Chemical Arithmetic
1
Chapter
1
Chemical Arithmetic
Chemistry is basically an experimental science. In it we study
physical and chemical properties of substance and measure it upto
possibility. The results of measurement can we reported in two steps,
(a) Arithmetic number, (b) Unit of measurement.
Every experimental measurement vary slightly from one another and
involves some error or uncertainty depending upon the skill of person
making the measurements and measuring instrument. The closeness of the
set of values obtained from identical measurement called precision and a
related term, refers to the closeness of a single measurement to its true
value called accuracy.
Significant figures
In the measured value of a physical quantity, the digits about the
correctness of which we are surplus the last digit which is doubtful, are
called the significant figures. Number of significant figures in a physical
quantity depends upon the least count of the instrument used for its
measurement.
(1) Common rules for counting significant figures Following are
some of the common rules for counting significant figures in a given
expression
Rule 1. All non zero digits are significant.
Rule 5. All zeros on the right of the non zero digit are not
significant.
Example : x  1000 has only one significant figure. Again
x  378000 has three significant figures.
Rule 6. All zeros on the right of the last non zero digit become
significant, when they come from a measurement.
Example : Suppose distance between two stations is measured to be
3050 m. It has four significant figures. The same distance can be expressed
as 3.050 km or 3.050  10 5 cm . In all these expressions, number of
significant figures continues to be four. Thus we conclude that change in
the units of measurement of a quantity does not change the number of
significant figures. By changing the position of the decimal point, the
number of significant digits in the results does not change. Larger the
number of significant figures obtained in a measurement, greater is the
accuracy of the measurement. The reverse is also true.
(2) Rounding off : While rounding off measurements, we use the
following rules by convention
Rule 1. If the digit to be dropped is less than 5, then the preceding
digit is left unchanged.
Example : x  1234 has four significant figures. Again x  189
has only three significant figures.
Rule 2. All zeros occurring between two non zero digits are
significant.
Example : x  7.82 is rounded off to 7.8, again x  3.94 is
rounded off to 3.9.
Example : x  1007 has four significant figures. Again
x  1.0809 has five significant figures.
Rule 3. In a number less than one, all zeros to the right of decimal
point and to the left of a non zero digit are not significant.
Example : x = 6.87 is rounded off to 6.9, again x = 12.78 is rounded
off to 12.8.
Example : x  0.0084 has only two significant digits. Again,
x  1.0084 has five significant figures. This is on account of rule 2.
Rule 4. All zeros on the right of the last non zero digit in the
decimal part are significant.
Example : x = 16.351 is rounded off to 16.4, again x  6.758 is
rounded off to 6.8.
Rule 4. If digit to be dropped is 5 or 5 followed by zeros, then preceding
digit is left unchanged, if it is even.
Example : x = 3.250 becomes 3.2 on rounding off, again
x  12.650 becomes 12.6 on rounding off.
Example : x  0.00800 has three significant figures 8, 0, 0. The
zeros before 8 are not significant again 1.00 has three significant figures.
Rule 2. If the digit to be dropped is more than 5, then the preceding
digit is raised by one.
Rule 3. If the digit to be dropped is 5 followed by digits other than
zero, then the preceding digit is raised by one.
2 Chemical Arithmetic
Rule 5. If digit to be dropped is 5 or 5 followed by zeros, then the
preceding digit is raised by one, if it is odd.
Example : x = 3.750 is rounded off to 3.8, again x  16.150 is
rounded off to 16.2.
(3) Significant figure in calculation
(i) Addition and subtraction : In addition and subtraction the
following points should be remembered
(a) Every quantity should be changed into same unit.
(b) If a quantity is expressed in the power of 10, then all the
quantities should be changed into power of 10.
(c) The result obtained after addition or subtraction, the number of
figure should be equal to that of least, after decimal point.
(ii) Multiplication and division
(a) The number of significant figures will be same if any number is
multiplied by a constant.
(b) The product or division of two significant figures, will contain
the significant figures equal to that of least.
Units for measurement
(1) C.G.S. System : Length (centimetre), Mass (gram), Time (second)
(2) M.K.S. System
: Length (metre), Mass (kilogram), Time
(second)
(3) F.P.S. System
: Length (foot), Mass (pound),
Time
(second)
(4) S.I. System : The 11th general conference of weights and measures
(October 1960) adopted International system of units, popularly known as the SI
units. The SI has seven basic units from which all other units are derived called
derived units. The standard prefixes which helps to reduce the basic units are now
widely used.
Dimensional analysis : The seven basic quantities lead to a number of
derived quantities such as pressure, volume, force, density, speed etc. The
units for such quantities can be obtained by defining the derived quantity in
terms of the base quantities using the base units. For example, speed
(velocity) is expressed in distance/time. So the unit is m / s or ms 1 . The
unit of force (mass  acceleration) is kg ms 2 and the unit for
acceleration is ms 2 .
The chosen standard of measurement of a quantity which has
essentially the same nature as that of the quantity is called the unit of the
quantity. Following are the important types of system for unit,
Table 1.1 Seven basic S.I. units
Length
Mass
Time
Temperature
Electric Current
Luminous Intensity
Amount of
substance
metre (m)
Kilogram (kg)
Second (s)
Kelvin (K)
Ampere (A)
Candela (Cd)
Mole (mol)
Table 1.2 Derived Units
Physical quantity
Unit
Symbol
Area
square metre
m
Volume
cubic metre
m
Velocity
metre per second
ms
Acceleration
metre per second square
ms
Density
kilogram per cubic metre
kg m
Molar mass
kilogram per mole
kg mol
Molar volume
cubic metre per mole
m mol
Molar concentration
mole per cubic metre
mol m
Force
newton (N)
kg m s
Pressure
pascal (Pa)
Nm
Energy work
joule (J)
kg m s , Nm
2
3
–1
–2
–3
–1
3
–1
–3
–2
–2
2
–2
Table 1.3 Standard prefixes use to reduce the basic units
Multiple
Prefix
Symbol
Submultiple
10
yotta
Y
10
zetta
Z
10
exa
E
10
peta
P
10
tera
T
24
21
18
15
12
Prefix
Symbol
10
deci
d
10
centi
c
10
milli
m
10
micro

10
nano
n
–1
–2
–3
–6
–9
Chemical Arithmetic
3
10
giga
G
10
pico
p
10
mega
M
10
femto
f
10
kilo
k
10
atto
a
10
hecto
h
10
zeto
z
10
deca
da
10
yocto
y
9
6
3
2
1
–12
–15
–18
–21
–24
Table 1.4 Conversion factors
1 e.s.u. = 3.3356  10 C
1 mole of a gas = 22.4 L at STP
1 dyne = 10 N
1 mole a substance = N molecules
1 atm = 101325 Pa
1 g atom = N atoms
1 m = 39.37 inch
1 cal = 4.184 J
1 inch = 2.54 cm
1 eV = 1.602  10 J
1 litre = 1000 mL
1 eV/atom =96.5 kJ mol
1 gallon (US) = 3.79 L
1 amu = 931.5016 MeV
1 bar = 1  10 N m
t ( F) = 9/5 t ( C) + 32
1 lb = 453.59237 g
1 kilo watt hour = 3600 kJ
1 litre atm = 101.3 J
1 g cm = 1000 kg m
1 horse power = 746 watt
1 year = 3.1536  10 s
1Å = 10 m
1 joule = 10 erg
1 debye (D) = 1  10 esu cm
1nm = 10 m
–10
–5
–19
1 newton =1 kg m s
–2
1 J = 1 Nm =1 kg m s
2
–2
7
–1
0
5
0
o
–2
–3
7
Laws of chemical combination
Various chemical reactions take place according to the certain laws,
known as the Laws of chemical combination.
(1) Law of conservation of mass : It was proposed by Lavoisier and
verified by Landolt. According to this law, Matter is neither created nor
destroyed in the course of chemical reaction though it may change from one
form to other. The total mass of materials after a chemical reaction is same
as the total mass before reaction.
(2) Law of constant or definite proportion : It was proposed by
Proust. According to this law, A pure chemical compound always contains
the same elements combined together in the fixed ratio of their weights
whatever its methods of preparation may be.
(3) Law of multiple proportion : It was proposed by Dalton and
verified by Berzelius. According to this law, When two elements A and B
combine to form more than one chemical compounds then different weights
of A, which combine with a fixed weight of B, are in proportion of simple
whole numbers.
(4) Law of equivalent proportion or law of reciprocal proportion : It
was proposed by Ritcher. According to this law, The weights of the two or
more elements which separately react with same weight of a third element
are also the weights of these elements which react with each other or in
simple multiple of them.
(5) Gay-Lussac’s law : It was proposed by Gay–Lussac and is
applicable only for gases. According to this law, When gases
combine, they do so in volumes, which bear a simple ratio to each
other and also to the product formed provided all gases are
measured under similar conditions. The Gay-Lussac’s law, was based
on experimental observation.
Important hypothesis
(1) Atomic hypothesis : Keeping in view various law of chemical
combinations, 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 is as followes,
(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 element.
o
–18
–3
–10
–9
(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 take part to form molecules i.e., compounds
are formed when atoms of more than one element combine.
(vi) In a given compound, the relative number and kinds of atoms
are constant.
(2) 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 Ca 40 and Ar 40 .
(iv) Atom is no longer indestructible. In many nuclear reactions, a
certain mass of the nucleus is converted into energy in the form of , 
and  rays.
(v) Atoms may not always combine in simple whole number ratios.
e.g., in sucrose (C12 H 22 O11 ) , 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.
(3) Berzelius hypothesis : “Equal volumes of all gases contain equal
number of atoms under same conditions of temperature and pressure”.
When applied to law of combining volumes, this hypothesis predicts that
atoms are divisible and hence it is contrary to Dalton's hypothesis.
(4) Avogadro’s hypothesis : “Equal volumes of all gases under similar
conditions of temperature and pressure contain equal number of
molecules.” Avogadro hypothesis has been found to explain as follows,
(i) Provides a method to determine the atomic weight of gaseous
elements.
(ii) Provides a relationship between vapour density (V.D.) and
molecular masses of substances.
Molecular mass  2  vapour density
4 Chemical Arithmetic
(iii) It helps in the determination of mass of fixed volume of a
particular gas.
Mass of 1 ml gas = V.D.  0.0000897 gm.
 V.D.  0.0000897 gm. gas has volume = 1 ml
2

V.D.(i.e.,
molecular
1  2  V .D.
volume 
ml  22400 ml
V .D.  0.0000897
Valency of the element =
gm.
mass)
=
S.T.P.
(v) It helps in determination of molecular formulae of gases and is
very useful in gas analysis. By knowing the molecular volumes of reactants
and products of reaction, molecular composition can be determined easily.
Atomic, Molecular and Equivalent masses
(1) Atomic mass : It is the average relative mass of atom of element
as compared with an atom of carbon –12 isotope taken as 12.
Average mass of an atom
1/12  Mass of an atom of C 12
Average atomic mass : If an elements exists in two isotopes having
atomic masses ‘a’ and ‘b’ in the ratio m : n, then average atomic mass
(m  a)  (n  b)
. Since the atomic mass is a ratio, it has no units and is
=
m n
expressed in amu, 1 amu = 1.66  10
equal to
24
Gram atomic mass (GAM) : Atomic mass of an element expressed in
grams is called Gram atomic mass or gram atom or mole atom.
Mass of an element
GAM
(ii) Mass of an element in gm. = No. of gm. atom
(iii) Number of atoms in 1 GAM = 6.02
 GAM
 10 23
 10 23
Atomic mass = Equivalent mass of metal
Mass
=
GAM
 6.02  10 23
6 .02  10
Atomic mass
23
(iv) Number of atoms in 1gm of element =
 Valency
(iii) Specific heat method : It is suitable only for gases. The two
types of specific heats of gases are C (at constant pressure) and C v (at
P
constant volume). Their ratio is known as  whose value is constant (1.66
for monoatomic, 1.40 for diatomic and 1.33 for triatomic gases).
Atomic mass of a gaseous element 
Molecular mass
Atomicity
(iv) Isomorphism method : It is based on law of isomorphism which
states that compounds having identical crystal structure have similar
constitution and chemical formulae.
Example : K 2 SO 4 , K 2 CrO 4 and K 2 SeO 4
(valency of S, Cr, Se = 6),
ZnSO 4 . 7 H 2 O, MgSO4 .7 H 2 O, FeSO 4 . 7 H 2 O
(valency of Zn, Mg, Fe = 2).
(2) Molecular mass : Molecular mass of a molecule, of an element or
a compound may be defined as a number which indicates how many times
1
heavier is a molecule of that element or compound as compared with
of
12
the mass of an atom of carbon–12. Molecular mass is a ratio and hence has
no units. It is expressed in a.m.u.
Molecular mass 
Mass of one molecule of the substance
1 / 12  Mass of one atom of C - 12
Actual mass of one molecule = Mol. mass 1.66 10 24 gm.
 Number of atoms in a given substance
= No. of GAM  6.02
2  Vapour density of chloride
Equivalent mass of metal  35.5
g . One atomic mass unit (amu) is
1
th of the mass of an atom of carbon-12 isotope.
12
(i) Number of gram atoms =
Molecular mass of chloride
Equivalent mass of chloride
has
 Molar mass of a gas or its 1 mole occupies 22.4 L volume at
Atomic mass 
Approximat e atomic mass
Equivalent mass
(ii) Vapour density method : It is suitable for elements whose
chlorides are volatile.
(iv) It also helps in the determination of molar volume at N.T.P.

Valency 
GAM
(v) Mass of one atom of the element (in gm.) 
6.02  10 23
Methods of determination of atomic mass
(i) Dulong and Pettit's method : According to Dulong and Pettit's
Molecular mass of a substances is the additive property and can be
calculated by adding the atomic masses present in one molecule.
Gram molecular mass (GMM) and Gram molar volume : Molecular
mass of an element or compound when expressed in gm. is called its gram
molecular mass, gram molecule or mole molecule.
Number of gm molecules =
Mass of substances
GMM
Mass of substances in gm = No. of gm. molecules  GMM
Volume occupied by one mole of any gas at STP is called Gram
molar volume. The value of gram molar volume is 22.4 litres. Volume of 1
mole of any gas at STP = 22.4 litres
law
Atomic mass  Specific heat = 6.4 (approx.)
Atomic mass (approx.) =
6.4
Specific heat (in cals.)
This law is applicable to solid elements only except Be, B, C and Si
because their specific heat is variable with temperature.
Atomic mass = Equivalent mass  Valency
Expression for mass and density
Mass of 11 .2 L of any gas at STP = V.D. of that gas in gm.
Density of a gas at NTP =
Mol.mass in gm.
22400 ml
Important generalisations
Number of atoms in a substance
Chemical Arithmetic
= Number of GMM
 6.02  10  Atomicity
23
(iv) EM of a salt =
Number of electrons in given substance
= Number of GMM  6.02  10  Number of electrons
23
(i) Diffusion method (For gases) : The ratio of rates of diffusion of
two gases is inversely proportional to the square root of their molecular
masses.
r1

r2

Formula mass
Number of electrons gained per molecule or Total change in O.N.
Equivalent mass of common oxidising agent changes with the
medium of the reaction.
Methods of determination of equivalent mass
M2
M1
(ii) Vapour density method (For gases only) : Mass of a fixed volume
of the vapour is compared with the mass of the same volume of hydrogen
under same conditions. The ratio of these masses is called Vapour density
or Relative density.
(i) Hydrogen displacement method : The mass of metal which
displaces 11200 ml of hydrogen at NTP from an acid, alkali or alcohol is the
equivalent mass of the metal.
(a) Equivalent mass of metal

Molecular mass  2  Vapour desity
(iii) Victor Meyer method (For volatile liquids or solids)
It is based on Dalton's law of partial pressure and Avogadro's hypothesis
(gram molar volume).
22400 ml of vapours of a substance = Molecular mass of that
substance
(iv) Colligative property method (For non-volatile solids)
Discussed in colligative properties of solutions.
Average atomic mass and molecular mass
A (Average atomic mass) =
 Ai X i
 X total
M (Average molecular mass) =
Where
Formula mass
Total positive or negative charge
(v) EM of an oxidising agent
Methods of determination of molecular mass
Following methods are used to determine molecular mass,
 Mi Xi
 X total
W
Mass of metal
 1 .008 g
 1 .008 
M
Mass of H 2 displaced
(b) Equivalent mass of metal

Mass of metal
W
11200
11200 =
V
Vol.(ml) of H 2 displaced at STP
This method is useful for metals which can displace hydrogen from
acids or can combine with hydrogen (Mg, Zn, Na, Ca etc.)
(ii) Oxide formation method : The mass of the element which
combines with 8 grams of oxygen is the equivalent mass of the element.
Mass of metal
8
Mass of oxygen
(a) Equivalent mass of metal =
(b) Equivalent mass of metal
=
A1 , A2 , A3 ..... are atomic mass of species 1, 2, 3,…. etc.
with % ratio as X1 , X 2 , X 3 ..... etc. Similar terms are for molecular masses.
(3) Equivalent mass : The number of parts by mass of a substance
that combines with or displaces 1.008 parts by mass of hydrogen or 8.0
parts of oxygen or 35.5 parts of chlorine is called its equivalent mass (EM).
On the other hand quantity of a substance in grams numerically equal to its
equivalent mass is called its gram equivalent mass (GEM) or gram
equivalent.
Expressions for equivalent mass (EM)
(ii) EM of an acid 
Atomic mass
Valency
Molecular mass
Basicity
(Basicity of acid is the number of replaceable hydrogen atoms in one
molecule of the acid).
Molecular mass
(iii) EM of a base 
Acidity
(Acidity of a base is the number of replaceable– OH groups in one
molecule of the base).
Mass of metal
 5600
Vol. of O 2 at S.T.P. in ml
(iii) Chloride formation method : The mass of an element which
reacts with 35.5 gm. of chlorine is the equivalent mass of that element.
(a) Equivalent mass of metal 
Mass of metal
 35.5
Mass of chlorine
(b) Equivalent mass of metal

Mass of the substance in grams
Number of GEM 
GEM of the substance
(i) EM of an element 
5
Mass of metal
 11200
Vol. of Cl 2 (in ml.) at STP
(iv) Neutralisation method : (For acids and bases).
Equivalent mass of acid (or base) 
W
VN
Where ,
W = Mass of acid or base in gm.,
V = Vol. of base or acid in litre required for neutralisation
N is Normality of base or acid
(v) Metal displacement method : It is based on the fact that one gm.
equivalent of a more electropositive metal displaces one gm equivalent of a
less electropositive metal from its salt solution.
Mass of metal added
Eq. mass of metal added

Mass of metal displaced Eq. mass of metal displaced
6 Chemical Arithmetic
Element
W1
E
 1
W2
E2
% Relative no. of atoms =
%/at. wt.
Simplest
Ratio
Empirical
Formula
(vi) Electrolytic method : The quantity of substance that reacts at
electrode when 1 faraday of electricity is passed is equal to its gram
equivalent mass.
Gram equivalent mass = Electrochemical equivalent  96500
The ratio of masses of two metals deposited by the same
quantity of electricity will be in the ratio of their equivalent masses.
Relative no. of atoms : Divide the percentage of each element
present in compound by its at. weight. This gives the relative no. of atoms
of element in molecule.
W1
E
 1
W2
E2
It the simplest ratio obtained are not complete integers, multiply
them by a common factor to get integer values of simplest ratio.
(vii) Double decomposition method
AB  CD  AD   CB
Empirical formula : Write all constituent atoms with their respective
no. of atoms derived in simplest ratio. This gives empirical formula of
compound.
Mass of compound AB Eq. mass of A  Eq. mass of B

Mass of compound AD Eq. mass of A  Eq. mass of D
Molecular formula : Molecular formula  n  empirical formula
where ' n' is the whole no. obtained by
or
Simplest ratio : Find out lowest value of relative no. of atoms and
divide each value of relative no. of atoms by this value to estimate simplest
ratio of elements.
Mass of salt taken (W1 )
Eq. mass of salt (E1 )

Mass of ppt. obtained (W2 ) Eq. mass of salt in ppt. (E 2 )
(viii) Conversion method : When one compound of a metal is
converted to another compound of the same metal, then
Mass of compound I (W1 ) E  Eq. mass of radical I

Mass of compound II (W2 ) E  Eq. mass of radical II
(E = Eq. mass of the metal)
n
Chemical stoichiometry
Stoichiometry (pronounced “stoy-key om-e-tree”) is the calculation
of the quantities of reactants and products involved in a chemical reaction.
That means quantitative calculations of chemical composition and reaction
are referred to as stoichiometry.
(ix) Volatile chloride method
Valency of metal 
E 
108  Mass of silversalt
 107
Mass of Ag metal
Molecular mass of acid = Equivalent mass of acid
 Basicity
The mole concept
mole
of
any
substance
(a) Simple calculations (gravimetric analysis) and
(b) More complex calculations involving concentration and volume
of solutions (volumetric analysis).
2  V.D. of Chloride
 35.5
Valency
(x) Silver salt method (For organic acids)
One
Basically, this topic involves two types of calculations.
2  V.D. of Chloride
2  V.D.

