gives
aldehyde. This specific type of reduction of nitrile is called stephen’s reduction.
(iv)
Clemmensen’s Reduction : [Zn-Hg/Conc. HCl]
It is used to prepare alkanes from carbonyl compounds (Aldehyde and ketones in absence of acid
sensitive groups).
Conc.
Zn – Hg / HCl
243
Mechanism :
⎯⎯
→
Zn − Hg / conc. HCl
→ RCH3 + H2O
(a) R – CHO ⎯⎯⎯⎯⎯⎯⎯⎯

O
||
Zn − Hg / conc. HCl
→ RCH2R + H2O
(b) R − C − R ⎯⎯⎯⎯⎯⎯⎯⎯

a
ri
Note : Clemmensen reduction is avoid to use for compounds which have acid sensitive group.
[Like: Alcohol, Alkene, Alkyne, Ether].
Section (B) : Reduction-2
1.3
Metal hydrides reduction
Certain complex metal and boron hydrides, are important reagents for reduction.
LiAlH4 (LAH) Lithium aluminium hydride [LiAlH4 / Ether or THF] :
uh
(i)
LAH is most common and versatile reagent. It is sensitive to protic solvent and therefore used in ether.
Reagent
Reactant
Product
LiAlH4
Aldehyde
1º alcohol
LiAlH4
Ketone
2º alcohol
LiAlH4/Excess
Acid
1º alcohol
LiAlH4/Excess
Acidanhydride
1º alcohol
LiAlH4
Acid chloride
1º alcohol
LiAlH4
Ester
1º alcohol
LiAlH4
Cyanide
1º amine
LiAlH4
Amide
1º amine
LiAlH4
Isocyanide
2º amine
LiAlH4
Nitro
1º amine
Ja
Note : Alkene, alkyne, benzene rings are not reduced by LiAlH4 in ether but it is reported that
(*) double bond can be reduced by LiAlH4 / THF in few cases like :
(i) LiAlH − THF
4
⎯
→ Ph–CH2–CH2–CH2OH
Ph–CH=CH–CHO ⎯⎯⎯⎯⎯⎯
(ii) H2O
[Cinnamaldehyde]
Sodium borohydride [NaBH4 / C2H5OH or Ether] :
lp
(ii)
Sa
nk
a
It is more specific than LAH as a reducing agent. It reduces ketones and aldehydes to the
corresponding alcohols without affecting other functional groups, reduces acid chlorides to 1º alcohols.
It does not reduce any other derivative of acid. It is effective even in protic solvent like alcohol.
Reagent
NaBH4
NaBH4
NaBH4
Reactant
Aldehyde
Ketone
Acid chloride
Product
1º alcohol
2º alcohol
1º alcohol
(iii)
NaBH
4
⎯⎯⎯⎯
→
Diisobutyl Aluminium Hydride [DIBAL-H / Inert solvent] :
Diisobutyl aluminium hydride is parallel to LAH (Lithium aluminium hydride) as a reducing agent but it is
more selective.
Reagent
DiBAL-H/(–78ºC)
DiBAL-H
Reactant
Ester
Cyanide
Product
Aldehyde
Aldehyde
(a)
(b)
By DIBAL at ordinary temperature esters are reduced to alcohols but at low temperature esters
are reduced to aldehyde.
244
(c)
LAH reduce RCN to amine but DIBAL is found to be reduce it to aldehyde.
1.4
Miscellaneous reductions :
(i)
By Red P & HI :
a
ri
CH3–CN
Used to prepare alkane from carbonyl compounds and alcohols generally.
uh
Re d P + HI
⎯⎯⎯⎯⎯⎯→ R–CH2–R

(a)
Re d P + HI
(c) R–CH2OH ⎯⎯⎯⎯⎯⎯→ R–CH3

Ja
Re d P + HI
(b) CH3CH=O ⎯⎯⎯⎯⎯⎯→ CH3CH3

Re d P + HI
(d) CH3 – CH – CH3 ⎯⎯⎯⎯⎯⎯→ CH3CH2CH3
|

(ii)
lp
OH
Meerwein-Pondorf-Verley reduction (MPV reduction) (Reduction by isopropyl
alcohol and aluminium isopropoxide): It is selective reduction of ketones to alcohol, even in
Sa
nk
a
presence of other functional groups using Aluminium isopropoxide in isopropyl alcohol.
eg.
(iii)
+
+
Wolff-kishner reduction [NH2NH2 / KOH] :
Used to prepane alkane from carbonyl compounds
;
NH2NH2 / KOH
→ RCH3
RCHO ⎯⎯⎯⎯⎯⎯⎯⎯

245
Mechanism :
Note : Wolff-kishner reduction is avoid to use for compounds which have base sensitive groups. [Like :
a
ri
Halogens, Acid halide, Esters, Anhydride]
Section (C) : Oxidation-1
Introduction
3.
uh
2.
oxidation is defined as the addition of oxygen (electronegative) element to a substance or removal of
hydrogen (electropositive) element from a substance.
or
Oxidation of an organic molecule usually corresponds to increasing its oxygen content or decreasing its
hydrogen content.
Oxidation of an organic compound may be more broadly defined as a reaction that increases its content
of any element more electronegative than carbon.
Replacing hydrogen atoms by chlorine atoms is an oxidation
Ar–CH3
Ja
1.
ArCH2Cl
ArCHCl2
ArCCl3
2.1
Oxidation of alkanes
Different products are formed by the use of different oxidising agents or different reaction conditions.
Chemical oxidation with KMnO4 or K2Cr2O7 : Alkanes are usually not affected by oxidising
agents like KMnO4 or K2Cr2O7. However, alkanes having tertiary hydrogen are oxidised by these
oxidising agents to an alcohol.
KMnO4
→ (CH3)3COH
(CH3)3CH ⎯⎯⎯⎯⎯
(Isobutane)
(Tertiary butyl alcohol)
Cu / 523K /1000atm
(i) 2CH4 + O2 ⎯⎯⎯⎯⎯⎯⎯
⎯
→ 2CH3OH (methanol)
Sa
nk
a
(i)
lp
When organic compound is oxidised, oxidising agent used is reduced. When an organic compound is
reduced, the reducing agent used must be oxidized.
Ex.
9: 1
Mo O
2 3
→ HCHO + H2O
(ii) CH4 + O2 ⎯⎯⎯⎯

(CH COO )Mn
3
2
→ 2CH3COOH + 2H2O
(iii) 2CH3–CH3 + 3O2 ⎯⎯⎯⎯⎯⎯

(iv) CH4 + O2 ⎯⎯⎯→
Burn
limited
C + 2H2O
carbon
black
2.2
Oxidation of alkenes and alkynes
(i)
Baeyer reagent [cold diluted to 1% alkaline KMnO4 solution] :
Baeyer Reagent
Reactant
Product
Baeyer Reagent
Alkene
Vicinal diol
Baeyer Reagent
Alkyne
Diketone
246
Stereochemistry : syn addition.
Both-OH groups are add from same stereochemical side.
General Reaction
Stereochemistry : syn addition.
Mn
O
OH
diol
HO
O¯
Osmium tetraoxide in alkaline medium [OsO4 / NaHSO3] :
Similar to Baeyer's reagent.
OsO4 /NaHSO3
OsO4 /NaHSO3
Reactant
Alkene
Product
Vicinal diol
OH
OsO4 , H2O2
⎯⎯⎯⎯⎯⎯→
(iii)
Oxidation with peroxyacids
Ja
OH
Cyclohexene
OsO4 /NaHSO3
Alkyne
Diketone
uh
Ex.
O
+ MnO2  + H2O
a
ri
O
(ii)
¯OH
H2O
⎯⎯⎯⎯
→
⎯⎯
→
Ex.
lp
An alkene is converted to an epoxide by a peroxyacid.
[a carboxylic acid that has an extra oxygen atom in a –O–O– (peroxy) linkage].
O
O
||
||
General Reaction
C=C
+ R – C – O–
–H
+ R – C – O – H (acid)
a
Some simple peroxyacids (sometimes called peracids) are shown below :
Sa
nk
(a)
(c)
(b)
(d)
Peroxyformic acid
General Reaction
Cl
Metachloro perbenzoic acid
(MCPBA)
H+ , H O
2 →
⎯⎯⎯⎯⎯
247
Stereochemistry : anti addition in diol formation.
OH
(i) per acid
⎯⎯⎯⎯⎯
→
+
Ex.
(ii) H3O
Cyclohexene
OH
trans+ Enantiomer
a
ri
Ex.
Ex.
(iv)
lp
Ja
HCO3H
⎯⎯⎯⎯⎯
→
–HCO2H
uh
Note : The more highly substituted olifinic bond is more nucleophilic and therefore reacts faster with the
peroxyacid than the less susbstituted double bond.
Oxidation with acidic KMnO4 [KMnO4/H+] : Stereochemistry : syn addition.
Sa
nk
a
When alkene & alkyne heated with KMnO4 in acidic or in alkaline medium; following changes takes
place.
KMnO4/H+
KMnO4/H+
KMnO4/H+
KMnO4/H+
KMnO4/H+
KMnO4/H+
Reactant .= CH2 group
.= CH R group
.= CR1R2 group
. CH group
. CR group
Product
CO2
RCOOH
O.= CR1R2 group
CO2
RCOOH
Carbon dioxide
Carboxylic acid
Ketone
Carbon dioxide
Carboxylic acid
Ex.
[O]
RCH=CH2 ⎯⎯
→ CO2 + H2O + RCOOH
Ex.
[O]
RCCH ⎯⎯
→ CO2 + H2O + RCOOH
Ex.
w
[O]
RCC–R ⎯⎯
→ 2RCOOH
Ex.
KMnO4
Warm
conc.
COOH
COOH
+ CO2 + H2O
O
O
O
||
||
(1) KMnO4 , NaOH, 
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
→
+
−
− CH2CH3
HO
C
−
−
CH
C
OH
3
+
Ex.
CH3–CC–CH2CH3
Ex.
O
||
(1) KMnO4 , NaOH, 
→ CH3CH2CH2 − C − OH + CO2 + H2O
CH3CH2CH2–CCH ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
+
(2) H
(2) H
248
(v).
Oxidation with ozone (ozonolysis) :
Alkene & Alkyne
O3/H2O2
Reactant
Product
O3/H2O2
=CH2 group
CO2
Carbon dioxide
a
ri
Like permanganate ozone cleaves double at very low temperature ( –78ºC) bonds to give Ketones and
aldehydes. However, ozonolysis is milder, and both Ketones and aldehydes can be recovered without
further oxidation.
Reductive Ozonolysis Products (O3/Zn, H2O)
O3/Zn, H2O
O3/Zn, H2O
O3/Zn, H2O
O3/Zn, H2O
O3/Zn, H2O
O3/Zn, H2O
Reactant
= CH2 group
= CH R group
= CR1 R2 group
 CH group
 CR group
O
||
Product
HCHO
O = CHR group
O = CR1R2 group
diketone
–C – CHO
Formaldehyde Aldehyde
Ketone
Keto aldehyde
diketone
Oxidative Ozonolysis Products
O3/H2O2
O3/H2O2
O3/H2O2
=CHR group
= CR1R2 group
CH group
RCOOH
O=CR1R2 group
CO2
Carboxylic acid
Ketone
Carbon dioxide
O3/H2O2
CR group
RCOOH
Carboxylic acid
O3 , H2O
C8H10 (A) ⎯⎯⎯⎯
→ C4H6O2 Acid (B). Identify (A) and (B) in the above reaction
Sol.
(A)
Ex.
A certain hydrocarbon has the formula C16H26. Ozonolysis followed by hydrolysis gives CH 3(CH2)4CO2H
and succinic acid as the only product. What is hydrocarbon
DU = 4

Molecular structure must be:
CH3(CH2)4CC–CH2–CH2–CC(CH2)4–CH3
COOH
Ja
Sol.
(B)
C C
uh
Ex.
2.3
lp
Section (D) : Oxidation-2
Oxidation reaction of alcohols :
Sa
nk
1º alcohol
or primary alcohol
2º alcohol
or secondary alcohol
3º alcohol
or tertiary alcohol
a
Oxidation Product of Alcohol-1
Weak oxidising agent
Strong oxidising agent
CrO3/Inert
CrO3 in water
PCC*
PDC**
KMnO4/H+
K2Cr2O7/H+
medium
or H2CrO4
(a)
Aldehyde
Carboxylic acid
Ketone
Ketone
Not oxidised
Not oxidised
Pyridinium Chloro Chromate (PCC)
Non aqueous solvent
+ HCl + CrO3 ⎯⎯⎯⎯⎯⎯⎯
→
(b)
(c)
(d)
(e)
.Cl–.CrO3
Pyridinium dichromate (PDC) = (2C6H5N.CrO3)
Jones reagent = dilute chromic acid + acetone
Collin's Reagent = CrO3 + pyridine, CH2Cl2
MnO2 = It is selectively oxidised reagent & oxidised allylic and benzylic alcohol into aldehyde
and ketone.
249
Copper & heat
as oxidising agent
Aldehyde
Ketone
Dehydrate to alkene
a
ri
Oxidation Product of Alcohol-2
Very-Very Strong oxidising agent
KMnO4/H+/Heat
1º alcohol or Primary alcohol
Carboxylic acid
2º alcohol or Secondary alcohol
Mixture of Carboxylic acid
3º alcohol or tertiary alcohol
Mixture of Carboxylic acid
Ex.
Write the product of following reactions.
(b) RCHO > RCH2OH
Ja
Rate of Oxidation :
(a) RCH2OH > R2CHOH > R3C–OH (inert)
uh
Oppenaur's oxidation
(i) This reaction involves the oxidation of a secondary alcohol with a ketone and base to the
corresponding ketone.
(ii) Commonly used bases are aluminium tert-butoxide.
lp
–
CH3–CH=CH– CH –CH2–CH2–OH
OH
(X)
O
K 2Cr2O7
X ⎯⎯⎯⎯⎯→ CH3 – C – OH + HOOC– C –CH2– C –OH
(1) H2SO4
O
O
PCC
(2)
a
→ CH3–CH=CH– C –CH2–CHO
X ⎯⎯⎯⎯
O
Oppenauer oxidation
X ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
→ CH3–CH=CH– C –CH2–CHO
Sa
nk
(3 )
O
CrO3 / aq.
→ CH3–CH=CH– C –CH2–COOH
X ⎯⎯⎯⎯⎯⎯
( 4) acetone
O
MnO2
→ CH3 − CH = CH − C − CH2 − CH2 − OH
X ⎯⎯⎯⎯
(5 )
||
O
2.4
Oxidation reaction of carbonyl compound
(i)
Acidic KMnO4 & K2Cr2O7 as oxidising agent :
Aldehydes are oxidised to carboxylic acid having same number of C atoms as aldehyde.
HCHO + [O] ⎯→ HCOOH
;
RCHO + [O] ⎯→ RCOOH
Ketones are oxidised with difficulty. They are oxidised only on heating with a strong oxidising agent.
2–
Cr2O7 + H+
→ CH3COOH + CO2 + H2O
+ [O] ⎯⎯⎯⎯⎯⎯⎯
or KMnO4 +H
In case of mixed or unsymmetrical ketones the >C = O group remains with the smaller alkyl group.
250
(Popoff’s rule)
[O]
[O]
→ 2CH3COOH ; CH3COCH2CH2CH3 ⎯⎯⎯
→ CH3COOH + CH3CH2COOH
CH3COCH2CH3 ⎯⎯⎯
Mechanism (with Cr+6 oxidising agents) :
(i) CrO3 + H2O
H2CrO4 =
O
H
–H2O
→
(ii) R – C – OH + H – O – Cr – OH ⎯⎯⎯⎯
O
OH
O
OH
H2CrO4
rds
R – C – OH ⎯⎯⎯⎯⎯
→
⎯⎯⎯
→ R – C = O + H2CrO3
R – C – O – Cr – OH
OH
H
H
O
Acid
Gemdiol
unstable
..
H2O
uh
–
– –
(iii) R – C = O
H
a
ri
H
rds
⎯⎯⎯
→ R – C = O + H2CrO3
H
(ii)
lp
Ja
Remarks :
(1) Primary alcohol forms a chromate ester with chromic acid.
(2) The chromate ester decomposes in 2nd slow step with the elimination of -hydrogen. So the first
oxidation product, an aldehyde is obtained.
(3) In aqueous medium, aldehyde forms a gemdiol (hydrated aldehyde). It is further oxidised to an acid
by similar mechanism.
(4) The following reactivity orders can be explained by this mechanism :
Rate of Oxidation
(a) R–CH2OH > R–CD2OH (Bond Energy : C–H < C–D ) (b) RCHO > RCDO
With Tollen's reagent :
Sa
nk
a
RCHO (aldehydes) can be easily oxidised to RCOOH (except HCHO that can be oxidised to CO 2) by
weak oxidising agents like ammonical AgNO 3 (Tollen’s reagent) hence they are better reducing
agents.
Aldehydes reduce Tollen’s reagent to Ag and appears in the form of silver mirror is called silver-mirror
test. It is given by all aldehydes and reducing sugars.

RCHO + 2[Ag(NH3)2+ ] + 3OH– ⎯⎯
→ RCOO– + 2Ag 
(iii)
+ 4NH3 + 2H2O
(Silver mirror )
With Fehling solution :
Aldehydes (except benzaldehyde) reduce Fehling’s solution (Cu2+ reduced to Cu+) which is an alkaline
solution of cupric (Cu2+) ion complexed with tartrate ion.
Cu2 O 

RCHO + 2Cu2+ + 3OH– ⎯⎯
+ 2H2O
→ RCOO– +
red ppt.
Aldehydes also reduce Benedict’s solution (Cu2+ complexed with citrate ion) to Cu+
(iv)
With Benedict's solution
Sodium citrate + NaOH + NaHCO3 + CuSO4
RCHO + Cu2+
HO
2
⎯⎯⎯
→

+
RCHO + HgCl2 + H2O ⎯⎯
→ RCOOH + 2HCl + Hg2Cl2

RCHO + Hg2Cl2 + H2O → RCOOH + 2HCl + 2Hg  grey ppt.
251
(v)
With Schiff's reagent
Schiff's Reagent is aq. solution of following base decolourised by passing SO 2.
Aldehyde restore pink colour of Schiff's reagent.
RCHO
SO2
→ RCOOH + Pink colour
⎯⎯⎯
→ Colourless solution (Schiff's Reagent) ⎯⎯⎯⎯
a
ri

(vi)
uh
p-Rosaniline Hydrochloride
Magenta colour (Fuschin)
Ketons are not easy to oxidize so they do not give these 5 tests. These five tests can be used to
distinguish aldehyde and ketones. Both gives 2, 4 DNP test.
Oxidation by using SeO2
SeO
2
⎯⎯⎯→
(vii)
Ja
SeO2 is a selective oxidizing agent with converts –CH2–group adjacent to carbonyl group into carbonyl
group. The reagent, in general, oxidises active methylene and methyl groups to ketonic and aldehydic
groups respectively.
SeO
2
⎯⎯⎯→
;
Baeyer-villiger oxidation
Ph COOOH
⎯⎯⎯⎯⎯→
(67 %)
O
a
CH2Cl2
lp
Baeyer-villiger oxidation is the oxidative cleavage of a carbon-carbon bond adjacent to a carbonyl which
converts ketones to esters and cyclic ketones to lactones.
It can be carried out with peracids such as MCBPA, or with hydrogen peroxide and a lewis acid.
O
O
O
HO
Sa
nk
2 2
⎯⎯⎯
→
BF3
Ether
O
(62%)
O
KETONES ARE DIFFICULT TO OXIDIZE :
Ketones can be oxidized from their enolic form at high temperature with very strong oxidizing agent.
Oxidation of ketones is sometimes governed by Popoff's rule. According to this rule carbonyl group
remains with the smaller alkyl group. More electron rich alkene will be easy to oxidized.
[O]
⎯⎯⎯
→ MeCOOH + CO2 + H2O

252
2.5
Oxidation reaction of diols
(i)
HIO4 oxidation : (Oxidation by lead acetate is similar to HIO4 oxidation)
Mechanism :
R
| +
R−C
||
O
R
|
C −R
||
O
Ex.
uh
a
ri
Remarks
(1) HIO4 (periodic acid) oxidises vicinal diols (1, 2-diols).
(2) It brings about oxidative cleavage of vicinal diol.
(3) It can also oxidise -hydroxy carbonyl compound and -dicarbonyl compound.
(4) HIO4 forms a cyclic periodate ester as an intermediate. So the two –OH groups should have synconformation.
(5) In cyclic diols only cis-vicinal diols are oxidised. Trans isomers are not oxidised.
General reaction
HIO
Ex.
HIO
4
⎯⎯⎯⎯
→
Section (E) : Hydrolysis
Ja
4
⎯⎯⎯⎯
→ R – CHO + HCOOH + R–CHO
lp
Introduction :
Hydrolysis of an ester :
Sa
nk
(i)
a
Hydrolysis is a chemical reaction or process in which a molecule splits into two parts by reacting
with a molecule of water, (H2O). One of the parts gets OH– from the water molecule and the other part
gets H+ from the water. Such reactions are endothermic.
This is distinct from a hydration reaction, in which water molecules are added to a substance, but no
fragmentation of molecule/species occurs. Such a process is exothermic.
Hydrolysis of an ester involves breaking off an ester link. It can takes place in
(b)
(a) Mild acidic medium : Dilute H2SO4, dilute HCl.
Strong alkaline medium : Aqueous NaOH or KOH and heat.
One hydrolysis product contains a hydroxyl functional group, while the other contains a carboxylic acid
functional group.
+
253
(ii)
Hydrolysis of an anhydride :
The hydrolysis of acid anhydride produces two carboxylic acids.
+
(iii)
OH
Hydrolysis of acid halide :
Cl
(iv)
a
ri
Hydrolysis of an acid halide results into a carboxylic acid and hydrogenhalide. Only the carboxylic acid
product has a hydroxyl group derived from the water. Hydrohalic acid product gains the remaining
hydrogen ion.
OH + HCl
Hydrolysis of acid amide :
NH2
uh
Hydrolysis of an amide results into a carboxylic acid and an amine product or ammonia, only the
carboxylic acid product has a hydroxyl group derived from the water. The amine product (or ammonia)
gains the remaining hydrogen ion.
OH + NH
3
Hydrolysis of cyanides :
Ja
(v)
Cyanide on hydrolysis produce ammonia and carboxylc acids. It is carried out in acidic medium
generally but hydrolyse in basic medium also.
OH + NH
(vi)
lp
3
Hydrolysis of isocyanides :
Isocyanides on hydrolysis produce primary amines and formic acids. It is carried out in acidic medium.
H
a
OH + RNH
2
Sa
nk
Note : Alkylisocyanide does not hydrolyse in basic medium.
(vii)
Hydrolysis of imine :
H O+
3
⎯⎯⎯→
R – CHO + NH
3
(viii) Hydrolysis of ethers :
Ethers has R–O–R group. In strong acidic medium (HI or HBr) in hydrolysis to produce 2 equivalent of
alcohols.
+
H3 O
C2H5–O–C(CH3)3 ⎯⎯⎯
→ C2H5OH + (CH3)3C–OH
(ix)
Hydrolysis of Vinyl ether :
H O+
3
⎯⎯⎯→
(x)
CH3–CHO + ROH
Hydrolysis of phenyl ether :
H O+
3
⎯⎯⎯→
Ph–OH + ROH
254
(xi)
Hydrolysis of hemiacetals and acetals :
Hemiacetals and Acetals has R–O–R group.
Hemiacetals are unstable and get hydrolysed to aldehyde/ketones even in aq medium. However
acetals/Ketals are stable and hydrolyse only in strong acidic medium (HI or HBr) to produce 2
equivalent of alcohols and one equivalent of aldehyde/ketone. Acetals are often used as protecting
groups.
C
+
H /H2O
OR
HO–H
OR
C
OR
HO–H
OH
C
ROH +
Unstable
(xii)
C=O
OH
Hydrolysis of epoxide :
H O+
3
⎯⎯⎯→
a
ri
OR
Ja
uh
Note : Three and four membered epoxides can also be hydrolysed in basic medium.
 Marked questions are recommended for Revision.
Section (A) : Reduction-1
A-1.
lp
PART - I : SUBJECTIVE QUESTIONS
Write the hydrogenation product of following species with H2/Pd.
H2 / Pd
(b) trans-2-butene ⎯⎯⎯⎯
→
CH2
H2 / Pd
C
⎯⎯⎯⎯
→
(d)
1 eq.
CH3
(Limonene)
a
H2 / Pd
(a) 1,2-Butadiene ⎯⎯⎯⎯
→
Sa
nk
H2 / Pd
(c) Benzaldehyde ⎯⎯⎯⎯
→
A-2. Write the hydrogenation product of following species
H (excess)
2
→
(a) CH2=CH–CH2–CH2–CC–CH3 ⎯⎯⎯⎯⎯
Pd
H2 /Pd−BaSO4
(b) CH2=CH–CH2–CH2–CC–CH3 ⎯⎯⎯⎯⎯⎯
→
H2 / Pd / BaSO 4
(c) Benzoylchloride ⎯⎯
⎯⎯⎯⎯→
A-3. Complete the following reactions :
(i) Na / NH (l)
3
→
(i) CH3–CH2–CC–CH2–(CH2)6–CH2OH ⎯⎯⎯⎯⎯⎯
(ii)
(ii) H2O
(i) Na / NH (l)
3
⎯⎯⎯⎯⎯⎯
→
(ii) H2O
255
A-4.
Give reaction conditions (reagents and/or catalyst) for effecting the following conversions :
(i) CH3–(CH2)7–CC–(CH2)7–CH3 ⎯⎯
→
(ii)
C=C
⎯⎯
→
What is the product of each reaction
(ii)
B-2. Identify a and b, in the following reactions :
NaBH4
a
CH3OH
(i)
a
(ii)
Complete the following reactions :
Sa
nk
B-3.
b
lp
LiAlH4
uh
(i)
Ja
B-1.
a
ri
Section (B) : Reduction-2
(a)
Re d P + HI
⎯⎯⎯⎯⎯⎯
→
(b)
Section (C) : Oxidation-1
C-1. Write the structural formulas for the products formed when 3-heptyne reacts with KMnO4 under ?
(i) neutral condition at room temp.
(ii) alkaline or acidic condition at higher temp.
C-2.
Complete the following reactions :
O
CH2–CH3
(a)
KMnO4
CH3
C
(b)
KMnO4
256
KMnO4
(c)
(d)
C-3. When t-Butanol and n-Butanol are separately treated with a few drops of dilute KMnO 4 in one case
only, the purple colour disappears and a brown precipitate is formed. Which of the two alcohols gives
the above reaction and which is the brown precipitate.
[IIT-JEE, 1994]
Section (D) : Oxidation-2
D-2.
Complete the following reactions :
D-3
Complete the following reactions :
(a) Ph–CH2–CH2–OH
Section (E) : Hydrolysis
; (c)
lp
; (b)
(c)
uh
(b)
Ja
(a)
a
ri
D-1. Complete the following reactions :
Write the products of following reaction
O
O
E-2.
Write the products of following reaction
a
E-1.
H O+
H O+
3
→
(b) CH3NC ⎯⎯⎯⎯
Sa
nk
3
(a) CH3–CN ⎯⎯⎯→
H O+
E-3.
3
→ CH3COOH + HCl
Reactant ⎯⎯⎯⎯
E-4.
3
→ Product is :
CH3CONH2 ⎯⎯⎯⎯
E-5.
The hydrolysis of acid anhydride produces ...................
H O+
NHCOCH3
H O+
3
⎯⎯⎯⎯
→ Product is :
E-6.
NO2
E-7.
H O+
3
→ CH3CHO + CH3OH
Reactant (C3H6O) ⎯⎯⎯⎯
257
PART - II : ONLY ONE OPTION CORRECT TYPE
Section (A) : Reduction-1
A-1.
The relative rates of hydrogenation is in the order of :
(A) CH2 = CH2 > RCH = CH2 > RCH = CHR > R2C = CHR
(B) R2C = CHR > RCH = CHR > RCH = CH2 > CH2 = CH2
(C) RCH = CHR > R2C = CHR > RCH = CH2 > CH2 = CH2
(D) R2C = CHR > CH2 = CH2 > RCH = CHR > RCH = CH2
A-2.
In which case the reaction is most exothermic with H2 / Ni.
(C)
(D)
An organic compound with molecular formula C6H10 is not reduced by H2 / Pd / BaSO4. From the given
options, the compounds may be :
(A) I, II
(B) I, III
A-4.
(C) II, IV
x, x is :
(B) Only (S,S) product
(D) Racemic mixture
Ja
(A) Only (R,R) product
(C) Meso compound
(D) II, II
uh
A-3.
(B)
a
ri
(A)
Na / NH3 ( )
⎯⎯⎯⎯⎯⎯
→ P, the product can be :
A-5.
(B)
a
(A)
lp
C2H5OH
Sa
nk
A-6.
(C)
(D)
product
(A)
(B)
(C) Both (A) and (B)
(D) None of these
A-7.
Which of the following reagents converts both acetaldehyde and acetone to alkanes ?
(A) Ni/H2
(B) LiAlH4
(C) I2/NaOH
(D) Zn–Hg/conc.HCl
A-8.
Stephen reduction (SnCl2 / HCl) converts cyanides to
(A) Aldehydes
(B) Ketones
(C) Amines
(D) Acids
258
Section (B) : Reduction-2
B-1. When benzoic acid is treated with LiAlH4, it forms
(A) Benzaldehyde
(B) Benzyl alcohol
(C) Benzene
Name the reaction
Me– 
Me– 
(A) Meervein-Ponndorf-verley reduction
(C) Bouveault-Blanc reduction
B-3.
(B) Wolff-kishner reduction
(D) Stephen's reduction
DIBAL −H, H O
2
→ Product
R–CN ⎯⎯⎯⎯⎯⎯
−40ºC
The product formed is :
(A) R–CO–NH2
(B) R–CH2–NH2
(C) R–CHO
CH2–CH2–CHO
B-4. B
A
CH2–CH2–CH2–OH
,
(B)
(C)
,
CH2–CH2–CH2–OH in both case
(D)
B-6.
in both case
In the following reaction C2H5OC2H5 + 4H
(A) Ethane
(B) Ethylene
diisobutyl
Red P + HI
2X + H2O, X is
(C) Butane
(D) Propane
(C) R–COOH
(D) R–CH3
⎯⎯⎯⎯⎯⎯⎯⎯→
lp
B-5.
CH2–CH2–CH2–OH
Ja
(A)
(D) R–CH2–NO2
uh
A and B are respectively :
a
ri
B-2.
(D) Toluene
→ Y + R' –OH
R–CO–O–R' ⎯⎯⎯⎯⎯⎯⎯⎯
(B) R–CHO
Sa
nk
The product Y is
(A) R–CH2–OH
a
aluminium hydride
– 78ºC
B-7. What are A and B in the following ?
Q
(A)
(B)
(C)
(D) P =
LiAlH
4
⎯⎯⎯⎯
→ P
in all cases
in all cases
in all cases
and Q=
259
Section (C) : Oxidation-1
C-1. Baeyer’s reagent decolourises which of the following :
(A) Alkane
(B) Alkene only
(C) Alkene and alkyne both
(D) Benzene
C-3.
C-4.
Ethanol on reaction with alkaline KMnO4 gives:
(A) Ethanal
(B) Glyoxal
(C) Acetic acid
(D) Acrolein.
KMnO / 
4
1-Butyne ⎯⎯⎯⎯⎯
→ X+Y
Identify X and Y ?
(A) CH3CH2CH2COOH + O2
(C) CH3CH2COOH + CO2 + H2O
(B) CH3CH2COOH
(D) CH3CH2COCH3 + HCOOH
a
ri
C-2.
An alkyne C7H12 when reacted with alkaline KMnO4 followed by acidification by HCl, yielded a
mixture of CH3 − CH − COOH & CH3CH2COOH. The alkyne is |
CH3
(B) 2-methyl-2-hexyne
(D) 3-methyl-2-hexyne
uh
(A) 3-hexyne
(C) 2-methyl-3-hexyne
; R1 and R2 are
(A) Cold alkaline KMnO4, OsO4/H2O2
(C) Cold alkaline KMnO4, C6H5CO3H
Section (D) : Oxidation-2
Ja
C-5.
(B) Cold alkaline KMnO4, HCO3H & H3O+
(D) C6H5CO3H, HCO3H
lp
D-1. Glucose as well as fructose are oxidized by periodic acid. The number of moles of HCOOH formed from
each mole of glucose and fructose are
(A) 5 and 5
(B) 5 and 4
(C) 5 and 3
(D) 4 and 3
D-2. Secondary alcohols on heating with copper at 300ºC give
(A) Alkenes
(B) Aldehydes
(C) Ketones
a
The reagent, with which both acetaldehyde and acetone react easily is :
(A) Tollens reagent
(B) Schiffs reagent
(C) H2 / Ni
(D) Fehling’s solution
Sa
nk
D-3.
(D) tert-alcohols
D-4. Which of the following compounds is resistant to periodic acid oxidation ?
CH2OH
CH2OH
CH2OH
(A)
CO
(B)
CH2
(C)
CH2OH
CH2OH
CH2OH
(D) CH OH
2
CHOH
CH2OH
Section (E) : Hydrolysis
E-1.
What product is obtained when Benzenecarbonitrile is hydrolysed.
(A) Benzoylchloride
(B) Benzenecarboxamide
(C) Benzaldehyde
(D) Benzoic acid
E-2.
The acid catalysed hydrolysis products in the following reaction are -
O
O
O
O
&
CHO
COOH
CHO
CHO
(A)

H3O
⎯⎯⎯
⎯
→ P+Q
(B)
CHO
&
COOH
CH 2–OH
CH 2–OH
CHO
(C)
&
CHO
CH 2–OH
CH 2–OH
COOH
(D)
CHO
&
COOH
CHO
260
OEt
E-3.
Product obtained in above reaction are :
OH
(A) EtOH,
, CH3–CHO
(B)
O
O
O
(C) CH3COOH,
(D) EtOH,
PART - III : MATCH THE COLUMN
(z)
HOH2C
Reagents-I
CrO3 / Pyridine / CH2Cl2
NaBH4
Na / C2H5OH
CrO3 / H+
(p)
(q)
(r)
(s)
CH3
O
Functional group oxidised / reduced-II
W
Z
X
Y
Match the following column:
Column-I
Reactant and reagents
COOMe
LiAlH4 / ether
(A)
Ph
⎯⎯⎯⎯⎯⎯
→
COOMe
(i)DIBAL −H ( −78C)
Ph
(B)
⎯⎯⎯⎯⎯⎯⎯⎯
→
Column-II
Products
Ph
(q)
Ph
NaBH
(r)
Me
(s)
Me
No reaction
lp
(p)
a
2.
O
(w)
Ja
(A)
(B)
(C)
(D)
(ii)H2O
(C)
COOMe
Me
4
⎯⎯⎯⎯
→
H +Pd / C
2
⎯⎯⎯⎯⎯
→
Sa
nk
(D)
COOMe
Ph
a
ri
Observe the following compound and match the reagents of List-I and List-II
(x)
O
(y)
HO
CHO
uh
1.
(t)
OH + MeOH
CHO
+ MeOH
CHO
+ MeOH
COOMe
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
1.
H / Ni
2
A ⎯⎯⎯
⎯
Heating with HI
B,
⎯⎯⎯⎯⎯⎯→
A and B can be :
(A) Both are n-Hexane
261
(B) Both are Hexan-1,2,3,4,5,6-hexaol
(C) A is n-Hexane B is Hexan-1,2,3,4,5,6-hexaol
(D) A is Hexan-1,2,3,4,5,6-hexaol and B is n-Hexane
2.
O3 / Zn / 
4
→ (C) Red colour ; (B) is :
→ (B) ⎯⎯⎯⎯⎯⎯⎯
⎯⎯⎯⎯⎯
→ (A) ⎯⎯⎯
Fehling solution
uh
HIO
3.
H2O
Consider reduction of 2-butanone.
A, B and C are respectively.
in all cases
a
(A)
in all case
Sa
nk
(C)
5.
(D) H–COOH
lp
B
(C) CH3–COOH
Ja
(B) CH3–CHO
(A)
4.
a
ri
In the above reaction the using reagents X and Y are ;
(A) Na/ liq. NH3 for X
(B) H2, Pd/ BaSO4 for Y
(C) BH3-THF + CH3COOH for Y
(D) All of these are correct.
(B)
,
(D)
,
,
,
Identify (P) and (Q) respectively in the given reaction :
KMnO 4
(A)
(C)
PCC,CH Cl
2 2
⎯⎯⎯⎯⎯
⎯
→ (Q)
Acidic
(P) ⎯⎯⎯
⎯
&
(B)
&
(D)
&
&
262
Which of the following sets of compounds cannot turn clear orange solution of CrO 3 / aq. H2SO4 to
greenish opaque solution
OH
OH
OH
|
|
CH3 − C − CH3
CH3 − C − CH3
CH3 CH2 − OH
|
|
CH3
H
( I)
,
,
,
( III )
(IV)
( II )
(A) I, IV
(B) II, III
(C) I, II
(D) III, IV
(i ) LiAlH4
PCC
N2H4 / glycol
⎯⎯
⎯⎯
⎯→ A ⎯⎯
⎯→ B ⎯⎯
⎯
⎯ ⎯→ C
(ii) H3 O +
KOH
7.
Product C is :
OH
(A)
(C)
(D)
The product which is not formed in the following reaction :
CH3
CH3
|
|
HIO4 (excess)
Ph − C − CH − CH − CH − CH2OH ⎯⎯⎯⎯⎯⎯⎯⎯
→
|
|
|
OH OH OH
Ja
8.
(B)
uh
COOH
a
ri
6.
(A) HCOOH
Reagent ''P'' in the given reaction is :
Sa
nk
9.
(A) LiAlH4
10.
CH3
|
(D) OHC − CH − CHO
a
CH3
|
(C) OHC − CH − CH2OH
lp
O
||
(B) Ph − C − CH3
(B) NaBH4
(C) DIBAl-H
(D) OH–
H O+
3
⎯⎯⎯→
P+Q
P and Q are respectively.
(A) Acetone and Hexane-1,2,5,6-tetraol.
(C) Acetaldehyde and Hexane-1,2,5,6-tetraol.
(B) Acetaldehyde and Acetone.
(D) Acetone and Formaldehyde.
263
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
1.
Find the value of x + y.
H / Pd
2
⎯⎯⎯⎯
→ -----------------
(a)
4
⎯⎯⎯⎯⎯
→ -----------------
(e)
Na / NH3 ( )
⎯⎯⎯⎯⎯
→ -----------------
H / Pd / BaSO
2
4
⎯⎯⎯⎯⎯⎯
→ -----------------
(b)
LiAlH / THF
(c)
a
ri
Number of reactions which give alcohol as product.
NaBH / EtOH
4
⎯⎯⎯⎯⎯⎯
→ -----------------
(d)
uh
2.
3.
How many reactions are correct ?
OH
dil. KMnO4 / HO
OH
lp
⎯⎯⎯→
Na / NH3 (  )
⎯⎯
⎯⎯
⎯→
CH3
Sa
nk
CH3
CH3
CH3
KMnO4
⎯⎯⎯⎯
→
(v)
OH
(i)
5HIO
4
⎯⎯⎯⎯
→

KMnO4 /HO / 
⎯⎯⎯⎯⎯⎯⎯
→
(iii)
a
(ii)
4.
O
NO2
NO2
(iv)
O
HO 4
⎯⎯⎯⎯⎯⎯⎯
→
(i)
Ja
C2H5OH
H
H
C == C
D2 / Ni
⎯⎯
⎯→
CH3
D
H
H
D
CH3
(ii)
5HIO
4
⎯⎯⎯⎯
→
Sum of moles of formaldehyde obtained in the reaction (i) and reaction (ii) ?
264
How many of following reactions are hydrolysis reactions ?
H+
→
(1) CH3–COOH + C2H5OH ⎯⎯⎯
H O+
H O+
3
(4) CH3–CN ⎯⎯⎯→
3
⎯⎯⎯→
(3)
H O+
H O+
3
⎯⎯⎯→
(5)
3
(6) CH3–CH=NH ⎯⎯⎯→
H O+
3
(7) CH2=CH–O–C2H5 ⎯⎯⎯→
H O+
3
⎯⎯⎯→
(8)
How many of the following may produce salicylic acid on hydrolysis under appropriate conditions ?
(ii)
(iv)
(v)
(iii)
uh
(i)
Ja
6.
H O+
3
⎯⎯⎯→
(2)
a
ri
5.
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
2.
lp
Which of the following catalysts is/are used for partial reduction of alkyne ?
(A) Na/NH3()
(B) Ni2B or P-2catalyst
(C) Lindlar catalyst
(D) Rossenmund catalyst
1. DiBAL −H
X ⎯⎯⎯⎯⎯
→
X is
Sa
nk
2. H2O
a
1.
(A)
3.
(B)
(C)
(D)
Identify the possible structure of X and Y
COOH
KMnO4 / OH / 
C9H12O
(X or Y)
resolution
K2Cr2O7 / H2SO4
d+
Blue green salt
+ other products
265
X
Y
(A)
(B)
a
ri
(C)
(D)
; (B) and (C) both give +ve iodoform test. Compound (A) is :
(B)
Ja
(A) CH3–CH=CH–O–CH2–CH3
uh
4.
(D) None of these
lp
(C)
Product is obtained in the above reaction is :
Sa
nk
a
5.
(A) R–CO2Na
(B)
(C) RCHO
(D) None of these
PART - IV : COMPREHENSION
Read the following passage carefully and answer the questions.
Comprehension # 1
266
1.
(P) and (Q) respectively are
OH
O
HO
O
DO
(A)
O
O
,
DO
OD
OH
O
OH
OH
O
(B)
,
a
ri
OH
OD
OH
OH
(C)
OH
HO
O
, HO
O
O
(D)
OH
OD
DO
,
uh
OD
O
HO
O
2.
Ja
OD
(R) is :
OH
O
OD
O
OH
D
HO
(C)
O
Sa
nk
D
O
OH
D
D
(B)
a
(A)
D
D
lp
OD
OH
OH
OD
OD
(D)
OD
OD
Comprehension # 2
H2 + Pd – BaSO 4
Na + liq. NH 3
+ EtOH
3.
(C)
alkaline
KMnO 4
(D)
(F)
alkaline
KMnO 4
(G)
HIO4
(E)
The compound (C) is :
(A)
(B)
(C) Et –  – Et
(D) Both (A) and (B)
267
The compound (F) is :
(A)
The compound (D) is :
(C)
(B)
(C)
(D) Both (B) and (C)
The compound (G) is :
(A)
7.
(B)
a
ri
(A)
6.
(D) Both (A) and (B)
The compound (E) is :
(A) Two moles of
.
(C) One mole of (A) and one mole of (B)
(D) Both (B) and (C)
uh
5.
(C) Et –  – Et
(B)
(B) Two moles of
Ja
4.
.
(D) No reaction.
CH3–C–O–C–CH3
O
O
OPh
Sa
nk
(II)
a
(I)
lp
Comprehension # 3
Answer Q.8, Q.9 and Q.10 by appropriately matching the information given in the three columns
of the following table.
Column-1, 2 and 3 contains starting material, reaction condition and type of reaction respectively.
Column-1
Column-2
Column-3
(i)
KMnO4
(P)
Oxidation
(ii)
Cu/
(Q)
Reduction
(iii)
H3O+
(R)
Hydrolysis
(iv)
LiAlH4
(S)
Dehydration
CH3
(III)
(IV)
CH3–C–OH
CH3
8.
Which of the following combination of reaction result in formation of an alkene.
(A) (I) (iv) Q
(B) (III) (i) S
(C) (III) (ii) S
(D) (II), (iii) R
9.
The only correct combination in which product gives position test with sodium bicarbonate is (A) (III) (ii) P
(B) (I) (iii) Q
(C) (II) (iii) R
(D) (IV) (i) P
10.
The reaction and condition which obey bimolecular nucleophilic substitution reaction with respect to
tetrahedral intermediate is (A) (IV) (i) P
(B) (I) (iv) Q
(C) (II) (iii) R
(D) (III) (ii) S
268
* Marked Questions may have more than one correct option.
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
1.
What would be the major product in the following reaction ?
[IIT-JEE 2000(M), 1/100]
H
2
⎯⎯⎯⎯⎯⎯
→
Lindlar catalyst
Hydrogenation of the adjoining compound in the presence of poisoned palladium catalyst gives.
a
ri
2.
[IIT-JEE
(B) an optically inactive compound
(D) a diastereomeric mixture
uh
2001(S), 1/35]
(A) an optically active compound
(C) a racemic mixture
1-Propanol and 2-Propanol can be best distinguished by :
[IIT-JEE 2001(S), 1/35]
(A) oxidation with alkaline KMnO4 followed by reaction with Fehling solution
(B) oxidation with acidic dichromate followed by reaction with Fehling solution
(C) oxidation by heating with copper followed by reaction with Fehling solution
(D) oxidation with concentrated H2SO4 followed by reaction with Fehling solution
4.
Assertion : Dimethylsulphide is commonly used for the reduction of an ozonide of an alkene to get the
carbonyl compounds
[IIT-JEE-2001(S), 1/35]
Reason : It reduces the ozonide giving water soluble dimethyl sulphoxide and excess of it evaporates
(A) Assertion is True, Reason is True; Reason is a correct explanation for Assertion.
(B) Assertion is True, Reason is True; Reason is NOT a correct explanation for Assertion.
(C) Assertion is True, Reason is False.
(D) Assertion is False, Reason is True.
5.
Five isomeric para-disubstituted aromatic compounds A to E with molecular formula C8H8O2 were given
for identification. Based on the following observations, give structures of the compounds.
(i) Both A and B form a silver mirror with Tollen’s reagent; also B gives a positive test with FeCl 3
solution.
(ii) C gives positive iodoform test.
(iii) D is readily extracted in aqueous NaHCO3 solution.
(iv) E on acid hydrolysis gives 1, 4–dihydroxybenzene.
[IIT-JEE-2002(M), 5/60]
Sa
nk
a
lp
Ja
3.
6.
The product of acid hydrolysis of P and Q can be distinguished by :
P=
(A) Lucas reagent
(C) Fehling’s solution
[IIT-JEE 2003(S), 3/84]
Q=
(B) 2,4–DNP
(D) NaHSO3
7.
Amongst the following the reagent that would convert 2-hexyne into trans-2-hexene is
[IIT-JEE 2004(S), 3/84]
(A) H2 .Pt / O2
(B) H2 .Pd / SO42–
(C) Li / NH3 / C2H5OH
(D) NaBH4
8.
A compound P(C5H10O) reacts with dilute sulfuric acid to give Q and R as the final products. This
reaction is about 1015 times faster than of ethylene. Both Q and R give positive iodoform test.
(a) Identify the structures of P, Q and R.
(b) Rationalize the extraordinary reactivity of P.
[IIT-JEE-2004(M), 2/60]
269
Match each of the compounds in Column I with its characteristic reaction(s) in Column II.
[IIT-JEE 2009, 8/160]
Column I
Column II
(A)
CH3CH2CH2CN
(p)
Reduction with Pd-C/H2
(B)
CH3CH2OCOCH3
(q)
Reduction with SnCl2/HCl
(C)
CH3–CH=CH–CH2OH
(r)
Development of foul smell on treatment with chloroform and
alcoholic KOH.
(D)
CH3CH2CH2CH2NH2
(s)
Reduction with diisobutylaluminium hydride (DIBAL-H)
(t)
Alkaline hydrolysis
10.*
With reference to the scheme given, which of the given statments(s) about T, U, V and W is (are)
correct?
a
ri
9.
uh
[IIT-JEE 2012, 4/136]
(A) T is soluble in hot aqueous NaOH
(B) U is optically active
(C) Molecular formula of W is C10 H18O4
(D) V gives effervescence on treatment with aqueous NaHCO3
Ja
The number of optically active products obtained from the complete ozonolysis of the given compound
is :
lp
11.
(A) 0
(B) 1
(C) 2
[JEE-2012, 3/70]
(D) 4
Consider all possible isomeric ketones, including stereoisomers of MW = 100. All these isomers are
idependently reacted with NaBH4 (Note: stereoisomers are also reacted separately). The total number
of ketones that give a racemic product(s) is/are
[JEE(Advance)-2014, 3/120]
13.*
Reagent(s) which can be used to bring about the following transformation is(are) :
[JEE(Advance)-2016, 3/124]
Sa
nk
a
12.
O
C
O
O
COOH
(A) LiAlH4 in (C2H5)2O
(C) NaBH4 in C2H5OH
O
O
C
O
H
O
OH
COOH
(B) BH3 in THF
(D) Raney Ni/H2 in THF
270
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) OFFLINE PROBLEMS
But-1-ene may be converted to butane by reaction with :
(1) Zn–HCl
(2) Sn–HCl
(3) Zn–Hg
[AIEEE-2003, 3/225]
(4) Pd/H2
2.
When CH2=CH–COOH is reduced with LiAlH4, the compound obtained will be: [AIEEE-2003, 3/225]
(1) CH3–CH2–COOH
(2) CH2=CH–CH2OH
(3) CH3–CH2–CH2OH (4) CH3–CH2–CHO.
3.
Which one of the following is reduced with Zn, Hg and HCl acid to give the corresponding hydrocarbon?
[AIEEE-2004, 3/225]
(1) Ethyl acetate
(2) Butan-2-one
(3) Acetamide
(4) Acetic acid
4.
The best reagent to convert pent-3-en-2-ol into pent-3-ene-2-one is
[AIEEE-2005, 3/225]
(1) Pyridinium chloro-chromate
(2) Chromic anhydride in glacial acetic acid
(3) Acidic dichromate
(4) Acidic permanganate
5.
The hydrocarbon which can react with sodium in liquid ammonia is:
[AIEEE-2008, 3/105]
(1) CH3CH2CCH
(2) CH3CH=CHCH3
(3) CH3CH2CCCH2CH3
(4) CH3CH2CH2CCCH2CH2CH3
6.
In the following sequence of reactions, the alkene affords the compound ‘B’
uh
a
ri
1.
H2O
[AIEEE-2008, 3/105]
→ B, The compound B is :
CH3CH=CHCH3 ⎯⎯⎯
→ A ⎯⎯⎯
O3
Zn
(1) CH3COCH3
(2) CH3CH2COCH3
(3) CH3CHO
(4) CH3CH2CHO
One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular
mass of 44 u. The alkene is :
[AIEEE-2010, 4/144]
(1) propane
(2) 1-butene
(3) 2-butene
(4) ethene
8.
Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the
presence of :
(1) two ethylenic double bonds
(2) a vinyl group
[AIEEE-2011, 4/120]
(3) an isopropyl group
(4) an acetylenic triple bond
9.
2-Hexyne gives trans-2-Hexene on treatment with :
(1) Pt/H2
(2) Li / NH3
(3) Pd/BaSO4
10.
In the given transformation, which the following is the most appropriate reagent ? [AIEEE-2012, 4/120]
[AIEEE-2012, 4/120]
(4) Li AlH4
a
lp
Ja
7.
Reagent
Sa
nk
⎯⎯⎯⎯→

(1) NH2NH2, O H
11.
(2) Zn–Hg/HCl
(3) Na, Liq, NH3
(4) NaBH4
Compound (A), C8H9Br, gives a white precipitate when warmed with alcoholic AgNO 3. Oxidation of (A)
gives an acid (B), C8H6O4. (B) easily forms anhydride on heating. Identify the compound (A).
[JEE(Main)-2013, 4/120]
CH2Br
CH2Br
C2H5
CH Br
2
(1)
(2)
(3)
(4)
CH3
Br
CH3
CH3
12.
The most suitable reagent for the conversion of R–CH2–OH → R–CHO is : [JEE(Main)-2014, 4/120]
(1) KMnO4
(2) K2Cr2O7
(3) CrO3
(4) PCC (Pyridinium Chlorochromate)
13.
In the following sequence of reactions :
271
KMnO
SOCl
H2 / Pd
2
4
Toluene ⎯⎯⎯⎯
→ B ⎯⎯⎯⎯
→ A ⎯⎯⎯⎯
→ C
BaSO4
the product C is :
(1) C6H5COOH
14.
(2) C6H5CH3
[JEE(Main)-2015, 4/120]
(4) C6H5CHO
(3) C6H5CH2OH
The correct sequence of reagents for the following conversion will be :
O
CH 3
HO
HO
CHO
CH3
CH3
[JEE(Main)-2017, 4/120]
uh
The major product obtained in the following reaction is :
O
O
DIBAL –H
⎯⎯
⎯⎯
⎯→
COOH
OH
CHO
(1)
Ja
OH
CHO
(2)
COOH
CHO
CHO
(3)
(4)
CHO
The trans-alkenes are formed by the reduction of alkynes with :
(1) Na/liq. NH3
(2) Sn - HCl
(3) H2–Pd/C, BaSO4
(4) NaBH4
CHO
COOH
[JEE(Main)-2018, 4/120]
lp
16.
a
ri
(1) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OH–
(2) CH3MgBr, [Ag(NH3)2]+OH–, H+/CH3OH
(3) [Ag(NH3)2]+OH–, CH3MgBr, H+/CH3OH
(4) [Ag(NH3)2]+OH– , H+/CH3OH, CH3MgBr
15.
[JEE(Main)-2017, 4/120]
JEE(MAIN) ONLINE PROBLEMS
The reagent needed for converting
a
1.
Ph
C=C
H
Sa
nk
Ph–CC–Ph
H is :
[JEE(Main) 2014 Online (11-04-14), 4/120]
Ph
(1) Cat. Hydrogenation
(3) Li/NH3
(2) H2/Lindlar Cat.
(4) LiAlH4
2.
The gas liberated by the electrolysis of Dipotassium succinate solution is :
[JEE(Main) 2014 Online (11-04-14), 4/120]
(1) Ethane
(2) Ethyne
(3) Ethene
(4) Propene
3.
Bouveault-Blanc reduction reaction involves:
[JEE(Main) 2016 Online (09-04-16), 4/120]
(1) Reduction of an anhydride with LiAlH4.
(2) Reduction of an ester with Na/C2H5OH.
(3) Reduction of a carbonyl compound with Na/Hg and HCl.
(4) Reduction of an acyl halide with H2/Pd.
4.
The reagent(s) required for the following conversion are: [JEE(Main) 2018 Online (15-04-18), 4/120]
OH
EtO2C
CO2H
⎯→
HO2C
CN
(1) (i) NaBH4 (ii) Raney Ni/H2 (iii) H3O+
(3) (i) B2H6 (ii) DIBAL-H (iii) H3O+
CHO
(2) (i) LiAlH4 (ii) H3O+
(4) (i) B2H6 (ii) SnCl2/HCl (iii) H3O+
272
5.
The main reduction product of the following compound with NaBH 4 in methanol is :
[JEE(Main) 2018 Online (15-04-18), 4/120]
O
O
NMe2
OH O
OH O
NMe2
(2)
OH
O
NMe2
(3)
O
NMe2
(4)
a
ri
NMe2
(1)
When 2-butyne is treated with H2/Lindlar's catalyst, compound X is produced as the major product and
when treated with Na/liq. NH3 it produces Y as the major product. Which of the following statements is
correct?
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) X will have higher dipole moment and higher boiling point than Y.
(2) Y will have higher dipole moment and higher boiling point than X.
(3) X will have lower dipole moment and lower boiling point than Y.
(4) Y will have higher dipole moment and lower boiling point than X.
7.
The major product formed in the following reaction is :
OCOCH3
CHCl3
HO
(1)
OH
O
lp
O
(2)
OH
H
O
Sa
nk
(3)
O
HO
OH
O
O
a
OCOCH3
[JEE(Main) 2018 Online (15-04-18), 4/120]
Ja
PCC
(Pyridinium
chlorochromate)
uh
6.
O
(4) HO
OH
OH
O
H
8.
The major product of following reaction is :
(1) AlH(i −Bu)2
→?
R–CN ⎯⎯⎯⎯⎯
[JEE(Main) 2019 Online (09-01-19), 4/120]
(2)H2 O
(1) RCONH2
9.
(2) RCH2NH2
(3) RCHO
(4) RCOOH
The major product ‘X’ formed in the following reaction is: [JEE(Main) 2019 Online (10-01-19), 4/120]
O
O
CH2–C–OCH3
NaBH 4
⎯⎯
⎯
⎯→ X
MeOH
OH
OH
CH2CH2OH
(1)
CH2CH2OH
(2)
273
O
OH
O
O
CH2–C–OCH3
CH2–C–H
(3)
The major product of the following reaction is:
O
CH3N
NaBH4
⎯⎯⎯⎯
→
O
(1) CH3NH
OH
(3) CH3N
11.
[JEE(Main) 2019 Online (10-01-19), 4/120]
OH
(2) CH3NH
OH
(4) CH3N
The major product of the following reaction is:
COCH3
a
ri
10.
(4)
[JEE(Main) 2019 Online (11-01-19), 4/120]
(i) KMnO / KOH,
4
⎯⎯⎯⎯⎯⎯⎯⎯
→
(ii) H2SO4 (dil)
CH3
COOH
(1)
(2)
OHC
HOOC
COOH
COCOOH
(4)
Ja
(3)
HOOC
HOOC
The major product obtained in the following reaction is : [JEE(Main) 2019 Online (11-01-19), 4/120]
O
OH
LiAlH4
CH3
(excess )
⎯⎯
⎯⎯→
NO2 OH
lp
12.
OH
a
OH
(1)
(2)
Sa
nk
CH3
NH2 OH
13.
uh
COCH3
CH3
OH
OH
(3)
NO2 OH
CH3
(4)
NH2 OH
NH2 OH
The major product of the following reaction is :
CN
CH3
[JEE(Main) 2019 Online (12-01-19), 4/120]
(i ) DIBAL −H
⎯⎯
⎯ ⎯⎯→
(ii ) H3O+
O
O
CHO
CH=NH
CHO
CHO
(1)
O
O
(2)
(3)
OH
O
CHO
(4)
OH
OH
274
14.
The major product of the following reaction is :
O
[JEE(Main) 2019 Online (12-01-19), 4/120]
NaBH
4
⎯⎯⎯⎯
→
EtOH
O
O
(1)
OH
OH
(2)
(3)
(4)
OEt
EXERCISE - 1
(c) Ph–CH2–OH
(a) Butane
(b) Butane
A-2.
(a) CH2=CH–CH2–CH2–CC–CH3
(b) CH2=CH–CH2–CH2–CC–CH3
(d)
CH3–CH 2–CH 2– CH2–CH2–CH2–CH3
Ja
A-1.
uh
PART - I
a
ri
OEt
lp
(c) Benzaldehyde (Benzene carbaldehyde)
A-3.
(i) CH3–CH2– CH = CH –(CH2)7–CH2–OH
A-4.
(i) H2/Pd-BaSO4
(ii) In this conversion –CO– converts into –CH2– which can be achieved by any one of the following
reagents :
(a) NH2–NH2/KOH/
(b) Zn-Hg, Conc HCl, 
(c) Red P + HI
(ii)
Sa
nk
a
trans
B-1.
LiAlH4 and NaBH4 both gives same products with carbonyl compounds.
(i) Ph–CH2–OH
(ii)
275
(i) a :
;b:
NH
HO
O
(ii) a :
NH
HO
;b:
HO
HO
CH3
CH 3
a
ri
B-2.
OH
Re d P + HI
⎯⎯⎯⎯⎯⎯
→
(a)
(b)
C-1.
(i) CH3CH2
(ii) CH3CH2COOH + HOOCCH2CH2CH3
(a)
KMnO4
O
CH3
(b)
KMnO4
OH
Cold dil. KMnO4
OH
(d)
OH
(1) Peracid
OH
Sa
nk
a
(c)
COOH
C
lp
C-2.
COOH
Ja
CH2–CH3
uh
B-3.
C-3.
t-Butanol is not oxidised by dilute KMnO4, so it does not give brown-black precipitate of MnO2. nButanol is oxidised with dil KMnO4 and MnO4– is converted to brown-black precipitate of MnO2.
Dil. KMnO
4
⎯⎯⎯⎯⎯
→ No reaction
Dil. KMnO
4
⎯⎯⎯⎯⎯
→ CH3CH2CH2COOH + MnO2  (Brown black ppt.) + 2KOH + 2H2O
D-1.
(a)
(b) 2 HCHO + 2 HCOOH
(c) CH3COOH + CH3CHO
276
D-2.
(P)
(Q)
D-3.
(a) Ph–CH2–CH2–OH
(R)
Ph–CH2–CH=O
(S)
(b)
a
ri
(c)
E-1.
+
CH3COOH
E-3.
CH3COCl
E-4.
CH3COOH + NH3
E-7.
CH2=CH–O–CH3
E-5.
Two carboxylic acids
Ja
NH2
uh
(a)
E-6.
CH3– NH3 + HCOOH
(b) CH3NC
E-2.
NO2
PART – II
A-2.
(B)
A-3.
(B)
A-4.
(D)
A-5.
(B)
A-6.
(B)
A-7.
(D)
A-8.
(A)
B-1.
(B)
B-2.
(A)
B-3.
(C)
B-4.
(B)
B-5.
(A)
B-6.
(B)
B-7.
(D)
C-1.
(C)
C-2.
(C)
C-3.
(C)
C-4.
(C)
C-5.
(B)
D-1.
(C)
D-2.
(C)
D-3.
(C)
D-4.
(B)
E-1.
(D)
E-2.
(C)
E-3.
(D)
lp
(A)
Sa
nk
a
A-1.
PART – III
1.
(A) – (q) ; (B) – (s) ; (C) – (p, s) ; (D) – (q, s)
2.
(A) – (p) ; (B) – (q) ; (C) – (t) ; (D) – (s).
EXERCISE - 2
PART - I
1.
(D)
2.
(D)
3.
(D)
4.
(B)
5.
(C)
6.
(D)
7.
(D)
8.
(D)
9.
(D)
10.
(A)
277
PART - II
1.
3
6.
4 (i, ii, iii, v)
1.
(ABCD)
2.
3
3.
2.
(AB)
3.
3 (i, ii, iv)
4.
3
5.
7
4.
(BC)
5.
(AB)
PART - III
(AD)
PART - IV
(D)
2.
(C)
3.
(A)
4.
(B)
5.
(A)
6.
(D)
7.
(A)
8.
(C)
9.
(D)
10.
(B)
a
ri
1.
EXERCISE - 3
2.
6.
(C)
8.
(a)
(B)
(C)
7.
P=
(D)
(C)
4.
(A)
(E)
(C)
;Q=
Sa
nk
P=
3.
lp
(A)
a
5.
(B)
Ja
1.
uh
PART - I
;R=
or
;Q=
;R=
(b) The greater stability of the oxonium ion intermediate is responsible for the extraordinary reactivity.
This is demonstrated for the two sets of P, Q, R (cases i and ii above) as follows.
(i)
(ii)
⎯⎯
→
⎯⎯
→
9.
(A) – p, q, s, t ; (B) – s, t ; (C) – p ; (D) – r
10.
(ACD)
11.
(A)
12.
5
13.
(CD)
278
PART - II
JEE(MAIN) OFFLINE PROBLEMS
1.
(4)
2.
(2)
3.
(2)
4.
(1)
5.
(1)
6.
(3)
7.
(3)
8.
(2)
9.
(2)
10.
(1)
11.
(4)
12.
(4)
13.
(4)
14.
(4)
15.
(4)
16.
(1)
(3)
2.
(3)
3.
(2)
4.
6.
(1)
7.
(3)
8.
(3)
9.
11.
(3)
12.
(1)
13.
(3)
14.
(4)
5.
(1)
(3)
10.
(2)
(3)
Ja
uh
1.
a
ri
JEE(MAIN) ONLINE PROBLEMS
This Section is not meant for classroom discussion. It is being given to promote self-study and
self testing amongst the students.
PART - I : PRACTICE TEST-1 (IIT-JEE (MAIN Pattern))
Important Instructions
The test is of 1 hour duration.
The Test Booklet consists of 30 questions. The maximum marks are 120.
Each question is allotted 4 (four) marks for correct response.
Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each
question.
¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction
from the total score will be made if no response is indicated for an item in the answer sheet.
5. There is only one correct response for each question. Filling up more than one response in any question
will be treated as wrong response and marks for wrong response will be deducted accordingly as per
instructions 4 above.
Sa
nk
a
1.
2.
3.
4.
Max. Marks : 120
lp
Max. Time : 1 Hr.
1.
Propyne and propene can be distinguished by :
(1) Conc. H2SO4
(2) Br2 in CCl4
(3) Dil. KMnO4
(4) AgNO3 in ammonia
2.
The reactivity order towards hydrogenation of the following compounds is
(I) CH3–CC–CH3
(III)
(1) I > II > III > IV
(II)
(IV)
(2) II > III > IV > I
(3) III > IV > II > I
(4) IV > III > II > I
279
Na / NH3 ( )
⎯⎯⎯⎯⎯⎯
→ P, the product can be :
3.
CN
4.
(2)
(3)
(4)
The product of following reaction is
(i) LiAlH4
→
PhCH=CHCH=O ⎯⎯⎯⎯⎯

(ii) H3O
(2) Ph(OH)C=CHCH3
(3) PhCH=CHCH2OH
(4) PhCH2CH2CH2–OH
LiAlD
4
→ is
The product of the reaction Ph2C=O ⎯⎯⎯⎯

H3O
(1) Ph2CD(OH)
6.
(1) NaBH4/EtOH
(2) LiAlH4/THF
(3) Na/EtOH
Hydrogenation of benzoyl chloride in the presence of Pd / BaSO 4 gives
(1) benzyl alcohol
(2) benzaldehyde
(3) benzoic acid
lp
8.
X can be
(4) None
(4) H2/Pd-BaSO4
(4) phenol
Which of the following reagent not convert carbonyl compound into alcohol ?
(1) DiBAL-H
(2) NH2–NH2/KOH
(3) Na–Hg/HCl
(4) LiAlH4
Sa
nk
9.
(1) NH2–NH2/ KOH
10.
(3) Ph2CD(OD)
a
7.
(2) Ph2CH(OD)
uh
(1) PhCH2CH=CHCH2OH
Ja
5.
a
ri
(1)
x
⎯⎯→
, X can be
(2) Zn–Hg/ HCl
(3) Red P + HI
(4) All
Identify a & b, in the following reaction :
NaBH4
CH3OH
(i) LiAlH4, ether
(ii) C2H5OH
a
b
(1) Both are
(2) Both are
(+ EtOH)
280
14.
15.
16.
(+ EtOH), b =
;
(1)
(2)
(3)
(4)
X is :
a
ri
+
CF CO H
alc. KOH
uh
An alkene on ozonolysis yields only ethanal. There is an isomer of this, which on ozonolysis yields :
(1) Propanone and methanal
(2) Propanone and ethanal
(3) Ethanal and methanal
(4) Only propanone
3
3
⎯
→ X ⎯⎯⎯⎯
(CH3)3CCl ⎯⎯⎯⎯
⎯
→Y ;

The product ‘Y’ is
(1) CH2=CH–CH2OH
(3) CH3COCH2CH3
(2)
Ja
13.
(4) a =
KMnO4
→
Alkene (X) ⎯⎯⎯⎯⎯

(+ EtOH)
Which of the following will decolorise alkaline KMnO4 solution ?
(1) C3H8
(2) CH4
(3) CCl4
Bayer's reagent is :
(1) alkaline permanganate solution
(2) acidified permanganate solution
(3) neutral permanganate solution
(4) aqueous bromine solution
(P)
(4) C2H4
O
||
1. LiAlH4
H3 O 
KMnO 4 / H
⎯⎯→ (B) ⎯⎯
⎯→ (A) + E(gas) ⎯⎯
⎯⎯⎯→ (C)
C—NH2 ⎯⎯
2. H O
Sa
nk
Select correct options, for identical pairs
(1) P, A
(2) A, C
OSO
HiO4
4
⎯⎯ ⎯
⎯
→ A ⎯⎯⎯
→ B+C;
H O
17.
(4) (CH3)2CHCH2OH
lp
12.
,b=
a
11.
(3) a =
2
2
(3) B, C
(4) P, C
Product B and C are respectively :
2
(1)
and CH3 – C– H
||
O
(3) H – C– H and CH3 − CH2 – C– CH3
||
||
O
O
(2) CH3 – CH2 – C– H and CH3 – C – H
||
||
O
O
(4) CH3 − C– CH3 and H − C– H
||
||
O
O
(1)
(2)
18.
281
(3)
19.
(4)
Fenton’s reagent is :
(1) FeSO4 + H2 O2
(2) HgSO4 + H2O2
(3) FeSO4 + H2O
(4) None of these
The reagent with which both acetaldehyde and acetone react easily is
(1) Tollen's reagent
(2) Schiff's reagent
(3) Grignard reagent
(4) Fehling reagent
21.
An organic compound (P) with molecular formula C5H8O4 is stable to heat but hydrolyse to (Q) and
MeOH by NaOH followed by acidification. (Q) on strong heating gives (R) which with Red P/HI gives
ethane. Compound (P) is :
(1) CH3–C–O–C–OC2H5
(2) CH3–C–C–C–OCH3
O
O O O
O
(4)
uh
(3)
a
ri
20.
When acetaldehyde is treated with Fehling's solution, it gives a precipitate of
(1) Cu
(2) CuO
(3) Cu2O
(4) Cu + Cu2O + CuO
23.
Identify the correct statement about MnO2/
(1) C6H5–CHOH–CH3 as well as CH3–CH=CH–CH2OH are oxidised
(2) C6H5–CH2–CH2–OH as well as CH2=CH–CH2–CH2–OH are oxidised
(3) C6H5–CHOH–CH3 is not oxidised but CH3–CH=CH–CH2–OH is oxidised.
(4) C6H5–CHOH–CH3 is oxidised but CH3–CH=CH–CH2OH is not oxidised.
24.
Which of the following reaction involves homogeneous reduction?
(A) CH2=CH2
lp
Ja
22.
CH3–CH3
Wilkinson' s catalyst
(C) CH3COCl
Sa
nk
(D) CH3CCH
a
(B) CH3–CC–CH3 ⎯⎯⎯⎯⎯⎯⎯⎯
→ CH3–CH=CH–CH3
CH3CH = CH2
X
⎯⎯→
25.
X is :
(1) NaBH4/EtOH
26.
CH3CHO
(2) LiAlH4/THF
(3) Al(OiPr)3 / CH3–CH–CH3 (4) All of these
OH
Which reducing agent, would you use to carry out the following transformation.
⎯⎯
→
(1) LiAlH4
(2) NaBH4
(3) Na/NH3
(4) DIBAL-H
282
OH
OH
27.
Product
OH
(1)
(2)
COOH
a
ri
O
COOH
(3)
(4)
O
O
(1)
29.
(2)
(3)
The reaction,
is known as :
(1) Wolff-kishner reduction
(3) Meerwein -Ponndorf reaction
(4)
(2) Oppenauer oxidation
(4) Clemmensen reduction
lp
30.
uh
An unknown compound decolorizes bromine in carbon tetrachloride, and it undergoes catalytic
reduction to give decalin. When treated with warm, conc. potassium permangate, this compound give
cis–cyclohexane-1,2-dicaboxylic acid and oxalic acid. Possible a structure for the unknown compound
is -
Ja
28.
The reagent used to convert RCOOH → RCH2OH is
(1) NaBH4
(2) Na/ Alcohol
(3) Zn/Hg-HCl
(4) LiAlH4
a
Practice Test-1 (IIT-JEE (Main Pattern))
OBJECTIVE RESPONSE SHEET (ORS)
1
2
3
Sa
nk
Que.
4
5
6
7
8
9
10
Ans.
Que.
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Ans.
Que.
Ans.
283
PART - II : NATIONAL STANDARD EXAMINATION IN CHEMISTRY (NSEC) STAGE-I
1.
If 2-pentanone is reacted with NaBH4 followed by hydrolysis with D2O the product will be [NSEC-2000]
(A) CH3CH(OD)CH2CH2CH3
(B) CH3CD(OH)CH2CH2CH3
(C) CH3CH(OH)CH2CH2CH3
(D) CH3CD(OD)CH2CH2CH3
2.
If 1 mole H2 is reacted with 1 mole of the following compound.
a
ri
Which double bond will be hydrogenated ?
(A) c
(B) b
[NSEC-2000]
(C) a
(D) d
3.
Which of the following can not be obtained when alkenes are oxidised with KMnO4 and then followed by
acid hydrolysis ?
[NSEC-2001]
(A) alkanoic acids
(B) alkanals
(C) alkanones
(D) carbon dioxide
4.
In the reaction
H2O
ether

CH3CN + 2H ⎯⎯⎯→ X ⎯⎯⎯→ Y, Y is
(A) acetaldehyde
(B) ethanamine
uh
[NSEC-2001]
HCl
(C) dimethylamine
(D) acetone
A compound is soluble in conc. H2SO4. It does not declourise bromine in carbon tetrachloride but
oxidised by chromic anhydride in aqueous sulphuric acid within two seconds, turning orange solution to
blue, green and then opaque. The original compound is :
[NSEC-2001]
(A) an alkane
(B) a tertiary alcohol
(C) a primary alcohol
(D) an ether
6.
If 3-hexanone is reacted with NaBH4 followed by hydrolysis with D2O, the product will be :
Ja
5.
8.
Hydrogenation of benzoyl chloride in the presence of Pd and BaSO 4 gives :
(A) benzyl alcohol
(B) benzoic acid
(C) benzaldehyde
(D) toluene
[NSEC-2001]
a
7.
[NSEC-2001]
(B) CH3CH2CD(OH)CH2CH2CH3
(D) CH3CH2CD(OD)CH2CH2CH3
lp
(A) CH3CH2CH(OD)CH2CH2CH3
(C) CH3CH2CH(OH)CH2CH2CH3
Zn −Hg
The reaction, R2CO + 4[H] ⎯⎯⎯⎯
→ R2CH2 + H2O
Sa
nk
Conc. HCl
9.
is well known as :
(A) Wurtz reaction
(C) Kolbe reaction
(B) Rosenmund reduction
(D) Clemmensen reduction
[NSEC-2001]
Acetone will be obtained on ozonolysis of
(A) 1-pentene
(B) 2-pentene
(C) isopentene
[NSEC-2002]
(D) 2-pentyne
10.
The reducing agent for conversion of O2NCH2CH2CH=CH2 to H2NCH2CH2CH=CH2 is :
(A) LiAIH4
(B) H2/Pd
(C) B2H6
(D) NaBH4
[NSEC-2002]
11.
Reduction of an isonitrile gives a
(A) primary amine
(C) tertiary amine
[NSEC-2002]
(B) secondary amine
(D) quaternary ammonium salt.
12.
Methane may be obtained from monochloromethane by
(A) reduction with nascent hydrogen (Zn + HCI) (B) reduction with hydrogen (H2)
(C) heating with sodium metal in dry ether
(D) hydrolysis with aqueous NaOH.
13.
The compound which does not react with lithium aluminium hydride is
(A) 3-penten-2-one
(B) methyl benzoate
(C) 2-pentanol
[NSEC-2002]
[NSEC-2003]
(D) propanenitrile
284
14.
The compound that would yield a 5-oxo-2-methylhexanal on ozonolysis is
(A)
(B)
(C)
[NSEC-2004]
(D)
15.
Reduction of methylbenzoate (C6H5COOCH3) to benzyl alcohol (C6H5CH2OH) can be accomplished
using
[NSEC-2005]
(A) H2/Pd
(B) LiAIH4
(C) NaBH4
(D) Zn-Hg/HCI
16.
Oxidation of cyclopentanol to cyclopentanone can be accomplished by using
[NSEC-2005]
(A) Tollen’s reagent
(B) chromic acid
(C) bromine water
(D) Fehling’s solution.
17.
Carbonyl compounds can generally be converted to hydrocarbons by
(A) H2/Pt
(B) LiAIH4
(C) N2H4-KOH
18.
To reduce a nitroaldehyde to a nitroalcohol the reducing agent of choice is
(A) LiAlH4
(B) NaBH4
(C) Molecular H2
(D) SnCl2
[NSEC-2008]
19.
Suggest the suitable reagent for the following transformation.
[NSEC-2008]
uh
a
ri
[NSEC-2006]
(D) K2Cr2O7-H2SO4
(A) meta-chloroperbenzoic acid
(C) potassium dichromate
(B) ozone
(D) alkaline hydrogen peroxide
An isocyanide on reduction with hydrogen in the presence of plantinum gives :
(A) amide
(B) primary amine
(C) secondary amine
(D) alcohol
[NSEC-2009]
21.
Compound X (C5H10O) is a chiral alcohol. It is catalytically hydrogenated to an achiral alcohol Y
(C5H12O) and oxidized by activated MnO2 to an achiral carbonyl compound Z (C5H8O). Compound X is
[NSEC-2009]
(A) 1-penten-3-ol
(B) 4-penten-2-ol
(C) 3-methyl-2-buten-1-ol
(D) 2-methyl-2-buten-1-ol
22.
4-Oxobutanoic acid is reduced with Na-borohydride and the product is treated with aqueous acid. The
final product is :
[NSEC-2009]
(B)
Sa
nk
(A)
a
lp
Ja
20.
(C)
(D)
23.
A solution of sodium metal in liquid ammonia is strongly reducing due to the presence of [NSEC-2013]
(A) sodium atoms
(B) sodium hydride
(C) sodium amide
(D) solvated electrons
24.
Which of the following statements is true for the reaction given below ?
alkaline KMnO
4⎯
⎯⎯⎯⎯⎯⎯⎯
→
0ºC
[NSEC-2013]
P
(A) P is a meso compound of 2,3-butanediol formed by syn addition.
(B) P is a meso compound of 2,3-butanediol formed by anti addition.
(C) P is a racemic mixture of d- and l-2,3-butanediol formed by anti addition.
(D) P is a racemic mixture of d- and l-2,3-butanediol formed by syn addition.
25.
Complete catalytic hydrogenation of naphthalene gives decalin (C 10H18). The number of isomers of
decalin formed and the total number of isomers of decalin possible are respectively.
[NSEC-2016]
(A) 1, 2
(B) 2, 2
(C) 2, 4
(D) 3, 4
285
26.
Which of the following on treatment with hot concentrated acidified KMnO 4 will give 2-methylhexane1,6-dioic acid as the only organic product?
[NSEC-2017]
Z
(A)
27.
(B)
(C)
(D)
An organic compound 'P' with molecular formula C9H8O2 on oxidation gives benzoic acid as one of the
products. The possible structure/s of 'P' is/are
[NSEC-2017]
COOH
(CH2)2COOH
a
ri
COOH
COOH
(I)
(A) I and III
(IV)
(III)
(C) I and II
(D) II only
The correct sequence of reagents from those listed below for the following conversion is [NSEC-2018]
uh
28.
(II)
(B) II and IV
⎯⎯⎯→
. Br2
V. H3O+
(C)  –  – 
Ja
. NaNH2
. H2/Pd-C, quinolone
(A) V –  – 
(B)  – V – 
(D)  –  – 
lp
PART - III : HIGH LEVEL PROBLEMS (HLP)
SUBJECTIVE QUESTIONS
An alkene (A) C16H16 on ozonolysis gives only one product (B) C8H8O. Compound (B) reaction with
NaOH / I2 yields sodium benzoate. Compound (B) reacts with KOH / NH 2NH2 yielding a hydrocarbon
(C) C8H10. Write the structures of compound (B) & (C). Based on this information two isomeric
structures can be proposed for alkene (A). Write their structure and identify the isomer which on
catalytic hydrogenation (H2/Pd–C) gives a racemic mixture.
[JEE-2001, 5/100]
Sa
nk
a
1.
2.
What is the product of each reaction
3._
Which alcohol is prepared from the following ketones via MPV reduction ?
(a)
4.
(b)
(c)
Complete the following
SeO
2
(a) CH3–CHO ⎯⎯⎯→
; (b)