Eq. mass of metal chloride E  35.5
Equivalent Mass of acid =
molecular weight of compound
empirical formula weight of compound
contains
a
fixed
number
(6.022  10 ) 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
23
correct to refer to a mole of helium, a mole of electrons, or a mole of Na  ,
meaning respectively Avogadro’s number of atoms, electrons or ions.
 Number of moles 
Weight(grams)
Weight of one mole (g/mole)

Weight
Atomic or molecular weight
Percentage composition & Molecular formula
(1) Percentage composition of a compound
Percentage composition of the compound is the relative mass of each of the
constituent element in 100 parts of it. If the molecular mass of a compound
is M and B is the mass of an element in the molecule, then
Mass of element
X
Percentage of element 
 100 
 100
Molecular mass
M
(2) Determination of empirical formula : The empirical formula of a
molecule is determined using the % of elements present in it. Following
method is adopted.
There is no borderline, which can distinguish the set of laws
applicable to gravimetric and volumetric analysis. All the laws used in one
are equally applicable to the other i.e., mole as well as equivalent concept.
But in actual practise, the problems on gravimetric involves simpler
reactions, thus mole concept is convenient to apply while volumetric
reactions being complex and unknown (unknown simple means that it is
not known to you, as it’s not possible for you to remember all possible
reactions), equivalent concept is easier to apply as it does not require the
knowledge of balanced equation.
(1) Gravimetric analysis : In gravimetric analysis we relate the
weights of two substances or a weight of a substance with a volume
of a gas or volumes of two or more gases.
Problems Involving Mass-Mass Relationship
Proceed for solving such problems according to the following
instructions,
(i) Write down the balanced equation to represent the chemical
change.
(ii) Write the number of moles below the formula of the reactants
and products. Also write the relative weights of the reactants and products
(calculated from the respective molecular formula), below the respective
formula.
(iii) Apply the unitary method to calculate the unknown factor (s).
Problems Involving Mass-Volume Relationship
For solving problems involving mass-volume relationship, proceed
according to the following instructions,
(i) Write down the relevant balanced chemical equations (s).
Chemical Arithmetic
(ii) Write the weights of various solid reactants and products.
(iii) Gases are usually expressed in terms of volumes. In case the
volume of the gas is measured at room temperature and pressure (or under
conditions other than N.T.P.), convert it into N.T.P. by applying gas
equation.
(iii) Iodiometric titrations : This is a simple titration involving free
iodine. This involves the titration of iodine solution with known sodium
thiosulphate solution whose normality is N . Let the volume of sodium
thiosulphate is V ml .
I 2  2 Na 2 S 2 O 3  2 NaI  Na 2 S 4 O6
(iv) Volume of a gas at any temperature and pressure can be
converted into its weight and vice-versa with the help of the relation, by
n  2, n 1
g
PV 
 RT where g is weight of gas, M is mole. wt. of gas, R is
M
gas constant.
Equivalents of I 2  Equivalent of Na 2 S 2 O 3
 Equivalents of I 2  N  V  10 3
Calculate the unknown factor by unitary method.
Moles of I 2 
Problems Based on Volume-Volume Relationship
Such problems can be solved according to chemical equation as,
(iii) In case volume of the gas is measured under particular (or
room) temperature, convert it to volume at NTP by using ideal gas
equation.
(iv) Iodometric titrations : This is an indirect method of estimation
of iodine. An oxidising agent is made to react with excess of solid KI . The
oxidising agent oxidises I  to I 2 . This iodine is then made to react with
Na 2 S 2 O 3 solution.
Oxidising Agent
Take the help of Avogadro’s hypothesis “Equal volume of different
2 Na 2 S 2 O3
( A)  KI  I 2 
 2 NaI  Na 2 S 4 O 6
gases under the similar conditions of temperature and pressure contain the
same number of molecules”.
(2) Volumetric analysis : It is a method which involves quantitative
determination of the amount of any substance present in a solution through
volume measurements. For the analysis a standard solution is required. (A
solution which contains a known weight of the solute present in known
volume of the solution is known as standard solution.)
Let the normality of Na 2 S 2 O 3 solution is N and the volume of
thiosulphate consumed to V ml .
Equivalent of A  Equivalent of I 2  Equivalents of Na 2 S 2 O 3
Equivalents of I 2 liberated from KI  N  V  10 3
To determine the strength of unknown solution with the help of
known (standard) solution is known as titration. Different types of titrations
are possible which are summerised as follows,
Moles of I 2 liberated from KI 
N  V  10 3
2
 N  V  10 3

 254  g .
Mass of I 2 liberated from KI  
2


(i) Redox titrations : To determine the strength of oxidising agents
or reducing agents by titration with the help of standard solution of
reducing agents or oxidising agents.
Examples:
N  V  10 3
2
 N  V  10 3

Mass of free I 2 in the solution  
 254  g .
2


(i) Write down the relevant balanced chemical equation.
(ii) Write down the volume of reactants and products below the
formula to each reactant and product with the help of the fact that 1 gm
molecule of every gaseous substance occupies 22.4 litres at N.T.P.
7
(v) Precipitation titrations : To determine the anions like

CN , AsO33  , PO43  , X  etc, by precipitating with AgNO3 provides
K 2 Cr2 O7  4 H 2 SO 4  K 2 SO 4  Cr2 (SO 4 )3  4 H 2 O  3[O]
[2 FeSO 4  H 2 SO 4  O  Fe2 (SO 4 )3  H 2 O]  3
6 FeSO 4  K 2 Cr2 O7  7 H 2 SO 4  3 Fe(SO 4 )3  K 2 SO 4  Cr2 (SO 4 )3 7 H 2 O
2 KMnO4  3 H 2 SO 4  K 2 SO 4  2 MnSO 4  3 H 2 O  5[O]
[2 FeSO 4  H 2 SO 4  O  Fe2 (SO 4 )3  H 2 O]  5
10 FeSO 4  2 KMnO4  8 H 2 SO 4  5 Fe2 (SO 4 )3  K 2 SO 4  2 MnSO 4  8 H 2 O
Similarly with H 2 C 2 O4
2 KMnO4  3 H 2 SO 4  5 H 2 C2 O4 
K 2 SO 4  2 MnSO 4  8 H 2 O  10CO 2 etc.
(ii) Acid-base titrations : To determine the strength of acid or base
with the help of standard solution of base or acid.
Example: NaOH  HCl  NaCl  H 2 O
and NaOH  CH 3 COOH  CH 3 COONa  H 2 O etc.
examples of precipitation titrations.
NaCl  AgNO3  AgCl   NaNO 3
KSCN  AgNO3  AgSCN   KNO 3
End point and equivalence point : The point at which
titration is stopped is known as end point, while the point at which
the acid and base ( or oxidising and reducing agents ) have been
added in equivalent quantities is known as equivalence point . Since
the purpose of the indicator is to stop the titration close to the
point at which the reacting substances were added in equivalent
quantities, it is important that the equivalent point and the end
point be as close as possible.
Normal solution : A solution containing one gram equivalent
weight of the solute dissolved per litre is called a normal solution ;
e.g. when 40 g of NaOH are present in one litre of NaOH solution,
the solution is known as normal ( N ) solution of NaOH . Similarly, a
solution containing a fraction of gram equivalent weight of the
8 Chemical Arithmetic
solute dissolved per litre is known as subnormal solution. For
example, a solution of NaOH containing 20 g (1/2 of g eq. wt.) of
NaOH dissolved per litre is a sub-normal solution. It is written as
N /2 or 0.5 N solution.
Formula used in solving numerical problems on volumetric analysis
(1) Strength of solution = Amount of substance in g litre1
(2) Strength of solution = Amount of substance in g moles litre1
(3) Strength of solution = Normality  Eq. wt. of the solute
= molarity  Mol. wt. of solute
Limiting reagent or reactant
In many situations, an excess of one or more substance is available
for chemical reaction. Some of these excess substances will therefore be left
over when the reaction is complete; the reaction stops immediately as soon
as one of the reactant is totally consumed.
The substance that is totally consumed in a reaction is called
limiting reagent because it determines or limits the amount of product. The
other reactant present in excess are called as excess reagents.
Let us consider a chemical reaction which is initiated by passing a
spark through a reaction vessel containing 10 mole of H and 7 mole of O .
2
(4) Molarity
Moles of solute
Volume in litre
(5) Number of moles 
Wt. in gm
 M  V(inl)
Mol.wt.
Volume in litres

at NTP (only for gases)
22.4
2 H 2 (g)  O2 (g)  2 H 2O (v)
Moles before reaction
10
7
0
Moles after reaction
0
2
10
The reaction stops only after consumption of 5 moles of O as no
further amount of H is left to react with unreacted O . Thus H is a limiting
reagent in this reaction.
2
2
(6) Number of millimoles 
Wt. in gm  1000
mol. wt.
2
2
2
 Molarity Volume in ml.
(7) Number of equivalents

Wt. in gm
 x  No. of moles  Normality  Volume in litre
Eq. wt.
(8) Number of milliequivalents (meq.)

Wt. in gm  1000
 normality  Volume in ml.
Eq. wt.
(9) Normality  x  No. of millimoles
 x  Molarity 
where x 
Strength in gm litre1
Eq. wt.
Mol.wt.
, x = valency or change in oxi. Number.
Eq. wt.
(10) Normality formula, N1 V1  N 2 V2
 Law of conservation of mass does not hold good for nuclear
reactions.
 Law of definite proportions do not hold good for nonstoichiometric compounds e.g., Wusitite Fe O.
0.95
 Law of definite proportions, law of multiple proportions and law of
reciprocal proportions do not hold good when same compounds is
obtained by using different isotopes of te same element e.g. H O
and D O.
2
2
 The term atom was introduced by Ostwald and the term molecule
was introduced by Avogadro.
 The concept of element was introduced by Robert Boyle.
 The number of atoms present in a molecule of a gaseous element is
called Atomicity.
Wt. of solvent
 100
(11) % by weight 
Wt. of solution
 Both atomic mass and molecular mass are just rations and hence
Wt. of solvent
 100
(12) % by volume 
Vol. of solution
 1 mol of H O  22400 cc of H O (because it is a liquid). Instead,
have no units.
2
2
1 mol of H O = 18cc of H O
2
Vol. of solvent
 100
(13) % by strength 
Vol. of solution
2
 1 M H SO = 2NH SO .
4
2
4
 Minimum molecular mass of a macromolecular substance can be

(15) Formality 
(because density of H O = 1 g/cc)
2
(14) Specific gravity
2
Wt. of solution
 Wt. of 1 ml. of solution
Vol. of solution
Wt. of ionic solute
Formula Wt. of solute Vin l
(16) Mol. Wt. = V.D  2 (For gases only)
calculated by analysing it for one of the minar components.
Minimum molecular mass is obtained when it is supposed that one
molecule of the macromolecule contains only one atom or molecule
of the minor component.
Minimum molecular mass
Chemical Arithmetic