SeO
2
⎯⎯⎯→

; (c)
SeO
2
⎯⎯⎯→

286
5.
Complete the following
H O
3
⎯⎯⎯→
(a)
H O
3
(b) Ph–O–CH=CH2 ⎯⎯⎯→
A+B
6.
Write the products P, Q, R and S in the given reaction sequence.
H O+
a
ri
H2 / Pd / BaSO4
3
(Q)
⎯⎯⎯⎯⎯⎯
→ (P) ⎯⎯⎯→
ONLY ONE OPTION CORRECT TYPE
Identify a reagent from the following list which can easily distinguish between 1-butyne and 2-butyne.
(A) bromine, CCl4
(B) H2, Lindlar catalyst
(C) dilute H2SO4, HgSO4
(D) ammonical Cu2Cl2 solution
[JEE-2002, 3/90]
uh
7.
O
O
O
Product obtained in above reaction is :
(A) 3CH3CHO
(B) 3HCHO
9.
O
(A)
, EtOH
(C)
OH
a
HO
, CH3OH
10.
(B)
, EtOH
HO
O
(D)
+ EtOH
O
Which of the following gives CH3–OH and
(A)
(D) 3CH3OH
(B)
O
Sa
nk
HO
(C) 3HCOOH
lp
OEt
Product of above reaction is :
Ja
8.
O
on hydrolysis with H3O+.
(C)
(D) CH3–O–CH2–CH2–CH2–CH3
287
SINGLE AND DOUBLE VALUE INTEGER TYPE
11.
In how many reaction CH3–CHO is obtained as major product ?
1. SnCl /dil. HCl
2
(P) CH3–CN ⎯⎯⎯⎯⎯⎯⎯
→
H O+
3
⎯⎯⎯⎯
→
(Q)
2. H3O+
O
OEt
2. H3O+
Dolastatin is an anti cancer compound isolated from Indian sea have Dobabella ausiculasia. One mole
of it on acidic hydrolysis yield how many products are formed.
H
H
H
O
N
N
N
N
N
O
OCH 3
CH 3
S
O
OCH3
O
uh
H
13.
(S) CH3–CN ⎯⎯⎯⎯⎯→
Low temperature
a
ri
12.
1. DBAL −H
DBAL −H
⎯⎯⎯⎯⎯⎯
→
(R) CH3–C
How many para substituted benzenoid isomers of C8H8O2 gives 1, 4-dihydroxy benzene on hydrolysis ?
14.
Ja
ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
The end product of following reaction is / are :
H / Pd / BaSO / Quinoline
D / Ni
2
2
4
→
→ ⎯⎯⎯⎯
Ph–CC–Ph ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
( 2)
(1)
Ph
(B)
(C)
Ph
(D)
D
H
D
H
D
H
D
H
D
Ph
a
H
Which of the following give only meso compound on catalytic reduction ?
Sa
nk
15.
Ph
Ph
lp
(A) Ph–CH2–CD2–Ph
H
Ph
D
(A)
(B)
(C)
(D)
(C)
(D) None of these
dil. H SO
2
4
⎯⎯⎯⎯⎯
→ A+B
16.
(A)
O
(B)
288
PART - IV : PRACTICE TEST-2 (IIT-JEE (ADVANCED Pattern))
Max. Time : 1 Hr.
Max. Marks : 66
Important Instructions
A. General %
7.
8.
C.
9.
10.
lp
11.
a
ri
6.
uh
5.
The test is of 1 hour duration.
The Test Booklet consists of 22 questions. The maximum marks are 66.
Question Paper Format
Each part consists of five sections.
Section-1 contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE is correct.
Section-2 contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE OR MORE THAN ONE are correct.
Section-3 contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from 0
to 9 (both inclusive).
Section-4 contains 1 paragraphs each describing theory, experiment and data etc. 2 questions relate to
paragraph. Each question pertaining to a partcular passage should have only one correct answer among
the four given choices (A), (B), (C) and (D).
Section-5 contains 1 multiple choice questions. Question has two lists (list-1 : P, Q, R and S; List-2 : 1, 2, 3
and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY ONE is
correct.
Marking Scheme %
For each question in Section 1, 4 and 5 you will be awarded 3 marks if you darken the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one
(–1) mark will be awarded.
For each question in Section 2, you will be awarded 3 marks. If you darken all the bubble(s) corresponding
to the correct answer(s) and zero mark. If no bubbles are darkened. No negative marks will be answered for
incorrect answer in this section.
For each question in Section 3, you will be awarded 3 marks if you darken only the bubble corresponding to
the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for incorrect
answer in this section.
Ja
1.
2.
B.
3.
4.
a
SECTION-1 : (Only One option correct Type)
This section contains 8 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which Only ONE option is correct.
1. H O+
3
⎯⎯⎯⎯
→ A is %
Sa
nk
1.
2. LiAlH4
(A)
(B)
(C)
(D) None
289
2.
(B) N is
(C) Y is
(D) All of these are correct
+ SeO2 ⎯⎯
→ (P). (P) will not :
3.
Q
(B) give Iodoform test.
(D) give cerric ammonium nitrate test.
uh
(A) reduce Tollens reagent.
(C) form dioxime
4.
a
ri
(A) M is
Na/NH3
P
(A)
in both cases
(C) P is I, Q is II
(B)
lp
Which of the following Reaction is not possible ?
alk. KMnO 4
⎯⎯
⎯
⎯
⎯→

(A)
CH3
CH3 –C–CH3
Sa
nk
alk. KMnO 4

(C)
alk. KMnO 4
⎯⎯
⎯
⎯
⎯→

(B)
NO 2 NH2
⎯⎯⎯⎯
⎯→
6.
in both cases
(D) P is II, Q is I
a
5.
Ja
P and Q are :
(D)
alk. KMnO 4
⎯⎯
⎯
⎯
⎯→

Hydrolysis of a compound C9H10ClBr (P) yields C9H10O(Q)
(Q) gives positive haloform test ?
Strong oxidation of (Q) yields a dibasic acid which gives only two mono-nitro derivative.
What is the structure of (P) ?
Br
Cl
CH2Cl
Br
Br
Cl
Cl
CH2Br
(A)
(B)
(C)
(D)
290
Section-2 : (One or More than one options correct Type)
This section contains 5 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.
Which of the following reaction is/are correct ?
O
OH
OH
O
LiAlH4
(A)
H2O
OH
O
LiAlH 4
(B)
CH3–CH2–NH2
H2O
a
ri
7.
O
(C)
O
CH2–OH
LiAlH4
H2O
CH2–OH
O
(D)
O
O
NaBH4
H2O
OH
O
Which of the following will give syn addition ?
(A)
Ja
8.
O
uh
O
(i) KMnO
4
⎯⎯⎯⎯⎯
→
–
(ii) Na2SO3
OsO 25ºC
4
⎯⎯⎯⎯⎯
→
lp
(C)
(i) OsO
4
⎯⎯⎯⎯
→
(B)
(ii) H2O / OH
(D)
(ii) H3 O
Periodic acid is generally used for the oxidation of vicinal diols or -hydroxycarboxyl compounds. Which
of the following statements are correct for this reaction
(A) oxidative cleavage takes place in the above reactions.
(B) final products are generally carbonyl compounds or carboxylic acids.
(C) HO4 reduced into HIO3
(D) Intermediate of this reaction for a vicinal diol is
Sa
nk
9.
(i) RCOOOH
⎯⎯⎯⎯⎯⎯
→
+
a
Na2SO3
10.
Mechanism of reductive ozonolysis is given below for an alkene.
R
R
O
C —C
R
R
R ⎯⎯⎯
R
H ⎯⎯
⎯
⎯→
⎯→
⎯⎯
⎯⎯→
C C
Step−
Step−
Step−
C == C
O
O
R
H
R
H
O
O —O
R
+ O == C
R
H
Which is correct for the above mechanism
(A) Ozone act as electrophile and as well as nucleophile in this reaction
291
(A) (A) and (B) are diastereomer’s of each other.
(B) upon catalytic hydrogenation (A) and (B) gives same product
(C) Product (C) and (D) are Identical
(D) Product (C) and (D) are separated by fractional distillation.
a
ri
11.
(B) First step of this reaction is an electrophilic addition
(C) ozonide is formed in the step-II
(D) When ozonide is cleaved in the presence of reducing agent such as Zn or Me 2S the products will be
aldehydes and/or ketones.
Observed the following reaction sequence and choose the correct options.
How many Stereoisomers are formed in following reaction ?
CH2
Pd
+ H2
Ja
12.
uh
Section-3 : (Single/ Double Integer Value Correct Type.)
This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive)
CH3
Ph
lp
Predict the product of following reaction.
(x moles of H2 used)
H2, Pd–C
(A)
H
C
C
H
a
13.
Sa
nk
H2, lindlar
(B)
catalyst
(y moles of H2 used)
Find the value of (x + y).
14.
Compound X gives smallest acid & smallest 2º amine on hydrolysis. What is the molecular weight of
compound X ?
15.
Na / NH ( )
3
(P) ⎯⎯⎯⎯⎯
→ Product(s)
The product(s) has/have X = degree of unsaturation and Y = number of isomeric product(s) formed.
Then X + Y = ?
292
SECTION-4 : Comprehension Type (Only One options correct)
This section contains 1 paragraphs, each describing theory, experiments, data etc. 3 questions
relate to the paragraph. Each question has only one correct answer among the four given
options (A), (B), (C) and (D)
Paragraph for Questions 16 to 17
(Read the paragraph carefully and give the answer of following questions.
Generally, during the clemmenson reduction >C=O group converts into >CH 2 after reacting with (conc.
HCl + Zn–Hg). But in case of -diketo compounds, its give unexpected products.
O
O
H3 C
CH3
Mechanism : Zn ⎯→ Zn2+ + 2e–
O
O
2e
H3 C
O¯

CH3
H3C
O
H3C

CH3
CH3
O
O
Ja
2e , HCl
O
O
HCl
H 3C
CH3
HO
OH
O
O
lp
(conc. HCl+ Zn–Hg)
⎯⎯⎯⎯⎯⎯⎯⎯
→ Product (X), (X) will be :

16.
O
(B)
O
O
(C)
a
(A)
(D)
OH
Zn–Hg + conc.HCl
CH3 —CH—C—CH2 —C—CH2 —CH3 ⎯⎯⎯⎯⎯⎯⎯
⎯
→ Product (X), (X) will be :

CH 3 O
O
Sa
nk
17.
CH3

uh
Ex :
(conc. HCl+ Zn–Hg)
⎯⎯⎯⎯⎯⎯⎯⎯
→

a
ri
O
(A)
C—CH2 —CH—CH2 —CH3
O
OH
(C) CH3 —CH—CH—CH2 —CH—CH 2—CH 3
CH 3 Cl
Cl
(B) CH3 —CH—CH 2—CH2—CH2—CH2—CH3
CH 3
(D) CH3 —CH—C—CH—CH2 —CH 3
CH 3 O CH3
293
SECTION-5 : Matching List Type (Only One options correct)
This section contains 1 questions, each having two matching lists. Choices for the correct
combination of elements from List-I and List-II are given as options (A), (B), (C) and (D) out of
which one is correct
Match the reducing agents of List-II with the reaction of List-I and select the correct answer using the
code given below the lists.
List-I
List-II
⎯⎯
→
⎯⎯
→
Code :
Q
3
1
R
1
2
Na / NH3 () / C2H5OH
4.
S
2
4
(B)
(D)
P
1
2
H2/Pd/BaSO4 / quinoline
Q
2
3
R
4
1
S
3
4
a
P
4
3
H2/Ni, (25ºC)
3.
S.
(A)
(C)
2.
Ja
R.
DIBAL-H
uh
⎯⎯
→
Q.
1.
a
ri
⎯⎯
→
P.
lp
18.
Practice Test-2 ((IIT-JEE (ADVANCED Pattern))
Sa
nk
OBJECTIVE RESPONSE SHEET (ORS)
Que.
1
2
3
4
5
6
7
8
11
12
13
14
15
16
17
18
9
10
Ans.
Que.
Ans.
294
PART - I
(4)
2.
(1)
3.
(1)
4.
(4)
5.
(1)
6.
(4)
7.
(2)
8.
(2)
9.
(4)
10.
(3)
11.
(1)
12.
(1)
13.
(2)
14.
(4)
15.
(1)
16.
(2)
17.
(1)
18.
(1)
19.
(1)
20.
(3)
21.
(3)
22.
(3)
23.
(1)
24.
26.
(2)
27.
(2)
28.
(4)
29.
(2)
25.
(4)
(2)
30.
(4)
(A)
5.
(C)
9.
(C)
10.
(A)
14.
(B)
15.
(B)
19.
(D)
20.
(C)
24.
(A)
25.
(A)
PART - II
2.
(D)
3.
(B)
6.
(A)
7.
(C)
8.
(D)
11.
(B)
12.
(A)
13.
(C)
16.
(B)
17.
(C)
18.
(B)
21.
(A)
22.
(A)
23.
(D)
26.
(C)
27.
(C)
28.
(C)
4.
uh
(A)
Ja
1.
a
ri
1.
lp
PART - III
Sa
nk
a
1.
A=
C6H5 – CH2 – CH3
or
B = C 6H5 − C − CH3
||
O
C = C6H5 – CH2 – CH3
Trans isomer give racemic mixture.
OH
2.
295
3.
(a)
(b)
(c)
In (a), the (C=C) bond is not reduced and in (c), the ester (–COOMe) is not reduced in MPV reduction.
(a)
5.
(a)
(b)
(c)
a
ri
4.
(b) A = PhOH, B = MeCHO
O
CH3
O
(P)
(Q)
O
(R)
OHC
(S)
C
uh
6.
HO
(D)
8.
(B)
9.
(B)
12.
6
13.
1
14.
10.
(C)
11.
4
15.
(AC)
16.
(AB)
4.
(D)
5.
(C)
Ja
7.
(BC)
PART - IV
(B)
2.
(D)
6.
(B)
7.
(ABCD)
8.
(ABC)
9.
(ABCD)
10. (ABCD)
11.
(ABC)
12.
2
13.
4
14.
73
15.
16.
(D)
17.
(D)
18.
(A)
a
Sa
nk
1.
3.
(D)
lp
1.
5
PART - I
Terminal alkyne gives white ppt with ammonical silver nitrate.
CH3—CCH + NH3 + AgNO3 → CH3—CC. Ag + NH4NO3
white ppt.
while propene does not give any reaction with ammonical AgNO 3 due to absence of acidic hydrogen.
2.
Rate of hydrogenation will decreases on increasing steric hinderance at -bond.
3.
It is birch reduction
6.
–COCl converts in –CHO by H2/Pd-BaSO4 (Rosenmund reduction)
8.
Wolf-kishwner reduction (NH2–NH2/KOH) give alkane after reduction of carbonyl compound.
296
All reagents are used to convert >C = O to >CH2
10.
Esters are not reduced by NaBH4, >C=O (carbonyl) change to –CH(OH) by use of NaBH4
11.
⎯⎯→
(1) O
3
Alkene ⎯⎯⎯⎯⎯
→ CH3CHO only
(2) Zn +H2 O
alco,KOH
⎯⎯⎯⎯⎯
→
13.

E2
uh
12.
a
ri
9.

H3 O
KMnO 4
LiAlH4
⎯→ PhCOOH ⎯⎯
Ph—C—NH 2 ⎯⎯
⎯
⎯→ Ph–CH2 OH ⎯⎯
⎯→ PhCOOH
(B)
(A)
(C)
O
17.
OSO4
H2 O 2
Ja
16.
HiO4
⎯⎯⎯
→
OH
OH
+
O
H
lp
Syn addition
O
18.
It is fact.
21.
(P) on hydrolysis gives propanedioic acid and methanol. Propanedioic acid on strong heating gives
acetic acid which when reduced with Red P/HI gives ethane.
Sa
nk
a
19.
24.
Reduction with Wilkinson's catalyst is homogeneous
25.
X can be NaBH4/EtOH or LiAlH4/THF or Al(OiPr)3 / CH3–CH–CH3
26.
OH
NaBH4 can not reduces ester.
KMnO
4
⎯⎯⎯⎯
→

28.
+
H
29.
Oppenauer's oxidation, oxidised secondary alcohol into ketone and there is no effect on double bond.
30.
It is fact.
297
PART - III
5.
H O
3
⎯⎯⎯→
(a)
+ NH3
H O
3
(b) Ph–O–CH=CH2 ⎯⎯⎯→
PhOH + MeCHO
7.
CH3–CH2–CCH
10.
Only C on hydrolysis gives CH3–OH and
11.
All four reactions gives CH3 –CHO as major product.
a
ri
+ NH4Cl + H2O
uh
.
Sa
nk
a
lp
Ja
12.
13.
14.
H O+
3
⎯⎯⎯→
+
D / Ni
H / Pd / BaSO / Quinoline
2
4
2
→ Ph–CH=CH–Ph ⎯⎯⎯⎯
→ Ph–CHD–CHD–Ph
Ph–CC–Ph ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
( 2)
(1)
trans
meso
298
meso
(B)
(C)
meso
(D)
and
Reaction is hydrolysis of acetal.
O and
are Products.
1. H O+
3
⎯⎯⎯⎯
→
1.
,
Y=
Sa
nk
a
, N=
lp
2. LiAlH4
M=
Both
Ja
PART - IV
2.
Not meso
a
ri
16.
(A)
uh
15.
3.
In (a), (P) reduces Tollens reagent, since it contains (–CHO) group.
In (b), (P) gives iodoform test, since it contains (MeCO–) group.
In (c), (P) forms dioxime, since it contains (–CHO) and
groups.
+ 2NH2OH ⎯⎯
→
In (d), (P) does not gives cerric ammonium nitrate test, since this test is given by alcohols and (P) does
not contain an alcoholic group. So the answer is (D).
299
4.
cis-alkene formed by lindlar catalyst and trans-alkene formed by Na/NH3.
COOH
5.
⎯⎯⎯⎯
⎯→
alk. KMnO 4

(A)
COOH
HOOC
COOH
⎯⎯⎯⎯
⎯→
(B)
a
ri
alk. KMnO 4

CH 3
CH3 –C–CH3
NO2
alk. KMnO 4
⎯⎯
⎯
⎯
⎯→

Cl
Br
+
6.
H3O
COCH3
CH3 oxidation
COOH
COOH
COOH
COOH
nitration
Two mononitro products
lp
(D)
Ja
NO2 NH2
uh
alk. KMnO 4
⎯⎯
⎯
⎯
⎯→ No reaction

(C)
Self explanatory.
9.
HIO4 is a mild oxidising agent.
a
8.
Sa
nk
Cyclic intermediate is formed with vicinal diols.
11.
300
CH3
CH3
Pd
+ H2
12.
CH3
+
CH3
(Trans)
CH3
(cis)
H2, Pd–C
Ph
13.
Ph–CH2–CH2–CH2–CH2–CH3
H
C
C
H
H2, lindlar
H
C
C
H
CH3
C
C
uh
catalyst
Ph
a
ri
CH2
H
H
Ja
lp
Na / NH ( )
3
(P) ⎯⎯⎯⎯⎯
→
Sa
nk
15.
m.w. is 73
O
O
OH

2e
⎯⎯⎯
→
16.
O
D.u. = X = 4 ; Y = 1
a
14.
⎯→
O

HCl
⎯⎯⎯
→
HCl
⎯⎯⎯
→

O
O

2e
⎯⎯⎯
→
HCl
OH
301
17.
CH3 —CH—C—CH2—C—CH2 —CH3
CH3 O
2e 
CH3 —CH—C——C—CH2 —CH3
O
CH3 O

O

HCl
CH3 —CH—C—C—CH2—CH3
CH3 O
HCl

CH3—CH—C——C—CH 2—CH 3
CH2
CH3 OH
OH
a
ri
2e, HCl
CH3 —CH—C—CH—CH2 —CH3
CH3
Sa
nk
a
lp
Ja
uh
CH3 O
302
CHEMICAL EQUILIBRIUM
JEE(Advanced) Syllabus
Law of mass action; Equilibrium constant, Le Chatelier’s principle (effect of concentration,
temperature and pressure); Significance of G and Gº in chemical equilibrium.
a
ri
JEE(Main) Syllabus
uh
Meaning of equilibrium, concept of dynamic equilibrium. Equilibria involving physical
processes:
Solid -liquid, liquid–gas and solid–gas equilibria, Henry’s law, general characterics of
equilibrium involving physical processes. Equilibria involving chemical processes: Law of
chemical equilibrium, equilibrium constants (Kp and KC) and their significance, significance
of G and Gº in chemical equilibria, factors affecting equilibrium concentration, pressure,
temperature, effect of catalyst; Le-Chatelier’s principle.
Ja
Introduction :
lp
Equilibrium is a state in which there are no observable changes as time goes by. When a chemical
reaction has reached the equilibrium state, the concentrations of reactants and products remain
constant over time and there are no visible changes in the system. However, there is much activity at
the molecular level because reactant molecules continue to from product molecules while product
molecules react to yield reactant molecules. This dynamic situation is the subject of this chapter. Here
we will discuss different types of equilibrium reactions, the meaning of the equilibrium constant and its
relationship to the rate constant and factors that can effect a system at equilibrium.
a
Section (A) : Properties of equilibrium, active mass
Types of chemical reactions
Sa
nk
The reaction which proceed
in one direction only
(a) Precipitation reactions e.g.
NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl–
(b) Neutralization reactions e.g.
HCl(aq) + NaOH(aq) → NaCl(s) + H2O
(c) Reactions in open vessels with
one of the gaseous product
1
2
3
4
5
(a)
Reactions which proceed in both
the direction. These are possible
only in closed vessel e.g.
(a) N2(g) + O2(g)
2NO(g)
(b) PCl5(g)
PCl3(g) + Cl2(g)
Types of chemical reactions
Irreversible reaction
The reaction which proceeds in one direction
(forward direction) only.
Reactants are almost completely converted
into products. Products do not react to form
reactants again.
Do not attain equilibrium state.
Such reactions are represented by single
arrow {→}
Examples –
Precipitation reactions e.g.
NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl 
1
2
3
4
5
(a)
Reversible reaction
The reaction which proceed in both the direction under the
same set of experimental conditions.
Reactants form products and products also react to form
reactants in backward direction. These are possible in closed
vessels .
Attain the equilibrium state and never go to completion.
)
Represented by double arrow (
) or (
Examples :–
Homogeneous reactions- only one phase is present
(i) Gaseous phase–
303
(b)
Neutralization reactions e.g
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O
(c)
2KClO3 (s) ⎯⎯
→ 2KCl(s) + 3O2(g)
(d)

Reactions in open vessel:
Even a reversible reaction will become
irreversible if it is carried out in open vessel.
Ex.
Open
CaCO3(s)
CaO(s) + CO2(g)
vessel
NH4HS(s)
NH3(g) + H2S(g)
(b)
H2(g) + I2(g)
2HI(g)
N2(g) + O2(g)
2NO(g) [Birkland eyde process
(HNO3)]
N2(g) + 3H2(g)
2NH3(g) (Haber’s process)
(ii) Liquid phase
CH3COOH(l) + C2H5OH(l)
CH3COOC2H5(l)+ H2O(l)
Heterogeneous reactions: More than one phases are
present
Closed
CaCO3(s)
CaO(s) + CO2(g)
vessel
NH4HS(s)
NH3(g) + H2S(g)
⚫
a
ri
State of Chemical equilibrium :
State of equilibrium means the balance of driving forces i.e. the factors taking the reaction in forward
direction and the backword direction are balancing each other.
The equilibrium state represents a compromise between two opposing tendencies.
Tendency to minimise energy.
⚫
Molecules try to maximise entropy.
Re ac tan ts
uh
In a reversible reaction like :
R1 + R 2
P1 + P2
Pr oducts
Sa
nk
a
lp
Ja
Initially only reactants are present. R1 and R2 combine to form P1 and P2. As soon as P1 and P2 are
formed, they start the backward reaction. As concentrations of R 1 and R2 decrease rate of forward
reaction decreases and rate of backward reaction increases. Ultimately a stage is reached when both
the rates become equal. Such a state is known as “Chemical Equilibrium” or “state of Equilibrium”.
At equilibrium :
(i) Rate of forward reaction (rf) = rate of backward reaction (rb) (dynamic nature)
(ii) All measurable parameters become constant with respect to time.
Types of equilibria on the basis of process
Physical Equilibrium
Equilibrium in physical process is called
physical equilibrium.
Chemical Equilibrium
Equilibrium in chemical process is called
chemical equilibrium.
For example phase changes like
H2O(l)
H2O(g) ;
Solvation like
For example
H2(g) + Cl2(g)
NaCl(s)
2HCl(g)
NaCl(aq)
304
Types of equilibria on basis of physical state
Homogeneous equilibrium
One phase in the system
H2(g) + Cl2(g)
2HCl(g)
SO2(g) + NO2(g)
SO3(g) + NO(g)
⚫
⚫
⚫
⚫
Ex-1.
a
ri
uh
⚫
The nature and the properties of the equilibrium state are the same regardless of the direction from
which it is achieved. It can be achieved in both directions.
Equilibrium is dynamic in nature.
It means that at microscopic level reaction has not stopped. It appears that no change is occuring but
both the opposing reactions are proceeding at the same rate. So there is no net change.Thus
equilibrium is not static in nature.
A catalyst can alter the rate of approach of equilibrium but does not change the state of equilibrium. By
using catalyst, the equilibrium can be achieved in different (more/less) time, but the relative
concentrations of reactants and products are same irrespective of the presence or absence of a
catalyst.
Equilibrium can be observed by constancy of some observable properties like colour, pressure,
concentration, density, temperature, refractive index etc.which may be suitable in a given reaction.
At equilibrium, free energy change G = 0
Equilibrium state can be affected by altering factors like pressure, volume, concentration and
temperature etc.(Le chateliers Principle).
System moves toward an equilibrium state spontaneously even if it is disturbed. It will return to original
state.
Ja
⚫
Characteristics of chemical equilibrium :
(1) Consider the following cases–
()
H2+I 2
()
2HI
lp
⚫
Heterogeneous equilibrium
More than one phase are present in system
3Fe(s) + 4H2O(g)
Fe3O4(s) + 4H2(g)
2Na2O2(s) + 2H2O()
4NaOH + O2(g)
Temperatures becomes
constant
T 1>T 2>T 3>T 4
Furnace
H2O(l)
Level of water
becomes
constant
a
Equilibrium state has
been attained
()
Metal Rod
Sa
nk
The nature of flow of energy in case () is same as that in–
(A) 
(B) 
(C)  and 
(D) None
(D) None, because in  and , the flow of energy or matter is taking place only in one direction. While
in equilibrium state, the flow of energy takes place in both directions equally. Thus () is a dynamic
equilibrium while states in  and  are called steady state (static equilibrium).
Sol.
Law of mass action : [By Guldberg and Waage]
Rate at which a substance reacts  [Active Mass of the substance]
Active Mass = Molar concentration i.e. Moles/Litres
Wt of substance (gram)
=
Molar wt.  Vol.(Litre)
It is represented in square brackets i.e. [ ] e.g. [A], [N2] etc.
Note: Active masses are dimensionless quantities but for our purposes we generally take them with
dimensions of molarity, partical pressure, etc.
The rate of a chemical reaction at a particular temperature is proportional to the product of active
masses of reactants raised to the powers of their stoichiometric coefficients.
Ex. aA + bB ⎯→ products
Rate of reaction  [A]a [B]b
Rate = k [A]a [B]b, where k is the rate constant of the reaction.
305
Section (B & C) : Homogeneous equilibrium: KC in gaseous system & KP in gaseous
system
Equilibrium constant (K) :
e.g.
N2(g) + 3H2(g)
e.g.
1
1
H2(g) + I2(g)
2
2

2NH3(g)

HI(g)
2
[NH3 ]eq
uh

Kc is a constant and is called the equilibrium constant in terms of concentration.
where all the concentrations are at equilibrium and are expressed in moles/litre.
[PCl3 ]eq [Cl2 ]eq
e.g.
PCl5 (g)
PCl3 (g) + Cl2(g)

KC =
[PCl5 ]eq
KC =
Kc =
[N2 ] [H2 ]3eq
[H]'eq
1/2
[H2 ]1/2
eq [ 2 ]eq
KP → Equilibrium constant in terms of partial pressure. It is defined for the equilibrium reaction
which contains at least one gaseous component.
e.g.
aA(g) + bB(g) → cC(g) + dD(g)
d
[PC ]ceq [PD ]eq
KP =
[PA ]aeq [PB ]beq
Ja

a
ri
For a general reaction
aA + bB
cC + dD,
Forward reaction rate rf = kf [A]a [B]b,
Backward reaction rate rb = kb [C]c[D]d,
At equilibrium rf = rb
kf [A]aeq [B]beq = kb [C]ceq [D]deq
The concentrations of reactants & products at equilibrium are related by
d
[C]ceq [D]eq
kf
= KC =
kb
[A]aeq [B]beq
lp
Where various pressures are the partial pressures of various gases substances.
Section (D) : Relation between KP and KC
Relation between Kp & KC
PV = nRT
or,
P=
a

n
RT
V
n
= (moles per litre)
V
PC = [C] RT ;
PD = [D] RT ; PA = [A] RT ; PB = [B] RT
[C]c (RT)c [D]d (RT)d
[C]c [D]d
KP =
=
(RT)(c+d) – (a+b)
[A]a (RT)a [B]b (RT)b
[A]a [B]b
Sa
nk
P = CRT where C =

Kp = Kc(RT)n
Where ng = (c + d) – (a + b), calculation of n involves only gaseous components.

ng = sum of the number of moles of gaseous products – sum of the number of moles of gaseous
reactants. ng can be positive, negative, zero or even fraction.
CaCO3(s)
CaO(s) + CO2(g)
ng = 1 (because there is only one gas component in the products and no gas component in the
reaction)
 Kp = Kc.(RT)
Unit of Equilibrium contants :
⚫
⚫
Unit of Kp is (atm)n
Unit of Kc is (mole/Lit)n = (conc.)n
306

In fact, equilibrium constant does not carry any unit because it is based upon the activities of
reactants and products and activities are unitless quantities. Under ordinary circumstances,
where activities are not known, above types of equilibrium constant and their units are
employed.
For pure solids and pure liquids, although they have their own active masses but they
remain constant during a chemical change (reaction). Therefore, these are taken to be
unity for the sake of convenience.
e.g.
CaCO3(s)
CaO(s) + CO2(g)
KC = [ CO2],
KP = PCO2
Section (E) : Reaction quotient and Its applications
Predicting the direction of the reaction
Reaction Quotient (Q)
At each point in a reaction, we can write a ratio of concentration terms having the same form as the
equilibrium constant expression. This ratio is called the reaction quotient denoted by symbol Q.
It helps in predicting the direction of a reaction.
[C]c [D]d
The expression Q =
at any time during reaction is called reaction quotient.The concentrations
[A]a [B]b
[C], [D], [A], [B] are not necessarily at equilibrium.
The reaction quotient is a variable quantity with time.
It helps in predicting the direction of a reaction.
if Q > Kc reaction will proceed in backward direction until equilibrium in reached.
if Q < Kc reaction will proceed in forward direction until equilibrium is established.
if Q = Kc Reaction is at equilibrium.
a



lp
⚫
⚫
Ja
uh
⚫
a
ri
Note : 
Sa
nk
eg.
2A(g) + B(g)
C(g) + D(g)
QC = Reaction quotient in terms of concentration
[C][D]
QC =
[A]2 [B]
[C]eq [D]eq
KC =
[Here all the conc. are at equilibrium]
[A]2eq [B]eq
Ex-2.
Sol.
1
Br2 (g)
2
KP = 0.15 atm at 90°C. If NOBr, NO and Br 2 are mixed at this temperature having partial pressures 0.5
atm, 0.4 atm and 0.2 atm respectively, will Br2 be consumed or formed ?
[PBr2 ]1/2 [PNO ]
[0.2]1/2 [0.4]
QP =
=
= 0.36
[0.50]
[PNOBr ]
KP = 0.15

QP > KP
Hence, reaction will shift in backward direction 
Br2 will be consumed
For the reaction NOBr (g)
NO(g) +
307
⚫
Predicting the extent of the reaction
K=
[Pr oduct]eq
[Reac tan t]eq
Case-I: If K is large (K > 103) then product concentration is very very larger than the reactant ([Product]
>>[Reactant]) Hence concentration of reactant can be neglected with respect to the product. In this
case, the reaction is product favourable and equilibrium will be more in forward direction than in
backward direction.
a
ri
Case-II : If K is very small (K < 10–3)
[Product] << [Reactant]
Hence concentration of Product can be neglected as compared to the reactant.
In this case, the reaction is reactant favourable.
The KP values for three reactions are 10–5, 20 and 300 then what will be the correct order of the
percentage composition of the products.
Sol.
Since Kp order is 10–5 < 20 < 300 so the percentage composition of products will be greatest for
Kp = 300.
Section (F) : Properties of equilibrium Constant
Sa
nk
a
lp
⚫
⚫
Equilibrium constant does not depend upon concentration of various reactants, presence of catalyst,
direction from which equilibrium is reached
The equilibrium constant does not give any idea about time taken to attain equilibrium.
K depends on the stoichiometry of the reaction.

If two chemical reactions at equilibrium having equilibrium constants K 1 and K2 are added then
the resulting equation has equilibrium constant K = K 1. K2
Equation constant
A (g)
B(g)
K1
B (g)
C(g)
K2
On adding
A(g)
C(g)
K = K1 . K2
1

If the reaction having eq. constant K1 is reversed then resulting equation has eq. constant
K1
A(g) + B(g)
C(g) + D(g)
1
On reversing, C(g) + D(g)
A(g) + B(g) K =
K1

If a chemical reaction having equilibrium constant K1 is multiplied by a factor n then the
resulting equation has equilibrium constant K = (K1)n , n can be fraction
eg.
D2(g)
2A(g)
K1
 1
1
Multiplying by   ,
D2
A
K = (K1)1/2 = K1
2
2
Ja
⚫
uh
Ex-3.
⚫
Equilibrium constant is dependent only on the temperature.
It means Kp and Kc will remain constant at constant temperature no matter how much changes are
made in pressure, concentration, volume or catalyst.

However if temperature is changed,
k
H  1 1 
log 2 =
 −  ; H = Enthalpy of reaction
2.303 R  T1 T2 
k1
K2 > K1 provided H = +ve (endothermic reaction)
K2 < K1 if H = –ve (exothermic reaction)
In the above equation, the unit of R and H/T should be same.
If T2 > T1 then
308
Section (G) : Homogenuous Equilibrium (liquid system)
The reaction between alcohol and acid to form ester is an example of homogeneous equilibrium in
liquid system.
CH3COOH() + C2H5OH()
CH3COOC2H5() + H2O()
KC =
Sol.
In an experiment starting with 1 mole of ethyl alcohol, 1 mole of acetic acid and 1 mole of water at TºC,
the equilibrium mixture on analysis shows that 54.3% of the acid is esterfied. Calculate the equilibrium
constant of this reaction.
CH3COOH() + C2H5OH()
CH3COOC2H5() + H2O()
Initial
1
1
0
1
At equilibrium 1 – x
1–x
x
1+x
1 – 0.543
1 – 0.543
0.543
1 + 0.543
Hence given x = 0.543 mole
Applying law of mass action : KC =
[ester][water]
0.543  1.543
=
= 4.0
0.457  0.457
[acid][alcohol]
EQUATION INVOLVING IONS :
Equilibrium involving ions always take place in aquous medium. In case of expression of K C
concentration of ion is taken.
Ex.
Ja

1  54.3
= 0.543 mole)
100
uh
(54.3% of 1 mole =
a
ri
Ex-4.
[CH3 COOC2H5 ][H2 O]
[CH3 COOH][C2H5 OH]
Ag+(aq.) + Cl–(aq.)
AgCl(s) Kc =
1
[Ag+ ][Cl– ]
lp
Section (H) : Heterogenuous equilibrium
a
For pure solid and pure liquid, active mass is taken to be unity i.e. 1 as they remain constant
throughout the reaction :
⚫
CaCO3 (s)
CaO (s) + CO2 (g)
KP = ( PCO2 )eq, KC = [CO2 (g)]eq
WCaCO3
moles
density CaCO3
=
=
= constant
MCaCO3
volume
MCaCO3
V
[CaO(s)]eq [CO2 (g)]eq
Sa
nk
[CaCO3(s)] =
K=
[CaCO3 (s)]eq
K.[CaCO3 (s)]eq
[CaO(s)]eq
⚫
= [CO2(g)]eq
KC = [CO2(g)]eq
H2O()
H2O(g)eq
KP = ( PH2O(g) )eq, KC = [H2O (g)]eq
[For pure solid and pure liquid active mass is taken as unity i.e. = 1]
Section (I) : Degree of dissociation () and vapour density
It is the fraction of one mole dissociated into the products. (Defined for one mole of substance)
So,  = no. of moles dissociated / initial no. of moles taken
= fraction of moles dissociated out of 1 mole.
Note : % dissociation =  × 100
Suppose 5 moles of PCl5 is taken and if 2 moles of PCl5 dissociated then  =
2
5
= 0.4
Let a gas An dissociates to give n moles of A as follows-
309
An(g)
a
t=0
nA(g)
0
a–x
t = teq
n.x
=
x
a

x = a.
a – a  = a(1–) n a 
Total no. of moles = a – a  + n a  = [1 + (n – 1)  ] a
Significance of n
sum of stoichiometric coefficient of product
n=
sum of coefficient of reactants
(i)
for PCl5(g)
PCl3(g) + Cl2(g)
(n = 2)
for 2HI(g)
H2(g) + 2(g)
(n =
3
1
+
= 2)
2
2
(n = 1)
Calculate the degree of dissociation and Kp for the following reaction.
PCl5(g)
PCl3(g) +
Cl2(g)
t=0
a
0
0
t=t
a –x
x
x
Since for a mole, x moles are dissociated
x

For 1 mole,
moles =  are dissociated
a

x=a
PCl5(g)
PCl3(g) +
Cl2(g)

At
t = teq a – a
a
a
Total no. of moles at equilibrium = a + a = a (1 +  )
a(1 −  ) P
a
a  .P
PPCl5 =
PCl2 =
, PPCl3 =
,
.P
a (1 +  )
a(1 +  )
a (1 +  )
a
ri
(iii)
N2(g) + 3H2(g)
2
KP =
 1−  
 1 +   P
2 . P
(Remember)
1− 2
a
KP =
 P  


 1 +   
lp
Ja
Sol.
for 2NH3(g)
uh
Ex-5.
(ii)
Sa
nk
Observed molecular weight and Observed Vapour Density of the mixture
Observed molecular weight of An(g) =

Mobs
=
a.Mth
molecular weight of A n (g)
=
total no. of moles at equilibrium
a(1 + (n − 1) )
Mth
[1 + (n − 1) ]
where Mth = theoritical molecular weight (n = atomicity)
Mmixture =
MAn
[1 + (n − 1) ]
, MAn = Molar mass of gas An
Vapour density (V.D). : Density of the gas divided by density of hydrogen under same temp &
pressure is called vapour density.

D = vapour density without dissociation =
MAn
2
d = vapour density of mixture = observed v.d. =
Mmix
2
D
= 1 + (n – 1) 
d

=
D−d
M − Mo
= T
(n − 1)  d
(n − 1)M0
310
Where MT = The oritical molecular wt., M0 = observed molecular wt. or molecular wt. of the
mixture at equilibrium.
Note : It is not applicable for n = 1 eg. Dissociation of HI & NO (as the total number of don’t change during the
dissociation.)
Sol.
The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 33°C calculate the no. of moles of
NO2 if 100g of N2O4 were taken initially.
N2O4(g)
2NO2(g)
Mmix = 2 × 38.3 = 76.6
M
92
Mmix = th =

 = 0.2
1+ 
1+ 
N2O4
2NO2
t=0
a
0
t=t
a – a
2a
2  100  0.2
no. of moles of NO2 = 2a =
= 0.435
92
a
ri
Ex-6.
uh
Section (J) : Thermodyanamics of equilibrium
lp
Ja
For a general reaction,
mA + nB
pC + qD,
G is given byG = Gº + 2.303 RT log10Q
where G = Gibb's Free energy change
Gº = Standard Gibb's Free energy change
Q = reaction quotient
Since, at equlibrium, Q = K
Here K is thermodynamic equilibrium constant replacing Kc or Kp
(a )p (a )q
K= C m D n ;
Here aX denotes the activity of X.
(a A ) (aB )
In fact, ‘ax’ is the ratio of the activity of substance at equilibrium and its activity in standard condition.
That is why it is unitless and K is also unitless.
a
Note : (i) Themodynamic equilibrium constant is unitless since activity is unitless.
(ii) For pure solids & pure liquids, activity is unity.
(iii) For gases (ideal behaviour), the activity is its partial pressure (in atm).
(iv) For components in solution, activity is molar concentration.
G = 0
Gº = – 2.303 RT log10K
Gº = Hº – TSº
Hº = Standard enthalpy change of the reaction
S° = Standard entropy change

– 2.303 RT log10K = Hº – TSº
1
Sº
Hº

log10K = –
.
+
2.303R
2.303 RT
1
Hº
Sº
If plot of ln k vs
is plotted then it is a straight line with slope = –
and intercept =
T
R
R
log K
log K
Sa
nk
At equilibrium,

Now since,
where
0
T – 1 or 1/T
Endothermic reaction
0
T – 1 or 1/T
Exothermic reaction
311
Slope =
−Hº
= tan 
2.303R
y intercept =
Sº
2.303R
If at temperature T1, equilibrium constant is K1 and at T2, it is K2 then ;
−Hº
Sº
1
log10K1 =
.
+
........... (i)
2.303R T1
2.303R
log10K2 =
−Hº 1
Sº
.
+
2.303R T2
2.303R
........... (ii)
[Assuming Hº and Sº remains constant in this temperature range.]
a
ri
Subtract eq. (ii) from (i) we get Vant Hoff equationK 
Hº  1
1
log  1  =
− 

2.303R  T2 T1 
 K2 
*
*
G > 0 for non-spontaneous process or reaction.
G = 0 for equilibrium.
Variation of equilibrium constant K with temperature T is given by
van’t
Hoff equation,
log K = log A –
lp
Ex-7.
Ja
uh
Note :  H should be substituted with sign.
 Unit of H/T and gas constant R should be same.
 For endothermic (H > 0) reaction value of the equilibrium constant increases with the rise in
temperature
 For exothermic (H < 0) reaction, value of the equilibrium constant decreases with increase in
temperature
Condition for spontaneity : G < 0 for spontaneous process or reaction.
Since, G = H – TS

H – TS < 0

T > H/S
Hº
2.303 RT
Sa
nk
a
A graph between log K and T–1 was a straight line as shown in the
figure and having  = tan–1 (0.5) and OP = 10. Calculate :
(a) H° (standard heat of reaction) when T = 300 K,
(b) A (pre-exponential factor),
(c) Equilibrium constant K, at 300 K,
(d) K at 900 K if H° is independent of temperature.
Sol.
(a)
log10 K = log10 A –
Hº
2.303 RT
It is an equation of a straight line of the type y = c + mx
H
Slope ‘m’ = tan = 2.303 R
Hº
2.303  8.314

H° =9.574 J mol–1
(b)
Intercept ‘c’ = log10 A = 10

A = 1010
(c)
log K = 10 –
0.5 =
(d)
9.574

2.303  8.314  298
K = 9.96 × 109
K 
1
H  1
log  2  =
 − 
2.303R  T1 T2 
 K1 
K2
1 
9.574
 1
−
log
=


2.303  8.314  298 798 
9.96  109
On solving
K2 = 9.98 × 109
312
Ans. (a) 9.574 J mol–1 ; (b) A = 1010 ; (c) 9.96 ×109 ; (d) 9.98 × 109
Section (K) : Le-chatelier's principle
⚫
Le Chatelier's Principle:
If a change is applied to the system at equilibrium, then equilibrium will be shifted in that direction in
which it can minimise the effect of change applied and the equilibrium is established again under new
conditions.
Effect of concentration : If the concentration of a component is increased, reaction shifts in a direction
which tends to decrease its concentration. e.g. In the following example.


a
ri
N2 (g) + 3H2(g)
2NH3(g)
[reactant] 
Forward shift
[Product] 
Backward shift
If concentration of reactant is increased at equilibrium then reaction shifts in the forward direction.
If concentration of product is increased then reaction shifts in the backward direction
Note : The addition of any solid component does not affect the equilibrium.
Effect of volume :

If volume is increased, pressure decreases hence reaction will shift in the direction in which
pressure increases that is in the direction in which number of moles of gases increases and
vice versa.

If volume is increased then,
for
ng > 0 reaction will shift in the forward direction
ng < 0 reaction will shift in the backward direction
ng = 0 reaction will not shift. eg. H2(g) + I2(g) 2HI(g)
(No effect)
Explanation :
(i)
ng > 0, eg.
QC =
V
PCl5(g)

PCl3(g) + Cl2 (g)
(nPCl3 )
V
(nPCl5 )
V
QC 

1
for ng > 0
V
lp
(nCl2 )
Ja
uh
⚫
[(n) = moles]
ng < 0, eg.
N2(g) + 3H2(g)
Sa
nk
(ii)
a
On incerasing V, QC, decreases.
Now, for QC < KC reaction will shift in forward direction.
Thus, if, Volume  QC  (Forward shift)
Volume  QC  (Backward shift)
QC =
 ( nNH3

 V
) 


 (nN2

 V
)   (nH2

  V
V  QC  (Backward shift)
⚫
2NH3(g)
2

3
) 


;
QC  V2
for ng < 0
V  QC  (Forward shift)
Effect of pressure :
On increasing pressure, equilibrium will shift in the direction in which pressure decreases i.e. no. of
moles in the reaction decreases and vice versa.

P  no. of moles
(i)
For ng = 0 → No. effects
(ii)
For ng > 0,
PCl5(g)
PCl3(g) + Cl2(g)
(XPCl3 P). (XCl2 P)
Qp =

QP  P
[( ) = mole fraction]
(XPCl5 . P)
P ; QP ; (Forward shift)
P ; QP; (Backward shift)
313
(iii)
For n < 0, eg.
N2(g) + 3H2(g)
2NH3(g)
2
QP =
(XNH )P)
3


[( XN2 ).P][(XH2 ) P]3
P ; QP ; (Forward shift)
1
P2

QP 
;
P ; QP ; (Backward shift)
Effect of catalyst :
Due to catalyst, the state of equilibrium is not affected i.e. no shift will occur as catalyst lowers the
activation energy of both the forward & reverse reaction by same amount, thus altering the forward &
reverse rate equally and hence, the equilibrium will be attained faster i.e time taken to reach the
equilibrium is less.
⚫
Effect of inert gas addition :
(a)
At constant volume : Inert gas addition has no effect at constant volume
(b)
At constant pressure : If inert gas is added then to maintain the pressure constant, volume is
increased. Hence equilibrium will shift in the direction in which larger no. of moles of gas is
formed
(i)
ng > 0, reaction will shift in the forward direction
(ii)
ng < 0, reaction will shift in the backward direction
(iii)
ng = 0, no effect
⚫
Effect of temperature :
(i)
Exothermic reaction : The reaction in which heat is evolved
A(g) + B(g)
C(g) + D(g) + Heat
H = – ve
eg.
N2(g) + 3H2(g)
2NH3(g) + Heat
Ja
uh
a
ri
⚫
 K' will decrease
K1
Hº  1 1 
−
log
=
(from vant’ hoff equation)
K2
2.303R  T2 T1 
T
a
Endothermic reaction : energy consumed.
A(g) + B(g)
C(g) + D(g) – Heat
T   K

Forward ;
Sa
nk
(ii)
lp
K1
<0

log K1 – log K2 > 0 
K2
Reaction will shift in backward direction.
T
 K will increases.
Reaction will shift in forward direction.
log
log K1 > log K2 
 H = + ve
T  K 

K1 > K2
Backward
Application of le chatelier’s principle : Practical equilibrium situations :
Section (L) : Vapour pressure and Relative Humidity
⚫
It is the pressure exerted by the vapours over it’s liquid when it is in equilibrium with the liquid. Vapour
pressure of water is also called aqueous tension.
H2O ()
H2O (g) ; KP = PH2O = constant at fixed temperature
Hence V.P. of liquid is independent of pressure, volume and concentration change.
e.g. at 25°C, vapour pressure of water  24 mm of Hg
Relative Humidity =
⚫
Partial pressure of H2O vapours
Vapour pressure of H2O at that temp.
Formation of diamond :
C (graphite)
C (diamond) – Heat;
H = + ve


Density Low
Density High
Volume High
Volume Low
Formation of diamond is favourable at high pressure and high temperature
314
⚫
Melting of ice :
H2O (s)

Density Low
Volume High
H2O () ;

Density High
Volume Low
H = + ve
Melting of ice is favourable at high temperature and high pressure.
Boiling of water :
H2O()
H2O(g)


Density High
Density Low
Volume Low
Volume High
On incerasing pressure, equilibrium will shift in the direction in which volume is decreasing i.e.
backward.
Hence, on incerasing pressure, the boiling point increases.
⚫
Formation of ammonia by Haber’s process :
N2 (g) + 3 H2 (g)
2NH3 (g) H = – 22.4 Kcal/mol.
(i) The reaction will shift in the forward direction at low temperature, but at very low temperature the rate
of reaction becomes very low; thus moderate temperature is used for this reaction.
(ii) At high pressure, reaction will shift in forward direction to form more product.
⚫
Manufacturing of SO3 by contact process
2SO2(g) + O2(g)
2SO3(g) + 45.2 kcal
High pressure (1.5 to 1.7 atm), Low temperature (500°C), higher quantity of SO2 and O2 are favourable
conditions for the formation of SO3.
⚫
Manufacturing of NO by Birkeland–Eyde process
N2(g) + O2(g)
2 NO(g) – 43.2 kcal

No effect on change of pressure

High temperature (1200°C to 2000°C), High concentration of N 2 and O2 are favourable
condition for the formation of NO.
Ex-8.
The equilibrium constant of the reaction at 25°C
CuSO4.5H2O (s)
CuSO4.3H2O(s) + 2H2O(g)
is 1.084 × 10–4 atm2. Find out under what conditions of relative humidity, CuSO 4.5H2O will start loosing
its water of crystallization according to above reaction. (Vapour pressure of water at 25°C is 24 mm of
Hg).
Sa
nk
a
lp
Ja
uh
a
ri
⚫
(
)
2
1.084  10 − 4 = 1.041 × 10–2 atm  8 mm of Hg
Sol.
KP = PH2O

If in a room, pressure of water is greater than 8 mm of Hg then CuSO 4.3H2O will absorb water from air
and will form CuSO4.5H2O & will keep absorbing until partial pressure of H 2O becomes 8 mm of Hg.
If PH2O < 8 mm of Hg then CuSO4.5H2O will loose water of crystallization and reaction will move in

so PH2O =
forward direction.
i.e. If relative humidity <
8
< 33.33% then CuSO4.5H2O will loose water of crystallization.
24
315
Section (M) : Simultaneous equilibria
If in any container there are two or more equilibria existing simultaneously involving one or more than
one common species. Then in both/all equilibrium the concentration of common species is the total
concentration of that species due to all the equilibria under consideration.
t=0
t = teq
A(s)
X(g) + Y(g)
a
0
0
a–t
t
t+u
B(s)
Z(g) + Y(g)
b
0
0
b–u
u
u +t
KC1 = t (u + t)
a
ri
e.g.
KC2 = (u + t) u
102 g of solid NH4HS is taken in the 2L evacuated flask at 57°C. Following two equilibrium exist
simultaneously
3
1
NH4HS (s)
NH3 (g) + H2S (g)
;
NH3 (g)
N2 (g) + H2 (g)
2
2
one mole of the solid decomposes to maintain both the equilibrium and 0.75 mole of H 2 was found at
the equilibrium then find the equilibrium concentration of all the species and K C for both the reaction.
102
Moles of NH4HS =
=2
51
NH4HS (s)
NH3 (g) + H2S (g)
KC1
Sol.
2
1
0
1–x
Ja
uh
Ex-9.
0
1
3
1
N2 (g) + H2 (g)
KC2
2
2
3x
x
1–x
2
2
3x
1
Given that moles of H2 =
= 0.75

x=
2
2
1 (1 − x) 1
=
[Since V = 2 L]
KC1 =
2 2
8
Sa
nk
a
lp
NH3 (g)
3/2
KC2
 3x   x 
 4   4 
=
 1− x 
 2 
½
=
 3
 8 
3/2
 1
 8 
1
4
½
3/2
= (3)
1 4 (3)3/2
 =
64 1
16
316
SUMMARY
Chemical equilibrium is a dynamic state in which the concentration of reactants and products remain
constant because the rates of the forward and the reverse reaction are equal. For the general reaction
a A(g) + b B (g)
c C(g) + d D(g)
concentrations in the equilibrium mixture are related by the equilibrium equation :
KC =
[C]ceq [D]deq
[A]aeq [B]beq
a
ri
The ratio on the right side of the equation is called the equilibrium constant expression. The equilibrium
constant Kc is the number obtained when equilibrium concentrations (in mol/L) are substituted into the
equilibrium constant expression. The value of Kc varies with temperature and depends upon the form of
the balanced chemical equation.
The equilibrium constant Kp can be used for gas phase reactions. It is defined in the same way as K c
uh
except that the equilibrium constant expression contains partial pressures (in atmospheres) instead of
molar concentrations. The constants Kp and Kc are related by the equation,
Kp = Kc (RT)ng, where ng = (c + d) – (a + b).
Ja
Homogeneous equilibrium are those in which all reactants and products are in a single phase;
heterogeneous equilibria are those in which reactants and products for heterogeneous equilibrium does
not include concentrations of pure solids or pure liquids.
The value of the equilibrium constant for a reaction makes it possible to judge the extent of reaction,
lp
predict the direction of reaction, and calculate equilibrium concentrations (or partial pressures) from
initial concentration (or partial pressures) . The farther the reaction proceed towards completion, the
larger the value of Kc. The direction of a reaction not at equilibrium depends on the relative values of K c
a
and the reaction quotient Qc which is defined in the same way as Kc except that the concentrations in
the equilibrium constant expression are not necessarily equilibrium concentrations. If Q c < Kc, net
reaction goes from left to right to attain equilibrium; if Qc = Kc, the system is at equilibrium.
Sa
nk
The composition of an equilibrium mixture can be altered by changes in concentration, pressure
(volume), or temperature. The qualitative effect of these changes is predicted by Le Chatelier's
principle, which says that if a stress is applied to a reaction mixture at equilibrium, net reaction occurs in
the direction that relieves the stress. Temperature changes affect equilibrium concentrations because
Kc is temperature-dependent. As the temperature increases, Kc for an exothermic reaction decreases,
and Kc for an endothermic reaction increases.
A catalyst increases the rate at which chemical equilibrium is reached, but it does not affect the
equilibrium constant or the equilibrium concentration. The equilibrium constant for a single-step reaction
equals the ratio of the rate constants for the forward and reverse reactions : Kc = kf / kb.
317
318
lp
a
Sa
nk
a
ri
uh
Ja
MISCELLANEOUS SOLVED PROBLEMS (MSPs)
Ans.
Sol.
2.
Ans.
Sol.
An example of a reversible reaction is :
(A) Pb(NO3)2(aq) + 2NaI(aq)
PbI2(s) + 2NaNO3(aq)
(B) AgNO3(aq) + HCl(aq)
AgCl(s) + HNO3(aq)
(C) 2Na(s) + H2O()
2NaOH (aq) + H2(g)
(D) KNO3(aq) + NaCl(aq)
KCl(aq) + NaNO3(aq)
(D)
Precipitation reactions, acid base reactions and reactions in which gases are liberated and are taking
place in open container will be irreversible reactions.
For the reaction, A + B
3C, if 'a' mol/litre of each 'A' & 'B' are taken initially then at equilibrium the
incorrect relation is :
(A) [A] − [B] = 0
(B) 3[B] + [C] = 3a
(C) 3[A] + [C] = 3 a
(D) [A] + [B] = 3 [C]
(D)
At equilibrium :
[A] = [B]
[A] +
1
[C] = a
3
The equilibrium constant for the decomposition of water H2O(g) → H2(g) +
1
O2(g) is given by :
2
uh
3.
a
ri
1.
(C) K =
Ja
( = degree of dissociation of H2O(g); p = Total equilibrium pressure )
 3/2p1/2
 2p1/ 2
(A) K =
(B)
K
=
(1 +  )(2 −  )1/ 2
(1 −  )(2 +  )1/2
 3p1/ 2
(D) K =
2
Ans.
(B)
Sol :
H2O(g) → H2(g) +
1
O2(g)
2
 3p3/ 2
(1 −  )(2 +  )1/ 2
1− 
p
1+  / 2

=
p
1+  / 2
/2
=
p
1+  / 2
pH2O =
Sa
nk
pH2
a
So,
lp
1
0
0
1–

/2
Total moles at equilibrium = 1 –  +  + /2 = 1 + /2
Let the total pressure at equilibrium be = p
pO2
So
Kp =
(pO2 )1/ 2 (pH2 )
(pH2O )
The reaction quotient Q for N2(g) + 3H2(g)
Ans.
Sol.
proceed in backward direction, when:
(A) Q = Kc
(B) Q < Kc
(C) Q > Kc
(D) Q = 0
(C)
When Q > K, reaction will favour backward direction and when Q < K, it will favour forward direction.
5.
2NH3(g) is given by Q =
[NH3 ]2
. The reaction will
[N2 ][H2 ]3
4.
Ans.
0.96 g of HI were heated to attain equilibrium 2HI(g)
H2(g) + I2(g). The reaction mixture on titration
requires 15.7 mL of N/10 hypo solution. Calculate degree of dissociation of HI.
20.9%
Sol.
meq of I2 =
w I2
127
× 1000 = 1.57
319
wI2 =
127  1.57
1000
2HI(g)
a
2
 = 0.209
a
2
1.57
0.96
×=
2000
256
Ans.
Would 1% CO2 by volume in air be sufficient to prevent any loss in weight when M2CO3 is heated at
120ºC ?
M2CO3(s)
M2O(s) + CO2(g)
Kp = 0.0095 atm at 120ºC. How long would the partial pressure of CO2 have to be to promote this
reaction at 120ºC ?
(No reactions)
For,
M2CO3(g) (s)
M2O (s) + CO2 (g)
1
1
× Pair =
× 1 atm. = 0.01 atm.
100
100
 CO2 is 1% in air ;

Also for equilibrium
KP = PCO2

PCO2 = 0.01 atm.
Given,
PCO2 =
a
ri
Ans.
Sol.
1.57
2000
H2 (g) + I2(g)
a – a
6.
nI2 =
= 0.0095 atm.
uh
Since decomposition is carried out in presence of PCO2 of 0.01 atm and KP = 0.0095 atm, thus,
more than 0.0095 atm.
Altrernate solution :
For
M2CO3 (s)
M2O (s) + CO2 (g)
Ja
practically no decomposition of M2CO3. Thus, 1% CO2 is sufficient to prevent any loss in weight.
If at all reaction is desired, the PCO2 must be lesser than 0.0095 atm as PCO2 at equilibrium cannot be
 1

 100 + P

KP = PCO2 and the pressure of CO2 already present in 1/100 atm. Let the decomposition of
M2CO3 produces the CO2 of pressure P, then
KP =
1
+P
100
or
0.0095 = P + 001
lp

or
P = – 0.0005.
The value of pressure comes negative and thus, it may be concluded that M2CO3 will not dissociate in
pressure of CO2 of pressure 0.01 atm.
For the chemical equilibrium, CaCO3(s)
CaO(s) + CO2(g)
Hfº can be determined from which one of the following plots ?
(B)
loge Pco2
(A)
loge Pco2
Sa
nk
a
7.
log T
1/T
(D)
loge Pco2
(C)
loge Pco2
1/T
logT
320
Sol.
CaCO3(s)
CaO(s) + CO2(g)
Kp = PCO2
H r0
2.303RT
log Kp = log A –
log PCO2 = log A –
Hrº
1
2.303 RT
......... (i)
Graph (a) represents (i) and its slope will be used to determine the heat of the reaction.
So, Ans. (A).
N2 (g)
+
9P – y – x
Total pressure
2H2 (g)
N2H4 (g)
13P – 3x – 2y
y
= PN2 + PH2 + PNH3 + PN2H4 = 3.5 atm
a
ri
Sol.
In a container of constant volume at a particular temparature N 2 and H2 are mixed in the molar ratio of
9:13. The following two equilibria are found to be coexisting in the container
N2(g) + 3H2(g)
2NH3(g)
N2(g) + 2H2(g)
N2H4(g)
The total equilibrium pressure is found to be 3.5 atm while partial pressure of NH 3(g) and H2(g) are 0.5
atm and 1 atm respectively. Calculate of equilibrium constants of the two reactions given above.
Let the initial partial pressures of N2 and H2 be 9P and 13 P respectively
N2 (g)
+ 3H2 (g)
2NH3 (g)
9P – y – x
13P – 3x – 2y
2x
uh
8.
PH2 = (13P – 3x – 2y) = 1 atm


so
9P = 2.25
P = 0.25 atm
from (3) equation
...(3)
(9P – x – y) + 1 atm + 0.5 + y = 3.5
(9P – x) = 2 atm
from (1)
2y = 1.5
y = 0.75 atm
PN2 = 9P – x – y = 1.25 atm
lp
so
...(1)
Ja
= (9P – x – y) + (13 P – 3x – 2y) + 2x + y = 3.5 atm
...(2)
PNH3 = 2x = 0.5 atm
PH2 = 1 atm
a
PNH3 = 0.5 atm
PN2H4 = 0.75 atm
KP1 =
2
PNH
3
=
0.5  0.5
= 0.2 atm–2
1  1  1  1.25
Sa
nk
So,
KP2 =
PH32 . PN2
PN2H4
PN2 .PH22
=
0.75
= 0.6 atm–2
1  1  1.25
321
 Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
Section (A) : Properties of equilibrium, active mass
A-1.
In a reaction A + B
C + D the rate constant of forward reaction & backward reaction is k f = 2 × 10–4
–1
–5
M sec and kb = 5 × 10 M sec–1 then the equilibrium constant (k) for reaction is expressed as :
a
ri
A-2. What is the active mass of 5.6 litres of O2 at S.T.P.?
Section (B) : Homogeneous equilibrium : KC in gaseous system
uh
B-1. A mixture of SO3, SO2 and O2 gases is maintained at equilibrium in 10 litre flask at a temperature at
which KC for the reaction, 2SO2(g) + O2(g)
2SO3(g) is 100 mol–1 litre. At equilibrium.
(a) If no. of mole of SO3 and SO2 in flask are same, how many mole of O2 are present ?
(b) If no. of mole of SO3 in flask are twice the no. of mole of SO2, how many mole of O2 are present ?
The equilibrium constant of the reaction, A2 (g) + B2 (g)
2 AB (g) at 100ºC is 16. Initially equal
moles of A2 & B2 are taken in 2L container. Then find mole % of A2 in equilibrium mixture.
B-3.
For the reaction 3A(g) + B(g)
2C(g) at a given temperature, Kc = 9.0. What must be the
concentration of (C) at equlibrium, if a mixture of 2.0 mol each of A, B and C exist in equilibrium ?
Ja
B-2.
lp
B-4.
The gas A2 in the left flask allowed to react with gas B2 present in right flask as
A2(g) + B2(g)
2AB(g) ; Kc = 4 at 27°C.
What is the concentration of AB when equilibrium is established?
a
Section (C) : Homogeneous equilibrium : KP in gaseous system
Sa
nk
C-1. n mole each of H2O(g), H2(g) and O2(g) are mixed at a suitable high temperature to attain the
equilibrium 2H2O(g)
2H2(g) + O2(g). If y mole of H2O(g) are the dissociated and the total pressure
maintained is P, calculate the KP.
C-2.
The moles of N2O4 and NO2 at equilirbrium are 1 and 2 respectively total pressure at equilibrium is 9
atm. Find KP for the reaction N2O4(g)
2 NO2(g).
C-3. 1 mole of N2 and 3 moles of H2 are placed in 1L vessel. Find the concentration of NH 3 at equilibrium, if
the equilibrium constant (KC) at 400 K is
N2 (g) + 3H2(g)
4
.
27
2NH3(g)
Section (D) : Relation between KP and KC
D-1.
Calculate the expression for Kc and Kp if initially a moles of N2 and b moles of H2 is taken for the
following reaction. N2 (g) + 3H2 (g)
2NH3 (g)
(n < 0) (P, T, V given)
D-2. 1 mole of a gas ‘A’ is taken in a vessel of volume 1L. It dissociates according to the reaction
A(g)
B(g) + C(g) at 27°C. Forward and backward reaction rate constants for the reaction are 1.5 ×
10–2 and 3 × 10–2 respectively. Find the concentrations of A, B and C at equilibrium. Also find K p and Kc.
322
D-3. 0.15 mole of CO taken in a 2.5 litre flask is maintained at 500 K along with a catalyst so that the
following reaction can take place;CO(g) + 2H2(g)
CH3OH(g).
Hydrogen is introduced until the total pressure of the system is 8.2 atm at equilibrium and 0.08 mole of
methanol is formed.Calculate :
(i)
Kp & Kc
;
(ii)
the final pressure if the same amount of CO and H2 as before are used, but with no catalyst so
that the reaction takes place on its own.
Section (E) : Reaction quotient and Its applications
At 460°C, KC = 81 for the reaction,
SO2 (g) + NO2 (g)
NO(g) + SO3 (g)
A mixture of these gases has the following concentrations of the reactants and products :
[SO2] = 0.04 M
[NO2] = 0.04 M
[NO] = 0.30 M
[SO3] = 0.3 M
Is the system at equilibrium? If not, in which direction must the reaction proceed to reach equilibrium.
What will be the molar concentrations of the four gases at equilibrium?
Section (F) : Properties of equilibrium Constant
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E-2.
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E-1. A mixture of 1.5 mol of N2, 2 mole of H2 and 8 mol of NH3 is introduced into a 20 L reaction vessel at
500 K. At this temperature, the equilibrium constant, Kc for the reaction
N2(g) + 3H2(g)
2NH3(g) is 1.7× 102.
Is the reaction mixture at equilibrium? If not what is the direction of the net reaction?
Explain the effect of the following on the equilibrium constant.
(i) Concentrations of the reactants are doubled
(ii) The reaction is reversed
(iii) Catalyst is added to the reaction
(iv) Temperature is increased.
F-2.
The equilibrium constant for the reactions N2 + O2
2NO and 2NO + O2
respectively, then what will be the equilibrium constant for the reaction N 2 + 2O2
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F-1.
H2O(g) + CO(g)
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F-3. Calculate the equilibrium constant for the reaction : H 2(g) + CO2(g)
at 1395 K, if the equilibrium constants at 1395 K for the following are
2H2O(g)
2H2 (g) + O2(g) K1 = 2.1 × 10–13
2CO2(g)
2CO(g) + O2(g) K2 = 1.4 × 10–12.
2NO2 are K1 and K2
2NO2?
a
Section (G) : Homogenuous Equilibrium (liquid system)
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G-1. The homogeneous reversible reaction, C2H5OH + CH3COOH
CH3COOC2H5 + H2O is studied at
various initial concentrations of the reactants at constant temperature. Calculate initial acid and alcohol
moles.
Moles of acid
Moles of alcohol
Moles of ester
per litre (initial)
per litre (initial)
per litre at equilibrium
(i)
1
1
0.667
(ii)
X
Y
8/3
Section (H) : Heterogenuous equilibrium
H-1. Write the expressions for equilibrium constant KC and KP and classify in Homogeneous and
Hetereogeneous equilibrium :
(i)
N2O4(g)
2NO2(g)
(ii)
3Fe(s) + 4H2O(g)
Fe3O4(s) + 4H2(g)
(iii)
NH4HS(s)
NH3(g) + H2S(g)
(iv)
CH3COOH() + C2H5OH()
CH3COOC2H5() + H2O()
(v)
MgCO3(s)
MgO(s) + CO2(g)
(vi)
2H2S(g)
2H2(g) + S2(g)
(vii)
SO2(g) + NO2(g)
SO3(g) + NO(g)
(viii)
NH4NO2(s)
N2(g) + 2H2O()
H-2.
For the reaction: CaCO3
CaO(s) + CO2(g); Kp = 1 atm at 927°C. If 20g of CaCO3 were kept in a
10 litre vessel at 927°C, then calculate percentage of CaCO 3 remaining at equilibrium :
323
H-3.
For the given reaction at equilibrium :
1
AgNO3(s)
Ag(s) + NO2(g) + O2(g)
2
If total pressure at equilibrium is P, then calculate KP for the given reaction.
Section (I) : Degree of dissociation () and vapour density
N2O4 is 25% dissociated at 37°C and one atmosphere pressure. Calculate (i) KP and (ii) the percentage
dissociation at 0.1 atmosphere and 37°C.
I-2.
At temperature T, the compound AB2 (g) dissociates according to the reaction; 2AB2 (g)
2 AB(g) +
B2(g).With a degree of dissociation x, which is small compared with unity.Deduce the expression for x in
terms of the equilibrium constant, Kp and the total pressure, P.
I-3.
Vapour density of the equilibrium mixture of NO2 and N2O4 is found to be 38.33. For the equilibrium
N2O4(g)
2NO2(g).
Calculate :
(i) abnormal molecular weight.
(ii) degree of dissociation.
(iii) percentage of NO2 in the mixture.
(iv) KP for the reaction if total pressure is 2 atm.
I-4.
When sulphur in the form of S8(g) is heated at 900 K, the initial partial pressure of S8(g) which was 1
atm falls by 29% at equilibrium. This is because of conversion of some S 8(g) to S2(g). Find the Kp for
reaction,
S8 (g)
4S2 (g).
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Section (J) : Thermodyanamics of equilibrium
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a
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I-1.
For the reaction,
SO2 (g) + 1/2O2(g)
SO3(g)
Hº298 = −98.32 kJ/mole, Sº298 = −95.0 J/mole-K. Find the Kp for this reaction at 298K. (Given that
10.27 = 1.86)
J-2.
From the following data :
(i) H2(g) + CO2(g)
H2O(g) + CO(g) ;
K2000K = 4.4
(ii) 2H2O(g)
2H2(g) + O2(g) ;
K2000K = 5.31 × 10–10
(iii) 2CO(g) + O2(g)
2CO2(g) ;
K1000K = 2.24 × 1022
State whether the reaction (iii) is exothermic or endothermic?
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J-1.
Which of the following reactions will get affected by increase of pressure ? Also mention, whether
change will cause the reaction to go into the right or left direction ?
(i)
CH4(g) + 2S2(g)
CS2(g) + 2H2S(g)
(ii)
CO2(g) + C(s)
2CO(g)
(iii)
4NH3(g) + 5O2(g)
4NO(g) + 6H2O(g)
(iv)
C2H4(g) + H2(g)
C2H6(g)
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K-1.
a
Section (K) : Le-chatelier's principle
K-2.
Using Le Chatelier's principle, predict the effect of
(i) decreasing the temperature and (ii) increasing the pressure on each of the following equilibria :
(a) N2(g) + 3H2(g)
2NH3(g) + Heat
(b) N2(g) + O2(g)
2NO(g) + Heat
(c) H2O(g) + Heat
K-3.
H2 (g) +
1
O2 (g)
2
(d) 2CO (g) + O2 (g)
2CO2 (g) + Heat
The decomposition of solid ammonium carbamate, (NH4)(NH2CO2), to gaseous ammonia and carbon
dioxide is an endothermic reaction.
(NH4)(NH2CO2) (s)
2NH3 (g) + CO2(g)
(a) When solid (NH4) (NH2CO2) is introduced into an evacuated flask at 25°C, the total pressure of gas
at equilibrium is 0.3 atm. What is the value of Kp at 25°C ?
(b) Given that the decomposition reaction is at equilibrium, how would the following changes affect the
total quantity of NH3 in the flask once equilibrium is re-established ?
(i)
Adding CO2
(ii)
Adding (NH4) (NH2CO2)
(iii)
Removing CO2
(iv)
Increasing the total volume
(v)
Adding neon (at constant volume)
(vi)
Increasing the temperature.
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K-4.
Following equilibrium is established at temperature T.
A(g)
B(g) + C (g)
at eq. 1M
2M
2M.
If volume of the vessel is doubled then find the equilibrium concentration of each species.
(Given that : 40 = 6.324)
Section (L) : Vapour pressure and Relative Humidity
L-2.
a
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L-1. Equilibrium constants is given (in atm) for the following reaction 0ºC :
Na2HPO4.12H2O(s)
Na2HPO4.7H2O(s) + 5H2O(g)
Kp = 2.43 × 10–13
The vapour pressure of water at 0ºC is 4.56 torr.
At what relative humidities will Na2HPO4.12H2O(s) be efflorescent when exposed to air at 0ºC ?
Equilibrium constant for the following equilibrium is given at 0ºC.
Na2HPO4 .12H2O (s)
Na2HPO4.7H2O(s) + 5H2O(g)
KP = 31.25 × 10–13
At equilibrium what will be partial pressure of water vapour :
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Section (M) : Simultaneous equilibria
When NO & NO2 are mixed, the following equilibria readily obtained;
2NO2
N2O4
Kp = 6.8 atm–1
NO + NO2
N2O3
Kp = ?
In an experiment when NO & NO2 are mixed in the ratio of 1 : 2, the total final pressure was 5.05 atm &
the partial pressure of N2O4 was 1.7 atm. Calculate
(a)
the equilibrium partial pressure of NO.
(b)
Kp for NO + NO2
N2O3
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M-2.
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M-1. Two solid compounds A and B dissociate into gaseous products at 20°C as
(i) A(s)
A'(g) + H2S(g)
(ii) B(s)
B'(g) + H2S(g)
At 20°C pressure over excess solid A is 50 mm and that over excess solid B is 60 mm find :
(a) The dissociation constant of A and B
(b) Relative no. of moles of A and B in the vapour phase over a mixture of solid A and B.
(c) Show that the total pressure of the gas over the solid mixture would be 39 mm
a
PART - II : ONLY ONE OPTION CORRECT TYPE
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Section (A) : Properties of equilibrium, active mass, homogeneous & heterogeneous
equilibrium (theoritical)
A-1.
A reversible reaction is one which
(A) Achieves equlibrium state
(C) Does not occurs at all
(B) Proceeds in both directions
(D) Both (A) and (B)
A-2. A chemical reaction is at equilibrium when
(A) Measurable properties becomes constant
(B) The rates of forward and backward reactions are equal
(C) Net rate of reaction is zero
(D) All are correct
A-3.
Molar concentration of 96 g of O2 contained in a 2 litre vessel is :
(A) 16 mol/litre
(B) 1.5 mol/litre
(C) 4 mol/litre
(D) 24 mol/litre
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A-4. Find correct graph reagarding equlibrium state :
(A)
(B)
(C)
(D) All of these
a
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A-5.
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Initially the reactions in the container a & b are at equilibrium when the products & reactants are put
together in a container c then at the equilibrium the total number of different chemical compounds are :
(A) 5
(B) 7
(C) 6
(D) 8
Section (B) : Homogeneous equilibrium : KC in gaseous system
B-1. In a reversible reaction A
B, the initial concentration of A and B are a and b in moles per litre, k 1
(A)
k1a − k 2b
k1 + k 2
(B)
k1a − k 2b
k1 − k 2
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and k2 are rate constants for forward & backward reactions respectively and the equilibrium
concentrations
are
(a − x) and (b + x) respectively; express x in terms of k 1, k2, a and b.
(C)
k1a − k 2b
k1 k 2
(D)
k1a + k 2b
k1 + k 2
a
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B-2. The reaction A(g) + B(g)
C(g) + D(g) is studied in a one litre vessel at 250°C. The initial
concentration of A was 3n and that of B was n. When equilibrium was attained, equilibrium
concentration of C was found to the equal to the equilibrium concentration of B. What is the
concentration of D at equilibrium?
(A) n/2
(B) (3n – 1/2)
(C) (n – n/3)
(D) n
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B-3. The figure show the change in concentration of species A and B as a fuctional of time. The equilibrium
constant KC for the reaction A(g)
2B (g) is :
(A) Kc > 1
B-4.
(B) Kc < 1
Kc = 9 for the reaction, A + B
equilibrium is :
(A) 1
(B) 0.25
(C) Kc = 1
(D) data insufficient
C + D, If one mole of each A and B are taken, then amount of C in
(C) 0.75
(D) None of these
B-5.
The equilibrium N2(g) + O2(g)
2NO(g) is estabilished in a reaction vessel of 2.5 L capacity. The
amounts of N2 and O2 taken at the start were respectively 2 moles and 4 moles. Half a mole of nitrogen
has been used up at equilibrium. The molar concentration of nitric oxide is :
(A) 0.2
(B) 0.4
(C) 0.6
(D) 0.1
B-6.
An equilibrium mixture for the reaction 2H2S(g)
2H2(g) + S2(g) had 1 mol of H2S, 0.2 mol of H2
and 0.8 mol of S2 in a 2 litre flask. The value of KC in mol lit–1 is :
(A) 0.08
(B) 0.016
(C) 0.004
(D) 0.160
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Section (C) : Homogeneous equilibrium : KP in gaseous system
C-1.
What is the unit of KP for the reaction ?
CS2 (g) + 4H2 (g)
CH4 (g) + 2H2S (g)
(A) atm
(B) atm–2
(C) atm2
(D) atm–1
C-2. N2 and H2 are taken in 1 : 3 molar ratio in a closed vessel to attained the following equilibrium
N2(g) + 3H2(g)
(A)
2NH3(g) . Find Kp for reaction at total pressure of 2P if PN2 at equilibrium is
1
3P
(B)
2
4
3P
(C)
2
4P2
3
P
3
(D) none
For the reaction
A2(g) + 2B2(g)
2C2(g)
the partial pressure of A2, B2 at equilibrium are 0.80 atm and 0.40 atm respectively. The pressure of the
system is 2.80 atm. The equilibrium constant Kp will be
(A) 20
(B) 5.0
(C) 0.02
(D) 0.2
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C-4.
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C-3. The equilibrium constant, Kp for the reaction 2SO2(g) + O2(g)
2SO3(g) is 4.0 atm–1 at 1000 K.
What would be the partial pressure of O2 if at equilibrium the amount of SO2 and SO3 is the same ?
(A) 16.0 atm
(B) 0.25 atm
(C) 1 atm
(D) 0.75 atm
(A)
bc
.RT
a
(B)
b
.P
(a + b + c)
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C-5. PCl5
PCl3 + Cl2 in the reversible reaction at equilibrium the moles of PCl5, PCl3 and Cl2 are a, b
and c respectively and total pressure is P then value of Kp is :
(C)
bc.P
a (a + b + c)
(D)
c
.P
(a + b + c)
The reaction, PCl5
PCl3 + Cl2 is started in a five litre container by taking one mole of PCl 5. If 0.3
mole of PCl5 is there at equilibrium, concentration of PCl3 and KC will respectively be :
(A) 0.14,
49
150
(B) 0.12,
23
100
a
C-7.
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C-6. A sample of pure NO2 gas heated to 1000 K decomposes : 2NO2(g)
2NO(g) + O2(g). The
equilibrium constant KP is 100 atm. Analysis shows that the partial pressure of O 2 is 0.25 atm. at
equilibrium. The partial pressure of NO2 at equilibrium is:
(A) 0.03
(B) 0.25
(C) 0.025
(D) 0.04
(C) 0.07,
23
100
(D) 20,
49
150
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Section (D) : Relation between KP and KC
D-1. At 527ºC, the reaction given below has KC = 4
NH3 (g)
3
1
N2(g) + H2(g)
2
2
What is the KP for the reaction ?
(A) 16 × (800 R)2
D-2.
 800 R 