Atomic/molecular mass of minor component
Mass of minor component per gram of macromolec ule
9
10 Chemical Arithmetic
13.
In known elements, the maximum number is of
[CPMT 1985]
Significant figures, Units for measurement,
Matter and Separation of mixture
1.
2.
3.
4.
One fermi is
14.
(a) Metals
(b) Non-metals
(c) Metalloids
(d) None of these
Which one of the following is not an element
(a) Diamond
(b) Graphite
(c) Silica
(d) Ozone
15.
A mixture of ZnCl 2 and PbCl 2 can be separated by
[AFMC 1989]
[Haryana CEET 1994; DPMT 2004]
(a) 10 13 cm
(b) 10 15 cm
(c) 10 10 cm
A picometre is written as
(d) 10 12 cm
(a) 10 9 m
(b) 10 10 m
(c) 10 11 m
One atmosphere is equal to
(a) 101.325 K pa
(d) 10 12 m
16.
(b) 1013.25 K pa
17.
(c) 10 5 Nm
(d) None of these
Dimensions of pressure are same as that of
[CBSE PMT 1995]
(a) Energy
(c) Energy per unit volume
5.
6.
(b) Force
(d) Force per unit volume
The prefix 1018 is
[Kerala MEE 2002]
(a) Giga
(b) Nano
(c) Mega
(d) Exa
Given the numbers : 161cm, 0.161cm, 0.0161 cm. The number
of significant figures for the three numbers are
18.
19.
[CBSE PMT 1998]
7.
8.
9.
10.
11.
(a) 3, 4 and 5 respectively
(b) 3, 3 and 3 respectively
(c) 3, 3 and 4 respectively
(d) 3, 4 and 4 respectively
Significant figures in 0.00051 are
(a) 5
(b) 3
(c) 2
(d) 4
Which of the following halogen can be purified by sublimation
(a)
F2
(b) Cl 2
(c)
Br2
(d) I2
Difference in density is the basis of
[Kerala MEE 2002]
(a) Ultrafiltration
(b) Molecular sieving
(c) Gravity Separation
(d) Molecular attraction
Which of the following elements of matter would best convey
that there is life on earth
(a) Oxygen
(b) Hydrogen
(c) Carbon
(d) Iron
The compound which is added to table salt for maintaining
proper health is
(a) KCl
(b) KBr
(c)
12.
NaI
(d) MgBr2
Which of the following contains only one element
(a) Marble
(b) Diamond
(c) Glass
(d) Sand
(a) Distillation
(b) Crystallization
(c) Sublimation
(d) Adding aceitic acid
A mixture of methyl alcohol and acetone can be separated by
(a) Distillation
(b) Fractional distillation
(c) Steam distillation
(d) Distillation under reduced pressure
In
the
final
answer
of
the
expression
(29.2  20.2) (1 .79  10 5 )
. The number of significant figures
1 .37
is
[CBSE PMT 1994]
(a) 1
(b) 2
(c) 3
(d) 4
81.4 g sample of ethyl alcohol contains 0.002 g of water. The
amount of pure ethyl alcohol to the proper number of
significant figures is
(a) 81.398 g
(b) 71.40 g
(c) 91.4 g
(d) 81 g
The unit J Pa 1 is equivalent to
(a) m 3
20.
(b) cm 3
(c) dm 3
(d) None of these
From the following masses, the one which is expressed nearest
to the milligram is
(a) 16 g
(b) 16.4 g
(c) 16.428 g
(d) 16.4284 g
[Manipal PMT 2001]
21.
22.
The number of significant figures in 6.02  10 23 is
(a) 23
(b) 3
(c) 4
(d) 26
The prefix zepto stands for
[DPMT 2004]
(a) 10 9
23.
24.
(b) 10 12
(c) 10 15
(d) 10 21
The significant figures in 3400 are
(a) 2
(b) 5
(c) 6
(d) 4
The number of significant figures in 6.0023 are
[BHU 2004]
[Pb.CET 2001]
(a) 5
(c) 3
25.
(b) 4
(d) 1
Given P  0.0030m , Q  2.40m , R  3000m , Significant
figures in P, Q and R are respectively
(a) 2, 2, 1
(c) 4, 2, 1
(b) 2, 3, 4
(d) 4, 2, 3
[Pb. CET 2002]
Chemical Arithmetic
26.
The number of significant figures in 60.0001 is
9.
[Pb. CET 2000]
27.
(a) 5
(b) 6
(c) 3
(d) 2
A sample was weighted using two different balances. The
result’s were (i) 3.929 g (ii) 4.0 g. How would the weight of the
sample be reported
(a) 3.929 g
(b) 3 g
(c) 3.9 g
(d) 3.93 g
10.
Laws of chemical combination
1.
Which of the following pairs of substances illustrate the law of
multiple proportions
[CPMT 1972, 78]
(a) CO and CO2
(c)
2.
3.
4.
5.
6.
NaCl and NaBr
(b) H 2 O and D2 O
(d) MgO and Mg(OH ) 2
1.0 g of an oxide of A contained 0.5 g of A. 4.0 g of another
oxide of A contained 1.6 g of A. The data indicate the law of
(a) Reciprocal proportions
(b) Constant proportions
(c) Conservation of energy
(d) Multiple proportions
Among the following pairs of compounds, the one that
illustrates the law of multiple proportions is
(a)
NH 3 and NCl 3
(b) H 2 S and SO 2
(c)
CuO and Cu 2 O
(d) CS 2 and FeSO 4
12.
The percentage of copper and oxygen in samples of CuO
obtained by different methods were found to be the same. This
illustrates the law of
[AMU 1982, 92]
(a) Constant proportions
(b) Conservation of mass
(c) Multiple proportions
(d) Reciprocal proportions
8.
13.
Two samples of lead oxide were separately reduced to metallic
lead by heating in a current of hydrogen. The weight of lead
from one oxide was half the weight of lead obtained from the
other oxide. The data illustrates
[AMU 1983]
(a) Law of reciprocal proportions
(b) Law of constant proportions
(c) Law of multiple proportions
(d) Law of equivalent proportions
Chemical equation is balanced according to the law of
[AMU 1984]
7.
11.
(a) Multiple proportion
(b) Reciprocal proportion
(c) Conservation of mass
(d) Definite proportions
Avogadro number is
(a) Number of atoms in one gram of element
(b) Number of millilitres which one mole of a gaseous
substances occupies at NTP
(c) Number of molecules present in one gram molecular mass
of a substance
(d) All of these
Different propartions of oxygen in the various oxides of
nitrogen prove the
[MP PMT 1985]
(a) Equivalent proportion
(b) Multiple proportion
(c) Constant proportion
(d) Conservation of matter
11
Two elements X and Y have atomic weights of 14 and 16. They
form a series of compounds A, B, C, D and E in which the same
amount of element X, Y is present in the ratio 1 : 2 : 3 : 4 : 5. If
the compound A has 28 parts by weight of X and 16 parts by
weight of Y, then the compound of C will have 28 parts weight
of X and
[NCERT 1971]
(a) 32 parts by weight of Y
(b) 48 parts by weight of Y
(c) 64 parts by weight of Y
(d) 80 parts by weight of Y
Carbon and oxygen combine to form two oxides, carbon
monoxide and carbon dioxide in which the ratio of the weights
of carbon and oxygen is respectively 12 : 16 and 12 : 32. These
figures illustrate the
(a) Law of multiple proportions
(b) Law of reciprocal proportions
(c) Law of conservation of mass
(d) Law of constant proportions
A sample of calcium carbonate (CaCO 3 ) has the following
percentage composition : Ca = 40%; C = 12%; O = 48%
If the law of constant proportions is true, then the weight of
calcium in 4 g of a sample of calcium carbonate obtained from
another source will be
(a) 0.016 g
(b) 0.16 g
(c) 1.6 g
(d) 16 g
n g of substance X reacts with m g of substance Y to form p g of
substance R and q g of substance S. This reaction can be
represented as, X  Y  R  S . The relation which can be
established in the amounts of the reactants and the products will be
(a) n  m  p  q
(b) n  m  p  q
(c) n  m
(d)
pq
Which of the following is the best example of law of
conservation of mass
[NCERT 1975]
(a) 12 g of carbon combines with 32 g of oxygen to form 44 g
of CO 2
(b) When 12 g of carbon is heated in a vacuum there is no
change in mass
(c) A sample of air increases in volume when heated at
constant pressure but its mass remains unaltered
(d) The weight of a piece of platinum is the same before and
after heating in air
14.
15.
The law of multiple proportions is illustrated by the two
compounds
[NCERT 1972]
(a) Sodium chloride and sodium bromide
(b) Ordinary water and heavy water
(c) Caustic soda and caustic potash
(d) Sulphur dioxide and sulphur trioxide
In compound A, 1.00 g nitrogen unites with 0.57 g oxygen. In
compound B, 2.00 g nitrogen combines with 2.24 g oxygen. In
compound C, 3.00 g nitrogen combines with 5.11 g oxygen.
These results obey the following law
[CPMT 1971]
(a) Law of constant proportion
(b) Law of multiple proportion
(c) Law of reciprocal proportion
(d) Dalton's law of partial pressure
12 Chemical Arithmetic
16.
Hydrogen combines with oxygen to form H 2 O in which 16 g
4.
of oxygen combine with 2 g of hydrogen. Hydrogen also
combines with carbon to form CH 4 in which 2 g of hydrogen
17.
18.
19.
20.
combine with 6 g of carbon. If carbon and oxygen combine
together then they will do show in the ratio of
(a) 6 : 16 or 12 : 32
(b) 6 : 18
(c) 1 : 2
(d) 12 : 24
2 g of hydrogen combine with 16 g of oxygen to form water
and with 6 g of carbon to form methane. In carbon dioxide 12 g
of carbon are combined with 32 g of oxygen. These figures
illustrate the law of
(a) Multiple proportions
(b) Constant proportions
(c) Reciprocal proportions
(d) Conservation of mass
An element forms two oxides containing respectively 53.33 and
36.36 percent of oxygen. These figures illustrate the law of
(a) Conservation of mass
(b) Constant proportions
(c) Reciprocal proportions
(d) Multiple proportions
After a chemical reaction, the total mass of reactants and
products
[MP PMT 1989]
(a) Is always increased
(b) Is always decreased
(c) Is not changed
(d) Is always less or more
A sample of pure carbon dioxide, irrespective of its source
contains 27.27% carbon and 72.73% oxygen. The data support
1 amu is equal to
(a)
1
of C  12
12
(c) 1g of H 2
5.
(b)
1
of O - 16
14
(d) 1.66  10 23 kg
Sulphur forms the chlorides S 2 Cl 2 and SCl 2 . The equivalent
mass of sulphur in SCl 2 is
[EAMCET 1985; Pb. CET 2001]
6.
(a) 8 g/mole
(b) 16 g/mole
(c) 64.8 g/mole
(d) 32 g/mole
The sulphate of a metal M contains 9.87% of M. This sulphate
is isomorphous with ZnSO 4 .7 H 2 O . The atomic weight of M
is
(a) 40.3
(c) 24.3
7.
[IIT 1991]
(b) 36.3
(d) 11.3
When 100 ml of 1 M NaOH solution and 10 ml of
10 N H 2 SO 4 solution are mixed together, the resulting solution
8.
will be
[DPMT 1982]
(a) Alkaline
(b) Acidic
(c) Strongly acidic
(d) Neutral
In chemical scale, the relative mass of the isotopic mixture of
oxygen atoms (O 16 , O 17 , O 18 ) is assumed to be equal to
[Bihar MADT 1981]
[AIIMS 1992]
21.
22.
(a) Law of constant composition
(b) Law of conservation of mass
(c) Law of reciprocal proportions
(d) Law of multiple proportions
The law of definite proportions is not applicable to nitrogen
oxide because
[EAMCET 1981]
(a) Nitrogen atomic weight is not constant
(b) Nitrogen molecular weight is variable
(c) Nitrogen equivalent weight is variable
(d) Oxygen atomic weight is variable
Which one of the following pairs of compounds illustrates the
law of multiple proportion
[EAMCET 1989]
(a)
H 2O, Na 2O
(b) MgO, Na 2O
(c)
Na 2O, BaO
(d) SnCl 2 , SnCl 4
9.
10.
(a) 6.02  10 23 atoms of H
(b) 4 g atom of Hydrogen
(c) 1.81  10 23 molecules of CH 4
11.
Atomic, Molecular and Equivalent masses
1.
[MP PMT 1986]
2.
3.
[MP PMT 2002]
12
(a)
C
(c)
H1
(b) O
16
(d) C 13
(d) 3.0 g of carbon
In the reaction
2 Na2 S 2O3  I2  Na2 S 4 O6  2 NaI , the
equivalent weight of Na 2 S 2 O 3 (mol. wt. = M) is equal to
(a) M
Which property of an element is always a whole number
(a) Atomic weight
(b) Equivalent weight
(c) Atomic number
(d) Atomic volume
Which one of the following properties of an element is not
variable
[Bihar MADT 1981]
(a) Valency
(b) Atomic weight
(c) Equivalent weight
(d) All of these
The modern atomic weight scale is based on
(a) 16.002
(b) 16.00
(c) 17.00
(d) 11.00
For preparing 0.1 N solution of a compound from its impure
sample of which the percentage purity is known, the weight of
the substance required will be
[MP PET 1996]
(a) More than the theoretical weight
(b) Less than the theoretical weight
(c) Same as the theoretical weight
(d) None of these
1 mol of CH 4 contains
12.
13.
(b) M / 2
(c) M / 3
(d) M / 4
When potassium permanganate is titrated against ferrous
ammonium sulphate, the equivalent weight of potassium
permanganate is
[CPMT 1988]
(a) Molecular weight /10
(b) Molecular weight /5
(c) Molecular weight /2
(d) Molecular weight
Boron has two stable isotopes, 10 B (19%) and 11 B (81%). The
atomic mass that should appear for boron in the periodic table
is
[CBSE PMT 1990]
(a) 10.8
(b) 10.2
(c) 11.2
(d) 10.0
Chemical Arithmetic
14.
What is the concentration of nitrate ions if equal volumes of 0.1
M AgNO 3 and 0.1 M NaCl are mixed together
27.
[CPMT 1983; NCERT 1985]
15.
16.
17.
(a) 0.1 M
(b) 0.2 M
(c) 0.05 M
(d) 0.25 M
Total number of atoms represented by the compound
CuSO4.5H2O is
[BHU 2005]
(a) 27
(b) 21
(c) 5
(d) 8
74.5 g of a metallic chloride contain 35.5 g of chlorine. The
equivalent weight of the metal is
[CPMT 1986]
(a) 19.5
(b) 35.5
(c) 39.0
(d) 78.0
7.5 grams of a gas occupy 5.8 litres of volume at STP the gas is
(a) NO
(b) N 2 O
(c)
18.
[CBSE PMT 1999; MH CET 2003]
19.
(b) 2  10 23
[AMU 1992]
20.
21.
C2 H 2
(b) CO
(c)
O2
(d) CH 4
29.
30.
23.
24.
25.
(a) 100
(b) 150
(c) 120
(d) 200
The oxide of a metal has 32% oxygen. Its equivalent weight
would be
[MP PMT 1985]
(a) 34
(b) 32
(c) 17
(d) 8
The mass of a molecule of water is
[Bihar CEE 1995]
(a)
3  10
26
kg
(b) 3  10
26
26.
25
(c) 1.5  10 kg
(d) 2.5  10
1.24 gm P is present in 2.2 gm
(a)
P4 S 3
(b) P2 S 2
(c)
PS 2
(d) P2 S 4
kg
26
kg
[CBSE PMT 2000]
(b) 2.24 L
(d) 1.12 L
If N A is Avogadro’s number then number of valence electrons
(a) 2.4 N A
(b) 4.2 N A
(c) 1.6 N A
(d) 3.2 N A
The weight of 1  10 22 molecules of CuSO 4 .5 H 2 O is
[IIT 1991]
(a) 41.59 g
(c) 4.159 g
31.
(b) 415.9 g
(d) None of these
Rearrange the following (I to IV) in the order of increasing
masses and choose the correct answer from (a), (b), (c) and (d)
(Atomic mass: N=14, O=16, Cu=63).
I.
1 molecule of oxygen
II. 1 atom of nitrogen
III. 1  10 10 g molecular weight of oxygen
IV. 1  10 10 g atomic weight of copper
[MP PMT 1995]
22.
(d) 2y
in 4.2 g of nitride ions ( N 3 )
The vapour density of a gas is 11.2. The volume occupied by
11.2 g of the gas at ATP will be
[Bihar CET 1995]
(a) 11.2 L
(b) 22.4 L
(c) 1 L
(d) 44.8 L
Equivalent weight of crystalline oxalic acid is
(a) 30
(b) 63
(c) 53
(d) 45
The equivalent weight of an element is 4. Its chloride has a V.D
59.25. Then the valency of the element is
[BHU 1997]
(a) 4
(b) 3
(c) 2
(d) 1
1.25 g of a solid dibasic acid is completely neutralised by 25 ml
of 0.25 molar Ba(OH ) 2 solution. Molecular mass of the acid is
y
4
Assuming fully decomposed, the volume of CO 2 released at
will be
(a) 0.84 L
(c) 4.06 L
(c) 4  10 23
(d) 6  10 23
One litre of a gas at STP weight 1.16 g it can possible be
(a)
(b)
STP on heating 9.85g of BaCO 3 (Atomic mass of Ba=137)
The number of atoms in 4.25 g of NH 3 is approximately
(a) 1  10 23
y
2
(c) y
(d) CO 2
CO
The atomic weights of two elements A and B are 40 and 80
respectively. If x g of A contains y atoms, how many atoms are
present in 2x g of B
(a)
28.
13
32.
(a) II<I<III<IV
(b) IV<III<II<I
(c) II<III<I<IV
(d) III<IV<I<II
1.520 g of the hydroxide of a metal on ignition gave
gm of oxide. The equivalent weight of metal is
0.995
[DPMT 1984]
(a) 1.520
(c) 19.00
33.
34.
(b) 0.995
(d) 9.00
How much coulomb charge is present on 1g ion of N 3 
(a)
5.2  10 6 Couloumb
(b) 2.894  10 5 Couloumb
(c)
6.6  10 6 Couloumb
(d) 8.2  10 6 Couloumb
Ratio of C p and C v of a gas X is 1.4, the number of atom of
the gas ‘X’ present in 11.2 litres of it at NTP will be
[CBSE 1999]
(a)
35.
6.02  10 23
(b) 1.2  10 23
(c) 3.01  10 23
(d) 2.01  10 23
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
[AIEEE 2005]
(a) Decrease twice
(b) Increase two fold
(c) Remain unchanged
(d) Be a function of the molecular mass of the substance
14 Chemical Arithmetic
36.
37.
What should be the equivalent weight of phosphorous acid, if
P=31; O=16; H=1
(a) 82
(b) 41
(c) 20.5
(d) None of these
The number of molecule at NTP in 1 ml of an ideal gas will be
(a)
38.
39.
6  10 23
(b) 2.69  10 19
(c) 2.69  10 23
(d) None of these
The specific heat of a metal is 0.16 its approximate atomic
weight would be
(a) 32
(b) 16
(c) 40
(d) 64
The weight of a molecule of the compound C 60 H 122 is
[AIIMS 2000]
(a) 1.4  10 21 g
40.
41.
42.
46.
48.
Mn 2 O 3
(b) MnO 2
(c)
MnO4
(d) MnO42 
100 mL of PH3 on decomposition produced phosphorus and
hydrogen. The change in volume is [MNR 1986]
52.
6.4 kg
96 kg
(a)
50 mL increase
(b) 500 mL decrease
(c)
900 mL decrease
(d) Nil.
12 g of Mg (at. mass 24) on reacting completely with acid
gives hydrogen gas, the volume of which at STP would be
[CPMT 1978]
(a)
22.4 L
(b) 11.2 L
(c)
44.8 L
(d) 6 .1 L
53.
6.02  10 23 cm 3
(b) 22400 cm 3
2 g atom of nitrogen
(c) 1 mole of S
54.
(d) 3.0  10 23 cm 3
Caffeine has a molecular weight of 194. If it contains 28.9% by
mass of nitrogen, number of atoms of nitrogen in one molecule
of caffeine is
(a) 4
(b) 6
(c) 2
(d) 3
A 400 mg iron capsule contains 100 mg of ferrous fumarate,
(CHCOO ) 2 Fe . The percentage of iron pasent in it is
Which of the following has least mass
(a)
If the density of water is 1 g cm 3 then the volume occupied
by one molecule of water is approximately
[Pb. PMT 2004]
55.
56.
(b) 3  10
[Pb. PET 1985]
23
atoms of C
(d) 7 .0 g of Ag
How many mole of helium gas occupy 22.4 L at 0 o C at 1
atm. pressure
(a) 0.11
(c) 1.0
[Kurukshetra CEE 1992; CET 1992]
(b) 0.90
(d) 1.11
Volume of a gas at STP is 1.12  10 7 cc. Calculate the number
of molecules in it
[BHU 1997]
(a)
3.01  10 20
(b) 3.01  1012
(c)
3.01  10 23
(d) 3.01  10 24
4 .4 g of an unknown gas occupies 2.24 L of volume at
standard temperature and pressure. The gas may be
(b) 25%
(d) 8%
[MP PMT 1995]
26
The element whose a atom has mass of 10.86  10 kg is
(a) Boron
(b) Calcium
(c) Silver
(d) Zinc
The number of gram atoms of oxygen present in 0.3 gram mole
of (COOH ) 2 .2 H 2 O is
(a) 0.6
(c) 1.2
47.
51.
[IIT 1988; CPMT 1994]
(a)
100ml of 2.5 molal (2.5m) ammonium hydroxide solution
(a) 0.056 litres
(b) 0.56 litres
(c) 5.6 litres
(d) 11.2 litres
approximately
(a) 33%
(c) 14%
45.
when it is converted to
16.023  10 23 g
required for the complete
[CBSE PMT 1989]
The equivalent weight of MnSO 4 is half its molecular weight
What volume of NH 3 gas at STP would be needed to prepare
(c)
44.
50.
(b) 1.09  10 21 g
(c) 5.025  10 23 g
(d)
What is the weight of oxygen
combustion of 2.8 kg of ethylene
(a) 2.8 kg
(b)
(c) 9.6 kg
(d)
(a) 18 cm 3
43.
49.
(a) 21
(b) 54
(c) 27.06
(d) 2.086
One gram of hydrogen is found to combine with 80g of
bromine one gram of calcium valency=2 combines with 4g of
bromine the equivalent weight of calcium is
(a) 10
(b) 20
(c) 40
(d) 80
57.
proportion. The weight of 2.24 litres of this mixture at NTP is
(a) 4.6 g
(b) 1.6 g
(c) 2.3 g
(d) 23 g
Vapour density of a metal chloride is 66. Its oxide contains
53% metal. The atomic weight of the metal is
[Bihar MADT 1982]
(b) Carbon monoxide
(c) Oxygen
(d) Sulphur dioxide
The number of moles of oxygen in 1 L of air containing 21%
oxygen by volume, in standard conditions, is
[CBSE PMT 1995; Pb. PMT 2004]
(b) 1.8
(d) 3.6
A gaseous mixture contains CH 4 and C 2 H 6 in equimolecular
(a) Carbon dioxide
58.
(a) 0.186 mol
(b) 0.21 mol
(c) 2.10 mol
(d) 0.0093 mol
The number of molecules in 8.96 L of a gas at 0 o C and 1
atmosphere pressure is approximately
(a)
6.02  10
23
(c) 18.06  10
23
(b) 12.04  10
[BHU 1993]
23
(d) 24.08  10 22
Chemical Arithmetic
59.
The equivalent weight of a metal is 9 and vapour density of its
chloride is 59.25. The atomic weight of metal is
(d) 12  10 23
6  10 23
The volume occupied by 4.4 g of CO 2 at STP is
(c)
6.
[Pb. CET 2002]
60.
(a) 23.9
(b) 27.3
(c) 36.3
(d) 48.3
The molecular weight of a gas is 45. Its density at STP is
[AFMC 1997, 2004; Pb. CET 1997, 2002]
7.
[Pb. PMT 2004]
61.
62.
(a) 22.4
(b) 11.2
(c) 5.7
(d) 2.0
Equivalent weight of a bivalent metal is 37.2. The molecular
weight of its chloride is
[MH CET 2003]
(a) 412.2
(b) 216
(c) 145.4
(d) 108.2
On reduction with hydrogen, 3.6 g of an oxide of metal left 3.2
g of metal. If the vapour density of metal is 32, the simplest
formula of the oxide would be
63.
MO
(b) M 2 O 3
(c)
M 2O
(d) M 2 O 5
8.
9.
10.
0.5  10
(c)
3.5  10 23
(b) 1.5  10
2.
(d) 1.8  10 32
11.
Which one of the following pairs of gases contains the same
number of molecules
[EAMCET 1987]
(a) 16 g of O 2 and 14 g of N 2
12.
(b) 2.0478  10 24
(c) 3.0115  10 24
(d) 4.0956  10 24
The number of molecules in 16 g of methane is
(b) 6.02  10 23
3.0  10 23
16
16
 10 23
 10 23
(c)
(d)
6 .02
3 .0
Number of molecules in 100 ml of each
[Bihar MADT 1985]
O 2 , NH 3 and CO 2 at STP are
(a) In the order CO 2  O2  NH 3
(c) 28 g of N 2 and 22 g of CO 2
(b) In the order NH 3  O 2  CO 2
(d) 32 g of O 2 and 32 g of N 2
(c) The same
(d) NH 3  CO 2  O 2
Number of gm of oxygen in 32.2 g Na 2 SO 4 .10 H 2 O is
(a) 20.8
(b) 22.4
(c) 2.24
(d) 2.08
250 ml of a sodium carbonate solution contains 2.65 grams of
Na 2 CO 3 . If 10 ml of this solution is diluted to one litre, what
(a) 0.1 M
(b) 0.001 M
(c) 0.01 M
(d) 10
4
M
13.
14.
g mol 1
(d) mol g 1
The number of water molecules in 1 litre of water is
[EAMCET 1990]
(b) 18  1000
(d) 55.55 N A
(a) 18
(c) N A
15.
The number of electrons in a mole of hydrogen molecule is
[CPMT 1987]
A molar solution is one that contains one mole of a solute in
(a) 1000 g of the solvent
(c) One litre of the solution
(b) One litre of the solvent
(d) 22.4 litres of the solution
The number of oxygen atoms in 4.4 g of CO 2 is approx.
[CBSE PMT 1990]
(a) 1.2  10 23
(b) 6  10 22
of
The molecular weight of hydrogen peroxide is 34. What is the
unit of molecular weight
[MP PMT 1986]
(a) g
(b) mol
(c)
[IIT 1986]
5.
(c) 6.02  10 24
(d) 6.02  10 25
The total number of protons in 10 g of calcium carbonate is
(b) 8 g of O 2 and 22 g of CO 2
is the concentration of the resultant solution (mol. wt. of
[EAMCET 2001]
Na 2 CO 3 =106)
4.
(b) 6.02  10 23
(a)
[Haryana PMT 2000]
3.
(c) 4.84  10 17
(d) 6.023  10 23
One mole of calcium phosphide on reaction with excess of
water gives
[IIT 1999]
(a) One mole of phosphine
(b) Two moles of phosphoric acid
(c) Two moles of phosphine
(d) One mole of phosphorus pentoxide
19.7 kg of gold was recovered from a smuggler. How many
atoms of gold were recovered (Au =197)
[Pb. CET 1985]
(a) 1.5057  10 24
23
The mole concept
1.
(b) 1.084  10 18
( N 0  6.023  10 23 )
[Pb. CET 2000]
(a)
6.023  10 19
(a) 100
The number of molecules in 4.25 g of ammonia are
23
(a) 22.4 L
(b) 2.24 L
(c) 0.224 L
(d) 0.1 L
The number of water molecules present in a drop of water
(volume 0.0018 ml) at room temperature is
[DCE 2000]
(a)
[DPMT 2004]
(a)
15
(a)
6.02  10
(b) 12.046  10
23
23
(c) 3.0115  10
(d) Indefinite
The numbers of moles of BaCO 3 which contain 1.5 moles of
23
16.
oxygen atoms is
(a) 0.5
(b) 1
(c) 3
(d) 6.02  10 23
[EAMCET 1991]
16 Chemical Arithmetic
17.
Which of the following is Loschmidt number
(a)
6  10
(b) 2.69  10
23
19
(c) 3  10
(d) None of these
How many molecules are present in one gram of hydrogen
23
18.
2.
[AIIMS 1982]
(a)
19.
20.
3.
(b) 3.01  10 23
(c) 2.5  10 23
(d) 1.5  10 23
The total number of gm-molecules of SO 2 Cl 2 in 13.5 g of
sulphuryl chloride is
[CPMT 1992]
(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.4
The largest number of molecules is in
[BHU 1997]
(a) 34 g of water
(b) 28 g of CO 2
(c)
21.
6.02  10 23
46 g of CH 3 OH
4.
The number of moles of sodium oxide in 620 g of it is
(a) 1 mol
(c) 18 moles
22.
[MP PMT 1986]
(d) 54 g of N 2 O5
[BHU 1992]
5.
(b) 10 moles
(d) 100 moles
[BHU 1992]
0 .5 g of hydrogen
(c) 7 g of nitrogen
23.
(b) 4 g of sulphur
(a) Different percentage composition
(b) Different molecular weights
(c) Same viscosity
(d) Same vapour density
A compound (80 g) on analysis gave C = 24 g, H = 4 g, O =
32 g. Its empirical formula is
[CPMT 1981]
(a)
C2 H 2 O2
(b) C 2 H 2 O
(c)
CH 2 O 2
(d) CH 2 O
The empirical formula of a compound is CH 2 O. 0.0835 moles
of the compound contains 1.0 g of hydrogen. Molecular
formula of the compound is
(a) C 2 H 12 O 6
(b) C 5 H 10 O 5
2 g of oxygen contains number of atoms equal to that in
(a)
(a) 40
(b) 60
(c) 8
(d) 10
The percentage of nitrogen in urea is about
[KCET 2001]
(a) 46
(b) 85
(c) 18
(d) 28
If two compounds have the same empirical formula but
different molecular formula, they must have
(c)
6.
(d) 2 .3 g of sodium
The empirical formula of an acid is CH 2O2 , the probable
(a)
CH 2 O
(b) CH 2O2
(c)
C 2 H 4 O2
(d) C3 H 6 O4
[CBSE PMT 2001]
24.
(a)
7.
(c) Fructose and sucrose
[BCECE 2005]
(a)
6.0  10
(c) 12  10 23
(b) 3  10
In which of the following pairs of compounds the ratio of C, H
and O is same
(b) Glucose and acetic acid
(c) 5  6.02  10 23 atoms/mole
(d) None of these
The number of molecules of CO2 present in 44g of CO2 is
23
[AFMC 2000]
(a) Acetic acid and methyl alcohol
45  6.02  10 23 atoms/mole
(b) 5  6.62  10 23 atoms/mole
25.
(d) C 3 H 6 O 3
molecular formula of acid may be
Molarity of liquid HCl with density equal to 1.17 g / cc is
(a) 36.5
(b) 18.25
(c) 32.05
(d) 4.65
How many atoms are contained in one mole of sucrose
[Pb. PMT 2002]
(C12 H 22 O11 )
C 4 H 8 O8
23
(d) All of these
Chemical stoichiometry
1.
(d) 3  10 10
How much of NaOH is required to neutralise 1500 cm 3 of 0.1
[KCET 2001]
N HCl (Na = 23)
(a) 40 g
26.
A sample of phosphorus trichloride (PCl 3 ) contains 1.4 moles
27.
(e) 2.409  10 24
The number of sodium atoms in 2 moles of sodium
ferrocyanide is
[BHU 2004]
(b) 4 g
(c) 6 g
(d) 60 g
of the substance. How many atoms are there in the sample[Kerala PMT 2004]
2.
How much water should be added to 200 c.c of semi normal
(a) 4
(b) 5.6
solution of NaOH to make it exactly deci normal
(c) 8.431  10 23
(d) 3.372  10 24
[AFMC 1983]
(a) 12  10 23
(b) 26  10 23
34  10 23
(d) 48  10 23
(c)
3.
Percentage composition & Molecular formula
1.
The percentage of oxygen in NaOH is
[CPMT 1979]
(a) 200 cc
(b) 400 cc
(c) 800 cc
(d) 600 cc
2.76 g of silver carbonate on being strongly heated yield a
residue weighing
[Pb. CET 2003]
(a) 2.16 g
(b) 2.48 g
(c) 2.64 g
(d) 2.32 g
Chemical Arithmetic
4.
In the reaction, 4 NH 3 (g)  5O 2 (g)  4 NO(g)  6 H 2 O(g) ,
When 1 mole of ammonia and 1 mole of O 2 are made to react
(c) 80 grams
13.
17
(d) 320 grams
What is the normality of a 1 M solution of H 3 PO4
to completion
[AIIMS 1982]
(a) 1.0 mole of H 2 O is produced
(b) 1.0 mole of NO will be produced
14.
(c) All the oxygen will be consumed
(a) 0.5 N
(b) 1.0 N
(c) 2.0 N
(d) 3.0 N
Normality of 2M sulphuric acid is
(a) 2N
(d) All the ammonia will be consumed
5.
6.
7.
8.
Haemoglobin contains 0.33% of iron by weight. The molecular
weight of haemoglobin is approximately 67200. The number of
iron atoms (At. wt. of Fe = 56) present in one molecule of
haemoglobin is
[CBSE PMT 1998]
15.
How many g of a dibasic acid (Mol. wt. = 200) should be
present in 100 ml of its aqueous solution to give decinormal
strength
[AIIMS 1992]
(a) 1 g
(b) 2 g
(d) 2
(c) 10 g
(d) 20 g
What quantity of ammonium sulphate is necessary for the
production of NH 3 gas sufficient to neutralize a solution
16.
The solution of sulphuric acid contains 80% by weight
H 2 SO 4 . Specific gravity of this solution is 1.71. Its normality
containing 292 g of HCl ? [HCl=36.5; (NH 4 )2 SO 4 =132;
is about
NH 3 =17]
(a) 18.0
(b) 27.9
(c) 1.0
(d) 10.0
[CPMT 1992]
(a) 272 g
(b) 403 g
(c) 528 g
(d) 1056 g
17.
[CBSE 1991]
Mohr's salt is dissolved in dil. H 2 SO 4 instead of distilled
The percentage of P2 O 5 in diammonium hydrogen phosphate
water to
(NH 4 ) 2 HPO4 is
(a) Enhance the rate of dissolution
[CPMT 1992]
(a) 23.48
(b) 46.96
(c) 53.78
(d) 71.00
(b) Prevent cationic hydrolysis
(c) Increase the rate of ionisation
1
moles of oxygen combine with Al to form Al2 O 3 the
2
weight of Al used in the reaction is (Al=27)
[EAMCET 1980]
If 1
(d) Increase its reducing strength
18.
(b) 54 g
19.
(d) 31 g
The percentage of Se in peroxidase anhydrous enzyme is 0.5%
by weight (atomic weight=78.4). Then minimum molecular
weight of peroxidase anhydrous enzyme is
(a) 1.568  10 4
(b) 1.568  10 3
(c) 15.68
(d) 3.136  10 4
20.
Acidified potassium permanganate solution is decolourised by
(a) Bleaching powder
(b) White vitriol
(c) Mohr's salt
(d) Microcosmic salt
Approximate atomic weight of an element is 26.89. If its
equivalent weight is 8.9, the exact atomic weight of element
would be
[DPMT 1984]
(a) 26.89
(b) 8.9
(c) 17.8
(d) 26.7
Vapour density of a gas is 22. What is its molecular mass
[AFMC 2000]
H 2 evolved at STP on complete reaction of 27 g of
Aluminium with excess of aqueous NaOH would be
21.
(a) 33
(b) 22
(c) 44
(d) 11
Equivalent weight of KMnO4 acting as an oxidant in acidic
medium is
[CPMT 1991]
12.
N
4
(c) 4
[CBSE PMT 2001]
11.
(d)
(b) 1
(c) 49.5 g
10.
N
2
(a) 6
(a) 27 g
9.
(c)
[AIIMS 1992]
(b) 4N
[CPMT 1990; MP PET 1999]
(a) 22.4
(b) 44.8
(a) The same as its molecular weight
(c) 67.2
(d) 33.6 litres
(b) Half of its molecular weight
(c) One-third of its molecular weight
What is the % of H 2 O in Fe(CNS )3 .3 H 2O
(a) 45
(b) 30
(c) 19
(d) 25
What weight of SO 2 can be made by burning sulphur in 5.0
moles of oxygen
(a) 640 grams
(b) 160 grams
(d) One-fifth of its molecular weight
22.
0.16 g of dibasic acid required 25 ml of decinormal NaOH
solution for complete neutralisation. The molecular weight of
the acid will be
[CPMT 1989]
(a) 32
(b) 64
(c) 128
(d) 256
18 Chemical Arithmetic
23.
24.
[CPMT 1992]
To neutralise 20 ml of M / 10 sodium hydroxide, the volume
of M / 20 hydrochloric acid required is
(a) 0.6 g
(b) 1.0 g
[Andhra MBBS 1980]
(c) 1.5 g
(d) 2.0 g
(a) 10 ml
(b) 15 ml
(c) 20 ml
(d) 40 ml
32.
In the preceeding question, the amount of Na 2CO 3 present in
the solution is
Hydrochloric acid solutions A and B have concentration of 0.5
N and 0.1 N respectively. The volume of solutions A and B
required to make 2 litres of 0.2 N hydrochloric are
[KCET 1993]
33.
(b) 1.060 g
(c) 0.530 g
(d) 0.265 g
How many ml of 1 (M) H 2 SO 4 is required to neutralise 10 ml
of 1 (M) NaOH solution
(a) 0.5 l of A + 1.5 l of B
[MP PET 1998; MNR 1982; MP PMT 1987]
(b) 1.5 l of A + 0.5 l of B
(c) 1.0 l of A + 1.0 l of B
(d) 0.75 l of A + 1.25 l of B
25.
5 ml of N HCl, 20 ml of N / 2 H 2 SO 4 and 30 ml of
34.
The normality of the resulting solution is
27.
(a)
N /5
(b)
N / 10
(c)
N / 20
(d)
N / 40
[MNR 1991]
35.
29.
(b) 5.0
(c) 10.0
(d) 20.0
Which of the following cannot give iodometric titrations
3
(a)
Fe
(c)
Pb2 
(b) Cu
(d)
2
Ag 
KMnO4 reacts with ferrous ammonium sulphate according to
Under similar conditions of pressure and temperature, 40 ml of
slightly moist hydrogen chloride gas is mixed with 20 ml of
ammonia gas, the final volume of gas at the same temperature
and pressure will be
[CBSE PMT 1993]
the equation
(a) 100 ml
(b) 20 ml
(a) 20 ml of 0.1 M FeSO 4
(c) 40 ml
(d) 60 ml
MnO4  5 Fe2   8 H   Mn2   5 Fe3   4 H 2O , here 10
ml of 0.1 M KMnO4 is equivalent to
2 MnO4  5 C2O42   16 H   2 Mn2   10CO 2  8 H 2O , here
(c) 40 ml of 0.1 M FeSO 4
20 ml of 0.1 M KMnO4 is equivalent to
(d) 50 ml of 0.1 M FeSO 4
[CBSE PMT 1996]
36.
Ca(OH ) 2  H 3 PO4  CaHPO4  2 H 2 O
(c) 50 ml of 0.5 M H 2C2O4 (d) 20 ml of 0.1 M H 2C2O4
weight of H 3 PO4 in the above reaction is
In order to prepare one litre normal solution of KMnO4 , how
(a) 21
(b) 27
many grams of KMnO4 are required if the solution is used in
(c) 38
(d) 49
acidic medium for oxidation
[MP PET 2002]
(a) 158 g
(b) 31.6 g
(c) 790 g
(d) 62 g
[CPMT 1999]
(b) 30 ml of 0.1 M FeSO 4
KMnO4 reacts with oxalic acid according to the equation,
(a) 20 ml of 0.5 M H 2C2O4 (b) 50 ml of 0.1 M H 2C2O4
28.
(a) 2.5
[AIIMS 1997]
N / 3 HNO3 are mixed together and volume made to one litre.
26.
[CPMT 1992]
(a) 2.650 g
37.
the
equivalent
[Pb. PMT 2004]
The mass of BaCO 3 produced when excess CO 2 is bubbled
through a solution of 0.205 mol Ba(OH ) 2 is
[UPSEAT 2004]
What is the concentration of nitrate ions if equal volumes of 0.1
M AgNO3 and 0.1 M NaCl are mixed together
(a) 81 g
(b) 40.5 g
(c) 20.25 g
(d) 162 g
[NCERT 1981; CPMT 1983]
30.
31.
(a) 0.1 N
(b) 0.2 M
(c) 0.05 M
(d) 0.25 M
38.
solution of NaOH to give a concentration of 10 mg per ml is
30 ml of acid solution is neutralized by 15 ml of a 0.2 N base.
The strength of acid solution is [CPMT 1986]
(a) 0.1 N
(b) 0.15 N
(c) 0.3 N
(d) 0.4 N
39.
(a) 100
(b) 200
(c) 250
(d) 500
Number of moles of KMnO 4 required to oxidize one mole of
Fe(C 2 O 4 ) in acidic medium is
A solution containing Na 2CO 3 and NaOH requires 300 ml of
0.1 N HCl using phenolpthalein as an indicator. Methyl orange
is then added to the above titrated solution when a further 25 ml
of 0.2 N HCl is required. The amount of NaOH present in
solution is (NaOH  40, Na 2CO3  106)
The amount of water that should be added to 500 ml of 0.5 N
40.
[Haryana CEE 1996]
(a) 0.6
(b) 0.167
(c) 0.2
(d) 0.4
A hydrocarbon contains 86% carbon, 488ml of the hydrocarbon
weight 1.68 g at STP. Then the hydrocarbon is an
Chemical Arithmetic
41.
(a) Alkane
(b) Alkene
(c) Alkyne
(d) Arene
(c) 18 g
50.
The ratio of amounts of H 2 S needed to precipitate all the metal
CuSO 4 will be
42.
43.
44.
45.
46.
47.
48.
49.
(a) 1:1
(b) 1:2
(c) 2:1
(d) None of these
(d) 19 g
in
M
FeSO 4 was titrated with
10
acidic medium. The amount of
KMnO4 used will be
[CPMT 1984]
A
solution
of
KMnO4 solution
ions from 100 ml of 1 M AgNO 3 and 100 ml of 1 M
51.
19
10
ml
(a) 5 ml of 0.1 M
(b) 10 ml of 1.1 M
(c) 10 ml of 0.5 M
(d) 10 ml of 0.02 M
c.c. of O 2 and 50 c.c. of H 2 . The volume of the gases formed
1.12 ml of a gas is produced at STP by the action of 4.12 mg of
alcohol, with methyl magnesium iodide. The molecular mass of
alcohol is
[Roorkee 1992; IIT 1993]
(i) at room temperature and (ii) at 1100C will be
(a) 16.0
(b) 41.2
(c) 82.4
(d) 156.0
An electric discharge is passed through a mixture containing 50
(a) (i) 25 c.c. (ii) 50 c.c.
(b) (i) 50 c.c. (ii) 75 c.c.
(c) (i) 25 c.c. (ii) 75 c.c.
(d) (i) 75 c.c. (ii) 75 c.c.
52.
100 ml of 0.1 N hypo decolourised iodine by the addition of x g
of crystalline copper sulphate to excess of KI. The value of ‘x’
is (molecular wt. of CuSO 4 .5 H 2 O is 250)
(a) 5.0 g
(b) 1.25 g
(c) 2.5 g
(d) 4 g
53.
How many grams of caustic potash required to completely
neutralise 12.6 gm HNO 3
(a) 22.4 KOH
(b) 1.01 KOH
(c) 6.02 KOH
(d) 11.2 KOH
(b) 9 kg
(c) 27 kg
(d) 1.8 kg
(a)
XY
(b)
X 2Y
(c)
XY 3
(d)
X 2Y3
A compound contains atoms of three elements in A, B and C. If
the oxidation number of A is +2, B is +5 and that of C is – 2, the
possible formula of the compound is
[CBSE PMT 2000]
If isobutane and n-butane are present in a gas, then how much
oxygen should be required for complete combustion of 5 kg of
this gas
(a) 17.9 kg
The simplest formula of a compound containing 50% of
element X (atomic mass 10) and 50% of element Y (atomic
mass 20) is
[Roorkee 1994]
54.
(a)
A 3 (BC 4 ) 2
(b)
A 3 (B 4 C)2
(c)
ABC 2
(d)
A 2 (BC 3 ) 2
What will be the volume of CO 2 at NTP obtained on heating
10 grams of (90% pure) limestone
16.8 litre gas containing H 2 and O 2 is formed at NTP on
(a) 22.4 litre
electrolysis of water. What should be the weight of electrolysed
water
(b) 2.016 litre
(a) 5 g
(b) 9 g
(d) 20.16 litre
(c) 10 g
(d) 12 g
[Pb. CET 2001]
(c) 2.24 litre
55.
The ratio of the molar amounts of H 2 S needed to precipitate
On electrical decomposition of 150 ml dry and pure O 2 , 10%
the metal ions from 20mL each of 1M Cd (NO 3 )2 and
of O 2 gets changed to O, then the volume of gaseous mixture
0.5 M CuSO 4 is
after reaction and volume of remaining gas left after passing in
turpentine oil will be
(a) 1 : 1
(a) 145 ml
(b) 149 ml
(c) 128 ml
(d) 125 ml
What should be the weight of 50% HCl which reacts with 100 g
of limestone
(a) 50% pure
(b) 25% pure
(c) 10% pure
(d) 8% pure
What should be the weight and moles of AgCl precipitate
obtained on adding 500ml of 0.20 M HCl in 30 g of
AgNO 3 solution? ( AgNO 3 = 170)
(a) 14.35 g
(b) 15 g
[CPMT 1997]
(b) 2 : 1
(c) 1 : 2
(d) Indefinite
56.
12 g of Mg (at. mass 24) will react completely with acid to
give
[MNR 1985]
(a) One mole of H 2
(b) 1/2 mole of H 2
(c) 2/3 mole of O 2
(d) Both 1/2 mol of H 2 and 1/2 mol of O 2
20 Chemical Arithmetic
57.
1 .5 mol of O 2 combine with Mg to form oxide MgO . The
(b) 22.4 L of CO 2 at STP
mass of Mg (at. mass 24) that has combined is
(c) 0.44 g of CO 2
[KCET 2001]
58.
(a) 72 g
(b) 36 g
(c)
(d) 24 g
48 g
8.
100 g CaCO3 reacts with 1litre 1 N HCl. On completion of
reaction how much weight of CO 2 will be obtain
[Kerala CET 2005]
(a)
5 .5 g
(b) 11 g
(c)
22 g
(d) 33 g
(e)
44 g
9.
(d) None of these
In a mole of water vapour at STP, the volume actually occupied
or taken by the molecules (i.e., Avogadro’s
No.  Volume
of one molecule) is
[Kerala EEE 2000]
(a) Zero
(b) Less than 1% of 22.4 litres
(c) About 10% of the volume of container
(d) 1% to 2% of 22.4 litres
(e) Between 2% to 5% of 22.4 litres
If 10 21 molecules are removed from 200mg of CO 2 , then the
number of moles of CO 2 left are
(a)
10.
2.85  10 3
[IIT 1983]
(b) 28.8  10 3
(c) 0.288  10 3
(d) 1.68  10 2
The set of numerical coefficient that balances
equation K 2 CrO4  HCl  K 2 Cr2 O7  KCl  H 2 O is
the
[Kerala CEE 2001]
1.
(a) 1, 1, 2, 2, 1
(c) 2, 1, 1, 2, 1
Mixture of sand and sulphur may best be separated by
[Kerala CET 2001]
(a) Fractional crystallisation from aqueous solution
(b) Magnetic method
(c) Fractional distillation
11.
(d) Dissolving in CS 2 and filtering
2.
3.
4.
13.
The molar heat capacity of water at constant pressure is 75
JK 1 mol 1 . When 1.0 kJ of heat is supplied to 100 g of water
6.
7.
[JIPMER 2000]
(a)
6.023  10
21
molecules of CO 2
of
Na 2 S 2O3
using
K2Cr2O7
(a)
MW / 2
(b) MW / 3
(c)
MW / 6
(d) MW / 1
by
[IIT 2000]
3.92 g of ferrous ammonium sulphate crystals are dissolved in
100 ml of water, 20 ml of this solution requires 18 ml of
KMnO4 during titration for complete oxidation. The weight of
[Tamilnadu CET 2002]
14.
which is free to expand, the increases in temperature of water is
(a) 6.6 K
(b) 1.2 K
(c) 2.4 K
(d) 4.8 K
A compound possesses 8% sulphur by mass. The least
molecular mass is
[AIIMS 2002]
(a) 200
(b) 400
(c) 155
(d) 355
Which of the following contains maximum number of atoms
standardization
KMnO4 present in one litre of the solution is
(d) NO
H2
In
(d) 12.15  10 3
iodometry, the equivalent weight of K2Cr2O7 is
3
contain the same number of molecules. The gas X is
(a) CO
(b) CO 2
5.
12.
10 dm of N 2 gas and 10 dm of gas X at the same temperature
(c)
One litre hard water contains 12.00 mg Mg 2  milli equivalent
of washing soda required to remove its hardness is
(a) 1
(b) 12.15
(c) 1  10 3
Irrespective of the source, pure sample of water always yields
88.89% mass of oxygen and 11.11% mass of hydrogen. This is
explained by the law of
[Kerala CEE 2002]
(a) Conservation of mass
(b) Constant composition
(c) Multiple proportions
(d) Constant volume
Zinc sulphate contains 22.65% of zinc and 43.9% of water of
crystallization. If the law of constant proportions is true, then
the weight of zinc required to produce 20 g of the crystals will
be
(a) 45.3 g
(b) 4.53 g
(c) 0.453 g
(d) 453 g
3
(b) 2, 2, 1, 1, 1
(d) 2, 2, 1, 2, 1
15.
16.
(a) 3.476 g
(b) 12.38 g
(c) 34.76 g
(d) 1.238 g
A 100 ml solution of 0.1 n HCl was titrated with 0.2 N NaOH
solution. The titration was discontinued after adding 30 ml of
NaOH solution. The remaining titration was completed by
adding 0.25 N KOH solution. The volume of KOH required for
completing the titration is
[DCE 1999]
(a) 70[CBSE
ml PMT 2003]
(b) 32 ml
(c) 35 ml
(d) 16 ml
What volume of Hydrogen gas, at 273 K and 1 atm pressure
will be consumed in obtaining 21.6 g of elemental boron
(atomic mass = 10.8) from the reduction of boron trichloride by
Hydrogen
[AIEEE 2003]
(a) 22.4 L
(b) 89.6 L
(c) 67.2 L
(d) 44.8 L
The mass of 112 cm 3 of CH 4 gas at STP is
[Karnataka CET 2001]
[
Chemical Arithmetic
17.
(a)
0.16 g
(b) 0 .8 g
(c)
0.08 g
(d) 1 .6 g
Complete combustion of 0.858 g of compound X gives
7.
Reason
:
Under similar conditions of temperature and
pressure all gases contain equal number of
molecules.
Assertion
:
Reason
:
One atomic mass unit (amu) is mass of an
atom equal to exactly one-twelfth the mass of
a carbon-12 atom.
Carbon-12 isotope was selected as standard.
Assertion
:
Molecular mass of A is
2.63 g of CO 2 and 1.28 g of H 2 O . The lowest molecular
mass X can have
(a)
[Kerala MEE 2000]
(b) 86 g
43 g
(c) 129 g
18.
8.
(d) 172 g
In the following reaction, which choice has value twice that of
the equivalent mass of the oxidising agent
SO 2  H 2O  3 S  2 H 2O
(a) 64
(c) 16
[DPMT 2000]
2.
Assertion
:
Volume of a gas is inversely proportional to
the number of moles of a gas.
Reason
:
The ratio by volume of gaseous reactants and
products is in agreement with their mole
ratio.
[AIIMS 1995]
Assertion
:
Molecular weight of oxygen is 16.
Reason
:
Atomic weight of oxygen is 16.
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Assertion
:
Atoms can neither be created nor destroyed.
Reason
:
Equivalent
weight
of
an
Atomic weight of the element