4 

(B) 
–2
(C) 4 × 800 R
(D) None of these
The value of Kp for the reaction, 2H2O(g) + 2Cl2(g)
4HCl(g) + O2(g) is 0.03 atm at 427º C, when
the partial pressure are expressed in atmosphere then the value of KC for the same reaction is :
(A) 5.23 × 10–4
(B) 7.34 × 10–4
(C) 3.2 × 10–3
(D) 5.43 × 10–5
D-3. log+
Kp
Kc
log RT = 0 is a relationship for the reaction :
(A) PCl5
(C) H2 + 2
PCl3 + Cl2
2H
(B) 2SO2 + O2
(D) N2 + 3H2
2SO3
2NH3
327
Section (E) : Reaction quotient and Its applications
E-1. 2 mole each of SO3, CO, SO2 and CO2 is taken in a one lit. vessel. If KC for
SO3(g) + CO(g)
SO2(g) + CO2(g) is 1/9 then
(A) total no. of moles at equilibrium are less than 8
(B) n(SO3) + n(CO2) = 4
(C) [n(SO2)/n(CO)] < 1
(D) both (B) and (C).
A reaction mixture containing H2, N2 and NH3 has partial pressure 2 atm, 1 atm and 3 atm respectively
at 725 K. If the value of KP for the reaction, N2 + 3H2
2NH3 is 4.28  10–5 atm–2 at 725 K, in which
direction the net reaction will go :
(A) Forward
(B) Backward
(C) No net reaction
(D) Direction of reaction cannot be predicted
a
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E-2.
E-3.
For the equilibium CH3–CH2–CH2–CH3(g)
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equilibrium
E-4.
Ja
constant is found to be 1.732 at 298 K. Now if in a vessel at 298 K, a mixture of
these two gases be taken as represented by the point P in the figure, predict
what will happen
(A) Immediately, above equilibrium will be setup
(B) Above reaction will go in the forward direction till it attains equilibrium
(C) Above reaction will go in the backward direction till it attains equilibrium
(D) Nothing can be said
The reaction quotient Q for N2(g) + 3H2(g)
(C) Q > KC
[NH3 ]2
[N2 ][H2 ]3
. The reaction will
(D) A = 0
For the reaction,
2A + B
3C
at 298 K,
KC = 49
A 3L vessel contains 2, 1 and 3 moles of A, B and C respectively. The reaction at the same
temperature
(A) must proceed in forward direction
(B) must proceed in backward direction
(C) must be equilibrium
(D) can not be predicted
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a
E-5.
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proceed in backward direction, when
(A) Q = KC
(B) Q < KC
2NH3(g) is given by Q =
Section (F) : Properties of equilibrium Constant
F-1.
At a certain temperature, the following reactions have the equilibrium constant as shown below :
S(s) + O2 (g)
SO2 (g); Kc1 = 5 × 1052
2S(s) + 3O2 (g)
2SO3(g); Kc2=1029
What is the equilibrium constant Kc for the reaction at the same temperature ?
2SO2(g) + O2 (g)
2SO3(g)
(A) 2.5 × 1076
(B) 4 × 1023
(C) 4 × 10–77
(D) None of these
F-2. The equilibrium constant of the reaction SO 2(g) + ½O2(g) SO3(g)
is 4 × 10–3 atm–1/2. The
equilibrium constant of the reaction 2SO3(g)
2SO2(g) + O2(g) would be :
(A) 250 atm
(B) 4 × 103 atm
(C) 0.25 × 104 atm
(D) 6.25 × 104 atm
F-3. Equilibrium constant for the reactions,
2 NO + O2
2 NO2
NO2 + SO2
2 SO3
SO3 + NO
2 SO2 + O2
is KC1 ;
is KC2 and
is KC3 then correct reaction is :
328
(A) KC3 = KC1  KC2
(B) KC3  KC1  K2C2 = 1
(A) KC3 = KC1  KC2 = 1
(D) KC3  K2C1  KC2 = 1
Section (G) : Homogenuous Equilibrium (liquid system)
G-1. When alcohol (C2H5OH ()) and acetic acid (CH3COOH ()) are mixed together in equimolar ratio at
27ºC , 33% of each is converted into ester. Then the K C for the equilibrium
C2H5OH() + CH3COOH ()
CH3COOC2H5 () + H2O() is :
(A) 4
(B) 1/4
(C) 9
(D) 1/9
a
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Section (H) : Heterogenuous equilibrium
H-1. What is the minimum mass of CaCO3 (s), below which it decomposes completely, required to establish
equilibrium in a 6.50 litre container for the reaction : CaCO 3(s)
CaO(s) + CO2(g); Kc = 0.05
mole/litre
(A) 32.5 g
(B) 24.6 g
(C) 40.9 g
(D) 8.0 g
In the reaction C(s) + CO2(g)
2CO(g), the equilibrium pressure is 12 atm. If 50% of CO 2 reacts
then Kp will be :
(A) 12 atm
(B) 16 atm
(C) 20 atm
(D) 24 atm
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H-2.
8M
NH2COONH4(s)
4M
CO2(g)
3M
1M
2 3 4 5
Time (sec.)
NH2COONH4(s)
Conc.(mol/Lt.)
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7M
6M
CO2(g)
H-4.
3M
2 3 4 5
Time(sec.)
NH2COONH 4(s)
(D)
6M
5M
4M
NH3(g)
3M
2M
1M
1M
2 3 4 5
Time (sec.)
CO2(g)
1
2M
1
NH3(g)
7M
NH3(g)
3M
4M
8M
5M
4M
5M
2M
Conc.(mol/Lt.)
8M
6M
1M
a
1
(B)
Conc.(mol/Lt.)
NH3(g)
5M
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Conc.(mol/Lt.)
7M
6M
2M
(C)
NH2COONH 4(s)
8M
7M
(A)
Ja
H-3. Solid ammonium carbamate dissociate to give ammonia and carbon dioxide as follows
NH2COONH4(s)
2NH3(g) + CO2(g)
which of the following graph correctly represents the equilibrium.
CO2(g)
1
2 3 4 5
Time(sec.)
For NH4HS(s)
NH3(g) + H2S(g) reaction started only with NH4HS(s), the observed pressure for
reaction mixture in equilibrium is 1.2 atm at 106°C. What is the value of Kp for the reaction ?
(A) 1.44 atm2
(B) 0.36 atm2
(C) 0.16 atm2
(D) 3.6 atm2
H-5. Consider the decomposition of solid NH4HS in a flask containing NH3(g) at a pressure of 2 atm. What
will be the partial pressure of NH3(g) and H2S(g) after the equilibrium has been attained?
KP for the reaction is 3.
(A) pNH3 = 6 atm, pH2S = ½ atm
(B) pNH3 = 1.732 atm, pH2S = 1.732 atm
(C) pNH3 = 3 atm, pH2S = 1 atm
(D) pNH3 = 1 atm, pH2S = 3 atm
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Section (I) : Degree of dissociation () and vapour density
I-1.
For the dissociation reaction N2O4(g)
total equilibrium pressure P is:
(A)  =
4p + Kp
Kp
(B)  =
Kp
4p + Kp
2NO2(g), the degree of dissociation () in terms of Kp and
(C)  =
Kp
4p
(D) None of these
I-2.
The degree of dissociation of SO3 is  at equilibrium pressure P0. Kp for 2SO3(g)
is:
(A) [(P03)/2(1 – )3]
(B) [(P03)/(2+)(1 – )2]
2
2
(C) [(P0 )/2(1 – ) ]
(D) None of these
I-3.
In the dissociation of N2O4 into NO2, (1 + ) values with the vapour densities ratio   is as given by :
d
a
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2SO2(g) + O2(g)
D


[-degree of dissociation, D-vapour density before dissociation, d-vapour density after dissociation]
(C)
In the above question,  varies with
(A)
D
according to :
d
(D)
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I-4.
(B)
uh
(A)
(B)
(C)
(D)
For the reaction N2O4(g)
2NO2(g), if percentage dissociation of N 2O4 are 20%, 45%, 65% & 80%,
then the sequence of observed vapour densities will be :
(A) d20 > d45 > d65 > d80
(B) d80 > d65 > d45 > d20
(C) d20 = d45 = d65 = d80
(D) (d20 = d45) > ( d65 = d80)
I-6.
The degree of dissociation of PCl5 () obeying the equilibrium, PCl5
related to the presure at equilibrium by (given  << 1) :
(B)  
Sa
nk
(A)   P
a
lp
I-5.
1
P
(C)  
1
P
PCl3 + Cl2, is approximately
(D)  
2
1
P4
I-7.
At 727°C and 1.23 atm of total equilibrium pressure, SO 3 is partially dissociated into SO2 and O2
according to SO3(g)
SO2(g) + 1/2O2(g). The density of equilibrium mixture is 0.9 g/litre. The
degree of dissociation is :
(A) 1/3
(B)2/3
(C) l/4
(D) 1/5.
I-8.
Consider the following hypothetical equilibrium 2B(g)
B2(g). If d is observed vapour density and D
is theoretical vapour density, then degree of association () will be :
D–d

 d 
(A)  = 2 
I-9.
(B)  =
2D – d
D
(C)  = 2 –
2D
d
(D)  =
2D
D–d
The degree of dissociation is 0.5 at 800 K and 2 atm for the gaseous reaction PCl 5
Assuming ideal behaviour of all the gases.
Calculate the density of equilibrium mixture at 800 K and 2 atm.
(A) 4.232 g/L
(B) 6.4 g/L
(C) 8.4 g/L
(D) 2.2 g/L
PCl3 + Cl2.
330
I-10.
SO3(g)
SO2(g) +
1
O2(g)
2
If observed vapour density of mixture at equilibrium is 35 then find out value of 
(A) 0.28
(B) 0.38
(C) 0.48
(D) 0.58
Section (J) : Thermodyanamics of equilibrium
The correct relationship between free energy change in a reaction and the corresponding equilibrium
constant K is
(A) – Go = RT ln K
(B) G = RT ln K
(C) – G = RT ln K
(D) Go = RT ln K
J-2.
For the reaction
H2(g) + I2(g)
2HI(g)
Kc = 66.9 at 350°C and Kc = 50.0 at 448°C. The reaction has
(A) H = + ve
(B) H = – ve
(C) H = zero
(D) H sign can not be determined
An exothermic reaction is represented by the graph :
(B)
(D) Slow
(C)
Ja
(A)
J-5.
(C) Fast
(D) None of these
An endothermic reaction is represented by the graph :
(B)
(C)
(D) None of these
a
(A)
lp
J-4.
2Br at 500 K and 700 K are 1  10–10 and 1  10–5
uh
J-3. The equilibrium constant for the reaction Br2
respectively. The reaction is :
(A) Endothermic
(B) Exothermic
a
ri
J-1.
Sa
nk
J-6. The value of G° for a reaction in aqueous phase having Kc= 1, would be :
(A) –RT
(B) –1
(C) 0
(D) + RT
J-7.
The effect of temperature on equilibrium constant is expressed as (T 2 > T1)
log
K2
1
−H  1
=
 −  . For endothermic reaction false statement is
K1
2.303  T2 T1 
 1
1
−  = positive
T
T
1
 2
(A) 
(B) H = positive
(C) log K2 > log K1
(D) K2 > K1
Section (K) : Le-chatelier's principle
K-1.
For the reaction CO(g) + H2O(g)
CO2(g) + H2(g) at a given temperature the equilibrium amount of
CO2(g) can be increased by :
(A) adding a suitable catalyst
(B) adding an inert gas
(C) decreasing the volume of container
(D) increasing the amount of CO(g)
K-2. Given the following reaction at equilibrium N2(g) + 3H2(g)
2NH3(g). Some inert gas at constant
pressure is added to the system. Predict which of the following facts will be affected.
(A) More NH3(g) is produced
(B) Less NH3(g) is produced
(C) No affect on the equilibrium
(D) Kp of the reaction is decreased
331
K-3.
The equilibrium, SO2Cl2(g)
SO2(g) + Cl2(g) is attained at 25°C in a closed container and an inert
gas, helium, is introduced. Which of the following statement(s) is/are correct.
(A) Concentrations of SO2, Cl2 and SO2Cl2 are changed
(B) No effect on equilibrium
(C) Concentration of SO2 is reduced
(D) Kp of reaction is increasing
K-5.
a
ri
K-4. Densities of diamond and graphite are 3.5 and 2.3 g/mL.
C (diamond)
C (graphite)
rH = – 1.9 kJ/mole
favourable conditions for formation of diamond are
(A) high pressure and low temperature
(B) low pressure and high temperature
(C) high pressure and high temperature
(D) low pressure and low temperature
Introduction of inert gas (at the same temperature) will affect the equilibrium if :
(A) volume is constant and ng  0
(B) pressure is constant and ng  0
(C) volume is constant and ng = 0
(D) pressure is constant and ng = 0
A reaction in equilibrium is represented by the following equation –
2A(s) + 3B(g)
3C(g) + D(g) + O2 if the pressure on the system is reduced to half of its original
value
(A) The amounts of C and D decreases
(B) The amounts of C and D increases
(C) The amount of B and D decreases
(D) All the amounts remain constant
Ja
K-7.
uh
K-6. For an equilibrium H2O(s)
H2O() which of the following statements is true.
(A) The pressure changes do not affect the equilibrium
(B) More of ice melts if pressure on the system is increased
(C) More of liquid freezes if pressure on the system is increased
(D) The degree of advancement of the reaction do not depend on pressure.
What is the relative humidity of air at 1 bar pressure and 313 K temperature if partial pressure of water
in air is 19.355 mmHg.for any data use the table given below :
(in mmHg)
(in K)
25.2
298
(B) 25%
31.8
303
(C) 75%
42.2
308
55.3
313
71.9
318
92.5
323
(D) 5%
Sa
nk
(A) 35%
V.P. of H2O
Temp.
a
L-1.
lp
Section (L) : Vapour pressure and Relative Humidity
L-2.
(a)
CuSO4.5H2O(s)
CuSO4.3H2O (s) + 2H2O (g)
KP = 4 × 10–4 atm2
(b)
Na2SO4.10H2O(s)
Na2SO4.5H2O (s) + 5H2O (g)
KP = 2.43 × 10–8 atm5
(c)
Na2S2O3.5H2O(s)
Na2S2O3.2H2O (s) + 3H2O (g)
KP = 6.4 × 10–5 atm3
What is order of partial pressure of water vapours at equilibrium and relative humidity respectively.
(A) c > b > a
Partial pressure
(B) c < b < a
Partial pressure
c>b>a
Relative humidity
c>b>a
Relative humidity
(C) a > c > b
Partial pressure
(D) a > c > b
Partial pressure
a>c>b
Relative humidity
a<c<b
Relative humidity
L-3. CuSO4.5H2O(s)
CuSO4.3H2O(s) + 2H2O(g)
KP = 4 × 10–4 atm2
and vapour pressure of water is 22.4 torr at 298 K. Then find out realative humidity
(A) 74.46%
(B) 78.46%
(C) 67.85%
(D) 70.46%
332
Section (M) : Simultaneous equilibria
M-1.
The two equilibria, AB(aq)
A+(aq) + B−(aq) and AB(aq) + B−(aq)
AB2−(aq) are simultaneously
maintained in a solution with equilibrium constants, K1 and K2 respectively. The ratio of concentration of
A+ to AB2− in the solution is :
(A) directly proportional to the concentration of B– (aq.).
(B) inversely proportional to the concentration of B– (aq.).
(C) directly proportional to the square of the concentration of B– (aq.).
(D) inversely proportional to the square of the concentration of B– (aq.).
M-2.
In the preceeding problem, if [A+] and [AB2−] are y and x respectively, under equilibrium produced by
adding the substance AB to the solvents, then K1/K2 is equal to
y
(y − x)2
x
(B)
y 2 (x + y)
x
(C)
y 2 (x + y)
x
[Note: Use the information of the preceeding problem]
(D)
The reactions PCl5(g)
PCl3(g) + Cl2(g) and COCl2(g)
CO(g) + Cl2(g) are simultaneously in
equilibrium at constant volume. A few moles of CO(g) are introduced into the vessel. After some time,
the new equilibrium concentration of
(A) PCl5 will remain unchanged
(B) Cl2 will be greater
(C) PCl5 will become less
(D) PCl5 will become greater
uh
M-3.
y
(x − y)
x
a
ri
(A)
PART - III : MATCH THE COLUMN
(A)
(B)
(C)
(D)
Column-
N2(g) + 3H2(g)
2NH3(g) (t = 300ºC)
PCl5(g)
PCl3(g) + Cl2(g) (t = 50ºC)
C(s) + H2O(g)
CO(g) + H2(g)
CH3COOH() + C2H3OH ()
CH3COOC2H5() + H2O()
Match the following :
Column-I
(Assume only reactant were present initially)
For the equilibrium NH4(s)
NH3(g) + H(g), if pressure
(A)
is increased at equilibrium
For the equilibrium N2(g) + 3H2(g)
2NH3(g), volume is
(B)
increased at equilibrium
For the equilibrium H2O(g) + CO(g)
H2(g) + CO2 (g),
(C)
inert gas is added at constant pressure at equilibrium
For the equilibrium PCl5(g)
PCl3(g) + Cl2(g), Cl2 is
(D)
removed at equilibrium.
Sa
nk
a
2.
Ja
Match the following : (Assume only reactants were present initially).
lp
1.
Column-
(p)
(q)
(r)
(s)
ng > 0
Kp < Kc
Kp not defined
Pinitial > Peq.
Column-II
(p)
Forward shift
(q)
No shift in equilibrium
(r)
Backward shift
(s)
Final pressure is more
than initial pressure
333
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
If K1, K2, K3 are equilibrium constant for formation of AD, AD2, AD3 respectively as follows A + D
AD,
AD + D
AD2, AD2 + D
AD3. Then equilibrium constant 'K' for A + 3D
AD3 is related as
(A) K1 + K2 + K3 = K
(B) logK1 + logK2 + logK3 = log K
(C) K1 + K2 = K3 + K
(D) log K1 + logK2 = logK3 + log K
2.
A 10 litre box contains O3 and O2 at equilibrium at 2000 K. KP = 4 × 1014 atm for 2O3(g)
Assume that PO2  PO3 and if total pressure is 8 atm, then partial pressure of O 3 will be :
(A) 8 × 10–5 atm
(B) 11.3 × 10–7 atm
(C) 9.71 × 10–6 atm
a
ri
1.
3O2(g).
(D) 9.71 × 10–2 atm
Sulfide ion in alkaline solution reacts with solid sulfur to form polysulfide ions having formulae S 22–, S32–,
S42– and so on. S (s) + S2– (aq)
S22– (aq) K1 = 12 & 2S (s) + S2– (aq)
S32– (aq) K2 = 132. What
is the equilibrium constant for the formation of S32– from S22– and S?
(A) 11
(B) 12
(C) 132
(D) None of these
4.
If for 2A2B(g)
2A2(g) + B2(g), Kp = TOTAL PRESSURE (at equilibrium) and starting the dissociation
from 4 mol of A2B then :
(A) degree of dissociation of A2B will be (2/3).
(B) total no. of moles at equilibrium will be (14/3).
(C) at equilibrium the no. of moles of A2B are not equal to the no. of moles of B2 .
(D) at equilibrium the no. of moles of A2B are equal to the no. of moles of A2 .
Ja
Attainment of the equilibrium A(g)
2C(g) + B(g)
gave the following graph. Find the correct option.
(% dissociation = fraction dissociated × 100)
lp
5.
uh
3.
Sa
nk
a
(A) At t = 5 sec equilibrium has been reached and Kc =
128 (mol/litre)2
(B) At t = 5 sec equilibrium has been reached and %
dissociation of A is 60%
(C) At t = 5 sec equilibrium has been reached and %
dissociation of A is 40%
(D) None of these
6.
A 10 L container at 300 K contains CO 2 gas at pressure of 0.2 atm and an excess solid CaO (neglect
the volume of solid CaO). The volume of container is now decreased by moving the movable piston
fitted in the container. What will be the maximum volume of container when pressure of CO 2 attains its
maximum value given that
CaCO3 (s)
CaO(s) + CO2(g)
Kp = 0.800 atm
(A) 5 L
(B) 2.5 L
(C) 1 L
(D)
The
information
is
insufficient.
7.
In the system, LaCl3(s) + H2O(g) + heat
LaClO(s) + 2HCl(g), equilibrium is established. More
water vapour is added to restablish the equlibrium. The pressure of water vapour is doubled. The factor
by which pressure of HCl is changed is:
(A) 2
(B) 2
(C) 3
(D) 5
8.
Some quantity of water is contained in a container as shown in figure. As neon is
added to this system at constant pressure, the amount of liquid water in the
vessel
(A) increases
(B) decreases
(C) remains same
(D) changes unpredictably
334
The equilibrium constant for, 2H2S(g)
2H2(g) + S2(g) is 0.0118 at 1300 K while the heat of
dissociation is 597.4 kJ. The standard equilibrium constant of the reaction at 1200 K is :
(A) 1.180 × 10–4
(B) 11.80
(C) 118.0
(D) cannot be calculated from given data
10.
For reaction, assuming large volume of water.
H2O()
H2O(g)
;
at temp. T K
Choose correct options :
(A) On introduction of an inert gas at constant temperature pressure in the container remains same at
equilibrium.
(B) For this system % relative humidity always remains 100% at constant temperature at equilibrium
(C) If steam at temperature ‘2T’ is passed into given system, after equilibrium is attained relative
humidity changes.
(D) This is a special case of equilibrium where pressure of H2O(g) remains same always due to unique
structural feature of H2O.
11.
In the Haber process for the industrial manufacturing of ammonia involving the reaction,
N2(g) + 3H2(g)
2NH3(g) at 200 atm pressure in the presence of a catalyst, a temperature of about
500ºC is used. This is considered as optimum temperature for the process because
(A) yield is maximum at this temperature
(B) catalyst is active only at this temperature
(C) energy needed for the reaction is easily obtained at this temperature
(D) rate of the catalytic reaction is fast enough while the yield is also appreciable for this exothermic
reaction at this temperature.
12.
Addition of water to which of the following equilibria causes it to shift in the backward direction?
(A) CH3NH2 (aq) + H2O ()
CH3NH3 (aq) + OH– (aq)
+
–
(B) AgCl (s)
Ag (aq) + Cl (aq)
(C) HCN (aq) + H2O ()
H3O+ (aq) + CN– (aq)
3+
(D) [Cr(dien)2] (aq) + 3H2O () + 3Cl– (aq)
[Cr (H2O)3Cl3] (aq) + 2 dien (aq)
13.
Consider the reactions
(i) PCl5(g)
PCl3(g) + Cl2(g)
(ii) N2O4(g)
2NO2(g)
The addition of an inert gas at constant pressure
(A) will increase the dissociation of PCl5 as well as N2O4
(B) observed molecular weight of PCl5 increases at equilibrium.
(C) Concentration NO2 increases at equilibrium.
(D) will not disturb the equilibrium of the reactions
14.
An equilibrium mixture [N2(g) + O2(g)
2NO(g)] in a vessel of capacity 100 litre contain 1 mol N 2, 2
mol O2 and 3 mol NO. Number of moles of O2 to be added so that at new equilibrium the conc. of NO is
found to be 0.04 mol/lit.:
(A) (101/18)
(B) (101/9)
(C) (202/9)
(D) None of these.
Sa
nk
a
lp
Ja
uh
a
ri
9.
15.
CaCl2.6H2O(s)
CaCl2(s) + 6H2O(g) Kp = 6.4 × 10–17 atm6
Excess solid CaCl2.6H2O & CaCl2 are taken in a container containing some water vapours at a
pressure of 1.14 torr at a particular temp.
(A) CaCl2(s) acts as drying agent under given condition.
(B) CaCl2(s) acts as hygroscopic substance given condition.
(C) CaCl2.6H2O(s) acts as effluoroscent substance.
(D) Mass of CaCl2.6H2O(s) increases dueto some reaction.
16.
A(s)
B(g) + C(g)
KP = 40 atm2
X(s)
B(g) + E(g)
Above equilibrium is allowed to attain in a closed container and pressure of B was found to be 10 atm.
Calculate standard Gibb’s free energy change for X(s)
B(g) + E(g) at 300 K (take R = 2 cal/K/mol)
(A) 3.5 Kcal/mol
(B) 3 Kcal/mol
(C) 2.5 Kcal/mol
(D) 2 Kcal/mol
335
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
How many of the following reactions are homogenous reversible reactions ?
(1) CH3COOH() + C2H5OH()
CH3COOC2H5() + H2O()
(2) H2(g) + CO2(g)
CO(g) + H2O(g)
(3) CO(g) + Cl2(g)
COCl2(g)
(4) NH4HS(s)
NH3(g) + H2S(g)
(5) CaCO3(s)
CaO(s) + CO2(g)
(6) N2(g) + O2(g)
2NO(g)
(7) CO2(g) + C(s)
2CO(g)
(8) SO2(g) + NO2(g)
SO3(g) + NO(g)
(9) NO(g) +
1
Br2()
2
2NOBr(g)
a
ri
1.
A(g) + B(g)
C(g) + D(g) above equilibrium is established by taking A & B in a closed container.
Initial concentration of A is twice of the initial concentration of B. At equilibrium concentrations of B and
C are equal. Then find the equilibrium constant for the reaction, C(g) + D(g)
A(g) + B(g).
3.
If 0.5 mole H2 is reacted with 0.5 mole I2 in a ten-litre container at 444°C and at same temperature
value of equilibrium constant KC is 49, the ratio of [HI] and [I2] will be :
4.
For the reaction, N2O5(g)
2NO2(g) + 1/2 O2(g), calculate the mole fraction of N2O5(g) decomposed
at a constant volume & temperature, if the initial pressure is 600 mm Hg & the pressure at any time is
960 mm Hg. Assume ideal gas behaviour. If answer is x then report 10x.
5.
Consider the equilibrium
Ni(s) + 4CO(g)
Ni(CO)4(g) ;
Kp = 0.125 atm–3.
If equal number of moles of CO and Ni(CO)4 (ideal gases) are mixed in a small container fitted with a
piston, find the maximum total pressure (in atm) to which this mixture must be brought in order to just
precipitate out metallic Ni ?
6.
Kp is 9 atm2 for the reaction: LiCl.3NH3(s)
LiCl.NH3(s) + 2NH3(g) at 40°C. How many moles of
ammonia must be added at this temperature to a 5 litre flask containing 0.1 mole of LiCl. NH 3 in order to
completely convert the solid to LiCl.3NH3? Multiply the obtained answer by 100. Round off the answer
to the nearest integer.
7.
Consider the reaction, 2Cl2(g) + 2H2O(g)
4HCl(g) + O2(g) ; Hº = + 113 kJ
The four gases, Cl2, H2O, HCl and O2, are mixed and the reaction is allowed to come to equilibrium.
Each operation is to be considered separately. Temperature and volume are constant unless stated
otherwise. Report the number of operations in the left column which lead to increase in the equilibrium
value of the quantity in the right column.
(a) Increasing the volume of the container
Number of moles of H2O
(b) Adding O2
Number of moles of H2O
(c) Adding O2
Number of moles of HCl
(d) Decreasing the volume of the container
Number of moles of Cl2
(e) Decreasing the volume of the container
Partial pressure of Cl2
(f) Decreasing the volume of the container
KC
(g) Raising the temperature
KC
(h) Raising the temperature
Concentration of HCl
(i) Adding He
Number of moles of HCl
(j) Adding catalyst
Number of moles of HCl
Sa
nk
a
lp
Ja
uh
2.
8.
For given simultaneous reaction :
X(s)
A(g) + B(s) + C(g)
KP1 = 500 atm2
Y(s)
D(g) + A(g) + E(s)
KP2 = 2000 atm2
If total pressure = x, then write your answer after dividing by 25.
336
For equilibrium N2O4(g)
2NO2(g) the observed vapour density of N2O4 is 40 at 350 K. Calculate
percentage dissociation of N2O4(g) at 350K.
10.
The vapour density of N2O4 at a certain temperature is 30.67. The % dissociation of N2O4 at this
temperature is :
11.
Solid ammonium carbamate dissociates to give ammonia and carbon dioxide as follows:
NH2 COONH4 (s)
2NH3 (g) + CO2 (g)
At equilibrium, ammonia is added such that partial pressures of NH 3 at new equilibrium equals the
original total pressure (at previous equilibrium). If the ratio of the total pressures now to the original total
pressure is a/b then report a + b.
a
ri
9.
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
Which of the following is correct about the chemical equilibrium ?
(A) (G)T,P = 0
(B) Equilibrium constant is independent of initial concentration of reactants
(C) Catalyst has no effect on equilibrium state
(D) Reaction stops at equilibrium
2.
For a reaction N2 + 3H2
2NH3, the value of KC does not depends upon :
(A) Initial concentration of the reactants
(B) Pressure
(C) Temperature
(D) catalyst
3.
Which of the following statement/s is/are correct :
(A) At equilibrium, vapour pressure of solution and refractive index of eq. mixture becomes constant.
(B) Equilibrium can be attained in both homogenous and heterogenous reaction.
(C) Approach to the equilibrium is fast in initial state but gradually it decreases.
(D) Equilibrium is dynamic in nature
4.
Equilibrium constant for following reactions respectively K1, K2 and K3
N2 + 3H2
2NH3
K1
N2 + O 2
2NO
K2
H2O
K3
a
1
O2
2
5
2NH3 + O2
2
H2 +
lp
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1.
2NO + 3H2O
K4
Sa
nk
Which of the following relation is correct.
(A) K1 =
5.
If log
K 2  (K 3 )3
K4
kc
− log
kp
(A) NH3(g)
(C) 2NO2(g)
6.
7.
(B) K4 = K1 × K2/(K3)3
(C) K2 =
K 4  K1
3
(K 3 )
(D) K4 =
K 2  (K 3 )3
K1
1
= 0, then above is true for the following equilibrium reaction
RT
1
3
N2(g) + H2(g)
2
2
N2O4 (g)
The reaction for which, KP = KC is satisfied
(A) A(g) + 2B(g)
3C(g)
(C) 2A(g)
B(g) + C(g)
(B) CaCO3(s)
(D) H2(g) + 2(g)
CaO(s) + CO2(g)
2H(g)
(B) A(s)
B(g)
(D) A(s) + B(g)
C(s) + 2D(g)
N2O4(g)
2NO2(g), Kc = 4. This reversible reaction is studied graphically as
shown in figure. Select the correct statements.
(A) Reaction quotient has maximum value at point A
(B) Reaction proceeds left to right at a point when [N2O4] = [NO2] = 0.1 M
(C) Kc = Q when point D or F is reached :
(D) None of these
337
8.
If reaction A + B
C + D, take place in 5 liter close vessel, the rate constant of forward reaction is
nine times of rate of backward reaction.
If initially one mole of each reactant present in the container, then find the correct option/s.
(A)
3
[C]
=
1
[B]
(B) log KP = log KC
(C) [D]eq = 15 × 10–2 mole L–1
(D) Keq = 9
9.
Consider the following equilibrium
2AB(g)
A2(g) + B2(g)
The vapour density of the equilibrium mixture does not depend upon
(A) Temperature
(B) Initial concentration
(C) Volume of contain
(D) Pressure of equilibrium mixture
10.
Vapour density of equilibrium PCl5(g)
(A) increasing temperature
(C) increasing pressure
11.
CuSO4.5H2O(s)
CuSO4(s) + 5H2O(g) Kp = 10–10 (atm). 10–2 moles of CuSO4.5H2O(s) is taken in a
2.5L container at 27°C then at equilibrium [Take : R =
1
litre atm mol–1 K–1]
12
uh
(A) Moles of CuSO4.5H2O left in the container is 9 × 10–3
(B) Moles of CuSO4.5H2O left in the container is 9.8 × 10–3
(C) Moles of CuSO4 left in the container is 10–3
(D) Moles of CuSO4 left in the container is 2 × 10–4
CuSO4.5H2O(s)
CuSO4.3H2O(s) + 2H2O(g)
KP = 0.4 × 10–3 atm2
Which of following statement are correct :
(A) Gº = – RT ln PH2O where PH2O = Partial pressure of H2O at equilibrium.
Ja
12.
a
ri
PCl3(g) + Cl2(g) is decreased by
(B) decreasing pressure
(D) decreasing temperature
1 mole each of H2(g) and 2(g) are introduced in a 1L evacuated vessel at 523K and equilibrium
H2(g) + 2(g)
2HI (g) is established. The concentration of HI(g) at equilibrium :
(A) Changes on changing pressure.
(B) Changes on changing temperature.
(C) Changes on changing volume of the vessel.
(D) Is same even if only 2 mol of HI (g) were introduced in the vessel in the begining.
(E) Is same even when a platinum gauze is introduced to catalyse the reaction.
Sa
nk
a
13.
lp
(B) At vapour pressure of H2O = 15.2 torr relative humidity of CuSO4.5H2O is 100%.
(C) In presence of aqueous tension of 24 torr, CuSO 4.5H2O can not loss moisture.
(D) In presence of dry atmosphere in open container CuSO 4.5H2O will completely convert into
CuSO4.3H2O
14.
For the reaction : PCl5 (g)
PCl3 (g) + Cl2 (g)
The forward reaction at constant temperature is favoured by
(A) introducing chlorine gas at constant volume
(B) introducing an inert gas at constant pressure
(C) increasing the volume of the container
(D) introducing PCl5 at constant volume
15.
Which of the following reaction will shift in forward direction. When the respective change is made at
equilibrium :
(A) N2(g) + 3H2(g)
2NH3(g)
increase in pressure at eq.
(B) H2O(s)
H2O()
addition of inert gas at constant volume
(C) PCl5(g)
PCl3(g) + Cl2(g)
addition of inert gas at constant pressure
(D) H2 + I2
2HI
increase in temperature
16.
2CaSO4(s)
2CaO(s) + 2SO2(g) + O2(g),
H > 0
Above equilibrium is established by taking some amount of CaSO 4(s) in a closed container at 1600 K.
Then which of the following may be correct option.
(A) moles of CaO(s) will increase with the increase in temperature
(B) If the volume of the container is doubled at equilibrium then partial pressure of SO2(g) will change at
new equilibrium.
338
(C) If the volume of the container is halved partial pressure of O2(g) at new equilibrium will remain same
(D) If two moles of the He gas is added at constant pressure then the moles of CaO(s) will increase.
2CaSO4(s)
2CaO(s) + 2SO2(g) + O2(g),
H > 0
17.
The dissociation of phosgene, which occurs according to the reaction
COCl2 (g)
CO(g) + Cl2(g)
Is an endothermic process. Which of the following will increase the degree of dissociation of COCl2?
(A) Adding Cl2 to the system
(B) Adding helium to the system at constant pressure
(C) Decreasing the temperature of the system
(D) Reducing the total pressure
Read the following passage carefully and answer the questions.
Comprehension # 1
Le chatelier's principle
a
ri
PART - IV : COMPREHENSION
(b)
Sa
nk
a
(c)
lp
(a)
Ja
uh
If a system at equilibrium is subjected to a change of any one of the factors such as concentration,
pressure or temperature, the system adjusts itself in such a way as to nulify the effect of that change.
Change of pressure : If a system in equilibrium consists of gases, then the concentrations of all the
components can be altered by changing the pressure. To increase the pressure on the system, the
volume has to be decreased proportionately. The total number of moles per unit volume will now be
more and the equilibirum will shift in the direction in which there is decrease in number of moles i.e.,
towards the direction in which there can be decrease in pressure .
Effect of pressure on melting point : There are two types of solids :
Solids whose volume decreases on melting, e.g., ice, diamond, carborundum, magnesium nitride and
quartz.
Solid (higher volume)
Liquid (lower volume)
The process of melting is facilitated at high pressure, thus melting point is lowered.
Solids whose volume increase on melting, e.g., Fe, Cu, Ag, Au, etc.
Solid (lower volume)
Liquid (higher volume)
In this case the process of melting become difficult at high pressure; thus melting point becomes high.
Solubility of substances : When solid substance are dissolved in water, either heat is evolved
(exothermic) or heat is absorbed (endothermic).
KCl + aq
KCl(aq) – heat
In such cases, solubility increase with increase in temperature. Consider the case of KOH; when this is
dissolved, heat is evolved.
KOH + aq
KOH(aq) + heat
In such cases, solubility decrease with increase in temperature.
Solubility of gases in liquids : When a gas dissolves in liquid, there is decrease in volume. Thus,
increase of pressure will favour the dissolution of gas in liquid.
(d)
1.
A gas 'X' when dissolved in water heat is evolved. Then solublity of 'X' will increase :
(A) Low pressure, high temperature
(B) Low pressure, low temperature
(C) high pressure, high temperature
(D) high pressure, low temperature
2.
Au(s)
Au()
Above equilibrium is favoured at :
(A) High pressure low temperature
(C) Low pressure, high temperature
3.*
For the reaction,
1
1
N2(g) + O2(g)
2
2
(B) High pressure high temperature
(D) Low pressure, low temperature
NO(g)
If pressure is increased by reducing the volume of the container then :
(A) Total pressure at equilibrium will change.
(B) Concentration of all the component at equilibrium will change.
(C) Concentration of all the component at equilibrium will remain same
(D) Equilibrium will shift in the forward direction
339
Comprehension # 2
Effect of temperature on the equilibrium process is analysed by using the thermodynamics
From the thermodynamics relation
Gº = – 2.30 RT logk
.......... (1)
Gº : Standard free energy change
Gº = Hº – TSº
.......... (2)
Hº : Standard heat of the reaction.
From (1) & (2)
– 2.3 RT logk = Hº – TSº
Sº : Standard entropy change

logK = −
Hº
Sº
+
2.3RT 2.3 R
.......... (3)
Clearly if a plot of log k vs 1/T is made then it is a straight line having slope =
S
2.3 R
a
ri
and Y intercept =
−Hº
2.3 R
If at temp. T1 equilibrium constant be k1 and at temperature T2 equilibrium constant be k2 then :
The above equation reduces to:
log K1 = −
Hº
Sº
+
2.3 R T1 2.3 R
.......... (4)

log K2 = −
Hº
Sº
+
2.3 R T2 2.3 R
.......... (5)
uh

Substracting (4) from (5) we get

log
K2
K
1
=
 Hº  1
1 
 −

2.30 R  T1 T2 
If standard heat of dissociation of PCl5 is 230 cal then slope of the graph of logk vs
5.
(B) – 50
(C) 10
lp
(A) +50
For exothermic reaction if S0 < 0 then the sketch of logk vs
(B)
1
is :
T
(D) None
1
may be :
T
(C) logk
(D)
1/T
Sa
nk
(A)
a
4.
Ja
From the above relation we can conclude that the value of equilibrium constant increases with increase
in temperature for endothermic reaction but value of equilibrium constant decreases with the increase in
temperature for exothermic reaction.
6.
If for a particular reversible reaction KC = 57 at 355ºC and KC = 69 at 450ºC then :
(A) H < 0
(B) H > 0
(C) H = 0
(D) H whose sign can’t be determined
340
Comprehension # 3
Answer Q.7, Q.8 and Q.9 by appropriately matching the information given in the three columns
of the following table.
Equilibrium is a state in which there are no observable changes as time goes by. When a chemical reaction has
reached the equilibrium state, the concentrations of reactants and products remain constant over time and there
are no visible changes in the system. However, there is much activity at the molecular level because reactant
molecules continue to from product molecules while product molecules react to yield reactant molecules. If a
change is applied to the system at equilibrium, then equilibrium will be shifted in that direction in which it can
minimise the effect of change applied and the equilibrium is established again under new conditions.
Column-2
Column-3
2NH3(g)
N2(g) + 3H2(g)
(i)
Homogeneous
(P)
(II)
N2O4(g)
2NO2(g)
(ii)
kP > kC (T = 298k)
(Q)
(III)
2O3(g)
3O2(g)
(iii)
degree of dissociation is
not affected by pressure
(R)
(IV)
2HI(g)
H2(g) + I2(g)
(iv)
Mtheoritical  Mexperimental
8.
(S)
(C) (III) (iii) (Q)
(D) (IV) (iv) (S)
Correct combination is
(A) (IV) (ii) (Q)
(B) (III) (i) (S)
Correct combination is
(A) (I) (iii) (R)
(B) (II) (iv) (S)
(C) (II) (iv) (R)
(D) (II) (iii) (S)
(C) (III) (iv) (Q)
(D) (IV) (iii) (P)
lp
9.*
Incorrect combination is
(A) (I) (i) (p)
(B) (II) (ii) (Q)
kP  1
On Increasing Temperature yield of
reaction increases
On increasing pressure vapour
density
of
equilibrium
mixture
decreases
Products are paramagnetic in nature
Ja
7.*
If 
uh
(I)
1
=
& Ptotal at equilibrium = 1 atm
2
a
ri
Column-1
a
* Marked Questions may have more than one correct option.
Sa
nk
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
1.
For a chemical reaction 3X(g) + Y(g)
(A) temperature and pressure
(C) pressure only
2/80]
2.
For the reversible reaction, N2 (g) + 3H2(g)
2NH3 at 500°C, the value of KP is 1.44 × 10–5 when
partial pressure is measured in atmospheres. The corresponding value of K C, with concentration in
mole litre–1, is
[JEE
2000,
1/35]
(A)
3.
1.44  10−5
(0.082  500)
−2
(B)
X3Y(g), the amount of X3Y at equilibrium is affected by
(B) temperature only
(D) temperature, pressure and catalyst [JEE-1999,
1.44  10−5
(8.314  773)
−2
(C)
1.44  10−5
2
(0.082  773)
(D)
1.44  10−5
(0.082  773)−2
When two reactants, A & B are mixed to give products C & D, the reaction quotient Q, at the initial
stages of the reaction.
[JEE2000, 1/35]
(A) is zero
(B) decrease with time
(C) is independent of time
(D) increases with time
341
4.
At constant temperature, the equilibrium constant (K P) for the decomposition reaction N2O4
is expressed by KP =
(4x
2
P)
(1 − x 2 )
2NO2
, where P = pressure, x = extent of decomposition. Which one of the
following statements is true?
1/35]
(A) KP increases with increase of P
(C) KP increases with decrease of x
[JEE
2001,
(B) KP increases with increase of x
(D) KP remains constant with change in P and x
Consider the following equilibrium in a closed container
[JEE 2002, 3/90]
N2 O4 (g)
2NO2 (g)
At a fixed temperature, the volume of the reaction container is halved. For this change, which of the
following statements holds true regarding the equilibrium constant (K P) and degree of dissociation ()?
(A) neither KP nor  changes
(B) both KP and  change
(C) KP changes, but  does not change
(D) KP does not change but  changes
6.
The value of log10K for a reaction A
B is : (Given : = rH298K –54.07 kJ mol–1, r S298K = 10 JK–1
mol–1 and R = 8.314 JK–1 mol–1; 2.303 × 8.314 × 298 = 5705)
[JEE
2007,
3/162]
(A) 5
(B) 10
(C) 95
(D) 100
7.*
The thermal dissociation equilibrium of CaCO3(s) is studied under different conditions.
CaCO3(s)
CaO(s) + CO2(g)
For this equilibrium, the correct statement(s) is (are) :
[JEE(Advanced)
3/120]
(A) H is dependent on T
(B) K is independent of the initial amount of CaCO 3
(C) K is dependent on the pressure of CO2 at a given T
(D) H is independent of the catalyst, if any
2013,
Ja
uh
a
ri
5.
Sa
nk
a
lp
Paragraph 1
Thermal decomposition of gaseous X2 to gaseous X at 298 K takes place according to the following
equation :
X2(g)
2X(g)
The standard reaction Gibbs energy, rGº, of this reaction is positive. At the start of the reaction, there
is one mole of X2 and no X. As the reaction proceeds, the number of moles of X formed is given by .
Thus. equilibrium is the number of moles of X formed at equilibrium. The reaction is carried out at a
constant total pressure of 2 bar. Consider the gases to behave ideally. (Given : R = 0.083 L bar K –1
mol–1)
8.
The equilibrium constant Kp for this reaction at 298 K, in terms of
(A)
9.
82equilibrium
2 − equilibrium
(B)
equilibrium , is
[JEE(Advanced)
2016,
3/124]
82equilibrium
4 − 2equilibrium
(C)
42equilibrium
2 − equilibrium
(D)
42equilibrium
4 − 2equilibrium
The INCORRECT statement among the following, for this reaction, is
[JEE(Advanced)
3/124]
(A) Decrease in the total pressure will result in formation of more moles of gaseous X
(B) At the start of the reaction, dissociation of gaseous X2 takes place spontaneously
(C) equilibrium = 0.7
(D) KC < 1
2016,
342
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
OFFLINE JEE-MAIN
Change in volume of the system does not alter the number of moles in which of the following
equilibriums:
[AIEEE 2002, 3/225]
(1) N2(g) + O2(g)
2NO(g)
(2) PCl5(g)
PCl3(g) + Cl2(g)
(3) N2(g) + 3H2(g)
2NH3(g)
(4) SO2Cl2(g)
SO2(g) + Cl2(g)
2.
In which of the following reactions, increase in the volume at constant temperature don’t effect the
number of moles of at equilibrium :
[AIEEE 2002,
3/225]
(1) 2NH3
N2 + 3H2
(2) C(g) + (1/2) O2(g)
CO (g)
(3) H2 (g) + O2 (g)
H2O2 (g)
(4) none of these.
3.
For the reaction CO (g) + (1/2) O2 (g)
(1) RT
(2) (RT)–1
4.
Consider the reaction equilibrium
2SO2 (g) + O2 (g)
2SO3 (g) ; H° = – 198 kJ.
On the basis of Le Chatelier’s principle, the condition favourable for the forward reaction is :
[AIEEE 2003, 3/225]
(1) lowering of temperature as well as pressure
(2) increasing temperature as well as pressure
(3) lowering the temperature and increasing the pressure
(4) any value of temperature and pressure.
5.
For the reaction equilibrium, N2O4(g)
2NO2(g) the concentrations of N2O4 and NO2 at equilibrium
are
4.8 × 10–2 and 1.2 × 10–2 mol L–1 respectively. The value of Kc for the reaction is [AIEEE 2003, 3/225]
(1) 3.3 × 102 mol L–1
(2) 3 × 10–1 mol L–1
(3) 3 × 10–3 mol L–1
(4) 3 × 103 mol L–1
6.
What is the equilibrium constant expression for the reaction :
P4 (s) + 5O2 (g)
P4 O10 (s) ?
(1) KC = [P4O10]/[P4] [O2]5
(2) KC = 1/[O2]5
5
(3) KC = [O2]
(4) KC = [P4O10] / 5[P4] [O2]
7.
For the reaction, CO(g) + Cl2 (g)
(1) 1/RT
(2) 1.0
8.
The equilibrium constant for the reaction, N2(g) + O2(g)
a
ri
1.
[AIEEE 2002, 3/225]
(4) (RT)1/2
Sa
nk
a
lp
Ja
uh
CO2 (g), Kc/Kp is :
(3) (RT)–1/2
value of Kc for the reaction, NO(g)
4/120]
(1) 2.5 × 102
(2) 0.02
[AIEEE 2004, 3/225]
COCl2 (g) then Kp / Kc is equal to :
[AIEEE 2004, 3/225]
(3) RT
(4) RT
2NO(g) at temperature T is 4 × 10–4 . The
1
1
N2 (g) + O2 (g) at the same temperature is :
2
2
[AIEEE 2004, 3/225 & JEE(Main) 2012,
(3) 4 × 10–4
(4) 50
9.
For the reaction,
2NO2 (g)
2 NO(g) + O2 (g),
(KC = 1.8 × 10–6 at 184°C)
(R = 0.0831 kJ/(mol.K))
When Kp and Kc are compared at 184°C it is found that :
[AIEEE 2005, 3/225]
(1) Whether Kp is greater than, less than or equal to Kc depends upon the total gas pressure
(2) Kp = Kc
(3) Kp is less than Kc
(4) Kp is greater than Kc
10.
The exothermic formation of ClF3 is represented by the equation Cl2(g)+3F2(g)
2ClF3(g);rH = –
329 J which of the following will increase the quantity of ClF 3 in an equilibrium mixture of Cl2, F2 and
ClF3.
[AIEEE 2005, 3/225]
(1) Adding F2
(2) Increasing the volume of container
343
(3) Removing Cl2
(4) Increasing the temperature
11.
An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature
at 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH 3 and H2S gases in the
flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84
atm? The equilibrium constant for NH4HS decomposition at this temperature is : [AIEEE
2005,
4½/225]
(1) 0.11
(2) 0.17
(3) 0.18
(4) 0.30
12.
Phosphorus pentachloride dissociates as follows in a closed reaction vessel.
PCl5(g)
PCl3(g) + Cl2(g)
If total pressure at equilibrium of the reaction mixture is P and degree of dissociation of PCl 5 is x, the
partial pressure of PCl3 will be :
[AIEEE 2006, 3/165]
The equilibrium constant for the reaction, SO 3(g)
of KC for the reaction 2SO2(g) + O2(g)
(1) 416
(2) 2.40 × 10–3
 x 
 x 
(4) 
 P
 1– x 
(3) 
 P
 x + 1
SO2(g) +
1
O2(g) is KC = 4.9 × 10–2. The value
2
2SO3(g) will be :
(3) 9.8 × 10–2
[AIEEE 2006, 3/165]
(4) 4.9 × 10–2
uh
13.
 2x 
(2) 
 P
 1– x 
a
ri
 x 
(1) 
P
 x + 1
For the following three reactions a, b and c, equilibrium constants are given:
(a) CO(g) + H2O(g)
CO2(g) + H2(g);
K1
(b) CH4(g) + H2O(g)
CO(g) + 3H2(g);
K2
(c) CH4(g) + 2H2O(g)
CO2(g) + 4H2(g);
K3
Which of the following relations is correct ?
[AIEEE 2008, 3/105]
(1) K2 K3 = K1
(2) K3 = K1K2
(3) K3 K23 = K12
(4) K1 K2 = K3
15.
The equilibrium constants K p1 and Kp for the reactions X
2
Ja
14.
2Y and Z
P + Q, respectively are in
A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on
the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is :
[AIEEE 2011, 4/120]
(1) 1.8 atm
(2) 3 atm
(3) 0.3 atm
(4) 0.18
Sa
nk
a
16.
lp
the ratio of 1 : 9. If the degree of dissociation of X and Z be equal then the ratio of total pressures at
these equilibria is
[AIEEE 2008,
3/105]
(1) 1 : 1
(2) 1 : 3
(3) 1 : 9
(4) 1 : 36
17.
The equilibrium constant (Kc) for the reaction N2(g) + O2(g)
The value of Kc for the reaction NO(g)
4/120]
(1) 0.02
18.
(2) 2.5 × 102
1
2
For the reaction SO2(g) + O2(g)
2NO(g) at temperature T is 4× 10–4.
1
1
N2(g) + O2(g) at the same temperature is:[AIEEE 2012,
2
2
(3) 4 × 10–4
(4) 50.0
SO3(g) ,if KP = KC(RT)x where the symbols have usual meaning then
the value of x is : (assuming ideality)
4/120]
(1) –1
19.
(2) –
1
2
[JEE(Main)
(3)
1
2
The standard Gibbs energy change at 300 K for the reaction 2A
2014,
(4) 1
B + C is 2494.2 J. At a given
1
1
time, the composition of the reaction mixture is [A] = , [B] = 2 and [C] = . The reaction proceeds in
2
2
the
[R = 8.314 J/K/mol, e = 2.718]
4/120]
[JEE(Main)
:
2015,
344
(1) forward direction because Q > KC
(3) forward direction because Q < KC
(2) reverse direction because Q > KC
(4) reverse direction because Q < KC
20.
The equilibrium constant at 298 K for a reaction A + B
C + D is 100. If the initial concentration of
all the four species were 1 M each, then equilibrium concentration of D (in mol L –1) will be :
[JEE(Main)
2016,
4/120]
(1) 0.818
(2) 1.818
(3) 1.182
(4) 0.182
21.
Which of the following lines correctly show
the temperature
ln K
dependence of equilibrium constant, K, for an exothermic reaction ?
A
B
[JEE(Main) 2018, 4/120]
a
ri
(0, 0 )
(1) C and D
(2) A and D

(3) A and B
(4) B and C
1
T (K )





C
D
uh
ONLINE JEE-MAIN
At a certain temperature, only 50% HI is dissociated into H2 and I2 at equilibrium. The equilibrium
constant is:
[JEE(Main) 2014 Online (09-04-14),
4/120]
(1) 1.0
(2) 3.0
(3) 0.5
(4) 0.25
2.
What happens when an inert gas is added to an equilibrium keeping volume unchanged ?
[JEE(Main) 2014 Online (12-04-14), 4/120]
(1) More product will form
(2) Less product will form
(3) More reactant will form
(4) Equilibrium will remain unchanged
3.
For the decomposition of the compound, represented as NH 2COONH4(s)
2NH3(g) + CO2(g) the
–5
3
Kp = 2.9 × 10 atm . If the reaction is started with 1 mol of the compounds, the total pressure at
equilibrium would be :
[JEE(Main) 2014 Online (19-04-14), 4/120]
(1) 1.94 × 10–2 atm
(2) 5.82 × 10–2 atm
(3) 7.66 × 10–2 atm
(4) 38.8 × 10–2 atm
4.
Gaseous N2O4 dissociates into gaseous NO2 according to the reaction N2O4(g)
2NO2(g) at 300 K
and 1 atm pressure, the degree of dissociation of N 2O4 is 0.2. If one mole of N2O4 gas is contained in a
vessel, then the density of the equilibrium mixture is :
[JEE(Main) 2015 Online (10-04-15),
4/120]
(1) 3.11 g/L
(2) 4.56 g/L
(3) 1.56 g/L
(4) 6.22 g/L
Sa
nk
a
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Ja
1.
5.
The increase of pressure on ice
water system at constant temperature will lead to :
[JEE(Main) 2015 Online (11-04-15), 4/120]
(1) a decrease in the entropy of the system
(2) an increase in the Gibbs energy of the system
(3) no effect on the equilibrium
(4) a shift of the equilibrium in the forward direction
6.
A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases
X and Y at temperature T. The equilibrium pressure is 10 bar in this vessel. K P for this reaction is :
[JEE(Main) 2016 Online (10-04-16), 4/120]
(1) 25
(2) 5
(3) 10
(4) 100
7.
The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal :
Fe2O3(s) + 3CO(g)
2Fe() + 3CO2(g)
Using the Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium ?
[JEE(Main) 2017 Online (09-04-17), 4/120]
(1) Addition of Fe2O3
(2) Removal of CO2
(3) Removal of CO
(4) Addition of CO2
345
In which of the following reactions, an increase in the volume of the container will favour the formation
of products ?
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 4NH3(g) + 5O2(g)
4NO(g) + 6H2O() (2) 2NO2(g)
2NO(g) + O2(g)
(3) 3O2(g)
2O3(g)
(4) H2(g) + I2(g)
2HI(g)
9.
At a certain temperature in a 5 L vessel, 2 moles of carbon monoxide and 3 moles of chlorine were
allowed to reach equilibrium according to the reaction,
CO + Cl2
COCl2
At equilibrium if one mole of CO is present then equilibrium constant KC for reaction is :
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 2
(2) 2.5
(3) 3
(4) 4
10.
At 320 K, a gas A2 is 20 % dissociated to A(g). The standard free energy change at 320 K and 1 atm in
J mol–1 is approximately : (R = 8.314 JK–1 mol–1 ; ln 2 = 0.693 ; ln 3 = 1.098)
[JEE(Main) 2018 Online (16-04-18), 4/120]
(1) 1844
(2) 2068
(3) 4281
(4) 4763
11.
The gas phase reaction 2NO2(g) → N2O4(g) is an exothermic reaction. The decomposition of N 2O4, in
equilibrium mixture of NO2(g) and N2O4(g), can be increased by :
[JEE(Main) 2018 Online (16-04-18), 4/120]
(1) addition of an inert gas at constant pressure. (2) lowering the temperature
(3) increasing the pressure
(4) addition of an inert gas at constant volume.
12.
Consider the following reversible chemical reactions :
K1
2AB(g)
K2
6AB(g)
3A2(g) + 3B2(g)
The relation between K1 and K2 is :
1
(1) K1K2 =
(2) K2 = K1–3
3
……(2)
[JEE(Main) 2019 Online (09-01-19), 4/120]
(3) K1K2 = 3
(4) K2 = K13
The values of KP/KC for the following reactions at 300 K are, respectively : (At 300 K, RT = 24.62 dm 3
atm mol–1)
N2(g) + O2(g)
2NO(g)
N2O4(g)
2NO2(g)
N2(g) + 3H2(g)
2NH3(g)
[JEE(Main) 2019 Online (10-01-19), 4/120]
(1) 1,4.1 × 10–2 dm–3 atm–1 mol, 606 dm6 atm2mol–2
(2) 1,24.62 dm3 atm–1 mol–1, 1.65 × 10–3 dm–6 atm2mol–2
(3) 24.62 dm3 atm mol–1 606.0 dm6 atm2 mol–2, 1.65 × 10–3 dm–6 atm–2 mol2
(4) 1,24.62 dm3 atm mol–1, 606.0 dm6 atm2mol–2
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13.
……(1)
Ja
A2(g) + B2(g)
uh
a
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8.
14.
5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327°C. 30% of the solid NH4SH decomposed to
NH3 and H2S as gases. The Kp of the reaction at 327°C is (R = 0.082 L atm mol–1K–1, molar mass of S =
32 g mol–1, molar mass of N = 14 g mol–1)
[JEE(Main) 2019 Online (10-01-19),
4/120]
(1) 4.9 × 10–3 atm2
(2) 0.242 × 10–4 atm2
(3) 1 × 10–4 atm2
(4) 0.242 atm2
15.
Consider the reaction, N2(g) + 3H2(g)
2NH3(g). The equilibrium constant of the above reaction is
KP. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by
(Assume that pNH  ptotal at equilibrium)
[JEE(Main) 2019 Online (11-01-19),
3
4/120]
(1)
2 2
33 / 2 K1/
p P
16
(2)
2 2
33 / 2 K1/
p P
4
(3)
2 2
K1/
p P
4
(4)
2 2
K1/
p P
16
346
16.
In a chemical reaction, A + 2B
K
2C+D, the initial concentration of B was 1.5 times of the
concentration of A, but the equilibrium concentrations of A and B were found to be equal. The
equilibrium constant (K) for the aforesaid chemical reaction is : [JEE(Main) 2019 Online (12-01-19),
4/120]
(1) 16
(2) 1
(3) 1/4
(4) 4
17.
Two solids dissociate as follows
A(s)
B(g) + C(g) ; K P1 = x atm 2
C(g) + E(g) ; K P2 = y atm 2
D(s)
The total pressure when both the solids dissociate simultaneously is :
[JEE(Main) 2019 Online (12-01-19), 4/120]
2
(1) x + y atm
(2) (x + y) atm
(3) x + y2 atm
(4) 2 x + y atm
(
uh
a
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)
EXERCISE - 1
Ja
PART - I
A-1.
4
A-2.
0.044 M.
B-3.
1/3M
B-4.
0.66
C-3.
[NH3] = 0.76 M
D-2.
[A]eq = [B]eq = [C]eq = 1/2 M, Kp = 12.3 atm, Kc = 0.5 M (unitless).
D-3.
(i) Kc =
lp
(a) 0.1 (b) 0.4
P(n + y / 2)(n + y)2
(3n + y / 2)(n − y)2
4x 2 V 2
(a − x)(b − 3x)3
100%
6
C-2.
KP = 12
; KP =
(a + b − 2x)2.4x 2
P2 (a − x)(b − 3x)3
a
KC =
B-2.
20000
58.3
= 58.3 mol-2 L2, KP =
= 0.035 atm-2 (ii) P = 8.2 atm
41 41
343
The reaction is N2(g) + 3H2(g)
Qc =
C-1.
D-1.
Sa
nk
E-1.
B-1.
[NH3 ]
2
3
[N2 ][H2 ]
=
2NH3(g)
8  8  20  20
64
=
× 102
3
3
 2 2 2
2
As Qc  Kc , the reaction mixture is not in equilibrium.
As Qc > Kc, the net reaction will be in the backward direction.
E-2.
[SO2] = 0.034 M ; [NO2] = 0.034 M ; [NO] = 0.306 M ; [SO3] = 0.306 M
F-1.
(i) No change
F-2.
For the required reaction, K = K1 × K2.
G-1.
X = 4, Y = 4
H-1.
(i)
(ii) 1/k
Homogeneous equilibrium
(iii) No change
F-3.
KC =
[NO2 ]2
[N2O4 ]
(iv) Change the equilibrium constant
2.58
KP =
(PNO2 )2
(PN2O4 )
347
[H2 ]4
(ii)
Hetereogeneous equilibrium
KC =
(iii)
Hetereogeneous equilibrium
KC = [NH3] [H2S]
Homogeneous equilibrium
system
(v)
Hetereogeneous equilibrium
KC
[H2O]4
KP = (PNH ) (PH S)
3
[CH3COOC2H5 ] [H2O]
=
[CH3COOH] [C2H5OH]
KC = [CO2]
2
Homogeneous equilibrium
KC =
[S2 ]
(vii)
Homogeneous equilibrium
KC =
[SO3 ] [NO]
[SO2 ][NO2 ]
(viii)
Hetereogeneous equilibrium
KC = [N2]
KP =
[H2S]2
I-1.
(i) 0.266 atm (ii) 63.25%
I-2.
I-3.
(i) 76.66, (ii) 0.2, (iii) 33.33 %, (iv) 1/3
atm−1/2
I-4.
J-2.
 reaction (iii) is exothermic.
K-1.
(i) unaffected; no shift
K-2.
(i) When decreasing temperature
(a) Forward
(b) Forward
(ii) Increasing the pressure
(a) Forward
(b) No change
KP =
(PH2 )2 (PS2 )
(PH2S )2
(PSO3 ) (PNO )
(PSO2 )(PNO2 )
KP = (PN )
2
KP =
2
33 / 2
P3 / 2
uh
H-3.
but KP is not define for liquid
2
(vi)
50%
2
KP = (PCO )
[H2 ]
H-2.
(PH2O )4
a
ri
(iv)
(PH2 )4
KP =
1/ 3
2 K 
P
x= 

 P 
J-1.
Ja
2.55 atm3
(ii) affected; left direction.
(iii) affected; left
KP = 1.86 x 1012
(iv) affected; right
(d) Forward
(c) Backward
(d) Forward
lp
(c) Backward
(a) 4 × 10–3 (b) (i) decrease (ii) no change (iii) increase (iv) increase (v) no change (vi) increase
K-4.
[A] = 0.34 M, [B] = 1.16 M, [C] = 1.16 M.
L-2.
5 × 10–3 atm.
M-2.
(a) 1.05 atm,
A-1.
(D)
A-2.
(D)
A-3.
(B)
A-4.
(D)
A-5.
(D)
B-1.
(A)
B-2.
(A)
B-3.
(A)
B-4.
(C)
B-5.
(B)
B-6.
(B)
C-1.
(B)
C-2.
(B)
C-3.
(B)
C-4.
(A)
C-5.
(C)
C-6.
(C)
C-7.
(A)
D-1.
(C)
D-2.
(A)
D-3.
(B)
E-1.
(D)
E-2.
(B)
E-3.
(C)
E-4.
(C)
E-5.
(A)
F-1.
(C)
F-2.
(D)
F-3.
(B)
G-1.
(B)
H-1.
(A)
H-2.
(B)
H-3.
(C)
H-4.
(B)
H-5.
(C)
I-1.
(B)
I-2.
(B)
I-3.
(A)
I-4.
(B)
I-5.
(A)
a
K-3.
Sa
nk
M-1.
L-1.
below 50%
(a) Kp1 = 625 mm2, Kp2 = 900 mm2
(b)
25
36
(b) 3.43 atm–1
PART – II
348
I-6.
(B)
I-7.
(B)
I-8.
(C)
I-9.
(A)
I-10.
(A)
J-1.
(A)
J-2.
(B)
J-3.
(A)
J-4.
(C)
J-5.
(B)
J-6.
(C)
J-7.
(A)
K-1.
(D)
K-2.
(B)
K-3.
(B)
K-4.
(C)
K-5.
(B)
K-6.
(B)
K-7.
(B)
L-1.
(A)
L-2.
(A)
L-3.
(C)
M-1.
(D)
M-2.
(A)
M-3.
(C)
PART – III
(A - q, s) ; (B - p) ; (C - p) ; (D - r)
2.
(A - r) ; (B - r) ; (C - q) ; (D - p)
EXERCISE – 2
(B)
3.
(A)
6.
(B)
7.
(B)
8.
(B)
11.
(D)
12.
(D)
13.
(A)
16.
(C)
1.
05
2.
3
3.
6.
78 mole
7.
5
8.
11.
58
1.
(ABC)
2.
(ABD)
3.
6.
(AC)
7.
(BC)
11.
(BD)
12.
(BCD)
17.
(BD)
4.
(A)
5.
(C)
9.
(A)
10.
(B)
14.
(A)
15.
(C)
uh
2.
Ja
(B)
a
PART – I
1.
a
ri
1.
PART – II
4.
4
5.
4
4
9.
15
10.
50
(ABCD)
4.
(ACD)
5.
(AB)
8.
(ABCD)
9.
(ABCD)
10.
(AB)
13.
(ABCDE)
14.
(BCD)
15.
5.
(B)
5.
(D)
lp
7
Sa
nk
PART – III
(ABC)
16.
(ACD)
PART – IV
1.
(D)
2.
(C)
3.*
(AB)
4.
(B)
6.
(B)
7.*
(CD)
8.
(B)
9.*
(BD)
EXERCISE – 3
PART - I
1.
(A)
2.
(D)
3.
(D)
4.
(D)
6.
(B)
7.*
(ABD)
8.
(B)
9.
(C)
349
PART – II
OFFLINE JEE-MAIN
1.
(1)
2.
(4)
3.
(4)
4.
(3)
5.
(3)
6.
(2)
7.
(1)
8.
(4)
9.
(4)
10.
(1)
11.
(1)
12.
(1)
13.
(1)
14.
(2)
15.
(4)
16.
(1)
17.
(4)
18.
(2)
19.
(2)
20.
(2)
21.
(3)
(1)
5.
(4)
(2)
10.
(3)
(4)
15.
(1)
(4)
2.
(4)
3.
(2)
4.
6.
(1)
7.
(1)
8.
(2)
9.
11.
(1)
12.
(2)
13.
(2)
14.
16.
(4)
17.
(4)
uh
1.
a
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ONLINE JEE-MAIN
 Marked questions are recommended for Revision.
Ja
This Section is not meant for classroom discussion. It is being given to promote selfstudy and self testing amongst the students.
PART - I : PRACTICE TEST-1 (IIT-JEE (MAIN Pattern))
Max. Time : 1 Hr.
Sa
nk
5.
The test is of 1 hour duration.
The Test Booklet consists of 30 questions. The maximum marks are 120.
Each question is allotted 4 (four) marks for correct response.
Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each
question.
¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No deduction
from the total score will be made if no response is indicated for an item in the answer sheet.
There is only one correct response for each question. Filling up more than one response in any question
will be treated as wrong response and marks for wrong response will be deducted accordingly as per
instructions 4 above.
a
1.
2.
3.
4.
lp
Important Instructions
Max. Marks : 120
1.
1 mole of N2 and 2 moles of H2 are allowed to react in a 1 dm3 vessel. At equilibrium, 0.8 mole of NH3 is
formed. The concentration of H2 in the vessel is :
(1) 0.6 mole
(2) 0.8 mole
(3) 0.2 mole
(4) 0.4 mole
2.
For the following mechanism , P + Q
PQ
R at equilibrium
[R]
is :
[P][Q]
[ k represents rate constant ]
(1)
3.
K A .KB
K C.K D
(2)
K A .KD
KB .K C
(3)
KB .K D
K A .K C
(4)
K A .K C
KB .KD
Select the reaction for which the equilibrium constant is written as [MX3]2 = Keq. [MX2]2 [X2]
(1) MX3
MX2 +
(3) 2MX2 + X2
1
X2
2
(2) 2MX3
2MX3
(4) MX2 +
2MX2+ X2
1
X2
2
MX3.
350
What should be the value of KC for the reaction 2SO2(g) + O2(g)
2SO3(g). If the amount are SO3 =
48g, SO2 = 12.8 and O2 = 9.6 at equilibrium and the volume of the container is one litre?
(1) 64
(2) 0.30
(3) 42
(4) 8.5
5.
The equilibrium constant (Kp) for the reaction PCl5(g)
PCl3(g) + Cl2(g) is 16. If the volume of the
container is reduced to one half its original volume, the value of K p for the reaction at the same
temperature will be :
(1) 32
(2) 64
(3) 16
(4) 4
6.
4.5 moles each of hydrogen and iodine heated in a sealed ten litre vessel. At equilibrium 3 moles of HI
were found. The equilibrium constant for H2(g) + I2(g)
2HI(g) is :
(1) 1
(2) 10
(3) 5
(4) 0.33
7.
In a 20 litre vessel initially each have 1 – 1 mole CO, H2O CO2 is present, then for the equilibrium of
CO + H2O
CO2 + H2 following is true :
(1) H2, more then 1 mole
(2) CO, H2O, H2 less then 1 mole
(3) CO2 & H2O both more than 1 mole
(4) All of these
8.
At 1000 K, the value of Kp for the reaction A(g) + 2B(g)
value of KC in terms of R would be :
(1) 20000 R
(2) 0.02 R
(3) 5  10–5 R
10.
(4) 5  10–5  R–1
uh
In which of the following reactions is Kp < Kc ?
(1) CO(g) + Cl2(g)
COCl2 (g)
(3) 2BrCl(g)
Cl2(g) + Br2(g)
3C(g) + D(g) is 0.05 atmosphere. The
(2) CH4(g) + H2O(g)
(4) I2(g)
2(g)
Ja
9.
a
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4.
K for the synthesis of HI is 50. K for dissociation of HI is :
(1) 50
(2) 5
(3) 0.2
CO(g) + 3H2(g)
(4) 0.02
The equilibrium constant of the reaction H2(g) + I2(g)
2HI(g) is 64. If the volume of the container
is reduced to one fourth of its original volume, the value of the equilibrium constant will be
(1) 16
(2) 32
(3) 64
(4) 128
12.
In equilibrium CH3COOH + H2O
CH3COO– + H3O+
The equilibrium constant may change when
(1) CH3COO– is added
(2) CH3COOH is added
(3) Catalyst is added
(4) Mixture is heated
13.
In the reaction, N2 + O2
2NO, the moles/litre of N2, O2 and NO respectively 0.25, 0.05 and 1.0 at
equilibrium, the initial concentration of N2 and O2 will be respectively :
(1) 0.75 mol/litre, 0.55 mole/litre
(2) 0.50 mole/litre, 0.75 mole/litre
(3) 0.25 mole/litre, 0.50 mole/ litre
(4) 0.25 mole/litre, 1.0 mole/litre
Sa
nk
a
lp
11.
14.
In the following reaction started only with A 8, 2A8(g)
2A3(g) + 3A2(g) + A4(g) mole fraction of A2 is
found to 0.36 at a total pressure of 100 atm at equilibrium. The mole fraction of A8(g) at equilibrium is :
(1) 0.28
(2) 0.72
(3) 0.18
(4) None of these
15.
In a 0.25 litre tube dissociation of 4 moles of NO is take place. If its degree of dissociation is 10%. The
value of Kp for reaction 2NO
N2 + O2 is :
(1)
16.
1
(18 )
2
(2)
1
(8)
2
(3)
1
16
(4)
1
32
For the given reaction at constant pressure,
n A (g)
An (g)
Initial moles
1
0
Moles at equilibrium 1 − 
/n
Then the correct relation between initial density (di) & final density (df) of the system is
 n − 1  df − di 
=
 
 n   df 
(1) 
(2)
n  df − di 
=
n−1
df
351
 n − 1  di − df 
=
 
 n   di 
(3) 
(4)
 di − df 
1

 =
(n − 1)  di 
On decomposition of NH4HS, the following equilibrium is established :
NH4HS(s)
NH3(g) + H2S (g)
If the total pressure is P atm, then the equilibrium constant K P is equal to
(1) P atm
(2) P2 atm2
(3) P2 / 4 atm2
(4) 2P atm
18.
At room temperature, the equilibrium constant for the reaction P + Q
R + S was calculated to be
4.32. At 425°C the equilibrium constant became 1.24 × 10–2. This indicates that the reaction
(1) is exothermic
(2) is endothermic
(3) is difficult to predict
(4) no relation between H and K
19.
Calculate G° for conversion of oxygen to ozone 3/2 O 2(g) ⎯⎯→ O3(g) at 298 K, if Kp for this conversion
is 2.47 × 10–29
(1) 163 kJ mol–1
(2) 2.4 × 102 kJ mol–1 (3) 1.63 kJ mol–1
(4) 2.38 × 106 kJ mol–1
20.
For the reaction, 4 NH3(g) + 5 O2(g)
4 NO(g) + 6 H2O() , H = positive .
At equilibrium which factor will not effect the concentration of NH3 is :
(1) change in pressure (2) change in volume (3) catalyst
(4) None of these
21.
The effect of adding krypton (Kr) gas on position of equilibrium, keeping the volume of the system
constant is
(1) If n = 0, backward reaction is favoured.
(2) If, n = +ve, forward reaction is favoured
(3) If n = –ve, forward reaction is favoured
(4) No effect whatever be the value of n
22.
Le-Chatelier’s principle is applicable only to a
(1) System in equilibrium
(3) Homogeneous reaction
Ja
uh
a
ri
17.
(2) Irreversible reaction
(4) Heterogeneous reaction
'a' moles of PCl5, undergoes, thermal dissociation as : PCl5
PCl3 + Cl2, the mole fraction of PCl3 at
equilibrium is 0.25 and the total pressure is 2.0 atmosphere. The partial pressure of Cl 2 at equilibrium
is:
(1) 2.5
(2) 1.0
(3) 0.5
(4) None
24.
The value of Gº for the phosphorylation of glucose in glycolysis is 15 kJ/mole. Find the value of K c at
300 K.
−
6
2.303
(3)
1
e−6
(4) 10
2.303
6
Which of the following statements is correct for a reversible process in a state of equilibrium ?
(1) G = 2.30 RT log K
(2) G° = –2.30 RT log K
(3) G° = 2.30 RT log K
(4) G = –2.30 RT log K
Sa
nk
25.
(2) 10
a
(1) e6
lp
23.
26.
For the following isomerisation reaction
cis-butene-2
trans-butene-2
KP = 1.732
A
P(trans-butene-2)
60º
P(cis-butene-2)
Which of the following statement is true at point ‘A’ ?
(1) Q > KP
(2) Q < KP
(3) Q = K = 1
27.
(4) Q = K = 1.732
The following equilibrium exists in a saturated solution of NH4CI.
NH4CI(S)
NH4+ (aq) + CI-(aq) ;
H25°C = 3.5 kcal mol-1
A change that will shift the equilibrium to the right is
(1) decrease in temperature
(2) increase in temperature
352
(3) addition of NH4CI crystals to the reaction mixture
(4) addition of NH4OH solution to the reaction mixture.
28.
For the reaction : PCl5 (g)
PCl3 (g) + Cl2 (g)
The backward reaction at constant temperature is favoured by
(1) introducing chlorine gas at constant volume
(2) introducing an inert gas at constant pressure
(3) increasing the volume of the container
(4) introducing PCl5 at constant volume
29.
Find out lnKeq for the formation of NO2 from NO and O2 at 298 K
1
O2
2
NO2 g
Gºf (NO2) = 52.0 KJ/mole
Given :
Gºf (NO) = 87.0 KJ/mole
Gºf (O2) = 0 KJ/mole
(2) –
35  103
8.314  298
(3)
35  103
2.303  8.314  298
35  103
2  298
(4)
If a reaction vessel at 400ºC is charged with equimolar mixture of CO and steam such that
PCO = PH2O = 4 bar what will be that partial pressure of H 2 at equilibrium
CO + H2O
(1) 0.3 bar
CO2 + H2
(2) 0.4 bar
KP = 9
(3) 0.2 bar
Ja
30.
35  103
8.314  298
uh
(1)
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NO(g) +
(4) 0.1 bar
Practice Test-1 (IIT-JEE (Main Pattern))
OBJECTIVE RESPONSE SHEET (ORS)
2
3
11
12
13
21
22
Ans.
Que.
Ans.
23
Sa
nk
Que.
4
5
6
7
8
9
10
15
16
17
18
19
20
25
26
27
28
29
30
lp
1
14
a
Que.
24
Ans.
PART - II : NATIONAL STANDARD EXAMINATION IN CHEMISTRY (NSEC) STAGE-I
1.
2.
3.
If the equilibrium constant for the reaction 0.125.
P4(g) + 6CI2(g)
4 PCI3(g)
The value of equilibrium for this reaction 4PCI3(g)
P4(g) + 6CI2(g).
(A) 0.25
(B) 8
(C) 0.125
[NSEC-2000]
(D) 6
The free energy change for a reversible reaction at equilibrium is :
(A) very large positive (B) positive
(C) zero
(D) negative
[NSEC-2000]
Pure ammonia is placed in a vessel at a temperature where its dissociation constant is appreciable. At
equilibrium :
[NSEC-2001]
(A) concentration of ammonia does not change with pressure.
(B) its degree of dissociation, a does not change with pressure.
(C) Kp does not change significantly with pressure.
(D) concentration of hydrogen is less than that of nitrogen.
353
4.
One mole of ethyl alcohol was treated with one mole of acetic acid at 25ºC. Two-third of the acid
changes into ester at equilibrium. The equilibrium constant for the reaction will be :
[NSEC-2001]
(A) 3
(B) 2
(C) 1
(D) 4
5.
The relationship between equilibrium constants Kp and Kc for a gaseous reaction is :
[NSEC-2001]
(A) Kp = Kc.R(T)n
(B) Kc = Kp.(RT)n
(C) Kp = Kc.(RT)n
(D) Kp = Kc/RTn
6.
For the gaseous reaction, C2H4 + H2
(A) mol2dm–3
(B) dm3mol–1
7.
The equilibrium constant for the reaction H 2 + Br2
constant for the dissociation of HBr is :
(A) 0.0147
(B) 67.80
(C) 33.90
8.
The equilibrium constant (K) for the reaction, A + 2B
(A)
(B)
[2C][D]
[A][2B]
(C)
2C + D is :
[NSEC-2001]
2
[C][D]
[A][B]
(D)
[C] [D]
[A][B]2
The following pictures represents the equilibrium state for three different reactions of the type
[NSEC-2002]
A2 + X2
2AX ( X = B, C or D)
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9.
[C] [D]
[A][2B]
2HBr is 67.8 at 300ºK. The equilibrium
[NSEC-2001]
(D) 8.349
a
ri
2
C6H6, the equilibrium constant has the units : [NSEC-2001]
(C) dm3mol–2
(D) mol.dm–3
Methanol (CH3OH) is manufactrued by reaction of carbon monoxide with hydrogen in the presence of
ZnO/Cr2O3 catalyst.
[NSEC-2002]
CO(g) + 2H2(g)
CH3OH (g) ; [Ho = – 91 kJ]
What happen to the amount of methanol when an equilibrium mixtrue of reactants and products is
subjected to rise in temperature ?
(A) Amount of methanol will increase
(B) Amount of methanol will decrease
(C) Amount of methanol remain the same
(D) None of these
Sa
nk
a
10.
2AD]
lp
[A2 + B2
2AB]
[A2 + C2
2AC] [A2 + D2
Which reaction has the largest equilibrium constant ?
(A) A2 + B2
2AB
(B) A2 + C2
2AC
(C) A2 + D2
2AD
(D) none of these
11.
For the reversible reaction, A + B
C, the specific reaction rates for forward and reverse reactions
are 1.25 × 10–3 and 2.75 ×10–5 respectively. The equilibrium constant for the reaction is : [NSEC-2002]
(A) 45.45
(B) 0.022
(C) 2.20
(D) 0.4545
12.
The equilibrium constant for the gaseous reaction H2 + CI2
(A) K =
[H2 ][CI2 ]
[HCI]
2
(B) K =
[H2 ][CI2 ]
2[HCI]
2HCI is given by
2
(C) K =
[HCI]
[H2 ][CI2 ]
(D) K =
[NSEC-2002]
2[HCI]
[H2 ][CI2 ]
13.
For the reaction, N2 + 3H2
2NH3, the units of Kc and Kp respectively are :
[NSEC-2003]
(A) mol–2 L2 and bar-2 (B) mol–2 L2 and bar–1 (C) mol–1 L and bar–2
(D) mol–1 L–1 and bar–1
14.
The equilibrium constant for the reaction N2 + 3H2
equilibrium constant for the reaction NH3
approximately
(A) 1.4×10–2
2NH3 is 70 at a certain temperature. Hence,
3
1
N2 +
H2 of the same temperature will be
2
2
[NSEC-2004]
(B) 1.2 × 10–1
(C) 2.0 × 10–4
(D) 2.9 × 10–2.
354
15.
For the reaction 4NH3 (g) + 7O2(g)
(A) Kp = Kc (RT)
(B) Kp = Kc
4NO2(g) + 6H2O(g), Kp is related to Kc by
[NSEC-2005]
(C) Kp = Kc (RT)3
(D) Kp = KcI (RT)–1.
16.
When Kc > 1 for a chemical reaction,
[NSEC-2005]
(A) the equilibrium would be achieved rapidly
(B) the equilibrium would be achieved slowly
(C) product concentrations would be much greater than reactant concentrations at equilibrium
(D) reactant concentrations would be much greater then product concentrations at equilibrium.
17.
Increased pressure shifts the equilibrium of the reaction : N2(g) + 3H2(g)
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ri
(A) form more ammonia gas
(B) produce more N2(g) and H2(g)
(C) keep the conversion to ammonia unaltered
(D) produce more H2(g).
2NH3(g) so as to
[NSEC-2006]
In which of the following reactions will an increase in volume of the reaction system favor the formation
of the products ?
[NSEC-2007]
(A) C(s) + H2O(g)
CO(g)+ H2(g)
(B) H2(g) + I2(g)
H2(g)
(C) 4NH3(g)+ 5O2(g)
4NO(g)+ 6H2O
(D) 3O2(g)
2O3(g)
19.
Which of the following changes the value of the equilibrium constant?
(A) change in concentration
(B) change in pressure
(C) change in volume
(D) none of these
[NSEC-2007]
20.
Consider the equilibrium reaction:
4NH3(g) + 3O2(g)
2N2(g) + 6H2O(g)
(H = –1268 KJ)
Which change will cause the reaction to shift to the right?
(A) Increase the temperature
(B) Decrease the volume of the container.
(C) Add a catalyst to speed up the reaction.
(D) Remove the gaseous water by allowing it to react and be absorbed by KOH.
[NSEC-2007]
21.
At a given temperature the equilibrium constants of the gaseous reactions
K1
NO(g) + 1/2 O2(g)
NO2(g)
K1
2NO2(g)
2NO(g) + O2(g) are related as :
a
lp
Ja
uh
18.
(A) K1 = 2K2
For the reaction 2HI(g)
H2(g) + I2(g)
(A) Kp = Kc
(B) Kp > Kc
Sa
nk
22.
23.
(D) K1 = K 2
(C) K1= 1 K 2
(B) K1 = 1/K2
[NSEC-2008]
[NSEC-2009]
(C) Kp < Kc
(D) Kc = Kp
If the equilibrium constants of the reactions,
2SO3
2SO2 + O2
and
[NSEC-2009]
1
SO2 + O2
2
SO3
are K1 and K2 respectively, the correct relation between the two equilibrium constant is,
(A) K2 = (K1)–1
24.
1
K1
 1 