Valency of the element
Reason
:
Under similar condition of temperature and
pressure, equal volume of gases does not
contain equal number of atoms.
Assertion
:
Reason
:
14.
Assertion
:
15.
Reason
Assertion
:
:
Reason
:
Assertion
:
Mass spectrometer is used for the
determination of isotopes.
Isotopes are the atoms of same element
differing in mass numbers.
Gases combine in simple ratio of their
volume but, not always.
Gases deviate from ideal behaviour.
Isomorphous substances form crystals of
same shape and can grow in saturated
solution of each other.
They have similar constitution and chemical
formulae.
Atomicity of oxygen is 2.
11.
12.
[AIIMS 1996]
3.
13.
[AIIMS 1994,2002]
4.
Assertion
:
One mole of SO 2 contains double the number of
molecules present in one mole of O 2 .
Reason
:
Molecular weight of SO 2 is double to that of
O2 .
5.
6.
Assertion
:
1.231 has three significant figures.
Reason
:
All numbers right to the decimal point are
significant.
Assertion
:
22.4 L of N 2 at NTP and 5.6 L O 2 at NTP
contain equal number of molecules.
mass of B is M.
Vapour density of A four times that of B.
Pure water obtained from different sources
such as, river, well, spring, sea etc. always
contains hydrogen and oxygen combined in
the ratio 1 : 8 by mass.
A chemical compound always contains
elements combined together in same
proportion by mass, it was discovered by
French chemist, Joseph Proust (1799).
As mole is the basic chemical unit, the
concentration of the dissolved solute is
usually specified in terms of number of moles
of solute.
The total number of molecules of reactants
involved in a balanced chemical equation is
known as molecularity of the reaction.
A certain element X, forms three binary
compounds
with
chlorine
containing
59.68%,68.95% and 74.75% chlorine
respectively. These data illustrate the law of
multiple proportions.
According to law of multiple proportions, the
relative amounts of an element combining
with some fixed amount of a second element
in a series of compounds are the ratios of
small whole numbers.
Equivalent weight of Cu in CuO is 63.6
and in Cu 2 O 31.8.
:
:
10.
1.
M
if the molecular
4
Reason
Assertion
9.
(b) 32
(d) 48
Read the assertion and reason carefully to mark the correct option out
of the options given below :
(a)
If both assertion and reason are true and the reason is the
correct explanation of the assertion.
(b)
If both assertion and reason are true but reason is not the
correct explanation of the assertion.
(c)
If assertion is true but reason is false.
(d)
If the assertion and reason both are false.
(e)
If assertion is false but reason is true.
21
16.
element
22 Chemical Arithmetic
17.
Reason
:
1 mole of an element contains 6.023  10 23
atoms.
Assertion
:
1 amu equals to 1.66  10 24 g .
Reason
1.66  10
:
24
1
th of mass of a
g equals to
12
C 12 atom.
1
a
2
b
3
b
4
c
5
a
6
b
7
a
8
c
9
d
10
c
11
b
12
c
13
c
14
d
15
a
16
a
17
b
18
b
19
a
20
a
21
b
22
b
23
c
24
a
25
a
26
c
27
d
Percentage composition & Molecular formula
1
a
2
a
6
b
7
b
Significant figures, Units for measurement,
Matter and Separation of mixture
3
b
4
d
5
a
Chemical stoichiometry
1
a
2
d
3
a
4
c
5
d
1
c
2
c
3
a
4
c
5
c
6
b
7
c
8
d
9
c
10
c
6
c
7
c
8
b
9
a
10
d
11
c
12
b
13
a
14
c
15
b
11
c
12
d
13
d
14
b
15
a
16
b
17
b
18
a
19
a
20
c
16
b
17
b
18
c
19
d
20
c
21
b
22
d
23
a
24
a
25
b
21
d
22
c
23
d
24
a
25
d
26
b
27
d
26
b
27
b
28
b
29
c
30
a
31
b
32
c
33
b
34
c
35
d
36
d
37
b
38
d
39
a
40
b
41
b
42
c
43
c
44
d
45
a
46
b
47
a
48
a
49
a
50
d
51
c
52
b
53
a
54
b
55
b
56
b
57
a
58
c
Laws of chemical combination
1
a
2
d
3
c
4
a
5
c
6
c
7
c
8
b
9
b
10
a
11
c
12
b
13
a
14
d
15
b
16
a
17
c
18
d
19
c
20
a
21
c
22
d
Critical Thinking Questions
Atomic, Molecular and Equivalent masses
1
d
2
b
3
b
4
a
5
c
1
c
2
b
3
a
4
a
5
b
6
b
7
b
8
b
9
a
10
d
6
c
7
d
8
b
9
a
10
b
11
a
12
c
13
a
14
d
15
c
11
a
12
b
13
a
14
c
15
b
16
c
17
a
18
b
16
c
17
a
18
d
19
a
20
a
21
b
22
b
23
d
24
c
25
a
26
a
27
c
28
d
29
a
30
c
1
e
2
e
3
c
4
e
5
d
31
a
32
d
33
b
34
a
35
c
6
d
7
a
8
c
9
a
10
b
36
b
37
b
38
c
39
a
40
b
11
a
12
e
13
e
14
a
15
a
41
c
42
d
43
a
44
d
45
d
16
b
17
a
46
b
47
c
48
c
49
b
50
b
51
a
52
b
53
b
54
c
55
b
56
a
57
d
58
d
59
a
60
d
61
c
62
d
63
b
The mole concept
Assertion & Reason
Chemical Arithmetic
Significant figures, Units of measurement,
Matter and Separation of mixture
6.
(c) As the given sulphate is isomorphous with ZnSO 4 .7 H 2 O
its formula would be MSO 4 .7 H 2 O .m is the atomic
weight of M, molecular weight of
4.
Force [MLT 2 ]
(c) Pressure 

 [ML1 T  2 ]
Area
[L2 ]
Energy per unit volume 
[ML2 T 2 ]
 [ML1 T  2 ]
[L3 ]
(29.2  20.2) (1 .79  10 5 ) 9 .0  1 .79  10 5

1 .37
1 .37
Least precise terms i.e., 9.0 has only two significant figures.
Hence, final answer will have two significant figures.
17.
(b)
18.
19.
(a) Pure ethyl alcohol  81.4  0.002  81.398 .
(a) JPa–1; Unit of work is Joule and unit of pressure is Pascal.
Dimension of Joule i.e. work  F  L  MLT 2  L

 ML2 T 2


1
1
1 1 A



 MLT 1
Pa Pressure F
F
A

 

  
So, JPa–1  ML2 T 2  L2  L  L3 .
22.
(d) 1 zepto  10 21
23.
(a) As we know that all non zero unit are significant number.
Therefore significant figure is 2.
24.
(a) Number of significant figures in 6.0023 are 5 because all
the zeroes stand between two non zero digit are counted
towards significant figures.
25.
(b) Given P  0.0030m , Q  2.40m & R  3000m In
P(0.0030) initial zeros after the decimal point are not
significant. Therefore, significant figures in P(0.0030) are 2.
Similarly in Q(2.40) significant figures are 3 as in this case
final zero is significant. In R  (3000) all the zeroes are
significant hence, in R significant figures are 4.
26.
(b) All the zeroes between two non zero digit are significatn.
Hence in 60.0001 significant figures is 6.
27.
(d) Round off the digit at 2nd position of decimal 3.929
= 3.93.
Laws of chemical combination
12.
(b)
X Y ⇌ R S
ng
mg
pg
qg
n  m  p  q by low of conservation of mass.
Atomic, Molecular and Equivalent masses
5.
(b) The atomic weight of sulphur =32
In SCl 2 valency of sulphur =2
So equivalent mass of sulphur 
32
 16 .
2
23
MSO 4 .7 H 2 O
 m  32  64  126  m  222
m
Hence % of M 
 100  9.87 (given) or
m  222
100m  9.87m  222  9.87 or 90.13m  222  9.87
222  9 .87
or m 
 24.3 .
90.13
24 Chemical Arithmetic
7.
 22.4gm of gas occupies 22.4L at S.T.P.
(d) For NaOH, M  N
N1 V1  100ml  1 N  100ml(N )
For H 2 SO 4 , N 2 V2  10ml  10 N  100ml(N )
21.
(b) Equivalent weight 
Hence, N 1 V1  N 2 V2 .
10 .
Molecular weight of
(b) 1 mole of CH 4 contains 4 mole of hydrogen atom i.e. 4g atom
of hydrogen.
2
11.
(a)
Na 2 SO 3  I 2  Na 2 S 4 O6  NaI
Molecular weight
Valency
COOH
|
C OOH
22.
2.5
(b) Valency of the element 

n  2  0. 5  1
E
M
M

M
n  factor
1
E
M
5
23.
(b)
13.
(a) Atomic mass 
10  19  81  11 190  891
1081


100
100
100
 10.81
14.
(c) 0.1M
(d) Molarity 
0 .25 
12.
AgNO3 will
react
0.1 M
with
24.
16.

NaCl to form
(c) wt. of metallic chloride  74.5
Equivalent weight of metal 
Molecular weight 
weight of metal  100  32  68 gm
"
"
wt. of metal
8
wt. of oxygen
=
26.
39
 35.5  39
35.5
6  10 23 molecules has mass  18 gm

27.
So molecular weight = 29
So, molecular formula of compound is NO
(d)  17gm NH 3 contains 6  10 23 molecules of NH 3
6  10 23
 4 .25
17
28.
 22.4 L of gas at S.T.P. weight  22.4  1.16
1.24gm P is present in =
220
 1 .24  2 .2 gm
124
x
40
x
 Avogadro no.  y (say)
Number of atoms of A 
40
40y
Or x 
Avogadro no.
(c) Number of moles of A 
2x
80
Number of atoms of B
2x
2
40 y

 Av.no. 

 Av.no.  y
80
80 Av.no.
(d) BaCO3  BaO  CO 2 
Molecular weight of BaCO3  137  12  3  16 =197
 197gm produces 22.4L at S.T.P.
22.4
 9 .85  1 .12 L at S.T.P.
 9.85 gm produces
197
 25.984  26
This molecular weight indicates that given compound is
C2 H 2 .
(a) Molecular weight  2  V.D  2  11.2  22.4
31  4
gm P is present in 220 gm P4 S 3
(124 )
Number of moles of B 
6  10 23  4 .25
 4  6  10 23 .
 No. of atoms 
17
(a)  1L of gas at S.T.P. weight 1.16g
18
 3  10 23 gm
6  10 23
(a) Choice (a) is P4 S 3

7.5
 22.4  28.96
5.8
 4.25 gm NH 3 contains =
(a)
68
 8  17 .
32
 3  10 26 kg .
(a)  5.8L of gas has mass  7.5 gm
 22.4L "
20 .
1 .25  1000
25  molecular weight
1 molecules has mass 
weight of metal
 35.5
weight of chlorine

19.
W (gm)  1000
V (ml )  molecular weight

 wt. of metal  74.5  35.5  39
18.
118.50
=3.
39.5
Equivalent weight of oxide 
25 .
126
 63 .
2
2  V .D
2  59.25

E  35.5
4  35.5
0.1
 0 .05 M .
2
wt. of chlorine = 35.5
17.
 2H 2O 
1 .25  1000
 200 .
0 .25  25
(c) Let weight of metal oxide = 100 gm
Weight of oxygen
= 32gm

0.1M NaNO 3 . But as the volume doubled, conc. of
NO 3 
22.4
 11.2  11.2 L .
22.4
 11.2gm of gas occupies
29.
(a) 14 gm N 3  ions have  8 N A valence electrons
Chemical Arithmetic
4.2gm of N 3  ions have 
30 .
Molecular weight of H 3 PO3  3  31  48  82
8 N A  4 .2
 2.4 N A
14
 Equivalent weight 
(c) [  Molecular weight of CuSO 4 .5 H 2O
 63.5  32  64  90  249.5 ]
6  10
23
37.
249.5  1  10
6  10 23
 1 ml at NTP has =
22
 41.58  10 1
 4.158
31.
38.
(a) (I) 1 molecule of oxygen

 1 molecule of O 2 has mass 
32
6  10 23
39.
(a) Molecular weight of C60 H122  12  60  122  1

6  10

1 molecule C60 H122 has mass
(II) 1 atom of nitrogen
2  6  10 23 atoms of N 2 has mass = 28gm
 1 atom of N 2 has mass
28

2  6  10 23
40 .
(III) 1  10 10 g molecular weight of oxygen
g atomic weight  2  1  10 10  2  10 10 g
(d)
41.
1 .520 x  17

0 .995
x 8
 1.520 x  1.520  8  0.995 x  0.995  17
42.
43.
(d) d 
M
M
; 1
or M = V; 18gm = 18ml
V
V
18
6  10 23
(a) 100gm caffeine has 28.9gm nitrogen

44.
28.9
 194  56.06 gm
100
No. of atoms in caffeine 
56.06
4.
14
(d) Molecular weight of (CHCOO )2 Fe  170
Fe present in 100 mg of (CHCOO )2 Fe
CP
 1 .4 so, given gas is diatomic
CV
56
 100mg  32.9 mg
170
This is present in 400 mg of capsule
32.9
 100  8 .2 .
% of Fe in capsule 
400

 No. of atoms  3.01  10 23  2  6.023  10 23 atoms
(b) The acid is dibasic.
64
 2 .8  10 3 gm
28
(c) 2.5 molal NH 4 OH means 2.5 moles of NH 3 in 1000 g
194gm caffeine has =
11.2L  3.01  10 23 molecules
36.
2.8  10 3 gm C 2 H 4 requires =
 3  10 23 cm 3 .
 2.89  10 5 coulomb
(a)

1 molecule of water has volume 
Then 1 gm ion N 3  (1 mole) carries
34.
28gm C 2 H 4 requires 64gm oxygen
6  10 23 molecule of water has volume =18 cc
(b) One ion carries 3  1.6  10 19 coulomb
 3  1.6  10 19  6.02  10 23

 0.25  22.4 L  5.6 L .
1.520 x  12.160  0.995 x  16.915
33.
(b) C2 H 4  2O2  2CO 2  2 H 2 O
Hence, 100 cc solution of NH 3 requires = 0.25 mole
or 0.525 x  4.755
4 .755
x
9.
0 .525
842
6  10 23
H 2 O (1000 cc of solution)
wt. of metal hydroxide
EM  EOH 

wt. of metal oxide
EM  EO 

molecule C60 H122 has mass = 842gm
 6.4  10 3 gm = 6.4kg.
So, order of increasing masses II  I  III  IV .
32.
23
 140.333  10 23 gm  1.4  10 21 gm .
 2.3  10 23 gm
(IV) 1  10 10 g atomic weight of copper
6 .4
 40 .
0 .16
 720  122  842
 5.3  10 23 gm

6 .023  10 23
22400
= 0.0002688  10 23  2.69  1019 .
(c) Sp. heat × atomic wt.= 6.4
0.16 × atomic wt.= 6.4
Atomic wt. 
6  10 23 molecule has mass  32 gm
Molecular weight 82
= 41.

Basicity
2
(b)  22400 ml at NTP has 6.023  10 23 molecule
molecules has weight  249.5 gm
1  10 22 molecules has weight 
25
45 .
26
(d) 1 atom has mass  10.86  10 kg
26 Chemical Arithmetic
 10.86  10 23 gm
3  10 23 atoms of C has mass 
6.023  10 23 atoms has mass
46.
= 10.86  10 23  6.023  10 23 = 65.40 gm
This is the atomic weight of Zn.
(b)  1mole (COOH )2 . 2 H 2 O has 96gm oxygen
 0.3 mole (COOH )2 . 2 H 2 O has 96  0.3  28.8 gm
47.
28.8
 1 .8 .
 No. of gram atoms of oxygen 
16
(c) Equimolecular proportion means both gases occupied equal
2 .24
volume 
 1 .12 L
2
For CH 4 :
5 4.
(c) 1mole of S has mass = 32gm
(d) 7.0 gm of Ag
So, lowest mass = 6gm of C.
(c) 1mole of any gas at STP occupies 22.4 L.
55.
(b) 

5 6. (a) 
16
 1 .12  0 .8 gm .
22.4
For C 2 H 6
5 7.
3 .0
30
gm  1 .5 gm
 1 .12 
2
22.4
Total mass  1.5 gm  0.8 gm  2.3 gm .
(c) Let wt. of metal oxide = 100 gm
wt. of metal = 53gm
wt. of oxygen = 47gm
wt. of metal
Equivalent weight of oxygen 
8
wt. of oxygen
53
 8  9 .02
47
2  V .D
2  66
132


 2.96  3
Valency 
E  35.5 9  35.5 44.5
 Atomic weight  Equivalentweight  Valency

(b)
5 2.
(a)
(b)
4.4
 22.4  44
2 .24
(d) 1L of air =210 cc O 2
1
 210  0.0093 .
22400
22.4L of a gas at STP has no. of molecules
(d) 
 6.023  10 23
 8.96L of a gas at STP has no. of molecules
6 .02  10 23  8 .96
 2.408  10 23  24.08  10 22 .
22.4
(a) Given equivalent weight of metal = 9
Vapour density of metal chloride = 59.25
 molecular weight of metal chloride

5 9.
 2  V.D  2  59.25  118.5
 valency of metal

molecular weight of metal chloride
equivalnet weight of metal  35.5
118.5
118.5

 2.66
9  35.5
44.5
Therefore atomic weight of the metal
=equivalent weight  valency
Valency of metal 
4
 9  2.66  23.9
2 PH 3  2 P  3 H 2
(solid)
2ml
3ml
100ml
150ml
Increase in volume  150ml  100ml  50ml increase.
molecular wt. of metal
volume
60 .
(d) The density of gas 
61.
45
 2 gm litre1
22.4
(c) Equivalent weight of bivalent metal = 37.2

 Atomic weight of metal  37.2  2  74.4
Mg  2 HCl  MgCl2  H 2
 Formula of chloride  MCl 2
 24g Mg evolves 22.4L H 2 at STP
Hence, molecular weight of chloride

5 3.
5 8.
Mn SO 4  Mn O2
Change of valency  4  2  2
M
.
 Equivalent weight 
2
5 1.
22.4L of gas has mass 
210 cc 
 9.02  3  27.06
(b) One gram of hydrogen combines with 80 gm of bromine.
So, equivalent weight of bromine = 80 gm
 4gm of bromine combines with 1 gm of Ca
1
 80gm of bromine combines with =  80  20 .
4
2
50 .
12g Mg evolves H 2 at STP
6  10 23  1 .12  10 7
22400
22400 cc = 1 mole
1.12L C 2 H 6 has mass 
49.
1.12  10 7 of gas at STP has
So given gas is CO 2 because CO 2 has molecular mass=44.
22.4L C 2 H 6 has mass = 30 gm
48.
22400 cc of gas at STP has 6  10 23 molecules
 .03  1014  3  1012 .
2.24L of gas has mass = 4.4gm

22.4L CH 4 has mass  16 gm
1.12L CH 4 has mass 
12  3  10 23
 6 gm
6  10 23
22.4
 12
24
=11.2L at STP.
(b) (a) 2gm atom of nitrogen = 28gm
(b) 6  10 23 atoms of C has mass  12 gm
(MCl2 )  74.4  2  35.5  145.4
62.
(c) As we know that
Equivalent weight 
weight of metal
8
weight of oxygen
Chemical Arithmetic

32
 8  64
0 .4
 1.2  10 23 atoms.
6.
(b) 44g CO 2 occupies 22.4L at STP
mol. wt
Vapour density 
2
4.4g CO 2 occupies 
Mol. wt  2  V.D  2  32  64
mol. wt 64
As we know that n 

1
eq. wt
64
7.
22.4
 4 .4 = 2.24L.
44
Mass
g
; 1
or g  ml
Volume
ml
0.0018ml = 0.0018gm
(a) D ensity 
Suppose, the formula of metal oxide be M 2 On . Hence the
No. of moles 
formula of metal oxide  M 2 O .
63.
(b) Molecular weight of NH 3 is 17
weight
0 .0018

 1  10  4
Molecular weight
18
No. of water molecules = 6.023  10 23  1  10 4

According to the mole concept
 6.023  1019 .
17 gm NH 3 has molecules  6.02  10 23
 1 gm NH 3 has molecules 
6 .02  10 23
17
Ca 3 P2  6 H 2 O  2 PH3  3Ca(OH)2
8.
(c)
9.
(d) Amount of gold  19.7kg  19.7  1000 gm =19700 gm
No. of moles 
 4.25 gm NH 3 has molecules
The mole concept
 6.023  10 25 atoms.
10 .
(c) 

(a) 16g O 2 has no. of moles 
16 1

32 2
14 1

28 2
No. of moles are same, so no. of molecules are same.
(b)
10 gm CaCO 3 =
6 .023  10 23
 10
100
1 molecule of CaCO 3 = 50 protons
6.023  10 22 molecule of CaCO 3  50  6.023  10 22
 3.0115  10 24
Na 2 SO 4 . 10 H 2 O  2  23  32  4  16  10  18
 46  32  64  180  322gm
322gm Na 2 SO 4 .10 H 2 O contains = 224 gm oxygen
11.
(b) 16gm of CH 4 = 1mole  6.023  10 23 molecules.
12.
(c)
According to avogadro's hypothesis equal volumes of all gases
under similar conditions of temperature and pressure contains
equal no. of molecules.
14.
(d)
d
32.2gm Na 2 SO 4 .10 H 2 O contains
=
3.
100 gm CaCO 3  6.023  10 23 molecules
 6.023  10 22 molecule
14g N 2 has no. of moles 
2.
19700
 100
197
 No. of atoms  100  6.023  10 23
6.02  10 23  4 .25

 1 .5  10 23 molecule
17
1.
32.2  224
 22.4 gm
322
(b) Molarity 
So, M  V
18gm = 18ml
18ml = N molecules (N = avogadro's no.)
A
1000ml 
10ml of this solution is diluted to 1000 ml N1 V1  N 2 V2
10  0.1  1000  x
x
0.1  10
 0 .001M .
1000
4.
(c) According to definition of molar solution  A molar solution is
one that contains one mole of a solute in one litre of the
solution.
5.
(a) 44g of CO has 2  6  10 23 atoms of oxygen
2
4.4g of CO has =
2
12  10 23
 4 .4
44
M
(d = density, M= mass, V =volume)
V
Since d = 1
W (gm)  1000
molecular wt.  V(ml.)
2.65  1000
 0 .1 M
106  250

27
15.
(a)
16.
(a) 
A
NA
 1000 = 55.555 N.
18
This is fact.

3 moles of oxygen is that in 1 mole of BaCO3
1.5 moles of oxygen is that in mole of BaCO3
1
1
 1 .5   0 .5 .
3
2
(b) The no. of molecules present in 1ml of gas at STP is known as
Laschmidt number.
22400 ml of gas has total no. of molecules

17.
A
 6.023  10 23
28 Chemical Arithmetic
1ml of gas has total no. of molecules 
6 .023  10 23
22400
25 .
(a) wt of CO 2  44
mol wt of CO 2  44
 2.69  1019 .
18.
(b) 

6 .02  10 23

 3 .01  10 23 molecule.
2
19.

13.5 gm of SO 2 Cl 2 
44
 6 .02  10 23  6.02  10 23
44
(c) No. of atoms in one molecule
26.
= no. of moles  6.022  10 23
27.
 1.4  6.022  10 23  8.432  10 23
(d) As we know that four sodium atom are present in sodium
ferrocyanide [ Na 4 Fe(CN )6 ]
1
 13.5  0 .1 .
135
Hence, number of Na atoms = No. of moles  number of
atom  Avogadro’s number
(a) (a) 34gm of water


2  4  6.023  10 23  48  10 23
18gm H 2 O = 6.023  10 23 molecule
6 .023  10 23
 34
18
34gm H 2 O =
 11.37  10 23 mole
Percentage composition & Molecular formula
1.
(a) 
 44gm CO 2  6  10 23 molecules
 32gm CH 3 OH  6  10 23 molecules
 46gm CH 3 OH 
3.
4.
23
molecules
(a) Urea- NH 2  CO  NH 2
5.
23.

24.
7.
(b) Glucose - C 6 H 12 O6
Ratio of C, H and O  1 : 2 : 1
In acetic acid CH 3  C  O  H
||
O
1170gm
Mol.wt.
1170
 32.05mole / litre
36.5
6.023  10 23 molecule of sucrose has
45  6.023  10 23 atoms/mole
Ratio of C, H and O 1 : 2 : 1 .
Chemical stoichiometry
(Mol. wt. of HCl =36.5)
(a) 1 mole of sucrose contains 6.023  10 23 molecules
 1 molecule of sucrose has 45 atoms

1
 11.97
0 .0835
(Empirical formula) n  Molecular formula
n = whole no. multiple i.e. 1,2,3,4..............
If n  1 molecular formula CH 2O2 .
(c) Molarity = mole/litre
 1cc contains 1.17gm
 1000 cc contains 1170 gm
28
 100  46.66 .
60
(b) Empirical formula of an acid is CH 2O2
2
1
 mole
16 8
4
1
 mole.
32 8
100 gm of urea contains
=12gm of hydrogen.
12 gm of H 2 is present in C 2 H12 O6
620gm of Na 2 O = 10 mole.
also 4g of sulphur 

 1gm mole of compound contain =
6.
(b) 2gm of oxygen contains atom 
60 gm of urea contains 28gm of nitrogen
So, Empirical formula  CH 2O
(Simplest formula).
(a)  0.0835 mole of compound contains 1 gm of hydrogen
Molecular weight = 46 +16 = 62
22.

(b) Based on facts.
(d) C  24 gm , H  4 gm , O  32 gm
6  10 23
 54  3  10 23 molecules.
 54gm of N 2 O5 
108
(b) Sodium oxide  Na 2 O
62gm of Na 2 O = 1 mole
16
 100 =40% oxygen.
40
So, Molecular formula  C2 H 4 O2
6  10 23
 46  8.625  10 23
32
(d)  108gm of N 2O5  6  10
21.
2.
6  10 23

 28  3 .8  10 23
44
(c) 46gm of CH 3 OH
40 gm NaOH contains 16gm of oxygen
 100 gm of NaOH contains
(b) 28gm of CO 2
 28gm CO 2
wt. of CO 2
 6 .02  10 23
mol wt of CO 2

(a) Molecular weight of SO 2 Cl 2
 32  32  2  35.5 = 135 gm
 135 gm of SO 2 Cl 2 = 1gm molecule
20 .
No. of molecule 
2gm of hydrogen  6.02  10 23 molecules
1gm of hydrogen
1.
(c)
W(gm)  1000
V  Eq.wt.
1500 ml of 0.1N HCl = 150 ml (N)
N 
Chemical Arithmetic
1
2.
(c)
W(gm)  1000
150  40
, W (gm ) 
 6 gm .
150  40
1000
1
1
 V2 ; V2  1000ml
N 1 V1  N 2 V2 ;  200 
2
10
10 .
27 gm
(a)
11.
(c) In Fe(CNS )3 . 3 H 2 O
12.
3  18
 100  19% .
284
(d) 5 S  5 O 2  5 SO 2 ; 5 O 2  5 SO 2 ; 5  64  320 gm .
13.
(d) H 3 PO4 is tribasic so N  3 M  3  1  3 .
14.
(b) H 2 SO 4 is dibasic N  2 M  2  2  4 .
15 .
(a) For Dibasic acid E 
% of H 2 O 

2 Ag2CO 3  4 Ag  2CO 2  O2
2  276 gm
4  108 gm
 2  276 gm of Ag2CO 3 gives 4  108 gm
 1 gm of Ag2CO 3 gives 
4  108
2  276
 2.76 gm of Ag2CO 3 gives
N
4  108  2.76
 2.16 gm
2  276
4.
16.
(b)
Oxygen is limiting reagent
So, X 
1
 0 .2 all oxygen consumed
5
18.
67200 gm Hb =
6.
(c)
(NH 4 )2 SO 4  2 NH 3 
132
N 
10  1.71  80  98
 27.9
98  49
2 KMnO4  10 FeSO 4  8 H 2 SO 4 
K 2 SO 4  2 MnSO 4  5 Fe2 (SO 4 )3  8 H 2 O
2 HCl
2(36 .5 )73 gm
10  sp. gr. of the solution wt.% of solute Mol.wt.
Molecular wt. of solute Eq. wt.
2 FeSO 4  H 2 SO 4  [O]  Fe2 (SO 4 )3  H 2 O]  5
[Mohrsalt]
67200  0 .33
gm Fe
100
672  0 .33
4.
56
gm atom of Fe =
N
K2 SO 4  2 MnSO4  3 H 2 O  [O]
(c)  100gm Hb contain = 0.33gm Fe

W  1000
E  V (in ml )
(c) 2 KMnO4  3 H 2 SO 4 
Left NH 3  1  4  0.2  0.2 .
5.
M 200

 100
2
2
1
W  1000

 W  1gm .
10 100  100
(c)
4 NH 3(g )  5 O2(g )  4 NO( g)  6 H 2 O( g )
t0
1
1
0
0
t  t 1  4x 1  5x
4x
6x
3
H2
2
3
22 .4 33 .6 L
2
Volume of water added  1000  200  800ml .
3.
(d) H 2 O  Al  NaOH  NaAlO2 
29
19.
Mohr-salt reducing agent KMnO4 / H   oxidising agent
(d) Atomic weight = Equivalent weight × Valency
73 g HCl  132 g(NH 4 )2 SO 4
292 g HCl  528 g(NH 4 )2 SO 4
7.
(c)
2(NH 4 )2 HPO4  P2 O5
2(36 1  31  64 ) 264
% of P2 O 5 
wt. of P2 O 5
 100
wt of salt

8.
(b)
2 Al 
62  80 142
26.89


 8.9  3  26.7  Valency 
 3 .
8.9


20 .
21.
MW  2  V.D.  2  22  44 .
(d) 2 KMnO4  3 H 2 SO 4  K 2 SO 4  2 MnSO 4  3 H 2 O  5[O]
(c)
Change by 5
142
 100  53.78% .
264
3
O 2  Al2 O 3
2
According to equation
Eq. wt. 
22.
0 .16
25
W
1
 1000 
 25 ;
 1000 
E
10
E
10
M  2  E  2  64  128 .
HCl
Al .
9.
100  78.4
 1.568  10 4
0 .5
Minimum m.w.  molecule at least contain one selenium.
Mol.wt.
5
(c) Dibasic acid NaOH; N 1 V1  N 2 V2
3
mole of O 2 combines with 2 mole
2
2 mole Al = 54gm
(a) 0.5 gm Se  100 gm peroxidase anhydrous enzyme
2
7
23.
(d) NaOH
N 1 V1  N 2 V2 ; 20 
78.4gm Se 
24.
(a)
1
1

 V ; V = 40 ml.
10 20
NV  N 1 V1  N 2 V2
0.2  2  0.5 x  0.1(2  x )
0. 4  0. 5 x  0. 2  0. 1 x
0.2  0.4 x
30 Chemical Arithmetic
x
25 .
(d)
35 .
1
L  0.5 L
2
M1 V1 M 2 V2 0 .1  10 M 2 V2
;
; M 2 V2  5 .


1
5
n1
n2
NV  N1V1  N 2 V2  N 3 V3
N  1000  1  5 
N  0 .025 
26.
(d) KMnO4  Mohr salt
(b)
t0
t t
1
1
 20   30  5  10  10  25
2
3
(d) The equivalent weight of H 3 PO4 
37.
98
 49
2
(b) Ba(OH)2  CO 2  BaCO3  H 2O
N
.
40
NH 3(g )  HCl( g )  NH 4 Cl(s)
20ml
40ml
0
0
20 ml
solid
molecular weight
2
 mole wt of H 3 PO4 = 3 + 31 + 64 = 98
36.

Atomic wt. of BaCO3 = 137  12  16  3 = 197
Final volume = 20 ml .
27.
(b)
No. of mole 
M 1 V1
M 2 V2 20  0 .1 M 2 V2
;
; M 2 V2  5 .


2
5
n1
n2
 205 mole of Ba(OH )2 will give .205 mole of BaCO3
(b) Acidic medium E 
29.
(c) 0.1 M AgNO3 will react with 0.1 M NaCl to form 0.1 M
NaNO 3 . But
 wt. of 0.205 mole of BaCO3 will be
M 158

 31.6 gm .
5
5
28.
30 .
wt. of substance
mol wt.
1
mole
of
Ba(OH )2 gives 1 mole of BaCO3

KMnO4 Oxalic acid
.205  197  40.385 gm  40.5 gm
38.
10  10 3 gm
 1000 =0.25 N
40  1
V1  500ml ,
V2  ?
N1V1  N 2 V2 ; 0.5  500  0.25  V2
(b) (I) Phenopthalein indicate partial neutralisation of
Na 2 CO 3  NaHCO 3
Meq. of Na 2 CO 3 + Meq. of NaOH = Meq. of HCl
W
W
 1000   1000  NV
E
E
39.
V2  1000mL final volume water added = 1000 – 500
= 500 mL.
(a) eq. of KMnO4 = eq. of Fe(C2 O4 )
x 5  1 3
x  0.6
40 .
(b)
(Suppose Na 2 CO 3  a gm , NaOH = b gm)
Element
At.wt.
Mole
Ratio
Empirical
formula
a
b
 1000 
 1000  300  0 .1 .....(1)
106
40
(II) Methyl orange indicate complete neutralisation
C =86%
H =14%
12
1
7.1
14
1
2
CH
HCl HCl
41.
N 1 V1  N 2 V2 , 25  0.2  0.1  V2 so V2  50ml excess
32.
N1  0.5 N  10mg per mL
N2 
0 .1
NO 3 
 0 .05 M
2
(a) Acid
base
N 1 V1 = N 2 V2 ; N 1  30  0.2  15 ; N1  0.1 N
31.
(d)
as the volume is doubled, conc. of
CuSO 4  Cu 2  S 2   CuS
(H 2 S )
[100×1=100 millimole]
 1 mole  1 mole
 100 millimole  100 millimole H 2 S required
a  Na 2 CO 3  0.53 gm .
M1V1 M 2 V2

( NaOH )
n1
n2
33.
34.
1  V1 1  10

; V1  5ml .
1
2
(c) Atom in highest oxidation state can oxidize iodide to liberate
I2 which is volumetrically measured by iodometric titration
using hypo.
2 I   I2
Pb2  Lowest oxidation state can not oxidise iodide to I2 .
AgNO3  2 Ag   S 2   Ag2 S
(H 2 S )
[100×1 =100 millimole]
 2 mole  1 mole
 100 miliimole  50 millimole H 2 S required
a
b
 1000 
 1000  350  0.1 .....(2)

53
40
From (1) and (2) b =1gm.
(c) From solution of (31)
From equation (1)
(b) (H 2 SO 4 )
(b)
2
Beleongs to
alkene Cn H 2n
50
1
 .
100 2
Ratio
42.
(c) At room temperature 2 H 2(g) 
O2(g)  2 H 2 O(l)
t =0
t =t
50 ml
50 – 2x
=0
In this case H 2 is limiting reagent
50 ml
0
50 – x 2x
25 gases (50)liquid
x = 25 ml
At 110°C
t =t
2 H 2(g)  O2(g)  2 H 2O(g)
0
25 ml
50 ml
Vgas  75 ml
Chemical Arithmetic
2
43.
(c)
2
2
CuSO 4  2 KI  K 2 SO 4  CuI 2 ;
1.12 mL is obtained from 4.12 mg
1
2 CuI 2  CuI 2  I2
 22400 mL will be obtained from
I2  2 Na 2 S 2O3  2 NaI  Na 2 S 4 O6
4 .12
 22400 mg  84.2 g
1.12
Eq. wt. Of CuSO 4 .5 H 2O  Mol.wt.  250
100 ml of 0.1 N hypo  100 ml of 0.1 N CuSO 4 .5 H 2O

44.
(d)
(b)
Element
X
Y
HNO3  KOH  KNO 3  H 2O
(a) Isobutane and n-butane
formula; C4 H10
46.
(b) n 
C4 H10 
have same molecular
5  208
O2 
 17.9 kg
58
(b) Cd 2 
Ratio = 2 : 1
5 6.
(b) Mg 2  H 2
n
3ml (O)  1ml O 3
5 7.
(a)
Turpentine oil absorb ozone.
(a) 50% HCl itself means 50 gm HCl react with 100 gm sample
50
 100  50% .
100
AgNO3  HCl  AgCl  HNO3
KMnO4
5 8.
1
5
ROH  CH 3 MgI  CH 4  Mg
1 mol.
(c)
CaCO 3  2 HCl  CaCl 2  CO 2  H 2 O
2N
100 g
44 g
100 g CaCO 3 with 1 N HCl gives 22 g CO 2
Critical Thinking Questions
3.
2
1
1

 10 
  0.2
10
10 5
(c)
1 .5
 3 mole
0 .5
24  3  72 gm .
2.
M 1 V1
n
M 2 V2
=
; M1 V1  1 M 2 V2
n1
n2
n2
5 1.
1
O 2  MgO
2
0 .5 mole
0.5 mole of oxygen react with 1 mole of Mg
Mg 
1mole
1.5 mole of oxygen react with
=14.345 gm
0.1mole
FeSO4
For (d), M1V1  0 .02  10 
12 gm 1
 mole of H 2
24 gm 2
100 g CaCO 3 with 2 N HCl gives 44 g CO 2
30
500  0 .2
170
1000
t =0 0.176 mole 0.1 mole limiting
t =t 0.076 mole 0
(d)
S 2   CdS
Cu 2  S 2   CuS
20×0.5 =10
O2
H2
0 .25
0 .25
9
mole
100
20×1= 20
% Purity 
50.
1
At NTP Vol. CO 2  0.09  22.4  2.016 L .
V of O2  V of O3  135  10  145ml
(a)
2.5
CaCO 3  CO 2  0.09 mole
55.
150  10
x
 15ml
100
49.
20
A3 (BC4 )2  3  2  [5  (2  4 )]2  0 .
90% pure 9gm 
30ml (O)  10 ml O 3
48.
50
(b) CaCO 3  CaO  CO 2
10 gm
16.8
 0 .75 mole of H 2 and O 2
22.4
(a) 
Ratio
2
5 4.
0.5 mole H 2 – 0.5 mole H 2 O = 9gm.
47.
a/b
5
(a)
13

O2  4 CO 2  5 H 2 O
2
2 H 2 O  2 H 2  O2 0 .75
2:1
2 mole H 2 – 2 mole H 2 O
At.wt.(b)
10
5 3.
For 58gm of C4 H10 208 gm O 2 is required then for 5 kg of
C4 H10
%(a)
50
Simplest formula  X 2 Y
12 . 6
 0.2 mole; HNO3  KOH
63
0.2 mole  0.2 mole
45 .
5 2.
250  0 .1  100
 2 .5 gm
100
0.2  56  11.2 gm .
31
(b) H 2 O contains H and O in a fixed ratio by mass. It illustrates
the law of constant composition.
(b) 100 g of ZnSO crystals are obtained from =22.65 g Zn
4
1g of ZnSO crystals will be obtained from 
4
20 g of ZnSO crystals obtained from 
4
4.
OR
I
1 mol  22400 cc
22.65
g Zn
100
22.65
 20  4 .53 g
100
(a) If same volume is occupied by the gas, the no. of molecules are
same, so no. of moles are same.
1 mole of N 2 gas  2  14  28 gm
32 Chemical Arithmetic
1 mole of CO gas  12  16  28 gm
5.
1000  100  4.17  t
6.
3 .92  1000 20
 W  1000 
 58  1000   18  392  1000  5 W=3.476gm/L.