 K1 
2
(C) K2 = 
(D)
K1
The equilibrium constant Kc for the reaction,
2NaHCO3 (s)
Na2CO3 (s) + CO2 (g) + H2O (l)
(A) Kc =
[Na2CO3 ][CO2 ][H2O]
[NaHCO3 ]2
(C) Kc = [CO2][H2O]
25.
(B) K2 =
(B) Kc =
[NSEC-2009]
[Na2CO3 ]
[NaHCO3 ]2
(D) Kc = pCO2 × pH2O
For the following reaction, the value of K changes with
N2(g) + O2(g)
2 NO(g)
H = + 180 kJ mol–1
(A) change in pressure
(B) change in concentration of oxygen
(C) introduction of NO(g)
(D) change in temperature
[NSEC-2010]
355
26.*
The formation of ammonia from nitrogen and hydrogen gases can he written by the following two
equations
(a)
3
1
N2(g) + H2(g)
2
2
NH3(g)
(b)
2
NH3(g)
3
1
N2 (g) + H2(g)
3
The two equations have equilibrium constants K1 and K2, respectively. The relationship between the
equilibrium constants is
[NSEC-2010]
(A) K1 = K22
(B) K13 = K22
(C) K12/3 = K2
(D) K1 = K23/2
For the reaction PCl3(g) + Cl2(g)
PCl5(g), Kc is 26 at 250°C. Kp at the same temperature is
(R = 8.314 JK–1mol–1)
[NSEC-2011]
–3
–3
–3
–3
(A) 4.6 × 10
(B) 5.7 × 10
(C) 6.0 × 10
(D) 8.3 × 10
28.
At 445ºC, Kc for the following reaction is 0.020.
2HI(g)
H2(g) + I2(g)
A mixture of H2, I2 and HI in a vessel at 445º C has the following concentrations :
[NSEC-2011]
[HI] = 2.0M, [H2] = 0.50M and [I2] = 0.10M. The statement that is true concerning the reaction quotient,
Qc is:
(A) Qc = Kc ; the system is at equilibrium
(B) Qc is less than Kc; more H2 and I2 will be produced
(C) Qc is less than Kc; more HI will be produced
(D) Qc is greater than Kc; more H2 and I2 will be produced
29.
The oxidation of SO2 by O2 is an exothermic reaction. The yield of SO3 can be maximized if :
(A) temperature is increased and pressure is kept constant
[NSEC-2012]
(B) temperature is decreased and pressure is increased
(C) both temperature and pressure are increased
(D) both temperature and pressure are decreased
30.
The KP/KC ratio for the reaction 4NH3(g) + 7O2(g)
4 NO(g) + 6 H2O(g), at 127ºC is: [NSEC-2013]
4NH3(g) + 7O2(g)
4 NO(g) + 6 H2O(g)
(A) 0.0301
(B) 0.0831
(C) 1.0001
(D) 33.26
31.
KP for the reaction given below is 1.36 at 499 K. Which of the following equaitons can be used to
calculate Kc for this reaction ?
[NSEC-2013]
N2O5(g) —→ N2O3(g) + O2(g)
[(0.0821)  (499)]
[1.36]
[1.36]
(C) Kc =
[(0.0821)  (499)]
At 700 K, for the reaction 2SO2(g) + O2(g)
2SO3(g) the Kp is 3.2 × 104. At the same temperature
the KP for the reaction SO3(g)
SO2(g) + 0.50O2(g) is :
[NSEC-2014]
(A) 3.125 × 10–5
(B) 5.59  10−3
(C) 1.79  104
(D) 1.79  10−2
Sa
nk
32.
[(1.36)  (0.0821)]
[499]
[(1.36)  (499)]
(D) Kc =
[0.0821]
(B) Kc =
a
(A) Kc =
lp
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uh
a
ri
27.
33.
For the following reaction, formation of the product is favored by
A2(g) + 4B2(g)
2AB4(g), H < 0
(A) Low temperature and high pressure
(B) High temperature and low pressure
(C) Low temperature and low pressure
(D) High temperature and high pressure
34.
The equilibrium constant of the following isomerisation reaction at 400K and 298 K are 2.07 and 3.42
respectively.
cis-butene
k1
[NSEC-2015]
trans-butene
k–1
Which of the following is/are correct ?
I. The reaction is exothermic
II. The reaction is endothermic
III. At 400K 50% of cis-butene and 50% of trans-butene are present of equilibrium
IV. Both at 298K and 400K, k1 =k–1
(A) I and IV
(B) II and IV
(C) I and III
(D) I only
[NSEC-2017]
356
35.
Acetic acid (CH3COOH) is partially dimerised to (CH3COOH)2 in the vapour phase. At a total pressure
of 0.200 atm, acetic acid is 92.0% dimerized at 298 K.
The value of equilibrium constant of dimerisation under these conditions is
[NSEC-2017]
(A) 57.5
(B) 9.7
(C) 97
(D) 194
PART - III : HIGH LEVEL PROBLEMS (HLP)
In a vessel, two equilibrium are simultaneously established at the same temperature as follows:
N2(g) + 3H2 (g)
2 NH3 (g)
...(1)
N2(g) + 2H2 (g)
N2H4 (g)
...(2)
Initially the vessel contains N2 and H2 in the molar ratio of 9 : 13. The equilibrium pressure is 7P 0, in
which pressure due to ammonia is P0 and due to hydrogen is 2P0. Find the values of equilibrium
constants (KP’s) for both the reactions
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1.
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SUBJECTIVE QUESTIONS
ONLY ONE OPTION CORRECT TYPE
3.
For the following gases equilibrium, N2O4(g)
attained when :
(A) 0ºC
(B) 273 K
(C) 1 K
(D) 12.19 K
Consider the following reversible gaseous reactions (at 298 K) :
(a) N2O4
2NO2
(b) 2SO2 + O2
2SO3
(c) 2H
H2 +  2
(d) X + Y
4Z
(A) d, b
Kp
Kc
will be shown by the equilibrium
lp
Highest and lowest value of
(B) a, c
(C) a, b
(D) b, c7
For a container containing A(g), B(g), C(g) & D(g) with rigid walls, an experiment is carried upon. This
experiment involves increase in temperature of container in steps of 1ºC and system is allowed to attain
equilibrium, followed by calculation of K1 & K2 at each step, where K1 & K2 are equilibrium constants for
reaction (1) & (2) respectively.
A(g) + 2B(g)
C(g) + D(g)
...(1)
C(g) + D(g)
A(g) + 2B(g)
...(2)
Select the graph showing correct relationship –
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nk
a
4.
2NO2(g), Kp is found to be equal to Kc. This is
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2.
(A)
(B)
(C)
(D)
5.
If CuSO4.5H2O(s)
CuSO4.3H2O(s) + 2H2O(g)
KP = 4 × 10–4 atm2
at 25ºC. The efflorescent nature of CuSO4.5H2O can be noticed when vapour pressure of H 2O in
atmosphere is
(A) > 15.2 mm
(B) < 15.2 mm
(C) < 15.2 mm
(D) = 15.2 mm
6.
At a temperature T, a compound AB4(g) dissociates as 2AB4(g)
A2(g) + 4B2(g) with a degree of
dissociation x, which is small compared with unity. The expression of K P in terms of x and total pressure
P is :
(A) 8P3x5
(B) 256P3x5
(C) 4Px2
(D) None of these.
357
The equilibrium, SO2Cl2(g) ⎯→ SO2(g) + Cl2(g) is attained at 25ºC in a closed container and inert gas
helium is introduced isothermally. Which of the following statement(s) is/are correct ?
I. Concentrations of SO2, Cl2 and SO2Cl2 change
II. More chlorine is formed
III. Concentration of SO2 is reduced
IV. More SO2Cl2 is formed.
(A) I, II, III
(B) II, III, IV
(C) III, IV
(D) None
8.
C(s) + CO2(g)
2CO(g)
KP = 1 atm
CaCO3(s)
CaO(s) + CO2(g)
KP = 4 × 10–2
Solid C, CaO and CaCO3 are mixed and allowed to attain equilibrium. Calculate final pressure of CO.
(A) 0.4 atm
(B) 0.2 atm
(C) 8 atm
(D) 0.01 atm
9.
Ammonia gas at 15 atm is introduced in a rigid vessel at 300 K. At equilibrium the total pressure of the
vessel is found to be 40.11 atm at 300ºC. The degree of dissociation of NH3 will be :
(A) 0.6
(B) 0.4
(C) Unpredictable
(D) None of these
10.
Two solid A and B are present in two different container having same volume and same temperature
following equilibrium are established :
In container (1) A(s)
D(g) + C(g) PT = 40 atm at equilibrium
In container (2) B(s)
E(g) + F(g) PT = 60 atm at equilibrium
If excess of A and B are added to a third container having double the volume and at same temperature
then, the total pressure of this container at equilibrium is :
(A) 50 atm
(B) 100 atm
(C) 200 atm
(D) 70 atm
11.
For equilibrium ZnSO4.7H2O(s)
ZnSO4.2H2O(s) + 5H2O(g), KP = 56.25 × 10–10 atm5 and vapour
pressure of water is 22.8 torr at 298 K. ZnSO 4.7H2O(s) is efflorescent (lose water) when relative
humidity is [ 5 56.25 = 2.23]
(A) more than 80.60%
(B) less than 74.60%
(C) Above than 74.60%
(D) Above than 70.60%
12.
Solid A and B are taken in a closed container at a certain temperature. These two solids decompose
and following equilibria are established simultaneously
A(s)
X(g) + Y(g)
KP1 = 250 atm2
Y(g) + Z(g)
KP2 = ?
a
B(s)
lp
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uh
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7.
Sa
nk
If the total pressure developed over the solid mixture is 50 atm. Then the value of K P for the 2nd
reaction.
(A) 375
(B) 625
(C) 225
(D) 250
13.
X(s)
Y(g) + 2Z(g)
A(s)
Y(g) + B(g)
Consider both these equilibrium to be established simultaneously in a closed container.
At equilibrium, pressure of Z and B were found to be same and sum of pressure of Z & B is 10 atm
more than that of species Y. Find ratio of standard gibb’s energy of two reactions.
(A) 20
14.
(B) 2.303 log1020
(C) log10 3 144
(D)
3 + log12
2 + log 6
In one experiment, certain amount of NH4I(s) was heated rapidly in a closed container at 375ºC. The
following equilibrium was established :
NH4I(s)
NH3(g) + HI(g)
Excess of NH4I(s) remained unreacted in the flask and equilibrium pressure was 304 mm of Hg. After
some time, the pressure started increasing further owing to the dissociation of H I.
2HI(g)
H2(g) + I2(g)
KC = 0.010 calculate final pressure.
(A) 328 mm Hg
(B) 331 mm Hg
(C) 662 mm Hg
(D) 151 mm Hg
358
15.
A(s)
B(g) + C(g)
KP1 = 36 atm2
E(s)
B(g) + D(g)
KP2 = 64 atm2
Both solids A & E were taken in a container of constant volume at a given temperature. Total pressure
in the container after equilibrium is
(A) 6 atm
(B) 5 atm
(C) 10 atm
(D) 20 atm
In a closed container following equilibrium will be attained –
A(s) + B(g)
AB(g)
B(g) + C(g)
BC(g)
On adding He gas (inert) to the above system at constant pressure & temperature
(A) Amount of AB(g) will be increased surely.
(B) Amount of B(g) will be decreased surely.
(C) Amount of C(g) will be decreased surely.
(D) Amount of BC(g) will be decreased surely.
17.
2NH3(g)
N2(g) + 3H2(g) in a V lit container total x mol at eq.
N2H4(g)
N2 + 2H2(g)
in V lit (other) container total y mol at eq.
If both are taken in same container (V lit) then at new equation total mols will be
(A) x + y
(B) > x + y
(C) < x + y
(D) No prediction is possible.
a
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16.
18.
The equilibrium SO2(g) +
uh
SINGLE AND DOUBLE VALUE INTEGER TYPE
1
O2(g)
2
SO3(g) is established in a container of 4L at a particular
19.
Ja
temperature. If the number of moles of SO2, O2 and SO3 at equilibrium are 2, 1 and 4 respectively then
find the value of equilibrium constant.
If the equilibrium constant of the reaction 2HI(g)
1
1
constant of the reaction H2 (g) +
2
2
I2 (g)
H2(g) + I2(g) is 0.25, find the equilibrium
HI(g).
A2(g) and B2(g) having partial pressures 60 mm of Hg & 42 mm of Hg respectively, are present in a
closed vessel. At equilibrium, partial pressure of AB(g) is 28 mm of Hg. If all measurements are made
under similar condition, then calculate percentage of dissociation of AB (g).
(Round of answers to nearest integer).
21.
NH4HS(s)
NH3(g) + H2S(g)
a
NH3(g)
lp
20.
3
1
N2(g) + H2(g)
2
2
KP1
KP2
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2 mol NH4HS(s) is taken & 50% of this is dissociated till at equilibrium in 1 litre container. Find
KP22
KP6
if
1
0.25 moles of N2 are found finally.
22.
Consider :
(I) C(s) + O2
(II) 2 C(s) + O2
CO2(g)
2 CO(g)
7
8
= 12.5 atm
Kp =
1
Kp
2
As 100 L of air (80 % N2, 20% O2 by volume) is pased over excess heated coke to establish these
equilibrium the equilibrium mixture is found to measure 105 L at constant temperature & pressure (105
atm). Assuming no other reaction, find the sum of partial pressure of CO and CO 2 in the final
equilibrium mixture.
23.
Two solid compounds A and C dissociate into gaseous product at temperature T as follows :
(i) A(s)
B(g) + D(g)
Kp1 = 625 (atm)2
(ii) C(s)
E(g) + D(g)
Kp2 = 975 (atm)2
Both solid are present in same container then calculate total pressure over the solid mixture.
359
24.
If a mixture 0.4 mole H2 and 0.2 mole Br2 is heated at 700 K at equilibrium, the value of equilibrium
constant is 0.25 × 1010 then find out the ratio of concentrations of (Br 2) and (HBr) (Report your answer
as
Br2
× 1011)
HBr
25.
2 mole of PCl5 were heated in a 5 liter vessel. It dissociated. 80% at equilibrium find out the value of
equilibrium constant. Report your answer as KC × 50.
26.
Two solids A and D dissociates into gaseous products as follows
A(s)
B(g) + C (g) ; KP1 = 300
;
D(s)
E(g) + C (g) KP2 = 600
a
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at 27ºC , then find the total pressure of the solid mixture.
ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
(i)
N2(g) + O2(g)
(ii)
 1
 1
  N2(g) +   O2(g)
2
 
2
(iii)
2NO(g)
(iv)
NO(g)
2NO(g),
K1
NO(g) ;
N2(g) + O2(g) ;
K2
K3
 1
 1
  N2(g) +   O2(g) ;
2
 
2
K4
K3  K 2 = 1
Ja
Correct relation between K1, K2, K3 and K4 is/are :
(A) K1 × K3 = 1
(B) K1  K 4 = 1
(C)
uh
27.
(D) None
A 2 lit vessel is filled by 1 mole of each gas A & B. If K C for reaction
A(g)
B(g) is 1.5 at temp. T. [Atomic mass of A is 40 & B is 20].
Which are correct option.
(A) [A] vs time is graph I
(B) [B] vs time is graph I
(C) [A] vs time is graph II
(D) [B] vs time is graph II
29.
The equilibrium constant for some reactions are given below against each of the reaction
(i) 2N2 + 5O2
2N2O5
;
K = 5 × 10–27
(ii) N2 + O2
2NO
;
K = 2 × 10–15
(iii) N2 + 2O2
2NO2
;
K = 1.5 × 10–29
Which of the following statement is correct
(A) The least stable oxide is NO2
(B) The most stable oxide is NO
(C) The stability order is N2O5 > NO2 > NO
(D) The stability order is NO2 > NO > N2O5
Sa
nk
a
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28.
30.
Sufficient amount of a solid X is taken in a rigid vessel at T°C where it attained the equilibrium :
X (s)
Y(g) + 2Z(g)
Total pressure was measured. Now the vessel is evacuated and filled with sufficient amount of another
solid V under same conditions where it attained theequilibrium :
V(s)
W (g) + 2Z(g)
Total pressure measured now is found to be double that of previous value. Now, if both X & V solids
are allowed to attain their respective equilibrium together in the same vessel at same temperature,
select the correct statement(s) :
(A) KP for decompostion reaction of V (s) = 8 × KP for decomposition reaction of X (s).
1
(B) In the 3rd case (when both solids are simultaneously estabilishing their equilibrium), P Y = PW.
8
1
(C) PY in 3rd case =
× PY in 1st case.
3 3
(D) In the 3rd case, Pw : PZ = 4:9
360
31.
CaCO3(s)
CO2(g)
CaO(s) + CO2(g)
CO(g) +
1
O2 (g)
2
For above simultaneous equilibrium if CO2 is added from out side at equilibrium then :
(A) PCO2 will increase
(B) PCO2 will decrease
(C) No shift in 2nd equilibrium
(D) Backward shift in 1st equilibrium
For the reaction SnO2(s) + 2H2(g)
2H2O(g) + Sn(s). If at 900 K, equilibrium mixture contains
45% H2 by volume and at 1100 K it contains 24% H2 by volume then which of the statements is/are
correct.
(A) Reaction is endothermic in nature
(B) At higher temperature, the efficiency of reduction of tin oxide will increase
(C) Reaction is exothermic in nature
(D) At lower temperature, the efficiency of reduction of tin oxide decreases.
33.
Consider equilibrium H2O()
relative humidity.
(A) R.H. > 1, rightward
(C) R.H. > 1, leftward
H2O(g). Choose the correct direction of shifting of equilibrium with
uh
(B) R.H. < 1, rightward
(D) R.H. < 1, leftward
An industrial fuel, ‘water gas’, which consists of a mixture of H 2 and CO can be made by passing steam
over red-hot carbon. The reaction is
C(s) + H2O(g)
CO(g) + H2(g), H = +131 kJ
The yield of CO and H2 at equilibrium would be shifted to the product side by :
(A) raising the relative pressure of the steam
(B) adding hot carbon
(C) raising the temperature
(D) reducing the volume of the system
Ja
34.
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32.
Max. Time : 1 Hr.
Important Instructions :
A. General %
Max. Marks : 60
a
The test is of 1 hour duration.
The Test Booklet consists of 20 questions. The maximum marks are 60.
Question Paper Format
Each part consists of five sections.
Section-1 contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE is correct.
Section-2 contains 4 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE OR MORE THAN ONE are correct.
Section-3 contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from
0 to 9 (both inclusive).
Section-4 contains 1 paragraphs each describing theory, experiment and data etc. 2 questions relate to
paragraph. Each question pertaining to a partcular passage should have only one correct answer among
the four given choices (A), (B), (C) and (D).
Section 5 contains 1 multiple choice questions. Question has two lists (list-1 : P, Q, R and S; List-2 : 1, 2,
3 and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY ONE
is correct.
Sa
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1.
2.
B.
3.
4.
lp
PART - IV : PRACTICE TEST-2 (IIT-JEE (ADVANCED Pattern))
5.
6.
7.
8.
361
C.
9.
10.
11.
Marking Scheme
For each question in Section 1, 4 and 5 you will be awarded 3 marks if you darken the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one
(– 1) mark will be awarded.
For each question in Section 2, you will be awarded 3 marks. If you darken all the bubble(s)
corresponding to the correct answer(s) and zero mark. If no bubbles are darkened. No negative marks will
be answered for incorrect answer in this section.
For each question in Section 3, you will be awarded 3 marks if you darken only the bubble corresponding
to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for
incorrect answer in this section.
Which of the following is correct for the equilibrium of the reaction
C(s) + H2O(g)
CO(g) + H2(g)
(B) pH2  pH2O
(A) pH2  pH2O
2.
(C) pH2  pH2 2O
(D) pH2 
uh
1.
a
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SECTION-1 : (Only One option correct Type)
This section contains 7 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which Only ONE option is correct.
2
pH
2O
pCO
For the reaction PCl5
PCl3 + Cl2 ; Supposing at constant temperature, if the volume is
increased 16 times the initial volume, the degree of dissociation for this reaction will becomes :
(B)
1
times
4
(C) 2 times
(D)
1
times
4
Ja
(A) 4 times
A vessel of 10 L was filled with 6 mole of Sb2S3 and 6 mole of H2 to attain the equilibrium at 440oC as :
Sb2S3(s) + 3H2(g)
2Sb(s) + 3H2S(g)
After equilibrium the H2S formed was analysed by dissolving it in water and treating with excess of Pb 2+
to give 708 g of PbS as precipitate. What is value of Kc of the reaction at 440ºC ? (At. weight of Pb =
206).
(A) 0.08
(B) 0.8
(C) 0.4
(D) 0.04
4.
Variation of log10 K with
a
1
is shown by the following graph in which straight line is at 45°, hence H° is :
T
(B) – 4.606 cal
Sa
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(A) + 4.606 cal
lp
3.
(C) 2 cal
(D) – 2 cal
5.
aA + bB
cC + dD
In above reaction low pressure and high temperature, conditions are shift equilibrium in back direction
so correct set :
(A) (a + b) > (c + d), H > 0
(B) (a + b) < (c + d), H > 0
(C) (a + b) < (c + d), H < 0
(D) (a + b) > (c + d), H < 0
6.
The value of kp for the reaction at 27ºC
Br2() + Cl2(g)
2BrCl(g)
is ‘1 atm’. At equilibrium in a closed container partial pressure of BrCl gas is 0.1 atm and at this
temperature the vapour pressure of Br2() is also 0.1 atm. Then what will be minimum moles of Br2() to
be added to 1 mole of Cl2, initially, to get above equilibrium situation :
(A)
7.
10
moles
6
(B)
5
moles
6
(C)
15
moles
6
(D) 2 moles
C(s)
2A(g) + B(s)
If the dissociation of C(s) is ‘’ and d is the density of the gaseous mixture in the container. Initially
container have only ‘C(s)’ and the reaction is carried at constant temperature and pressure.
362
(A)
(B)
(C)
(D)
Section-2 : (One or More than one options correct Type)
This section contains 4 multipole choice questions. Each questions has four choices (A), (B),
(C) and (D) out of which ONE or MORE THAN ONE are correct.
For which reaction at 298 K, the value of
(A) N2O4
(C) X + Y
2NO2
4Z
Kp
Kc
is maximum and minimum respectively :
(B) 2SO2 + O2
(D) A + 3B
2SO3
7C
a
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8.
For the equilibrium 2SO2(g) + O2(g)
2SO3(g), H = –198 kJ, the equilibrium concentration of SO 3
will be affected by
(A) doubling the volume of the reaction vessel
(B) increasing the temperature at constant volume
(C) adding more oxygen to the reaction vessel
(D) adding helium to the reaction vessel at constant volume
10.
AB(s)
A(g) + B(g) Kp = 4, H = +ve
In a container, A (g) and B (g) are filled to partial pressure of 1 atm each. Now AB (s) is added (in
excess quantity). Which of the following is CORRECT ? (No other gas is present in container) :
(A) At equilibrium, the total pressure in the container is 4 atm.
(B) Equilibrium pressure decreases uniformly on increasing the volume by container.
(C) At equilibrium, the total pressure in the container is more than 4 atm, if temperature is increased.
(D) None of these
11.
Following two equilibria are established seperately in 2 different containers of unequal volume.
PCl5(g)
PCl3(g) + Cl2(g)
COCl2(g)
CO(g) + Cl2(g)
Now the containers are connected together by a thin tube of negligible volume.
Select incorrect statements. (Assume T constant)
(A) Degree of dissociation of both PCl5(g) & COCl2(g) will decrease
(B) Degree of dissociation of both PCl5(g) & COCl2(g) will increase
(C) Degree of dissociation of PCl5(g)may increase ; decrease or remain the same irrespective of effect
on degree of dissociation of COCl2(g).
(D) Degree of dissociation of PCl5(g) may increase, decrease or remain the same, but the effect would
be same as that on degree of dissociation of COCl2.
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Ja
uh
9.
Section-3 : (One Integer Value Correct Type.)
This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive)
12.
If 1 mole of CH3COOH and 1 mole of C2H5OH are taken in 1 litre flask, 50% of CH 3COOH is converted
into ester as ;
CH3COOH() + C2H5OH()
CH3COOC2H5() + H2O()
There is 33% conversion of CH3COOH into ester, if CH3COOH and C2H5OH have been taken initially in
molar ratio x : 1, find x.
13.
Solid ammonium carbamate (NH2COONH4) was taken in excess in closed container of volume 5 Litre
according to the following reaction NH2COONH4(s)
2NH3(g) + CO2(g). If equilibrium partial
pressure of ammonia is 4 atm, it’s equilibrium constant K p is x atm3. If the above equilibrium mixture is
transferred to another vessel of volume 10 Litre, total equilibrium pressure is y atm. Calculate the value
of (x + y). Assuming temperature is constant during the whole experiment.
363
14.
Find the percentage dissociation of ammonia into N 2 and H2 if the dissociation is carried out at constant
pressure and the volume at equilibrium is 20% greater than initial volume. (Initially, equal moles of NH 3
and N2 are present with no hydrogen)
15.
A2B(g) is introduced in a vessel at 1000 K. If partial pressure of A2B(g) is 1 atm initially and KP for
reaction A2B(g)
2A(g) + B(g) is 81 × 10–6 then calculate percentage of dissociation of A2B.
16.
Consider the following two equilibrium established together in a closed container
A(s)
2B(g) + 3C(g)
;
KP1
A(s)
3D(g)
;
KP2
17.
a
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Starting with only A(s), molar ratio of B(g) & D(g) at equilibrium is found to be in a ratio 1 : 6 determine
KP2
.
8KP1
0.1 mol each of ethyl alcohol and acetic acid are allowed to react and at equilibrium the acid was
exactly neutralised by 100 ml of 0.75 N NaOH. If no hydrolysis of ester is supposed to have undergo
uh
–1
find KC in terms (x)10
.
Ja
SECTION-4 : Comprehension Type (Only One options correct)
This section contains 1 paragraphs, each describing theory, experiments, data etc. 2 questions
relate to the paragraph. Each question has only one correct answer among the four given
options (A), (B), (C) and (D)
Paragraph for Question Nos. 18 to 19
05 moles of NH4HS(s) are ta1ken in a container having air at 1 atm. On warming the closed container
a
18.
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There are different types of equilibria depending on the composition of equilibrium mixture.
Homogenous equilibrium contains all the species in same phase and heterogeneous equilibrium
contains different phases. If heterogeneous equilibrium involves pure solid or liquid, then their
concentration or pressure term is not included in KC or KP respectively.
to 50ºC the pressure attained a constant value of 1.5 atm, with some NH 4HS (s) remaining unreacted.
Sa
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The KP of reaction NH4HS(s)
(A) 0.25
19.
(B) 0.625
NH3(g) + H2S(g) at 50°C is :
(C) 0.025
(D) 0.0625
How many moles of water are in vapour phase present inside the vessel containing 1 L water after
sufficient time? (Vapour pressure of water at 27ºC = 3000 Pa, R =
(A) 5 × 10–4
(B) 120
(C) 1.2 × 10–3
25
J/mol–K)
3
(D) None of these
SECTION-5 : Matching List Type (Only One options correct)
This section contains 1 questions, each having two matching lists. Choices for the correct
combination of elements from List-I and List-II are given as options (A), (B), (C) and (D) out of
which one is correct
20.
In given setup, container I has double the volume to that of container II. Container I & II are connected
by a narrow tube with two knobs.
364
Closed – No gas is allowed to pass through
Open – All gases can pass
Knob B :
Closed – A thin filter of Pd is introduced on complete cross section of tube
Open – All gases can pass.
Initially both knobs are closed. In container I, some amount of NH 3 gas is introduced which sets up
equilibrium according to following reaction :
2NH3(g)
N2(g) + H2(g)
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Knob A :
Match the actions in column I to corresponding value in column II and select the correct answer using
the code given below the column. Assume each action from initial stage.
I
nH
2
Q.
A is open & B is closed,
=?
R.
A is open & B is open, nNI 2 / nHII2 = ?
S.
A & B are left open for long time;
now B is closed & volume of container II is halved.
( nNI 2 + nNII2 )/( nHI 2 + nHII2 ) = ?
Code :
P
4
2
(A)
(C)
Q
2
1
R
3
3
1.
2.
3
3.
2
4.
2/3
Ja
/
I
nN
2
Column II
1/3
uh
P.
Column I
A & B are closed, pH2 / pN2 = ?
S
1
4
(B)
(D)
P
2
1
Q
3
2
R
4
3
S
1
4
1
2
11
12
Ans.
13
Sa
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Que.
3
4
a
Que.
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Practice Test-2 (IIT-JEE (ADVANCED Pattern))
OBJECTIVE RESPONSE SHEET (ORS)
14
5
6
7
8
9
10
15
16
17
18
19
20
Ans.
PART - I
1.
(2)
2.
(4)
3.
(3)
4.
(2)
5.
(3)
6.
(1)
7.
(2)
8.
(4)
9.
(1)
10.
(4)
11.
(3)
12.
(4)
13.
(1)
14.
(1)
15.
(1)
16.
(2)
17.
(3)
18.
(1)
19.
(1)
20.
(3)
21.
(4)
22.
(1)
23.
(3)
24.
(2)
25.
(2)
26.
(4)
27.
(2)
28.
(1)
29.
(1)
30.
(1)
365
PART - II
(B)
2.
(C)
3.
(C)
4.
(D)
5.
(C)
6.
(B)
7.
(A)
8.
(D)
9.
(B)
10.
(B)
11.
(A)
12.
(C)
13.
(A)
14.
(B)
15.
(D)
16.
(C)
17.
(A)
18.
(A)
19.
(D)
20.
(D)
21.
(D)
22.
(A)
23.
(B)
24.
(A)
25.
(D)
26.
(BC)
27.
(C)
28.
(B)
29.
(B)
30.
(A)
31.
(C)
32.
(B)
33.
(A)
34.
(D)
35.
(D)
PART - III
1
(D)
3.
5.
(B)
6.
10.
(B)
15.
(A)
4.
(C)
(A)
7.
(D)
8.
(B)
9.
(B)
11.
(B)
12.
(A)
13.
(D)
14.
(A)
(D)
16.
(D)
17.
(C)
18.
4
19.
2
20.
72 %
21.
27
22.
15
23.
80 atm
24.
80
25.
64
26.
80 atm.
27.
(ABC)
28.
(BC)
29.
(AB)
30.
(ABCD)
31.
(CD)
32.
(AB)
33.
(BC)
34.
(AC)
(A)
4.
(B)
5.
(D)
uh
2.
20 P02
Ja
KP1 =
20P02
, KP2 =
3
1.
a
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1.
PART - IV
(B)
2.
(A)
3.
6.
(C)
7.
(D)
8.
(BD)
9.
(ABC)
10.
(ACD)
11.
(ABD)
12.
2
13.
38
14.
40
15.
3
16.
8
17.
1
18.
(D)
19.
(A)
20.
(B)
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1.
1.
t=0
t = eq
2.
N2 + 3H2
1 mole 2 mole
1–x
2–3x
x = 0.4
mole of N2 = 0.6
mole of H2 = 0.8
KA
[PQ]
=
KB [P] [Q]
..... (i)
KC
[R]
=
KD [PQ]
..... (ii)
PART – I
2 NH3
0
2x = 0.8
On multiply equation (i) and (ii) we get
K A .K C
[R]
=
KB .KD
[P][Q]
366
4.
2SO2(g) + O2(g)
2SO3(g)
2
KC =
[SO3 ]
Concentration in gram mole/litre, therefore
[SO2 ]2 [O2 ]
48
80  1
128
[SO2] =
64  1
9.6
[O2] =
32  1
[SO3] =
(Where 80 is molecular weight of SO3)
(Where 64 is molecular weight of SO2)
(Where 32 is molecular weight of O2)
Since, Kp is temperature dependent only.
6.
H2(g) + I2(g)
2HI(g)
t=0
4.5
4.5
0
t = teq. 4.5 – x 4.5 – x
2x
put
x = 1.5
4.5 – 1.5
4.5 – 1.5


3
3
KC =
2 × 1.5 = 3

3
(3)2
=1
33
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5.
a
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Thus, KC = = 0.30
Co + H2O
CO2 + H2
t=0
1
1
1
0
t = teq 1 – x 1 – x 1 + x x
at equilibrium, only CO2 has (1 + x) moles.
8.
Kp = Kc (RT)n, n = 4 – 3 = 1
0.05 = Kc R × 1000
Kc = 5 × 10–5 × R–1
13.
N2
+
O2
2NO
2x = 1.0 mole/litre
a
b
0
x = 1.0/2 mole/litre = 0.50 mole/litre
(a – x)
(b–x)
2x
If a – x = 0.25, b – x = 0.05
[N2] = a = a – x + x = 0.25 + 0.50 = 0.75 mole/litre
lp
a
14.
Ja
7.
2A8
2 A3
+
2
0
2 – 2
2
nT = 2 + 4
given mole fraction of A2 is = 0.36.
Sa
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t=0
t = teq
3 A2
0
3
+
A4
0

3

 = 0.46
2 + 4
2 − 2 2 − 2  0.46
Mole fraction of A8 =
=
= 0.28
2 + 4 2 + 4  0.46
0.36 =
15.
t=0
2NO
4 – .4
3.6
 = 10%
N2 + O 2
.2
.2
0.2
0.2
n = 0,
 KP = KC,
16.
KC =
(.2 / V)2
2
(3.6 / V)
=
4
36  36
1
n

Total moles at equilibrium = 1 −  + /n = 1 +  − 1 

1 
d
So using i = 1 +  − 1  .
df
n 
367
17.
NH4HS(s)
NH3(g) + H2S(g)
Total pressure is P
So,
PNH3 = PH2S =
P
2
Kp = PNH3 × PH2S =
P2
.
4
At room temperature, K = 4.32 and at 425°C, equilibrium constant become 1.24 × 10 –4 i.e. it is
decreases with increase in temperature. So, it is exothermic reaction.
20.
Catalyst can't disturb the state of the equilibrium.
23.
PCl5
a
a–xx
at t = 0
t = teq
PCl3 + Cl2
0
0
x
PPCl3 = XPCl3  PT = .25 × 2 = .5 atm
PPCl3 = PCl2 = .5 atm
lnKeq = –6
25
15000 = –
× 300 ln Keq
3
15000
lnKeq = –
=6
2500
Keq = e–6
G° = –2.303 RT log K
26.
At point A,
28.
According to KP =
Q = tan 60ºC
Q = 1.732
 Q = K = 1.732
PPCl3
(g)  PCl2 (g)
PPCl5
=
(nPCl3
Ja
25.
uh
Gº = – RT ln Keq
(g) )eq.  (nCl2 (g) )eq.
V  (nPCl5
(g) )eq.
lp
24.
a
ri
18.
(g)
and on adding inert gas at constant pressure effect on equilibrium will be similar to as if volume of
container has been increased.
NO +
1
O2
2
NO2
a
29.
º
GR
H = 52 – 87 = –35 kJ
Sa
nk
Gº = –RTlnKeq
lnKeq =
30.
35000
.
8.314  298
CO + H2O
0.4
0.4
0.4 – x 0.4 – x
CO2 + H2
x
x
x
=3
0.4 − x
1.2 – 3x = x
1.2 = 4x
x=
1.2
= 0.3
4
x = 0.3
PART - III
1.
N2(g)
9P – x – y
N2(g)
9P – x – y
+
3H2(g)
13P – 3x – 2Y
+
2H2(g)
13p – 3x – 2Y
2NH3 (g)
2x
N2H4(g)
Y
368
9P – x – y + 13P – 3x –2y + 2x + y = 7Po
22P
– 2x – 2y = 7Po
........(1)
2x = Po
..........(2)
13P – 3x – 2y = 2Po
.........(3)

then
and
P=
Po
2
9Po
3P
5P
P
– o – o = o
2
2
2
2
6Po
13Po 3Po
13P – 3x – 2y =
–
–
= 2Po
2
2
2
9P – x – y =
(2x)2
K1 =
3
(9p − x − y)(13p − 3x − 2y)
=
2y =
y=
Po2
5
Po .(2Po )3
2
=
13 − 7
Po = 3Po
2
3
Po
2
1
a
ri
given
20Po2
1 = RT, T =
4.
K1 =

5.
1
= 12.19 K
.0821
1
K2
Ja
N2O4
2NO2(g)
n = 2 – 1 = 1, KP = KC (given)
We know, KP = KC (RT)n
for same value of T.
K1K2 = 1
similar to yx = const.
An efflorescent salt is one that loss of H2O to atmosphere.
CuSO4.5H2O(s)
CuSO4.3H2O(s) + 2H2O(g)
lp
2.
uh
3
Po
3
2
K2 =
=
2
2
5 
20P
o
 2 Po  ( 2Po )


KP = ( PH' 2O )2 = 4 × 10–4
a
PH' 2O = 2 × 10–2 atm = 15.2 mm Hg
If PH' 2O at 25ºC < 15.2 mm only then reaction will proceed in forward direction.
2AB4(g)
A2(g) + 4B2(g)
Sa
nk
6.

2
1–
2x

n =1+
3
−
2
1
1- − 1
4
    2

 2 P 1 P

 = 8P35
KP = 
( P )2
8.
C(s) + CO2(g)
2CO(g)
2
KP =
(PCO )
'
PCO
2
= 1.0
At equilibrium, PCO2 = KP = 4 × 10–2 remain constant
(PCO )2
4  10 –2
=1
PCO = 4  10 –2 = 0.2 atm
369
9.
P1 = 15 atm
;
T1 = 300 K.
Equilibrium temperature is 300ºC that is 573 K.
So first of all we have to calculate pressure of NH3 at 573 K.
P
P1
15
P
= 2 =
= 2
300
573
T2
T1
P2 = 28.65 atm at 300ºC.
t=0
28.65 atm
t = teq.
3
H2(g).
2
1
N2(g) +
2
NH3 (g)
0
0
3
x
2
x
atm
2
[28.65–x]
But according to question.
Or
+
x
2
+
3
x
2
a
ri
Ptotal = 28.65 – x
28.65 + x = 40.11.
x = 11.46.
Degree of dissociation of NH3 =
11.46
= 0.4.
28.65
D(g) + C(g) Kp = (20)2
20 atm 20 atm
(2) B(s)
E(g) + F(g) Kp = (30)2
30 atm 30 atm
(3) container Kp is same so on doubling the volume partial pressure does not change but moles of C, D,
E, F will change to maintain their partial pressure. So total pressure = 40 + 60 = 100 atm.
(1) A(s)
11.
KP = ( PH2O )5 = 56.25 × 10–10
Ja
uh
10.
( PH2O ) = (56.25 × 10–10)1/5 = (56.25)1/5 × 10–2 = 2.23 × 10–2 × 760 = 17.01 torr
A(s)
X + Y

+
B(s)
Y + Z
+

 KP1 =  ( + )
KP2 =  ( + )
(less than 74.60%)
a
12.
Parital pressure
17.01
× 100 =
× 100 = 74.60%
22.8
Vapour pressure
lp
% Relative humidity =
Sa
nk
Ptotal = ( + ) +  +  = 2 ( + )

2 ( + ) = 50 
 +  = 25

250 = 25 

 = 10 ,  = 15

KP2 =  ( + ) = 15 × 25 = 375
13.
X(s)
Y(g) + 2Z(g)
A(s)
(a+b) 2a
t = eq.
(a+b)
pZ = pB
2a = b
and
pZ + pB = pY + 10
(2a + b) = (a + b) + 10
a = 10 atm

b = 20 atm
= (a + b) (2a)2 = (30) (400) = 12000
t = eq.
Given
KP1
Y(g) + B(g)
b
KP2 = (a + b) (b) = (30) (20) = 600
G = –RTlnK
G1
=
G2
nK1
log12000
3 + log12
=
=
nK 2
log 600
2 + log 6
370
14.
NH4I(s)
NH3(g) + HI(g)
at eq. total pressure = 304 mm of Hg
304
pNH3 = pHI =
2
= 152 mm of Hg = 0.2 atm
2HI(g)
0.2
t=0
t = eq.
H2(g) + I2(g)
0
0
x
2
0.2+y–x
eq.
NH4I(s)
–
NH3(g) + HI(g)
0.2+y 0.2+y–x
x
2
KC = KP =
(x / 2)(x / 2)
(0.2 + y – x)2
= 0.010
a
ri
We assumed that x atm of HI is used for dissociation which results formation of y atm of H I from
NH4I(s)
for
NH4I(s)
NH3(g) + HI(g)
KC = 0.2 × 0.2
in other case KP = (0.2 + y) (0.2 + y – x)
(0.2 + y) (0.2 + y – x) = 0.2 × 0.2
...(i)
For
2HI(g)
H2(g) + I2(g)
...(ii)
uh
on solving the equation :
x = 0.036 & y = 0.016
total pressure at eq. = pNH3 + pHI + pH2 + pI2
= 0.432 atm = 328 mm of Hg
A(s)
B(g) + C(g) ,
E(s)
B(g) + D(g)
(P1 + P2) P1
(P1 + P2) P2
Total pressure = 2(P1 + P2) = 2 Ksp1 + Ksp2 = 20 atm
16.
On addign inert gas at content temperature & pressure 2nd reaction will be shifted in backward direction.
17.
On mixing, P(N2) & P(H2) will add up. This will shift both the equilibrium backwards thereby decreasing
the number of moles.
Ja
15.
No. of mole
conc.
KC =
2
[SO2 ][O2 ]1/ 2
For reaction, H2(g) + I2(g)
For reaction,
20.
1
O2 (g)
2
SO3 (g)
1
2
4
[SO3 ]
=
1
1
H2 +
2
2
I2
4
4
4
1
4
1
(1/ 2) (1/ 4)1/ 2
Sa
nk
19.
+
lp
SO2 (g)
a
18.
=
1
= 4 Ans.
(1/ 2)  (1/ 2)
2HI(g)
K’ =
1
=4
0.25
K’’ = 4
HI(g)
A2(g) + B2(g)
at t = 0
60
42
at eq.
(60–x) (42–x)
(Partial pressures)
 2x = 28

x = 14
14
(28)2
(PAB )2
Kp =
=
=
23
(PA 2 )(PB2 ) (46)(28)
.
2AB(g)
0
(2x)
14
23
Let degree of dissociation for AB is ‘x’ , then

1
23 
=

2AB(g)
A2(g) + B2(g)  K C2 =

K C1 14 

1
0
0
For ng = 0
;
Kp = KC1 =
371
at t = 0 (1 – x)
x
x
 x  x 
 2  2 
   = 23 →
 =0.719
14
(1– x)2
Hence percentage of dissociatin = 0.719 × 100 = 72 %
21.
NH4HS(s)
NH3(g) + H2S(g)
1–0.5
1
3
1
N2(g) + H2(g)
2
2
NH3(g)
0.5
0.25
0.75
(0.251/ 2 ) (0.75)3/2
0.5
KP2 =
2
/ K p2 =
 P(O2) = PCO
Also, KP1 =
PCO2
7
× solution = 7 atm
8
=
PO2
102 atm2
= solutions
125atm
uh
Since (I) causes no change in volume due to reaction change from 100 L is due to (II) only.
 v (CO) finally = 10 L
 10 
P(O) finally = 
  105 = 10 L
 105 
 P(CO2) = KP1. P(O2)
Ja
22.
a
ri
KP1 = 0.5
 P(CO) + P(CO2) finally is 8 + 7 = 15 atm
A(s)
B(g) + D(g)
C(s)
Kp1 = 625
(P1 +P2 )
P1
E(g) +
Kp2 = 975
D(g)
(P1 +P2 )
P2
lp
23.
Kp1 + Kp2 = ( p12 + p 22 + 2p1p2) = (p1 + p2)2 = (625 + 975)  (p1 + p2) = 40
so total pressure = 2[p1 + p2] = 80 atm.
t=0
t = teq.
H2 + Br2
2HBr
0.4
0.2
–
0.2
y
0.4
= negligible = y
a
24.
Sa
nk
0.4  0.4
1
× 1010 =
4
0.2  y

y = 3.2 × 10–10
3.2
Br2
× 1011 =
× 10–10 × 1011 = 80
HBr
0.4
25.
PCl5
2
2–1.6
= 0.4
0.4
5
KC =
PCl3 + Cl2
–
–
1.6
5
1.6
5
1.6
1.6
5
64
4  1.6
=
50
5
KC = a = 2 × 0.8 = 1.6
1.6
1.6 1.6

5
KC = 5
0.4
5
64
Ans. is
× 50 = 64
50
372
26.
A(s)
–
B(g) + C(g)
KP1 = 300
P1
KP1 = P1(P1+P2)
D(s)
(P1+P2)
E(g) + C(g)
P2
KP2 = 600
P1+P2
KP2 = P2(P1+P2)
 Kp1 
P1(P1+P2 )

=
 Kp2  P2 (P1 + P2 )
P 
300
=  1
600
 P2 
P1
1
=
2
P2

KP1 = P1(P1+P2)
KP1 = P1 (P1+2P1)

a
ri
P2 = 2P1
300 = P1 (3P1)
P12
27.
uh
= 100
(P1 = 10)
P2 = 20 atm
Now Total pressure
PB + PE + PC
(P1+P2) + (P1+P2) = (10 + 20) + (30) = 60 atm.
From given reactions.
(i) = – (iii)
1
(i) = – (iv)
2
;
1
(iii) = – (ii)
2
;
Higher is the value of equilibrium const, higher will be the stability of products.
31.
Second equilibrium will not be affected by CO2 addition only first will shift backward.
32.
Since  in temperature results in shifting of equilibria in forward direction hence reaction is endothermic
in nature. Endothermic reaction is driven in forward direction upon increase in temperature.
33.
R.H. > 1 
R.H. < 1 
34.
Addition of solids have no effect on equilibrium and temperature favours endothermic direction while
increasing pressure will shift equilibrium in backward direction as ng is +ve.
PCO(g).PH2 (g)
PH2O(g)
=
Q>K
Q<K
lp
a
KP =


> V.P.
< V.P.
(PH2 (g) )2
PH2O(g)
leftward shift
Rightward shift
PART - IV
(as PCO(g) = PH2 (g) )
Sa
nk
1.
Ja
29.
3.
Mole of PbS = 708 / 236 = 3 mole = mole of H2S
Sb2S3(s) + 3H2(g)
2Sb(s) + 3H2S(g)
initial
6
6
0
0
at eq.
5
3
2
3
KC =
4.
(3 / 10)3  (2 / 10)2
3
(5 / 10)  (3 / 10)
=
4
= 0.08
50
K = A e–H/RT
H
.
2.303RT
H
1
log K = log A –
×
.
T
2.303 R
log K = log A –

H

1
log K =  –
+ log A.
 ×
T
 2.303 R 
– H
= 1.
2.303 R
H = – 2.303 R = – 4.606 cal.
373
aA + bB
cC + dD
At high temp. & low pressure equilibrium is shifting in backward direction. It means (a + b) > (c + d) &
heat will reaction in the formation of producers is H < 0.
6.
Br2() + Cl2(g)
1
(1 – x)
t=0
2BrCl(g)
0
2x
kp =
0.1
2x
nBrCl
=
= 10 =
1− x
0.01
nCl2
then at equilibrium,
10 – 10x = 2x
So,
(PBrCl )2
= 1 so, PCl2 = (PBrCl )2 = 0.01 atm
PCl2
or
x=
10
5
=
12
6
moles
Moles of Br2() required for maintaining vapour pressure of 0.1 atm
a
ri
5.
5
10
moles =
moles = moles of BrCl(g).
6
6
5
Moles required for taking part in reaction = moles of Cl2 used up =
moles.
6
5 10 15
Hence total moles required = +
=
moles.
6 6
6
uh
=2×
As in the gaseous mixture only A will be present so the molecular weight of the gaseous mixture will be
‘MA’

PMA = dRT
where R, P, T are const.
so
d  MA
and hence it does not depend on ‘’
so ‘d’ will remain constant.
Option (D) is correct.
12.
CH3COOH() + C2H5OH()
1 – 0.5
1 – 0.5
KC =
0.5  0.5
=1
0.5  0.5
CH3COOC2H5() + H2O()
0.5
0.5
lp
So,
Ja
7.
Now let a moles of CH3COOH and b moles of C2H5OH are taken :
b–
KC =
a
3
(a / 3)  (a / 3)
a

2a / 3   b − 
3

Sa
nk
So,
a
3
a
a–
13.
NH2COONH4(s)
At eq.
or
a
3
a
3

a


a
2 b −  =
3
3
or
2b = a
or
a
2
=
b
1
2NH3(g) + CO2(g).
PNH3 = 4 atm
PCO2 = 2 atm
2
Kp = PNH
× PCO2 = 32 atm3.
3
Therefore x = 32 & y = 6  x + y = 38.
14.
N2 + 3H2
a
0
a
3a
+a
2
2
teq
a(1 – )
3a
nT = 2a + a = a(2 + ) = 2a + 2 × 2a
2 +  = 2.4

 = 0.4
Ans. 40%
t =0
2NH3
a
374
15.
A2B(g)
1
(1–x)
at t = 0
at eq.
2A(g) + B(g)
0
0
2

(partial pressure)
Hence, (1 – ) ; 1
(2x)2  ()
= 81 × 10–6
(1)
 = 3 × 10–2
percentage of dissociation = 3 ×10–2 × 100 = 3%
PB : PD = 1 : 6
Let the partial pressure of Beq. be P0
=
KP1
KP2
8KP1
3
2
 
=
64
8
=8
CH3COOH + C2H5OH
CH3COC2H5 + H2O
At t = 0
0.1
0.1
0
0
At equilibrium 0.1–x
0.1–x
x
x
Meq of acetic acid left = Meq. of NaOH used = 100 × 0.75 = 75
Milimoles of acetic acid left = 75 ( monobasic)
Moles of acetic acid left = 0.075
0.1 – x = 0.075

x = 0.025
KC =
x2
2
(0.1– x)
=
(0.025)2
(0.075)2
NH4HS(s)
NH3(g) + H2S(g)
p
p
2p = 1.5 – 1
2p = 0.5
p = 0.25
Kp = 0.25 × 0.25 = 0.0625
Sa
nk

= 0.111 = 1.11 × 10–2
a
18.
= 64
3
Ja
17.
63
uh
KP2
a
ri
 3P 
KP1 = (P0)2  0 
 2 
KP2 = (6P0)3
3
lp
16.
19.
n=
20.
(P)
PV
3000  10−3
=
= 1.2 × 10–3 moles.
25
RT
 300
3
Initially, only NH3 was present & according to reaction
2NH3(g)
N2(g) + 3H2(g)
nN2
1
=
3
nH2
pN2
Same volume for both gases 
(Q)
pH2
=
pH
1
 2 =3
3
pN2
When B is closed, only H2 diffuses through filter until partial pressure of H2 becomes equal.
nH2
total
nN2
=3
pHI 2 = pHII2
nHI 2 + nHII2 = nH2
...(1)
&

total

VI = 2VII
nH2
I
total
/ nH2 =
3
2
nHI 2 = 2 nHII2
...(2)
375
from eq.(1) & (2)
=
nN2
(R)
nHI 2
nH2
total
nH2
total
nN2
=
2
×3=2
3
Again same volume for N2 & H2
1
nN2 / nH2 =
total
total
3
I
II
pN2 = pN2
&
VI = 2VII
nN2
total
nH2
total
= nNI 2 + nNII2 = 3 nNII2 =
...(1)

nNI 2 = 2 nNII2
1
3 I
nN / 3 nHII2 =
2 2
3
nHI 2 = 2 nHII2

nNI 2
nHII2
=
2
3
1
3
Sa
nk
a
lp
Ja
uh
I
II
I
II
( nN2 + nN2 )/( nH2 + nH2 ) =
&
3 I
nN
2 2
I
II
II
= nH2 + nH2 = 3 nH2
from eq.(1)
(S)
×
a
ri
nHI 2
376
IONIC EQUILIBRIUM (ELEMENTARY)
JEE(Advanced) Syllabus
Solubility product, common ion effect, pH and buffer solutions; Acids and bases (Bronsted and
Lewis concepts); Hydrolysis of salts.
a
ri
JEE(Main) Syllabus
Ionic equilibrium: Weak and strong electrolytes, ionization of electrolytes, various concepts of
acids and bases (Arrhenius, Bronsted – Lowry and Lewis) and their ionization, acid – base
equilibria (including multistage ionization) and ionization constants, ionization of water, pH
Ja
soluble salts and solubility products, buffer solutions.
uh
scale, common ion effect, hydrolysis of salts and pH of their solutions, solubility of sparingly
Introduction :
Here we deal with the equilibria of species wich are ionic in nature. Generally the concentration of H +
and OH– ions is of importance. Also the concentration of ions forming precipitates is focused upon. As
like dissolves like, the solvent for production of these ions should be polar. It is generally water.
Sa
nk
a
lp
Table-1
377
SA : Strong Acid ; SB : Strong Base ; WA : Weak Acid ; WB : Weak Base
Acid-Base concepts
Arrhenius Concept
D1 :
Arrhenius Acid : Substance which gives H+ ion from its own molecule on dissolving in an ionising
solvent .
Eg : HCl, H2SO4, CH3COOH etc.
Table-2
a
ri
(A)
;
Arrhenius Base : Substance which gives OH– ion from its own molecule on dissolving in an ionising
solvent.
Eg : NaOH, Ba(OH)2, NH4OH etc.
lp
D2 :
;
Ja
Figure-1
uh
 H3BO3 is not Arrhenius acid as it does not give H+ from its own molecule.
 H+ ion in water is extremely hydrated (in form of H3O+, H7O2+, H7O3+) due to its high charge density.
 The structure of solid HClO4 is studied by X-ray. It is found to consist of H3O+ & ClO4– as :
HClO4 + H2O
H3O+ + ClO4–
Table-3
Sa
nk
a
Types of bases
NH3.H2O

ion also exists in hydrated form of H3O2–, H2O4–, H5O3–.
 First group elements of modern periodic table (except Li) form strong bases.
 Insoluble hydroxides - like Fe(OH)3, Mg(OH)2, Cr(OH)3.
OH–
D3 :
D4 :
Bronsted-Lowry concept (Conjugate acid-base concept / Protonic concept)
Bronsted Lowry Acid : Species which donate H+ are Bronsted Lowry acids (H+ donor).
Bronsted Lowry Base : Species which accept H+ are Bronsted Lowry bases (H+ acceptor).
HB+
+
X–
Conjugated acid-base pairs : In a typical acid-base reaction,
Forward reaction : Here HX, being a proton donor is an acid. Here HB, being a proton acceptor is a
base.
378
Backward reaction : Here HB+, being a proton donor is an acid. Here X–, being a proton acceptor is a
base.
Table-4
a
ri
Acid
Base
Conjugate acid Conjugate base
HCl
+
H2O
H3O+
+
Cl–
HSO4– +
NH3
NH4+
+
SO42–
3+
+
[Fe(H2O)6] + H2O
H3O
+
[Fe(H2O)5(OH)]2+
+
 Conjugate acid-base pair differ by only one proton (H ).
 Strong acid will have weak conjugate base. This can be explained as the strong electrolyte will move
the equilibrium where it is in dissociated form.
 Weak acid/base will also have weak conjugate base/acid, because weak electrolyte has tendency to
be in undissociated form.
 Reaction will always proceed in a direction from stronger acid to weaker acid or from stronger base
to weak base.
Some Acid-Base Conjugate Pairs
NH4+
R–NH3+
a
lp
Ja
uh
NH3 or NH3.H2O
Sa
nk
Note : In polyprotic acids, the tendency of release of subsequent H + decreases.
Eg : Order of acidic strength: H3PO4 > H2PO4– > HPO42–.
D5 :
Amphiprotic/Amphoteric Species : Species which can act as an acid as well as a base. Eg : H2O,
NH3.
HCl + H2O
H3O+ + Cl– (H2O acting as base)
NH3 + H2O
NH3 + CH3–
+
–
NH4 + OH
NH2– + CH4
(H2O acting as acid; NH3 acting as base)
(NH3 acting as acid)
Lewis Concept (Electronic Concept)
D6 :
D7 :
Lewis Acid : An Lewis acid is a species which can accept an electron pair with the formation of
coordinate bond.
Acid  Electron pair donor

Acid : Electron pair acceptor.
eg : Electron deficient molecules : BF3, AlCl3, H3BO3 , BeCl2
Cations : H+, Fe3+, Na+
Molecules with vacant orbitals : SiCl4, SO2.
Lewis Base : A Lewis base is a species wich has a lone pair of electrons available for donation.
Base → Electron pair acceptor 
Base : Electron pair donor
..
..
..
eg : Molecules with donatable lone pairs : NH3 , H2 O
, CH3 OH
..
..
Ions : X–, CN–, OH–.
379
Note :
Here, Boric acid [i.e. B(OH) 3] accepts a lone pair. So, it is a lewis acid.
Sol.
In which direction will the following equilibria I and II proceed ?
I : H2SO4 (aq) + NH3 (aq)
NH4+ (aq) + HSO4– (aq)
II : HCO3– (aq) + SO42– (aq)
HSO4– (aq) + CO32– (aq)
(A) I forward & II backward
(B) I backward & II forward
(C) Both forward
(D) Both backward
Equilibrium proceeds in the direction from strong (Acid/Base) to weak (Acid/Base).
I:
a
ri
Ex-1.
Here, acidic strength is compared among the 2 acids in the reaction & basic strength is compared
among 2 bases in that reaction only.
II :
Ex-3.
Sol.
uh
Ja
In the above question, identify the true (T) & false (F) statements if HI is replaced with C 2H5NH2 in each
statement.
(A) T F F F T T
(B) T F F F F T
(C) F F T T F F
(D) F T T T F F
Lewis base donates a lone pair to an electron deficient species. Arrhenious base releases OH –.
Bronsted base accepts H+ from a species.
Ans. (B)
Ammonium ion is :
(A) Lewis acid
(B) Lewis base
NH4+ ⎯⎯
→ NH3 + H+  Bronsted Acid
Sa
nk
Ex-4.
lp
Sol.
For the given reaction, identify the true (T) & false (F) statements.
C2H5 NH2 + H
(C2H5NH3)+ –
S1 : HI is bronsted base.
S2 : HI is bronsted acid.
S3 : HI is arrhenius acid.
S4 : HI is lewis acid.
S5 : HI is arrhenius base.
S6 : HI is lewis base.
(A) T F F F T T
(B) F T T T F F
(C) F T T F F F
(D) T F F F T F
H has donated H+ in the above reaction. So it acts as an Arrhenius acid as well as a Bronsted acid.
Ans. (C)
a
Ex-2.
Ans. (A)
Sol.
Ex-5.
Sol.
Ex-6.
Sol.
(C) Bronsted acid
Ans. (C)
(D) Bronsted base
In which of the following reactions, does NH3 act as an acid ?
(A) NH3 + H+ ⎯⎯
(B) NH3 + H– ⎯⎯
→ NH4+
→ NH2– + H2
(C) NH3 + HCl ⎯⎯
(D) None, as NH3 is a base
→ NH4Cl
In the reaction, NH3 changes to NH2– . So, NH3 has donated a proton (H+) and hence acts as an acid.
Ans. (B)
Sulphanilic acid is a/an :
(A) Arrhenius acid
(B) Lewis base
Sulphanilic acid is
(C) Neither (A) or (B)
(D) Both (A) & (B)
. Its –SO3H group is capable of donating H+, and hence it acts as arrhenius
acid, while –NH2 group’s nitrogen has lone pair of electron which can be donated as :
380
Ans. (D)
Properties of Water, ph scale, autoprotolysis
Properties of water :
a
ri
(B)
Amphoteric Acid / Base Nature :
Water acts as an acid as well as base according to Arrhenius & Bronsted - Lowry theory, but according
to Lewis concept it is a base, not an acid.
H2O + H2O
H3O+ + OH–
+
–
In pure water, [H ] = [OH ], so it is always neutral
F1 :
Ionic product of water :
According to arrhenius concept,
H2O
H+ + OH–
Ionic product of water is defined as :
Kw = [H+] [OH–] = 10–14 at 25°C (experimental data)
 at 25°C pure water contains [H+] = [OH–] = 10–7 M
uh
Molar concentration / Molarity of water :
1000 /18
= 55.55 moles / litre
Molarity = No. of moles / litre =
1
= 55.55 M (taking density = 1 g / cc)
Ja

lp
Dissociation of water is endothermic, so on incresing temperature, K eq increases. So, Kw increases with
increase in temperature & decreases with decrease in temperature.
For eg. at 25° C, Kw = 1 × 10–14 ; at 40°C, Kw = 2.916 × 10–14 ; at 90°C , Kw = 10–13
a
Ionic product of water is always a constant whatever may be dissolved in water. As it is an equilibrium
constant, it will depend only on temperature.
Sa
nk
Der1 : Degree of dissociation of water :
H2O
H+ + OH–   =
no. of moles dissociated
10 −7
=
no. of moles initially taken 55.55
 = 1.8 × 10–9 or 1.8 × 10–7 % (at 25ºC)
Der2 : Absolute dissociation constant of water :
[H+ ][OH− ] 10−7  10−7
H2O
H+ + OH–  K a = K b =
=
= 1.8 × 10–16
[H2O]
55.55
So, pKa of H2O = pKb of H2O = – log(1.8 × 10–16) = 15.74 (at 25ºC)

pH Scale :
Acidic strength means the tendency of an acid to give H3O+ or H+ ions in water.
So greater the tendency to give H+, more will be the acidic strength of the substance.

Basic strength means the tendency of a base to give OH – ions in water.
So greater the tendency to give OH– ions, more will be basic strength of the substance.
D8 :
The concentration of H+ ions is written in a simplified form introduced by Sorenson known as pH scale.
pH is
defined as negative logarithm of activity of H+ ions.

pH = – log aH+ (where aH+ is the activity of H+ ions)
381

Activity of H+ ions is the concentration of free H+ ions or H3O+ ions in a dilute solution.

The pH scale was marked from 0 to 14 with central point at 7 at 25°C taking water as solvent.
1
pH = – log [H+] or pH = log + or pH = – log[H3O+]
[H ]
F2 :
1
[OH− ]
Der3 : From, Kw = [H+] [OH–] = 10–14 (at 25°C)
Taking negative log both sides,
– log [H+] – log[OH–] = – log (10–14) = – log kw = 14

pH + pOH = pKw = 14 (for an aqueous solution at 25°C)
pOH = – log [OH–] or pOH = log

If the temperature and the solvent are changed, the pH range of the scale will also change.
For example :
0 – 14
at 25°C
Neutral point pH = 7
0 –13
at 80°C (Kw = 10–13)
Neutral point pH = 6.5
pH can also be negative or greater than 14.
Now, pH = – log [H+] = 7 and pOH = – log [OH–] = 7,
for water at 25°C (experimental)
pH = 7 = pOH

neutral
pH < 7 or pOH > 7

acidic
pH > 7 or pOH < 7

basic


 at 25 C only


uh


a
ri
F3 :
Autoprotolysis :
Some substances like HCOOH, NH3 etc are observed to be self ionised in pure liquid state as follows :
2HCOOH ()
HCOO– + HCOOH2+
This phenomena is called Self-Ionisation or Autoprotolysis and equilibrium constant corresponding to
above is called Autoprotolysis constant defined as follows :
Keq = [HCOO–] [HCOOH2+]
(C)
Relation between Ka and kb for conjugate acid-base pair
lp
Ja
D9 :
Der4 : For any conjugate acid-base pair (HA & A–) in aqueous solution :
[H O+ ][A − ]
HA + H2O
A– + H3O–
:
Ka = 3
[HA]
Now,
HA + OH–
a
A– + H2O
:
Kb =
[HA][OH− ]
[A − ]
Ka × Kb = Kw  pKa + pKb = pKw = 14 (at 25°C)
Sa
nk
Eg. pKa (CH3COOH) + pKb (CH3COO–) = pKw = 14 ; pKa(NH4+) + pKb(NH3) = pKw = 14.
Degree of dissociation ()


When an electrolyte is dissolved in a solvent (H2O), it spontaneously dissociates into ions.
It may dissociate partially ( < 1) or sometimes completely (  1)
Eg.
NaCl + aq
Na+ (aq) + Cl– (aq)
(  1)
CH3COOH + aq
CH3COO– (aq) + H+ (aq)
( < 1)
D10 :
The degree of dissociation of an electrolyte () is the fraction of one mole of the electrolyte that has
dissociated under the given conditions.
F4 :
=
No. of moles dissociated
No. of moles taken initially
The value of  depends on :
(a)
Nature of electrolyte: Strong electrolytes dissociate completely ( = 1) whereas weak
electrolytes dissociate partially. (  1)
(b)
Nature of solvent: A solvent having high value of dielectric constant will favour dissociation.
(Generally polar solvents)
(c)
Dilution: For weak electrolytes, degree of dissociation will increase with dilution (Ostwald’s
dilution law)
382
(d)
(e)
F6 :
Ex-7.
Sol.
(a) Strong acid solution :
(i)
If [H+]from strong acid is greater than 10–6 M
In this case, H+ ions coming from water can be neglected.
So, [H+] = Molarity of strong acid solution × number of H+ ions per acid molecule.
(ii)
If [H+]from strong acid is less than 10–6 M
In this case, H+ ions coming from water cannot be neglected.
So, [H+] = [H+] from strong acid + [H+] coming from water in presence of this strong acid.
a
ri
F5 :
pH Calculation : Strong acid SolutioNs, strong base solutions, solutions
containing mixture of two or more strong acids, solutions containing mixture of
two or more strong bases, solutions containing mixture of strong acid and
strong base
Find the pH of :
(a) 10–3 M HNO3 solution
(b) 10–4 M H2SO4 solution (Take log 2 = 0.3)
(a) pH = − log[H+ ]HNO3 = – log (10–3) = 3
Ja
(b) pH = − log[H+ ]H2SO4 = – log (2 × 10–4) = 4 – log 2 = 3.7
uh
(D)
Temperature: On increasing temperature, generally degree of dissociation increases. (For
endothermic dissociations)
Presence of other solute: When a substance is present in a solution, it may effect the
dissociation of another substance. Generally, presence of common ion supresses degree of
dissociation of weak electrolyte. (Common ion effect)
In both solutions, [H+]from strong acid > 10–6 M. So H+ from water has not been considered.
Calculate pH of 10–8 M HCl solution at 25ºC. (Take log 1.05 = 0.02)
Here, [H+]HCl = 10–8 M (< 10–6 M). So [H+] from water has to be considered. But,
[H+]from H2O  10–7 M because of common ion effect exerted on it by H+ ions of HCl. So, considering
dissociation of H2O :
H2O
H+
+
OH¯
10–8 + x x
Kw = [H+] [OH¯]
10–14 = x(x + 10–8)
 x2 + x × 10–8 – 10–14 = 0
1
−10−8 + 10−7 4 +
−8
−16
−14
−8
−10  10 + 4x10
100 = ( 401 − 1)10 = 0.95 x 10–7
x=
=
2
2
2
[H+] = 10.5 x 10–8 = 1.05 x 10–7 M
pH = 7 – log 1.05  6.98
Note : For 10–9 M HCl pH  7, For 10–12 M HCl pH  7
Sa
nk
a
lp
Ex-8.
Sol.
F7 :
F8 :
(b) Strong base solution :
(i)
If [OH–]from strong base is greater than 10–6 M
In this case, OH– ions coming from water can be neglected.
So, [OH–] = Molarity of strong base solution × number of OH– ions per base molecule.
(ii)
If [OH–]from strong base is less than 10–6 M
In this case OH– ions coming from water cannot be neglected.
So, [OH–] = [OH–] from strong base + [OH–] coming from water in presence of this strong base.
Example-9.
Solution.
What will be the pH of 5 × 10–6 M Ba(OH)2 solution at 25ºC ?
[OH–]from strong base = 2(5 × 10–6) = 10–5 M
 pH = 14 – p(OH) = 14 – (–log [OH–]) = 14–(–log 10–5) = 14 – 5 = 9
383
Example-10.
Solution.

Calculate pH of 10–7 M of NaOH solution at 25ºC. (Take log 0.618= 0.21)
[OH–] from NaOH = 10–7 M (< 10–6 M)
[OH–] from water = x (< 10–7 M ; due to common ion effect)
H2O
OH– + H+
–
(x + 10–7) x
+
Kw = [H ] [OH–] = 10–14 = x (x + 10–7)
x2 + 10–7x – 10–14 = 0
5 −1
x=
× 10–7 = 0.618 × 10–7 M = [H+]
( 5 = 2.236)
2
 pH = 7.21
pH OF ACIDS/BASES MIXTURES :
(B)
F10 :
a
ri
uh
F9 :
Mixture of two strong acids :
If V1 volume of a strong acid solution with H+ concentration [H+]1 is mixed with V2 volume of another
strong acid solution with H+ concentration [H+]2, then
moles of H+ ions from -solution = M1V1
moles of H+ ions from -solution = M2 V2
If final H+ ion concentration is [H+]f and final volume is Vf (= V1 + V2), then :
[H+]fVf = [H+]1 V1 + [H+]2 V2
[Dissociation equilibrium of none of these acids will be disturbed as both are strong acids]
[H+ ]1 V1 + [H+ ]2 V2

[H+]f =
V1 + V2
Mixture of two strong bases :
Similar to above calculation,
[OH– ]1 V1 + [OH– ]2 V2
10−14
[OH–]f =
&
[H+]f =
V1 + V2
[OH– ]f
Ja
(A)
Take log 2 = 0.3
Sa
nk

1
400
4
4

=
, [H+]2 V2 =
, H+ ions from water can be neglected
100 1000 1000
1000
[H+]1V1+ [H+]2V2 = 8 × 10–3 and Vf = 0.4 + 0.4 + 0.2 = 1 L
8  10 −3
[H+]f =
= 8 × 10–3 M

pH = 3 – log 8 = 2.1.
1
[H+]1V1 =
a
Sol.
1
1
M H2SO4) + (400 mL,
M HCl) + (200 mL of water).
200
100
lp
Ex-11. Calculate pH of mixture of (400 mL,
Ex-12. 500 mL of 10–5 M NaOH is mixed with 500 mL of 2.5 x 10–5 M of Ba(OH)2. To the resulting solution, 99
L water is added. Calculate pH of final solution. Take log 0.303 = – 0.52.
(500  10 −5 ) + (500  2  2.5  10 −5 )
Sol.
[OH–]f =
= 3 × 10–5 M
1000
Vi = 1 L &
Vf = 100 L
no. of moles of [OH–] in resulting solution = no. of moles of [OH–] in final
3 × 10–5 = [OH–]f × 100
[OH–]f = 3 × 10–7 M (< 10–6 M)
So, OH– ions coming from H2O should also be considered.
H2O
H+
+
OH–
x
(x + 3 × 10–7)
–7
Kw = x (x + 3 × 10 ) = 10–14
 13 − 3 
x= 
 × 10–7 M = [H+]

2


So,
pH = 7 – log 0.303 = 7.52.
384
(C)
Mixture of a strong acid and a strong base :


Der5:
Acid Base neutralisation reaction will take place.
The solution will be acidic or basic, depending on which component has been taken in excess.
If V1 volume of a strong acid solution with H+ concentration [H+] is mixed with V2 volume of a strong
base solution with OH– concentration [OH–], then
Number of moles H+ ions from -solution = [H+]1V1
Number of moles OH– ions from -solution = [OH–]2V2
Ex-13. Calculate pH of mixture of (400 mL,
1 
1



 0.4  50  −  0.4  200  2 


 = 4 × 10–3 M.
[H+]f = 
0.4 + 0.4 + 0.2
So,
pH = 3 – 2log 2 = 2.4.
Ostwald’s Dilution Law, ph calculation: Solutions of weak monoprotic acid,
solutions of weak monoacidic base
Ostwald’s Dilution Law
lp
(E)
1
1
M Ba(OH)2) + (400 mL,
M HCl) + (200 mL of Water)
200
50
Ja
Sol.
uh
a
ri
F11 :
Sa
nk
a
Der6 : For a weak electrolyte A+B– dissolved is water, if  is the degree of dissociation then :
AB (aq)
A+ (aq) + B– (aq)
initial conc
C
0
0
conc. at eq.
C(1 – )
C
C
Then,
[A + ][B – ] C . C
C 2
=
=
F12 : Keq =
= dissociation constant of the weak electrolyte
[AB]
C(1 −  )
(1 − )
(Keq = Ka for weak acid; Kb for weak base)
K eq
F13 : If  is negligible in comparison to unity, then 1 –   1. So Keq = 2 C   =
C
1

concentration

As concentration increases   decreases
F14 :
[A+] = [B–] = C  = CK eq  Upon dilution, C  & Keq remains same, so [ions]  and moles of ions 

At infinite dilution  reaches its maximum value, unity (1). Here, weak electrolyte also starts
behaving like a strong electrolyte.
385
Figure-2
pH calculation : Solutions of weak monoprotic acid, solutions of weak
monoacidic base
uh
a
ri
Weak Acid (monoprotic) solution :

Weak acid does not dissociate 100% therefore we have to calculate the percentage
dissociation using Ka (dissociation constant of the acid).
Der7 : We have to use Ostwald’s Dilution law (as we have been derived earlier)
HA
H+ + OH–
t=0
C
0
0
[H+ ] [OH− ] C  2
t = teq
C(1–)
C C
Ka =
=
...(1)
[HA]
1− 
Assume  << 1  (1 –  )  1  Ka  C2   =
Ka
=
C
on dilution
 So, pH =

C
Ka
is greater than 0.1, solve quadratic equation (1) and get accurate .
C
[H+] = C  & now pH calculation can be done.
lp
Then,
Ex-14. Calculate pH of (a) 10–1 M CH3COOH
Take Ka = 2 ×10–5, at 25ºC.
(a)  =
So,
Ka
=
C
2  10 −5
=
10 −1
2  10−4
(b) 10–3 M CH3COOH (c) 10–6 M CH3COOH
( << 0.1)
a
Sol.
1
(pK a – log C ) (valid if  < 0.1 or 10%)
2
 
and [H+]   pH 
Ka  C
If  obtained from
Note :
(valid if  < 0.1 or 10%)
Ja
[H+] = C = C
F15 :
Ka
C
[H+] = 10 –1 ×
2 ×10 –2  pH = 3 –
1
log 2 = 2.85 Ans.
2
Sa
nk
Ka
2  10 −5
=
= 2  10−2 ( > 0.1)
C
10 −3
So, we have to do the exact calculations
10 −3   2
C 2
Ka =
 2×10–5 =
  = 13.14%
1− 
1− 
[H+] = 10–3 × 0.1314 =1.314 ×10–4
 pH = 4 – log (1.314) 3.8 Ans.
(b)  =
(c) If approximation is used, then  =
2  10 −5
=
10 −6
2
  0.95 or 95%
1− 
[H+] = 0.95 ×10–6 = 9.5 × 10–7 
pH = 7– log (9.5) = 6.022 Ans.
At very low concentration (at infinite dilution), weak electrolyte will be almost 100% dissociated,
so it will behave as strong electrolyte
pH of 10–6 M HCl  pH of 10–6 M CH3COOH  6)
So, we have to do the exact calculations,

20 (> 1; not possible)
2×10–5 = 10–6
386
F16 :
Weak base (monoacidic) solution :
Proceed similarly as done for weak monoprotic acid.
1
 pOH =
(pKb – log C)
(if  < 0.1 or 10%)
2
& then pH = 14 – pOH.
Isohydric solutions:
If the concentration of the common ions in the solution of two electrolytes, for example H+ ion
concentration in two acid solutions HA1 and HA2 or OH– ion concentration in two base solutions B1OH
and B2OH is same, then on mixing them, there is no change in the degree of dissociation of either of
the electrolytes (common ion effect is not exerted by one on other). Such solutions are called isohydric
solutions.
Consider two isohydric solutions of weak acids HA1 and HA2. Let C1 and C2 be their concentration and
1 and 2 be their degree of dissociation. Then,
F17 :
C11 = C22 (on equating the H+ concentrations from both acids).
Relative strength of acids and bases :
a
ri
D11 :
(F)
Similarly, relative strengths of any two weak bases at the same concentration are given by the ratio of
the square-roots of their dissociation constants i.e.,
K b1

Basic strength of BOH1
= 1 =
Basic strength of BOH2 2
K b2
Ja
F19 :
K a1
Acid strength of HA1 1
=
=
Acid strength of HA 2 2
K a2
Salt Hydrolysis, pH calculation : solutions of salt of monoprotic acid and
monoacidic base
Salt Hydrolysis
Hydrolysis : The reaction of an ion with water in which either H 3O+ or OH– is produced, by dissociation
of water molecule.
Salt + Water
acid +base
When acids and bases are mixed so that none of the two is left, then we will have salt solution in water
and we have to calculate pH of salt solution.
When a salt is added to water, the solid salt first dissolves and breaks into ions completely (unless
otherwise specified). The ions of the salt may or may not react with water. The cations on reaction with
water will produce H3O+ ions and the anions on reaction with water will produce OH – ions. Depending
on the extent of hydrolysis and on the amounts of H 3O+ and OH– ions, the solution can be acidic, basic
or neutral. If salt is BA, then :
BA(s) ⎯⎯
→ BA(aq) ⎯⎯
→ B+(aq) + A–(aq)
–
A (aq) + H2O(l)
HA(aq) + OH–(aq)
(anionic hydrolysis)
+
B (aq) + 2H2O(l)
BOH(aq) + H3O+(aq)
(cationic hydrolysis)
Sa
nk
a
D12 :
lp
F18 :
uh
In practice, Ka is used to define the strength only of those acids that are weaker than H 3O+ and Kb is
used to define the strength of only those bases that are weaker than OH –. For two weak acids HA1 and
HA2 of ionisation constant Ka1 and Ka2 respectively at the same concentration C, we have :
ANIONIC HYDROLYSIS
Anions can function as a base on reaction with water and hydrolyse as follows :
A–(aq) + H2O(l)
HA(aq) + OH–(aq)
The extent of hydrolysis of a given anion depends on its basic strength.
(a)
Complete hydrolysis
The anions, which are stronger base than OH– and have conjugate acids weaker than H2O, will show
complete hydrolysis in aqueous medium.
For example : H– + H2O ⎯⎯
NH2– + H2O ⎯⎯
→ H2 + OH– ;
→ NH3 + OH–
(b)
Hydrolysis to a limited extent
The anions, which are weaker base than OH– and have conjugate acids stronger than H 2O but weaker
acid than H3O+, will hydrolyse to a limited extent in aqueous medium.
387
For example :
CN– + H2O
HCN + OH–
–
Other examples are CH3COO–, NO2 , S2– etc.
(c)
No hydrolysis
The anions, which are weaker base than OH– and have conjugate acids stronger than both H2O and
H3O+, do not hydrolyse at all.
For example : Cl– + H2O
HCl + OH–
–
Other examples include HSO4 , NO3–, ClO4– etc.
CATIONIC HYDROLYSIS
Cations can function as acid on reaction with water and hydrolyse as follows :
B+(aq) + 2H2O(l)
BOH(aq) + H3O+(aq)
The extent of hydrolysis of a given cation depends on its acidic strength.
a
ri
(a)
Complete hydrolysis
The cations, which are stronger acids than H3O+ and their conjugate bases are very much weaker than
H2O will show complete hydrolysis.
For example : PH+4 + H2O ⎯⎯
→ H3O+ + PH3
uh
(b)
Hydrolysis to a limited extent
The cations, which weaker acid than H3O+ ion and their conjugate bases are stonger than H 2O but
weaker than OH–, show hydrolysis to a limited extent.
For example : NH4+ + 2H2O
NH4OH + H3O+
+
Other examples are C6H5NH3 , CH3NH3+ etc.
Ja
(c)
No hydrolysis
The cations, which are weaker acid than H3O+ and their conjugate bases are stronger than both H 2O
and
OH–, do not hydrolyze at all. Examples are alkali and alkaline earth metal ions.
For example : Na+ + 2H2O
NaOH + H3O+
pH calculation : Solutions of salt of monoprotic acid and monoacidic base.
lp
There are four types of salts :
(A)
Salt of strong acid and strong base
(C)
Salt of weak acid and strong base
(B)
(D)
Salt of strong acid and weak base
Salt of weak acid and weak base
Salts of first type does not undergo hydrolysis and rest three types undergo hydrolysis.
Currently considering only monoprotic acids & monoacidic bases,
Sa
nk
a
(A)
Salt of strong acid and strong base
Neither of the ions will undergo hydrolysis, so the solution involves only the equilibrium of ionization of
water.
2H2O(l)
H3O+ + OH–
Thus, the pH of solution will be 7 (neutral solution at 25ºC).
(B)
Salt of strong acid and weak base
Examples can be NH4Cl, (NH4)2SO4, C6H5NH3+Cl–
Only the cation will undergo hydrolysis and the solution will be acidic in nature. Cation is considered
responsible for the acidic nature of solution.
Der8 : For example, in the solution of NH4Cl of concentration c, we will have :
NH4+ + H2O
NH4OH + H+
t= 0
c
–
0
0
at eq.
c(1– h)
ch
ch
(h - degree of hydrolysis)
[NH4 OH][H+ ]
Kh=
= hydrolysis constant of the salt
+
[NH4 ]
+
NH4OH
H2O
NH4+
+
OH– ,
H+ + OH–,
[NH4 ][OH− ]
Kb=
[NH4 OH]
Kw = [H+] [OH–]
From above equations, we can get :
Kh × Kb = Kw
388
ch2
ch .ch
=
c(1 − h) (1 − h)
Kh =
F20 :
...(2)
Generally, h << 1  1 – h 1. So we get 

[H+] = ch = Kh  c =

pH = – log [H+] = –
h=
Kh
c
Kw
c
Kb
1
[log Kw – log Kb + log c]
2
1
[pKw – pKb – log c]
(valid if h < 0.1 or 10%)
2
Note : (1) c is the concentration of ion undergoing hydrolysis, not the concentration of salt.
Kh
(2) If h obtained from
is greater than 0.1, solve quadratic equation (2) and get accurate h. Then,
c
[H+] = ch & now pH calculation can be done.

pH =
a
ri
F21 :
Now
or

Ja
NH2CONH3Cl is a salt of SA + WB
10−14
K
Kh = w =
= 6.667 x 10–1
K b 1.5 x 10 −14
10−14
Kh
=
C
1.5 x 10 −14 x 1
h = 0.816 (> 0.1)
So we use actual relation
Ch2
1
Kh =
=
1 − h 1.5
1.5 h2 + h – 1 = 0

h = 0.55
[H+] = ch = 0.55 M
pH = 0.26.
h=
lp
Sol.
uh
Ex-15. Calculate degree of hydrolysis, Kh and pH of 1 M urea hydrochloride solution in water, K b (Urea)
= 1.5 x 10–14 at 25ºC. Consider urea as a monoacidic base. Take log 0.55 = – 0.26.
C = [NH4+] = 0.1 M ( volume got doubled, so concentration must have been halved)
10 −14
1.8  10 −5  0.1
Sa
nk
Sol.
a
Ex-16. Equal volume of 0.2 M NH4OH (or ammonia) and 0.1 M H2SO4 are mixed. Calculate pH of final solution.
Given : Kb of NH3 = 1.8 × 10–5 at 25ºC.
h=
Kh
=
C
 pH = 1/2 {14 – 4.74 + 1} =
(C)
(< 0.1)
10.26
= 5.13
2
Der9 :
Salt of weak acid and strong base
The examples can be CH3COONa, KCN etc.
Proceeding similar to above analysis of salt of weak base & strong acid, we will get :
ch2
ch .ch
Kh × Ka = Kw
&
Kh =
=
c(1 − h) (1 − h)
F22 :
So, h =
Kh
c
[OH–] = ch = Kh  c =
 [H+] =
Kw
=
[OH− ]
Kw
c
Ka
K w  Ka
C
389
pH = – log [H+] = –
F23 :

1
[ log K w + log K a
2
− log c]
1
[pK w + pK a + logc]
(valid if h < 0.1 or 10%)
2
Solution will be basic in nature due to hydrolysis of anion.

pH =
Ex-17. If the equilibrium constant for reaction of HCN with NaOH is 10 10, then calculate pH of 10–3 M NaCN
solution at 25ºC.
HCN + NaOH
H2O + NaCN
CN– + H2O
HCN + OH–
10–3 M
0
0
10–3 (1 – h)
10–3h 10–3h
10 −3 h  10 −3 h
Kh = 10–10 =

10 −3 (1 − h)

t=0
at eq.

h=
Kh
=
c
10−7 (< 0.1)
3
1
1
log 10–10 + log 10–3 = 7 + 5 – = 10.5.
2
2
2
uh
pH = 7 –
K = 1010
Kh = 10–10
a
ri
Sol.
Kh =
Kw
10−14
=
= 5  10 −10
K a 2  10−5
 pH =
Ja
Ex-18. Calculate degree of hydrolysis(h) and pH of solution obtanied by dissolving 0.1 mole of CH 3 COONa in
water to get 100 L of solution. Take Ka of acetic acid = 2 ×10–5 at 25ºC.
0.1
= 1  10−3 M
Sol.
c=
100
h=

Kh
=
c
5  10−10
= 5 × 10–3 = 0.5%
2  10−5
1
1
1
[pKw + pKa + log c] = [14 + 5 – log 2 + log 10–3] = [15.7] = 7.85.
2
2
2
CH3COO– + H+,
Sa
nk
CH3COOH
a
lp
(D)
Salt of weak acid and weak base
Examples can be CH3COONH4, NH4CN etc.
Der10 :
CH3COO– + NH4+ + H2O
CH3COOH + NH4OH
t=0
c
c
0
0
at eq. c – ch
c – ch
ch
ch
[CH3 COOH] [NH4 OH]
Kh =
+
[CH3 COO− ] [NH4 ]
NH4OH
So,
F24 :
[CH3 COO− ][H+ ]
[CH3 COOH]
.....(ii)
+
[NH4 ][OH− ]
[NH4 OH]
Kw = [H+] [OH–]
NH4+ + OH–,
Kb =
H2O
H+ + OH–,
Kh × Ka × Kb = Kw ,
ch . ch
 h 
=
c(1 − h) . c(1 − h)  1 − h 
 h 

 1– h  = Kh


From (ii) equation,

Ka =
.....(i)
......(iii)
......(iv)
2
Kh =
[H+] = Ka
Kw
[CH3 COOH]
h
ch
= Ka
= Ka ×
= Ka × Kh = Ka ×
=
−
1− h
K a K b
c(1 − h)
[CH3 COO ]
K w  Ka
Kb
390
F25 :



1
[pKw + pKa – pKb ]
2
This formula is always valid for any Ka and Kb at any temperature, for any h.
pH is independent of concentration of salt solution.
Even if Ka of weak acid  Kb of week base, degree of hydrolysis of cations & anions are very close to
each other when they are getting hydrolysed in presence of each other. So, for numerical analysis, they
are taken same.
pH = – log [H+] =


Kw
=
Ka  Kb
10−14
=1
5  10−10  2  10 −5
2h = 1
1
h = = 0.5
2
Table-5
Κh =
Kw
Ka
Kw
Kb
lp
(i) Salt of weak acid and strong base
(ii) Salt of strong acid and weak base
Κh =
Kh =
Kw
K aK b
K 
h =  h
 C
Expression for pH
pH =
C]
(h < 0.1)
K 
h =  h
 C
pH =
1
[pKw – pKb– log
2
C]
(h < 0.1)
h
= (K h )
1− h
1
[pKw + pKa + log
2
pH =
1
[pKw+ pKa– pKb]
2
Sa
nk
a
(iii) Salt of weak acid and weak base
Expression for h
Ja
Expression for
Kh
Types of salt
uh
h
= Kh =
1− h
a
ri
Ex-19. Calcluate pH and degree of hydrolysis of 10–2 M NH4CN solution.
Given that Ka of HCN = 5 × 10–10 and Kb of (aq.NH3) = 2 × 10–5 at 25ºC.
1
1
1
Sol.
pH =
[14+ pKa – pKb] =
[14 + 10 – log 5 – 5 + log 2 ] =
[18.6] = 9.3
2
2
2
(A)
Buffer Solution: Definition and identification
D13 :
Buffer solutions are those solutions which resist a change in pH upon addition of small amount
of small amount of acid or base.
This does not mean that the pH will not change, all it means is that the pH change would be less than
the change that would have occurred had it not been a buffer.
There are various types of buffers :
(i)
Buffer of a weak acid and its salt with a strong base : Can be prepared by
(a) Mixing weak acid solution and solution of its salt with a strong base.
(b) Mixing weak acid solution and lesser amount of strong base solution than that required for
neutralization.
(c) Mixing salt solution of a weak acid and strong base with lesser amount of strong acid
solution than that required for complete reaction.
Eg. A solution containing CH3COOH & CH3COONa.
(ii)
Buffer of a weak base and its salt with a strong acid : Can be prepared by
(a) Mixing weak base solution and solution of its salt with a strong acid.
391
(iii)
(B)
(b) Mixing weak base solution and lesser amount of strong acid solution than that required for
neutralization
(c) Mixing salt solution of a weak base and strong acid with lesser amount of strong base
solution than that required for complete reaction.
Eg. A solution containing NH4OH & NH4Cl.
Solution of salt of a weak acid and a weak base :
Eg. A solution of CH3COONH4.
pH calculation: Buffer solutions generated from monobasic acid /
monoacidic base
CH3COOH
CH3COO¯ + H+
t=0
C1
C2
0
t = eq C1(1 – )
C2 + C1
C1
–
+
[CH3COO ] [H ] (C2 + C1 ) C1
Ka =
=
[CH3CO2H]
C1(1 −  )
...(1)
a
ri
Der11: To calculate the pH of a buffer solution consisting of a weak acid (CH 3COOH; C1 concentration) and its
salt with a strong base (CH3COONa; C2 concentration of anion), we have :
[H+] = C1 = Ka ×
C1
C2
Taking log of both sides,
[Acid]
[Anion of Salt]
[Anion of Salt]
[Acid]
This is known as the Henderson’s equation of a buffer.
NOTE : If  from (2) comes greater than 0.1, calculate exact  by solving quadratic (1) & then [H+] =
C1.
For a buffer made up of weak base and its salt with a strong acid, the Henderson’s equation looks like
this:
[Cation of Salt]
pOH = pKb + log
[Base]

pH = pKa + log
a
F27 :
log [H+] = log Ka + log
lp
F26 :
Ja

uh
Expecting  << 1 (due to common ion effect exerted by CH3COO– on dissociation of CH3COOH),
K
= a
...(2)
C2
Sa
nk
Ex-20. Calculate the amount of (NH4)2SO4 in grams which must be added to 500 ml of 0.2 M NH3 to give a
solution of pH = 9.3. Given pKb for NH3 = 4.7
Sol.
This is a buffer solution made up of weak base and its salt with a strong acid. On checking  (refer
derivation), it comes less than 0.1 (can be considered negligible).
[Conjugate acid]

pOH = pKb + log
(Cation of salt here is same as conjugate acid)
[Base]
x
 x = [NH+4 ] = 0.2, so concentration of (NH4)2SO4 required = 0.1 M
0.2
moles of (NH4)2SO4 needed = 0.1 × 0.5 = 0.05
weight of (NH4)2SO4 needed = 132 × 0.05 = 6.6 g
4.7 = 4.7 + log


392
Ex-21. Calculate [H+] in a 0.20 M solution of dichloroacetic acid (Ka= 5
dichloroacetate. Neglect hydrolysis of sodium salt.
Sol.
CHCl2COOH
CHCl2COO– +
Before dissociation
0.2
–
After dissociation
(0.2 – x)
x
CHCl2COONa ⎯⎯
→ CHCl2COO– +
0.1
For the dissociation of acid
K a = 5  10−2 =
[CHCl2COO− ] [H+ ]
[CHCH2COOH]
or
× 10–2) that also contains 0.1 M sodium
H+
–
x
Na+
0.1
0.05 =
[0.1 + x] [x]
[0.2 − x]
a
ri
x = 0.05 or [H+] = 0.05 M.
Solubility, Solubility Product and simple solubility calculations
F28 :
Solubility product (Ksp) is a type of equilibrium constant, so will be dependent only on
temperature for a particular salt.
Following examples will illustrate the different type of solubilities and the effects of different factors or
situations on solubility of a salt.
Simple solubility
Let salt AxBy be dissolved in water. Let its solubility in H2O = ‘s’ M. Then :
AxBy
xAy+ + yB–x
–
xs
ys

Ksp = (xs)x (ys)y = xx.yy.(s)x+y
Ja
uh
(A)
Ex-22. Calculate Ksp of Fe4[Fe(CN)6]3 at a particular temperature, where solubility in water = s mol/L
Ksp = 44.33.(s)3+4 = 6912 s7
(B)
Condition of precipitation, common ion effect on solubility
Condition of precipitation


Consider ionic product (KIP) similar to reaction quotient Q in chemical equilibrium.
For precipitation, ionic product (KIP) should be greater than solubility product Ksp. This will make the
equilibrium of undissolved salt and dissolved salt shift in backward direction leading to precipitation.
After precipitation, solution will become saturated and KIP = Ksp.
Remember to modify the concentration of the precipitating ions because of volume change occouring
upon mixing both solutions.
a
Sa
nk


lp
Sol.
Ex-23. You are given 10–5 M NaCl solution and 10–8 M AgNO3 solution. They are mixed in 1:1 volume ratio.
Predict whether AgCl will be precipitated or not, if solubility product (Ksp) of AgCl = 10 –10.
Sol.
Upon mixing equal volumes, volume of solution will get doubled. So concentration of each ion will get
halved from original value.
10−5 10−8
Ionic product KIP = ([Ag+] [Cl–])upon mixing =
×
= 25 × 10–15 (< Ksp)
2
2
Hence, no precipitation will take place.
Common Ion effect on solubility


Because of the presence of common ion in solution, the solubility of the sparingly soluble salt generally
decreases.
Neglect the concentration of common ion coming from sparingly soluble salt with respect to that coming
from completely soluble salt.
393
Ex-24. Calculate solubility of silver oxalate in 10–2 M potassium oxalate solution. Given that Ksp of silver oxalate
= 10–10.
Sol.
Let the solubility be x mol/L
Ag2C2O4
t = teq
–
2Ag+ + C2O42–
2x
x+10–2 (  10–2)
10−8

= x2
22

x = 5 × 10–5 M
Sa
nk
a
lp
Ja
uh
a
ri
Ksp = 10–10 = 10–2 × (2x)2
394
Check List
F10
F11
F12
F13
Formulae (F)
Ionic Product of Water (Kw)
pH
pOH

[H+] in strong acid solution (H+ from H2O negligible)
[H+] in strong acid solution (H+ from H2O significant)
[OH–] in strong base solution (OH– from H2O negligible)
[OH–] in strong base solution (OH– from H2O significant)
[H+] in a solution containing mixture of two
strong acids
[OH–] in a solution containing mixture of two strong bases
[H+]/[OH–] in a solution containing mixture of a strong acid & a
strong base
Keq for a weak electrolyte (in terms of C & )
 for weak electrolyte (if < 0.1)
F15
F16
F17
F18
F19
F20
F21
F22
F23
F24
F25
F26
F27
F28
Concentrations of ions produced by weak electrolyte
(with  < 0.1)
[H+] & pH of a weak monoprotic acid solution
pOH of a weak monoacidic base solution
Condition for two solution to be isohydric
Relative strength of two weak acids
Relative strength of two weak bases
Degree of hydrolysis (h) of a salt of WB & SA (< 0.1)
pH of a solution of a salt of WB & SA (h < 0.1)
Degree of hydrolysis (h) of a salt of WA & SB (< 0.1)
pH of a solution of a salt of WA & SB (h < 0.1)
Degree of hydrolysis (h) of a salt of WA & WB
pH of a solution of a salt of WA & WB
pH of a buffer solution of WA & its conjugate base
pH of a buffer solution of WB & its conjugate acid
Relation between solubility(s) & solubility product (Ksp) for
sparingly soluble salts
a
ri
F14
Derivation (Der)
Der1
Der2
Der3
Der4
Der5
Der6
Der7
Der8
Der9
Der10
Der11
Absolute dissociation constant of water (Ka or Kb)
Relation between pH & pOH for an aqueous solution
Relation between pKa & pKb for a conjugate acid base pair
[H+]/[OH–] in a solution containing mixture of SA & SB
Ostwald’s Dilution Law
Equilibrium of a weak monoprotic acid
Hydrolysis of cation in a salt solution of WB & SA
Hydrolysis of anion in a salt solution of WA & SB
Hydrolysis of cation & anion in a salt solution of WA & WB
pH calculation of a buffer solution consisting of a weak acid &
its salt with SB
uh
F1
F2
F3
F4
F5
F6
F7
F8
F9
Definitions (D)
Arrhenius Acid
Arrhenius Base
Brönsted–Lowry Acid
Brönsted–Lowry Base
Amphiprotic species
Lewis Acid
Lewis Base
pH & pH scale
Autoprotolysis
Degree of dissociation ()
Isohydric Solutions
Hydrolysis
Buffer solution
Ja
D1
D2
D3
D4
D5
D6
D7
D8
D9
D10
D11
D12
D13
MISCELLANEOUS SOLVED PROBLEMS (MSPs)
Ans.
Sol.
Which of the following is the strongest base?
–
(A) C2H5–
(B) C2H5COO
(C) C2H5O–
(A)
Acidic strength, C2H6 < C2H5OH < H2O < C2H5COOH
Weakest acid will have strongest conjugate base.
(D) OH–
lp
1.
A solution of HCI has a pH = 5. If one mL of it is diluted to 1 litre, what will be pH of resulting solution.
[HCI]i = 10–5 M
since pH = 5
Since volume of original solution has been made 1000 times, so concentration of solution will decrease
by 1000 times.

[HCl]f = 10–8 M. So H+ from water should also be considered (as done in solved example-8)
Then, pH of resulting solution = 6.96
3.
Calculate the pH of 0.001 M HOCl having 25% dissociation. Also calculate dissociation constant of the
acid. Take log 2 = 0.3
HOCl
H+ + OCl–
t=0
a
0
0
t=eq
a – a
a
a
25
So,
[H+] = a = 10–3 ×
= 2.5 × 10–4
100
So,
pH = 3.6
1
(a ) (a)
a 2
Now, Ka =
=
=
× 10–3
a (1 −  )
1 −  12
Sa
nk
a
2.
Sol.
Sol.
4.
Sol.
The solubility product of SrF2 in water is 8×10–10. Calculate its solubility in 0.1 M NaF aqueous solution.
Ksp = [Sr2+] [F–]2
8 × 10–10 = s[2s + 0.1]2 = s[0.1]2 (neglecting the F– coming from sparingly soluble salt SrF2)
8  10 −10

s=
= 8 × 10–8 M
(0.1)2
395
IONIC EQUILIBRIUM-I
 Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
Section (A) : Acid-Base Concepts
Commit to memory :
uh
(a) Select Polyprotic Arrhenius acids from the following : H3PO2, H3PO3, H3BO3, HCOOH, (COOH)2.
(b) Write conjugate acids of
SO42–, RNH2, NH2–, C2H5OC2 H5, F–
(c) Write conjugate base of
HNO2, OH–, H2CO3, HClO4.
(d) Write conjugate acid and conjugate base of following amphoteric species :
HS–, NH3, C2H5OH, H2O
Ja
A-1.
a
ri
Polyprotic Arrhenius acid : 2 or more replaceable H per acid molecule.
Polyprotic Arrhenius base : 2 or more replaceable OH per base molecule.
Acid – 1 H+ = its Conjugate Base
Base + 1 H+ = its Conjugate Acid
Amphiprotic species : H+ donor as well as acceptor.
Lewis acid : Lone pair acceptor
Lewis base : Lone pair donor
..
(e) Classify the following into Lewis acid & Lewis base : H+, FeCl3, (CH3)3N, F–, CH2
Classify the following into acid, base and amphiprotic species on the basis of protonic concept :
(i) H2PO2–
(ii) H2PO3–
(iii) H2PO4–
(iv) HPO32–
(v) HPO42–
(vi) NH4+
(vii) CH3COOH2+
A-3.
Comment upon H2O as an Arrhenius acid/base, Bronsted–Lowry acid/base and Lewis acid/base.
lp
A-2.
Section (B) : Properties of water, pH scale, Autoprotolysis
Commit to memory :
Sa
nk
a
Kw = [H+] [OH–] = 10–14 at 25°C
[H+] = [OH–] (in water / a neutral solution)
pH = –log [H+] ; pOH = –log [OH–]
For H2O, pH & pOH  with  in temperature
pH + pOH = pKw = 14 (for an aqueous solution at 25°C)
Kself ionisation of HA = [A–] [H2A+]
B-1.
At –50°C, liquid NH3 has ionic product is 10–30. How many amide (NH2–) ions are present per mm3 in
pure liquid NH3 ? (Take NA = 6 × 1023)
Section (C) : Relation between Ka and Kb for conjugate acid - base pair
Commit to memory :
Ka (acid) × Kb (conjugate base) = Kw
pKa (acid) + pKb (conjugate base) = pKw = 14 (at 25°C)
C-1.
What is ionisation constant of HOCl, if Kb of OCl– = 4 × 10–10 ? Also find its pKa.
C-2.
Ka1 , K a2 and K a3 values for H3PO4 are 10–3, 10–8 and 10–12 respectively. If Kw (H2O) = 10–14, then :
(i) What is dissociation constant of HPO42– ?
(ii) What is Kb of HPO42– ?
(iii) What is Kb of H2PO4– ?
(iv) What is order of Kb of PO43– ( Kb 3 ), HPO42–( Kb 2 ) and H2PO4–( K b 1 ) ?
396
Section (D) : pH calculation : Strong acid solutions, Strong base solutions, Solutions
containing mixture of two or more strong acids, Solutions containing mixture of two or
more strong bases, Solutions containing mixture of strong acid and strong base
Commit to memory :
Strong acid solution: [H+] = Molarity of strong acid solution × number of H+ ions per acid molecule.
Strong base solution: [OH–] = Molarity of strong base solution × number of OH– ions per base molecule.
[H+ ]1 V1 + [H+ ]2 V2
Mixture of two strong acids :
[H+]f =
V1 + V2
[OH–]f =
[OH– ]1 V1 + [OH– ]2 V2
V1 + V2
a
ri
Mixture of two strong bases :
D-1.
Calculate pH of following solutions :
(i) 0.001 M HNO3,
(ii) 0.005 M H2SO4 ,
(iv) 10–8 M NaOH,
(v) 0.0008 M Ba(OH)2 .
uh
Mixture of a strong acid and a strong base :
(iii) 0.01 M KOH
Calculate the pH of the following solutions :
(i) 2.21 g of TOH dissolved in water to give 2 litre of solution. (Assume TOH to be a strong base)
(ii) 0.49% w/v H2SO4 solution
M
(iii)
Sr(OH)2 solution is diluted to quadruple volume.
1000
(iv) 1 mL of 12 M HCl is diluted with water to obtain 1 litre of solution.
D-3.
Calculate the pH of solution obtained by mixing 10 mL of 0.2 M HCl and 40 mL of 0.1 M H 2SO4.
D-4.
Calculate the pH of the resulting solution formed by mixing the following solutions :
(a) 20 mL of 0.2 M Ba(OH)2 + 30 mL of 0.1 M HCl
(b) 2 mL of 0.1 M HCl + 10 mL of 0.01 M Sr(OH)2
(c) 10 mL of 0.1 M H2SO4 + 10 mL of 0.1 M KOH.
a
lp
Ja
D-2.
Sa
nk
Section (E) : Ostwald dilution law, pH calculation : Solutions of weak monoprotic acid,
Solutions of weak monoacidic base
Commit to memory :
AB (aq)
=
K eq
C
A+ (aq) + B– (aq) ; Keq =
; [A+] = [B–] = C  =
CK eq
Weak Acid (monoprotic) solution :  =
Weak base (monoacidic) solution :  =
[A + ][B – ] C . C
C 2
=
=
[AB]
C(1 −  )
(1 − )
(both valid if  < 0.1 or 10%)
Ka
1
; pH = (pKa – log C) (both valid if  < 0.1 or 10%)
2
C
Kb
1
; pOH = (pKb – log C) (both valid if  < 0.1 or 10%)
2
C
K a1 Basic strength of BOH1 1
K b1
Acid strength of HA1 1
=
=
=
=
;
Acid strength of HA 2 2
K a2 Basic strength of BOH2 2
K b2
E-1.
Acetic acid gets 1.3% ionised in its decimolar solution. What is be the ionisation constant of acetic acid?
397
E-2.
Prove that degree of dissociation of a weak monoprotic acid is given by :
1
=
1 + 10(pKa −pH)
where Ka is its dissociation constant.
E-3. Calculate the pH of a 500 mL solution of 1 M BOH. (Kb = 2.5 × 10–5)
E-4.
Whose pH increases by greater value on dilution from initial pH = 2 ?
(a) CH3COOH solution (b) HCl solution.
Section (F) : Salt hydrolysis, pH calculation : Solutions of salt of monoprotic acid and
Salt of strong acid and weak base : Kh × Kb = Kw ; h =
a
ri
monoacidic base
Commit to memory :
1
Kh
; pH = [pKw–pKb–log c] (valid if h < 0.1 or 10%)
2
c
1
Kh
; pH = [pKw+pKa+log c] (valid if h < 0.1 or 10%)
2
c
1
 h 
= K h ; pH = [pKw+pKa–pKb]
Salt of weak acid and weak base : Kh × Ka × Kb = Kw ; 

 1– h 
2
uh
Salt of strong base and weak acid : Kh × Ka = Kw ; h =
Which of the following ions or compounds in a solution tend to produce an acidic, a basic or a neutral
solution ?
(a) C2H5O–
(b) Cu+2
(c) SO32–
(d) F–
(e) NH4+
(f) CH3COONa
(g) KNO3
(h) NaOCl
(i) Na2CO3
(j) ZnCl2
F-2.
Calculate pH of 0.2 M aqueous solution of sodium butyrate. Given : Ka of butyric acid = 2 × 10–5.
F-3.
A 0.25 M solution of pyridinium chloride C5H5NH+Cl− was found to have a pH of 2.75. What is Kb for
pyridine, C5H5N ?
Ja
F-1.
lp
F-4. Calculate the percentage hydrolysis & the pH of 0.02 M CH 3COONH4. Kb(NH3) = 1.6 × 10–5,
Ka(CH3COOH) = 1.6 × 10–5.
PART - II : ONLY ONE OPTION CORRECT TYPE
Commit to memory :
a
Section (A) : Acid-Base Concepts
Sa
nk
Polyprotic Arrhenius acid : 2 or more replaceable H per acid molecule.
Polyprotic Arrhenius base : 2 or more replaceable OH per base molecule.
Acid – 1 H+ = its Conjugate Base
Base + 1 H+ = its Conjugate Acid
Amphiprotic species : H+ donor as well as acceptor.
Lewis acid : Lone pair acceptor
Lewis base : Lone pair donor
A-1. An acid with molelcular formula C7H6O3 forms three types of sodium salts. i.e. C 7H5O3Na, C7H4O3Na2
and C7H3O3Na3. The basicity of the acid is :
(A) One
(B) Two
(C) Three
(D) Six
A-2.
A-3.
Select the incorrect option :
(A) H3PO4 is a tribasic acid.
(C) Sr(OH)2 is a strong diacidic base.
In the reaction HC2O4– (aq) + PO43–(aq)
bases ?
(A) HC2O4– and PO43–
(C) HC2O4– and HPO42–
(B) H3BO3 is not an Arhenius acid.
(D) NH3.H2O is a strong monoacidic base.
HPO42–(aq) + C2O42–(aq), which are the two Bronsted
(B) HPO42– and C2O42–
(D) PO43– and C2O42–
398
A-4.
The following equilibrium is established when HClO4 is dissolved in weak acid HF solvent :
HF + HClO4
ClO4– + H2F+
Which of the following is correct set of conjugate acid base pair ?
(A) HF and HClO4
(B) HF and ClO4–
(C) HF and H2F+
(D) HClO4 & H2F+
A-5.
Which of the following correctly explains the nature of boric acid in aqueous medium ?
2H2O
H2O
(A) H3BO3 ⎯⎯⎯
(B) H3BO3 ⎯⎯⎯
→ H3O+ + H2BO3–
→ 2H3O+ + HBO32−
3H2O
(C) H3BO3 ⎯⎯⎯
→ 3H3O+ + BO33−
H2O
(D) H3BO3 ⎯⎯⎯
→ B(OH)4– + H+
Section (B) : Properties of water, pH scale, Autoprotolysis
Kw = [H+] [OH–] = 10–14 at 25°C
[H+] = [OH–] (in water / a neutral solution)
pH = –log [H+] ; pOH = –log [OH–]
For H2O, pH & pOH  with  in temperature
pH + pOH = pKw = 14 (for an aqueous solution at 25°C)
Kself ionisation of HA = [A–] [H2A+]
(B) [H+] >
K W & [OH−] < K W for an acidic solution.
(C) [H+] <
K W & [OH−] >
−
K W for an alkaline solution.
−7
= [OH ] = 10 M for a neutral solution at all temperatures .
Ja
(D)
[H+]
uh
B-1. Which of the following expression is not true ?
(A) [H+] = [OH−] = K W for a neutral solution at all temperatures.
a
ri
Commit to memory :
pOH of H2O is 7 at 298 K. If water is heated to 350 K, which of the following statement should be true ?
(A) pOH will decrease.
(B) pH will increase.
(C) pOH will remain 7.
(D) Both (A) and (B).
B-3.
Kw of H2O at 373 K is 1 × 10–12. Identify, which of the following is incorrect :
(A) pH + pOH = 12, for every aqueous solutions.
(B) pH of H2O is 6.
(C) H2O has increased from its value at 298 K.
lp
B-2.
(D) H2O is acidic.
In pure HCOOH liquid, concentration of HCOO– = 10–3 M at 27ºC. What is the self ionisation constant at
27ºC (K = [HCOOH2+] [HCOO–]) ?
(A) 10–3
(B) 103
(C) 106
(D) 10–6
a
B-4.
Sa
nk
Section (C) : Relation between Ka and Kb for conjugate acid-base pair
Commit to memory :
Ka (acid) × Kb (conjugate base) = Kw
pKa (acid) + pKb (conjugate base) = pKw = 14 (at 25°C)
C-1.
Given %
HF + H2O
Ka
H3O+ + F–
Kb
F– + H2O
HF + OH–
Which relation is correct.
1
(A) Kb =
(B) Ka . Kb = Kw
Ka
C-2.
(C) Ka.Kb.Kw = 1
(D)
Ka
= Kw
Kb
Which of the following is incorrect ?
(A) Ka (weak acid). Kb (conjugate weak base) = Kw
(B) Ka (strong acid). Kb (conjugate weak base) = Kw
(C) Ka (weak acid). Kb (weak base) = Kw
(D) Ka (weak acid). Kb (conjugate strong base) = Kw
399
Section (D) : pH calculation : Strong acid solutions, Strong base solutions, Solutions
containing mixture of two or more strong acids, Solutions containing mixture of
two or more strong bases, Solutions containing mixture of strong acid and
strong base
Commit to memory :
Mixture of two strong bases :
[OH–]f =
[OH– ]1 V1 + [OH– ]2 V2
V1 + V2
The [OH–] in 100 mL of 0.016 M HCl (aq) is :
(A) 6.25 × 10–12 M
(B) 3 × 10–10 M
(C) 6.25 × 10–13 M
Ja
D-1.
uh
Mixture of a strong acid and a strong base :
a
ri
Strong acid solution : [H+] = Molarity of strong acid solution × number of H+ ions per acid molecule.
Strong base solution : [OH–] = Molarity of strong base solution × number of OH – ions per base
molecule.
[H+ ]1 V1 + [H+ ]2 V2
Mixture of two strong acids :
[H+]f =
V1 + V2
(D) 1.6 × 10–3 M
D-2. How many moles of NaOH must be removed from one litre of its aqueous solution to change its pH
from 12 to 11 ?
(A) 0.009
(B) 0.01
(C) 0.09
(D) 0.1
Which statement/relationship is correct ?
(A) pH of aqueous solutions of 0.1 M HNO3, 0.1M HCl, 0.1M H at 25ºC is not equal.
1
(B) For a dilute solution, pH = – log +
[H ]
(C) At 25°C, the pH of pure water is 7.
(D) The value of pKw at 25°C is 7.
D-4.
Upon mixing equal volume of two solutions of strong acids having pH values 2 & 4, resulting solution
will have pH :
(A) equal to 3
(B) closer to 2 than 4
(C) closer to 4 than 2
(D) closer to 3 than 2
Sa
nk
a
lp
D-3.
D-5.
On adding 0.04 g solid NaOH to a 100 mL,
(A) 0
D-6.
(B) +0.3
M
Ba(OH)2 solution, determine change in pH :
200
(C) –0.3
(D) +0.7
Upon mixing equal volume of a strong acid solution (HA) and a strong base (BOH) solution, pH of
resulting solution :
(A) may be less than 7
(B) may be greater than 7
(C) will be equal to 7
(D) Both (A) & (B)
D-7. Which of the following solutions will have pH close to 1 ?
(A) 100 mL of M/10 HCl + 100 mL of M/10 NaOH
(B) 55 mL of M/10 HCl + 45 mL of M/10 NaOH
(C) 10 mL of M/10 HCl + 90 mL of M/10 NaOH
(D) 75 mL of M/5 HCl + 25 mL of M/5 NaOH.
400
Section (E) : Ostwald dilution law, pH calculation : Solutions of weak monoprotic acid,
Solutions of weak monoacidic base
Commit to memory :
AB (aq)
=
K eq
C
A+ (aq) + B– (aq) ; Keq =
[A + ][B – ] C . C
C 2
=
=
[AB]
C(1 −  )
(1 − )
; [A+] = [B–] = C  = CK eq (both valid if  < 0.1 or 10%)
Ka
1
; pH =
(pKa – log C) (both valid if  < 0.1 or 10%)
2
C
Weak base (monoacidic) solution :  =
Kb
1
; pOH = (pKb – log C) (both valid if  < 0.1 or 10%)
2
C
a
ri
Weak Acid (monoprotic) solution :  =
K a1 Basic strength of BOH1 1
K b1
Acid strength of HA1 1
=
=
=
=
;
Acid strength of HA 2 2
K a2 Basic strength of BOH2 2
K b2
Which of the following has the maximum degree of ionisation ?
(A) 1 M NH3
(B) 0.001 M NH3
(C) 0.1 M NH3
(D) 0.0001 M NH3.
uh
E-1.
Ka for formic acid and acetic acid are 1.8 × 10–4 and 1.8 × 10–5 respectively. The relative strength of
acids is :
(A) 10 : 1
(B) 1 : 10
(C) 1 : 10
(D) 10 : 1
E-3.
Ka for a monobasic acid, whose 0.1 M solution has pH of 4.5, is :
(A) 10–10
(B) 10–8
(C) 10 × 10–4
Ja
E-2.
(D)
10 × 10–6
Section (F) : Salt hydrolysis, pH calculation : Solutions of salt of monoprotic acid and
monoacidic base.
1
Kh
; pH = [pKw–pKb–log c] (valid if h < 0.1 or 10%)
2
c
lp
Commit to memory :
Salt of strong acid and weak base : Kh × Kb = Kw ; h =
1
Kh
; pH = [pKw+pKa+log c] (valid if h < 0.1 or 10%)
2
c
1
 h 
= K h ; pH = [pKw+pKa–pKb]
Salt of weak acid and weak base : Kh × Ka × Kb = Kw ; 
 1– h 
2
Sa
nk
a
Salt of strong base and weak acid : Kh × Ka = Kw ; h =
F-1.
Aqueous solution of NH4Cl is ____ in nature due to behaviour of ____ ion in solution :
(A) acidic ; NH4+
(B) alkalline ; NH4+
(C) acidic ; Cl–
(D) alkalline ; Cl–
F-2. The chloride salt of a certain weak monoacidic organic base is hydrolysed to an extent of 3% in its 0.1M
solution at 25ºC. Given that the ionic product of water is 10 −14 at this temperature, what is the
dissociation constant of the base?
(A)  1 x 10−10
(B)  1 x 10−9
(C) 3.33 x 10−9
(D) 3.33 x 10−10
F-3.
The pH of 0.1 M solution of the following salts increases in the order :
(A) NaCl < NH4Cl < NaCN < HCl
(B) HCl < NH4Cl < NaCl < NaCN
(C) NaCN < NH4Cl < NaCl < HCl
(D) HCl < NaCl < NaCN < NH4Cl
F-4.
The degree of hydrolysis of a salt of weak monobasic acid and weak monoacidic base in its 0.1 M
solution is found to be 50%. If the molarity of the solution is 0.2 M, the percentage hydrolysis of the salt
should be :
(A) 100 %
(B) 50 %
(C) 25 %
(D) None of these
401
PART - III : MATCH THE COLUMN
1.
Match the Column.
Column-I
(A)
HCl
(B)
KOH
(C)
NH3
(D)
BF3
(p)
(q)
(r)
(s)
Column-II
Lewis acid
Arrhenius acid
Lewis base
Arrhenius base
a
ri
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
The self ionisation constant for pure formic acid, K = [HCOO H2+ ][HCOO–] has been estimated as 10–6 at
room temperature. The density of formic acid is 1.15 g/cm 3. What percentage of formic acid molecules
in pure fomic acid are converted to formate ion ?
(A) 0.002%
(B) 0.004%
(C) 0.006%
(D) 0.008%
2.
pKa for an acid HA is 6. The value of K for the reaction A– + H3O+
(A) 1 × 10–6
(B) 1 × 108
(C) 1 × 10–8
HA + H2O is :
(D) 1 × 106
3.
10–6 M HCI is diluted to 100 times. Its pH is :
(A) 6
(B) 8
(D) 7.02
uh
1.
Ja
(C) 6.98
Which of the following solutions has a pH exactly equal to 8 ?
(A) 10–8 M HCl solution
(B) Solution containing 10–8 M H+
–6
(C) 2 × 10 M Ba(OH)2 solution
(D) 10–8 M NaOH solution
5.
10 mL of a strong acid solution of pH = 2 are mixed with 990 mL of another strong acid solution of pH =
4. The pH of the resulting solution will be :
(A) 3
(B) 3.3
(C) 3.7
(D) Molecular formula of both strong acids should be known to answer above question.
6.
% dissociation of a 0.024 M solution of a weak acid HA (Ka = 2 × 10–3) is :
(A) 0.25%
(B)  29%
(C) 25%
(D)  0.29%
7.
For a weak base BOH, Kb = 10–4. Calculate pH of 10–4 M BOH solution. (Take log 6.2 = 0.79)
(A) 10
(B) 9.79
(C) 8
(D) None of these
Sa
nk
a
lp
4.
8.
Four separate solutions of sodium salts NaW, NaX, NaY and NaZ have pH 7, 9, 10 and 11 respectively,
when each solution has concentration 0.1 M. Then the strongest acid is :
(A) HW
(B) HX
(C) HY
(D) HZ
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
1.
2.
If an acid-base reaction HA(aq) + B– (aq)
HB(aq) + A– (aq) has Keq = 10–4, how many of the
following statements are true ?
(i) HB is stronger acid than HA
(ii) HA is stronger acid than HB
(iii) HA and HB have the same acidic strength (iv) B– is stronger base than A–
(v) A– is stronger base than B–
(vi) B– and HB are conjugate acid-base pair
–
(vii) A is the conjugate base of acid HA.
(viii) HA can be HSO4– and HB can be HCOOH.
–
–
–
–
(ix) A can be F and B can be CN .
What is pOH of an aqueous solution with [H+] = 10–2 M and Kw = 2 × 10–12 ? Report your answer after
dividing by 2 and round it off to the nearest whole number.
3.
Percentage ionisation of water as follows at certain temperature is 3.6 × 10 –7 %. Calculate Kw and pH of
water at this temperature. 2H2O
H3O+ + OH–
4.
0.1 mole HCl is dissolved in distilled water of volume V. Then, at lim , (pH)solution is equal to .....
V →
402
Determine pKa (H2O) + pKa (H3O+).
6.
What volume (in L) of water must be added to 1 L of 0.1 M solution of B (weak organic monoacidic
base; ionisation constant = 10–5) to triple the % ionisation of base ?
7.
If pHx% is the pH of a 1 M weak monoprotic acid which is x % ionised, then find the value of
 pH50% 

 × 100.
 pH10% 
8.
If the equilibrium constant for the reaction of weak acid HA with a strong base is 10 9, then determine pH
of 0.1 M NaA solution.
9.
What is the concentration of CH3COOH(aq) in a solution prepared by dissolving 0.01 mole of
NH+4 CH3COO– in 1 L H2O ? Report your answer after multiplying by 9 × 105.
[Ka(CH3COOH) = 1.8 × 10–5; Kb(NH3.H2O) = 1.8 × 10–5]
a
ri
5.
1.
Which statement(s) is/are correct ?
(A) All Bronsted bases are also Lewis bases.
(B) All Bronsted acids are not Lewis acids.
(C) Most cations are acids and most anions are bases.
(D) All Bronsted bases are also Arrhenius bases.
2.
Select the incorrect statement(s) :
(A) NH4+ (aq) is a strong acid.
(C) H– is a weak base.
(B) CH3COO– (aq) is a weak base.
(D) HS– is a weak acid as well as a weak base.
Ja
3.
uh
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
For pure water :
(A) pKw increases with decrease in temperature.
(B) Absolute dissociation constant decreases with decrease in temperature.
(C) H2O increases with decrease in temperature.
lp
(D) Both pH and pOH decrease with rise in temperature.
One litre of a strong acid solution contains 10–5 moles of H+ ions. Then :
(A) pOH = 9.
(B) Percentage ionisation of water in solution is 1.8 × 10–9 %.
(C) Number of OH– ions per mL of solution = 6.022 × 1015.
(D) [H+ ]from H2O = 10–9 M
5.
Equal volumes of 0.2 M HCl and 0.2 M Sr(OH)2 are mixed. Which of the following statement is/are
correct ?
(A) [Sr2+] = 0.1 M
(B) [Cl–] = 0.1 M
(C) pH of resulting solution = 13
(D) Solution is neutral.
6.
If 0.1 M CH3COOH (Ka = 1.8 × 10–5) is diluted at 25°C, then which of the following will be correct ?
(A) [H+] will increase.
(B) pH will increase.
(C) number of H+ ions will increase.
(D) Ka will increase.
7.
Degree of hydrolysis for a salt of strong acid and weak base :
(A) is independent of dilution
(B) increases with dilution
(C) increases with decrease in Kb of the bases
(D) decreases with decrease in temperature.
8.
Equal volumes of following solutions are mixed, in which case the pH of resulting solution will be
average value of pH of two solutions.
(A) Aqueous HCl of pH = 2, aqueous NaOH of pH = 12
(B) Aqueous HCl of pH = 2, aqueous HCl of pH = 4
(C) Aqueous NaOH of pH = 12, aqueous NaOH of pH = 10
(D) Aqueous CH3COOH of pH = 5, aqueous NH3 of pH = 9. [Ka (CH3COOH) = Kb (NH3)]
Sa
nk
a
4.
403
PART - IV : COMPREHENSION
Read the following passages carefully and answer the questions.
Comprehension # 1
pH calculation upon dilution of a strong acid solution is generally done by equating nH+ in original
solution & diluted solution. However, if strong acid solution is very dilute, then H + from water are also to
be considered.
Take log 3.7 = 0.568 and answer the following questions.
A 1 litre solution of pH = 4 (solution of a strong acid) is added to the 7/3 litre of water. What is the pH of
resulting solution ?
(A) 4.52
(B) 4.365
(C) 4.4
(D) 4.432
2.
A 1 litre solution of pH = 6 (solution of a strong acid) is added to the 7/3 litre of water. What is the pH of
resulting solution ? Neglect the common ion effect on H2O.
(A) 6.4
(B) 6.52
(C) 6.365
(D) 6.432
a
ri
1.
Sa
nk
a
lp
Ja
uh
Comprehension # 2
Consider a solution of CH3COONH4 which is a salt of weak acid & weak base.
The equilibrium involved in the solutions are :
CH3COO– + H2O
CH3COOH + OH–
..........(1)
NH4– + H2O
NH3.H2O + H+
..........(2)
H+ + OH–
H2O
..........(3)
If we add these three reactions, then the net reaction is :
CH3COO– + NH4+ + H2O
CH3COOH + NH3.H2O
..........(4)
Both CH3COO– and NH4+ get hydrolysed independently and their hydrolysis depends on :
(i)
their initial concentration
K
K
(ii)
the value of Kh which is w for CH3COO– and w for NH4+.
Ka
Kb
Since both of the ions were produced from the same salt, their initial concentrations are same.
K
K
Therefore unless & until the value of w and w or Ka and Kb is same, the degree of hydrolysis of ion
Ka
Kb
can't be same.
To explain why we assume that degree of hydrolysis of cation and anion is same, we need to now look
at the third reaction i.e., combination of H+ and OH– ions. It is obvious that this reaction happens only
because one reaction produced H+ ion and the other produced OH– ions. We can also note that this
reaction causes both the hydrolysis reaction to occur more since their product ions are being
consumed. Keep this thing in mind that the equilibrium which has smaller value of equilibrium constant
is affected more by the common ion effect. For the same reason, if for any reason a reaction is made to
occur to a greater extent by the comsumption of any one of the product ion, the reaction with the
smaller value of equilibrium constant tends to get affected more.
Therefore we conclude that firstly the hydrolysis of both the ions ocurs more in the presence of each
other (due to consumption of the product ions) than in each other's absence. Secondly, the hydrolysis
of the ion which occurs to a lesser extent (due to smaller value of K h) is affected more than the one
whose Kh is greater. Hence, we can see that the degree of hydrolysis of both the ions would be close to
each other when they are getting hydrolysed in the presence of each other.
Now answer the following questions :
3.
In the hydrolysis of salt of weak acid & weak base :
(A) degree of hydrolysis of cation and anion is different
(B) degree of hydrolysis of cation and anion is same
(C) degree of hydrolysis of cation and anion is different and they can never be assumed same.
(D) degree of hydrolysis of cation and anion is different but they are very close to each other when they
are getting hydrolysed in the presence of each other.
404
4.
For 0.1 M CH3COONH4 salt solution given, Ka (CH3COOH) = Kb (NH3.H2O) = 2 × 10–5.
In this case, degree of hydrolysis of cation and anion are :
(A) exactly same
(B) slightly different
(C) can't say
(D) different but can be take approximatly same
JEE(MAIN) OFFLINE PROBLEMS
1.
2.
The conjugate base of H2PO4– is :
(1) PO43–
(2) P2O5
(3) H3PO4
What is the conjugate base of OH– ?
(1) O2
(2) H2O
(3) O–
a
ri
JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
[AIEEE-2004, 3/225]
(4) HPO42–
[AIEEE-2005, 3/225]
(4) O2–
Hydrogen ion concentration in mol/L in a solution of pH = 5.4 will be :
(1) 3.98 × 108
(2) 3.88 × 106
(3) 3.68 × 10–6
[AIEEE-2005, 3/225]
(4) 3.98 × 10–6
4.
The first and second dissociation constants of an acid H 2A are 1.0 × 10–5 and 5.0 × 10–10 respectively.
The overall dissociation constant of the acid will be :
[AIEEE-2007, 3/120]
(1) 5.0 × 10–15
(2) 0.2 × 105
(3) 5.0 × 10–5
(4) 5.0 × 1015
5.
The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous
solution of the corresponding salt, BA, will be :
[AIEEE-2008, 3/105]
(1) 4.79
(2) 7.01
(3) 9.22
(4) 9.58
6.
Three reactions involving H2PO4– are given below :
(i) H3PO4 + H2O → H3O+ + H2PO4–
(ii) H2PO4– + H2O → HPO42– + H3O+
(iii) H2PO4– + OH– → H3PO4 + O2–
In which of the above, does H2PO4– act as an acid ?
(1) (ii) only
(2) (i) and (ii)
(3) (iii) only
lp
Ja
uh
3.
[AIEEE-2010, 4/144]
(4) (i) only
The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, K a of the acid
is:
[AIEEE-2012, 4/120]
(1) 3 × 10–1
(2) 1 ×10–3
(3) 1 × 10–5
(4) 1 × 10–7
8.
How many litres of water must be added to 1 litre an aqueous solution of HCl with a pH of 1 to create
an aqueous solution with pH of 2 ?
[JEE(Main) 2013, 4/120]
(1) 0.1 L
(2) 0.9 L
(3) 2.0 L
(4) 9.0 L
9.
pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their
salt (AB) solution is :
[JEE(Main) 2017, 4/120]
(1) 6.9
(2) 7.0
(3) 1.0
(4) 7.2
10.
Which of the following salts is the most basic in aqueous solution?
(1) FeCl3
(2) Pb(CH3COO)2
(3) Al(CN)3
Sa
nk
a
7.
[JEE(Main) 2018, 4/120]
(4) CH3COOK
JEE(MAIN) ONLINE PROBLEMS
1.
Assuming that the degree of hydrolysis is small, the pH of 0.1 M solution of sodium acetate
(Ka = 1.0 × 10–5) will be :
[JEE(Main) 2014 Online (11-04-14), 4/120]
(1) 5.0
(2) 6.0
(3) 8.0
(4) 9.0
2.
The conjugate base of hydrazoic acid is :
(1) N–3
(2) N3–
(3) N2–
[JEE(Main) 2014 Online (12-04-14), 4/120]
(4) HN3–
405
What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20%
solution of the same acid to produce 800 mL of a 29.875% acid solution ?
[JEE(Main) 2017 Online (09-04-17), 4/120]
(1) 330
(2) 316
(3) 320
(4) 325
4.
Following four solutions are prepared by mixing different volumes of NaOH and HCl of different
concentrations, pH of which one of the them will be equal to 1?
[JEE(Main) 2018 Online (15-04-18), 4/120]
M
M
M
M
(1) 100 mL
HCl + 100 mL
NaOH
(2) 75 mL
HCl + 25 mL
NaOH
10
10
5
5
M
M
M
M
(3) 60 mL
HCl + 40 mL
NaOH
(4) 55 mL
HCl + 45 mL
NaOH
10
10
10
10
a
ri
3.
PART - I
(a) H3PO3, (COOH)2
(b) HSO4–, RNH3+ NH3, (C2H5)2OH+, HF
(c) NO2–, O2–, HCO3–, ClO4–
(d) Conjugate acids: H2S, NH4+, C2H5OH2+, H3O+
Conjugate base: S2–, NH2–, C2H5O–, OH–
Ja
A-1.
uh
EXERCISE - 1
..
(e) Lewis acid: H+, FeCl3, CH2
Lewis base: (CH3)3N, F–
Acidic – (vi), (vii) ;
Basic – (i), (iv) ;
Amphiprotic – (ii), (iii), (v)
A-3.
H2O – Arrhenius acid, Arrhenius base, Bronsted–Lowry acid, Bronsted–Lowry base, Lewis base but not
Lewis acid.
B-1.
600 ions / mm3
C-2.
(i) 10–12
(ii) 10–6
D-1.
(i) 3,
(ii) 2,
lp
A-2.
C-1.
2.5 × 10–5 ; 4.6
(iv) K b 1 < Kb 2 < Kb 3
(iii) 12,
(iv) 7.02,
Sa
nk
a
(iii) 10–11
D-2.
(i) 11.7, (ii) 1, (iii) 10.7, (iv) 1.92
D-3.
0.7
E-1.
1.69  10–5
E-2.
Refer class notes / sheet theory.
E-3.
11.7
E-4.
HCl
F-1.
(a) Basic
(g) neutral
F-2.
9
(b) acidic
(h) basic
(c) basic
(i) basic
F-3.
D-4.
(v) 11.2
(a) 13 ; (b) 7 ; (c) 1.3.
(d) basic
(j) acidic
(e) acidic
(f) basic
Kb = 8 × 10–10
F-4.
0.625%, pH = 7
PART - II
A-1.
(C)
A-2.
(D)
A-3.
(D)
A-4.
(C)
A-5.
(D)
B-1.
(D)
B-2.
(A)
B-3.
(D)
B-4.
(D)
C-1.
(B)
C-2.
(C)
D-1.
(C)
D-2.
(A)
D-3.
(C)
D-4.
(B)
D-5.
(B)
D-6.
(D)
D-7.
(D)
E-1.
(D)
E-2.
(D)
E-3.
(B)
F-1.
(A)
F-2.
(A)
F-3.
(B)
F-4.
(B)
406
PART - III
1.
(A) → r; (B) → s; (C) → r; (D) → p
EXERCISE – 2
PART - I
(B)
2.
(D)
3.
(C)
6.
(C)
7.
(B)
8.
(A)
4.
PART - II
1.
4 (i, v, vi, vii)
2.
5 (Actual answer = 9.7)
4.
7
5.
14
8.
9
9.
50
8
PART - III
(ABC)
2.
(AC)
3.
(ABD)
6.
(BC)
7.
(BCD)
8.
(AD)
4.
Ja
1.
7.
5.
(C)
Kw = 4 × 10–14, pH = 6.7
30
uh
6.
3.
(B)
a
ri
1.
(ABD)
5.
(ABC)
PART - IV
(A)
2.
(D)
3.
(D)
4.
(A)
lp
1.
EXERCISE – 3
(4)
2.
6.
(1)
7.
(4)
3.
(4)
4.
(1)
5.
(2)
(3)
8.
(4)
9.
(1)
10.
(4)
Sa
nk
1.
a
JEE(MAIN) OFFLINE PROBLEMS
JEE(MAIN) ONLINE PROBLEMS
1.
(4)
2.
(2)
3.
(2)
4.
(2)
407
IONIC EQUILIBRIUM-II
Note : Take water as solvent and temperature as 25ºC, if not specified.
Take log 2 = 0.3, log 3 = 0.48, log 5 = 0.7, log 7 = 0.845, if not specified.
 Marked questions are recommended for Revision.
Section (A) : Buffer Solutions : Definition and Identification
Commit to memory :
a
ri
PART - I : SUBJECTIVE QUESTIONS
uh
Buffer Solutions : Solution containing weak acid and it's conjugate base, solution containing weak
base and it's conjugate acid, solution containing salt of weak acid and weak base.
Preparation :
(i) Solution of weak acid (or weak base) + Solution of it's conjugate base (or it's conjugate acid)
(ii) Solution of weak acid (or weak base) + Solution of strong base (or strong acid) (n 1 > n2)
(iii) Solution of salt of weak acid and strong base (or salt of weak base and strong acid) + Solution of
strong acid (or strong base) (n1 > n2)
Ja
A-1. V1 mL of a CH3COONa solution (of molarity M1) and V2 mL of a HCl solution (of molarity M2) are
available. Can the two be mixed to obtain a buffer solution ? If yes, what should be the mathematical
condition relating M1, M2, V1 & V2 for this ?
A-2. Select pair(s) of solutions from below which could be mixed to produce a buffer solution :
NH4OH solution (S1), (NH4)2 SO4 solution (S2), HCl solution (S3), KOH solution (S4).
Commit to memory :
lp
Section (B) : pH Calculation : Buffer solutions generated from Monobasic acid /
Monoacidic base
Sa
nk
a
pH Calculation : Buffer solutions generated from Monobasic acid / Monoacidic base :
(i) pH of a buffer solution consisting of a weak acid (HA ; C 1 concentration) and its salt with a strong
[Anion of Salt]
base (NaA ; C2 concentration of anion) : pH = pKa + log
[Acid]
(ii) pH of a buffer solution consisting of a weak base (B ; C1 concentration) and its salt with a strong acid
[Cation of Salt]
(BH+Cl– ; C2 concentration of cation) : pOH = pKb + log
[Base]
B-1.
Calculate pH of following solutions :
(a) (4 g CH3COOH + 4.1 g CH3COONa) in 100 mL aqueous solution ; Ka for CH3COOH = 1.8 × 10–5
(b) 5 mL of 0.1 M BOH + 25 mL of 0.1 M BCl
; Kb for BOH = 1.8 × 10–5
B-2. 50 mL of 0.2 M solution of an acid HA (Ka = 10−5) & 50 mL of a NaA solution are given. What should be
the concentration of NaA solution to make a buffer solution with pH = 4 upon mixing the two ?
B-3.
Calculate the pH of 0.5 L of a 0.2 M NH4CI – 0.2 M NH3 buffer before and after addition of (a) 0.05 mole
of NaOH and (b) 0.05 mole of HCl. Assume that the volume remains constant.
[Given : pKb of NH3 = 4.74]
408
PART - II : ONLY ONE OPTION CORRECT TYPE
Section (A) : Buffer Solutions : Definition and Identification
Commit to memory :
Buffer Solutions : Solution containing weak acid and it's conjugate base, solution containing weak
base and it's conjugate acid, solution containing salt of weak acid and weak base.
Preparation :
(i) Solution of weak acid (or weak base) + Solution of it's conjugate base (or it's conjugate acid)
(ii) Solution of weak acid (or weak base) + Solution of strong base (or strong acid) (n 1 > n2)
(iii) Solution of salt of weak acid and strong base (or salt of weak base and strong acid) + Solution of
strong acid (or strong base) (n1 > n2)
A solution is 0.1 M in CH3COOH and 0.1 M in CH3COONa. Which of the following will change its pH
significantly?
(A) Addition of small amount of water
(B) Addition of small amount of HCl
(C) Addition of small amount of NaOH
(D) None will change the pH significantly.
A-2.
Which of the following may be added to one litre of water to act a buffer ?
(A) One mole of CH3COOH and one mole of HCI
(B) One mole of NH4OH and one mole of NaOH
(C) One mole of NH4CI and one mole of HCI
(D) One mole of CH3COOH and 0.5 mole of NaOH
A-3.
In which of the following respective volume ratios should 0.1 M NH 4OH solution & 0.1 M HCl solution be
mixed, so that the resulting solution behaves like a buffer solution ?
(A) 1 : 1
(B) 2 : 1
(C) 1 : 2
(D) No such volume ratio is possible
Ja
uh
a
ri
A-1.
Section (B) : pH Calculation : Buffer solutions generated from Monobasic acid /
Monoacidic base
lp
Commit to memory :
Sa
nk
a
pH Calculation : Buffer solutions generated from Monobasic acid / Monoacidic base :
(i) pH of a buffer solution consisting of a weak acid (HA ; C 1 concentration) and its salt with a strong
[Anion of Salt]
base (NaA ; C2 concentration of anion) : pH = pKa + log
[Acid]
(ii) pH of a buffer solution consisting of a weak base (B ; C1 concentration) and its salt with a strong acid
[Cation of Salt]
(BH+Cl– ; C2 concentration of cation) : pOH = pKb + log
[Base]
B-1. Fear or excitement generally cause one to breathe rapidly and it results in the decrease of
concentration of CO2 in blood. In what way, it will change pH of blood ?
(A) pH will significantly increase
(B) pH will significantly decrease
(C) No significant change in pH
(D) pH will be 7
B-2.
B-3.
pH of a mixture containing 0.1 M X– and 0.2 M HX is : [pKb (X–) = 4]
(A) 4 + log 2
(B) 4 – log 2
(C) 10 + log 2
(D) 10 – log 2
10–10.
Ka for HCN is 5 ×
For maintaining a constant pH of 9, the volume of 5 M KCN solution required to
be added to 10 mL of 2 M HCN solution is :
(A) 4 mL
(B) 8 mL
(C) 2 mL
(D) 10 mL
B-4. A buffer solution made up of BOH and BCl of total molarity 0.29 M has pH = 9.6 and Kb = 1.8 × 10–5.
Concentration of salt and base respectively is :
(A) 0.09 M and 0.2 M
(B) 0.2 M and 0.09 M
(C) 0.1 M and 0.19 M
(D) 0.19 M and 0.1 M
409
PART - III : MATCH THE COLUMN
1.
At the equivalence point of titration of (equivalence point = the point at which reaction is just complete) :
(A) a strong acid with a strong base
(p) pH < 7
(B) a weak acid with a strong base
(q) pH > 7
(C) a weak base with a strong acid
(r) pH = 7
(D) a weak acid with a weak base
(s) pH may be less than or greater than 7
 Marked questions are recommended for Revision.
a
ri
PART - I : ONLY ONE OPTION CORRECT TYPE
To prepare a buffer of pH 8.26 amount of (NH 4)2 SO4 to be added to 500 mL of 0.01 M NH 4OH solution
is : [pKa (NH4+) = 9.26]
(A) 0.05 mole
(B) 0.025 mole
(C) 0.10 mole
(D) 0.005 mole
2.
A weak acid (HA) after treatment with 12 mL of 0.1 M strong base (BOH) solution has a pH of 5. At the
end point, the volume of same base solution required is 27 mL. Ka of acid is :
(A) 1.8 × 10–5
(B) 8 × 10–6
(C) 1.8 × 10–6
(D) 8 × 10–5
uh
1.
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
How many of the following statement(s) is/are correct for making a buffer solution ?
(i)
It can be formed by mixing equal concentrations of HCl and CH3COONa
(ii)
It can be formed by mixing equal concentrations of HNO 3 and NH3
(iii)
It can be formed by mixing equal concentrations of HCOOH and Aniline.
(iv)
It can be formed by mixing equal volumes of NH4OH and HClO4.
(v)
It can be formed by mixing equal volumes of HCN and KOH.
(vi)
There is no change in the pH of a buffer solution on adding small amount of a strong acid/base.
(vii)
The concentrations of acid and base being mixed must be different to form a buffer.
(viii)
The volumes of acid and base being mixed must be different to form a buffer.
(ix)
The concentrations and volumes of acid and base being mixed must be different to form a
buffer.
2.
1 M benzoic acid (pKa = 4.2) and 1M C6H5 COONa solutions are given separately. What is the volume
of benzoic acid required to prepare a 93 mL buffer solution of pH = 4.5 ?
a
lp
Ja
1.
Sa
nk
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
1.
A buffer solution can be prepared from a mixture of :
(A) Sodium acetate and acetic acid in water
(B) Sodium acetate and hydrochloric acid in water
(C) Ammonia and ammonium chloride in water
(D) Ammonia and sodium hydroxide in water
[JEE-1999, 3/80]
410
JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) OFFLINE PROBLEMS
1.
The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA, in which 50% of
the acid is ionized, is :
[AIEEE-2007, 3/120]
(1) 9.5
(2) 7.0
(3) 4.5
(4) 2.5
JEE(MAIN) ONLINE PROBLEMS
In some solutions, the concentration of H3O+ remains constant even when small amounts of strong acid
or strong base are added to them. These solutions are known as :
[JEE(Main) 2014 Online (11-04-14), 4/120]
(1) Ideal solutions
(2) Colloideal solutins (3) true solutions
(4) Buffer solutions
2.
Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisation
constant of HA is 10–5, the ratio of salt to acid concentration in the buffer solution will be :
[JEE(Main) 2017 Online (08-04-17), 4/120]
(1) 10 : 1
(2) 4 : 5
(3) 1 : 10
(4) 5 : 4
3.
50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl . If pK b of ammonia solution is
4.75, the pH of the mixture will be :
[JEE(Main) 2017 Online (09-04-17), 4/120]
(1) 4.75
(2) 3.75
(3) 9.25
(4) 8.25
4.
20 mL of 0.1 M H2SO4 solution is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant
mixture is : (pKb of NH4OH = 4.7)
[JEE(Main) 2019 Online (09-01-19), 4/120]
(1) 9.0
(2) 5.2
(3) 9.4
(4) 5.0
lp
Ja
uh
a
ri
1.
a
EXERCISE – 1
Yes, M1 V1 > M2 V2.
B-1.
(a) 4.62
Sa
nk
A-1.
(b) 8.56
A-1.
(D)
A-2.
(D)
B-3.
(C)
B-4.
(A)
1.
B-2.
PART – I
0.02 M
A-2.
S1 & S2 ; S1 & S3 ; S2 & S4.
B-3.
pH = 9.26 ; (a) pH = 9.74 ; (b) 8.78
PART - II
A-3.
(B)
B-1.
(C)
B-2.
(D)
PART - III
(A) → R; (B) → Q; (C) → P; (D) → S
EXERCISE – 2
PART - I
1.
(B)
2.
(B)
PART - II
1.
5 [(i) to (v)]
2.
31
411
PART - III
1.
(A) (B)(C)
EXERCISE – 3
JEE(MAIN) OFFLINE PROBLEMS
1.
(1)
JEE(MAIN) ONLINE PROBLEMS
(4)
2.
(1)
3.
(3)
4.
 Marked questions are recommended for Revision.
uh
IONIC EQUILIBRIUM-III
(1)
a
ri
1.
Ja
PART - I : SUBJECTIVE QUESTIONS
Section (A) : Solubility, Solubility product and Simple solubility calculations
Commit to memory :
Solubility (s ; in mol/L) of a sparingly soluble salt AxBy : Ksp = xx.yy.(s)x+y
If the solubility product of a salt MX is 3 × 10–10 at 0ºC, determine its solubility.
A-2.
A saturated solution of PbCl2 contains 2 × 10–3 mol of PbCl2 per litre. What is the Ksp of PbCl2 ?
A-3.
Ksp at 25ºC of AgCl, AgBr and AgI are respectively 3 × 10–10, 7.7 × 10–13, 1.5 × 10–16. Write decreasing
order of solubility of these salts.
lp
A-1.
a
A-4. Write solubility product expression for Hg2SO4.
Section (B) : Condition for precipitation, Common ion effect on solubility
Commit to memory :
Sa
nk
Condition for precipitation of AxBy : Ionic product or KIP  Ksp ; [Ay+]x [Bx–]y > Ksp of AxBy
Common ion effect on solubility : Solubility (s') of AxBy (sparingly soluble ; Ksp) in a solution of 'c' M
ApDq/ErBp (both readily soluble) : (pc)x(ys')y = Ksp ; s' = .....
B-1. Determine solubility of PbI2 in a 0.1 M solution of NaI. (Ksp of PbI2 = 8 × 10–9)
B-2. Determine solubility of PbI2 in a 0.1 M solution of Pb(NO3)2. (Ksp of PbI2 = 8 × 10–9)
B-3.
At 25ºC, the solubility of Ag2CO3 (Ksp = 4.3 × 10–13) would be in what order in the following solutions ?
(a) 0.01 M AgNO3
(b) 0.04 M K2CO3
(c) pure water
(d) in a buffer (pH = 4)
412
PART - II : ONLY ONE OPTION CORRECT TYPE
Section (A) : Solubility, Solubility product and Simple solubility calculations
Commit to memory :
Solubility (s ; in mol/L) of a sparingly soluble salt AxBy : Ksp = xx.yy.(s)x+y
A-1. M(OH)x (producing Mx+ and OH– ions) has Ksp 4 × 10–12 and solubility 10–4 M. The value of x is :
(A) 1
(B) 2
(C) 3
(D) 4
A-2.
If the solubility of Lithium sodium hexafluoridoaluminate, Li3Na3[AlF6 ]2 is ‘s’ mol L–1, its solubility product
3−
is: (Assume no ionisation of  AlF6  )
(B) 186624 s8
(C) 1458 s8
(D) 2916 s8
a
ri
(A) 18 s3
Which of the following is most soluble in water ? Assume no reaction of cation/anion.
(A) MnS (Ksp = 2.5 × 10–13 )
(B) ZnS (Ksp = 1.6 × 10–24)
–72
(C) Bi2S3 (Ksp =1.6 × 10 )
(D) Ag2S (Ksp = 10–51)
A-4.
When different types of salts have nearly same solubility product constants K sp, but less than one, the
most soluble salt is that :
(A) Which produces maximum number of ions per formula unit
(B) Which produces minimum number of ions per formula unit
(C) Which produces ions with maximum charge
(D) Which produces ions with miniumum charge
A-5.
In a saturated solution of Ag2SO4, silver ion concentration is 3 × 10–2 M. Its solubility product is :
Assume no reaction of cation/anion.
(A) 1.35 × 10–5
(B) 1.08 × 10–4
(C) 2.7 × 10–5
(D) 4.5 × 10–4
Ja
uh
A-3.
A-6. The minimum volume of the water needed to dissolve 1 g of BaSO 4 (Ksp = 10–10) is about: Assume no
reaction of cation/anion. [Mol. mass (BaSO4) = 233 u]
(A) 105 litres
(B) 430 litres
(C) 43 litres
(D) 4300 litres
Commit to memory :
lp
Section (B) : Condition for precipitation, Common ion effect on solubility
a
Condition for precipitation of AxBy : Ionic product or KIP  Ksp ; [Ay+]x [Bx–]y > Ksp of AxBy
Common ion effect on solubility : Solubility (s') of AxBy (sparingly soluble ; Ksp) in a solution of 'c' M
ApDq/ErBp (both readily soluble) : (pc)x(ys')y = Ksp ; s' = .....
Sa
nk
B-1. The solubility product of BaCrO4 is 2.4 × 10–10 M2. The maximum concentration of Ba(NO3)2 possible
without precipitation in a 6 × 10–4 M K2CrO4 solution is :
(A) 4 × 10–7 M
(B) 1.44 × 10–13 M
(C) 2 × 10–7 M
(D) 2.5 × 106 M
B-2. The solubility product of AgCl is 1.8 × 10–10. Precipitation of AgCl will occur only when equal volumes of
solutions of :
(A) 2 × 10–5 M Ag+ and 2 × 10–5 M Cl– are mixed.
(B) 10–7 M Ag+ and 10–7 M Cl– are mixed.
(C) 10–5 M Ag+ and 10–5 M Cl– are mixed.
(D) 10–4 M Ag+ and 10–4 M Cl– are mixe
B-3. The solubility of CaF2 (Ksp = 5.3 × 10–9) in 0.1 M solution of NaF would be : Assume no reaction of
cation/anion.
(A) 5.3 × 10–10 M
(B) 5.3 × 10–8 M
(C) 5.3 × 10–7 M
(D) 5.3 × 10–11 M
B-4. Let the solubilities of AgCl in pure water, 0.01 M CaCl2, 0.01 M NaCl & 0.05 M AgNO3 be s1, s2, s3 & s4
respectively. What is the correct order of these quantities ? Neglect any complexation.
(A) s1 > s4 > s3 > s2
(B) s1 > s2 = s3 > s4
(C) s1 > s3 > s2 > s4
(D) s4 > s2 > s3 > s1
B-5.
Solubility of BaF2 in a solution of Ba(NO3)2 will be represented by which concentration term ? Assume
no reaction of cation/anion.
(A) [Ba2+]
(B) [F–]
(C) [F–]/2
(D) 2[]
413
PART - III : MATCH THE COLUMN
Match the correct Ksp expression in terms of solubility (s) for given salts :
(Dont assume hydrolysis of any ion)
Column-I
Column-II
(A)
Ca3(PO4)2
(p)
4s3
(B)
(q)
27s4
Hg2I2
(C)
Cr(OH)3
(r)
108s5
(D)
CaF2
(s)
16s4
a
ri
1.
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
Slaked lime, Ca(OH)2 is used extensively in sewage treatment. What can be the maximum pH of
Ca(OH)2 (aq) ? (Take log11 = 1.04)
Ca(OH)2(s)
Ca2+(aq) + 2OH–(aq)
;
Ksp = 5.324 × 10–6
(A) 12.04
(B) 12.34
(C) 10.68
(D) 14
2.
The solubility of Ag2CO3 in water is 1.26 × 10−4 mole/litre. What is its solubility in 0.02 M Na2CO3
solution ? Assume no hydrolysis of CO32− ion. (Take 3 2 = 1.26)
(A) 5 × 10−6 M
uh
1.
50 × 10−6 M
(C) 10−5 M
(D) 2 × 10−5 M
Ja
(B)
PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
1.
How many of the following relations are correct for the solubility product (K sp) & solubility (s g/litre) of
sparingly soluble salt A3B2 (producing A2+ & B3– ions ; mol. wt. M) in water ? (Assume no hydrolysis of
any ion).
3
lp
 3s   2s 
2. Ksp =    
M M
1. Ksp = 108s5
[B3 − ]
1 M4
=
5.
K sp
54 s4
2s
=
M
108s5
3− 2
[B ]
8.
K sp
[A 2 + ]
= 36s4
3. Ksp = (3[A2+])3 (2[B3–])2
1/ 2
 K sp 
6. [A ] =  3 − 3 
 [B ] 
2+
9. Ksp = [A2+]2 [B3–]3
Sa
nk
7. [A 2+ ]3 M5 =
a
4.
[B3–]
2
2.
8 × 10–6 M AgNO3 solution is gradually added in 1 L of 10–4 M KCl solution. Upto what volume of AgNO3
solution being added (in L), precipitation of AgCl will not take place? (K sp of AgCl = 2 × 10–10)
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
1.
The solubility of a sparingly soluble salt AxBy in water is 1.4 × 10–4 M. The solubility product is
1.1 × 10–11. The possibilities are :
(A) x = 1, y = 2
(B) x = 2, y = 1
(C) x = 1, y = 3
(D) x = 3, y = 1
JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
1.
The solubility of Mg(OH)2 is s moles/litre. The solubility product under the same condition is :
[AIEEE-2002, 3/225]
(1) 4s3
(2) 3s4
(3) 4s2
(4) s3
2.
The solubility in water of a sparingly soluble salt AB 2 is 1.0 × 10–5 mol L–1. Its solubility product will be :
[AIEEE-2003, 3/225]
(1) 4 × 10–15
(2) 4 × 10–10
(3) 1 × 10–15
(4) 1 × 10–10
414
The molar solubility (in mol L–1) of a sparingly soluble salt MX4 is s. The corresponding solubility product
is Ksp.. s is given in terms of Ksp by the relation :
[AIEEE-2004, 3/225]
(1) s = (Ksp/128)1/4
(2) s = (128Ksp)1/4
(3) s = (256Ksp)1/5
(4) s = (Ksp/256)1/5
4.
The solubility product of a salt having general formula MX2, in water is : 4 × 10–12. The concentration of
M2+ ions in the saturated aqueous solution of the salt is :
[AIEEE-2005, 3/225]
(1) 2.0 × 10–6 M
(2) 1.0 × 10–4 M
(3) 1.6 × 10–4 M
(4) 4.0 × 10–10 M
5.
In a saturated solution of the sparingly soluble strong electrolyte AgIO 3 (Molecular mass = 283), the
equilibrium which sets in is :
AgIO3(s)
Ag+(aq) + IO3– (aq)
If the solubility product constant Ksp of AgIO3 at a given temperature is 1.0 × 10–8, what is the mass of
AgO3 contained in 100 mL of its saturated solution?
[AIEEE-2007, 3/120]
(1) 1.0 × 10–7 g
(2) 1.0 × 10–4 g
(3) 28.3 × 10–2 g
(4) 2.83 × 10–3 g
6.
Solid Ba(NO3)2 is gradually dissolved in 1.0 × 10–4 M Na2CO3 solution. At what concentration of Ba2+
will a precipitate begin to form ? (Ksp for BaCO3 = 5.1 × 10–9)
[AIEEE-2009, 4/144]
(1) 5.1 × 10–5 M
(2) 8.1 × 10–8 M
(3) 8.1 × 10–7 M
(4) 4.1 × 10–5 M
7.
Solubility product of silver bromide is 5.0 × 10–13 . The quantity of potassium bromide (molar mass
taken as 120 g mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitation of
AgBr is :
[AIEEE-2010, 4/144]
(1) 1.2 × 10–10 g
(2) 1.2 × 10–9 g
(3) 6.2 × 10–5 g
(4) 5.0 × 10–8 g
8.
At 25°C, the solubility product of Mg(OH)2 is 1.0  10–11. At what pH, will Mg2+ ions start precipitating in
the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions ?
[AIEEE-2010, 4/144]
(1) 9
(2) 10
(3) 11
(4) 8
9.
An aqueous solution contains an unknown concentration of Ba 2+. When 50 mL of a 1M solution of
Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of
BaSO4 is 1 × 10–10. What is the original concentration of Ba2+.
[JEE(Main) 2018, 4/120]
(1) 1.1 × 10–9 M
(2) 1.0 × 10–10 M
(3) 5 × 10–9 M
(4) 2 × 10–9 M
Ja
uh
a
ri
3.
lp
JEE(MAIN) ONLINE PROBLEMS
Zirconium phosphate [Zr3(PO4)4] dissociates into three zirconium cations of charge +4 and four
phosphate anions of charge –3. If molar solubility of zirconium phosphate is denoted by S and its
solubility product by Ksp then which of the following relationship between S and Ksp is correct ?
[JEE(Main) 2014 Online (19-04-14), 4/120]
(1) S = {Ksp /(6912)1/7} (2) S = {Ksp /144}1/7
(3) S = (Ksp /6912)1/7
(4) S = {Ksp /6912}7
2.
The minimum volume of water required to dissolve 0.1 g lead(II) chloride to get a saturated solution (Ksp
of PbCl2 = 3.2 × 10–8 ; atomic mass of Pb = 207 u) is : [JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 1.798 L
(2) 0.36 L
(3) 17.98 L
(4) 0.18 L
Sa
nk
a
1.
3.
A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume
was made up to 100 mL. The mass of calcium sulphate formed and the concentration of OH – in
resulting solution, respectively, are : (Molar mass of Ca(OH) 2, Na2SO4 and CaSO4 are 74, 143 and 136
g mol–1, respectively; Ksp of Ca(OH)2 is 5.5 × 10–6)
[JEE(Main) 2019 Online (10-01-19), 4/120]
(1) 13.6 g, 0.14 mol L–1
(2) 13.6 g, 0.28 mol L–1
(3) 1.9 g, 0.28 mol L–1
(4) 1.9 g, 0.14 mol L–1
4.
If Ksp of Ag2CO3 is 8 × 10–12, the molar solubility of Ag2CO3 in 0.1 M AgNO3 is :
[JEE(Main) 2019 Online (12-01-19), 4/120]
(1) 8 × 10–10 M
(2) 8 × 10–12 M
(3) 8 × 10–13 M
(4) 8 × 10–11 M
415
EXERCISE - 1
PART – I
1.73 × 10–5 mol/L
A-2.
3.2 × 10–8
A-3.
AgCl > AgBr > AgI
A-4.
[ Hg22+ ][SO42–]
B-1.
8 × 10–7 M
B-2.
1.414 × 10–4 M
B-3.
(d) > (c) > (b) > (a)
PART - II
(B)
A-2.
(D)
A-3.
(A)
A-4.
A-6.
(B)
B-1.
(A)
B-2.
(D)
B-3.
B-5.
(C)
PART - III
1.
(A → r) ; (B → p) ; (C → q) ; (D → p)
(A)
A-5.
(A)
(C)
B-4.
(C)
5.
(4)
uh
A-1.
a
ri
A-1.
EXERCISE - 2
1.
(B)
2.
(C)
Ja
PART - I
PART - II
4 (2, 4, 5 and 7)
2.
1
lp
1.
PART - III
1.
(A) B)
a
EXERCISE – 3
JEE(MAIN) OFFLINE PROBLEMS
(1)
2.
6.
(1)
7.
(1)
3.
(4)
4.
(2)
(2)
8.
(2)
9.
(1)
Sa
nk
1.
JEE(MAIN) ONLINE PROBLEMS
1.
(3)
2.
(4)
3.
(3)
4.
(1)
416
SURFACE CHEMISTRY
JEE(Advanced) Syllabus
JEE(Main) Syllabus
a
ri
Concepts : Elementary concepts of adsorption (excluding adsorption isotherms);
Colloids : Types, methods of preparation and general properties; Elementary ideas of emulsions,
surfactants and micelles (only definitions and examples).
Ja
uh
Adsorption - Physisorption and chemisorption and their characteristics, factors affecting
adsorption of gases on solids - Freundlich and Langmuir adsorption isotherms, adsorption from
solutions.
Catalysis - Homogeneous and heterogeneous, activity and selectivity of solid catalysts, enzyme
catalysis and its mechanism.
Colloidal state-distinction among true solutions, colloids and suspensions, classification of
colloids- lyophilic, lyophobic; multi molecular, macromolecular and associated colloids
(micelles), preparation and properties of colloids Tyndall effect, Brownian movement,
electrophoresis, dialysis, coagulation and flocculation; Emulsions and their characteristics.
INTRODUCTION :
Section (A) : Adsorption
lp
Surface chemistry is that branch of chemistry which deals with study of the phenomena occurring at the
surface or interface, i.e. at the boundary separating two bulk phases. In this chapter our main emphasis
will be on three important topics related to surface chemistry, viz., adsorption, colloids and emulsions.
⚫ Adsorption : The phenomenon of attracting and retaining the molecules of a substance on the
Sa
nk
a
surface of a liquid or a solid resulting into a higher concentration of the molecules on the surface is
called adsorption. As a result of adsorption, there is a decrease of surface energy. The process of
removal of an adsorbed substance from the surface on which it is adsorbed is called desorption. It is
the reverse of adsorption and can be brought about by heating or by reducing the pressure.
⚫ Adsorbent and adsorbate : The substance on the surface of which adsorption occurs is known
as adsorbent. The substances that get adsorbed on the solid surface due to intermolecular attractions
are called adsorbate. Charcoal, silica, gel, alumina gel are good adsorbents because they have highly
porous structures and have large surface area. Colloids on account of their extremely small dimensions
possess enoromous surface area per unit mass and are, therefore, also good adsorbents.
Examples of adsorption :
⚫ Adsorption of a gas by charcoal : Finely divided activated charcoal has a tendency to adsorb a
number of gases like ammonia, sulphur dioxide, chlorine, phosgene, etc. In this case, charcoal acts as
an adsorbent while gas molecules act as adsorbate.
⚫ Adsorption of a dye by charcoal : Animal charcoal is used for decolourising a number of organic
substances in the form of their solutions. The discharge of the colour is due to the fact that the coloured
component (generally an organic dye) gets adsorbed on the surface of the adsorbent (animal charcoal).
⚫
Sorption : When both adsorption and absorption take place simultaneously.
Eg : Dyes get adsorbed as well as absorbed in the cotton fibre i.e. sorption takes place.
417
Difference between adsorption and absorption :
⚫
a
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The important points of distinction between adsorption and absorption
Absorption
Adsorption
It is the phenomenon in which the particles of gas
It is the phenomenon of higher concentration of gas
or liquid get uniformly distributed throughout the
or liquid on the surface than in the bulk of the solid.
body of the solid.
The concentration on the surface of the adsorbent is
The concentration is the same throughout the
different (has higher concentration) from that in the
material.
bulk.
It is a bulk phenomenon.
It is a surface phenomenon.
Adsorption is rapid in the beginning and its rate
Absorption occurs at uniform rate.
slowly decreases
It is a slow process
It is a fast process
Thermodynamics of adsorption
uh
1. Adsorption involves attracting molecules of adsorbate on surface of the adsorbent. Due to this
energy is released and thus heat of adsorption is negative i.e. adsorption is always exothemic
2. The molecules of adsorbate are held on surface of the solid adsorbent due to this entropy decreases
i.e. S is negative.
⚫
Ja
3. G = H – TS, Therefore adsorption will occur only when G is negative, and this is possible only if
H| > |TS|
4. As adsorption proceeds H becomes less and less negative hence H becomes equal to TS and
G becomes zero. This is the state at which equilibrium is attained.
Enthalpy of adsorption Hadsorption : It is the amount of the heat released when 1 mole of an
lp
adsorbate gets adsorbed on a particular adsorbent at adsorption equilibrium. It depends upon the
nature of both the adsorbate as well as adsorbent.
Types of adsorption : The adsorption is classified into two types:
Physical adsorption (i.e. physisorption) : When the particles of the adsorbate are held to the surface
of the adsorbent by the physical forces such as van der Waal’s forces, the adsorption is called physical
adsorption or vanderwaals adsorption.
Sa
nk
a
(i)
(ii)
Chemical adsorption (i.e. chemisorption) :
When the molecules of the adsorbate are held to the surface of the adsorbent by the chemical forces,
the adsorption is called chemical adsorption.
418
Differences between Physical and Chemical Adsorption
physical Adsorption
Chemical Adsorption
Nature of adsorption
Weak
Strong
Enthalpy of adsorption
Low
High
Reversibility of adsorption
Reversible and occur rapidly
Irreversible and occurs slowly
Temp. at which adsorption is
more pronounced
Low temp.
High temp.
Effect of change in temp.
Decreases with rise in temp.
Increases with rise in temp.
Not specific, generally take
place on all surface
Multi-layered
(at
high
pressure)
Highly specific,
specific surface
Nature of adsorbate layer
Very low
Ease of desorption
Easy, since Vander-Waal's
forces are involved
on
significantly high
Not easy, since chemical forces are
involved
x
m
Graph
Temperature
lp
Temperature
Competitive adsorption : When an adsorbent is in contact with more than one species (adsorbate).
There will be competition among them to get adsorbed on to the surface of the adsorbent. The one that
is more strongly adsorbed gets deposited first in preference to the others. Further a strongly adsorbed
substance may displace a weakly adsorbed substance.
Ex. NH3 can displace O2 or N2 from the surface of charcoal.
a
⚫
place
Mono-layered
Ja
Energy of activation
x
m
take
uh
Specificity of adsorption
a
ri
Property
Adsorption of gases on solids :
Sa
nk
The extent of adsorption of a gas on a solid surface is affected by the following factors:
⚫
The nature of the gas (i.e. nature of the adsorbate). The easily liquefiable gases such as HCl, NH 3,
Cl2 etc. are adsorbed more than the permanent gases such as H2, N2 and O2.
extent of adsorption  critical temperature of gas  ease of liquification
Gas
H2
CH4
CO2
SO2
TC
33
190 K 304 K 330K
⚫
Nature of adsorbent. The extent of adsorption of a gas depends upon the nature of adsorbent.
Activated charcoal (i.e. activated carbon), metal oxides (silica gel and aluminium oxide) and clay can
adsorb gases which are easily liquified. Gases such as H 2, N2 and O2 are generally adsorbed on finely
divided transition metals Ni and Co.
Extent of adsorption  surface area of solid.
⚫
Activation of adsorbent :
(a) Metallic adsorbents are activated by mechanical rubbing or by subjecting it to some chemical
reactions.
(b) To increase the adsorbing power of adsorbents, they are sub-divided into smaller pieces. As a
results, the surface area is increased and therefore, the adsorbing power increases.
419
⚫
Effect of temperature : Mostly the process of adsorption is exothermic and the reverse process or
desorption is endothermic. If the above equilibrium is subjected to increase in temperature, then
according to Le-Chaterlier’s principle, with increase in temperature, the desorption will be favoured.
Physical adsorption decreases continuously with increase in temperature whereas chemisorption
increases initially, shows a maximum in the curve and then it decreases continuously.
x
m
x
m
Effect of pressure. The extent of adsorption of a gas per unit mass of
adsorbent depends upon the pressure of the gas. The variation of extent of
adsorption expressed as x/m (where x is the mole of adsorbate and m is the
mass of the adsorbent) and the pressure is given as below. A graph between
the amount of adsorption and gas pressure keeping the temperature
constant is called an adsorption isotherm.
x
P
m
Where x = mass of adsorbate
M = mass of adsorbent
Ja
uh
⚫
a
ri
Temperature
Temperature
The initial increase in chemisorption with increase in temperature is because of activation energy
required.
This is why the chemical adsorption is also known as “Activated adsorption”.
A graph between degree of adsorption (x/m) and temperature ‘t’ at a constant pressure of adsorbate
gas is known as adsorption isobar.
It is clear from the figure-1 that extent of adsorption (x/m) increases with pressure and becomes
maximum corresponding to pressure Ps called saturation pressure.
Freundlich Adsorption isotherm
Sa
nk
a
lp
The variation of extent of adsorption (x/m) with pressure (P) was given mathematically by Freundlich.
Where n can take any whole number value which depends upon the nature of adsorbate and
adsorbent. The above relationship is also called Freundlich’s adsorption isotherm.
x
1/n
 m  = kp
 