14.
(d) Volume m of HCl neutralised by NaOH = (Caustic soda) = V1
1000
t
 2 .4 K .
100  4 .17
(b)  8gm sulphur is present in 100 gm of substance

7.
M1 V1 M 2 V2  W
 V


 1000   2
1
5
M V
 5
75
(c) Heat capacity of water per gram 
 4 .17
18
Q = mST
32gm sulphur will present =
(b) (a) 6.023  10
23
N1V1  N 2 V2 ; 0.1  V1  0.2  30 ; V1  60ml
V total (HCl ) = 100 ml
40 ml
40ml 0.1N HCl is now neutralised by KOH (0.25 N) 
molecules of CO 2
(HCl ) N1V1  N 2 V2 (KOH)
No. of atoms  3  6.023  10 21 = 18.069  10 21 atoms
0.1  40  0.25  V2 ;
(b) 22.4L of CO 2
No. of atoms = 6.023  10 23  3  18.069  10 23 atoms
15 .
(c)
(c) 0.44gm of CO 2
No. of moles 
0.44
1

 6.023  10 23 moles
44
100
B
 6.023  10 moles  3  6.023  10 atoms
21
8.
9.
(a) 200mg of CO 2 = 200  10 3  0.2 gm
= 6  10 23 molecules
0.2gm of CO 2 =
6  10 23
 0 .2  0 .0272  10 23
44
17.
(a) %C 
2 WH 2 O
2

 100 
18
W
18
Element
%(a)
At.wt.(b)
C
83.6
12
H
16.4
1
18.
2 K 2 Cr2 O 4  2 HCl  K 2 Cr2 O7  2 KCl  H 2 O
W
24
1000  Mg 2 ; EW 
 12
E
2
12  10
12
12.
(c)
3
Mol.wt.
Eq. wt. 
6
13.
(a)
KMnO4 = Mohr salt
M 64

 16 ;
4
4
Twice 16  2  32
Assertion & Reason
1.
(e) We know that from the reaction H 2  Cl2  2 HCl that the
ratio of the volume of gaseous reactants and products is in
agreement with their molar ratio. The ratio of
H 2 : Cl2 : HCl volumes is 1 : 1 : 2 which is the same as
 1000  1 .
K 2 Cr2 O7  4 H 2 SO 4  K 2 SO 4 Cr2 (SO 4 )3
 6/two atom
 12/two atom
Change by 6
4 H 2 O  3[O]
7
(b) SO 2  2 H 2 O  S  2 H 2 O2
0
4
EW 
 2.85  10 3 .
(a) Meq of Mg 2  Meq of washing soda
1 .28
 100  16.4%
.858
a/b
Ratio
6.96
1
×3
16.4
2.3

C3 H7  12  3  7  43 gm .
Now, 6.023  10 23 molecules = 1mole
11.
12 WCO 2
12
2.63

 100 

 100  83.6%
44
W
44 0.858
%H 
 10 21 (2.72  1) = 1.72  10 21 molecules
(d)
W
V
W
112


;
; W  0.08 gm .
M 22400 16 22400
(c)
So remaining molecules  2.72  10 21  10 21
10 .
n
16.
Now 10 21 molecule are removed.
1  1 .72  10 21
 0 .285  10  2
6 .023  10 23
3
mole ; 2 mole – 3 mole
2
V  3  22.4  67.2 L .
 2.72  10 21 molecule
1.72  10 21 molecules 
3
21.6
H 2  B  3 HCl ; B 
 2 mole
2
10.8
3
H2
2
1mole 
18.069  10 21 atoms
(b) It is about 22.4L.
V2  16ml .
BCl3  3[H ]  B  3 HCl
BCl3 
21
44gm of CO 2
= 60 ml
V1
100
 32  400 .
8
their molar ratio. Thus volume of gas is directly related to the
number of moles. Therefore, the assertion is false but reason is
true.
2.
(e) We know that molecular weight of substance is calculated by
adding the atomic weight of atoms present in one molecules.
We also know that molecular weight of oxygen (O2 ) =2x
(Atomic weight of oxygen)  2  16  32 a.m.u. Atomic
Chemical Arithmetic
weight of oxygen is 16, because it is 16 times heavier than1/12 of
carbon atom. Therefore assertion is false but reason is true.
2 HCl  H 2  I2 (Bimolecular)
th
3.
4.
(c) According to Dalton's atomic theory atoms can neither be
created nor destroyed and according to berzelius hypothesis,
under similar condition of temperature and pressure equal
volumes of all gases contain equal no. of atom. Therefore
assertion is true but reason is false.

(a) Both assertion and reason are true and reason is the correct
explanation of assertion.
12.
(e) Equivalent wt. of Cu in CuO =
5.
6.
Molecular weight of SO 2 is double to that of O 2 .
(d) 1.231 has four significant figures all no. from left to right are
counted, starting with the first digit that is not zero for
calculating the no. of significant figure.
13.
(e) Mass spectrometer is the instrument used for the
determination of accurate atomic mass and the relative
abundance of the isotopes.
14.
(a) Both assertion and reason are true and reason is the correct
explanation of assertion.
15 .
(a) Example
16.
17.
6 .023  10 23  5 .6
22.4
1
 6 .023  10 23
4
According to avagadro's hypothesis equal volume of all gases
contain equal no. of molecules under similar condition of
temperature and pressure.
7.
(a) For universally accepted atomic mass unit in 1961,
C-12
was selected as standard. However the new symbol used is ' v'
(unified mass) in place of amu.
8.
(c) Vapour density of B 
M
,
2
Vapour density of A  4 
M
 2M
4
Molecular mass of A  2  2 M  4 M .
9.
(a) Pure water always contains hydrogen and oxygen in the ratio
1 : 8 by mass. This is in accordance with the law of constant
composition.
10 .
(b) The number of moles of a solute present in litre of solution is
known is as molarity ( M).
The total no. of molecules of reactants present in a balanced
chemical equation is known as molecularity. For example,
PCl5  PCl3  Cl2 (Unimolecular)
compounds
are
(b) No. of atoms present in a molecules of a gaseous element is
called atomicity.
For example, O 2 has two atoms and hence its atomicity is 2.
32 gm  6.023  10 23 molecules = 22.4L

isomorphous
ZnSO 4 . 7 H 2O, MgSO4 . 7 H 2O, FeSO4 . 7 H 2O (valency
of Zn, Mg, Fe =2).
Similarly, O2  2  16  32 gm ,
22.4 L  6.023  10 23 or 5 .6 L 
of
K2 SO 3 , K2CrO4 , K2 SeO 4 (valency of S, Cr, Se = 6) and
Now 22.4L of N 2  volume occupied by one mole of

63.6
=63.6
1
(Valency of Cu =1).
(d) Molar volume (at NTP) = 22.4 L
N 2  28 gm  6.023  10 23 molecules.
63.6
At.wt.
=31.8

2
Valency
Equivalent wt. of Cu in Cu 2 O 
Molecular weight of SO 2  32  2  16  64 gm .

Molarity and molecularity are used in different sense.
11.
(e) One mole of any substance corresponding to 6.023  10 23
entities is respective of its weight.
Molecular weight of O2  16  2  32 gm .
33
(a) 12gm of C-12 contain 6.023  10 23 atom

12
 10  23  1.66  10  24 .
6.023
34 Chemical
1.
Arithmetic
A m ixture of sand and iodine can be separated by
8.
[Kerala CEE 2002]
2.
2.
(a) Cry stallisation
(b) Sublim ation
(c) Distillation
(d) Fractional distillation
The elem ent sim ilar to carbon is
(a) Mg
(b) Mn
(c) Sn
(d) Po
The law of m ultiple proportions was proposed by
(a) Lav oisier
(b) Dalton
(c) Proust
(d) Gay -Lussac
10.
11.
(d) PbSO4 , NiSO4
M is the m olecular weight of KMnO4 . The equiv alent
(a) M
(b) M / 3
(c) M / 5
(d) M / 7
An aqueous solution of 6.3 g of oxalic acid dihy drate is
m ade up of to 2 50 ml. The v olum e of 0.1
N NaOH required to com pletely neutralise 1 0 ml of this
[IIT 2001]
(a) 4 0 ml
(b) 2 0 ml
(c) 1 0 ml
(d) 4 ml
(b) Law of m ultiple proportions
(c) Law of reciprocal proportions
(a) 1 1 N
(b) 2 2 N
(c) 3 3 N
(d) 4 4 N
(d) Gay -Lussac’s law of gaseous v olum es
One sam ple of atmospheric air is found to hav e 0.03 % of
carbon dioxide and another sam ple 0.04 %. This is
ev idence that
12.
The equivalent weight of phosphoric acid (H 3 PO4 ) in the
reaction, NaOH  H 3 PO4  NaH 2 PO4  H 2O is
[A IIMS 1999; BHU 2005]
(a) The law of constant com position is not alway s true
(a) 2 5
(b) 49
(b) The law of m ultiple proportions is true
(c) 59
(d) 9 8
(c) Air is a com pound
13.
(d) Air is a m ixture
Volume of 0.6 M NaOH required to neutralize 30 cm 3 of
0.4 M HCl is
One part of an elem ent A com bines with two parts of
another B. Six parts of the element C com bine with four
parts of the element B. if A and C com bine together the
ratio of their weights will be governed by
[A MU 1984]
(a) 30 cm
3
(c) 50 cm 3
14.
(a) Law of definite proportion
(b) Law of m ultiple proportion
7.
(c) ZnSO 4 , MgSO4
The normality of orthophosphoric acid hav ing purity of
70% by weight and specific gravity 1.54 would be[CPMT 1992]
(a) Law of constant composition
6.
(b) MgSO4 , CaSO 4
solution is
1 L of N 2 com bines with 3 L of H 2 to form 2 L of
NH 3 under the sam e conditions. This illustrates the
5.
(a) ZnSO 4 , SnSO 4
weight of KMnO 4 when it is conv erted into K 2 MnO4 is
[IIT 1992]
4.
9.
Crystals of which pair are isom orphous [MP PMT 1985]
[KCET 1995]
(b) 20 cm
(d) 45 cm 3
One m ole of potassium dichrom ate com pletely oxidises
the following num ber of m oles of ferrous sulphate in
acidic m edium
[MP PET 1998]
(c) Law of reciprocal proportion
(a) 1
(b) 3
(d) Law of conserv ation of m ass
(c) 5
(d) 6
The m aximum amount of BaSO 4 precipitated on m ixing
equal v olum es of BaCl2 (0.5 M) with H 2 SO 4 (1 M) will
correspond to
15.
3
The number of equivalents of Na 2 S 2O3 required for the
volumetric estimation of one equivalent of Cu 2  is
[A IIMS 1997]
[Kerala MEE 2000]
(a) 0.5 M
(b) 1 .0 M
(a) 1
(b) 2
(c) 1 .5 M
(d) 2 .0 M
(c) 3 /2
(d) 3
(SET -1)
Chemical Arithmetic
1.
(b) Iodine shows sublimation and hence v olatalizes on
heating, the vapour condenses on cooling to give pure
iodine.
2.
(c) Carbon and tin both are same group elements so have
sim ilarities in properties.
3.
(b) Law of m ultiple proportions was proposed by Dalton
and v erified by Berzelius.
4.
(d) Gay - Lussac's law : The volumes of the reacting gases
and those of the gaseous products bear the sim ple
ratio (also called the law of gaseous v olum es).
5.
(d)
6.
(c) The weights of two elements combining with a fixed
am ount of the third elem ent will bear the sam e
ratio(or sim ple m ultiple of it) in which they
them selv es react.
7.
10.
 W 1000 
 E  V   V1  N 2 V2


6.3 1000

 10  0 .1  V V= 4 0ml.
63 250
11.
solution/sam ple
V
12.
13.
Hence 0.5 mole will react with 0.5 mole of H 2 SO 4
8.
(c) Isom orphous substance m olecules contain the sam e
num ber of atom s bonded in sim ilar fashion.
9.
(a) KMnO4  K 2 MnO4

14.
(d) NaOH  H 3 PO4  NaH 2 PO4
(PO43 )
( NaPO42 )
MW
98

.
1
no. of ionisableH 
(b) NaOH HCl
(d) Cr2 O7  Cr 3  ; Fe   Fe  
n 1
n6
eq. of K2Cr2O7 = eq. of FeSO 4
6
 Equivalent weight of KMnO4
W 100
70  1000

N
 11 N .
d 1 .54
98  100 / 1.54
N1 V1  N 2 V2 ; 0.6  V1  0.4  30 ; V1  20ml .
i.e. BaCl2 is the limiting reagent.
Change in 0.5 per atom  7  6  1
(a) 7 0% by weight 70 gm H 3 PO4  100 gm
EW 
One m ole of BaCl2 reacts with one m ole of H 2 SO 4 .
NaOH
N 1 V1  N 2 V2
(a) BaCl2  H 2 SO 4  BaSO4  2 HCl
7
(a) Oxalic acid
35
1  6  x 1
15.
(b) Cu 2   2 I   CuI 2 2CuI 2  Cu 2 I2  I2
I2  2 Na 2S 2 O3  2 NaI  Na 2 S 4 O6
M
Molecular weight of KMnO4

M.
1
Change of 0.5 per atom
Cu 2   2 Na 2 S 2O3
***
34 Chemical
1.
Arithmetic
A m ixture of sand and iodine can be separated by
8.
[Kerala CEE 2002]
2.
2.
(a) Cry stallisation
(b) Sublim ation
(c) Distillation
(d) Fractional distillation
The elem ent sim ilar to carbon is
(a) Mg
(b) Mn
(c) Sn
(d) Po
The law of m ultiple proportions was proposed by
(a) Lav oisier
(b) Dalton
(c) Proust
(d) Gay -Lussac
10.
11.
(d) PbSO4 , NiSO4
M is the m olecular weight of KMnO4 . The equiv alent
(a) M
(b) M / 3
(c) M / 5
(d) M / 7
An aqueous solution of 6.3 g of oxalic acid dihy drate is
m ade up of to 2 50 ml. The v olum e of 0.1
N NaOH required to com pletely neutralise 1 0 ml of this
[IIT 2001]
(a) 4 0 ml
(b) 2 0 ml
(c) 1 0 ml
(d) 4 ml
(b) Law of m ultiple proportions
(c) Law of reciprocal proportions
(a) 1 1 N
(b) 2 2 N
(c) 3 3 N
(d) 4 4 N
(d) Gay -Lussac’s law of gaseous v olum es
One sam ple of atmospheric air is found to hav e 0.03 % of
carbon dioxide and another sam ple 0.04 %. This is
ev idence that
12.
The equivalent weight of phosphoric acid (H 3 PO4 ) in the
reaction, NaOH  H 3 PO4  NaH 2 PO4  H 2O is
[A IIMS 1999; BHU 2005]
(a) The law of constant com position is not alway s true
(a) 2 5
(b) 49
(b) The law of m ultiple proportions is true
(c) 59
(d) 9 8
(c) Air is a com pound
13.
(d) Air is a m ixture
Volume of 0.6 M NaOH required to neutralize 30 cm 3 of
0.4 M HCl is
One part of an elem ent A com bines with two parts of
another B. Six parts of the element C com bine with four
parts of the element B. if A and C com bine together the
ratio of their weights will be governed by
[A MU 1984]
(a) 30 cm
3
(c) 50 cm 3
14.
(a) Law of definite proportion
(b) Law of m ultiple proportion
7.
(c) ZnSO 4 , MgSO4
The normality of orthophosphoric acid hav ing purity of
70% by weight and specific gravity 1.54 would be[CPMT 1992]
(a) Law of constant composition
6.
(b) MgSO4 , CaSO 4
solution is
1 L of N 2 com bines with 3 L of H 2 to form 2 L of
NH 3 under the sam e conditions. This illustrates the
5.
(a) ZnSO 4 , SnSO 4
weight of KMnO 4 when it is conv erted into K 2 MnO4 is
[IIT 1992]
4.
9.
Crystals of which pair are isom orphous [MP PMT 1985]
[KCET 1995]
(b) 20 cm
(d) 45 cm 3
One m ole of potassium dichrom ate com pletely oxidises
the following num ber of m oles of ferrous sulphate in
acidic m edium
[MP PET 1998]
(c) Law of reciprocal proportion
(a) 1
(b) 3
(d) Law of conserv ation of m ass
(c) 5
(d) 6
The m aximum amount of BaSO 4 precipitated on m ixing
equal v olum es of BaCl2 (0.5 M) with H 2 SO 4 (1 M) will
correspond to
15.
3
The number of equivalents of Na 2 S 2O3 required for the
volumetric estimation of one equivalent of Cu 2  is
[A IIMS 1997]
[Kerala MEE 2000]
(a) 0.5 M
(b) 1 .0 M
(a) 1
(b) 2
(c) 1 .5 M
(d) 2 .0 M
(c) 3 /2
(d) 3
(SET -1)
Chemical Arithmetic
1.
(b) Iodine shows sublimation and hence v olatalizes on
heating, the vapour condenses on cooling to give pure
iodine.
2.
(c) Carbon and tin both are same group elements so have
sim ilarities in properties.
3.
(b) Law of m ultiple proportions was proposed by Dalton
and v erified by Berzelius.
4.
(d) Gay - Lussac's law : The volumes of the reacting gases
and those of the gaseous products bear the sim ple
ratio (also called the law of gaseous v olum es).
5.
(d)
6.
(c) The weights of two elements combining with a fixed
am ount of the third elem ent will bear the sam e
ratio(or sim ple m ultiple of it) in which they
them selv es react.
7.
(a) BaCl2  H 2 SO 4  BaSO4  2 HCl
14.
(d) Cr2 O7  Cr 3  ; Fe   Fe  
n 1
n6
eq. of K2Cr2O7 = eq. of FeSO 4
1  6  x 1
15.
(b) Cu
i.e. BaCl2 is the limiting reagent.
9.
(a) KMnO4  K 2 MnO4
7
6
Change in 0.5 per atom  7  6  1
 Equivalent weight of KMnO4

10.
M
Molecular weight of KMnO4

M.
1
Change of 0.5 per atom
(a) Oxalic acid
NaOH
N 1 V1  N 2 V2
 W 1000 
 E  V   V1  N 2 V2


6.3 1000

 10  0 .1  V V= 4 0ml.
63 250
11.
(a) 7 0% by weight 70 gm H 3 PO4  100 gm
solution/sam ple
V
12.
W 100
70  1000

N
 11 N .
d 1 .54
98  100 / 1.54
(d) NaOH  H 3 PO4  NaH 2 PO4
(PO43 )
( NaPO42 )
EW 
13.
 2 I  CuI 2 2CuI 2  Cu 2 I2  I2
Cu 2   2 Na 2 S 2O3
Hence 0.5 mole will react with 0.5 mole of H 2 SO 4
(c) Isom orphous substance m olecules contain the sam e
num ber of atom s bonded in sim ilar fashion.

I2  2 Na 2S 2 O3  2 NaI  Na 2 S 4 O6
One m ole of BaCl2 reacts with one m ole of H 2 SO 4 .
8.
2
MW
98

.
1
no. of ionisableH 
(b) NaOH HCl
N1 V1  N 2 V2 ; 0.6  V1  0.4  30 ; V1  20ml .
***
35
36 Structure of atom
Chapter
2
Structure of atom
John Dalton 1808, believed that matter is made up of extremely
minute indivisible particles, called atom which can takes part in chemical
reactions. These can neither be created nor be destroyed. However, modern
researches have conclusively proved that atom is no longer an indivisible
particle. Modern structure of atom is based on Rutherford’s scattering
experiment on atoms and on the concepts of quantization of energy.
Composition of atom
The works of J.J. Thomson and Ernst Rutherford actually laid the
foundation of the modern picture of the atom. It is now believed that the atom
consists of several sub-atomic particles like electron, proton, neutron, positron,
neutrino, meson etc. Out of these particles, the electron, proton and the neutron
are called fundamental subatomic particles and others are non-fundamental
particles.
(vi) Cathode rays heat the object on which they fall due to transfer
of kinetic energy to the object.
(vii) When cathode rays fall on solids such as Cu, X  rays are
produced.
(viii) Cathode rays possess ionizing power i.e., they ionize the gas
through which they pass.
(ix) The cathode rays produce scintillation on the photographic
plates.
(x) They can penetrate through thin metallic sheets.
(xi) The nature of these rays does not depend upon the nature of gas
or the cathode material used in discharge tube.
(xii) The e/m (charge to mass ratio) for cathode rays was found to

Electron ( e )
(1) It was discovered by J.J. Thomson (1897) and is negatively
charged particle. Electron is a component particle of cathode rays.
(2) Cathode rays were discovered by William Crooke's & J.J.
Thomson (1880) using a cylindrical hard glass tube fitted with two metallic
be the same as that for an e (1.76  10 8 coloumb per
gm). Thus, the cathode rays are a stream of electrons.
(xiii) According to Einstein’s theory of relativity, mass of electron in
Rest mass of electron(m)
motion is, m  
[1  (u / c)2 ]
electrodes. The tube has a side tube with a stop cock. This tube was known
as discharge tube. They passed electricity (10,000V) through a discharge
Where u = velocity of electron, c= velocity of light.
When u=c than mass of moving electron =.
o
–1
tube at very low pressure ( 10 2 to 10 3 mm Hg) . Blue rays were
emerged from the cathode. These rays were termed as Cathode rays.
(3) Properties of Cathode rays
(i)
Cathode rays travel in straight line.
(ii) Cathode rays produce mechanical effect, as they can rotate the
wheel placed in their path.
(iii) Cathode rays consist of negatively charged particles known as
electron.
(iv) Cathode rays travel with high speed approaching that of light
(v)
Proton ( H , H , P)
(1) Proton was discovered by Goldstein and is positively charged
1
+
1
particle. It is a component particle of anode rays.
(2) Goldstein (1886) used perforated cathode in the discharge tube
and repeated Thomson's experiment and observed the formation of anode
rays. These rays also termed as positive or canal rays.
(3) Properties of anode rays
(i) Anode rays travel in straight line.
(ii) Anode rays are material particles.
(iii) Anode rays are positively charged.
(ranging between 10 9 to 10 11 cm/sec)
Cathode rays can cause fluorescence.
Table : 2.1 Comparison of mass, charge and specific charge of electron, proton and neutron
Structure of atom
Name of constant
Unit
Electron(e )
0.000546
9.109 × 10
1/1837
Proton(p )
1.00728
1.673 × 10
1
Neutron(n)
1.00899
1.675 × 10
1
– 1.602 × 10
– 4.8 × 10
–1
1.76 × 10
+1.602 × 10
+4.8 × 10
+1
9.58 × 10
Zero
Zero
Zero
Zero
2.17  10 17
1.114  1014
1.5  10 14
–
Amu
Mass (m)
Kg
Charge(e)
Specific charge (e/m)
Density
+
–31
Relative
Coulomb (C)
Esu
Relative
C/g
Gram / cc
37
–27
–19
–27
–19
–10
–10
8
4
 The atomic mass unit (amu) is 1/12 of the mass of an individual atom of 6 C 12 , i.e. 1.660  10 27 kg .
Table : 2.2 Other non fundamental particles
Particle
Symbol
Nature
Charge esu
Mass
10–10
(amu)
Discovered by
Positron
e  , 1e 0 ,  
+
+ 4.8029
0.0005486
Anderson (1932)
Neutrino

0
0
< 0.00002
Pauli (1933) and Fermi (1934)
Anti-proton
p
–
– 4.8029
1.00787
Positive mu meson

+
+ 4.8029
0.1152
Yukawa (1935)
Negative mu meson

–
– 4.8029
0.1152
Anderson (1937)
Positive pi meson

+
+ 4.8029
0.1514
Negative pi meson

–
– 4.8029
0.1514
0
0
0
0.1454
Neutral pi meson
(1) Atomic number or Nuclear charge
(v) Anode rays also affect the photographic plate.
(i) The number of protons present in the nucleus of the atom is
called atomic number (Z).
(vi) The e/m ratio of these rays is smaller than that of electrons.
(ii) It was determined by Moseley as,
(vii) Unlike cathode rays, their e/m value is dependent upon the
nature of the gas taken in the tube. It is maximum when gas present in the
tube is hydrogen.
Where,   X  ray’s frequency
Neutron ( n , N)
o
(1) Neutron was discovered by James Chadwick (1932) according to
the following nuclear reaction,
or 5 B 11  2 He 4  7 N 14  o n 1
(2) Neutron is an unstable particle. It decays as follows,
 1 H 1 
Proton
0
0
1 e 
0
electon antinutrino
Z Fig. 2.1
Z= atomic number of the metal a & b are constant.
1
Be 9  2 He 4  6 C 12  o n 1
 s 1
  a(Z  b) or aZ  ab
(viii) These rays produce flashes of light on ZnS screen.
1
0n
neutron
Powell (1947)
Atomic number, Mass number and Atomic species
(iv) Anode rays may get deflected by external magnetic field.
4
Chamberlain Sugri (1956) and Weighland (1955)
(iii) Atomic number = Number of positive charge on nucleus =
Number of protons in nucleus = Number of electrons in nutral atom.
(iv) Two different elements can never have identical atomic number.
(2) Mass number
Mass number (A) = Number of protons or Atomic number (Z) +
Number of neutrons or Number of neutrons = A – Z .
(i) Since mass of a proton or a neutron is not a whole number (on
atomic weight scale), weight is not necessarily a whole number.
(ii) The atom of an element X having mass number (A) and atomic
number (Z) may be represented by a symbol,
Z
XA.
Table: 2.3 Different types of atomic species
Atomic species
Similarities
Differences
Isotopes
(i) Atomic No. (Z)
(i) Mass No. (A)
(Soddy)
(ii) No. of protons
(ii) No. of neutrons
Examples
(i) 11 H , 12 H , 13 H
38 Structure of atom
(iii) No. of electrons
(iii) Physical properties
(ii)
16
17
18
8 O, 8 O, 8 O
(iv) Electronic configuration
(iii)
(v) Chemical properties
35
17
37
Cl, 17
Cl
(vi) Position in the periodic table
(i) Mass No. (A)
(i) Atomic No. (Z)
(i)
(ii) No. of nucleons
(ii) No. of protons, electrons and
neutrons
(ii)
Isobars
40
18
40
Ar, 19
K, 40
20 Ca
130
52
130
Te , 130
54 Xe, 56 Ba
(iii)Electronic configuration
(iv) Chemical properties
(v) Position in the perodic table.
No. of neutrons
(i) Atomic No.
(i)
30
14
31
32
Si, 15
P, 16
S
(ii) Mass No., protons and electrons.
Isotones
(iii) Electronic configuration
(ii)
39
19
(iv) Physical and chemical properties
(iii)
3
1
(iv)
13
14
6 C, 7
(v) Position in the periodic table.
Isotopic No.
(N – Z) or (A – 2Z)
Isodiaphers
H , 42 He
N
(i) At No., mass No., electrons, protons,
neutrons.
(i) 92 U 235 , 90 Th 231
(ii) Physical and chemical properties.
(ii)
(iii)
(i) No. of electrons
K, 40
20 Ca
At. No., mass No.
19
K 39 , 9 F19
29 Cu
65
, 24 Cr 55
(i) N 2 O, CO 2 , CNO  (22e  )
(ii) Electronic configuration
(ii) CO , CN  , N 2 (14 e  )
Isoelectronic species
(iii) H  , He, Li  , Be 2  (2e  )
(iv) P 3  , S 2 , Cl  , Ar, K  and Ca 2 (18 e  )
(i) No. of atoms
(i) N 2 and CO
(ii) No. of electrons
(ii) CO 2 and N 2 O
(iii) Physical and chemical properties.
Isosters
(iii) HCl and F2
(iv) CaO and MgS
(v) C 6 H 6 and B3 N 3 H 6
Electromagnetic radiations
(1) Light and other forms of radiant energy propagate without any
medium in the space in the form of waves are known as electromagnetic
radiations. These waves can be produced by a charged body moving in a
magnetic field or a magnet in a electric field. e.g.   rays,   rays,
cosmic rays, ordinary light rays etc.
(2) Characteristics
(i) All electromagnetic radiations travel with the velocity of light.
(ii) These consist of electric and magnetic fields components
that oscillate in directions perpendicular to each other and
perpendicular to the direction in which the wave is travelling.
(3) A wave is always characterized by the following five
characteristics,
(i) Wavelength : The distance between two nearest crests or nearest
troughs is called the wavelength. It is denoted by  (lambda) and is
measured is terms of centimeter(cm), angstrom(Å), micron(  ) or
nanometre (nm).
Crest
Wavelength
Vibrating
source
Energy
Trough
Fig. 2.2
Structure of atom
1 Å  10 8 cm  10 10 m ; 1  10 4 cm  10 6 m ;
1nm  10 7 cm  10 9 m ; 1cm  10 8 Å  10 4   10 7 nm
(ii) Frequency : It is defined as the number of waves which pass
through a point in one second. It is denoted by the symbol  (nu) and is
expressed in terms of cycles (or waves) per second (cps) or hertz (Hz).
  distance travelled in one second = velocity =c
 
c

(iii) Velocity : It is defined as the distance covered in one second by
the wave. It is denoted by the letter ‘c’. All electromagnetic waves travel
with the same velocity, i.e., 3  1010 cm / sec .
c    3  1010 cm / sec
(iv) Wave number : This is the reciprocal of wavelength, i.e., the
number of wavelengths per centimetre. It is denoted by the symbol  (nu
bar). It is expressed in cm 1 or m 1 .
 
incandescent object resolved through prism or spectroscope, it also gives
continuous spectrum of colours.
(ii) Line spectrum : If the radiation’s obtained by the excitation of a
substance are analysed with help of a spectroscope a series of thin bright
lines of specific colours are obtained. There is dark space in between two
consecutive lines. This type of spectrum is called line spectrum or atomic
spectrum..
(2) Absorption spectrum : Spectrum produced by the absorbed
radiations is called absorption spectrum.
Hydrogen spectrum
(1) Hydrogen spectrum is an example of line emission spectrum or
atomic emission spectrum.
(2) When an electric discharge is passed through hydrogen gas at
low pressure, a bluish light is emitted.
(3) This light shows discontinuous line spectrum of several isolated
sharp lines through prism.
(4) All these lines of H-spectrum have Lyman, Balmer, Paschen,
Barckett, Pfund and Humphrey series. These spectral series were named by
the name of scientist discovered them.
(5) To evaluate wavelength of various H-lines Ritz introduced the
following expression,
 
1

(v) Amplitude : It is defined as the height of the crest or depth of
the trough of a wave. It is denoted by the letter ‘A’. It determines the
intensity of the radiation.
The arrangement of various types of electromagnetic radiations in
the order of their increasing or decreasing wavelengths or frequencies is
known as electromagnetic spectrum.
Table: 2.4
Name
Radio wave
Microwave
Wavelength (Å)
Frequency (Hz)
3  10 14  3  10 7
1  10 5  1  10 9
3  10  6  10
1  10  5  10
7
6
9


1
1 
 R 2  2 
c
 n1 n 2 

Where R is universal constant known as Rydberg’s constant its value
is 109, 678 cm 1 .
Plum pudding model of Thomson
(1) He suggected that atom is a positively charged sphere having
electrons embedded uniformly giving an overall picture of plum pudding.
–
11
6  10 6  7600
Visible
7600  3800
Ultraviolet (UV)
3800  150
X-Rays
150  0.1
  Rays
2  10 16  3  10 19
0.1  0.01
0.01- zero
3  10 19  3  10 20
Cosmic Rays
1
3.95  10
7.9  10
14
 7.9  10
–
3  10 20  infinity
+
Electron unifromly
embedded
+
Positive charge spreaded throughout the sphere
14
 2  10 16
–
+ – +
–
Positively charged
sphere
+
+
5  10 11  3.95  10 16
16
–
+
Infrared (IR)
39
2.3
(2) This model failed toFig.explain
the line spectrum of an element and
the scattering experiment of Rutherford.
Rutherford's nuclear model
(1) Rutherford carried out experiment on the bombardment of thin
(10 mm) Au foil with high speed positively charged   particles emitted
from Ra and gave the following observations based on this experiment,
–4
Atomic spectrum - Hydrogen spectrum
Atomic spectrum
Spectrum is the impression produced on a photographic film when
the radiation (s) of particular wavelength (s) is (are) analysed through a
prism or diffraction grating.
Types of spectrum
(1) Emission spectrum : Spectrum produced by the emitted radiation
is known as emission spectrum. This spectrum corresponds to the radiation
emitted (energy evolved) when an excited electron returns back to the
ground state.
(i) Continuous spectrum : When sunlight is passed through a prism, it
gets dispersed into continuous bands of different colours. If the light of an
(i) Most of the

particles passed without any deflection.
(ii) Some of them were deflected away from their path.
(iii) Only a few (one in about 10,000) were returned back to their
original direction of propagation.
Deflected
-particles
-rays
+ve Nucleus
Fig. 2.4
ZnS screen
40 Structure of atom
(2) From the above observations he concluded that, an atom
consists of
(i) Nucleus which is small in size but carries the entire mass i.e.
contains all the neutrons and protons.
(ii) Extra nuclear part which contains electrons. This model was
similar to the solar system.
(3) Properties of the nucleus
(i) Nucleus is a small, heavy, positively charged portion of the atom
and located at the centre of the atom.
(ii) All the positive charge of atom (i.e. protons) are present in
nucleus.
(iii) Nucleus contains neutrons and protons, and hence these
particles collectively are also referred to as nucleons.
(iv) The size of nucleus is measured in Fermi (1 Fermi = 10 cm).
–13
(v) The radius of nucleus is of the order of 1.5  10 13 cm. to
6.5  10 13 cm. i.e. 1 .5 to 6 .5 Fermi. Generally the radius of the
nucleus ( rn ) is given by the following relation,
rn  ro ( 1.4  10 13 cm)  A1 / 3
This exhibited that nucleus is 10 5 times small in size as
compared to the total size of atom.
(vi) The Volume of the nucleus is about 10 39 cm 3 and that of
atom is 10
atom.
24
3
cm , i.e., volume of the nucleus is 10
15
times that of an
(vii) The density of the nucleus is of the order of 10
15
g cm
3
or
10 8 tonnes cm 3 or 10 12 kg / cc . If nucleus is spherical than,
Density =
mass of the nucleus

volume of the nucleus
mass number
4
6 .023  10 23  r 3
3
(4) Drawbacks of Rutherford's model
(i) It does not obey the Maxwell theory of electrodynamics,
according to it “A small charged particle moving around an oppositely
charged centre continuously loses its energy”. If an electron does so, it
should also continuously lose its energy and should set up spiral motion
ultimately failing into the nucleus.
(ii) It could not explain the line spectra of H  atom and
discontinuous spectrum nature.
Planck's quantum theory
Where,
h
hc

Planck's constant = 6.62×10
6.62  10 34 Joules sec .
Photoelectric effect
(1) When radiations with certain minimum frequency ( 0 ) strike
the surface of a metal, the electrons are ejected from the surface of the
metal. This phenomenon is called photoelectric effect and the electrons
emitted are called photo-electrons. The current constituted by
photoelectrons is known as photoelectric current.
(2) The electrons are ejected only if the radiation striking the
surface of the metal has at least a minimum frequency ( 0 ) called
Threshold frequency. The minimum potential at which the plate
photoelectric current becomes zero is called stopping potential.
(3) The velocity or kinetic energy of the electron ejected depend
upon the frequency of the incident radiation and is independent of its
intensity.
(4) The number of photoelectrons ejected is proportional to the
intensity of incident radiation.
(5) Einstein’s photoelectric effect equation
According to Einstein,
Maximum kinetic energy of the ejected electron = absorbed energy
– threshold energy
1
1
1 
2
mv max
 h   h  0  hc  

2


0 

Where,  0
wavelength.
–27
erg. sec. or
and 0
are threshold frequency and threshold
Bohr’s atomic model
Bohr retained the essential features of the Rutherford model of the
atom. However, in order to account for the stability of the atom he
introduced the concept of the stationary orbits. The Bohr postulates are,
(1) An atom consists of positively charged nucleus responsible for
almost the entire mass of the atom (This assumption is retention of
Rutherford model).
(2) The electrons revolve around the nucleus in certain permitted
circular orbits of definite radii.
(3) The permitted orbits are those for which the angular
momentum of an electron is an intergral multiple of h / 2 where h is
the Planck’s constant. If m is the mass and v is the velocity of the
electron in a permitted orbit of radius r, then
L  mvr 
When black body is heated, it emits thermal radiation’s of different
wavelengths or frequency. To explain these radiations, max planck put
forward a theory known as planck’s quantum theory.
(i) The radiant energy which is emitted or absorbed by the black
body is not continuous but discontinuous in the form of small discrete
packets of energy, each such packet of energy is called a 'quantum'. In case
of light, the quantum of energy is called a 'photon'.
(ii) The energy of each quantum is directly proportional to the
frequency (  ) of the radiation, i.e.
E   or E  hv 
(iii) The total amount of energy emitted or absorbed by a body
will be some whole number quanta. Hence E  nh , where n is an
integer.
nh
; n  1 , 2, 3, …… 
2
Where L is the orbital angular momentum and n is the number
of orbit. The integer n is called the principal quantum number. This
equation is known as the Bohr quantization postulate.
(4) When electrons move in permitted discrete orbits they do not
radiate or lose energy. Such orbits are called stationary or non-radiating
orbits. In this manner, Bohr overcame Rutherford’s difficulty to account for
the stability of the atom. Greater the distance of energy level from the
nucleus, the more is the energy associated with it. The different energy
levels were numbered as 1,2,3,4 .. and called as K, L, M , N , …. etc.
(5) Ordinarily an electron continues to move in a particular
stationary state or orbit. Such a state of atom is called ground state.
When energy is given to the electron it jumps to any higher energy
level and is said to be in the excited state. When the electron jumps
from higher to lower energy state, the energy is radiated.
Structure of atom
Advantages of Bohr’s theory
(i) Bohr’s theory satisfactorily explains the spectra of species having
one electron, viz. hydrogen atom, He  , Li 2  etc.
(ii) Calculation of radius of Bohr’s orbit : According to Bohr, radius
of n orbit in which electron moves is

2 2k 2me 4 Z 2
ch3
 1
1

 n2  n2
2
 1

th

 n2
h2
rn   2 2 .
 4 me k  Z
9.1  10
n  Orbit
Where,
31

number,
Where, R 

m  Mass
kg , e  Charge on the electron 1.6  10 19


number of element, k = Coulombic constant 9  10 9 Nm 2c 2
number

Z  Atomic
After putting the values of m,e,k,h, we get.
rn 
n2
 0 .529 Å
Z
 Ze 2 
2e 2 ZK
, Vn  

nh
 mr 

kZe
kZe

2r
r
2

kZe
2r
1/2
; Vn 
2 .188  10 8 Z
cm. sec 1
n
of
electron
 2 2 mZ 2 e 4 k 2
Where, n=1, 2,
n 2h2
Putting the value of m, e, k, h,  we get
E  21.8  10 12 
(i) The light absorbed or emitted as a result of an electron changing
orbits produces characteristic absorption or emission spectra which can be
recorded on the photographic plates as a series of lines, the optical
spectrum of hydrogen consists of several series of lines called Lyman,
Balmar, Paschen, Brackett, Pfund and Humphrey. These spectral series were
named by the name of scientist who discovered them.
1
1 
1
  R 2  2 
 c
 n1 n 2 
2 2me 4
 Rydberg's constant
ch 3
It's theoritical value = 109,737 cm
Z2
J per atom(1 J  10 7 erg)
n2
2
Z
eV per atom(1eV  1.6  10 -19 J )
n2
Z2
k .cal / mole (1 cal = 4.18J)
n2
109,677.581cm
–1
and It's experimental value =
1
This remarkable agreement between the theoretical
experimental value was great achievment of the Bohr model.
and
(iii) Although H-atom consists of only one electron yet it's spectra
consist of many spectral lines.
(iv) Comparative study of important spectral series of Hydrogen is
shown in following table.
(v) If an electron from n excited state comes to various energy
th
1312 2
Z kJmol 1
or
n2
states, the maximum spectral lines obtained will be
When an electron jumps from an outer orbit (higher energy) n 2 to
an inner orbit (lower energy) n1 , then the energy emitted in form of
radiation is given by
2 2 k 2 me 4 Z 2
h2
 
Where, R is =
Z2
erg per atom
n2
 21.8  10 19 
E  En2  En1 
The negative sign in the above equations shows that the electron
and nucleus form a bound system, i.e., the electron is attracted towards the
nucleus. Thus, if electron is to be taken away from the nucleus, energy has
to be supplied. The energy of the electron in n  1 orbit is called the
ground state energy; that in the n  2 orbit is called the first excited state
energy, etc. When n   then E  0 which corresponds to ionized atom
(ii) To evaluate wavelength of various H-lines Ritz introduced the
following expression,
3………. 
 13.6 
value to be used is 109678cm 1 .
(6) Spectral evidence for quantisation (Explanation for hydrogen spectrum
on the basisof bohr atomic model)
2
Substituting of r, gives us E 
E  13.6 
2 2 k 2 me 4
; R is known as Rydberg constant. Its
ch 3
(ionization).
(iv) Calculation of energy of electron in Bohr’s orbit
Total energy of electron = K.E. + P.E.
2
 1
1 
   RZ 2  2  2 
n
n
2 
 1
i.e., the electron and nucleus are infinitely separated H  H   e 
(iii) Calculation of velocity of electron
Vn 




1
This can be represented as
41
 1
1 

 n2  n2 
2 
 1
 1
1 
 E  13.6 Z 2  2  2 eV / atom
n
n
2 
 1
As we know that E  h , c   and  
1


E
,
hc
=
n(n  1)
. n=
2
principal quantum number.
As n=6 than total number of spectral lines
=
6(6  1) 30

 15.
2
2
(vi) Thus, at least for the hydrogen atom, the Bohr theory accurately
describes the origin of atomic spectral lines.
(7) Failure of Bohr model
(i) Bohr theory was very successful in predicting and accounting the
energies of line spectra of hydrogen i.e. one electron system. It could not
explain the line spectra of atoms containing more than one electron.
(ii) This theory could not explain the presence of multiple spectral
lines.
42 Structure of atom
(iii) This theory could not explain the splitting of spectral lines in
magnetic field (Zeeman effect) and in electric field (Stark effect). The
intensity of these spectral lines was also not explained by the Bohr atomic
model.
(iv) This theory was unable to explain of dual nature of matter as
explained on the basis of De broglies concept.
(v) This theory could not explain uncertainty principle. (vi) No
conclusion was given for the concept of quantisation of energy.
Table: 2.5
S.No.
Spectral series
(1)
Lymen series
(2)
Balmer series
Lies in the
region
Transition
Ultraviolet
region
n1  1
n2  n1
max 
n1  2
 max 
Paschen series
(4)
Brackett series
(5)
Pfund series
(6)
Humphrey
series
n=3
Infra red
region
Infra red
region
1
n 2  4,5,6.... 
n1  4
n 2  5,6,7.... 
Infra red
region
Far infrared
region
n1  5
n 2  6,7,8.... 
n1  6
n 2  7,8.... 
 max 
max 
 max 
Dual nature of electron
(1) In 1924, the French physicist, Louis de Broglie suggested that if
light has both particle and wave like nature, the similar duality must be true
for matter. Thus an electron, behaves both as a material particle and as a
wave.
(2) This presented a new wave mechanical theory of matter.
According to this theory, small particles like electrons when in motion
possess wave properties.
(3) According to de-broglie, the wavelength associated with a
particle of mass m, moving with velocity v is given by the relation
h
, where h = Planck’s constant.
mv
(4) This can be derived as follows according to Planck’s equation,
h.c

c

   


36
5R
 min 
9
5
4
R
n1  3 and n 2  
144
7R
 min 
16
7
9
R
n1  4 and n 2  
16  25
9R
 min 
25
9
16
R
n1  5 and n 2  
25  36
11R
 min 
n1  6 and n 2  7
 max
4
3
1
R
n1  2 and n2  
n1  5 and n 2  6
It is an extension of Bohr’s model. The electrons in an atom revolve
around the nuclei in elliptical orbit. The circular path is a special case of
ellipse. Association of elliptical orbits with circular orbit explains the fine
line spectrum of atoms.
E  h 
 min 
n1  4 and n2  5
Bohr–Sommerfeild’s model

4
3R
n1  3 and n 2  4
 max 
 max
n2
 2 2 2
 min n 2  n1
n12
R
n1  1 and n 2  
n1  2 and n 2  3
n 2  3,4,5.... 
(3)
 min 
n1  1 and n 2  2
n 2  2,3,4.... 
Visible region
n12n22
 n12 )R
(n22
36
11
25
R
n1  6 and n 2  
36  49