Pressure
Graph
Relation
x
= kP
At low pressure
straight line
m
(x/m) = kp1/n
1
At intermediate pressure dependent on power of pressure
(0< <1)
n
x
=k
At high pressure
Independent of pressure
m
where, x = Amount of gas adsorbed, m = Mass of adsorbent, K and n
are adsorption constant, p is adsorption equilibrium pressure.
Slope=(1/n)
log(x/m)
The constant k and n can be determined as explained below :
Taking logarithms on both sides
Eq. (x/m) = kp1/n
intercept = log k
we get log (x/m) = logk + (1/n) log p.
log p

One of the drawbacks of Freundlich isotherm is that it fails at high pressure of the gas.
This equation applicable only when adsorbate substance form unimolecular layer on
adsorbent surface.

420
Langmuir Adsorption Isotherm :
Assumptions :
1.
2.
3.
4.
5.
Gas is considered to behave ideally.
A solid surface is considered as homogenous but contains a fixed number of adsorption sites on surface.
Each site adsorb a single molecule; means adsorption is monomolecular.
Rate of adsorption = Rate of desorption.
There is no lateral overlap between adsorbed molecules.
a
ri
Derivation
Rate of adsorption  fraction of surface available for adsorption next line
 Pressure of gas.
If  is covered fraction of surface covered. (1 – ) is the free surface area.
Rate of adsorption = KaP(1 – )
Rate of desorption = Kd
KaP(1 – ) = Kd
KaP – KaP = Kd
K aP
K
KP
= 
=
where K = a
1 + KP
K aP + K d
Kd
....(a)
Ja
x
aP
=
m
1 + bP
uh
Amount of gas adsorbed ‘x’ by given mass of adsorbent ‘m’ is proportional to 
x

m
KK a / K dP
x
=
m
1 + K a / K dP
Ka
where ‘a’ and ‘b’ are Langmuir parameter.
Kb
At very high pressure : x/m =a/b ..........(b)
At very low pressure : x/m = ap ........ (c)
For determination of the parameters ‘a’ and ‘b’, Eq. (a) may be written in its inverse form.
m 1 + bp b 1
=
= +
............ (d)
x
ap
a ap
A plot of m/x against 1/p gives a straight line with slope and intercept equal to 1/a and b/a respectively.
At low pressure according to Eq. (c) x/m increases linearly with p. At high pressure according to Eq. (b)
x/m becomes constant i.e. the surface is fully covered and change in pressure has no effect and no
further adsorption takes place which is clear from the Figure-1.
Sa
nk
a
lp
a = Kb, b =
1.
2.
3.
Adsorption from solutions :
The extent of adsorption increases with increase in the surface area of adsorbent.
The extent of adsorption decreases with increase in temp.
The extent of adsorption is related to concentration of solution through this equation
Freundlich isotherm :
(x/m) = k(c)1/n (n  1) where c is the equilibrium concentration of the solute in solution.
Temperature dependence here also is similar to that for adsorption of gases and in place of equilibrium
pressure, we use equilibrium concentrations of the adsorbates in the solution.
Applications of adsorption :
1.
2.
3.
4.
In gas masks : Activated charcoal is generally used in gas masks to adsorb poisonous and toxic
gases from air. These masks are commonly used by the miners because there are poisonous gases
like CO, CH4 etc. in the atmosphere in coal mines.
In dyeing of cloths : Mordants such as alums are used in dyeing of cloths. They adsorb the dye
particles which, otherwise, do not stick to the cloths.
In dehumidizers : Silica gel is commonly used to adsorb humidity or moisture from air.
Removal of colouring matter : Many substances such as sugar, juice and vegetable oils are coloured
due to the presence of impurities. They can be decolourised by placing them in contact with adsorbents
like activated charcoal or fuller’s earth.
421
10.
11.
Ex-1.
Sol.
a
ri
8.
9.
uh
6.
7.
Heterogeneous catalysis : The phenomenon of adsorption is useful in the heterogeneous catalysis.
The metals such as Fe, Ni, Pt, Pd, etc. are used in the manufacturing processes such as Contact
process, Haber process and the hydrogenation of oils. Their use is based upon the phenomenon of
adsorption.
Refining Petroleum : Silica gel is used as adsorbent in petroleum refining.
Chromatography : It is a method for separation of component and is based on preferential adsorption
column is very common device used.
Creating vacuum : High vacuum can be created by removing gas by adsorption.
Adsorption Indicators : In volumetric analysis, adsorption indicator is used Surface of certain
precipitates such as silver halide have the property of adsorbing some dye like eosin, fluorescein, etc In
the case of precipitation titration (AgNO3 vs NaCI) of the indicator is adsorbed at the end point
producing a characteristic colour on the precipitate.
In froth floatation process : (in metallurgy).
Softening of hard water : Ion exchange resins used for softening of hard water is based upon
selective and competive adsorption of ions on resins.
Na2Z + Ca+2 ⎯⎯
→ CaZ + 2 Na+
The organic polymers containing groups like –COOH, –SO3H and –NH2 etc. possess the property of
selective adsorption of ions from solution. These are quite useful in the softening of water.
A sample of charcoal weighing 6 g was brought into contact with a gas contained in a vessel of one litre
capacity at 27ºC. The pressure of the gas was found to fall from 700 to 400 mm. Calculate the volume
of the gas (reduced to STP) that is adsorbed per gram of the adsorbent under the condition of the
experiment (density of charcoal sample is 1.5 g cm3).
The adsorption is taking place in a closed vessel hence if pressure falls there is correspondingly
increase in volume constant, excess of the volume of the gas would be adsorbed.
P1V1 = P2V2
V
1000
V2 = P1 1 = 700 ×
= 1750 mL.
P2
400
Ja
5.
lp
Actual volume of the flask = 1000 – volume of charcoal = 1000 –
6.00
= 996 mL.
1.50
a
Volume of the gas adsorbed = 1750 – 996 = 754 mL.

P V P V  125.67  400  273
Volume of the gas adsorbed per gram at STP  Using 1 1 = 2 2  =
= 60.19 mL.
300  760
T1
T2 

Sa
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Section (B) : Catalysis
⚫
Catalysts : Berzillus in 1835 used the word catalyst first time for some substance which alter rate of
chemical reaction and themselves remain chemically and quantitatively unchanged after the reaction
and the phenomenon is known as catalysis.
Eg : Potassium chlorate when heated at 653K to 873K, it gives O2, When MnO2 is used in this reaction
the O2 is quickly released at the low temperature hence MnO2 is a catalyst
2KCIO3 → 2KCI + 3O2
⚫
Homogeneous Catalysis : When catalysts and reactants are in same phase then the process is said
to be homogeneous catalysis and
NO(g)
Eg :
(i) 2SO2 (g) + O2 (g) ⎯⎯⎯
→ 2SO3 (g)
HCl( )
(ii) CH3COOCH3 ( ) ⎯⎯⎯
→CH3COOH(aq)
H2SO4 ( )
(iii) C12H22O11(aq.) + H2O( ) ⎯⎯⎯⎯
→C6H12O6 (aq.) + C6H12O6 (aq.)
Glucose Fructose
⚫
Heterogenous Catalysis : When catalysts and reactants are in different phases, then process is know
as heterogenous catalysis and catalyst is called heterogeneous catalyst
Pt(s)
Eg :
(i) 2SO3 (g) + O2 (g) ⎯⎯⎯
→ 2SO3 (g)
Fe(s)
(ii) N2 (g) + 3H2 (g) ⎯⎯⎯
→ 2NH3 (g)
Pt(s)
(iii) 4NH3 (g) + 5O2 (g) ⎯⎯⎯
→ 4NO(g) + 6H2O(g)
422
Ni(s)
(iv) Vegetable oils ( ) + H2 (g) ⎯⎯⎯
→ Vegetable ghee (s).
Types of Catalysis
⚫
Negative
Substance which decrease the
rate of chemical reaction
H2O2
2H2O2 ⎯⎯⎯⎯⎯⎯⎯
→ 2H2O+
negative catalyse
Auto
One
of
Product
behave as catalyst
for that reaction and
increase the rate of
reaction, then this
phenomena is called
autocatalysis
CH3COOC2H5 + H2O

CH3COOH+C2H5OH
Induced
When one reaction
catalyses
another
reaction, then this
phenomena is called
as induced catalysis
Na2SO3 → Na2SO4
a
ri
⚫
Positive
Substance
which
increase the rate of
chemical reaction
Pt
2SO2 + O2 ⎯⎯→
2SO3
Promotors/Activators : Substance which are not catalyst themselves but its presence can increase
the catalytic activity of catalyst. Promotors increase the number of active sites on the surface Eg :
(i)
Fe(catalyst)
N2 + 3H2 ⎯⎯⎯⎯⎯
→ 2NH3
Mo(promotor )
Ni(catalyst)
→ Vegetable ghee.
Vegetable Oil + H2 ⎯⎯⎯⎯⎯
(iii)
CO + 2H2 ⎯⎯⎯⎯⎯⎯
→ CH3OH
uh
(ii)
Cu(promotor )
Cr2O3 (promotor )
Catalytic Poisons/Anti catalysts : Substance which
are not catalyst themselves but whose
presence decrease the activity of the catalyst. Poisoning is due to preferential adsorption of poison on
the surface of the catalyst.
Fe(catalyst )
N2 + 3H2 ⎯⎯⎯⎯⎯⎯⎯
→ 2NH3
(i)
CO / H2 S(catalytic poisons)
(iii)
As2 S3 (catalytic poisons)
Pd(catalyst )
→ RCHO + HCI
Rosenmund Reactions : RCOCl + H2 ⎯⎯⎯⎯⎯⎯⎯
BaSO4 ( poisons catalytst )
Characteristics of Catalysis :
(i)
(ii)
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(iii)
A Catalyst remains unchanged in mass and chemical compositions at the end of reactions.
However its physical state can be changed.
Eg : Granular MnO2 during decomposition of KClO3 is left as powder at the end of the reaction.
Highly efficient: Finely devided state of catalyst is more efficient for the reactions because
surface area increases and more adsorption take place.
A catalyst cannot initiate the reaction. But some times the activation energy is so large that
practically a reaction may not start until a catalyst lowers the activation energy significantly. For
example, mixture of hydrogen and oxygen do not react at room temperature but the reaction
occurs very rapid in presence of Pt black.
room temperature
H2 + O2 ⎯⎯⎯⎯⎯⎯
→ No reaction
Pt black
H2 + O2 ⎯⎯⎯⎯
→ H2O.
Highly specific: Catalyst are generally specific in nature. A substance which act as a catalyst
in a particular reaction, fails to catalyse other reaction.
Catalyst cannot change equilibrium state but it helps to attain equilibrium quickly.
A catalyst does not change the enthalpy, entropy and free energy of a reaction.
Highly active under optimum temperature: There is a particular temperature at which the
efficiency of a catalyst is maximum, this temperature is known as optimum temperature. On
either side of the optimum temperature, the activity of catalyst decreases. (Optimum range :
298 to 310 K).
Highly active under optimum pH: Range is pH 5 to 7.
Influence of Inhibitor or poison.
a
⚫
Platinised asbestos(catalyst )
2SO2 + O2 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
→ 2SO3
lp
(ii)
Ja
⚫
ZnO(catalyst)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
⚫
Adsorption Theory of Heterogeneous Catalyst : This theory explains the mechanism of
heterogeneous catalyst. This theory is combination of two theory, intermediate compound formation
theory and the old adsorption theory, the catalytic activity is localised on the surface on the catalyst.
The mechanism involves 5 steps.
(i) Diffusion of reactant to the surface of the catalyst of.
423
(ii) Adsorption of reactant molecules on the surface of the catalyst.
(iii) Formation of activated intermediate.
(iv) Formation of reactions product on the catalyst surface.
(v) Diffusion of reactions product from the catalyst surface or desorption.
Ja
uh
a
ri
Examples : Let us consider addition of H2 gas to ethlene in presence of Ni catalyst, the reaction takes
places as follows.
lp
Factors Supporting Theory :
(i)
This theory explains the role of active centre, more free valency which provides the more space
for more adsorption and concentration increases as a result of increase in rate of reaction.
(ii)
Rough surface has more active centres and pores, there will be more free valency so more will
be rate of reaction.
The theory explains the centre action of promoters which
occupied the interstial void as a result surface area for the
adsorption increases therefore rate of reaction increases.
.
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(iii)
a
⚫
(iv)
The theory explains the function of poisons or inhibitors. In
poisoning preferential adsorption of poisons takes place on the
catalyst, surface area for the adsorption on the catalyst
decreases hence rate of reaction decreases.
more strain
H
H
Ni
Ni surface area increase
promoters
inhibitors
catalyst
424
Some Industrial Catalytic reactions
Process
1
Haber's Process for the manufacture of ammonia
N2 + 3H2 → 2NH3
3
4
⚫
Ostwald's process for the manufacture of nitric acid.
4NH3 + 5O2 → 4NO + 6H2O ;
2NO + O2 → 2NO2
4NO + 2H2O + O2 → 4HNO3.
Lead chamber process for the manufacture of sulphuric acid
2SO2 + O2 → 2SO3;
SO3 + H2O → H2SO4.
Contact process for the manufacture of sulphuric acid
2SO2 + O2 → 2SO3;
SO3 + H2SO4 → H2S2O7
Shape selective catalysis
Platinised asbestos
Temperature 300ºC
Nitric oxide.
Platinised
asbestos
or
vanadium pentoxide(V2O5).
Temperature 400–450ºC.
uh
2
Catalyst
Finely
divided
iron,
Molybdenum as promoter.
Conditions: 200 atmospheric
pressure
and
450-500ºC
temperature.
a
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S.N.
2.
ZSM−5
R – OH ⎯⎯⎯⎯
→ gasoline
It convert alcohol directly into gasoline by dehydrating them to give a mixture of hydrocarbon.
3.
4.
a
Enzyme catalyst :
lp
1.
2.
Ja
1.
Catalysis action depends upon.
Shape of the catalyst (Pore structure or cavities)
Size of reactant and product.
Zeolites
They are honey comb like structure i.e. shape selective catalysis.
They are microporous aluminium silicate with 3D silicate in which silicon atom are replaced by
aluminum atoms i.e. Al–O–Si
The pore size is generally in range 260-740 pm.
General formula of zeolite is Nax/n[(AlO2)x(SiO2)y].zH2O
Use of Zeolite
It is used as catalyst in petrolium industries for cracking hydrocarbon and isomerism i. e. ZSM-5.
1.
2.
Sa
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Enzyme are complex nitrogenous organic compound which are produced by living plants and animal
the chracterstics of enzyme catalyst are.
1. Most highly efficient
2. Highly specific in nature
3. Highly active under optimum temperature
4. Highly active under optimum Ph.
5. Increase activity in presence of coenzyme
6. Influence of induce and poison
Section (C) : Classification and Preparation of Colloid
⚫
Colloid Solution :
Colloid State : A substance is said to be in colloidal state when the size of the particle of disperse
phase is greater than particle of true solution and less than that of suspension solution particle, their
range of diameters lie between 1 and 1000 nm (10–9 to 10–6 m).
Colloid solution : It is a heterogeneous system consisting of 2 phase :
(1) Disperse Phase (D.P) : The phase which is dispersed through the medium is called dispersed
phase or discontinuous phase or internal phase.
(2) Dispersion Medium (D.M) : A medium in which colloidal particles are dispersed is called dispersion
medium. It is also known as continuous phase or outer phase or external phase.
Colloidal solution = D.P. + D.M.
Ex. In Gold sol, Gold is D.P and water is D.M.
425
⚫
Differentiating point of colloids :
(1) A colloid is a heterogeneous system in which one substance is dispersed (dispersed phase) as
very fine particles in another substance called dispersion medium.
(2) The solution and colloid essentially differ from one another by particle size.
* In a solution, the particles are ions or small molecules.
* In a colloid, the dispersed phase may consist of particles of a single macromolecule (such as
protein or synthetic polymer) or an aggregate of many atoms, ions or molecules.
(3) Colloidal particles are larger than simple molecules but small enough to remain suspended. They
have a range of diameters between 1 and 1000 nm (10–9 to 10–6 m).
Classification of colloids :
DP
Solid
Solid
Solid
Liquid
Liquid
Liquid
Gas
Gas
DM
Solid
Liquid
Gas
Solid
Liquid
Gas
Solid
Liquid
Type of colloid
Solid Sol
Sol
Aerosol
Gel
Emulsion
Liquid Aerosol
Solid Sol
Foam
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On the basis of physical state of D.P. and D.M.
On the bases of physical state of D.P. and D.M. colloidal solution may be divided into eight system.
Examples
Some coloured glasses, and gem stones
Paints, cell fluids
Smoke, dust
Cheese, butter, jellies
Milk, hair cream
Fog, mist, cloud, insecticide sprays
Pumice stone, foam rubber
Froth, whipped cream, soap lather.
uh
1.
Ja
* Solution of gas in gas is not a colloidal system because it forms homogeneous mixture.
On the basis of D.M. : Colloidal solution are classified as
D.M.
Name of colloidal system
Water
Hydro sol or aqua sol
Alcohol
Alco sols
Benzene
Benzo sols
Air
Aero sols
*
*
Aquadag & oildag are colloidal solution of graphite in water & oil respectively.
Colloidal solution are often termed as sol.
On the Basis of interaction of D.P. for D.M. : There are two types(i) Lyophilic colloids / liquid loving sols / intrinsic colloid. The colloidal solution in which the
particles of the dispersed phase have a great affinity (or love) for the dispersion medium, are called
lyophilic colloids. These solutions are easily formed and the lyophilic colloids are reversible in nature. In
case when water acts as the dispersion medium, the lyophilic colloid is called hydrophilic colloid. The
common examples of lyophilic colloids are glue, gelatin, starch, proteins, egg albumin, rubber, etc.
Sa
nk
a
3.
lp
2.
*
(ii) Lyophobic colloids / solvent hating colloid / extrinsic colloid. The colloidal solutions in which
there is no affinity between particles of the dispersed phase and the dispersion medium are called
lyophobic colloids. Such solutions are formed with difficulty only by special methods. These sols are
readily precipitated (or coagulated) upon addition of small amounts of electrolytes, by heating or by
shaking and hence are not stable. Further, once precipitated, they do not give back the colloidal sol by
simple addition of the dispersion medium. Hence these sols are also called irreversible sols. They need
stabilising agents for their preservation. In case the dispersion medium is water, the lyophobic sol is
called hydrophobic colloid. For example, the solution of metals like Ag and Au, hydroxides like Al (OH) 3,
Fe(OH)3, metal sulphides like As2S3 etc.
Lyophilic sols are more stable than lyophobic sols, the additional stability is due presence of an
envelope of the solvent layer (say water) around the colloidal particle, the process is known as
hydration, To coagulate a hydrophilic sols we have to add a dehydrating agent in addition to electrolyte.
426
2
3
4
5
6
The particles of colloids are heavily
hydrated due to the attraction for
the solvent.
Hydration
8
Viscosity
9
Tyndall effect
It is higher than that of dispersion
medium
They do not show tyndall effect
10
Surface tension
Lower than dispersion medium
Coagulation
of
precipitation
Migration in electric
field
Precipitated by high concentration
of electrolyte
May or may not migrate as they
may not carry charge
Mostly organic nature; Starch and
Gelatin
On the basis of chemical composition :
Inorganic Colloids
(i) Metal sols : Cu, Ag, Au, Pt Sols.
(ii) Non Metal sols : S, I2, Graphite
(iii) Sol of oxide and hydroxide : SnO2, TiO2, Fe2O3, Fe(OH)3, AI(OH)3, Cr(OH)3
(iv) Salt Sol - AgBr, AgI, As2 S3, etc.
Sa
nk
4.
Example
lp
13
It is nearly same as that of dispersion
medium
They show tyndall effect.
It is nearly same as that of dispersion
medium
Precipitated by low concentration of
electrolyte
Migrate toward anode or cathode as
these particles carry charge
Inorganic nature; Transition metal salt
in water, gold etc., Metal solution
a
12
The particles of colloids are not
appreciably hydrated.
Ja
7
11
a
ri
1
DISTINCTION BETWEEN LYOPHILIC AND LYOPHOBIC COLLOIDS
Property
Lyophilic colloids
Lyophobic colloids
There are easily formed by direct These are formed only by special
Ease of preparation
mixing.
methods
Reversible
or
These are reversible in nature
These are irreversible in nature.
irreversible nature
The particles of colloids are true The particles are aggregates of many
Particles nature
molecules but are big in size
molecules
These are unstable and require traces
Stability
These are very stable
of stabilizers
The addition of small amount of
Action
of
electrolytes
causes
precipitation
No effect
electrolytes
(called
coagulation)
of
colloidal
solution.
The particles move in a specific
The particles do not carry any direction either towards anode or
Charge on particles
charge.
cathode depending upon their charge
in an electric field
uh
S.No.
Organic Colloids
(i) Homopolar sol - In this type of colloid, particles carry similar type of charge. eg. Sol of rubber in
benzene which contain - ve charge colloidal particle of latex.
(ii) Hydroxy Sol- Starch sol
5.
On the basis of charge on particles
(i) Positive Sol
(a) Metal Oxide & Hydroxide - SnO2, TiO2, Fe2O3, AI(OH)3, Fe(OH)3, Cr(OH)3.
(b) Basic Dyes Methylene blue, vismark brown.
(ii) Negative Sol
(a) Metal sol- Ag, Au, Pt, Cu
(b) Acidic dye - congo red, eosin
(c) Sulphide sol- CdS, HgS, As2S3, Sb2S3.
(d) Natural sol- Blood, clay, charcoal, latex rubber, dust particle in water, starch carbon particle in
smoke, gum.
427
On the basis of type of particles of dispersed phase :
6.
Multimolecular, macromolecular and associated colloids
Multimolecular colloids : In this type, the particles consist of an aggregate of atoms or small
molecules size less than 1 nm. For example, sols of gold atoms and sulphur (S 8) molecules. In these
colloids, the particles are held together by van der Waal’s forces.
⚫
Macromolecular colloids: In this types, the particles of the dispersed phase are sufficiently big in size
(macro) to be of colloidal dimensions. These macromolecules forming the dispersed phase are
generally polymers having very high molecular masses. These colloids are quite stable and resemble
true solutions in many respects. Naturally occurring macromolecules are starch, cellulose, proteins,
enzymes, gelatin, etc.
⚫
Associated colloids (Micelles): These are the substances which behave as normal strong electrolytes
at low concentration but behave as colloidal particles at higher concentration. These associated
particles are also called micelles. Ex. Soap.
a
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⚫
Micelles :
Ja
uh

Micelles are relatively small, spherical structures composed of any where for a few to few
thousand molecule that attract one another to reduce surface tension within a membrane of cell.
The formation of micelles takes place only at particular temp, that temperature is called Kraft
temperature.
The concentration above which micelle formation becomes appreciable is termed is critical micelles
concentration. Its value depend upon natures of D.P. and D.M. eg. Surface active agent (surfactants,
which decrease the surface tension) like soaps and detergents form micelle beyond CMC (~10 –3
mol/litre for soaps).
* Usually longer the hydrophobic chain, smaller is its CMC.
* Also CMC increase with decreasing polarity of the D.M.
* The micelles ‘formation takes place only above a particular temperature called as Kraft Temperature
(Tk ).
*At CMC, the micelles are spherical in shape, but that start flattening with increase in concentration and
ultimately form sheet or film like structures which have a thickness of two molecules. These are called
lamelar micelles or McBain Micelles.
Sa
nk
a
lp
Mechanism of micelle formation: Let us take the example of soap solutions. Soap is sodium salt of a
higher fatty acid and may be represented as RCOO –Na+ e.g., sodium stearate viz. CH3(CH2)16COO–
Na+ which is a major component of many bar soaps. When dissolved in water, it dissociates into
RCOO- and Na+ ions. The RCOO– ions, however, consist of two parts i.e., long hydrocarbon chain R
(also called non-polar 'tail') which is hydrophobic (water repelling) and the polar group COO – (also
called polar-ionic 'head') which is hydrophilic (water loving). The RCOO – ions are, therefore, present on
the surface with their COO– groups in water and the hydrocarbon chains R staying away from it, and
remain at the surface, but at higher concentration these are pulled into the bulk of the solution and
aggregate in a spherical form with their hydrocarbon chains pointing towards the centre with COO – part
remaining outward on the surface. An aggregate thus formed is known as ‘Ionic micelle'. These
micelles may contain as many as upto 100 such ions.
Aggregation of RCOO– ions to form an ionic micelle.
Similarly, in case of detergents, e.g., sodium lauryl sulphate
viz. CH3(CH2)11SO4–Na+, the polar group is –SO4– along with
the long hydrocarbon chain. Hence, the mechanism of micelle
formation is same as that of soaps.
Example of micelles :
(i) Sodium stearate C17H35COO–Na+(Soap).
(ii) Sodium lauryl sulphate CH3 [CH2]11 SO4– Na+ (Detergent).
(iii) Cetyl trimethyl ammonium bromide (Detergent). CH3(CH2)15N+(CH3)3Br–.
The Cleansing Action of Soaps: It has been mentioned earlier that a micelle consists of a
hydrophobic hydrocarbon like central core. The cleansing action of soap is due to these micelles,
because oil and grease can be solubilised in their hydrocarbon, like centres which are not otherwise
soluble in water. This is shown diagrammatically in Figure. The dirt goes out along with the soap
micelles.
428

Sodium stearate (C17H35COO– Na+)
(a)
Hydrophobic tail
(b)
(c)
(d)
Fig. : The cleansing action of soap.
a
ri
a
(e)
[A]
Ja
uh
(a) A sodium stearate molecule
(b) The simplified representation of the molecule that shows a hydrophilic head and a hydrophobic tail
(c) Grease (oily substance) is not soluble in water
(d) When soap is added to water, the non-polar tails of soap molecules dissolve in grease
(e) Finally, the grease is removed in the form of micelles containing grease.
*Surfactants : They can be ionic as well as non-ionic. The ionic are soaps and detergent. The
surfactant gets adsorbed at the interface between the dispersed droplets and dispersion medium in the
form of mono molecular layer and lowers the interfacial tension between oil and water so as to facilitate
the mixing of two liquids.
Preparation of lyophobic colloidal sols :
Condensation methods :
Sa
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a
lp
In these methods particles of atomic or molecular size are induced to combine to form aggregates
having colloidal dimensions. For this purpose chemical as well as physical methods can be applied.
(a) Chemical methods. Colloidal solutions can be prepared by chemical reactions leading to formation
of molecules by double decomposition, oxidation, reduction or hydrolysis. These molecules then
aggregate leading to formation of sols.
(i) Double decomposition : When a hot aqueous dilute solution of arsenous oxide (As2O3) is mixed
with a saturated solution of H2S in water, a colloidal sol of arsenous sulphide (As 2S3) is obtained.
Double decomposition
As2O3(in hot water) + 3H2S (saturated solution in H2O) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→ As2S3(sol) + 3H2O
CH ( OH) COOK
CH ( OH) COOK
+ 3H2S → |
+ Sb2S3 (orange sol) + 2H2O
|
CH ( OH) COO ( SbO )
CH ( OH) COOH
(ii) Oxidation : A colloidal sol of sulphur is obtained by passing H 2S into a solution of sulphur dioxide.
Oxidation → 3S(sol) + 2H2O
SO2 + 2H2S(saturated solution in H2O) ⎯⎯⎯⎯⎯⎯
Sulphur sol can also be obtained when H2S is bubbled through Br2 water or nitric acid (oxidizing agent).
2H2S (aq.) + Br2 (aq.) ⎯⎯
→ 2HBr (aq.) + S (sol).
or
by bubbling O2 (g) through a solution of H2S : 2H2S (aq.) + O2 (g) ⎯⎯
→ 2H2O (l) + 2S (sol).
(iii) Reduction : Colloidal sol of metals like gold, silver solution are obtained by following method.
2 AuCl3 + 3 HCHO + 3H2O ⎯⎯
→ 2Au(sol) + 3HCOOH + 6HCl.
(purple of cassius)
Reduction
2 AuCl3 + 3 SnCl2 ⎯⎯⎯⎯
→ 3 SnCl4 + 2Au(sol)

Reduction
AgNO3 + tannic acid ⎯⎯⎯⎯
→ silver sol.
NH2NH2 can also be used as reducing agent.
*Sol of gold is also known as purple of cassius.
(iv) Hydrolysis : A colloidal sol of metal hydroxides like Al(OH) 3 or Cr(OH)3 is obtained by boiling a
dilute solution of FeCl3 , AlCl3 or CrCl3 .
FeCl3 + 3H2O ⎯⎯
AlCl3 + 3H2O ⎯⎯
→ Fe(OH)3 (sol) + 3HCl ;
→ Al(OH)3 (sol) + 3HCl
The colloidal sol of sillicic acid is also obtained by hydrolysis of dilute solution of sodium silicate with
hydrochloric acid. Na4SiO4 + 4HCl ⎯⎯
→ Si(OH)4 (sol) + 4NaCl.
(b) Physical methods : The following physical methods are used to prepare the colloidal solutions.
429
(i) By Exchange of solvent : When a true solution is mixed with an excess of the other solvent in
which the solute is insoluble but solvent is miscible, a colloidal sol is obtained. For example,
⚫ when a solution of sulphur in alcohol is poured in excess of water, a colloidal sol of sulphur is
obtained.
⚫ when a solution of phenolphthalein in alcohol is poured in excess of water
a white sol of
phenolphthalein is found.
⚫ Phenolphthalein, I2 , sulphur sol can be prepared by this methods.
(ii) Excessive cooling : Molecules of certain substance condense together on excess cooling to form
colloidal size particle. The colloidal sol of ice in an organic solvent such as CHCl 3 or ether can be
obtained by freezing a solution of water in the solvent. The molecules of water which can no longer be
held in solution separately combine to form particles of colloidal size.
[B]
a
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(iii) By condensing vapours of a substance into solvent : Substance like sulphur and Hg in water
are prepared by passing their vapours in cold water containing small amount of stabilising agent like
ammonium nitrate.
Dispersion Methods : In these methods large particles of the substance are broken into particles of
lp
Ja
(a) Mechanical dispersion (By colloidal mill):
Substance is first finely powdered.

It is shaken with the D.M. to form a suspension.

This suspension is passed through a colloidal mill.

The simplest type of colloidal mill is disc mill which
consists of two metal discs nearly touching each other
& rotating in opposite.

Direction at a high speed (7,000 revolutions per min.).

The suspended particles are broken to produce
colloidal size particle.
* This method is used to prepare printing ink.
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colloidal dimensions in the presence of dispersion medium. These are stabilized by adding some
suitable stabilizer. Some of the methods employed are given below :
Sa
nk
a
(b) Electrical disintegration or Bredig's Arc method : This
process involves dispersion as well as condensation. Colloidal
sols of less reactive metals such as gold, silver, platinum,
copper, lead etc., can be prepared by this method. In this
method, electric arc is struck between electrodes of the metal
immersed in the dispersion medium as shown in fig. The
intense heat produced vaporises the metal, which then
condenses to form particles of colloidal size by surrounding
cooling mixture (ice).
*A slight trace of KOH is added in water to stabilized colloidal solutions.
(c) Ultrasonic dispersion : Ultrasonic vibration (having frequency larger than audible range) can bring
about the transformation of coarse suspension or liquids like oil, mercury etc. into colloidal range.
*This is the latest method for preparation of metal oxides and metal sulphide sols from their coarse
suspension.
*It is a suitable technique for oils also. This method also comprises both dispersion and condensation.
(d) Peptization: The term has originated from the digestion of proteins by the enzyme pepsin.
Peptization may be defined as (the process of converting a precipitate into colloidal sol by shaking it
with dispersion medium in the presence of a small amount of electrolyte).
The electrolyte used for this purpose is called peptizing agent. This method is applied, generally, to
convert a freshly prepared precipitate into a colloidal sol. During peptization, the precipitate adsorbs
one of the ions of the electrolyte on its surface.
430
The ion adsorbed on the surface is common either with
the anion or cation of the electrolyte. This causes the
development of positive or negative charge on
precipitates which ultimately break up into smaller
particles having the dimensions of colloids.
For example :
(i) When freshly precipitated Fe(OH)3 is shaken with aqueous solution of FeCl3 (peptizing agent) it
adsorbs Fe3+ ions and thereby breaks up into small-sized particles.
FeCl3 ⎯⎯
Fe(OH)3 + Fe3+ ⎯⎯
→ Fe3+ + 3Cl– ;
→ Fe(OH)3 | Fe3+
+3
+3
+3
Fe
Peptizing agent
ppt of Fe (OH)3
+3
+3
+3
+3
+3
a
ri
+
+3
+3
+3
+3
+3
+3
+3
+3
+3
+3
+3 +3
+3
+3 +3
+3
+3
+3
+3
+3
+3
+3
+3
+3
+3
+3
+3
3+
+3
uh
Charge Colloidal particals of Fe (OH)3
Ja
(ii) Freshly prepared stannic oxide on treatment with a small amount of dilute hydrochloric acid forms a
stable colloidal sol of stannic oxide, SnO2 ; Sn4+ .
SnO2 + 4HCl → Sn4+ + 2H2O + 4Cl– ; SnO2 + Sn4+ → SnO2 / Sn4+ .
(iii) Freshly precipitated silver chloride can be converted into a colloidal sol by adding a small amount of
hydrochloric acid, AgCl : Cl– .
(iv) Cadmium sulphide can be peptised with the help of hydrogen sulphide, CdS : S 2– .
Section (D) : Purification and Properties of Colloid
Purification of Colloidal Sols : The colloidal sols obtained by various methods are impure and
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contain impurities of electrolytes and other soluble substances. These impurities may destabilise the
sol. Hence, they have to be removed. A very important method of removal of soluble impurities from
sols by a semipermeable membrane is known as dialysis.
Sa
nk
a
A. Dialysis : It is a process of removing a dissolved substance from a colloidal solution by means
diffusion through suitable membrane. Since particles in true solution (ions or smaller molecules) can
pass through animal membrane or parchment paper or cellophane sheet but colloidal particle do not,
the appratus used for this purpose is called Dialyser.
A bag of suitable membrane containing the colloidal solutions is suspended in a vessel through which
fresh water continously flow. The molecules and ions (crystalloids) diffuse through membrane into the
outer water & pure colloidal solution is left behind.
Movement of ions across the membrane can be expedited by applying electric potential through two
electrodes as shown in fig.
This method is faster than simple dialysis and is known as
Electrodialysis.
*The most important applications of dialysis is in the purification of blood in the artificial kidney machine.
In case of kidney failure, blood cannot be purified. Under such condition, the blood is separated from
dissolved toxic impurities by dialysis and re-introduced in the bloods stream.
*Dialysis is not applicable for non-electrolytes like glucose, sugar, etc.
431
B. Ultra Filtration : In this method, colloidal sols are purified by carrying out filtration through special
type of graded filters called ultra-filters. These filter papers allow only the electrolytes to pass through.
These filter papers are made of particular pore size by impregnating with colloidal solution and
subsequently hardened by soaking in formaldehyde collodion. In order to accelerate the filtration
through such filter papers, increased pressure or suction is employed.
When SO2 is bubbled into H2S gas, colloidal sol is formed. What type of colloidal sol is it ?
2H2S + SO2 ⎯⎯
→ 2H2O + 3S (colloidal).
Lyophobic colloidal sol of sulphur is formed.
Ex-3.
A reddish brown positively charged sol is obtained by adding small quantity of FeCl 3 solution to freshly
prepared and well washed Fe(OH)3 precipitate. How does it take place ?
It is due to adsorptions of Fe3+ ions on the surface of Fe(OH)3 which gives colloidal sol.
Fe(OH)3 (ppt.) + Fe3+ (ions adsorbed) ⎯⎯
→ [Fe(OH)3]Fe3+ (colloidal sol).
Ex-4.
Sol.
Suppose we have a cube of 1.00 cm length. It is cut in all three directions, so as to produce eight
cubes, each 0.50 cm on edge length. Then suppose these 0.50 cm cubes are each subdivided into
eight cubes 0.25 cm on edge length, and so on. How many of these successive subdivisions are
required before the cubes are reduced in size to colloidal dimensions of 100 nm.
We find that every division in two equal halves also reduces the size of edge lengths to one half.
1
In first subdivision 1 cm is reduceds to 0.5 cm =
cm.
2
uh
Sol.
a
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Ex-2.
Sol.
2
n
1
 1
cm =   cm
4
2
Ja
In second subdivision 0.5 cm is reduced to 0.25 cm =
 1
In n subdivision 1 cm is reduced to   . Size of colloidal particles lies between 1 to 1000 mm.
2
Thus to make n subdivision required particle size may be attained.
n
lp
a

 1
–9
–7
 2  = 100 nm = 100 × 10 m = 100 × 10 cm.
 
n log 2 = 5
n × 0.3010 = 5.
5
n=
= 16.61 = 17 subdivisions are required for dimension of 100 nm.
0.3010
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Important properties of colloidal sols :
⚫ Heterogeneous character :
Colloidal sols are heterogeneous in character as they consist of two phases.
(a) dispersed phase and
(b) dispersion medium.
⚫ Visibility : Due to scattering caused by the colloidal particles, it will appear as a bright spot moving
randomly.
⚫
Filterability : Colloidal particles pass through an ordinary filter paper. However, the particle do not
pass through other fine membranes.
⚫
Colligative Properties : Colloidal sols show the colligative properties viz. relative lowering of
vapour pressure, elevation in boiling point, depression in freezing point and osmotic pressure. However,
due to high average molecular masses of colloidal particles, mole fraction of the dispersed phase is
very low. Hence, the values of the colligative properties observed experimentally are very small. Only
osmotic pressure measurements are used in determining the molecular mass of polymers.
⚫
Optical Properties-Tyndall effect : Tyndall, in 1869, observed that if a strong beam of light is
passed through a colloidal sol placed in a dark place, the path of the beam gets illuminated. This
phenomenon is called Tyndall effect, which is due to the scattering of light by the colloidal particles. The
illuminated path of beam is called Tyndall cone. This phenomenon is due to scattering of light from the
surface of colloidal particles. In a true solution there are no particles of sufficiently large diameter to
scatter light & hence the beam is invisible.
432
Eye
microscope
Tyndall cone
Light source
Scattered light
True solution
Colloidal solution
⚫
Application of Tyndall Effect :
In making ultramicroscopes.
In finding heterogenity of solution.
Ja
(i)
(ii)
uh
Example of Tyndall Effect
→
Blue colour of sky and sea water.
→
Visibility of tail of comets.
→
Light thrown from a projector in cinema hall.
→
Appearance of dust particle in a semi darked room.
a
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*The intensity of scattered light depends on the difference between the refractive indice of the D.P and
D.M., In lyophobic colloids, this difference is appreciable and therefore the tyndal effect is quite well
defined but in lyophilic sols the difference is very small and the tyndal effect is very weak. Thus in sols
of silicic acid, blood serum, albumin, etc. there is little or no tyndal effect.
Under what conditions is Tyndal effect observed ?
Tydalls effect is applicable when :
(a) The diameter of the dispersed particles is not much smaller than the wavelength of the light used.
(b) The refractive indices of the dispersed phase and the dispersion medium must differ greatly in
magnitude.
Ex-6.
In the lower layer of the atmosphere, there is a great deal of dust. When the weather is fine, it is
possible to see the magnificent red colour of the setting sun. What have these observation to do with
colloids ?
Dust in the atmosphere is often colloidal. When the sun is low down on the horizon, light from it has to
pass through a great deal of dust to reach your eyes. The blue part of the light is scattered away from
your eyes. You see the red part of the spectrum, which remains. Red sunsets are the Tyndall effect on
a large scale.
a
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Sol.
lp
Ex-5.
Sol.
⚫
Mechanical Properties :
Brownian movement: Robert Brown, a botanist, discovered in 1827 that
pollen grains placed in water do not remain at rest but move about
continuously and randomly. Later on, this phenomenon was observed in
case of colloidal particles when they were seen under an ultramicroscope.
The particles were seen to be in constant zig-zag motion as shown in fig.
This zig-zag motion is called Brownian movement.
Brownian movement arises because of the impact of the molecules of the
dispersion medium with the colloidal particles.
Dependence :
(i) Size of colloidal particles : Mobility 
(ii) Viscosity of solution : Mobility 
1
Size of the particle
1
Viscosity
(iii) Temperature : Mobility  Temperature
433
⚫
Factors Affecting Brownian Movement :
(i) If particles is large then brownian movement becomes less.
(ii) Brownian movement increases with increasing temperature.
(iii) The brownian movement does not change with time & remains same for months or even for a year.
⚫
Important :
(i) In confirmation of kinetic energy.
(ii) Determination of Avogadro numbers.
(iii) Stability of colloidal solution: Brownian movement does not allow the colloidal particles to settle
down to gravity & thus is responsible for their stability.
⚫
Electrical Properties (Electrophoresis) :
Ja
uh
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The existence of charge on colloidal particles is confirmed by electrophoresis experiment. When electric
potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles
move towards one or the other electrode. The movement of colloidal particles under on applied electric
potential is called electrophoresis.
Positively charged particles move towards the cathode while negatively charged particles move towards
the anode. Depending upon the direction a movement of particles towards cathode or anode
electrophoresis can be called ’cataphoresis‘ or ‘Anaphoresis’
Arsenious sulphide, gold, silver and platinum particles in their respective colloidal sole are negatively
charged while particles of Fe(OH)3, Al(OH)3 are positively charged.
(Fig. : A set up for electrophoresis.)
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Sa
nk
⚫
a
⚫
The colloidal solution is placed in a U-tube fitted with platinum electrodes. On passing an electric
current, the charged colloidal particles move towards the oppositely charged electrode. Thus, if arsenic
sulphide sol is taken in the U-tube, in which negatively charge particle of arsenic sulphide move
towards the anode.
*Earlier this process was called cataphoresis because most of the colloidal sols studied at that time
were positively charged and moved towards cathode.
Electro osmosis: When movement of colloidal particles is prevented by some suitable means (porous
diaphragm or semi permeable membranes), it is observed that the D.M. begins to move in an electric
field. This phenomenon is termed electrosmosis.
Sedimentation potential or Dorn potential : When the charged colloidal particles are made to settle
down under centrifugal field, there occurs a charge separation and a potential difference is developed.
This effect is called Dorn effect and the potential difference thus developed is called Dorn potential or
sedimentation potential. This process is reverse of electrophoresis.
Isoelectric point : The H+ concentration at which the colloidal particles have no charge is known as the
isoelectric point. At this point stability of colloidal particles becomes very less & do not move under
influence of electric field.
Streaming potential : A potential difference is developed across a porous partition when the
dispersion medium of a charged colloid is forced through it. This is called Streaming potential. This
process is reverse of electro-osmosis.
Charge on colloidal particles : Colloidal particles are either positively charged or negatively
charged. This charge is due to preferential adsorption of either positive or negative ions on their
surface. There is adsorption of common ion present in excess.
⚫ Fe(OH)3 sol prepared by the hydrolysis of FeCl3 solution adsorbs Fe3+ and this is positively charged.
FeCl3 + 3H2O
Fe(OH3) + 3HCl ;
Fe(OH)3 + FeCl3 → Fe(OH)3 Fe3+ :
3Cl–
Fixed part
Diffused part.
Positive charge on colloidal sol is due to adsorption of Fe 3+ ion (common ion between Fe(OH)3 and
FeCl3).
⚫ As2S3 colloidal sol is obtained when As2O3 is saturated with H2S :
As2O3 + 3H2S → As2S3 + 3H2O.
As2S3 adsorbs S2– ions (common between H2S and As2S3 and thus is negatively charged).
As2S3 + H2S → As2S3 S2– : 2H+.
⚫
⚫
⚫
434
⚫
AgI in contact with AgNO3 forms positively charged colloidal sol due to adsorption of Ag+ ion.
AgI + AgNO3 → [AgI]Ag+ : NO3– , AgI in contact with KI forms negatively charged colloidal sol due to
adsorption of I– ion
AgI + KI → AgI I– : K+.
⚫
SnO2 in acidic medium forms positively charged colloidal sol due to adsorption of Sn 4+ formed.
SnO2 + 4H+ → Sn4+ + 2H2O
SnO2 + Sn4+ → SnO2 Sn4+
SnO2 in alkaline medium forms negatively charged colloidal sol due to adsorption of SnO 32– formed.
SnO2 + 2OH– → SnO32– + H2O
SnO2 + SnO32– → SnO2 SnO32–
Charge on colloidal particle may be due to some other reasons also e.g.
*Due to electron capture by collidal particle during electro dispersion of metal.
*Due to self dissociation.
Electric Double Layer Theory or Helm-holtz Electric double layer :
Electric double layer Theory
–
–
+
+ +
+
–
+
–
Colloidal
Particle
+
–
+
–
+
–
❖
a
Sa
nk
K
Sol.
Mobile layer
–
–
+
K
K
+
+
–
I–
I
–
K+
I
–
I
I–
+
K
–
AgI
I– I– I–
Zetapotential K+
Ex-8.
–
–
Fixed layer
Sol.
+
+
The combination of two layer of opposite charges (+ve and –ve charge) around colloidal particle is
called Helm-Holtz electrical double layer.
First layer of ions is firmly held and is termed as fixed layer.
Secondary layer is mobile and is termed as diffused layer.
The charges of opposite sign on fixed and diffused parts of double layer result in difference in potential
between these layers.
This potential difference between the fixed layer and diffused layer of opposite charges is called
electrokinetic potential or zeta potential.
4
Z=
n = Viscosity coefficient
D
D = Dielectric constant of medium
 = Velocity of colloidal particle when an electric field is applied.
e.g.
AgNO3 + K I → AgI + KNO3
(excess)
Ex-7.
–
Ja
❖
❖
❖
+
+
+
–
Fixed layer
–
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❖
–
uh
–
–
a
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⚫
I
I–
K+
+
K+
K
Classify the following sols according to theirs charges :
(a) gold sol
(b) ferric hydroxide sol
(e) sulphur
(f) arsenious sulphide
Negatively charged colloidal sol : (a), (c), (d), (e), (f).
Positively charged colloidal sol : (b), (g).
Diffused layer
(c) gelatine
(g) titanium oxide.
(d) blood
SnO2 forms positively charged colloidal sol in acidic medium and negatively charged colloidal sol in
basic medium. Explain ?
SnO2 is amphoteric reacting with acid and base both. In acidic medium (say HCl) Sn 4+ ion is formed
which is preferentially adsorbed on SnO2 giving positively charged colloidal sol :
SnO2 + 4HCl ⎯⎯
;
→ SnCl4 + 2H2O
SnO2 + SnCl4 ⎯⎯
→ [SnO2]Sn4+(positively charged) + 4Cl–.
435
Section (E) : Coagulation, Protection And application of colloid
⚫
Coagulation/Flocculation : This process of aggregation of colloidal particles into an insoluble
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+ ++
+
+ + + + ++ + +
+
+ +
+ +
+ ++ + +
+ ++ +
+ ++
+ + + + + ++ + + + +
+ ++ +++ + ++
+ ++ + + + + +
++
+ ++ + + +
+ ++ + +
+ ++ +
+ ++
+
+ + +
+
+
+ + + + ++ +
+ ++ + +
+ ++ +
+ ++
+ + ++
+ ++ + + +
++++
+++ + + + + + +
+ + ++ + +
+
+ + + + + + ++
+++
+
+
+ + +
+
+
+ + + + ++ +
++
+ + ++ + ++ +
+ +
+ +
+ ++ ++
+
+ + + + +
+ ++ +
+ ++
+ + + + ++ +
+ + + + +
+ ++ +
+ ++
Neutralised
sol particles
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precipitate by the addition of some suitable electrolyte is known as coagulation. At lower
concentration of electrolytes, the aggregation of particles is called flocculation that can be reversed on
shaking while at higher concentration of electrolyte, coagulation takes place and the same cannot be
reversed simply by shaking. The stability of the lyophobic colloids is due to presence of charge on
colloidal particles. If, somehow, the charge is removed, the particles will come near to each other to
form aggregates and settle down under the force of gravity.
Coagulation value or Flocculation value: It needs to be noted that the coagulation of a colloidal
solution by an electrolyte does not take place until the added electrolyte has certain minimum
concentration in the solution. The minimum concentration of electrolyte in millimoles required to
cause coagulation of one litre of colloidal solution is called coagulation value. It is express in
terms of millimoles/litre.
millimoles of electrolyte
Coagulation value =
volumeof sol in litre
Coagulated sol
Sa
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a
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Ja
Coagulation of lyophobic sols can be carried out by the following methods.
(i) By electrophoresis
(ii) By mutual precipitation : It is a process in which oppositely charged sol are mixed in proper
proportion to neutralise the charge of each other causing coagulation of both the sol.
Example : Positively charged Fe(OH)3 and negatively charged As2S3 colloidal particle containing sol on
mixing get coagulated.
(iii) By Prolonged Dialysis : On prolonged dialysis, traces of the electrolyte present in the sol are
removed almost completely and the colloid becomes unstable and ultimately coagulate.
(iv) By Boiling : Sols such as sulphur and silver halides dispersed in water may be coagulated by
boiling because increased collisions between sol particle and the water molecule removed the
adsorbed electrolytes. This takes away the charge from the particles and helps them to coagulate.
(v) By cooling : Certain sol can also be coagulated by lowering temperature. For example,
accumulation of cream on the surface of milk on cooling. This is because at lower temperature the
dispersion medium molecules do not exert sufficient force on to the dispersed particles and hence the
Brownian motion becomes less effective.
(vi) By the addition of electrolyte : When excess of an electrolyte is added, the colloidal particles are
precipitated.
⚫
⚫
⚫
Comparision of relative coagulating power of two electrolyte for the same colloidal solution :
The coagulation value decrease with increase in charge of the coagulating ion.
1
Coagulating power 
.
coagulation value
coagulation valueB
coagulating power of electrolyte A
=
.
coagulation value A
coagulating power of electrolyteB
Factor-Affecting Coagulations : (i) Nature of sols : The lyophobic colloid can easily coagulate
because it is a less stable colloid, but lyophilic colloids coagulate hardly by the addition of electrolyte
due to protective layer of D.M. surrounding the colloidal particle.
(ii) Nature of electrolyte : In equimolar electrolyte, strong electrolyte have greater coagulating power
than weak electrolyte. Example : 0.1M NaCl > 0.1M CH3COOH.
Hardy-Schulze Rule : According to this rule greater is the valency of coagulating ion, greater its power
to cause precipitation. This is known as Hardy-Schulze rule.
In case of positive charged sol, the coagulating power of anion is in the order of [Fe(CN) 6]4– > PO43– >
SO42– > Cl–
In case of negative charged sol, the coagulating power of cation is in the order of Al 3+ > Ba2+ > Na+.
436
⚫
The coagulating power of bivalent ion is 20-80 times higher than monovalent ion and coagulating power
of trivalents is many times more than bivalent.
Ex-9
The particles of a particular colloidal solution of arsenic trisulphide (As2S3) are negatively charged.
Which 0.0005 M solution would be most effective in coagulating this colloidal solution. KCl, MgCl 2, AlCl3
or Na3PO4? Explain.
Since As2S3 is a negatively charged colloidal sol hence positively charged ion will cause its coagulation.
By Hardy-Schulze rule “greater the charge on ion, greater the coagulating power to coagulate
oppositely charged colloidal sol”, hence out of K+, Mg2+, Al3+ and Na+, Al3+ would be most effective.
Sol.
Protection of colloidal sols : Lyophilic colloidal sols are much more stable than lyophobic colloidal
weight of lyophilic sol in mg 10
volume of gold sol in mL
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Gold Number =
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a
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sols. This is due to the extensive solvation of lyophilic colloidal sols, which forms a protective layer
outside it and thus prevents it from forming associated colloids. Lyophobic sols can easily precipitate by
addition of small amount of an electrolyte. They can be prevented from coagulation by previous addition
of some lyophilic colloid. This is due to formation of a protective layer by lyophilic sols outside lyophobic
sols. Process of protecting the lyophobic colloid solution from precipitation by an electrolyte due to
previous addition of some lyophilic colloid is called protection of colloid and lyophilic colloidal sols are
called protective sols.
Eg : Gelatin, Sodium caseinate, Egg albumin, Gum arabic, Potato starch etc.,
Gelatin (lyophilic) protects gold sol (lyophobic) colloids is expressed in terms of gold number.
Gold Number : Zpsigmondy (1901) introduce a term called gold number it is defined as ‘’the
minimum amount of the protective colloid in milligrams which when added to 10 ml of a standard gold
sol is just sufficient to prevent a colour change from red to blue on the addition of 1 ml of 10% sodium
chloride solution. It may be noted that smaller of the gold number, greater will be protecting power of
the protective colloid.
1
Protecting power 
.
gold number
Uses of protective action :
(i) Gelatin is added in the preparation of ice cream to protect the particle of ice.
(ii) Protargol and Argyrol, is a silver sol protected by organic material used as eye drop.
Applications of Colloids : Colloids including emulsions find a number of uses in our daily life and
1.
a
industry. Some of the uses are given below.
In medicines : A wide variety of medicinal and pharmaceutical preparations are emulsions. Colloidial
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medicines are easily adsorbed by the body tissue because of large surface area.
* Colloidal antimony is used in curing kalaazar.
* Milk of magnesia, an emulsion, is used for stomach disorder.
* Colloidal gold is used for intramuscular injection.
* Colloidal sulphur are used as Germicides.
* Argyrol is a silver sol used as an eye lotion.
* Colloidal Fe(OH)3 is given to arsenic poisoning patients as it adsorbs arsenic and then gets omited
out.
2.
3.
4.
5.
Tanning
Photographic plate & Film
Rubber plating
Sewage disposal
437
6.
Cottrell smoke precipitator :
Smoke is a dispersion of negatively charged colloidal particles of
carbon in air and can be made free of these colloidal particles by
passing it through cottrell precipitator as shown in fig. installed in
the chimney of an industrial plant. It consists of two metal discs
charged to a high potential. The carbon particles get discharged
and precipitate, while gases come out from the chimney.
7.
Formation of deltas: The river water contains colloidal particles of sand and clay which carry
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negative charge. The sea water contains +ve ions such as Na +, Mg2+, Ca2+ etc. As the river water
meets sea water, these ions discharge the sand or clay particle which are precipitated in the form of
delta.
Artificial rain
Stop bleeding from a cut
Stop Screen
Preparation of nano-materials
In disinfectants : The disinfectants such as dettol and lysol give emulsions of the oil-in-water type
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8.
9.
10.
11.
12.
when mixed with water.
STEM TECHNOLOGY : The size and shape of the colloidal particles is determined with the help of
Section (F) : Emulsion and Gel
⚫
Ja
an electron microscope which has much more resolving power (of the order of 10 –12m.) The different
techniques used to study the colloidal particles are :
(i) Scanning electron microscope (SEM)
(ii) Transmission electron microscope (TEM) and
(iii) Scanning transmission electron microscope (STEM).
Emulsions : Pair of immiscible liquid is called emulsion. Emulsion are unstable and some time they
a
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are separated into two layers on keeping still, for the stabilising of an emulsion, a third component is
added called emulsifying-Agent form an interfacial film between D.P. and D.M.
Emulsion droplets are bigger than sol particles and can be seen under an ordinary microscope and
sometimes even with a magnifying glass.
Example : Milk is an emulsion in which liquid fat is D.P. and liquid water is D.M. and casein is
emulsifying agent.
 Demulsification : The separation of an emulsion into its constituent liquids is called demulsification.
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Various techniques employed for this are freezing, boiling, centrifugation, electrostatic precipitation or
chemical methods which destroys the emulsifying agents.
Demulsification can be brought about by :
(i) Freezing
(ii) Heating
(iii) Centrifugal action (Separation of cream of milk done by centrifugation).
(iv) Removal of emulsifiers by adding a better solvent for them like alcohol, phenol etc, called
demulsifiers.
⚫
Types of emulsions : Depending on the nature of the dispersed phase, the emulsions are classified
as:
(a) Oil in water emulsions
(b) Water in oil emulsions
Inversion of phase : The conversion of emulsion of oil in water (o/w) into water in oil (w/o) or vice
versa is called the inversion of phase.
[A] Identification of the type of emulsion : These two types may be identified by :
⚫
⚫
Dilution test : An emulsion can be diluted with any amount of the dispersion medium, while the
dispersed liquid, if added, forms a separate layer. Thus if a few drops of water added to the emulsion
are soluble in it, it is oil in water type and if immiscible, it is water in oil type.
Dye test : If a small amount of oil soluble dye gives a uniform colour to the emulsion, it is water in oil
type otherwise it is oil in water type.
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⚫
Electrical conductivity test : If conductivity of emulsion increases significantly by adding a very small
amount of electrolyte, it is oil in water type and if there is no significant increase in conductivity, it is
water in oil type.
[B] Applications of emulsions :
⚫
⚫
⚫
⚫
⚫
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Disinfactants like phenyl, dettol when mixed with water form emulsion.
Digestion of fat in small intestine occurs easily due to emulsion.
In metallurgical process the concentration of ore by froath floatation method is based upon emulsion.
Milk is an emulsion of liquid fat in water in which casein emulsifying agent.
Cleansing action of soap is due to formation of emulsions. Soaps and detergents emulsify the grease
along with the adhering dirt and carry them away in the wash water.
⚫ For concentrating ores, the finely powdered ore is treated with an oil. Oil forms emulsion with the ore
particles. When air is bubbled into the mixture, emulsion containing the particles of the mineral are
carried to the surface.
Gel :
Ex :
Gelatin dissolved in water forming a colloidal. Sol which when cooled sets into a gelly.
*Gel have honey-comb structure :
Sillicic acid, Gum arabic, Sodium oleate, Gelatin, Solid alcohol, etc.
Types of Gel :
(i) Elastic gel : Those gel which have elastic properties.
(ii) Non- elastic gel : Those gel which are rigid.
2.
Properties of Gel :
Syneresis/weeping of gel : The spontaneous liberation of liquid from a gel is called syneresis or
weeping of gels. It is reverse of swelling.
Eg : Gelatin, Agar-Agar show syneresis at low concentration while sillicic acid shows it at high
concentration.
Imbibition or swelling of gel : When gel is kept in a suitable liquid (water) it absorb large volume of
liquid. The phenomenon is called, imbibition or swelling of gel.
Thixotropic : Some gels when shaken to form a sol, on keeping changes into gel are termed as
thixotropic gel and phenomenon is called thixotropy.
Eg : Gelatin and silica liquify on shaking changing into corresponding sol and the sol on keeping
changes back into gel.
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3.
Eg : Gelatin, Starch, Agar-Agar etc.
Eg : Silica gel.
Ja
1.
uh
Ex :
Which of the following is (are) lyophobic colloids ?
(A) Gold sol
(B) As2S3 sol
(C) Starch sol
(D) Fe(OH)3 sol
(ABD) Gold sol, As2S3 and Fe(OH)3 are lyophobic colloid. Therefore, (A, B, D) are correct options.
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1.*
a
MISCELLANEOUS SOLVED PROBLEMS (MSPs)
Sol.
2.
Sol.
3.
Ans.
Sol.
4.
The presence of colloidal particles of dust in air imparts blue colour to the sky. This is due to
(A) Absorption of the light
(B) Scattering of the light
(C) Reflection of the light
(D) None of these
(B) Due to scattering of the light. Therefore, (B) is correct option.
The volume of nitrogen gas Um (measured at STP) required to cover a sample of silica gel with a monomolecular layer is 129 cm3 g–1 of gel. Calculate the surface area per gram of the gel if each nitrogen
molecule occupies 16.2 × 10–20 m2.
561.8 cm3
22400 cm3 of N2 at STP contain = 6.022 × 1023 molecules
6.022  1023  129
 129 cm3 of N2 at STP will contain =
= 3.468 × 1021 molecules
22400
Area occupied by a single molecule = 16.2 × 10 –20 m2
 Area occupied by 3.468 × 1021 molecules of nitrogen
= (16.2 × 10–20) × (3.468 × 1021) m2 = 561.8 m2.
Which of the following has minimum gold number ?
(A) Potato starch
(B) Gum arabic
(C) Gelatin
(D) Albumin
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Sol.
(C) Gelatin has minimum gold number. Therefore, (C) is correct option.
5.*
Which of the following are correctly matched ?
(A) Butter-gel
(B) Milk-emulsion
(ABC) are correct matches.
Sol.
(C) Fog-aerosol
(D) Dust-solid sol
Explain the adsorption of nitrogen on iron.
When nitrogen gas is brought in contact with iron at 83 K, it is physisorbed on iron surface as nitrogen
molecules, N2. As the temperature is increased the amount of nitrogen adsorbed decreases rapidly and
at room temperature, practically there is no adsorption of nitrogen on iron. At 773 K and above, nitrogen
is chemisorbed on the iron surface as nitrogen atoms.
7.
How do size of particles of adsorbent, pressure of gas and prevailing temperature influence the extent
of adsorption of a gas on a solid ?
(a) Smaller the size of the particles of the adsorbent, greater is the surface area and hence greater is
the adsorption
(b) At constant temperature, adsorption first increases with increase of pressure and then attains
equilibrium.
(c) In physical adsorption, it decreases with increase of temperature but in chemisorption, first it
increases and then decreases.
Sol.
a
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6.
Sol.
How is adsorption of a gas is related to its critical temperature ?
Higher is the critical temperature of a gas, greater the van der Waal’s forces of attraction and hence
greater is the adsorption.
9.
Physical adsorption is essentially quite appreciable :
(A) at room temperature
(B) at higher temperature
(C) at lower temperature
(D) none of these
(C) Rate of physical adsorption decreases with increase in temperature (exothermic process).
Therefore, (C) is correct option.
Sol.
Ja
What happens when persistent dialysis of a colloidal solution is carried out.
The stability of a colloidal sol is due to the presence of a small amount of the electrolyte. On persistent
dialysis, the electrolyte is completely removed. As a result, the colloidal sol becomes unstable and gets
coagulated.
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11.
Sol.
What type of colloidal sols are formed in the following ?
(i) Through cooled water, vapours of sulphur are passed.
(ii) White of an egg is mixed with water.
(i) Sulphur molecules associate together to form molecular sols.
(ii) Macromolecular sol because protein molecules present in the white of the egg are macromolecules
& are soluble in water.
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10.
a
Sol.
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8.
Sol.
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 Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
Section (A) : Adsorption
A-1. Why adsorption is always exothermic ?
A-2. What is the difference between physical adsorption and chemisorption ?
What are the factors which influence the adsorption of a gas on a solid ?
A-4.
What is an adsorption isotherm ?
A-5.
What do you understand by activation of adsorbent ? How is it achieved ?
A-6.
Which will be adsorbed more readily on the surface of charcoal and why– NH3 or CO2 ?
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A-3.
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A-7. In an Adsorption experiment a graph between log x/m versus log P was found to be linear with
a slope of 45º the intercept of the log x/m was found to be 0.3010. Calculate the amount of gas
adsorbed per gram of charcoal under a pressure of 0.6 bar.
A-9.
Ja
A-8. 1 gm of charcoal adsorbs 100 mL of 0.5 M CH3COOH to form mono layer and there by the
molarity of CH3COOH reduces to 0.49 M. Calculate the surface area of the charcoal adsorbed
by each molecule of CH3COOH. Surface area of charcoal = 3.01 × 102 m2/g.
What role does adsorption play in heterogeneous catalysis ?
A-10.
How many grams of gas would be adsorbed per gram of a substance at 8 atm by
assuming Freundlich adsorption isotherm.
= kp1/n
k = 10–2 atm–1/3
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x
m
and
&
n = 3.
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A-11. 10 mg of an adsorbate gets adsorbed on a surface. This causes the release of 3J of heat at
constant pressure and at 27ºC. [Molar mass of adsorbate = 100 g/mol].
(i) Find Hadsorption.
(ii) Argue whether the adsorption is physical or chemical ?
(iii) If 20 mg of adsorbate is adsorbed at temperature T0. Then compare T0 and 27ºC :
Section (B) : Catalysis
B-1. Give two examples of heterogeneous catalysis.
B-2.
Identify the correct order of steps in hetereogeneous catalysis.
(i) Adsorption of reactant molecules on the surface of the catalyst.
(ii) Diffusion of reactant to the surface of the catalyst.
(iii) Formation of reactions product on the catalyst surface.
(iv) Diffusion of reactions product from the catalyst surface or desorption.
(v) Formation of activated intermediate.
Section (C) : Classification and Preparation of Colloid
C-1. How are the colloidal solutions classified, on the basis of physical states of the dispersed phase
and dispersion medium ?
C-2.
Explain the following terms with suitable examples.
(a) Gel
(b) Liquid Aerosol
(c) Hydrosol
C-3. How are associated colloids different from multimolecular and macromolecular colloids ?
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C-4.
Give one example of multimolecular and macromolecular colloids.
C-5.
Describe a method each for the preparation of sols of sulphur and platinum in water
Section (D) : Purification and Properties of Colloid
D-1. Explain the following terms :
(a) Peptization
(b) Electrophoresis
movement
D-2.
(c) Dialysis
(d)
Brownian
Why the sun looks red at the time of setting ?
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D-3. Why is osmotic pressure of a colloidal solution less than that of true solution ?
Section (E) : Coagulation, Protection And application of colloid
E-1. Which one of the following electrolytes is most effective for the coagulation of Fe(OH) 3 sol
and why ? NaCl, Na2SO4, Na3PO4.
E-3. Rivers form delta on meeting with ocean, why ?
Artificial rain is made by spraying salt over clouds, why ?
Section (F) : Emulsion and Gel
F-1. Name two demulsifier.
Ja
E-4.
uh
E-2. What do you understand by "isoelectric point" of a colloid ?
F-2. What is the difference between sols and emulsions.
What is demulsification ?
F-4.
What is phase inversion in emulsion ?
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F-3.
PART - II : ONLY ONE OPTION CORRECT TYPE
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Section (A) : Adsorption
A-1. Which of the following statements about chemisorption is not applicable?
(A) It involves chemical forces between adsorbent and absorbate
(B) It is irreversible in nature
(C) It involves high heat of adsorption
(D) It does not require activation energy
A-2. Following is the variation of physical adsorption with temperature:
(A)
(B)
(C)
(D)
A-3. Adsorption is the phenomenon in which a substance:
(A) accumulates on the surface of the other substance
(B) goes into the body of the other substances
(C) remains close to the other substance
(D) none of these
442
A-4. Finely divided catalyst has greater surface area and has greater catalytic activity than the
compact solid. If a total surface area of 6291456 cm2 is required for adsorption in a catalysed
gaseous reaction, then how many splits should be made to a cube of exactly 1 cm in length to
achieve required surface area. (Given : One split of a cube gives eight cubes of same size)
(A) 60
(B) 80
(C) 20
(D) 22
A-5. Volume of N2 at NTP required to form a mono layer on the surface of iron catalyst is 8.15
ml/gram of the adsorbent. What will be the surface area of the adsorbent per gram if each
nitrogen molecule occupies 16 × 10–22 m2.
(A) 16 × 10–16 cm2 (B) 0.35 m2/g
(C) 39 m2/g
(D) 22400 cm2
Softening of hard water is done using sodium aluminium silicate (zeolite). This causes :
(A) adsorption of Ca2+ and Mg2+ ions of hard water replacing Na+ ions.
(B) adsorption of Ca2+ and Mg2+ ions of hard water replacing Al3+ ions
(C) both (A) and (B)
(D) none of these
Ja
A-8.
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A-7.
There is desorption of physical adsorption when :
(A) temperature is increased
(B) temperature is decreased
(C) pressure is increased
(D) concentration is increased
The rate of chemisorption :
(A) decreases with increase of pressure
(B) increases with increase of pressure
(C) is independent of pressure
(D) is independent of temperature
uh
A-6.
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Section (B) : Catalysis
B-1. Which one is false in the following statement ?
(A) A catalyst is specific in its action
(B) A very small amount of the catalyst can alter the rate of a reaction.
(C) The number of free valencies on the surface of the catalyst increases on sub-division
(D) Ni is used as a catalyst in the manufacture of ammonia
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B-2. A catalyst increases rate of reaction by :
(A) Decreasing enthalpy
(C) Decreasing activation energy
(B) Decreasing internal energy
(D) Increasing activation energy
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Section (C) : Classification and Preparation of Colloid
C-1. Colloidal solution of gold prepared by different methods of different colours because of :
(A) different diameters of colloidal gold particles
(B) variable valency of gold
(C) different concentration of gold particles
(D) impurities produced by different methods
C-2.
At CMC, the surfactant molecules :
(A) Decomposes
(C) Associate
(B) Become completely soluble
(D) Dissociate
Section (D) : Purification and Properties of Colloid
D-1.  A colloidal solution can be purified by the following method :
(A) dialysis
(B) peptization
(C) filtration
D-2.
(D) oxidation
Peptisation is :
(A) conversion of a colloidal into precipitate form
(B) conversion of precipitate into colloidal sol
(C) conversion of metal into colloidal sol by passage of electric current
(D) conversion of colloidal sol into macromolecules
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D-3. Bleeding is stopped by the application of ferric chloride. This is because :
(A) the blood starts flowing in opposite direction
(B) the blood reacts and forms a solid, which seals the blood vessel
(C) the blood is coagulated and thus the blood vessel is sealed
(D) the ferric chloride seals the blood vessel.
E-2. Protective sols are :
(A) lyophilic
(B) lyophobic
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Section (E) : Coagulation, Protection And application of colloid
E-1. Gold number of a lyophilic sol is such property that:
(A) the larger its value, the greater is the peptising power
(B) the lower its value, the greater is the peptising power
(C) the lower its value, the greater is the protecting power
(D) the larger its value, the greater is the protecting power
(C) both (A) and (B) (D) none of (A) and (B)
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E-3. For the coagulation of 200 mL of As2S3 solution, 10 mL of 1 M NaCl is required. What is the
coagulating value (number of milli moles of solute needed for coagulation of 1 liter of
solution) of NaCl.
(A) 200
(B) 100
(C) 50
(D) 25
Which of the following ions is most effective in the coagulation of an arsenious sulphide
solution ?
(A) K+
(B) Mg2+
(C) Al3+
(D) C
E-5.
Which of the following ions is most effective in the coagulation of ferric hydroxide solution ?
(A) Cl¯
(B) Br –
(C) NO2¯
(D) SO42–
Ja
E-4.
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Section (F) : Emulsion and Gel
F-1. Small liquid droplets dispersd in another liquid is called :
(A) Suspension
(B) Emulsion
(C) Gel
solution
(D)
True
Match list I with list II and select the correct answer :
List-I
List-II
(P) Mechanical property of (1) Dialysis
colloid
(Q) Purification
(2) Peptization
(R) Gold number
(3) Brownian
movement
(S) Formation of a sol
(4) Protection
Code:
P
Q
R
S
P
(A)
3
4
1
2
(B)
1
(C)
3
1
4
2
(D)
2
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1.
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PART - III : MATCH THE COLUMN
Q
2
3
R
4
1
S
3
4
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Match list I with list II and give the correct answer :
List-I
List-II
(A) Gold sol
(p) Bredig's Arc method
(B) Purification of colloidal (q) Negatively charged
solution
(C) As2S3 sol
(r) Dialysis
(D) Zeta potential
(s) Electro kinetic potential
(E) Casein
(t) Double
decomposition
reaction
(u) Protective colloid
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2.
 Marked questions are recommended for Revision
uh
PART - I : ONLY ONE OPTION CORRECT TYPE
Which of the following statements is correct for a lyophilic solution ?
(A) It is not easily solvated
(B) The coagulation of this sol is irreversible in
nature
(C) It is unstable
(D) It is quite stable in a solvent
2.
Liquid-liquid colloidal system is known as
(A) aerosol
(B) foam
(C) emulsion
Ja
1. 
(D) gel
The colloidal system consisting of a liquid dispersed in a solid dispersion medium is termed as
:
(A) aerosol
(B) foam
(C) emulsion
(D) gel
4.
Which of the following statements is not correct ?
(A) A colloidal solution is a heterogeneous two-phase system
(B) Silver sol in water is an example of lyophilic solution.
(C) Metal hydroxides in water are examples of lyophobic solution
(D) Liquid-liquid colloidal solution is not a stable system
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3.
5.
6.
Size of colloidal particles may range from :
(A) 1 to 1000 nm
(B) 10 to 100 pm
(C) 1 to 100 µm
(D) 1 to 10 mm
Which of the following represents a multimolecular colloidal particles?
(A) Starch
(B) A sol of gold
(C) Proteins
(D) Soaps
7.
Which of the following anions will have minimum flocculation value for the ferric oxide
solution ?
(A) Cl–
(B) Br–
(C) SO42–
(D) [Fe(CN)6l3-
8.
Which of the following represents a macromolecular colloidal particles ?
(A) Solution of gold (B) Cellulose
(C) Soaps
(D) Synthetic detergents
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9.
Gold number of some lyophilic sols are :
Casein
0.01

Haemoglobin 0.03

Gum arabic
0.15

Sodium oleate 0.40
V
Which has maximum protective power :
(A) 
(B) 
(C) 
(D) V
Arsenic () sulphide forms a sol with a negative charge. Which of the following ionic
substances should be most effective in coagulating the sol ?
(A) KCl
(B) MgCl2
(C) Al2(SO4)3
(D) Na3PO4
11.
Smoke is a dispersion of :
(A) gas in gas
(B) gas in solid
gas
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10.
(C) solid in gas
12. Smoke has generally blue tinge. It is due to :
(A) scattering
(B) coagulation
uh
(C) Brownian motion (D) electro-osmosis
Which one of the following statements is false for hydrophilic sols ?
(A) they do not require electrolytes for stability
(B) their viscosity is of the order of that of water
(C) their surface tension is usually lower than that of dispersion medium.
(D) none of these
Ja
13.
(D) liquid in
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14. Soaking of sponge by water is an example of :
(A) Simple adsorption
(B) Physical adsorption
(C) Chemisoption
(D) Absorption
15. Identify the appropriate graph between enthalpy and progress of physical adsorption.
(B)
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(A)
(C)
16.
(D)
Hydrolysis of ester in catalysed by acid. Rate of hydrolysis of ester were obtained initially and
after some ester has been hydrolysed as R0 and Rt then (same temp.)
(A) R0 = Rt
(B) R0 < Rt
(C) R0 > Rt
(D) Cannot be determined
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PART - II : SINGLE OR DOUBLE INTEGER TYPE
When a graph is plotted between log x/m and log p, it is straight line with an angle 45° and
intercept 0.3010 on y-axis. If initial pressure is 0.3 atm, what will be the amount of gas
adsorbed per gram of adsorbent : (Report your answer after multiplying by 10)
2.
The volume of nitrogen gas (measured at STP) required to cover a sample of silica gel with a
mono-molecular layer is 129cm3/g of gel. Calculate the surface area per gram of the gel if each
nitrogen molecule occupies 16.2 × 10–20m2. (Report your answer after dividing by 10).
3.
How many of these reactions are homogeneously catalyzed?
NO(g)
(i) 2SO2 (g) + O2 (g) ⎯⎯⎯
→ 2SO3 (g)
H SO ( )
(ii) C12H22O11(aq.) + H2O( ) ⎯⎯⎯⎯
→C6H12O6 (aq.) + C6H12O6 (aq.)
Glucose
Fructose
Pt(s)
(iii) 2SO3 (g) + O2 (g) ⎯⎯⎯→ 2SO3 (g)
Fe(s)
(iv) N2 (g) + 3H2 (g) ⎯⎯⎯
→ 2NH3 (g)
Pt(s)
(v) 4NH3 (g) + 5O2 (g) ⎯⎯⎯→ 4NO(g) + 6H2O(g)
HCl( )
(vi) CH3COOCH3 ( ) ⎯⎯⎯
→CH3COOH(aq)
Ni(s)
(vii) Vegetable oils ( ) + H2 (g) ⎯⎯⎯
→ Vegetable ghee (s).
4
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2
a
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1.
Coagulation value of the electrolytes AlCl3 and NaCl for As2S3 sol are 0.093 and 52
repectively. How many times AlCl3 has greater coagulating power than NaCl.
5.
Among the following number of correct statements are :
(i) Stability of lyophilic colloids is mainly due to the strong interaction between dispersed
particle and dispersion medium.
(ii) Entropy change for adsorption of gases over solid is positive.
(iii) Gelatin has considerably low value of gold number and is effective protective colloid.
(iv) Zeta potential is also responsible for stability of lyophobic colloid solution.
(v) Surface tension of lyophilic colloidal solution is lesser than that of dispersion medium.
6.
For the just coagulation of 250 mL of Fe(OH)3 sol, 2 mL of 1 M Na2SO4 electrolyte is
required. What is the coagulating value of Na2SO4 electrolyte.
7.
The minimum concentration of an electrolyte required to cause coagulation of a sol is called its
flocculation value. It is expressed in millimoles per litre. If the flocculation value of MgSO4 for
standard As2S3 sol is 3.33. How many milligrams of MgSO4 is to be added to 20 ml standard
As2S3 sol so that flocculation just starts ?
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4.
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
1.
Which of the following statements about physical adsorption is correct ?
(A) It is always monolayer
(B) It is reversible in nature
(C) It involves van der Waals interactions between adsorbent and adsorbate
(D) It involves small enthalpy of adsorption as compared to chemisorption.
2.
Which of the following statements regarding adsorption is correct ?
(A) Extent of adsorption of gases on charcoal increases with increase in pressure of the gas
(B) Extent of adsorption is independent of temperature
(C) Extent of chemisorption by a given mass of adsorbent is limited
(D) Extent of adsorption is dependent on the nature of adsorbent
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3.
Which of the following is characteristic of chemisorption?
(A) it is irreversible
(B) it is specific
(C) it is multilayer phenomenon
(D) heat of adsorption is generally around
80 – 240 kJ
4.
Which is/are a purely surface phenomena :
(A) surface tension
(B) adsorption
7.
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6.
(D) none of these
Which of the following are correct statements ?
(A) Spontaneous adsorption of gases on solid surface is an exothermic process as entropy
decreases during adsorption
(B) Formation of micelles takes place when temperature is below Kraft Temperature (Tk) and
concentration is above critical micelle concentration (CMC)
(C) Longer the length of hydrophobic chain, smaller is the value of critical micelle
concentration (CMC)
(D) According to Hardy-Schulze rule the coagulation (flocculating) value of Fe3+ ion will be
more than Ba2+ or Na+.
Which of the following statements are true for physisorption?
(A) Extent of adsorption increases with increase in pressure.
(B) It needs activation energy
(C) It can be reversed easily
(D) It occurs at high temperature.
Ja
5.
(C) absorption
Identify the reactions that includes inhibtors in the reactions mixture.
Fe
Ni
⎯
→ 2NH3
(A) N2 + 3H2 ⎯⎯
(B) Vegetable Oil + H2 ⎯⎯→
Vegetable ghee.
Mo
(C) N2 + 3H2 ⎯⎯⎯⎯→ 2NH3
Fe
(D) RCOCl + H2 ⎯⎯⎯→ RCHO + HCI
BaSO4
lp
CO / H2S
Cu
Pd
Which of the following are the correct :
(A) A Catalyst remains unchanged in mass and chemical compositions at the end of reactions.
(B) Finely devided state of catalyst is more efficient for the reactions.
(C) Catalyst change equilibrium state of the reaction.
(D) A catalyst changes the entropy and the free energy of a reaction.
9.
The diameter of colloidal particle is of the order :
(A) 10–3 m
(B) 10–6 m
(C) 10–15 m
10.
Which of the following are examples of aerosols?
(A) Whipped cream (B) Cloud
(C) Fog
Sa
nk
a
8.
11.
Which of the following are hydrophobic sols ?
(A) Protein sol
(B) Gold sol
(C) Gum sol
(D) 10–7 m
(D) Soap lather
(D) Fe(OH)3 sol.
12.
Which of the following are multimolecular colloids ?
(A) Sulphur
(B) Egg albumin in water
(C) Gold sol
(D) Soap solution
13.
The origin of charge on colloidal solution is
(A) Self dissociation (in soaps and detergents)
method
(C) Selective adsorption of ion on their surface
colloids
(B) Electron capture during Bredig's arc
(D) It is due to addition of protective
448
14. Which of the following is/are not true for lyophilic colloid ?
(A) These are prepared by special indirect methods.
(B) The particles must travel towards the anode or cathode under the influence of an electric
field.
(C) These are called on intrinsic colloid
(D) Small quantity of electrolyte is sufficient to cause precipitation of these.
15. Which of the following are based on Tyndall effect.
(A) Tail of comets
(B) Deltas
(C) Blue colour of sky
(D) Coagulation
Which is an example of coagulation?
(A) curdling of milk
(C) formation of deltas at the river beds
uh
17.
a
ri
16. Which of the following statements is correct?
(A) Peptization is the process by which some fresh precipitates are converted into the colloidal
state by addition of little suitable electrolyte.
(B) Metal sols of gold, silver and platinum can be prepared by Bredig's arc method.
(C) Impurities present in a solution makes it more stable.
(D) Dialysis is a process to remove impurities of ions and molecules from a solution.
(B) purification of water by addition of alum
(D) formation of ice
When negatively charged colloids like As2S3 sol is added to positively charged Fe(OH)3 sol in
suitable amounts
(A) Both the sols are precipitated simultaneously .
(B) This process is called mutual coagulation.
(C) They become positively charged colloids.
(D) They become negatively charged colloids.
19.
Which of the following are incorrect statements ?
(A) Hardy schulz rule is related to coagulation
(B) Brownian movement and Tyndall effect are the characteristic of colloids.
(C) In gel, the liquid is dispersed in liquid
(D) Higher the gold number, more is the protective power of lyophillic sols.
20.
Which of the following sols is positively charged?
(A) Arsenious sulphide
(B) Aluminium hydroxide
(C) Ferric hydroxide
(D) Silver iodide in silver nitrate solution
Sa
nk
a
lp
Ja
18.
PART - IV : COMPREHENSION
Read the following passage carefully and answer the questions.
Comprehension # 1
Many lyophilic sols and few lyophobic sols when coagulated under some special conditions
changes into semi rigid mass, enclosing whole amount of liquid within itself, it is called gel
and the process is called gelation. Gelatin Agar-agar, gum-Arabic can be converted into gels by
cooling them under moderate concentration conditions. Hydrophobic sols like silicic acid.
Al(OH)3 are prepared by double decomposition and exchange of solvent method.
(i)
Types of Gel :
Elastic gel : Those gel which have elastic properties.
Ex : Gelatin, Strach, Agar-Agar etc.
449
1.
2.
3.
1.
Non- elastic gel : Those gel which are rigid.
Ex: Silica gel.
Properties of Gel :
Syneresis/weeping of gel : The spontaneous liberation of liquid from a gel is called syneresis
or weeping of gels. It is reverse of swelling.
Ex: Gelatin, Agar-Agar show syneresis at low concentration while sillicic acid shows it at high
concentration.
Imbibition or swelling of gel : When gel is kept in a suitable liquid (water) it absorb large
volume of liquid. The phenomenon is called imbibition or swelling of gel.
Thixotropic : Some gels when shaken to form a sol, on keeping changes into gel are termed as
thixotropic gel and phenomenon is called thixotropy.
Ex: Gelatin and silica liquify on shaking changing into corresponding sol and the sol on
keeping changes back into gel.
Which of the following is used to adsorb water ?
(A) Silica gel
(B) Calcium acetate (C) Hair gel
a
ri
(ii)
(D) Cheese
The process of imbibing water when elastic gel are placed in water is called :
(A) imbibition
(B) synerisis
(C) coagulation
(D) thixotropy
3.
Some types of gels like gelatin and silica liquify on shaking thereby changing into sols. The
sols on standing change back into gels. This process is know as
(A) syneresis
(B) thixotropy
(C) double decompostion
(D) peptization
Ja
uh
2.
When excess of AgNO3 is treated with KI solution, AgI forms
(A) +ve charged sol (B) –ve charged sol (C) neutral sol
Sa
nk
4.
a
lp
Comprehension # 2
The clouds consist of charged particles of water dispersed in air. Some of them are +vely
charged, others are –vely charged. When +vely charged clouds come closer they cause
lightening and thundering whereas when +ve and –ve charged colloids come closer they cause
heavy rain by aggregation of minute particles. It is possible to cause artificial rain by throwing
electrified sand or silver iodide from an aeroplane and thus coagulating the mist hanging in air.
Smoke screen is a cloud of smoke used to hide military, naval police etc. it consists of fine
particles of TiO2.
5.
6.
AgI helps in artificial rain because :
(A) it helps in ionisation of water
(C) it helps in coagulation
(D) true solution
(B) it helps in dispersion process
(D) all of them
Smoke screens consist of
(A) fine particles of TiO2 dispersed in air by aeroplanes
(B) fine particles of AgI dispersed in air by aeroplanes
(C) fine particles of Al2O3 dispersed in air by aeroplanes
(D) None of these
450
(C) (II) (iii) (S)
Select the only incorrect option for gold sol.
(A) (II) (ii) (P)
(B) (II) (iii) (S)
(Q)
(C) (I) (iii) (R)
(C) (III) (iv) (S)
(D) (IV) (iii)
(D) (IV) (iv)
(D) (II) (iii)
lp
9.
Select the only correct option Fe(OH)3 sol.
(A) (I) (iii) (Q)
(B) (IV) (i) (Q)
(S)
uh
8.
Select the only incorrect option for AgI / I– sol.
(A) (II) (ii) (P)
(B) (IV) (iv) (P)
(Q)
Ja
7.
a
ri
Comprehension # 3
Answer Q.7, Q.8 and Q.9 by appropriately matching the information given in the three
columns of the following table.
Column-1
Column-2
Column-3
During
electrophoresis
Positively charged
Can be coagulated by
(I)
(i)
(P) coagulation will take place at
colloid
adding metal sulphide sol
anode
During electro-osmosis level
Negatively charged
Can be coagulated by
(II)
(ii)
(Q) of dispersion medium will
colloid
adding metal oxide sol
increase on anode side.
Can be prepared by
Coagulation value of
During
Electro-phoresis
(III
Bredig’s
Arc (iii) Na2SO4 > MgCl2 for this (R) coagulation will take place at
)
method
colloid
cathode.
(I
Coagulating power of
During electro-osmosis level
Can be prepared by
V
(iv) MgSO4 > NaCl for this (S) of dispersion medium will
peptisation
)
colloid.
increase on cathode side.
a
* Marked Questions are having more than one correct option.
Sa
nk
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
1.
Rate of physisorption increases with
(A) decrease in temperature
(C) decrease in pressure
[JEE 2003, 3/60]
(B) increase in temperature
(D) decrease in surface area
2.
Adsorption of gases on solid surface is generally exothermic because
3/84]
(A) enthalpy is positive
(B) entropy decreases
(C) entropy increases
(D) free energy increases
[JEE 2004,
3.
Lyophilic sols are
(A) Irreversible sols
(C) Coagulated by adding electrolytes
4.
Among the following, the surfactant that will form micelles in aqueous solution at the lowest
molar concentration at ambinent condition is :
[JEE
2008, 3/163]
(A) CH3(CH2)15N+(CH3)3Br–
(B) CH3(CH2)11OSO3–Na+
(C) CH3(CH2)6COO–Na+
(D) CH3(CH2)11N+(CH3)3Br–
[JEE 2005, 3/84]
(B) They are prepared from inorganic compound
(D) Self-stabilizing
451
Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating
agent for Sb2S3 sol is :
[JEE 2009,
3/160]
(A) Na2SO4
(B) CaCl2
(C) Al2(SO4)3
(D) NH4Cl
6.
Silver (atomic weight = 108 gm mol–1) has a density of 10.5 gm cm–3. The number of silver
atoms on a surface of area 10–12 m2 can be expressed in scientific notation as y × 10x. The
value of x is :
[JEE 2010, 3/163]
7.*
The correct statement(s) pertiaining to the adsorption of a gas on a solid surface is (are)
[JEE 2011, 4/180]
(A) Adsorption is always exothermic
(B) Physisorption may transform into chemisorption at high temperature
(C) Physisorption increases with increasing temperature but chemisorption decreases with
increasing temperature
(D) Chemisorption is more exothermic than physisorption, however it is very slow due to
higher energy of activation.
8.*
Choose the correct reason(s) for the stability of the lyophobic colloidal particles. [JEE 2012,
4/168]
(A) Preferential adsorption of ions on their surface from the solution.
(B) Preferential adsorption of solvent on their surface from the solution.
(C) Attraction between different particles having opposite charges on their surface.
(D) Potential difference between the fixed layer and the diffused layer of opposite charges
around the colloidal particles.
9.*
The given graph / data I, II, III and IV represent general trends observed for different
physisorption and chemisorption processes under mild conditions of temperature and pressure.
Which of the following choice (s) about I, II, III and IV is (are) correct.
[JEE 2012, 4/168]
Sa
nk
(i)
a
lp
Ja
uh
a
ri
5.
(iii)
(ii)
(iv)
(A) I is physisorption and II is chemisorption
chemisorption
(C) IV is chemisorption and II is chemisorption
chemisorption
(B)
I
is
physisorption
and
III
is
(D) IV is chemisorption and III is
452
Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25° C. For this
process, the correct statement is
[JEE(Advanced)-2013, 2/120]
(A) The adsorption requires activation at 25°C.
(B) The adsorption is accompanied by a decreases in enthalpy.
(C) The adsorption increases with increase of temperature.
(D) The adsorption is irreversible.
11.*
When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The
TRUE statement(s) regarding this adsorption is(are)
[JEE(Advanced)2015, 4/168]
(A) O2 is physisorbed
(B) heat is released
*
(C) occupancy of 2p of O2 is increased
(D) bond length of O2 is increased
12.
The qualitative sketches I, II and III given below show the variation of surface tension with
molar concentration of three different aqueous solution of KCl, CH3OH and CH3(CH2)11
OSO3– Na+ at room temperature. The correct assignment of the sketches is :
[JEE(Advanced)-2016, 3/124]
Concentration
uh
II
Concentration
lp
(A) I : KCl
II : CH3OH
(B) I : CH3(CH2)11 OSO3– Na+
II : CH3OH
(C) I : KCl
II : CH3(CH2)11 OSO3– Na+
(D) I : CH3OH
II : KCl
+
Na
III
Concentration
III : CH3(CH2)11 OSO3– Na+
III : KCl
III : CH3OH
III : CH3(CH2)11 OSO3–
The correct statement(s) about surface properties is(are)
[JEE(Advanced)2017, 4/122]
(A) The critical temperatures of ethane and nitrogen are 563 K and 126 K, respectively. The
adsorption of ethane will be more than that of nitrogen of same amount of activated charcoal at
a given temperature.
(B) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is
dispersion medium.
(C) Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system.
Sa
nk
a
13.*
Surface tension
I
Ja
Surface tension
Surface tension
a
ri
10.
(D) Brownian motion of colloidal particles does not depend on the size of the particles but
depends on viscosity of the solution.
453
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) ONLINE PROBLEMS
Which one of the following characteristics is not correct for physical adsorption? [AIEEE
2003, 3/225]
(1) Adsorption on solids is reversible
(2) Adsorption increases with increase in temperature
(3) Adsorption is spontaneous
(4) Both enthalpy and entropy of adsorption are negative.
2.
The disperse phase in colloidal iron () hydroxide and colloidal gold is positively and
negatively charged, respectively. Which of the following statements is NOT correct ?
[AIEEE 2005, 3/225]
(1) Coagulation in both sols can be brought about by electrophoresis
(2) Mixing the sols has no effect
(3) Sodium sulphate solution causes coagulation in both sols
(4) Magnesium chloride solution coagulates, the gold sol more readily than the iron ()
hydroxide sol.
3.
The volume of collodial particle VC as compared to the volume of a solute particle in a true
solution VS could be :
[AIEEE 2005, 3/225]
(1) ~ 1
(2) ~ 1023
(3) ~ 10–3
(4) ~ 103
4.
In langmuir’s model of adsorption of a gas on a solid surface :
[AIEEE
2006, 3/165]
(1) the rate of dissociation of adsorbed molecules from the surface does not depend on the surface
covered
(2) the adsorption at a single site on the surface may involve multiple molecules at the same
time
(3) the mass of gas striking a given area of surface is proportional to the pressure of the gas
(4) the mass of gas striking a given area of surface is independent of the pressure of the gas
5.
Gold numbers of protective colloids A, B, C and D are 0.50, 0.01, 0.10 and 0.005, respectively.
The correct order of their protective powers is
[AIEEE
2008, 3/105]
(1) C < B < D < A
(2) A < C < B < D
(3) B < D < A < C
(4) D < A < C < B
Sa
nk
a
lp
Ja
uh
a
ri
1.
6.
Which of the following statements is incorrect regarding physiosorptions ? [AIEEE
4/144]
(1) More easily liquefiable gases are adsorbed readily.
(2) Under high pressure it results into multi molecular layer on adsorbent surface.
(3) Enthalpy of adsorption (Hadsorption) is low and positive.
(4) It occurs because of van der Waal’s forces.
7.
According to Freundlich adsorption isotherm which of the following is correct?
2012, 4/120]
(1)
x
 p0
m
(2)
x
m
 p1
(3)
x
m
2009,
[AIEEE
 p1/n
(4) All the above are correct for different ranges of pressure
454
The coagulating power of electrolytes having ions Na+, Al3+ and Ba2+ for arsenic sulphide sol
increases in the order :
[JEE(Main) 2013, 4/120]
(1) Al3+ < Ba2+ <
Na+ (2) Na+ < Ba2+ < Al3+ (3) Ba2+ < Na+ < Al3+ (4) Al3+ < Na+ <
2+
Ba
9.
3 gram of activated charocoal was added to 50 mL of acetic acid solution (0.06N) in a flask.
After an hour it was filtered and the strength of the fitrate was found to be 0.042 N. The
amount of acetic acid adsorbed (per gram of charcoal) is :
[JEE(Main)-2015, 4/120]
(1) 18 mg
(2) 36 mg
(3) 42 mg
(4) 54 mg
10.
For a linear plot of log(x/m) versus log p in a Freundlich adsorption isotherm, which of the
following statements is correct? (k and n are constants)
[JEE(Main)-2016, 4/120]
(1) 1/n appears as the intercept
(2) Only 1/n appears as the slope.
(3) log(1/n) appears as the intercept.
(4) Both k and 1/n appear in the slope term.
11.
The Tyndall effect is observed only when following conditions are satisfied :
[JEE(Main)2017, 4/120]
(a) The diameter of the dispersed particles is much smaller than the wavelength of the light
used.
(b) The diameter of the dispersed particles is not much smaller than the wavelength of the light
used
(c) The refractive indices of the dispersed phase and dispersion medium are almost similar in
magnitude.
(d) The refractive indices of the dispersed phase and dispersion medium differ greatly in
magnitude.
(1) (b) and (d)
(2) (a) and (c)
(3) (b) and (c)
(4) (a) and (d)
lp
Ja
uh
a
ri
8.
JEE(MAIN) ONLINE PROBLEMS
a
The following statements relate to the adsorption of gases on a solid surface. Identify the
incorrect statement among them :
[JEE(Main)
2015
Online (10-04-15), 4/120]
(1) On adsorption decrease in surface energy appears as heat
(2) Enthalpy of adsorption is negative
(3) On adsorption, the residual forces on the surface are increased
(4) Entropy of adsorption is negative
Sa
nk
1.
2.
3.
Under ambient conditions, which among the following surfactants will form micelles in
aqueous solution at lowest molar concentration ?
[JEE(Main) 2015
Online (11-04-15), 4/120]

(1) CH3–(CH2)8 –COO– Na+
(2) CH3(CH2)11 N (CH3)3Br–
(3) CH3–(CH2)13–OSO3– Na+
(4) CH3(CH2)15 N (CH3)3Br–

The most appropriate method of making egg-albumin sol is:[JEE(Main) 2016 Online (09-0416), 4/120]
(1) Keep the egg in boiling water for 10 minutes. After removing the shell, transfer the yellow
part of the content to 100 mL of 5% w/V saline solution and homogenize with a
mechanical shaker.
455
(2) Break an egg carefully and transfer the transparent part of the content to 100 mL of 5%
w/V saline solution and stir well.
(3) Keep the egg in boiling water for 10 minutes. After removing the shell, transfer the white
part of the content to 100 mL of 5% w/V saline solution and homogenize with a
mechanical shaker.
(4) Break an egg carefully and transfer only the yellow part of the content to 100 mL of 5%
w/V saline solution and stir well.
A particular adsorption process has the following characteristics: (i) It arises due to vander
Waals forces and (ii) it is reversible. Identify the correct statement that describes the above
adsorption process:
[JEE(Main) 2016 Online (09-04-16),
4/120]
(1) Enthalpy of adsorption is greater than 100 kJ mol–1.
(2) Adsorption is monolayer.
(3) Adsorption increases with increase in temperature.
(4) Energy of activation is low.
5.
Gold numbers of some colloids are :
Gelatin : 0.005 - 0.01; Gum Arabic : 0.15 - 0.25; Oleate : 0.04 - 1.0; Starch : 15 - 25.
Which among these is a better protective colloid ?
[JEE(Main) 2016 Online (10-04-16),
4/120]
(1) Gelatin
(2) Starch
(3) Gum Arabic
(4) Oleate
6.
Among the following, correct statement is :
[JEE(Main) 2017 Online (08-04-17),
4/120]
(1) One would expect charcoal to adsorb chlorine more than hydrogen sulphide
(2) Sols metal sulphides are lyophilic
(3) Hardy Schulze law states that bigger the size of the ions, the greater is its coagulating
power.
(4) Brownian movement is more pronounced for smaller particles than for bigger-particles.
7.
Adsorption of a gas on a surface follows Freundlich adsorption isotherm. Plot of log
a
lp
Ja
uh
a
ri
4.
Sa
nk
log p gives a straight line with slope equal to 0.5, then :
04-17), 4/120]
(
x
m
x
m
versus
[JEE(Main) 2017 Online (09-
is the mass of the gas adsorbed per gram of adsorbent)
(1) Adsorption is proportional to the pressure.
(2) Adsorption is proportional to the square root of pressure.
(3) Adsorption is proportional to the square of pressure.
(4) Adsorption is independent of pressure.
8.
Which of the following statements about colloids is False ? [JEE(Main) 2018 Online (15-0418), 4/120]
(1) When silver nitrate solution is added to potassium iodide solution a negatively charged
colloidal solution is formed.
(2) Freezing point of colloidal solution is lower than true solution at same concentration of a
solute.
(3) Colloidal particles can pass through ordinary filter paper.
(4) When excess of electrolyte is added to colloidal solution, colloidal particle will be
precipitated.
456
9.
If x gram of gas is adsorbed by m gram of adsorbent at pressure P, the plot of log
x
m
versus log
P is linear. The slope of the plot is : (n and k are constants and n > 1)
[JEE(Main) 2018 Online (15-04-18), 4/120]
(1) 2 k
(2) log k
(3) n
(4)
1
n
Two compounds I and II are eluted by column chromatography (adsorption of I > II). Which
one of the following is a correct statement?
[JEE(Main) 2018 Online
(15-04-18), 4/120]
(1) I moves faster and has higher Rf value than II
(2) II moves faster and has higher Rf value than I
(3) I moves slower and has higher Rf value than II
(4) II moves slower and has higher Rf value than I
11.
Which one of the following is not a property of physical adsorption ?
[JEE(Main) 2018 Online (16-04-18),
4/120]
(1) Higher the pressure, more the adsorption (2) Greater the surface area, more the adsorption
(3) Lower the temperature, more the adsorption
(4) Unilayer adsorption occurs
12.
Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the mass of
uh
a
ri
10.
x
m
Ja
the gas adsorbed on mass m of the adsorbent at pressure p.
is proportional to :
[JEE(Main) 2019 Online (09-01-19),
4/120]
x
m
2 unit
lp
Log
4 unit
Log P
(3) p1/2
(4) p1/4
For coagulation of arsenious sulphide sol, which one of the following salt solution will be most
effective?
[JEE(Main) 2019 Online (09-01-19),
4/120]
(1) Na3PO4
(2) AlCl3
(3) BaCl2
(4) NaCl
Sa
nk
13.
(2) p
a
(1) p2
14.
Which of the following is not an example of heterogeneous catalytic reaction ?
[JEE(Main) 2019 Online (10-01-19),
4/120]
(1) Ostwald’s process
(2) Haber’s process
(3) Combustion of coal
(4) Hydrogenation of vegetable oils
15.
Haemoglobin and gold sol are examples of:
[JEE(Main) 2019 Online (10-01-19),
4/120]
(1) negatively charged sols
(2) positively charged sols
(3) positively and negatively charged sols, respectively
(4) negatively and positively charged sols, respectively
457
16.
An example of solid sol is:
4/120]
(1) Butter
(2) Hair cream
[JEE(Main) 2019 Online (11-01-19),
(3) Paint
(4) Gem stones
Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the
dispered phase and dispersion medium, respectively is :
[JEE(Main) 2019 Online
(11-01-19), 4/120]
(1) C : liquid in solid; M : liquid in liquid; S : solid in gas
(2) C : solid in liquid; M : liquid in liquid; S : gas in solid
(3) C : solid in liquid; M : solid in liquid; S : solid in gas
(4) C : liquid in solid; M : liquid in solid; S : solid in gas
18.
Given
Gas
H2
CH4 CO2 SO2
Critical
33
190
304 630
Temperature / K
On the basis of data given above, predict which of the following gases shows least adsorption
on a definite amount of charcoal ?
[JEE(Main) 2019 Online (12-0119), 4/120]
(1) CH4
(2) H2
(3) CO2
(4) SO2
19.
Among the following, the false statement is :
[JEE(Main) 2019 Online (12-0119), 4/120]
(1) It is possible to cause artificial rain by throwing electrified sand carrying charge opposite to
the one on clouds from an aeroplane.
(2) Lyophilic solution can be coagulated by adding an electrolyte
(3) Latex is a colloidal solution of rubber particles which are positively charged
(4) Tyndall effect can be used to distinguish between a colloidal solution and a true solution
Sa
nk
a
lp
Ja
uh
a
ri
17.
458
EXERCISE - 1
PART - I
Adsorption is accompanied by decrease of randomness, i.e. this factor opposes the process, i.e.
S is –ve. For the process to be spontaneous, G must be –ve. Hence, according to eqn, G =
H – TS, G can be –ve only if H is –ve.
A-2.
Difference between physical adsorption and chemical adsorption :
Physical Adsorption
Chemical Adsorption
The forces between the adsorbate molecules and the
adsorbent are strong chemical forces.
Low heat of adsorption of the order of 20-40 kJ mol–1
High heat of adsorption of the order 80-240 kJ mol–1
Usually occurs at low temperature and decreases with
increasing temperature.
It is reversible.
The extent of adsorption depends upon the ease of
liquefication of the gas.
It occurs at high temperature decreases with increasing
temperature.
It is irreversible.
There is no correlation between extent of adsorption and the
ease of liquefication of gas.
It is less specific in nature, all gases are adsorbed on the
surface of a solid to some extent adsorbent and adsorbate
molecules.
It is highly specific in nature and occurs only when there is
bond formation between extents.
Ja
uh
The forces between the adsorbate molecules and the
adsorbent are weak van der Waal’s forces.
It forms multimolecular layers.
It forms mono-molecular layer.
(i) The nature of the gas (i.e. nature of the adsorbate). The easily liquefiable gases such as
HCl, NH3, Cl2 etc. are adsorbed more than the permanent gases such as H2, N2 and O2. The
ease with which a gas can be liquefied is primarily determined by its critical temperature.
Higher the critical temperature (Tc) of a gas, the more easily it will be liquefied and, therefore,
more readily it will be adsorbed on the solid.
Gas SO2
CH4
H2
TC
330K
190 K
33 K
(ii) Nature of adsorbent. The extent of adsorption of a gas depends upon the nature of
adsorbent. Activated charcoal (i.e. activated carbon), metal oxides (silica gel and aluminium
oxide) and clay can adsorb gases which are easily liquified. Gases such as H2, N2 and O2 are
generally adsorbed on finely divided transition metals Ni and Co.
(iii) Activation of adsorbent.
(a) Metallic adsorbents are activated by mechanical rubbing or by subjecting it to some
chemical reactions.
(b) To increase the adsorbing power of adsorbents, they are sub-divided into smaller pieces. As
a result, the surface area is increased and therefore, the adsorbing power increases.
(iv) Effect of temperature.
Mostly the process of adsorption is exothermic and
the reverse process or desorption is endothermic. If
the above equilibrium is subjected to increase in
temperature, then according to Le-Chaterlier’s
x
x
principle, with increase in temperature, the m
m
desorption will be favoured. Physical adsorption
decreases continuously with increase in temperature
Temperature
Temperature
whereas chemisorption increases initially, shows a
maximum in the curve and then it decreases
continuously.
Sa
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a
lp
A-3.
a
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A-1.
459
uh
a
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The initial increase in chemisorption with increase in temperature is because of activation
energy required.
This is why the chemical adsorption is also known as “Activated adsorption”.
A graph between degree of adsorption (x/m) and temperature ‘t’ at a constant pressure of
adsorbate gas is known as adsorption isobar.
(v) Effect of pressure. The extent of adsorption of a gas per unit
mass of adsorbent depends upon the pressure of the gas. The
variation of extent of adsorption expressed as x/m (where x is the
mole of adsorbate and m is the mass of the adsorbent) and the
pressure is given as below. A graph between the amount of
adsorption and gas pressure keeping the temperature constant is
called an adsorption isotherm.
It is clear from the figure that extent of adsorption (x/m) increases
with pressure and becomes maximum corresponding to pressure Ps
called equilibrium pressure. Since adsorption is a reversible process,
the
desorption also takes place simultaneously. At this pressure (Ps) the amount of gas adsorbed
becomes equal to the amount of gas desorbed.
It represents the variation of the mass of the gas adsorbed per gram of the adsorbent with
pressure at constant temperature.
A-5.
It means increasing the adsorption power of an adsorbent and is done by increasing the surface
area of the adsorbent by a suitable method.
A-6.
NH3 has higher critical temperature than that of CO2, i.e. NH3 is more easily liquefiable than
CO2 because, NH3 has greater intermolecular forces of attraction and hence will be adsorbed
more readily.
A-7.
1.2
A-8.
5 × 10–19 m2
A-9.
In heterogeneous catalysis, generally the reactants are gaseous where as catalyst is a solid. The
reactant molecules are adsorbed on the surface of the catalyst. As a result, the concentration of
the reactant molecules on the surface increases and hence the rate of reaction increases.
Sa
nk
a
lp
Ja
A-4.
A-10. 0.02 g
A-11. T0 < 27ºC
B-1.
(i) Mfg. of NH3 (Haber's process) – using iron as catalyst
(ii) Mfg. of H2SO4 – using platinised asbestos or V2O5 as catalyst.
B-2.
(ii) → (i) → (v) → (iii) → (iv)
C-1.
On the bases of physical state of D.P. and D.M. colloidal solution may be divided into eight
system.
460
a
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Table : Type of Colloidal Systems
DP
DM
Type of colloid Examples
Solid Solid Solid Sol
Some coloured glasses and gem
stones
Solid Liquid Sol
Paints, cell fluids
Solid Gas
Aerosol
Smoke, dust
Liquid Solid Gel
Cheese, butter, jellies
Liquid Liquid Emulsion
Milk, hair cream
Liquid Gas
Liquid Aerosol Fog, mist, cloud, insecticide sprays
Gas
Solid Solid Sol
Pumice stone, foam rubber
Gas
Liquid Foam
Froath, whipped cream, soap lather.
(a) Gel – a colloidal dispersion of a liquid in a soild, e.g., butter
(b) Liquid aerosol – a colloidal dispersion of a liquid in a gas, e.g., fog
(c) Hydrosol – a colloidal sol of a solid in water as the dispersion medium, e.g, starch sol or
gold sol.
C-3.
Associated colloids are formed by electrolytes so that they are dissociated into ions and these
ions associate together to form ionic micelles whose size lies in the colloidal range, e.g. soaps.
Multimolecular colloids–formed by the aggregation of a large number of simple molecules.
Macromolecular colloids – due to large size of the molecules themselves.
C-4.
Multimolecular – S8 ; Macromolecular – starch
C-5.
Sol of sulphur – oxidation method or by exchange of solvent.
Sol of platinum – Bredig's electro–disintegration method.
D-1.
(a) Peptization: The term has originated from the digestion of proteins by the enzyme pepsin.
Peptization may be defined as (the process of converting a precipitate into colloidal sol by
shaking it with dispersion medium in the presence of a small amount of electrolyte). The
electrolyte used for this purpose is called peptizing agent. This method is applied, generally, to
convert a freshly prepared precipitate into a colloidal sol. During peptization, the precipitate
adsorbs one of the ions of the electrolyte on its surface. The ion adsorbed on the surface is
common either with the anion or cation of the electrolyte. This causes the development of
positive or negative charge on precipitates which ultimately break up into smaller particles
having the dimensions of colloids.
Sa
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a
lp
Ja
uh
C-2.
(b) Electrical Properties (Electrophoresis) : The particles of the colloids are electrically
charged and carry positive or negative charge. The dispersion medium has an equal and
opposite charge making the system neutral as a whole. Due to similar nature of the charge
carried by the particles, they repel each other and do not combine to form bigger particles. That
is why, a sol is stable and particles do not
461
Sa
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a
lp
Ja
uh
a
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settle down. Arsenious sulphide, gold, silver
and platinum particles in their respective
colloidal sols are negatively charged while
particles of ferric hydroxide, aluminium
hydroxide are positively charged. The
existence of the electric charge is shown by
the phenomenon of electrophoresis. It
involves the 'movement of colloidal particles
either towards the cathode or anode, under
(Fig. : A set up for electrophoresis.)
the influence of the electric field'. The
apparatus used for electrophoresis as shown
in fig.
The colloidal solution is placed in a U-tube fitted with platinum electrodes. On passing an
electric current, the charged colloidal particles move towards the oppositely charged electrode.
Thus, if arsenic sulphide sol is taken in the U-tube, in which negatively charge particle of
arsenic sulphide move towards the anode.
*Earlier this process was called cataphoresis because most of the colloidal sols studied at that
time were positively charged and moved towards cathode.
(c) Dialysis : It is a process of removing a dissolved substance from a colloidal solution by
means diffusion through suitable membrane. Since particles in true solution (ions or smaller
molecules) can pass through animal membrane or parchment paper or cellophane sheet but
colloidal particle do not, the appratus used for this purpose is called Dialyser.
A bag of suitable membrane containing the colloidal solutions is suspended in a vessel through
which fresh water continously flow. The molecules and ions (crystalloids) diffuse through
membrane into the outer water & pure colloidal solution is left behind.
(d) Mechanical Properties :
Brownian movement: Robert Brown, a botanist, discovered in 1827
that pollen grains placed in water do not remain at rest but move about
continuously and randomly. Later on, this phenomenon was observed in
case of colloidal particles when they were seen under an
ultramicroscope. The particles were seen to be in constant zig-zag
motion as shown in fig. This zig-zag motion is called Brownian
Figure
movement
Brownian movement
D-2.
At the time of settings, the sun is at the horizon. The light emitted by the sun has to travel a
longer distance through the atmosphere. As a result, blue part of the light is scattered away by
the dust particles in the atmosphere. Hence, the red part is visible.
D-3.
Because colloidal solutions being bigger aggregate of a large number of molecule, the effective
number of particles in colloidal solution is relative much smaller.
E-1.
According to Hardy – schulze rule, greater the charge on the oppositely charged ion of the
electrolyte added, more effective it is in bringing about coagulation. Hence Na3PO4 (PO4–3) is
most effective.
E-2.
Isoelectric point : The H+ concentration at which the colloidal particles have no charge is
known as the isoelectric point. At this point stability of colloidal particles becomes very less &
do not move under influence of electric field.
462
Formation of deltas : The river water contains colloidal particles of sand and clay which carry
negative charge. The sea water contains +ve ions such as Na+, Mg2+, Ca2+, etc. As the river
water meets sea water, these ions discharge the sand or clay particle which are precipitated in
the form of delta.
E-4.
Artificial rain : Cloud consists of charged particle of water dispersed in air. Rain is caused by
aggregation of these minute particles. Artificial rain can be done by throwing electrified sand
of Agl from aeroplanes, colloidal H2O particle present in cloud will get coagulated by these
sand or Agl particles to form bigger water drops causing rain.
F-1.
Alcohol, phenol.
F-2.
(i) Sols are dispersions of solids in liquids while emulsions are dispersions of liquids in liquids.
(ii) Sols are quite stable wheres as emulsions are less stable.
F-3.
The process of separation of the constituent liquids of an emulsion is called demulsification.
F-4.
Changing of W/O emulsion to O/W emulsion and vise-versa is known as phase inversion.
(D)
A-2.
(B)
A-3.
(A)
A-6.
(A)
A-7.
(B)
A-8.
(A)
C-1.
(A)
C-2.
(C)
D-1.
(A)
E-1.
(C)
E-2.
(A)
E-3.
(C)
F-1.
(B)
A-4.
(C)
A-5.
(B)
B-1.
(D)
B-2.
(C)
D-2.
(B)
D-3.
(C)
E-4.
(C)
E-5.
(D)
Ja
A-1.
uh
PART - II
a
ri
E-3.
1.
(C)
lp
PART - III
(A – p, q) ; (B–r) ; (C – q, t) ; (D – s) ; (E – u)
2.
(D)
2.
(C)
Sa
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1.
a
EXERCISE - 2
PART - I
3.
(D)
4.
(B)
5.
(A)
6.
(B)
7.
(D)
8.
(B)
9.
(A)
10.
(C)
11.
(C)
12.
(A)
13.
(B)
14.
(D)
15.
(A)
16.
(B)
PART - II
3.
3 (i, ii, vi)
4.
560
5.
4
1.
6
2.
56
6.
8
7.
8
1.
(BCD)
2.
(ACD)
PART - III
3.
(ABD)
4.
(AB)
5.
(AC)
6.
(AC)
7.
(CD)
8.
(AB)
9.
(BD)
10.
(BC)
11.
(BD)
12.
(AC)
13.
(ABC)
14.
(ABD)
15.
(AC)
16.
(ABD)
(BCD)
17.
(ABC)
18.
(AB)
19.
(CD)
20.
463
PART - IV
1.
(A)
2.
(A)
3.
(B)
4.
(A)
6.
(A)
7.
(D)
8.
(B)
9.
(D)
5.
(C)
EXERCISE - 3
PART - I
(A)
2.
(B)
3.
(D)
4.
(A)
5.
(C)
6.
7
7.*
(ABD)
8.*
(AD)
9.*
(AC)
10.
(B)
11.*
(BCD)
12.
(D)
13.*
(AC)
PART - II
JEE(MAIN) OFFLINE PROBLEMS
(2)
2.
(2)
3.
(4)
6.
(3)
7.
(4)
8.
(2)
11.
(1)
4.
(3)
5.
(2)
(1)
10.
(2)
uh
1.
a
ri
1.
9.
Ja
JEE(MAIN) ONLINE PROBLEMS
(3)
2.
(4)
3.
6.
(4)
7.
(2)
8.
11.
(4)
12.
(3)
13.
16.
(4)
17.
(1)
(2)
4.
(4)
5.
(1)
(2)
9.
(4)
10.
(2)
(2)
14.
(3)
15.
(3)
19.
(3)
lp
1.
(2)
Sa
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a
18.
464
 Marked questions are recommended for Revision.
This Section is not meant for classroom discussion. It is being given to promote selfstudy and self testing amongst the students.
PART - I : PRACTICE TEST-1 (IIT-JEE (MAIN Pattern))
Max. Time : 1 Hr.
Max. Marks : 120
5.
The test is of 1 hour duration.
The Test Booklet consists of 30 questions. The maximum marks are 120.
Each question is allotted 4 (four) marks for correct response.
Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each
question. ¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No
deduction from the total score will be made if no response is indicated for an item in the answer sheet.
There is only one correct response for each question. Filling up more than one response in any question
will be treated as wrong response and marks for wrong response will be deducted accordingly as per
instructions 4 above.
uh
1.
2.
3.
4.
a
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Important Instructions
For adsorption of a gas on a solid, the plot of log (x/m) Vs log P is linear with a slope equal to :
[n being a whole number]
(1) K
(2) log K
(3) n
(4) 1/n
2.
Surface tension of lyophilic sols is :
(1) Lower than that of H2O
(3) Equal to that of H2O
Ja
1.
(2) More than that of H2O
(4) None of the above
On passing light from collidal solution, the effect due to scattering of light is known as :
(1) Electrophoresis
(2) Tyndall effect
(3) Electro osmosis
(4) Coagulation
4.
Tyndall effect is shown by :
(1) Colloid
(2) True Solution
(3) Suspension
(4) all of these
Milk is an example of :
(1) True solution
(2) Gel
(3) Suspension
(4) Emulsion
Most effective ion to coagulate a negative sol is :
(1) PO43–
(2) Al3+
(3) Ba2+
(4) K+
Sa
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6.
a
5.
lp
3.
7.
Which of the following electrolytes will be most effective in the coagulation of gold sol :
(1) NaNO3
(2) K4[Fe(CN)6]
(3) Na3PO4
(4) MgCl2
8.
The stability of lyophilic colloid is due to which of the following :
(1) Charge on their particles
(2) Large size of their particles
(3) Small size of their particles
(4) Solvation by dispersion medium
9.
A colloidal solution is subjected to an electrical field. The particles move towards anode. The
coagulation of the same solution is studied using NaCl, BaCl2 and AlCl3 solutions. Their coagulating
power should be
(1) NaCl > BaCl2 > AlCl3
(2) BaCl2 > AlCl3 > NaCl
(3) AlCl3 > BaCl2 > NaCl
(4) BaCl2 > NaCl > AlCl3
10.
Which of the following is most effective in coagulating a ferric hydroxide sol :
(1) KCl
(2) KNO2
(3) K2SO4
(4) K3[Fe(CN)6]
11.
Fog is an example of colloidal system of :
(1) Liquid dispersed in gas
(3) Solid dispersed in gas
12.
(2) Gas dispersed in gas
(4) Solid dispersed in liquid
The charge on As2S3 sol is due to the adsorbed :
(1) H+
(2) OH–
(3) O–2
(4) S–2
465
13.
14.
The sky looks blue due to :
(1) Dispersion
(2) Reflection
(3) Transmission
(4) Scattering
Tyndall effect will be mainly observed in :
(1) Lyophilic colloid
(2) Lyophobic colloid
(3) True solution
(4) Vapour
The Brownian motion is due to :
(1) Temperature fluctuation within the liquid phase
(2) Attraction and repulsion between charge on the colloidal particles
(3) Impact of molecules of the dispersion medium on the colloidal particles
(4) All of these
16.
In coagulating the colloidal solution of As2S3 which has the minimum coagulating value :
(1) NaCl
(2) KCl
(3) BaCl2
(4) AlCl3
17.
Positive sol is :
(1) Gold
19.
(3) As2S3
(4) None
Which one is a lyophobic colloid :
(1) Gelatin
(2) Starch
(3) Sulphur
Smoke is an example of :
(1) Gas dispersed in liquid
(3) Solid dispersed in gas
(2) Gas dispersed in solid
(4) Solid dispersed in solid
(4) Gum arabic
uh
18.
(2) Gelatin
a
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15.
A colloidal solution of arsenious sulphide is most readily coagulated by the addition of a normal
solution?
(1) NaCl
(2) CaCl2
(3) Na3PO4
(4) Al2(SO4)3
21.
A colloid always :
(1) Contains two phases
(3) Contains three phases
(2) Is a true solution
(4) Contains only water soluble particles
Which of the following ions has maximum flocculation value :
(1) [Fe(CN)6)]4–
(2) Cl–
(3) SO2–
4
lp
22.
Ja
20.
(4) PO3–
4
Which of the following gases, will be adsorbed maximum on a solid surface :
(1) CO2
(2) O2
(3) N2
(4) H2
24.
Which of the following is a mismatch :
(1) Lyophilic colloids
–
reversible sols
(2) Associated colloids –
micelles
(3) Tyndall effect
–
scattering of light by colloidal particle
(4) Electrophoresis
–
movement of dispersion medium under the influence of electric field
Sa
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a
23.
25.
A negative catalyst will
(1) raise the energy of activation for a given reaction
(2) take away the internal energy of reactants and deactivate them
(3) catalyse the backward reaction more than the forward one, thereby shifting equilibrium backward.
(4) none of these
26.
A liquid is found to scatter a beam of light but leaves no residue when passed through the filter paper.
The liquid can be described as
(1) a suspension
(2) Oil
(3) a colloidal sol
(4) a true solution
27.
Which of the following kinds of catalysis can be explained by the adsorption theory?
(1) heterogeneous catalysis
(2) enzyme catalysis
(3) homogeneous catalysis
(4) acid base catalysis
466
28.
Which of the following relations is (are) correct according to Freundlich ?
(i) x/m = constant
(ii) x/m = constant × p1/n (n > 1)
(iii) x/m = constant × pn (n > 1)
(1) All are correct
(2) All are wrong
(3) (ii) is correct (4) (iii) is correct
29.
The physical adsorption of gases on the solid surface is due to
(1) vander Waals forces
(2) covalent bonding
(3) hydrogen bonding
(4) All of these
30.
Correct equation of Freundlich isotherm is
1
x
(1) log   = log K +
log C
n
m
1
x
(3) log   = log C +
log C
K
m
 
1
log C
m
a
ri
x
(2) log   = log m +
m
x
(4) log   = log C +
m
1
log K
n
Practice Test-1 (IIT-JEE (Main Pattern))
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Que.
Ans.
lp
Ans.
Ja
Ans.
Que.
uh
OBJECTIVE RESPONSE SHEET (ORS)
Smoke is a colloidal sol of
(A) gas dispersed in a solid
(C) solid dispersed in a liquid
Sa
nk
1.
a
PART - II : NATIONAL STANDARD EXAMINATION IN CHEMISTRY (NSEC) STAGE-I
[NSEC-2000]
(B) solid dispersed in a gas
(D) gas dispersed in a liquid
2.
A catalyst :
(A) alters the reaction mechanism
(B) decreases the activation energy
(C) increases the average kinetic energy of reacting molecule
(D) increases the frequency of collisions of reacting species
[NSEC-2001]
3.
The colloidal system in which the disperse phase and dispersion medium are both liquids is known as :
[NSEC-2001]
(A) an emulsion
(B) an areosol
(C) gel
(D) a foam
4.
Soaps essentially form a colloidal solution in water and remove the greasy matters by : [NSEC-2001]
(A) adsorption
(B) emulsification
(C) coagulation
(D) absorption
5.
Swimming for a long time in salt water makes the skin of one’s finger tips wrinkled. Which one of the
following properties is responsible for this observation ?
[NSEC-2002]
(A) osmosis
(B) dialysis
(C) electrodialysis
(D) coagulation.
6.
Tyndal effect in a colloid is due to
(A) interference of light
(C) reflection of light
[NSEC-2002]
(B) defraction of light
(D) scattering of light.
467
7.
Ferric chloride is used to stop bleeding in cuts, because
(A) Fe3+ coagulates blood which is a positively charged sol
(B) Fe3+ coagulates blood which is a negatively charged sol
(C) CI– coagulates blood which is a positively charged sol
(D) CI– coagulates blood which is negatively charged sol
8.
A catalyst is a substance which
(A) accelerates the rate of reaction
(C) changes the equilibrium position
9.
In nature, ammonia is synthesisd by nitrifying bacteria using enzymes while in industry it is
manufactured from N2 and H2 using iron oxide catalyst at 550ºC. Under the same industrial conditions,
enzymes cannot be used because
[NSEC-2003]
(A) enzymes get deactivated at high temperature
(B) enzymes catalyze reactions only in living systems
(C) the reaction becomes vigorous and uncontrollable
(D) the enzymes use nitrates in place of N 2.
10.
A catalyst increases the
[NSEC-2003]
(A) rate of forward reaction only
(B) free energy change in the reaction
(C) rates of both forward and reverse reactions (D) equilibrium constant of the reaction.
11.
A soap solution in water removes greasy substances by
(A) adsorption
(B) peptization
(C) coagulation
[NSEC-2004]
(D) emulsifiction.
12.
Smoke is an example of
(A) sol
(B) aerosol
(D) gel.
a
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uh
Ja
(B)
[NSEC-2008]
(C)
[NSEC-2009]
(D)
Frendlich adsorption isotherms are properly represented as in
(B)
Sa
nk
(A)
[NSEC-2005]
(B) increasing the activation energy
(D) decreasing energy of activation
The plot representing Langmuir's adsorption isotherm is :
(A)
15.
(C) emulsion
lp
14.
[NSEC-2002]
(B) does not influence the rate of reaction
(D) does not alter the energy of activation of the reaction
A catalyst speeds up a chemical reraction by
(A) shifting the equilibrium
(C) initiating the reaction
a
13.
[NSEC-2002]
(C)
[NSEC-2010]
(D)
16.
A gold sol is prepared by :
(A) dissolving gold in Aqua-regia and precipitation by dilution
(B) double decomposition of AuCI3
(C) Bredy's arc method
(D) grinding in a colloidal mill
[NSEC-2010]
17.
Effective electrolyte to cause the flocculation of a negatively charged arsenium sulphide colloid is:
[NSEC-2012]
(A) NaCl
(B) BaCl2
(C) K3Fe(CN)6
(D) AICI3
18.
A catalyst is a substance that :
(A) undergoes chemical change to accelerate the rate of the reaction
(B) decreases the energy of activation of the reaction
(C) increases the kinetic energy of the reaction
(D) lowers the potential energy of the products with respect to that of the reactants.
[NSEC-2013]
468
19.
Which of the following reaction parameters will change due to addition of a catalyst ?
[NSEC-2013]
(A) Free energy
(B) Only equilibrium constant
(C) Only rate constant
(D) Both equilibrium constant and rate constant
20.
In electrophoresis,
(A) the colloidal particles migrate in an applied electric field.
(B) the medium migrates in an applied electric field
(C) both colloidal particles and the medium migrate.
(D) neither the particles nor the medium migrate.
21.
100 mL of 0.3 M acetic acid is shaken with 0.8 g wood charcoal. The final concentration of acetic acid in
the solution after adsorption is 0.125 M. The mass of “acetic acid adsorbed per gram of charcoal is:
[NSEC-2015]
(A) 1.05 g
(B) 0.0131 g
(C) 1.31 g
(D) 0.131 g
22.
The mass of argon adsorbed per unit mass of carbon surface is plotted against pressure. Which of the
following plots is correct if x and m represent the masses of argon and carbon respectively ? (........
represents extrapolated data)
[NSEC-2016]
uh
(D) x/m
P
P
V0bPO
where V is the
(1 + bP )
volume of gas adsorbed at pressure P. For a given adsorbate/adsorbent system, V 0 and b are
constants. The dependence of V on P can be depicted as
[NSEC-2018]
(A)
Ja
An adsorption isotherm equation proposed by Langmuir is of the form V =
1/V
(B) 1/V
a
1/P
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23.
(C) x/m
P
P
a
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(B) x/m
(A) x/m
[NSEC-2014]
V
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(C)
P
(D)
1/P
1/V
P
PART - III : HIGH LEVEL PROBLEMS (HLP)
SUBJECTIVE QUESTIONS
1.
A 1 cm × 1 cm square paper coated with a suitable adsorbent on both sides. The paper is dipped in a
aqueous solution of glucose of volume 20 mL and concentration 20 ppm. Final concentration glucose was
dropped to 19 ppm due to adsorption. Find the number of glucose particles per unit area of the paper.
2.
When 9.0 ml of arsenius sulphide sol and 1.0 ml of 1.0 × 10 –4 M BaCl2 are mixed, turbidity due to
precipitation just appears after 2 hours. Find the coagulating value of the effective ion.
469
ONLY ONE OPTION CORRECT TYPE
3.
4.
Some type of gels like gelatin loose water slowly. The process is known as :
(A) Synerisis
(B) Thixotropy
(C) Peptisation
(D) Imbibition
Select correct statement (s) :
(A) hydrophilic colloid is a colloid in which there is a strong attraction between the dispersed phase and
water
(B) hydrophobic colloid is a colloid in which there is a lack of attraction between the dispersed
phase and water
(C) hydrophobic sols are often formed when a solid crystallises rapidly from a chemical reaction
or a supersaturated solution
(D) all of the above
A reddish brown sol (containing Fe3+) is obtained by:
(A) the addition of small amount of FeCl3 solution to freshly prepared Fe(OH)3 precipitate
(B) the addition of Fe(OH)3 to freshly prepared FeCl3 solution
(C) the addition of NH4OH to FeCl3 solution dropwise
(D) the addition of NaOH to FeCl3 solution dropwise
6.
The stabilisation of a lyophobic colloid is due to :
(A) preferential adsorption of similiar charged particle on colloids surface.
(B) interaction between dispersed phase and dispersion medium
(C) the formation of a covalent bond between two phases.
(D) the viscosity of the medium.
7.
Compared to common colloidal sols, micelles have :
(A) higher colligative properties
(B) lower colligative properties
(C) same colligative properties
(D) none of these
8.
Graph between log x/m and log p is a straight line inclined at an angle of 45º. When pressure is 0.5 atm
and ln k = 0.693, the amount of solute adsorbed per gram of adsorbent will be :
(A) 1
(B) 1.5
(C) 0.25
(D) 2.5
9.
The coagulation of 200 mL of a positive colloid took place when 0.73 g HCl was added to it without
changing the volume much. The flocculation value of HCl for the colloid is :
(A) 0.36
(B) 36.5
(C) 100
(D) 150
lp
Ja
uh
a
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5.
10.
a
SINGLE AND DOUBLE VALUE INTEGER TYPE
0.02 g of gelatin is required to protect 10 mL of gold sol from 10% NaCl, then find the gold number for
gelatin. Report your answer by multiplying by 100.
Sa
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ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
11.
The incorrect statements are (A) For coagulation of As2S3 sol, +ve ions are effective.
(B) For coagulation of aluminium hydroxide sol Ba2+ ions are more effective than Na+
(C) Cellulose solution is an example of multimolecular colloid system
(D) Colloidal sol of metals such as gold, silver etc are prepared by Bredig's arc method.
12.
An example of extrinsic colloid (lyophobic colloids) is :
(A) As2S3 sol
(B) Fe(OH)3 sol
(C) Egg albumin
(D) Au sol
13.
Which of the following are the characteristic of chemisorption :
(A) Multilayer adsorption
(B) Exothermic nature
(C) Strong adsorption by adsorption sites
(D) Irreversible
14.
If Cl2 gas is enclosed in presence of powdered charcoal in a closed vessel, the pressure of the gas
decreases. It is because
(A) the gas molecules are absorbed at the surface
(B) the gas molecules concentrate at the surface of the charcoal
(C) the gas molecules are adsorbed at the surface
(D) the gas molecules are desorbed by the surface
470
16.
17.
Which of the following colloidal solutions contain negatively charged colloidal particles?
(A) Fe(OH)3 sol
(B) As2 S3 sol
(C) Blood
(D) Gold sol
Which of the following statements is correct for a lyophobic solution ?
(A) It can be easily solvated
(B) It carries charges
(C) The coagulation of this sol is irreversible in nature
(D) It is less stable in a solvent
Which one of the following statements is/are not correct ?
(A) Brownian movement is more pronounced for smaller particles than for bigger ones
(B) Sols of metal sulphides are lyophilic
(C) Schulze-Hardy law states, the bigger the size of the ion, the greater is its coagulating power
(D) One would expect charcoal to adsorb hydrogen gas more strongly than chlorine.
a
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15.
PART - IV : PRACTICE TEST-2 (IIT-JEE (ADVANCED Pattern))
Max. Time : 1 Hr.
Max. Marks : 66
Important Instructions
General :
The test is of 1 hour duration.
The Test Booklet consists of 22 questions. The maximum marks are 66.
Question Paper Format :
Each part consists of five sections.
Section-1 contains 7 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE is correct.
5.
Section-2 contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE OR MORE THAN ONE are correct.
6.
Section-3 contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from
0 to 9 (both inclusive).
7.
Section-4 contains 1 paragraphs each describing theory, experiment and data etc. 3 questions relate to
paragraph. Each question pertaining to a partcular passage should have only one correct answer among
the four given choices (A), (B), (C) and (D).
8.
Section-5 contains 1 multiple choice questions. Question has two lists (list-1 : P, Q, R and S; List-2 : 1, 2,
3 and 4). The options for the correct match are provided as (A), (B), (C) and (D) out of which ONLY ONE
is correct.
C. Marking Scheme :
9.
For each question in Section-1, 4 and 5 you will be awarded 3 marks if you darken the bubble
corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus one
(– 1) mark will be awarded.
10. For each question in Section-2, you will be awarded 3 marks. If you darken all the bubble(s)
corresponding to the correct answer(s) and zero mark. If no bubbles are darkened. No negative marks will
be answered for incorrect answer in this section.
11. For each question in Section-3, you will be awarded 3 marks if you darken only the bubble corresponding
to the correct answer and zero mark if no bubble is darkened. No negative marks will be awarded for
incorrect answer in this section.
Sa
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Ja
uh
A.
1.
2.
B.
3.
4.
SECTION-1 : (Only One option correct Type)
This section contains 7 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which Only ONE option is correct.
1.
An arsenious sulphide sol carries a negative charge. The maximum precipitating power of this sol is :
possessed by :
(A) K2SO4
(B) CaCl2
(C) Na3PO4
(D) AlCl3
2.
According to Hardy-Schulze rule, the coagulating power of cation follows the order :
(A) Na+ > Ba2+ > Al3+
(B) Al3+ > Ba2+ > Na+
2+
3+
+
(C) Ba > Al > Na
(D) Al3+ > Na+ > Ba2+
3.
Which of the following electrolytes is least effective in causing flocculation of ferric hydroxide sol :
(A) K4[Fe(CN)6]
(B) K2CrO4
(C) KBr
(D)K2SO4
471
4.
Lyophobic colloids are :
(A) Reversible colloids (B) Irreversible colloids (C) Protective colloids
(D) Gum proteins
Which among the following statements is false ?
(A) Increase of pressure increases the amount of adsorption.
(B) Increase of temperature may decrease the amount of adsorption.
(C) The adsorption may be monolayered or multilayered.
(D) Particle size of the adsorbent will not affect the amount of adsorption.
6.
(i) At 298 K the volume of NH3 adsorbed by 1g of charcoal is higher than that of H 2 under similar
conditions.
(ii) The movement of collodial particles towards the oppositely charged electrodes on passing electric
current is known as Brownian movement.
(A) T, T
(B) T, F
(C) F, T
(D) F, F
7.
Identify the correct statement regarding enzymes.
(A) Enzymes are specific biological catalysts that can normally function at very high tempt. (T  1000 K)
(B) Enzymes are normally heterogeneous catalysts that are very specific in action
(C) Enzymes are specific biological catalysts that can not be poisoned
(D) Enzymes are non-biological catalysts.
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a
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5.
Ja
Section-2 : (One or More than one options correct Type)
This section contains 5 multiple choice questions. Each questions has four choices (A), (B), (C)
and (D) out of which ONE or MORE THAN ONE are correct.
Which of the following statements is correct?
(A) The efficiency of a heterogeneous catalyst depends upon its surface area.
(B) Catalyst operates by providing alternate path for the reaction that involves a lower activation energy.
(C) Catalyst lowers the energy of activation of the forward direction without affecting the energy of
activation of the backward direction.
(D) Catalyst does not affect the overall enthalpy change of the reaction.
9.
Which of the following statements is (are) true?
(A) The concentration of a homogeneous catalyst may appear in the rate expression.
(B) A catalyst is always consumed in the reaction.
(C) A catalyst must always be in the same phase as the reactants.
(D) None of these
10.
The correct statement(s) pertiaining to the adsorption of a gas on a solid surface is (are)
(A) Adsorption is always exothermic
(B) Physisorption may transform into chemisorption at high temperature
(C) Physisorption increases with increasing temperature but chemisorption decreases with increasing
temperature
(D) Chemisorption is more exothermic than physisorption, however it is very slow due to higher energy
of activation.
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8.
11.
The incorrect statements are (A) For coagulation of As2S3 sol, +ve ions are effective.
(B) For coagulation of aluminium hydroxide sol Ba 2+ ions are more effective than Na+
(C) Cellulose solution is an example of multimolecular colloid system
(D) Colloidal sol of metals such as gold, silver etc are prepared by Bredig's arc method.
12.
Select the CORRECT statements :
(A) Langmuir adsorption isotherm is expected to be applicable at low gas pressure and moderately high
temperature.
(B) The rate of an enzyme catalysed reaction depends upon enzyme concentration.
(C) A negative catalyst raises the activation energy barrier of a reaction and thus reduces its rate.
(D) In physisorption gases having higher critical temperature are adsorbed to a greater extent than
those with lower critical temperature.
472
Section-3 : (One Integer Value Correct Type.)
This section contains 6 questions. Each question, when worked out will result in one integer
from 0 to 9 (both inclusive)
13.
On addition of one ml solution of 10% NaCl to 10 ml gold sol in the presence of 0.0250 g of starch the
coagulation is just prevented. Starch has the following gold number :
14.
At 2 atm pressure the value of
15.
1 L of 0.6 M acetic acid is shaken with 2 g activated carbon. Activated carbon absorbs some acetic acid
on its surface only. This process is called adsorption. The final concentration of the solution after
adsorption is 0.5 M. What is the amount of acetic acid adsorbed per gram of carbon.
16.
A detergent (C12H25SO4Na) solution becomes a colloidal solution at a concentration of 10 –3 mol/lit. On
an average 1013 colloidal particles are present in 1 mm3 what is average number of C12H25SO4Na in one
colloidal particle ?
17.
For the coagulation of 200 mL of As2S3 solution, 10 mL of 1 M NaCl is required. What is the coagulating
value of NaCl.
18.
A solution of palmitic acid (Molar mass = 256) in Benzene contain 5.12 g of acid per litre of solution.
When this solution is dropped on a water surface, the Benzene evaporates and acid forms a monolayer
V
film of solid type. If 500 cm2 are is to be covered by a monolayer, then find X, where X =
, when V
100
is volume required of solution. The area covered by 1 molecule = 0.2 nm2.
lp
Ja
uh
a
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x
will be : (log 2 = 0.3010)
m
a
SECTION-4 : Comprehension Type (Only One options correct)
This section contains 1 paragraphs, each describing theory, experiments, data etc. 3 questions
relate to the paragraph. Each question has only one correct answer among the four given
options (A), (B), (C) and (D).
Paragraph for Questions 19 to 21
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Read the following passage carefully and answer the questions.
The Colloidal particles are electrically charged as is indictated by their migration towards cathode or
anode under the applied electric field. In a particular colloidal system, all particles carry either positive
charge or negative charge.
The electric charge on colloidal particles originate in several ways. According to preferential adsorption
theory, the freshly obtained precipitate particles adsorb ions from the dispersion medium, which are
common to their lattice and acquire the charge of adsorbed ions. For example, freshly obtained
Fe(OH)3 precipitated is dispersed, by a little FeCI 3, into colloidal solution owing to the adsorptions of
Fe3+ ions in preference. Thus sol particles will be positively charged.
In some cases the colloidal particles are aggregates of cations or anions having amphiphilic character.
When the ions possess hydrophobic part (hydrocarbon end) as well as hydrophilic part (polar end
group), they undergo association in aqueous solution to form particles having colloidal size. The
formation of such particles, called micelles plays a very important role in the solubilization of water
insoluble substances, (hydrocarbon, oils, fats, grease etc.). In micelles, the polar end groups are
directed towards water and the hydrocarbon ends into the centre.
The charge on sol particles of proteins depends on the pH. At low pH, the basic group of protein
molecule is ionized (protonated) and at higher pH (alkaline medium), the acidic group in ionized. At
Isoelectric pH, characteristic to the protein, both basic and acidic groups are equally ionized.
The stability of colloidal solution is attributed largely to the electric charge of the dispersed particles.
This charge causes them to be coagulated or precipitated. On addition of small amount of electrolytes,
the ions carrying opposite charge are adsorbed by sol particles resulting in the neutralization of their
charge.When the sol particles either with no charge or reduced charge, come closer due to Brownian
473
movement, they aggregate to form bigger particles resulting in their separation from the dispersion
medium. This is what is called coagulation or precipitation of the colloidal solution. The coagulating
power of the effective ion, which depend on its charge, is expressed in terms of its coagulating value,
defined as its minimum concentration (m mol/L) needed to precipitate a given sol.
How would you obtain a sol of AgI, the particles of which migrate toward cathode under the electric
field?
(A) By adding little excess of K to AgNO3 solution
(B) By adding little excess of AgNO3 to K solution
(C) By mixing equal volumes of 0.010 M AgNO3 and 0.010 M K
(D) None of these
20.
100 ml each of two sols of Ag, one obtained by adding AgNO3 to slight excess of K and another
obtained by adding K to slight excess of AgNO3, are mixed together. Then
(A) The two sols will stabilize each other
(B) The sol particles will acquire more electric charge
(C) The sols will coagulate each other mutually (D) A true solution will be obtained
21.
Under the influence of an electric field, the particles in a sol migrate towards cathode. The coagulation
of the same sol is studied using NaCl, Na2SO4 and Na3PO4 solutions. Their coagulating values will be in
the order
(A) NaCl > Na2 SO4 > Na3 PO4
(B) Na2 SO4 > Na3 PO4 > NaCl
(C) Na3 PO4 > Na2 SO4 > NaCl
(D) Na2 SO4 > NaCl > Na3 PO4
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19.
Match list I with list II and select the correct answer :
List I
List II
P
Coagulation
1.
Scattering of light
Q
Dialysis
2.
Formation of colloidal solution from precipitates.
R
Peptization
3.
Purification of colloids
S
Tyndall effect
4.
Accmulation of collidal sols
Code:
P
Q
R
S
P
Q
R
(A)
3
4
1
2
(B)
1
2
4
(C)
3
3
2
1
(D)
2
3
1
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22.
Ja
SECTION-5 : Matching List Type (Only One options correct)
This section contains 1 questions, each having two matching lists. Choices for the correct
combination of elements from List-I and List-II are given as options (A), (B), (C) and (D) out of
which one is correct.
S
3
4
Practice Test-2 ( (IIT-JEE (ADVANCED Pattern))
OBJECTIVE RESPONSE SHEET (ORS)
Que.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
Ans.
Que.
Ans.
Que.
Ans.
474
PART - I
(4)
2.
(1)
3.
(2)
4.
(1)
5.
(4)
6.
(2)
7.
(4)
8.
(4)
9.
(3)
10.
(4)
11.
(1)
12.
(4)
13.
(4)
14.
(2)
15.
(3)
16.
(4)
17.
(2)
18.
(3)
19.
(3)
20.
(4)
21.
(1)
22.
(2)
23.
(1)
24.
(4)
25.
(2)
26.
(3)
27.
(1)
28.
(3)
29.
(1)
30.
(1)
(B)
2.
(B)
3.
(A)
6.
(D)
7.
(B)
8.
(A)
11.
(D)
12.
(B)
13.
(D)
16.
(C)
17.
(D)
18.
(B)
21.
(C)
22.
(D)
23.
(B)
4.
(B)
5.
(A)
9.
(A)
10.
(C)
14.
(B)
15.
(C)
19.
(C)
20.
(A)
3.
(A)
Ja
1.
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PART - II
a
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1.
3.33 × 1016 cm–2
4.
(D)
5.
9.
(C)
10.
(A)
a
1.
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PART - III
Sa
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20
14.
(BC)
15.
(BCD)
2.
Ba2+, 10 m mol/L
6.
(A)
7.
(B)
8.
(A)
11.
(BC)
12.
(ABD)
13.
(BCD)
16.
(BCD)
17.
(BCD)
PART - IV
1.
(D)
2.
(B)
3.
(C)
4.
(B)
5.
(D)
6.
(B)
7.
(B)
8.
(ABD)
9.
(A)
10.
(ABD)
11.
(BC)
12.
(ABD)
13.
25
14.
4
15.
03
16.
60
17.
50
18.
2
19.
(B)
20.
(C)
21.
(A)
22.
(C)
475
PART - I
1
x
= log K + log P
m
n
x
= KP1/n,
m
2.
That's why lyophilic colloid has affinity for water.
3.
Scattering of light by colloidal particles is known as Tyndall effect.
4.
Tyndall effect is shown by colloidal solution.
5.
In milk, liquid fat particles are dispersed in water.
6.
Higher the charge on coagulating ion, higher the coagulating power.
7.
Gold sol is – ve sol, so coagulating ion is cation.
8.
Lyophilic colloid is solvated by dispersion medium and becomes more stable.
9.
Coagulating power  charge on coagulating ion.
10.
Effectiveness of coagulation by electrolyte  charge on coagulating ion.
11.
Fog is an example of liquid dispersed in gas.
12.
As2S3 colloidal sol is obtained when As2O3 is saturated with H2S :
uh
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log
a
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1.
lp
As2O3 + 3H2S → As2S3 + 3H2O.
As2S3 adsorbs S2– ions (common between H2S and As2S3 and thus is negatively charged).
As2S3 + H2S → As2S3
S2– : 2H+.
Light is scattered by colloidal particles present in environment so sky looks blue.
14.
Colloidal particle shows Tyndall effect because of it's larger size.
15.
Brownian motion is due to impact of molecules of the dispersion medium on the colloidal particles.
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a
13.
16.
As2S3 is negatively charged sol so more positively charged ion will have minimum coagulating value.
17.
Gelatin is positive sol.
18.
Sulphur is a lyophobic colloid.
19.
Smoke is an example of solid dispersed in gas.
20.
Arsenious sulphide is negatively charged sol so more the charge on cation of electrolyte, more the
efficiency of electrolyte for coagulation.
21.
Colloid is heterogeneous, biphasic solution.
22.
Smaller the charge on coagulating ion, higher the flocculation value.
23.
Easily liquefiable gases like CO2 are adsorbed to a greater extent than gases like O2, N2 and H2
24.
Electrophoresis means movement of colloidal particles under the influence of electric field.
476
25.
Negative catalyst provides a path of higher activation energy
26.
These are the properties of colloidal solution.
27.
Adsorption theory is given for heterogeneous catalyst. Example : adsorption of gas on solid surface.
28.
According to Freundlich adsorption isotherm,
29.
Physical adsorption is due to vander waals forces.
30.
According to Freundlich isotherm :
x
m
 kp1/n (n > 1).
PART - III
Mass of glucose adsorbed = (20 – 19) ×
Moles of glucose adsorbed =
20
= 2 × 10–5 g.
106
2  10 −5
10−6
=
.
180
9
10−6
20
× 6 × 1023 =
× 1016
9
3
Ja
Number of glucose molecules =
uh
1.
a
ri
1
x
x
= Kp1/n or log
= log K + logP (For solution, P = C).
m
m
n
Number of glucose molecules per unit area =
1
20
×
× 1016 = 3.33 × 1016 cm–2.
2
3
10 −4
M = 10–2 M = 10 mmole/L.
10  10 −3
Conc. of Ba2+ =
3.
Loss of water from gel is known as synerisis.
4.
All are facts to remember.
5.
Reddish brown sol is prepared by adding FeCl3 in Fe(OH)3 precipitate.
6.
Similar charged colloidal particles will repel each other so colloidal system will not be suspended.
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2.
7.
Micelles have large molar mass so less colligative property.
8.
log
1
x
= log k +
log P
M
n
1
= tan 45°
n
n=1
x
= 2 x (0.5)1
m
x=1
9.
ln k = 0.69
k=2
200 mL of the sol require = 0.73 g HCl
0.73
=
mol = 0.02 mol = 20 m mol.
36.5
So, flocculation value of HCl =
20mmole
= 100
0.2lit
477
Gold number is minimum amount of protective colloid which can protect 10 mL standard gold sol from
coagulation when 1 mL of 10% NaCl is added.
11.
(B) Aluminium hydroxide is a +ve sol, so – ve ions are effective in coagulation.
(C) Cellulose solution is an example of macromolecular colloid.
12.
Egg albumin is lyophlic colloid.
13.
Monolayer is formed during chemisorption.
14.
This is because of absorption.
15.
Fe(OH)3 is postive sol, remaining all three are negative sol.
16.
Lyophobic colloid are solvent hating.
17.
Small particles has more random behaviour.
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10.
PART - IV
–ve charged sol is coagulated by cation of electrolyte. More the charge on cation, higher the
precipitating power.
2.
More the charge on cation, higher the coagulating power.
3.
Ferric hydroxide sol is + vely charged sol.
4.
Lyophobic colloids are irreversible colloids.
5.
Particle size of the adsorbent affects the amount of adsorption.
6.
(i) NH3 is easily liquifiable, so ordered more than H2.
(ii) This phenomenon is known as electrophoresis, not Brownian movement.
7.
Enzymes are highly specific heterogeneous catalyst.
8.
Efficiency of a heterogeneous catalyst increases with its surface area.
Catalyst provides a path of lower activation energy but enthalpy of reaction is not affected.
9.
Catalyst may appear in rate expression. But it is not consumed in the reaction.
Sa
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Ja
uh
1.
10.
(A) H = –ve for adsorption
(B) fact
(D) chemical bonds are stronger than vander waal’s forces so chemical adsorption is more exothermic.
11.
(B) Aluminium hydroxide is a +ve sol, so – ve ions are effective in coagulation.
(C) Cellulose solution is an example of macromolecular colloid.
12.
A negative catalyst does not raises the activation energy barrier of a reaction; it actually functions to
remove the active intermediate from the reaction and thus, it retards the rate of a reaction.
13.
Gold number is minimum amount of protective colloid which can protect 10 ml standard gold sol from
coagulation when 1 ml of 10% NaCl is added.
15.
Mass of acetic acid adsorbed = (0.6 – 0.5) × 1 × 60 = 6 g
Mass of acetic acid adsorbed per gram of carbon =
16.
6
=3g
2
1 litre colloidal solution contains = 10–3 × NA molecule of C12H25SO4Na.
478
10−3  NA  1
= 6 × 1014 molecule of C12H25SO4Na.
106
1013 colloidal particles = 6 × 1014 molecule of C12H25SO4Na.
1 mm3 colloidal solution contains =
1 colloidal particle =
17.
6  1014
molecule of C12H25SO4Na = 60 molecules.
1013
10 ml of 1 M NaCl contains NaCl = 10 × 1 = 10 milli mole
200 ml of As2S3 required NaCl for the coagulation = 10 milli mole
 1000 ml of As2S3 required NaCl for the coagulation = 10 × 1000/ 200 = 50 milli mole
Let V litre of palmitic acid is required.
5.12 V
Number of molecules =
× 6.023 × 1023 = 1.2 × 1022 V
256
Area covered = (1.2 × 1022 V) (0.2 × 10–14 cm2) = 2.4 × 107 V cm2
2.4 × 107 V cm2 = 500 cm2
2
V = 0.02 cm3 =
cm3
100
X=2
19.
We want to prepare sol of Ag having positively charged particles, so a little excess of Ag + should be
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18.
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added to K.
20.
The sols will neutralise each other so will coagulate each other because of opposite charge.
21.
Greater the charge on negative ions of salt used (since sol is positively charged) smaller will be its
coagulating value.
22.
P.
Q.
R.
S.
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Coagulation is known as accomulation of collidal sols.
Dialysis is purification of colloids.
Peptization is formation of colloidal solution from precipitates.
Tyndall effect is scattering of light by colloidal particle.
479
EQUIVALENT CONCEPTS & TITRATIONS
JEE(Advanced) Syllabus
Concept of oxidation and reduction, redox reactions, oxidation number, balancing redox
reactions and normality, Law of Equivalence, titration, Application of redox titration, hardness
JEE(Main) Syllabus
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of water, parts per million (PPM), Bleaching powder, Hydrogen peroxide (H2O2), Oleum.
Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for
assigning oxidation number, balancing of redox reactions, concept of equivalents, titration,
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hardness of water.
Section (A) : Classical Concept of Equivalent weight / Mass, Equivalent weight,
n-factor and Normality for Acid, Base and Precipitate
Concept of equivalents :
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Equivalent mass of element : Number of parts by mass of an element which reacts or displaces from
a compound 1.008 parts by mass of hydrogen, 8 parts by mass of oxygen and 35.5 parts by mass of
chlorine, is known as the equivalent weight of that element.
e.g.
2Mg + O2 ⎯⎯
→ 2MgO
48 g 32 g
12 g 8 g
32 g of O2 reacts with 48 g of Mg
48  8

8 g of O2 =
= 12 g
32

Equivalent weight of Mg = 12
Similarly,
Zn + H2SO4 ⎯⎯
→ ZnSO4 + H2
65.5 g
32.75
65.5

Equivalent weight of Zn =
= 32.75 g
2
3
Al
+
Cl2 ⎯⎯
→ AlCl3
2
3
27 g
× 71 g
2
111.5 g chlorine reacts with 27 g of Al.
27  35.5

35.5 chlorine reacts with =
9.0 g of Al
111.5
27
= 9.0
3
As we can see from the above examples that equivalent weight is the ratio of atomic weight and a
factor (say n-factor or valency factor) which in the above three cases is their respective valencies.

Equivalent weight of aluminium =
480
Equivalent weight (E) :
In general,
Eq. wt. (E) =
Mol. wt.
M
Atomic weight or Molecular weight
=
=
n − factor
x
valency factor(v.f )
Number of equivalents =
⚫
⚫
⚫
mass of species
eq. wt. of that species
For a solution, Number of equivalents = N1V1, where N is the normality and V is the volume in litres
Equivalent mass is a pure number which, when expressed in gram, is called gram equivalent mass.
The equivalent mass of a substance may have different values under different conditions.
There is no hard and fast rule that equivalent weight will be always less than the molecular mass.
Valency factor calculation :
For Elements : Valency factor = valency of the element.
For Acids : Valency factor = number of replaceable H + ions per acid molecule.
Ex.-1
HCl ,
H3PO4
1
2
H3PO3
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Valency factor →
H2SO4
{see there are only two replaceable H+ions}
2
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Sol.
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⚫
⚫
3
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(Assume 100% dissicoiation)
Eq. wt. (E)
→
M/1
M/2
M/3
M/2

Replaceable hydrogen atoms are those hydrogen atoms which are attached with the atoms of
group VI and group VII i.e. O, S, Se, Te & F, Cl, Br, I.
For Bases :
Valency factor = number of replacable OH– ions per base molecule.
Ex-2.
Sol.
NaOH,
KOH
v .f. →
1
1
Eq. wt. →
M/1
M/1

Bases may be defined as the substances in which OH group is/are directly attached with group
I elements (Li,Na, K,Rb,Cs), group II elements (Be, Mg,Ca,Ba ) or group III elements (Al, Ga,In,Tl),
transition metals, non-metallic cations like PH4+ , NH4+ etc.
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⚫
⚫
Acid-base reaction :
In case of acid base reaction, the valence factor is the actual number of H + or OH– replaced in the
reaction. The acid or base may contain more number of replaceble H + or OH– than actually replaced in
reaction.

v. f. for base is the number of H+ ion from the acid replaced by each molecule of the base.
Ex-3.
Sol.
2NaOH + H2SO4 ⎯⎯
→ Na2 SO4 + 2H2O
Base
Acid
Valency factor of base = 1
Here, two molecule of NaOH replaced 2H+ ion from the H2 SO4. Therefore, each molecule of NaOH
replaced only one H+ ion of acid, so v.f. = 1.

v. f. for acid is the number of OH– replaced from the base by each molecule of acid.
481
⚫
Ex-5.
Sol.
Salts :
(a) In non-reacting condition

Valency factor = Total number of positive charge or negative charge present in the compound.
V.f.
Eq.wt.
Na2 CO3 ,
2
M/2
Fe2(SO4)3 ,
2×3 = 6
M/6
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Sol.
NaOH + H2SO4 ⎯⎯
→ NaHSO4 + H2O
Base
Acid
Valency factor of acid = 1
Here, one of molecule of H2SO4 replaced one OH– from NaOH. Therefore, valency factor for H2SO4 is
one
Mol.wt

Eq. wt. of H2SO4 =
1
FeSO4.7H2O
2
M/2
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Ex-4.
Note : In case of hydrated salt, positive/negative charge of water molecule is not counted.
Ex-6.
Sol.
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(b) In reacting condition
Na2CO3 + HCl ⎯⎯
→ NaHCO3 + NaCl
Base
Acid
It is an acid base reaction, therefore valency factor for Na2CO3 is one while in non-reacting condition, it
will be two.
Section (B) : Equivalent weight, n-factor and Normality for Oxidant and Reductant
KMnO4 + H2O2 ⎯⎯
→ Mn2+ + O2
Mn in KMnO4 is going from +7 to +2, so change in oxidation number per molecule of KMnO 4 is 5. So the
a
Ex-7
Sol.
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(c) Equivalent weight of oxidising / reducing agents in a redox reaction
In case of redox change , v.f. = Total change in oxidation number per molecule .
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valency factor of KMnO4 is 5 and equivalent weight is
M
.
5
Normality :
Normality of a solution is defined as the number of equivalents of solute present in one litre (1000 mL)
solution.
Let V mL of a solution is prepared by dissolving W g of solute of equivalent weight E in water.
⚫
Number of equivalents of solute =
W
E
W
equivalents of solute
E
W  1000

1000 mL solution will contain
equivalents of solute.
EV
W  1000
Normality (N) =
EV
VmL of solution contain
⚫
⚫
Relations between Normality and Molarity :
Normality (N) = Molarity x Valency factor
or
N × V (in mL) = M × V (in mL) × n
or
milliequivalents = millimoles × n
482
Ex-8.
Calculate the normality of a solution containing 15.8 g of KMnO 4 in 50 mL acidic solution.
So.
Normality (N) =
Here
W  1000
EV
W = 15.8 g ,
V = 50 mL
E=
molar mass of KMnO 4
= 158/5 = 31.6
Valency factor
So, normality = 10 N
Calculate the normality of a solution containing 50 mL of 5 M solution of K2Cr2O7 in acidic medium.
Normality (N) = Molarity × valency factor = 5 x 6 = 30 N
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Ex-9.
Sol.
Section (C) : Equivalent Concept for Acid Base Titration and Precipitation Reactions
Law of Equivalence :
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The law states that one equivalent of an element combine with one equivalent of the other. In a
chemical reaction, equivalents and milli equivalents of reactants react in equal amount to give same
number of equivalents or milli equivalents of products separately.
Accordingly
(i) aA + bB → mM + nN ;
meq of A = meq of B = meq of M = m.eq. of N
(ii) In a compound MxNy ;
meq of MxNy = meq of M = meq of N
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Ex-10. Find the number of moles of KMnO4 needed to oxidise one mole Cu2S in acidic medium.
The reaction is KMnO4 + Cu2S ⎯⎯
→ Mn2+ + Cu2+ + SO2
Sol.
From law of equivalence,
equivalents of Cu2S = equivalents of KMnO4
moles of Cu2S × v.f. = moles of kMnO4 × v.f.
1 × 8 = moles of KMnO4 × 5

moles of KMnO4 = 8/5
( v.f. of Cu2S = 2 (2 – 1) + 1 (4 – (–2))) = 8 and v.f. of KMnO4 = 1 (7 –2) = 5)
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Ex-11. The number of moles of oxalate ions oxidized by one mole of MnO 4– ion in acidic medium are :
3
5
2
5
(A)
(B)
(C)
(D)
2
5
5
3
Sol.
Equivalents of C2O42– = equivalents of MnO4–
x(mole) × 2 = 1 × 5
( v.f. of C2O42– = 2 (4 – 3) = 2 and v.f. of MnO4– = 1 (7 – 2) = 5).
5
x = mole of C2O42– ions.
2
Drawbacks of Equivalent concept :
⚫
Since equivalent weight of a substance (for example oxidising or reducing agent) may be variable
hence it is better to use mole concept.
e.g.
5e– + 8H+ + MnO4– ⎯→ Mn2+ + 2H2O
e.g.
3e– + 2H2O + MnO4– ⎯→ MnO2 + 4OH–

Mol. wt. of MnO 4 –
5
Mol. wt. of MnO 4 −
=
3
Eq.wt of MnO4– =
Eq.wt of MnO4–
Thus, the number of equivalents of MnO 4– will be different in the above two cases but number of
moles will be same.
⚫
Normality of any solution depends on reaction while molarity does not.
For example : Consider 0.1mol KMnO4 dissolved in water to make 1L solution. Molarity of this solution
is 0.1 M. However, its normality is NOT fixed. It will depend upon the reaction in which KMnO4
participates. e.g. if KMnO4 forms Mn2+, normality = 0.1 x 5 = 0.5 N. This same sample of KMnO4, if
employed in a reaction giving MnO2 as product (Mn in +4 state), will have normality 0.1 × 3 = 0.3 N.
483
⚫
The concept of equivalents is handy, but it should be used with care. One must never equate
equivalents in a sequence which involves same element in more than two oxidation states. Consider an
example, KIO3 reacts with KI to liberate iodine and liberated Iodine is titrated with standard hypo
solution. The reactions are :
(i)
O3– + ¯ ⎯→ 2
(ii) 2 + S2O32– ⎯→ S4O62– + ¯
−
meq of hypo = meq of I2 = meq of IO3 = meq of I−

meq of hypo = meq of IO3−.
This is wrong. Note that I2 formed by equation (i) has v.f. = 5/3 & in equation (ii) has v.f. = 2.
v.f. of I2 in both the equation are different, therefore we cannot equate milli equivalents in
sequence. In this type of case, students are advised to use mole concept.
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
V = 50 mL.
Method -2 : Equivalent Method :
At the equivalence point,
milliequivalents of MnO4¯ = milliequivalents of Fe2+
M1 × vf1 × V1 = M2 × vf2 × V2
0.02 × 5 × V1 = 0.2 × 1 × 25
( MnO4– ⎯⎯
→ Mn2+ ; v.f. = 5, Fe2+ ⎯⎯
→ Fe3+ ; v.f. = 1)

V1 = 50 mL.
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
m.moles of MnO4–
m.moles of Fe 2 +
=
5
1
(25)(0.2)
V(0.02)
=
(from (1) & (2))
1
5
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
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Ex-12 How many millilitres of 0.02 M KMnO4 solution would be required to exactly titrate 25 mL of 0.2 M
Fe(NO3)2 solution in acidic medium ?
Sol.
Method -1 : Mole concept method
Starting with 25 mL of 0.2 M Fe2+, we can write :
Millimoles of Fe2+ = 25 x 0.2
........(1)
and in volume V (in milliliters) of the KMnO4,
Millimoles of MnO4¯ = V (0.02)
........(2)
The balanced reaction is :
MnO4¯ + 5Fe2+ + 8H+ ⎯⎯
→ Mn2+ + 5Fe3+ + 4H2O
This requires that at the equivalent point,
Section (D) : Equivalent Concept for Redox reactions, KMnO4 / K2Cr2O7 v/s Reducing
Agents & their Redox Titration
Titrations :
⚫
⚫
Titration is a procedure for determining the concentration of a solution by allowing its carefully
measured volume to react with a standard solution of another substance, whose concentration is
known.
Standard solution - It is a solution whose concentration is known and is taken in burette. It is also
called Titrant.
There are two type of titrants :
Primary titrants/standard : These reagents can be accurately weighed and their solutions are not to
be standardised before use.
Ex : Oxalic acid, K2Cr2O7, AgNO3, CuSO4, ferrous ammonium sulphate, hypo etc.
Secondary titrants/standard : These reagents cannot be accurately weighed and their solutions are to
be standardised before use.
Ex : NaOH, KOH, HCl, H2SO4, 2, KMnO4 etc.
Titrate : Solution consisting of substance to be estimated, its generally taken in a beaker .
484
Equivalence point : It is the point when number of equivalents of titrant added becomes equal to
number of equivalents of titrate.
At equivalence point :
n1V1M1 = n2V2M2
Indicator : An auxiliary substance added for physical detection of completion of titration at equivalence
point. It generally show colour change on completion of titration.
Type of Titrations :
Acid-base titrations (to be studided in Ionic equilibrium)
Redox Titrations
Some Common Redox Titrations
Table of Redox Titrations : (Excluding Iodometric / Iodimetric titrations)
Estimation
By titrating
Relation*between
S.No.
Reactions
of
with
OA and RA
2+
3+
–
Fe
⎯→
Fe
+
e
5Fe2+  MnO4–
1
Fe2+
MnO4–
Eq. wt. of Fe2+ = M/1
MnO4– + 8H+ + 5e– ⎯→ Mn2+ + 4H2O
Fe2+ ⎯→ Fe3+ + e–
6Fe2+  Cr2O72–
2
Fe2+
Cr2O72–
Eq.wt. of Cr2O72– = M/6
Cr2O72– + 14H+ + 6e– ⎯→ 2Cr3+ + 7H2O
2–
–
C
O
⎯→
2CO
+
2e
5C2O42–  2MnO4¯
2 4
2
3
C2O42–
MnO4–
–
+
–
2+
Eq. wt. of C2O42– = M/2
MnO4 + 8H + 5e ⎯→ Mn + 4H2O
+
–
H2O2 ⎯→ 2H + O2 + 2e
5H2O2  2MnO4–
4
H2O2
MnO4–
–
+
–
2+
Eq.wt. of H2O2 = M/2
MnO4 + 8H + 5e ⎯→ Mn + 4H2O
As2O3 + 5H2O ⎯→ 2AsO43– + 10H+ + 4e–
–
5
As2O3
MnO4
Eq. wt. of As2O3 = M/4
MnO4– + 8H+ + 5e– ⎯→ Mn2+ + 4H2O
Eq. wt. of AsO33– = M/2
AsO33– + H2O ⎯→ AsO43– + 2H+ + 2e–
6
AsO33–
BrO3–
–
+
–
–
Eq.wt. of BrO3– = M/6
BrO3 + 6H + 6e ⎯→ Br + 3H2O
Permanganate Titrations :
KMnO4 is generally used as an oxidising agent in acidic medium generally provided by dilute H 2SO4 .
KMnO4 works as self indicator persistent pink color is the indication of end point.
Mainly used for estimation of Fe2+, oxalic acid ,oxalates, H2O2 etc.
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⚫
⚫
⚫
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⚫
⚫
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Ex-13 Write the balanced reaction of titration of KMnO 4 Vs oxalic acid in presence of H2SO4.
So.
Reaction :
2KMnO4 + 3H2SO4 + 5H2C2O4 ⎯→ K2SO4 + 2MnSO4 + 8H2O + 10CO2
M

Redox Changes :
C23+ ⎯→ 2C4+ +2e
 EH2C2O4 = 2 


M

5e + Mn7+ ⎯→ Mn2+
 EKMnO4 = 5 


Indicator :
KMnO4 acts as self indicator.
Ex-14 Write the balanced reaction of titration of KMnO4 vs ferrous ammonium sulphate in presence of H2SO4.
Sol.
Reaction :
2KMnO4 + 10[FeSO4(NH4)2SO4.6H2O] + 8H2SO4 ⎯→
5Fe2(SO4)3 + 10(NH4)2SO4 + K2SO4 + 2MnSO4 + 68H2O
M

Redox Changes :
Fe2+ ⎯→ Fe3+ + e
 EFeSO4 = 1 


M

Mn7+ + 5e ⎯→ Mn2+
 EKMnO4 = 5 


Indicator : KMnO4 acts as self indicator
485
Section (E) : Iodometric/Iodimetric Titration, Calculation of Available Chlorine from a
sample of bleaching powder
Iodometric/Iodimetric Titrations :
Compound containing iodine are widely used in titrations.
(i) Iodide ions can be oxidised to 2 by suitable oxidising agent.
2− (aq) ⎯→ 2 (s) + 2e−
−
(ii) Iodine (V) ions, O3 , will oxidise − to 2.
O− (aq) + 5− (aq) + 6H+ (aq) ⎯→ 32 (s) + 3H2O ()
(iii) Thiosulphate ions, S2O32–, can reduce iodine to iodide ions.
2S2O− (aq) +  (s) ⎯→ S4 O62– + 2–
colourless
black
colourless
Estimation of
1.
2
2.
CuSO4
3.
CaOCl2
Reaction
2 + 2Na2S2O3 ⎯→ 2Na + Na2S4O6
or 2 + 2S2O32– ⎯→ 2– + S4O62–
2CuSO4 + 4K ⎯→ 2Cu + 2K2SO4 + 2
or 2Cu2+ + 4– ⎯→ 2Cu + 2
white ppt
CaOCl2 + H2O ⎯→ Ca(OH)2 + Cl2
Cl2 + 2K ⎯→ 2KCl + 2
Cl2 + 2– ⎯→ 2Cl– + 2

MnO2 + 4HCl(conc.) ⎯⎯
→ MnCl2 + Cl2 + 2H2O
Cl2 + 2K ⎯→ 2KCl + 2
or MnO2 + 4H+ + 2Cl– ⎯→ Mn2+ + 2H2O + Cl2
Cl2 + 2– ⎯→ 2 + 2Cl–
Relation between
O.A. and R.A.
222Na2S2O3
Eq.wt. of Na2S2O3 = M/1
2CuSO422=2Na2S2O3
Eq.wt.of CuSO4 = M/1
CaOCl2Cl2222Na2S2O3
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S.No.
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Iodometric Titrations (Titration Solution is of Na2S2O3.5H2O)
MnO2
5.
O3–
O3– + 5– + 6H+ ⎯→ 32 + 3H2O
6.
H2O2
H2O2 + 2– + 2H+ ⎯→ 2 + 2H2O
7.
Cl2
Cl2 + 2– ⎯→ 2Cl– + 2
8.
O3
O3 + 6– + 6H+ ⎯→ 32 + 3H2O
9.
ClO¯
10.
Cr2O72–
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4.
ClO¯ + 2– + 2H+ ⎯→ H2O + Cl– + 2
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Cr2O72– + 14H+ + 6– ⎯→ 32 + 2Cr3+ + 7H2O
11.
MnO4–
2MnO4– + 10–+ 16H+ ⎯→ 2Mn2+ + 52 + 8H2O
12.
BrO3–
BrO3– + 6– + 6H+ ⎯→ Br– + 32 + 3H2O
13.
As(V)
H2AsO4 + 2– + 3H+ ⎯→ H3AsO3 + H2O + 2
14.
HNO2
2HNO2 + 2– ⎯→ 2 + 2NO + H2O
15.
HClO
HClO + 2– + H+ ⎯→ Cl– + 2 + H2O
Eq.wt. of CaOCl2 = M/2
MnO2Cl222Na2S2O3
Eq.wt. of MnO2 = M/2
O3–  32  6  6Na2S2O3
Eq.wt. of O3– = M/6
H2O2  2  2  2Na2S2O3
Eq.wt. of H2O2 = M/2
Cl2  2  2  2Na2S2O3
Eq.wt. of Cl2 = M/2
O3  32  6  6Na2S2O3
Eq.wt. of O3 = M/6
ClO–  2  2  2Na2S2O3
Eq.wt. of OCl– = M/2
Cr2O72–  32  6
Eq.wt. of Cr2O72– = M/6
2MnO4¯  52  10
Eq.wt. of MnO4¯ = M/5
BrO3–  32  6
Eq.wt. of BrO3– = M/6
H3AsO4  2  2
Eq.wt. of H3AsO4 = M/2
2HNO2  2  2
Eq.wt. of HNO2 = M/1
HClOI22Na2S2O3
Eq.wt. of HClO = M/2
486
Iodimetric Titrations
S.No.
Relation between
O.A. and R.A.
Estimation of
Reaction
H2S
(in acidic medium)
SO32–
(in acidic medium)
Sn2+
(in acidic medium)
H2S + 2 ⎯⎯
→ S + 2¯ + 2H+
4.
As(III) (at pH 8)
H2AsO3¯ + 2 + H2O ⎯⎯
→ HAsO42– + 2¯ + 3H+
5.
N2H4
N2H4 + 22 ⎯⎯
→ N2 + 4H+ + 4¯
1.
2.
Sn2+ + 2 ⎯⎯
→ Sn4+ + 2¯
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3.
SO32– + 2 + H2O ⎯⎯
→ SO42– + 2¯ + 2H+
H2S  2  2
Eq.wt. of H2S = M/2
SO32–  2  2
Eq.wt. of SO32– = M/2
Sn2+  2  2
Eq.wt. of Sn2+ = M/2
H2AsO3–  2  2
Eq.wt. of H2AsO3¯ = M/2
N2H4 = 22  4
Eq.wt. of N2H4 = M/4
Ja
uh
Ex-15. The sulphur content of a steel sample is determined by converting it to H 2S gas, absorbing the H2S in
10 mL of 0.005 M I2 and then back titrating the excess I2 with 0.002 M Na2S2O3. If 10 mL Na2 S2O3 is
required for the titration, how many milligrams of sulphur are contained in the sample?
Reactions :
H2S + I2 ⎯→ S + 2I– + 2H+
;
I2 + 2S2O32– ⎯→ 2I– + S4O62–
m.moles of hypo used
Sol.
Used millimoles of I2 = (m.moles of I2 taken initially) –
2
10
= 0.005 × 10 – 0.002 ×
2
= 0.04 = millimoles of H2S

Weight of sulphur = 0.04 × 10–3 × 32 × 103 mg = 1.28 mg.
Calculation of available chlorine from a sample of bleaching powder :
a
lp
The weight of available Cl2 released from the given sample of bleaching powder on reaction with dilute
acids or CO2 is called available chlorine.
CaOCl2 + H2SO4 ⎯→ CaSO4 + H2O + Cl2
CaOCl2 + 2HCl ⎯→ CaCl2 + H2O + Cl2
CaOCl2 + 2CH3COOH ⎯→ Ca(CH3COO)2 + H2O + Cl2
CaOCl2 + CO2 ⎯→ CaCO3 + Cl2
Sa
nk
Method of determination :
CaOCl2
+ 2CH3COOH
(Sample of bleaching powder)
⎯→ Ca(CH3COO)2 + H2O + Cl2
Cl2 + 2KI ⎯→ 2KCl + I2
I2
v.f. = 2
+ 2Na2S2O3 ⎯⎯⎯⎯⎯⎯⎯
→ Na2S4O6 + 2Nal
Starch as indicator
v.f. = 1
End point is indicated by disappearance of blue colour.
Let
M = Molarity of hypo (Na2S2O3) solution

millimoles of Cl2 produced = m.moles of I2 used by hypo
M V
=
where V = vol of hypo solution used in ml.
2
mass of Cl2 produced

M  V  10 −3
× 71
2
= 35.5 × M × V × 10–3
=
35.5  M  V  10 −3
× 100
W
where W = amount of belaching powder taken in g.
% of available chlorine
=
487
or
% of available Cl2 =
3.55  M  V
W
Section (F) : Volume strength of H2O2 , Hardness of water
Hydrogen peroxide (H2O2)
a
ri
Ex-16 3.55 g sample of bleaching powder suspended in H 2O was treated with enough acetic acid and K
solution. Iodine thus liberated required 80 mL of 0.2 M hypo for titration. Calculate the % of available
chlorine.
3.55  0.2  80
So.
% of Cl2 =
= 16%
3.55
Sa
nk
a
⚫
Oxidising agent : (H2O2 → H2O)
(a) Acidic medium :
2e– + 2H+ + H2O2 ⎯→ 2H2O
v.f. = 2
(b) Basic medium :
2e– + H2O2 ⎯→ 2OH–
v.f = 2
Reducing agent : (H2O2 → O2)
(a) Acidic medium :
H2O2 ⎯→ O2 + 2H+ + 2e–
v.f = 2
(b) Basic medium :
2OH– + H2O2 ⎯→O2 + 2H2O + 2e–
v.f = 2
Volume strength of H2O2 : Strength of H2O2 is represented as 10V , 20 V , 30 V etc.
20V H2O2 means one litre of this sample of H2O2 on decomposition gives 20L of O2 gas at STP.
Decomposition of H2O2 is given as :
1
H2O2 ⎯→ H2O + O2
2
1
1 mole
× 22.4 L O2 at STP
2
= 34g
= 11.2 L O2 at STP
lp
⚫
Ja
uh
H2O2 can behave both like oxidising and reducing agent in both the mediums (acidic and basic).
⚫
Molarity of H2O2 (M) =
Volume strength of H2O2
11.2
Strength (in g/L) : Denoted by S
Strength = Molarity × Mol. wt = Molarity × 34
Hardness of water (Hard water does not give lather with soap)
Temporary hardness - due to bicarbonates of Ca & Mg
Permanent hardness - due to chlorides & sulphates of Ca & Mg. There are some method by which we
can soften the water sample.
(a)
By boiling
:
2HCO3– ⎯→ H2O + CO2 + CO32–
or
488
Ca(HCO3)2 + Ca(OH)2 ⎯→ CaCO3 + 2H2O
Ca2+ + CO32– ⎯→ CaCO3
By Washing Soda
:
CaCl2 + Na2CO3 ⎯→ CaCO3 + 2NaCl
By ion exchange resins :
Na2R + Ca2+ ⎯→ CaR + 2Na+
By adding chelating agents like (PO3–)3 etc.
By Slaked lime
(b)
(c)
(d)
:
a
ri
Measurement of Hardness :
Hardness is measured in terms of ppm (parts per million) of CaCO 3 or equivalent to it.
mass of CaCO3
Hardness in ppm =
× 106
Total mass of solution
Sa
nk
a
lp
Ja
uh
Ex-17 0.00012% MgSO4 and 0.000111% CaCl2 is present in water. What is the measured hardness of water
and millimoles of washing soda required to purify water 1000 L water ?
Sol.
Basis of calculation = 100 g hard water
0.00012
MgSO4 = 0.00012 g =
mole
120
0.000111
CaCl2 = 0.000111 g =
mole
111
 0.00012 0.000111 

equivalent moles of CaCO3 = 
mole
+
111 
 120
 0.00012 0.000111 

mass of CaCO3 = 
× 100 = 2 × 10–4 g
+
111 
 120
2  10 −4
Hardness (in terms of ppm of CaCO3) =
 106 = 2 ppm
100
CaCl2 + Na2CO3 ⎯→ CaCO3 + 2NaCl
NaSO4 + Na2CO3 ⎯→ MgCO3 + Na2SO4
 0.00012 0.000111 

Required Na2CO3 for 100g of water = 
mole
+
111 
 120
= 2 × 10–6 mole
2  10 −6
2

Required Na2CO3 for 1000 litre water =
mole ( d = 1g/mL)
 106 =
100
100
20
=
mole = 20 m mole
1000
Strength of Oleum :
Oleum is SO3 dissolved in 100% H2SO4. Sometimes, oleum is reported as more than 100% by weight,
say y% (where y > 100). This means that (y − 100) grams of water, when added to 100 g of given
oleum sample, will combine with all the free SO3 in the oleum to give 100% sulphuric acid.
Hence, weight % of free SO3 in oleum = 80(y −100)/18
Ex-18 What volume of water is required (in mL) to prepare 1 L of 1 M solution of H 2SO4 (density = 1.5g/mL) by
using 109% oleum and water only (Take density of pure water = 1 g/mL).
Sol.
1 mole H2SO4 in 1L solution = 98 g H2SO4 in 1500 g solution = 98 g H2SO4 in 1402 g water.
Also, in 109% oleum, 9 g H 2O is required to form 109 g pure H2SO4 & so, to prepare 98 g H2SO4, water
needed is 9/109 × 98 = 8.09 g.
Total water needed = 1402 + 8.09 = 1410.09 g = 1410.09 mL
489
MISCELLANEOUS SOLVED PROBLEMS (MSPS)
Ans.
3.
Ans.
4.
Ans.
5.
Sol.
Find the valency factor for following bases :
(i) Ca(OH)2
(ii) CsOH
(i) 2
(ii) 1
(iii) Al(OH)3
(iii) 3
Find the valence factor for following salts :
(i) K2SO4.Al2(SO4)3.24H2O
(ii) CaCO3
(i) 8
(ii) 2
Find the valency factor for following redox reactions :
acidic
a
ri
2.
(iii) H3BO3
(iii) 1
(i) KMnO4
(ii) K2Cr2O7 ⎯⎯⎯→ Cr3+
(iii) C2O42− ⎯→ CO2
(i) 5, 3, 1;
(ii) 6 ;
(iv) Fe2+ ⎯→ Fe3+
(iv) 1
(iii) 2 ;
uh
Ans.
Find the valency factor for following acids
(i) CH3COOH
(ii) NaH2PO4
(i) 1
(ii) 2
Calculate the normality of a solution obtained by mixing 50 mL of 5 M solution of K 2Cr2O7 and 50 mL of
2 M K2Cr2O7 in acidic medium.
v.f. of K2Cr2O7 = 6
so
Nf =
5  6  50 + 2  6  50
N1V1 + N2 V2
=
= 21 N
50 + 50
V1 + V2
Ja
1.
6.
Calculate the normality of a solution containing 13.4 g of Sodium oxalate in 100 mL Sol.
Sol.
Normality =
wt. in g / eq. wt
vol. of solution in litre
Here, eq. wt. of Na2C2O4 = 134/2 = 67
Sol.
The number of moles of ferrous oxalate oxidised by one mole of KMnO 4 in acidic medium is :
(A) 5/2
(B) 2/5
(C) 3/5
(D) 5/3
Eq. of FeC2O4 = Eq. of KMnO4
moles of FeC2O4 × 3 = moles of KMnO4 × 5
so,
moles of FeC2O4 = 5/3 Ans. (D)
How many moles of KMnO4 are needed to oxidise a mixture of 1 mole of each FeSO 4 & FeC2O4 in
acidic medium ?
(A) 4/5
(B) 5/4
(C) 3/4
(D) 5/3
Eq. of KMnO4 = Eq. of FeSO4 + Eq. of FeC2O4
moles of KMnO4 × 5 = moles of FeSO4 × 1 + moles of FeC2O4 × 3
 moles of KMnO4 = 4/5
Ans. (A)
Sa
nk
8.
13.4 / 67
= 2N
100 / 1000
a
7.
N=
lp
so
Sol.
9.
Sol.
A sample of hydrazine sulphate [N2H6SO4] was dissolved in 100 mL water. 10 mL of this solution was
treated with excess of FeCl3 Sol. Ferrous ions formed were estimated and it required 20 mL of M/50
KMnO4 solution in acidic medium.
Fe3+ + N2H4 ⎯→ N2 + Fe2+ + H+
MnO4– + Fe2+ + H+ ⎯→ Mn2+ + Fe3+ + H2O
(a) Write the balanced redox reactions.
(b) Estimate the amount of hydrazine sulphate in one litre of Sol.
(a) Given
4Fe3+ + N2H4 ⎯→ N2 + 4Fe2+ + 4H+
MnO4– +5Fe2+ + 8H+ ⎯→ Mn2+ + 5Fe3+ + 4H2O
(b) In 10 mL solution, eq. of N2H6SO4 = Eq. of Fe2+ = Eq. of KMnO4
= 20 ×
1
× 5 × 10–3 = 2 × 10–3
50
v.f. of N2H6SO4 = 4
490
so,
Ans.
11.
Ans.
Sol.
2  10 −3  1000
× 130 = 6.5 g.
4  10
Write the balanced redox reaction and calculate the equivalent weight of oxidising agent and reducing
agent for titration of K2Cr2O7 Vs Ferrous ammonium sulphate.
The reaction : 6[FeSO4(NH4)2SO4.6H2O] + K2Cr2O7 + 7H2SO4 ⎯→
3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 6(NH4)2SO4 + 43H2O
M 
M

Redox changes :  EFeSO4 =  ;  EK 2Cr2O7 = 
1 
6

One litre of acidified KMnO 4 solution containing 15.8 g KMnO4 is decolorized by passing sufficient SO 2.
If SO2 is produced by FeS2, what is the amount of FeS2 required to give desired SO2 ?
15 g.
v.f. of KMnO4 = 5 & v.f. of SO2 = 2
Now, Eq. of KMnO4 = Eq. of SO2
a
ri
10.
weight of N2H6SO4 in 1 L solution =
15.8
= moles of SO2 × 2
158 / 5
moles of FeS2 =
1 1 1
 =
4 2 8
so,
weight of FeS2 =
1
× 120 = 15 g.
8
uh
so,
An aqueous solution containing 0.1 g KIO 3 (formula weight = 214) and an excess of K was acidified
with HCl. The liberated 2 consumed 45 mL of thiosulphate. The molarity of sodium thiosulphate
solution is :
The reaction involved is : O3– + – + H+ ⎯→ 2 + H2O
(A) 0.0623 M
(B) 0.0313 M
(C) 0.126 M
(D) 0.252 M
O3– + 5– + H+ ⎯→ 32 + H2O
2Na2S2O3 + 2 ⎯→ 2Na + Na2S4O6
0.1
Now, Moles of KO3 =
214
0.1
So,
Moles of 2 = 3 ×
214
0.1
Now, Moles of Na2S2O3 = 2 × 3 ×
214
45
0.1
0.1

M × VL = 2 × 3 ×

M×
=2×3×
214
214
1000
0.1 1000

Now, Molarity of hypo solution = 2 × 3 ×
= 0.0623 M
Ans. (A)
214
45
Calculate the percentage of available chlorine in a sample of 3.55 g of bleaching powder which was
dissolved in 100 mL of water. 25 mL of this solution, on treatment with KI and dilute acid, required
20 mL of 0.125 N sodium thiosulphate Sol.
10 %
CaOCl2 + H2O ⎯→ Ca(OH)2 + Cl2
Cl2 + 2K ⎯→ 2KCl + 2
2 + 2Na2S2O3 ⎯→ Na2S2O6 + 2Na
20
0.125
In 25 mL solution,
moles of Na2S2O3 =
×
= 25 × 10–4
1
1000
1
So,
moles of 2 = × moles of Na2S2O3
2
1
= × 25 × 10–4 = 12.5 × 10–4
2
So,
in 100 mL solution, moles of Cl2 = 4 × 12.5 × 10–4 = 50 × 10–4
So,
weight of Cl2 = 50 × 10–4 × 71 g
Sa
nk
a
lp
Sol.
moles of SO2 = 1/4
applying POAC on S, we get :
2 × mole of FeS2 = 1 × moles of SO2
Ja
12.
so,
Now,
13.
Ans.
Sol.
491
% of available Cl2 =
50  10 −4  71
× 100 = 10%
3.55
 Marked questions are recommended for Revision.
PART - I : SUBJECTIVE QUESTIONS
A-2.
Determine the equivalent weight of the following ions :
(a) Na+
(b) Al3+
(c) NO+
(e) CO32–
(f) SO42–
(g) PO43–
Determine the equivalent weights of the following salts :
(a) NaCl
(b) K2SO4
(c) Ca3(PO4)2
(d) Cl–
uh
A-1.
a
ri
Section (A) : Classical Concept of Equivalent weight / Mass, Equivalent weight,
n-factor and Normality for Acid, Base and Precipitate
A-3. 1.12 litre dry chlorine gas at STP was passed over a heated metal when 5.56 g of chloride of the metal
was formed. What is the equivalent weight of the metal?
Ja
Section (B) : Equivalent weight, n-factor and Normality for Oxidant and Reductant
B-1. A mixture of CuS (molecular weight = M1) and Cu2S (molecular weight = M2) is oxidised by KMnO4
(molecular weight = M3) in acidic medium, where the product obtained are Cu2+, Mn2+ and SO2. Find the
equivalent weight of CuS, Cu2S and KMnO4 respectively.
Determine the equivalent weight of the following oxidising and reducing agents :
(a) KMnO4 (reacting in acidic medium MnO4– ⎯→ Mn2+)
(b) KMnO4 (reacting in neutral medium MnO4– ⎯→ MnO2)
lp
B-2.
0.98 g of the metal sulphate was dissolved in water and excess of barium chloride was added. The
precipitated barium sulphate weighted 0.95 g. Calculate the equivalent weight of the metal.
Sa
nk
C-1.
a
Section (C) : Equivalent Concept for Acid Base Titration and Precipitation Reactions
C-2.
A dilute solution of H2SO4 is made by adding 5 mL of 3N H2SO4 to 245 mL of water. Find the normality
and molarity of the diluted solution.
C-3. What volume at NTP of gaseous ammonia will be required to be passed into 30 cm 3 of 1 N H2SO4
solution to bring down the acid strength of the latter to 0.2 N ?
Section (D) : Equivalent Concept for Redox reactions, KMnO4 / K2Cr2O7 v/s Reducing
Agents & their Redox Titration
D-1. 1.60 g of a metal A and 0.96 g of a metal B when treated with excess of dilute acid, separately,
produced the same amount of hydrogen. Calculate the equivalent weight of A if the equivalent weight of
B is 12.
D-2.
It requires 40 mL of 1 M Ce4+ to titrate 20 mL of 1M Sn2+ to Sn4+. What is the oxidation state of the
Cerium in the product ?
D-3.
25 mL of a solution of Fe2+ ions was titrated with a solution of the oxidizing agent Cr 2O72–. 50 mL of 0.01
M K2Cr2O7 solution was required. What is the molarity of the Fe2+ solution ?
492
D-4. How many mL of 0.3M K2Cr2O7 (acidic) is required for complete oxidation of 5 mL of 0.2 M SnC 2O4
solution.
Section (E) : Iodometric/Iodimetric Titration, Calculation of Available Chlorine from a
sample of Bleaching Powder
a
ri
E-1. 10 g sample of bleaching powder was dissolved into water to make the solution one litre. To this
solution 35 mL of 1.0 M Mohr salt solution was added containing enough H 2SO4. After the reaction was
complete, the excess Mohr salt required 30 mL of 0.1 M KMnO 4 for oxidation. The % of available Cl2
approximately is (mol wt = 71)
E-2. A mixture containing As2O3 and As2O5 required 20 mL of 0.05 N iodine solution for titration. The
resulting solution is then acidified and excess of KI was added. The liberated iodine required 1.116 g
hypo (Na2S2O3.5H2O) for complete reaction. Calculate the mass of the mixture. The reactions are:
As2O3 + 2I2 + 2H2O ⎯⎯
→ As2O5 + 4H+ +4I–
As2O5 + 4H+ + 4I– ⎯⎯
→ As2O3 + 2I2 + 2H2O
uh
(Atomic weight : As = 75)
Section (F) : Volume strength of H2O2, Hardness of water
20 ml of H2O2 after acidification with dil H2SO4 required 30 ml of
Detemine the strength of H2O2 solution.
N
KMnO4 for complete oxidation.
12
Ja
F-1.
lp
F-2. A 100 mL sample of water was treated to convert any iron present to Fe 2+. Addition of 25 mL of 0.002 M
K2Cr2O7 resulted in the reaction :
6Fe2+ + Cr2O72– + 14H+ ⎯→ 6Fe3+ + 2Cr3+ + 7H2O
The excess K2Cr2O7 was back-titrated with 7.5 mL of 0.01 M Fe2+ solution. Calculate the parts per
million (ppm) of iron in the water sample.
By which reason temporary and permanent hardness occur ?
F-4.
Define two method by which we can soften the water sample.
a
F-3.
PART - II : ONLY ONE OPTION CORRECT TYPE
Sa
nk
Section (A) : Classical Concept of Equivalent weight / Mass, Equivalent weight,
n-factor and Normality for Acid, Base and Precipitate
A-1. x g of the metal gave y g of its oxide. Hence equivalent weight of the metal
y−x
x+y
x
x
(A)
×8
(B)
×8
(C)
×8
(D)
×8
x
x
(y − x)
y
A-2.
A-3.
Equivalent wt. of H3PO4 in each of the reaction will be respectively H3PO4 + OH– → H2PO4– + H2O
H3PO4 + 2OH– → HPO42– + 2H2O
H3PO4 + 3OH– → PO43– + 3H2O
(A) 98, 49, 32.67
(B) 49, 98, 32, 67
(C) 98, 32.67, 49
(D) 32.67, 49, 98
3 g of an oxide of a metal is converted to chloride completely and it yielded 5 g of chloride. Equivalent
weidht of the metal is :
(A) 33.25
(B) 3.325
(C) 12
(D) 20
493
Section (B) : Equivalent weight, n-factor and Normality for Oxidant and Reductant
B-1. An ion is reduced to the element when it absorbs 6 × 10 20 electrons. The number of equivalents of the
ion is:
(A) 0.1
(B) 0.01
(C) 0.001
(D) 0.0001
B-2. When N2 is converted into NH3, the equivalent weight of nitrogen will be :
(A) 1.67
(B) 2.67
(C) 3.67
(D) 4.67
In the ionic equation 2K+BrO3– + 12H+ + 10e– ⎯→ Br2 + 6H2O + 2K+, the equivalent weight of KBrO3
will be:
(A) M/5
(B) M/2
(C) M/6
(D) M/4
(where M = molecular weight of KBrO3)
a
ri
B-3.
Section (C) : Equivalent Concept for Acid Base Titration and Precipitation Reactions
If one mole of H2SO4 reacts with one mole of NaOH, equivalent weight of H2SO4 will be :
(A) 98
(B) 49
(C) 96
(D) 48
C-2.
How many millilitres of 0.1N H2SO4 solution will be required for complete reaction with a solution
containing 0.125 g of pure Na2CO3 :
(A) 23.6 mL
(B) 25.6 mL
(C) 26.3 mL
(D) 32.6 mL
uh
C-1.
Ja
C-3. One litre of a solution contains 18.9 g of HNO3 and one litre of another solution contains 3.2 g of NaOH.
In what volume ratio must these solution be mixed to obtain a neutral solution?
(A) 3 : 8
(B) 8 : 3
(C) 15 : 4
(D) 4 : 15
Section (D) : Equivalent Concept for Redox reactions, KMnO4 / K2Cr2O7 v/s Reducing
Agents & their Redox Titration
lp
If equal volumes of 0.1 M KMnO4 and 0.1 M K2Cr2O7 solutions are allowed to oxidise Fe2+ to Fe3+ in
acidic medium, then Fe2+ oxidised will be :
(A) more by KMnO4
(B) more by K2Cr2O7
(C) equal in both cases
(D) cannot be determined.
a
D-1.
Sa
nk
D-2. Which of the following solutions will exactly oxidize 25 mL of an acid solution of 0.1 M iron () oxalate:
(A) 25 mL of 0.1 M KMnO4
(B) 25 mL of 0.2 M KMnO4
(C) 25 mL of 0.6 M KMnO4
(D) 15 mL of 0.1 M KMnO4
D-3. An element A in a compound ABD has oxidation number –n. It is oxidised by Cr2O72– in acid medium. In
the experiment, 1.68 × 10–3 moles of K2Cr2O7 were used for 3.36 × 10–3 moles of ABD. The new
oxidation number of A after oxidation is :
(A) 3
(B) 3 – n
(C) n – 3
(D) +n
D-4.
The number of moles of oxalate ions oxidized by one mole of MnO 4– ion in acidic medium is :
(A) 5/2
(B) 2/5
(C) 3/5
(D) 5/3
Section (E) : Iodometric/Iodimetric Titration, Calculation of Available Chlorine from a
sample of Bleaching Powder
E-1.
What can be the maximum percentage of available chlorine possible in a given bleaching powder
sample (Take formula of bleaching powder as CaOCl2) ?
(A) 52.9%
(B) 55.9 %
(C) 58%
(D) 60%
E-2. A 0.2 g sample containing copper () was analysed iodometrically, where copper() is reduced to
copper () by iodide ions. 2Cu2+ + 4– ⎯⎯
→ 2 Cu + 2
494
If 20 mL of 0.1 M Na2S2O3 solution is required for titration of the liberated iodine, then the percentage of
copper in the sample will be :
(A) 31.75 %
(B) 63.5 %
(C) 53 %
(D) 37 %
Section (F) : Volume strength of H2O2, Hardness of water
F-1.
A substance which participates readily in both acid-base and oxidation-reduction reactions is :
(A) Na2 CO3
(B) KOH
(C) KMnO4
(D) H2 C2 O4
F-2._
A fresh H2O2 solution is labeled as 11.2 V. Calculate its concentration in wt/vol percent.
(A) 3.4
(B) 6.8
(C) 1.7
(D) 13.6
F-4.
a
ri
F-3. The amount of lime, Ca(OH)2 required to remove the hardness in 60 L of pond water containing 1.62
mg of calcium bicarbonate per 100 ml of water, will be :
(A) 4.44 g
(B) 0.222 g
(C) 2.22 g
(D) 0.444 g
What will the concentration of [Ca+2] in a sample of 1 litre hard water if after treatment with washing
soda 10 g insoluable CaCO3 is precipitated.
(A) 0.2 M
(B) 0.1 M
(C) 0.3 M
(D) 0.4 M
Column 
(A) 4.1 g H2SO3
(B) 4.9 g H3PO4
(C) 4.5 g oxalic acid (H2C2O4)
(D) 5.3 g Na2CO3
(p) 200 mL of 0.5 N base is used for complete neutralization
(q) 200 millimoles of oxygen atoms
(r) Central atom is in its highest oxidation number
(s) May react with an oxidising agent
Ja
Column 
lp
1.
uh
PART - III : MATCH THE COLUMN
 Marked questions are recommended for Revision.
PART - I : ONLY ONE OPTION CORRECT TYPE
a
The equivalent weight of a metal is double that of oxygen. How many times is the weight of its oxide
greater than weight of the metal?
(A) 1.5
(B) 2
(C) 0.5
(D) 3
Sa
nk
1.
2.
Oxalic acid, H2C2O4, reacts with paramagnet ion according to the balanced equation 5H 2C2O4 (aq) +
2MnO4– (aq)
2 Mn2+ (aq) + 10 CO2 (g) + 8 H2O (l). The volume in mL of 0.0162 M KMnO4 solution
required to react with 25.0 mL of 0.022 M H2C2O4 solution is :
(A) 13.6
(B) 18.5
(C) 33.8
(D) 84.4
3.
x mmol of KMnO4 react completely with y mmol of MnSO4 in presence of fluoride ions to give MnF4
quantitatively. Then :
(A) x = y
(B) 4x = y
(C) x > y
(D) x < y
4.
1 mol each of H3PO2, H3PO3 and H3PO4 will neutralise respectively x mol of NaOH, y mol of Ca(OH) 2
and z mol of Al(OH)3 (assuming all as strong electrolytes). x, y, z are in the ratio of :
(A) 3 : 1.5 : 1
(B) 1 : 2 : 3
(C) 3 : 2 : 1
(D) 1 : 1 : 1
5.
The amount of wet NaOH containing 15% water required to prepare 70 litres of 0.5 N solution is :
(A) 1.65 kg
(B) 1.4 kg
(C) 16.5 kg
(D) 140 kg
6.
28 NO3– + 3As2S3 + 4H2O ⎯→ 6AsO43– + 28NO + 9SO42– + 8H+.
What will be the equivalent mass of As2S3 in above reaction : (Molecular mass of As2S3 = M)
M
M
M
M
(A)
(B)
(C)
(D)
2
4
24
28
495
If 25 mL of a H2SO4 solution reacts completely with 1.06 g of pure Na 2CO3, what is the normality of this
acid solution :
(A) 1 N
(B) 0.5 N
(C) 1.8 N
(D) 0.8 N
8.
125 mL of 63% (w/v) H2C2O4.2H2O solution is made to react with 125 mL of a 40%(w/v) NaOH solution.
The resulting solution is: (ignoring hydrolysis of ions)
(A) neutral
(B) acidic
(C) strongly acidic
(D) alkaline
9.
25 mL of a 0.1 M solution of a stable cation of transition metal Z reacts exactly with 25 mL of
0.04 M acidified KMnO4 solution. Which of the following is most likely to represent the change in
oxidation state of Z correctly :
(A) Z+ → Z2+
(B) Z2+ → Z3+
(C) Z3+ → Z4+
(D) Z2+ → Z4+
10.
How many litres of Cl2 at STP will be liberated by the oxidation of NaCl with 10 g KMnO 4 in acidic
medium: (Atomic weight : Mn = 55 and K = 39)
(A) 3.54
(B) 7.08
(C) 1.77
(D) None of these
11.
One gram of Na3AsO4 is boiled with excess of solid KI in presence of strong HCl. The iodine evolved is
absorbed in KI solution and titrated against 0.2 N hypo solution. Assuming the reaction to be
AsO43– + 2H+ + 2– ⎯→ AsO33– + H2O + 2
calculate the volume of hypo consumed. [Atomic weight of As = 75]
(A) 48.1 mL
(B) 38.4 mL
(C) 24.7 mL
(D) 30.3 mL
12.
If 10 g of V2O5 is dissolved in acid and is reduced to V 2+ by zinc metal, how many mole of 2 could be
reduced by the resulting solution, if it is further oxidised to VO 2+ ions :
[Assume no change in state of Zn2+ions] (Atomic masses : V = 51, O = 16,  = 127)
(A) 0.11
(B) 0.22
(C) 0.055
(D) 0.44
13.
During the disproportionation of Iodine to iodide and iodate ions, the ratio of iodate and iodide ions
formed in alkaline medium is :
(A) 1 : 5
(B) 5 : 1
(C) 3 : 1
(D) 1 : 3
14.
If 1 mL of a KMnO4 solution react with 0.140 g Fe2+ and if 1 mL of KHC2O4. H2C2O4 solution react with
0.1mL of previous KMnO4 solution, how many millilitres of 0.20 M NaOH will react with 1 mL of previous
KHC2O4. H2C2O4 solution in which all the protons (H+) are ionisable ?
(A)15/16 mL
(B) 13/16
(C) 11/14
(D) None of these
a
lp
Ja
uh
a
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7.
Sa
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PART - II : SINGLE AND DOUBLE VALUE INTEGER TYPE
1.
How many equivalents of Mg would have to react in order to liberate 4 N A electrons? (Mg–2e– → Mg2+)
2.
A certain weight of pure CaCO3 is made to react completely with 20 mL of a HCl solution to give 224
mL of CO2 gas at STP. The normality of the HCl solution is:
3._
The volume of 3 M Ba(OH)2 solution required to neutralize completly 120 mL of 1.5M H3PO4 solution is:
4.
In an experiment, 50 mL of 0.1 M solution of a salt reacted with 25 mL of 0.1 M solution of sodium
sulphite. The half equation for the oxidation of sulphite ion is :
SO32– (aq) + H2O ⎯→ SO42– (aq) + 2H+ + 2e–
If the oxidation number of metal in the salt was 3, what would be the new oxidation number of metal :
5.
When tetracarbonylnickel(0) is heated, it dissociates into its components. If 5 moles of this compound is
heated and the resulting gaseous component is absorbed by sufficient amount of 2O5, liberating 2.

What volume of 4M Hypo solution will be required to react with this 2 : Ni(CO)4 ⎯⎯
→ Ni + 4CO
6.
1 mole of OH– ions is obtained from 85 g of hydroxide of a metal. What is the equivalent weight of the
metal?
7.
An oxide of a metal contains 40% oxygen, by weight. What is the equivalent weight of the metal?
496
In the following reaction, 3Fe + 4H2O ⎯→ Fe3O4 + 4H2, if the atomic weight of iron is 56, then its
equivalent weight will be :
9.
What volume of 0.05 M Ca(OH)2 solution is needed for complete conversion of 10 mL of 0.1 M H 3PO4
into Ca(H2PO4)2?
10.
Potassium acid oxalate K2C2O4.3H2C2O4.4H2O can be oxidized by MnO4– in acid medium. Calculate the
volume of (in mL) 1 M KMnO4 reacting in acid solution with 5.08 gram of the acid oxalate.
11.
In the following reaction, SO2 acts as a reducing agent :
SO2 + Cl2 + 2H2O ⎯→ H2SO4 + 2HCl
Find the equivalent weight of SO2.
a
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8.
PART - III : ONE OR MORE THAN ONE OPTIONS CORRECT TYPE
In the titration of K2Cr2O7 and ferrous sulphate, following data is obtained :
V1 mL of K2Cr2O7 solution of molarity M1 requires V2 mL of FeSO4 solution of molarity M2.
Which of the following relations is/are true for the above titration :
(A) 6 M1V1 = M2V2
(B) M1V1 = 6 M2V2
(C) N1V1 = N2V2
(D) M1V1 = M2V2
2.
Choose the correct statement(s) :
(A) 1 mole of MnO4– ion can oxidise 5 moles of Fe2+ ion in acidic medium.
(B) 1 mole of Cr2 O72– ion can oxidise 6 moles of Fe2+ ion in acidic medium.
(C) 1 mole of Cu2S can be oxidised by 1.6 moles of MnO4– ion in acidic medium.
(D) 1 mole of Cu2S can be oxidised by 1.33 moles of Cr2O72– ion in acidic medium.
3.
Which of the following samples of reducing agents is /are chemically equivalent to 25 mL of 0.2 N
KMnO4 to be reduced to Mn2+ and water :
(A) 25 mL of 0.2 M FeSO4 to be oxidized to Fe3+
(B) 50 mL of 0.1 M H3AsO3 to be oxidized to H3AsO4
(C) 25 mL of 0.1 M H2O2 to be oxidized to H+ and O2
(D) 25 mL of 0.1 M SnCl2 to be oxidized to Sn4+
4.
To a 25 ml H2O2 solution excess acidified solution of KI was added. The iodine liberated 20 ml of 0.3 N
sodium thiosulphate solution. Use these data to choose the correct statements from the following :
(A) The weight of H2O2 present in 25 ml solution is 0.102 g
(B) The molarity of H2O2 solution is 0.12 M
(C) The weight of H2O2 present in 1 L of the solution is 0.816 g
(D) The volume strength of H2O2 is 1.344 L
Sa
nk
a
lp
Ja
uh
1.
5.
There are two sample of HCl having molarity 1N and 0.25 N. Find volume of these sample taken in
order to prepare 0.75 N HCl solution. (Assume no water is used) :
(A) 20 mL, 10 mL
(B) 100 mL, 50 mL
(C) 40 mL, 20 mL
(D) 50 mL, 25 mL
6.
If mass of KHC2O4 (potassium acid oxalate) required to reduce 100 mL of 0.02 M KMnO 4 in acidic
medium is x g and to neutralise 100 mL of 0.05 M Ca(OH) 2 is y g, then which of the following options
may be correct :
(A) If x is 1 g then y is 2 g
(B) If x is 5.5g then y is 11 g
(C) If x is 2 g then y is 1 g
(D) If x is 11 g then y is 5.5 g
497
PART - IV : COMPREHENSION
Read the following passage carefully and answer the questions.
a
ri
Comprehension # 1
Equivalent Mass :
The equivalent mass of a substance is defined as the number of parts by mass of it which combine with
or displace 1.0078 parts by mass of hydrogen, 8 parts by mass of oxygen and 35.5 parts by mass of
chlorine.
The equivalent mass of a substance expressed in grams is called gram equivalent mass.
The equivalent mass of a substance is not constant. It depends upon the reaction in which the
substance is participating. A compound may have different equivalent mass in different chemical
reactions and under different experimental conditions.
(A) Equivalent mass of an acid : It is the mass of an acid in grams which contains 1.0078 g of
replaceable H+ ions or it is the mass of acid which contains one mole of replaceable H + ions. It may be
calculated as :
Molecular mass of acid
Equivalent mass of acid =
Basicity of acid
uh
Basicity of acid = number of replaceable hydrogen atoms present in one molecule of acid
(B) Equivalent mass of a base : It is the mass of the base which contains one mole of replaceable
OH– ions in molecule.
Molecular mass of base
Equivalent mass of base =
Aciditiy of base
Equivalent mass of an oxidising agent :
Ja
Acidity of base = Number of replaceable OH – ions present in one molecule of the base
(a) Electron concept : Equivalent mass of oxidising agent =
Molecular mass of oxidising agent
Number of electrons gained by one molecule
lp
(b) Oxidation number concept : Equivalent mass of oxidising agent =
Molecular mass of oxidising agent
Total change in oxidation number
per molecule of oxidising agent
Equivalent mass of Ba(MnO4)2 in acidic medium is : (where M stands for molar mass)
(A) M/5
(B) M/6
(C) M/10
(D) M/2
2.
Equivalent mass of Fe0.9O in reaction with acidic K2Cr2O7 is : (M = Molar mass)
(A) 7 M/10
(B) 10 M/7
(C) 7 M/9
(D) 9 M/7
Sa
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a
1.
3.
Equivalent weight of oxalic acid salt in following reaction is : (Atomic masses : O = 16, C = 12, K = 39)
H2C2O4 + Ca(OH)2 ⎯→ CaC2O4 + H2O
(A) 90
(B) 45
(C) 64
(D) 128
Comprehension # 2
Some amount of “20V” H2O2 is mixed with excess of acidified solution of KI. The iodine so liberated
required 200 mL of 0.1 N Na2S2O3 for titration.
4.
The volume of H2O2 solution is :
(A) 11.2 mL
(B) 37.2 mL
(C) 5.6 mL
(D) 22.4 mL
5.
The mass of K2Cr2O7 needed to oxidise the above volume of H2O2 solution is :
(A) 3.6 g
(B) 0.8 g
(C) 4.2 g
(D) 0.98 g
6.
The volume of O2 at STP that would be liberated by above H2O2 solution on disproportionation is :
(A) 56 mL
(B) 112 mL
(C) 168 mL
(D) 224 mL
498
Comprehension # 3
Answer Q.7, Q.8 and Q.9 by appropriately matching the information given in the three columns
of the following table.
Equivalent weight =
Molecular weight / Atomic weight
n-factor is very important in redox as well as nonn − factor
redox reactions.
In general n-factor of acid/base is number of moles of H+/OH– furnished per mole of acid/base. n-factor
of reactions is number of moles of electrons lost or gained per mole of reactant columns 1, 2, 3 contain
reactions, n-factor & equivalent weight respectively.
(i)
1
(II)
MnO4– ⎯→ MnO4–2
(ii)
10
6
(III)
Br2 + OH– ⎯→ BrO3– + Br–
(iii)
3
(IV)
H2O2 ⎯→ O2 + H2O
(iv)
2
(P)
158
(Q)
96
(R)
34
(S)
52.6
For KMnO4 in strong basic medium correct combination is (A) (I) (ii) (R)
(B) (II) (i) (P)
(C) (II) (iii) (S)
(D) (I) (iv) (Q)
For KMnO4 in neutral medium correct combination is (A) (I) (iii) (Q)
(B) (II) (i) (R)
(C) (I) (iii) (S)
(D) (II) (iii) (R)
For a disproportionation reaction the only correct combination is (A) (I) (ii) (R)
(B) (II) (ii) (Q)
(C) (IV) (i) (S)
(D) (III) (ii) (Q)
lp
9.
MnO4– + 2H2O ⎯→ MnO2+4OH–
Column-3
uh
8.
(I)
Ja
7.
Column-2
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Column-1
* Marked Questions may have more than one correct option.
In basic medium, I– is oxidised by MnO4–. In this process, I– changes to :
[JEE 2004, 3/144]
–
–
–
(A) IO3
(B) I2
(C) IO4
(D) IO
Sa
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1.
a
PART - I : JEE (ADVANCED) / IIT-JEE PROBLEMS (PREVIOUS YEARS)
2.
Consider a titration of potassium dichromate solution with acidified Mohr's salt solution using
diphenylamine as indicator. The number of moles of Mohr's salt required per mole of dichromate is :
[JEE 2007, 3/162]
(A) 3
(B) 4
(C) 5
(D) 6
3.
25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4N acetic acid. In
the titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The
molarity of the household bleach solution is :
[JEE 2012, 3/136]
(A) 0.48 M
(B) 0.96 M
(C) 0.24 M
(D) 0.024 M
4.*
For the reaction : I¯ + ClO3– + H2SO4 ⎯→ Cl¯ + HSO4– + I2
The correct statement(s) in the balanced equation is/are :
[JEE (Advanced) 2014, 3/120]
–
(A) Stoichiometric coefficient of HSO4 is 6.
(B) Iodide is oxidized.
(C) Sulphur is reduced.
5.
(D) H2O is one of the products.
To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to
KMnO4 using the reaction.
MnCl2 + K2S2O8 + H2O ࢮ KMnO4 + H2SO4 + HCl (equation not balanced).
499
Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid
(225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of
MnCl2 (in mg) present in the initial solution is ____.
(Atomic weights in g mol–1 : Mn = 55, Cl = 35.5)
[JEE (Advanced) 2018, 3/120]
PART - II : JEE (MAIN) / AIEEE PROBLEMS (PREVIOUS YEARS)
JEE(MAIN) OFFLINE PROBLEMS
When KMnO4 acts as an oxidising agent and ultimately forms Mn O24− , MnO2, Mn2O3 and Mn2+, then the
number of electrons transferred in each case is :
(1) 4, 3, 1, 5
(2) 1, 5, 3, 7
(3) 1, 3, 4, 5
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1.
[AIEEE 2002, 3/225]
(4) 3, 5, 7, 1
What will happen if the solution of potassium chromate reacts with excess amount of nitric acid
(1) Cr reduces in the oxidation state +3 from CrO42–.
[AIEEE 2003, 3/225]
(2) Cr oxidises in the oxidation state +7 from CrO42–.
(3) Cr+3 and Cr2O72– will be formed.
(4) Cr2O72– and H2O will be formed.
3.
The oxidation state of chromium in the final product formed by the reaction between KI and acidified
potassium dichromate solution is :
[AIEEE 2005, 3/225]
(1) + 4
(2) + 6
(3) + 2
(4) + 3
4.
Amount of oxalic acid present in a solution can be determined by its titration with KMnO4 solution in the
presence of H2SO4. The titration gives unsatisfactory result when carried out in the presence of HCl,
because HCl :
[AIEEE 2008, 3/105]
(1) furnishes H+ ions in addition to those from oxalic acid.
(2) reduces permanganate to Mn2+.
(3) oxidises oxalic acid to carbon dioxide and water.
(4) gets oxidised by oxalic acid to chlorine.
5.
29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method and
the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15
mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is :
[AIEEE 2010, 4/144]
(1) 59.0
(2) 47.4
(3) 23.7
(4) 29.5
Sa
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Ja
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2.
6.
The values of x, y and z in the reaction are, respectively :
(1) 5, 2 and 16
(2) 2, 5 and 8
(3) 2, 5 and 16
7.
z
H2O
2
[JEE(Main) 2013, 4/120]
(4) 5, 2 and 8
Consider the following reaction : xMnO4– + yC2O42– + zH+ ⎯→ xMn2+ + 2yCO2 +
For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the
M
evolved ammonia was absorbed in 60 mL of
sulphuric acid. The unreacted acid required 20 mL of
10
M
sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is :
10
[JEE(Main) 2014, 4/120]
(1) 6%
(2) 10%
(3) 3%
(4) 5%
500
JEE(MAIN) ONLINE PROBLEMS
Hydorgen peroxide acts both as an oxidising and as a reducing agent depending upon the nature of the
reacting species. In which of the following cases H2O2 acts as a reducing agent in acid medium ?
[JEE(Main) 2014 Online (12-04-14), 4/120]
(1) MnO4–
(2) Cr2O72–
(3) SO32–
(4) KI
2.
Permanent hardness in water cannot be cured by :
[JEE(Main) 2015 Online (10-04-15), 4/120]
(1) Treatment with washing soda
(2) Boiling
(3) Ion exchange method
(4) Calgon's method
3.
1.4 g of an organic compound was digested according to Kjeldahl's method and the ammonia evolved
was absorbed in 60 mL of M/10 H2SO4 solution. The excess sulphuric acid required 20 mL of M/10
NaOH solution for neutralization. The percentage of nitrogen in the compound is :
[JEE(Main) 2015 Online (10-04-15), 4/120]
(1) 24
(2) 5
(3) 10
(4) 3
4.
The volume of 0.1 N dibasic acid sufficient to neutralize 1 g of a base that furnishes 0.04 mole of OH – in
aqueous solution is :
[JEE(Main) 2016 Online (10-04-16), 4/120]
(1) 400 mL
(2) 600 mL
(3) 200 mL
(4) 80 mL
5.
For standardizing NaOH solution, which of the following is used as a primary standard ?
[JEE(Main) 2018 Online (16-04-18), 4/120]
(1) Sodium tetraborate
(2) Ferrous Ammonium Sulfate
(3) Oxalic acid
(4) dil. HCl
6.
The temporary hardness of water is due to :
(1) CaCl2
(2) Ca(HCO3)2
Ja
uh
a
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1.
(3) NaCl
[JEE(Main) 2019 Online (09-01-19), 4/120]
(4) Na2SO4
In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in
producing one molecule of CO2 is :
[JEE(Main) 2019 Online (10-01-19), 4/120]
(1) 5
(2) 1
(3) 2
(4) 10
8.
25 ml of the given HCl solution requires 30 mL of 0.1 M sodium carbonate solution. What is the volume
of this HCl solution required to titrate 30 mL of 0.2 M aqqueous NaOH solution?
[JEE(Main) 2019 Online (11-01-19), 4/120]
(1) 12.5 mL
(2) 75 mL
(3) 50 mL
(4) 25 mL
Sa
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7.
9.
The hardness of water sample (in terms of equivalents of CaCO 3) containing 10–3 M CaSO4 is :
(molar mass of CaSO4 = 136 g mol–1)
[JEE(Main) 2019 Online (12-01-19), 4/120]
(1) 10 ppm
(2) 50 ppm
(3) 90 ppm
(4) 100 ppm
10.
The volume strength of 1M H2O2 is: (Molar mass of H2O2 = 34 g mol–1)
[JEE(Main) 2019 Online (12-01-19), 4/120]
(1) 11.35
(2) 22.4
(3) 16. 8
(4) 5.6
501
EXERCISE - 1
PART - I
A-1.
(a) 23 ; (b) E = 9 ; (c) E = 30 ; (d) E =
A-2.
(a) E = 58.5 or E = 58.5 ;
A-3.
20.1
C-2.
35.5
; (e) E = 30 ; (f) E = 48 ; (g) E = 31.67
1
(b) E = 87 or E = 87 ; (c) E = 51.67 or E = 51.67
M1 M2 M3
,
,
6
8
5
B-2.
(a) 31.6 ; (b) 52.67
0.06 N, 0.03 M C-3.
537.6 mL
D-1.
20
D-3.
0.12 M.
D-4.
2.22 mL.
E-1.
7.1%
F-1.
2.12 g/L
F-2.
126 ppm
F-3.
Temporary hardness - due to bicarbonates of Ca & Mg
Permanent hardness - due to chlorides & sulphates of Ca & Mg.
F-4.
There are some method by which we can soften the water sample.
(a)
By boiling :
2HCO3– ⎯→ H2O + CO2 + CO32–
or
By Slaked lime :
Ca(HCO3)2 + Ca(OH)2 ⎯→ CaCO3 + 2H2O
Ca2+ + CO32– ⎯→ CaCO3
(b)
By Washing Soda :
CaCl2 + Na2CO3 ⎯→ CaCO3 + 2NaCl
(c)
By ion exchange resins :
Na2R + Ca2+ ⎯→ CaR + 2Na+
(d)
By adding chelating agents like (PO3–)3 etc.
72.61
a
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C-1.
D-2.
+3
E-2.
0.25075 g
Ja
uh
B-1.
(B)
A-2.
(A)
B-3.
(A)
C-1.
(A)
D-2.
(D)
D-3.
(B)
F-1.
(D)
F-2.
a
A-1.
lp
PART - II
Sa
nk
(A)
1.
A-3.
(A)
B-1.
(C)
B-2.
(D)
C-2.
(A)
C-3.
(D)
D-1.
(B)
D-4.
(A)
E-1.
(B)
E-2.
(B)
F-3.
(D)
F-4.
(B)
PART - III
(A – p,s) ; (B – q,r) ; (C – p,q,s) ; (D – r)
EXERCISE - 2
PART - I
1.
(A)
2.
(A)
3.
(D)
4.
(D)
5.
(A)
6.
(D)
7.
(D)
8.
(A)
9.
(D)
10.
(A)
11.
(A)
12.
(A)
13.
(A)
14.
(A)
PART - II
1.
4
2.
1
3.
90
4.
2
5.
2
6.
68
7.
12
8.
21
9.
10 mL
10.
16
11.
32
4.
(ABD)
5.
PART - III
1.
(AC)
2.
(ABCD)
3.
(ACD)
(ABCD)
502
6.
(AB)
PART - IV
1.
(C)
2.
(B)
3.
(C)
4.
(C)
6.
(B)
7.
(B)
8.
(C)
9.
(D)
4.*
(ABD)
5.
(D)
5.
126 mg
EXERCISE - 3
PART - I
1.
(A)
2.
(D)
3.
(C)
JEE(MAIN) OFFLINE PROBLEMS
1.
6.
(3)
(3)
2.
7.
(4)
(2)
3.
(4)
4.
(4)
2.
(2)
3.
(3)
6.
(2)
7.
(2)
8.
(4)
5.
(3)
4.
(3)
5.
(3)
9.
(4)
10.
(1)
Ja
1.
(2)
uh
JEE(MAIN) ONLINE PROBLEMS
a
ri
PART - II
lp
This Section is not meant for classroom discussion. It is being given to promote self-study
and self testing amongst the students.
PART - I : PRACTICE TEST-1 (IIT-JEE (MAIN Pattern))
Marks : 120
a
Max. Time : 1 Hr. Max.
Important Instructions
The test is of 1 hour duration.
The Test Booklet consists of 30 questions. The maximum marks are 120.
Each question is allotted 4 (four) marks for correct response.
Candidates will be awarded marks as stated above in Instructions No. 3 for correct response of each
question. ¼ (one fourth) marks will be deducted for indicating incorrect response of each question. No
deduction from the total score will be made if no response is indicated for an item in the answer sheet.
There is only one correct response for each question. Filling up more than one response in any
question will be treated as wrong response and marks for wrong response will be deducted accordingly
as per instructions 4 above.
Sa
nk
1.
2.
3.
4.
5.
1.
In the reaction : Na2S2O3 + 4Cl2 + 5H2O ⎯→ Na2SO4 + H2SO4 + 8HCl,
the equivalent weight of Na2 S2 O3 will be : (M = molecular weight of Na2S2O3)
(1) M/4
(2) M/8
(3) M/1
(4) M/2
2.
In the reaction, 2CuSO4 + 4K ⎯→ 2Cu22 + 2 + 2K2SO4
the equivalent weight of CuSO4 will be :
(1) 79.75
(2) 159.5
(3) 329
3.
(4) None of these
100 milli moles of dichloroacetic acid (CHCl2 COOH) can neutralize how many moles of ammonia to
form ammonium dichloroacetate :
(1) 0.0167
(2) 0.1
(3) 0.3
(4) 0.6
503
The number of moles of ferrous oxalate oxidised by one mole of KMnO 4 in acidic medium is :
(1) 5/2
(2) 2/5
(3) 3/5
(4) 5/3
5.
How many moles of KMnO4 are needed to oxidise a mixture of 1 mole of each FeSO 4 & FeC2O4 in
acidic medium :
(1) 4/5
(2) 5/4
(3) 3/4
(4) 5/3
6.
22.7 mL of (N/10) Na2CO3 solution neutralises 10.2 mL of a dilute H2SO4 solution. The volume of water
that must be added to 400 mL of this H2SO4 solution in order to make it exactly N/10.
(1) 490.2 mL
(2) 890.2 mL
(3) 90.2 mL
(4) 290.2 mL
7.
HNO3 oxidises NH4+ ions to nitrogen and itself gets reduced to NO2. The moles of HNO3 required by 1
mole of (NH4)2SO4 is :
(1) 4
(2) 5
(3) 6
(4) 2
8.
The mass of oxalic acid crystals (H2C2O4.2H2O) required to prepare 50 mL of a 0.2 N solution is :
(1) 4.5 g
(2) 6.3 g
(3) 0.63 g
(4) 0.45 g
9.
When HNO3 is converted into NH3, the equivalent weight of HNO3 will be :
(1) M/2
(2) M/1
(3) M/6
(4) M/8
(M = molecular weight of HNO3)
10.
In the conversion NH2OH ⎯→ N2O, the equivalent weight of NH2OH will be :
(1) M/4
(2) M/2
(3) M/5
(4) M/1
(M = molecular weight of NH2OH)
11.
Number of moles of CaO required to remove hardness from 1000 litre water having 324 ppm of calcium
bicarbonate and 7