13 R
 min
energy of
49
13
36

R
photon (on the basis of Einstein’s mass energy
relationship), E  mc 2
Equating both
Broglie relation.

hc

 mc 2 or  
mc  p 
h
mc
which is same as de-
(5) This was experimentally verified by Davisson and Germer by
observing diffraction effects with an electron beam. Let the electron is
accelerated with a potential of V than the Kinetic energy is
1
mv 2  eV ; m 2 v 2  2eVm
2
mv  2eVm  P ;  
h
2eVm
(6) If Bohr’s theory is associated with de-Broglie’s equation then
wave length of an electron can be determined in bohr’s orbit and relate it
with circumference and multiply with a whole number
2r  n  or  
2r
n
From de-Broglie equation,  
h
.
mv
Structure of atom
h
2r
nh
or mvr 

2
mv
n
(iii) If  2 is maximum than probability of finding e  is maximum
(7) The de-Broglie equation is applicable to all material objects but
it has significance only in case of microscopic particles. Since, we come
across macroscopic objects in our everyday life, de-broglie relationship has
no significance in everyday life.
Heisenberg’s uncertainty principle
the
particle,
Now since p  m v
h
h
or x  v 
4
4m
In terms of uncertainty in energy, E and uncertainty in time
h
t, this principle is written as, E . t 
4
Schrödinger wave equation
(1) Schrodinger wave equation is given by Erwin Schrödinger in
1926 and based on dual nature of electron.
(2) In it electron is described as a three dimensional wave in the electric
field of a positively charged nucleus.
(3) The probability of finding an electron at any point around the
nucleus can be determined by the help of Schrodinger wave equation which
is,
2
x
2

2
y
2

2
z
2

8 2m
h2
Where x, y and z are the 3 space co-ordinates, m = mass of
(4) The Schrodinger wave equation can also be written as,
8 2m
h
2
14
12
10
8
6
4
2
0
2 4 6 8
0.53Å
r(Å)
1s
5
4
3
2
1
0
0.53Å
Quantum numbers
2 4 6 8
2.7Å r(Å)
2s
Fig. 2.5
5
4
3
2
1
0
2
4
2.1Å
2s
6
8
r(Å)
Each orbital in an atom is specified by a set of three quantum
numbers (n, l, m) and each electron is designated by a set of four quantum
numbers (n, l, m and s).
(1) Principle quantum number (n)
(i) It was proposed by Bohr and denoted by ‘n’.
(ii) It determines the average distance between electron and nucleus,
means it denotes the size of atom.
(iii) It determine the energy of the electron in an orbit where
electron is present.
(iv) The maximum number of an electron in an orbit
(E  V )   0
electron, h = Planck’s constant, E = Total energy, V = potential energy of
electron,  = amplitude of wave also called as wave function,  = for an
infinitesimal change.
2  
R  4r 2 dr 2 . The plats of R distance from nucleus as follows
4r2 dr 2
of
Radial probability distribution curves : Radial probability is
4r2 dr 2
h
Mathematically it is represented as , x . p 
4
So equation becomes, x . m v 
(iv) The solution of this equation provides a set of number called
quantum numbers which describe specific or definite energy state of the
electron in atom and information about the shapes and orientations of the
most probable distribution of electrons around the nucleus.
This principle states “It is impossible to specify at any given
moment both the position and momentum (velocity) of an electron”.
Where
x  uncertainty is position
p  uncertainty in the momentum of the particle
around nucleus and the place where probability of finding e  is maximum
is called electron density, electron cloud or an atomic orbital. It is different
from the Bohr’s orbit.
4r2 dr 2
Thus
43
represented by this quantum number as 2n 2 . No energy shell in
atoms of known elements possess more than 32 electrons.
(v) It gives the information of orbit K, L, M, N------------.
(vi) Angular momentum can also be calculated using principle
quantum number
(2) Azimuthal quantum number (l)
(i) Azimuthal quantum number is also known as angular
quantum number. Proposed by Sommerfield and denoted by ‘ l ’.
(E  V )   0
(ii) It determines the number of sub shells or sublevels to which the
electron belongs.
Where  = laplacian operator.
(5) Physical significance of  and  2
(iii) It tells about the shape of subshells.
(i) The wave function  represents the amplitude of the
electron wave. The amplitude  is thus a function of space co-ordinates
and time i.e.   (x , y, z...... times)
(ii) For a single particle, the square of the wave function ( ) at
any point is proportional to the probability of finding the particle at that
point.
2
(iv) It also expresses the energies of subshells s  p  d  f
(increasing energy).
(v) The value of l  (n  1) always. Where ‘n’ is the number of
principle shell.
(vi)
Value of l
=
0
1
2
3…..(n-1)
Name of subshell
=
s
p
d
f
Shape of subshell
=
Spheric
al
Dumbbell
Double
dumbbell
Complex
44 Structure of atom
(vii) It represent the orbital angular momentum. Which is equal to
h
2
l(l  1)
(vii) Degenerate orbitals : Orbitals having the same energy are
known as degenerate orbitals. e.g. for p subshell p x p y p z
(viii) The number of degenerate orbitals of s subshell =0.
(viii) The maximum number of electrons in subshell  2(2l  1)
s  subshell 2 electrons d  subshell 10 electrons
p  subshell 6 electrons f  subshell 14 electrons.
(4) Spin quantum numbers (s)
(i) It was proposed by Goldshmidt & Ulen Back and denoted by the
symbol of ‘s’.
The value of ' s' is  1/2 and - 1/2, which signifies the spin
(ii)
(ix) For a given value of ‘n’ the total values of ‘l’ is always equal to
the value of ‘n’.
or rotation or direction of electron on it’s axis during movement.
(iii) The spin may be clockwise or anticlockwise.
(3) Magnetic quantum number (m)
(i) It was proposed by Zeeman and denoted by ‘m’.
(ii) It gives the number of permitted orientation of subshells.
(iii) The value of m varies from –l to +l through zero.
(iv) It tells about the splitting of spectral lines in the magnetic field
i.e. this quantum number proves the Zeeman effect.
(v) For a given value of ‘n’ the total value of ’m’ is equal to n 2 .
(iv) It represents the value of spin angular momentum is equal to
h
2
s(s  1).
(v) Maximum spin of an atom  1 / 2  number of unpaired
electron.
(vi) This quantum number is not the result of solution of
schrodinger equation as solved for H-atom.
(vi) For a given value of ‘l’ the total value of ‘m’ is equal to (2l  1).
Table : 2.6 Distribution of electrons among the quantum levels
n
l
m
Designation of
orbitals
Number of Orbitals in the subshell
1
0
0
1s
1
2
0
0
2s
1
2
1
–1, 0, +1
2p
3
3
0
0
3s
1
3
1
–1, 0, +1
3p
3
3
2
–2, –1, 0, +1, +2
3d
5
4
0
0
4s
1
4
1
–1, 0, +1
4p
3
4
2
–2, –1, 0, +1, +2
4d
5
4
3
–3, –2, –1, 0, +1, +2, +3
4f
7
Shape of orbitals
(1) Shape of ‘s’ orbital
(2) Shape of ‘p’ orbitals
(i) For ‘s’ orbital l=0 & m=0 so ‘s’ orbital have only one
unidirectional orientation i.e. the probability of finding the electrons is same
in all directions.
(i) For ‘p’ orbital l=1, & m=+1,0,–1 means there are three ‘p’ orbitals,
which is symbolised as p x , p y , p z .
(ii) The size and energy of ‘s’ orbital with increasing ‘n’ will be
1s  2s  3 s  4 s.
(ii) Shape of ‘p’ orbital is dumb bell in which the two lobes on
opposite side separated by the nodal plane.
(iii) s-orbitals known as radial node or modal surface. But there is
no radial node for 1s orbital since it is starting from the nucleus.
(iii) p-orbital has directional properties.
Z
Z
Z
Y
X
PY
Fig. 2.7
Fig. 2.6
2S
X
X
Px
1S
Y
Y
Pz
Structure of atom
(3) Shape of ‘d’ orbital
(i) For the ‘d’ orbital l =2 then the values of ‘m’ are –2, –1, 0, +1, +2.
It shows that the ‘d’ orbitals has five orbitals as d xy , d yz , d zx , d x 2 y 2 , d z 2 .
(ii) Each ‘d’ orbital identical in shape, size and energy.
(iii) The shape of d orbital is double dumb bell .
(iv) It has directional properties.
Z
Z
Y
Z
“Electron pairing in p, d and f orbitals cannot occur until each
orbitals of a given subshell contains one electron each or is singly occupied”.
X
This is due to the fact that electrons being identical in charge, repel
each other when present in the same orbital. This repulsion can however be
minimised if two electrons move as far apart as possible by occupying
different degenerate orbitals. All the unpaired electrons in a degenerate set
of orbitals will have same spin.
Y
Y
Z
X
X
Z
Y
dYZ
1s
Because there are only two possible values of s, an orbital can hold
not more than two electrons.
This rule deals with the filling of electrons in the orbitals having
equal energy (degenerate orbitals). According to this rule,
Y
dXY
Z
does not represent a possible
(4) Hund’s Rule of maximum multiplicity
X
dZX
The orbital diagram
arrangement of electrons
dX2–Y2
As we now know the Hund’s rule, let us see how the three electrons
are arranged in p orbitals.
The important point ot be remembered is that all the singly
occupied orbitals should have electrons with parallel spins i.e in the same
direction either-clockwise or anticlockwise.
2 py
2 px
2 pz
2 px
X
d
45
2 py
2 pz
or
Electronic configurations of elements
z2
Fig. 2.8
(4) Shape of ‘f’ orbital
(i) For the ‘f’ orbital l=3 then the values of ‘m’ are –3, –2, –
1,0,+1,+2,+3. It shows that the ‘f’ orbitals have seven orientation as
fx ( x 2  y 2 ) , fy( x 2  y 2 ) , fz ( x 2  y 2 ), fxyz , fz 3 , fyz 2 and fxz 2 .
On the basis of the elecronic configuration principles the electronic
configuration of various elements are given in the following table :
The above method of writing the electronic configurations is quite
cumbersome. Hence, usually the electronic configuration of the atom of any
element is simply represented by the notation.
nlx
(ii) The ‘f’ orbital is complicated in shape.
number of
principal
shell
Rules for filling of electrons in various orbitals
The atom is built up by filling electrons in various orbitals according
to the following rules,
(1) Aufbau’s principle
This principle states that the electrons are added one by one to the
various orbitals in order of their increasing energy starting with the orbital
of lowest energy. The increasing order of energy of various orbitals is
1s  2s  2 p  3 s  3 p  4 s  3d  4 p  5 s  4 d  5 p  6 s  4 f
 5d  6 p  7 s  5 f  6 d  7 p.........
(2) (n+l) Rule
In neutral isolated atom, the lower the value of (n + l) for an orbital,
lower is its energy. However, if the two different types of orbitals have the
same value of (n + l), the orbitals with lower value of n has lower energy.
(3) Pauli’s exclusion principle
According to this principle “no two electrons in an atom will have
same value of all the four quantum numbers”.
If one electron in an atom has the quantum numbers n  1 , l  0 ,
m  0 and s  1 / 2 , no other electron can have the same four
quantum numbers. In other words, we cannot place two electrons with the
same value of s in a 1s orbital.
Number of
electrons
Present
symbol of
subshell
Some Unexpected Electronic Configuration
Some of the exceptions are important though, because they occur
with common elements, notably chromium and copper.
has 29 electrons. Its excepted electronic configuration is
Cu
1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 9
2
2
6
2
6
1
1s 2 s 2 p 3 s 3 p 4 s 3 d
10
but
in
reality
the
configuration
is
as this configuration is more stable. Similarly
Cr has the configuration of 1s 2 2 s 2 sp 6 3 s 2 3 p 6 4 s 1 3d 5 instead of
1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 4 .
Factors responsible for the extra stability of half-filled and
completely filled subshells,
(i) Symmetrical distribution : It is well known fact that symmetry
leads to stability. Thus the electronic configuration in which all the orbitals
of the same subshell are either completely filled or are exactly half filled are
more stable because of symmetrical distribution of electrons.
(ii) Exchange energy : The electrons with parallel spins present in
the degenerate orbitals tend to exchange their position. The energy released
during this exchange is called exchange energy. The number of exchanges
that can take place is maximum when the degenerate orbtials (orbitals of
same subshell having equal energy) are exactly half-filled or completely. As a
result, the exchange energy is maximum and so it the stability.
46 Structure of atom
 When energy or frequency of scattered ray is lesser than the
incident ray, it is known as Compton effect.
 The instrument used to record solar spectrum is called
spectrometer or spectrograph developed by Bunsen and Kirchoff in 1859.
 The intensities of spectral lines decreases with increase in the value
 All lines in the visible region are of Balmer series but reverse is not
of n . For example, the intensity of first Lyman line (2  1) is greater
true i.e., all Balmer lines will not fall in visible region.
 A part of an atom up to penultimate shell is a kernel or atomic
core.
 If the energy supplied to hydrogen atom is less than 13.6 eV it will
awpt or absorb only those quanta which can take it to a certain higher
energy level i.e., all those photons having energy less than or more than a
particular energy level will not be absorbed by hydrogen atom, but if
energy supplied to hydrogen atom is more than 13.6eV then all photons
are absorbed and excess energy appear as kinetic energy of emitted
photo electron.
 No of nodes in any orbital  (n  l  1)
than second line (3  1) and so on.
 No of nodal planes in an orbitals  l
 The d orbital which does not have four lobes is d z 2
 The d orbital whose lobes lie along the axis is d x 2 y 2
 Spin angular momentum  s(s  1)
h
2
 In Balmer series of hydrogen spectrum the first line (3  2) is
also known as L line. The second line (4  2) is L  line. The line
from infinity energy shell is called limiting line.
Discovery and Properties of anode, cathode rays
neutron and Nuclear structure
1.
A neutral atom (Atomic no. > 1) consists of
[CPMT 1982]
n
2
 Total spin   ; where n is no of unpaired e 
(a) Only protons
(b) Neutrons + protons
 Magnetic moment  n(n  2) B.M. (Bohr magnetron) of n
(c) Neutrons + electrons
unpaired e 
(d) Neutron + proton + electron
 Ion with unpaired electron in d or f orbital will be coloured.
 Exception of E.C. are Cr(24) , Cu(29) , Mo(42) , Ag(47) ,
2.
The nucleus of the atom consists of
[CPMT 1973, 74, 78, 83, 84; MADT Bihar 1980;
DPMT 1982, 85; MP PMT 1999]
W (74 ) , Au(79) .
 No. of waves n 
2r

(where  
(a) Proton and neutron
h
)
mv
 No. of revolutions of e  per second is 
(b) Proton and electron
(c) Neutron and electron
v
.
2r
 The solution of schrodinger wave equation gives principal,
(d) Proton, neutron and electron
3.
azimuthal and magnetic quantum numbers but not the spin quantum
number.
[CPMT 1982; MP PMT 1991]
 In the Rydberg formula, when n 2   the line produced is called
the limiting line of that series.
 Among various forms of visible light, violet colour has shortest
The size of nucleus is of the order of
4.
12
(a)
10
(c)
10 15 m
8
(b) 10 m
m
(d) 10 10 m
Positive ions are formed from the neutral atom by the
[CPMT 1976]
wavelength, highest frequency and highest energy.
(a) Increase of nuclear charge
 Red coloured light has largest wavelength, least frequency and
lowest energy in visible light.
(b) Gain of protons
 Elements give line spectra. The line spectrum is characteristic of
(c) Loss of electrons
the excited atom producing it. No two elements have identical line
spectrum.
(d) Loss of protons
5.
The electron is
 The line spectrum results from the emission of radiations from the
[DPMT 1982; MADT Bihar 1980]
atoms of the elements and is therefore called as atomic spectrum.
 Atoms give line spectra (known as atomic spectrum) and the molecules
give band spectra (known as molecular spectrum).
 The negative potential at which the photoelectric current becomes
zero is called cut off potential or stopping potential.
6.
 -ray particle
(a)  -ray particle
(b)
(c) Hydrogen ion
(d) Positron
Who discovered neutron
[IIT 1982; BITS 1988;CPMT 1977; NCERT 1974;
MP PMT 1992; MP PET 2002]
Structure of atom
7.
(a) James Chadwick
(b) William Crooks
(c) J.J. Thomson
(d) Rutherford
The ratio of charge and mass would be greater for
[BHU 2005]
(a) Proton
(b) Electron
(c) Neutron
(d) Alpha
47
48 Structure
8.
9.
10.
of atom
Magnitude of K.E. in an orbit is equal to [BCECE 2005]
(a) Half of the potential energy
(b) Twice of the potential energy
(c) One fourth of the potential energy
(d) None of these
The density of neutrons is of the order
[NCERT 1980]
(a) 10 3 kg / cc
(b) 10 6 kg / cc
(c) 10 9 kg / cc
(d) 1011 kg / cc
The discov ery of neutron becom es v ery late because
[CPMT 1987; AIIMS 1998]
11.
(a) Neutrons are present in nucleus
(b) Neutrons are highly unstable particles
(c) Neutrons are chargeless
(d) Neutrons do not m ov e
The fundamental particles present in the nucleus of an
atom are
(a) Alpha particles and electrons
(b) Neutrons and protons
(a) Proton is nucleus of deuterium
(b) Proton is ionized hy drogen m olecule
(c) Proton is ionized hy drogen atom
20.
(d) Electrons, neutrons and protons
(d) None of these
Anode rays were discov ered by
[DPMT 1985]
(a) Goldstein
(b) J. Stoney
(c) Rutherford
(d) J.J. Thom son
21.
22.
(a) 10
13.
(b) 10 8 kg / cc
(c) 10 9 kg / cc
(d) 10 12 kg / cc
Cathode rays are
(a) Protons
23.
14.
Number of neutron in C 12 is
(a) 6
(b) 7
(c) 8
15.
16.
Heaviest particle is
(a) Meson
[BCECE 2005]
(d) A charg of –1 and a m ass of 1 unit
Cathode rays have
26.
(a) Mass only
(b) Charge only
(c) No m ass and charge (d) Mass and charge both
The size of nucleus is m easured in
27 .
(a) am u
(b) Angstrom
(c) Ferm i
(d) cm
Which phrase would be incorrect to use
[DPMT 1983; MP PET 1999]
(b) Neutron
(c) Proton
(d) Electron
Penetration power of prot on is
[A MU (Engg.) 1999]
(a) A m olecular of a com pound
(a) More than electron
(b) Less than electron
(c) More than neutron
An elementary particle is
(d) None
[CPMT 1973]
(b) A m olecule of an elem ent
(c) An atom of an elem ent
28.
[MP PET 2002]
The nucleus of helium contains
[CPMT 1972; DPMT 1982]
29.
(a) Four protons
(b) Four neutrons
(c) Two neutrons and two protons
(d) Four protons and two electrons
Which is correct statem ent about proton
[CPMT 1979; MP PMT 1985; NCERT 1985; MP PET 1999]
(d) None of these
Which one of the following pairs is not correctly m atched
(a) Rutherford-Proton
(c) A sub-atom ic particle
(d) A fragm ent of an atom
19.
[CPMT 1982]
[EA MCET 1988; CPMT 1994]
(d) 9
(a) An elem ent present in a com pound
(b) An atom present in an elem ent
18.
(c) 10 15 cm
(d) 10 8 cm
Neutron possesses
[CPMT 1982]
(a) Positiv e charge
(b) Negativ e charge
(c) No charge
(d) All are correct
Neutron is a fundamental particle carrying
25.
[BHU 1985; CPMT 1982, 88]
17.
(b) 10 13 cm
[CPMT 1990]
(d)  -particles
(c) Neutrons
cm
(a) A charge of + 1 unit and a m ass of 1 unit
(b) No charge and a m ass of 1 unit
(c) No charge and no m ass
[JIPMER 1991; NCERT 1976]
(b) Electrons
10
[CPMT 1983, 84]
[NCERT 1981, CPMT 1981, 2003]
(a) 10 8 kg / cc
The radius of an atom is of the order of
[A MU 1982; IIT 1985; MP PMT 1995]
24.
The order of density in nucleus is
[A MU 1983]
(a) Positiv ely charged particles
(b) Negativ ely charged particles
(c) Neutral particles
(c) Neutrons and electrons
12.
(d) Proton is  -particle
Cathode rays are made up of
(b) J.J. Thom som -Electron
(c) J.H. Chadwick-Neutron
(d) Bohr-Isotope
Proton was discov ered by
(a) Chadwick
(c) Goldstein
30.
[A FMC 2004]
(b) Thom son
(d) Bohr
The m inimum real charge on any particle which can exist
is
[RPMT 2000]
(a) 1.6  10 19 Coulomb
(b) 1.6  10 10 Coulomb
Structure of atom
(c) 4.8  10 10 Coulomb
31.
(d) Zero
The nature of anode ray s depends upon
[MP PET 2004]
(a) Nature of electrode
[CPMT 1986; MP PMT 1987]
All the abov e
One would expect proton to hav e v ery large
[Pb. CET 2004]
33.
34.
(a) Ionization potential
(b) Radius
(c) Charge
(d) Hy dration energy
(a) Num ber of e 1  W  N
(a) 6.023  10 23 g
(b) 1.008 g and 0.55mg
(b) Num ber of 0 n 1  W  N
(c) 9.1  10 28 kg
(d) 2 gm
(c) Num ber of 1 H 1  W  N
The average distance of an electron in an atom from its
nucleus is of the order of
[MP PET 1996]
(b) 10 6 m
(c) 10 10 m
(d) 10 15 m
The mass of 1 mole of electrons is
[Pb. CET 2004]
(a) 9.1  10 28 g
(c) 0.55 mg
36.
3.
(b) 1.008 mg
27
(d) 9.1  10
(d) Num ber of 0 n 1  N
4.
5.
(a) Na
g
The ratio of specific charge of a proton and an  -particle
is
38.
39.
(a) 2 : 1
(b) 1 : 2
(c) 1 : 4
(d) 1 : 1
Ratio of masses of proton and electron is
[BHU 1998]
(a) Infinite
(b) 1.8  10
(c) 1 .8
(d) None of these
Splitting of signals is caused by
[Pb. PMT 2000]
(b) Neutron
(c) Positron
(d) Electron
40.
(a) Deutron
(b) Positron
(c) Meson
(d) Nucleon
Which of the following has the sam e m ass as that of an
electron
[A FMC 2002]
(a) Photon
(c) Positron
41.
(b) Neutron
7.
[IIT 1979; MP PMT 1994; RPMT 1999]
[NCERT 1972; MP PMT 1995]
(a) Proton
(c) Electrons
8.
[UPSEAT 2004]
The number of electrons in an atom of an element is equal
to its
[BHU 1979]
CO has sam e electrons as or the ion that is isoelectronic
with CO is
[CPMT 1984; IIT 1982;
(a)
N 2
(c) O 2
(b) CN 
(d) O 2
9.
The m ass of an atom is constituted m ainly by
10.
(a) Neutron and neutrino (b) Neutron and electron
(c) Neutron and proton (d) Proton and electron
The atom ic num ber of an elem ent represents
[DPMT 1984, 91; AFMC 1990]
[CPMT 1983; CBSE PMT 1990; NCERT 1973; AMU 1984]
11.
(a) Num ber of neutrons in the nucleus
(b) Num ber of protons in the nucleus
(c) Atom ic weight of elem ent
(d) Valency of elem ent
An atom has 2 6 electrons and its atomic weight is 56. The
num ber of neutrons in the nucleus of the atom will be
[CPMT 1980]
Atomic number, Mass number, Atomic species
1.
(b) Neutron
(d) Protons and electrons
EA MCET 1990; CBSE PMT 1997]
(d) Proton
(b) 1 : 1
(d) 1 : 3
(b) K  and O
(a) 2 2
(b) 4 4
(c) 6 6
(d) 88
Chlorine atom differs from chloride ion in the num ber of
What is the ratio of m ass of an electron to the m ass of a
proton
(a) 1 : 2
(c) 1 : 1 83 7
and Ne
6.
The proton and neutron are collectiv ely called as
[MP PET 2001]

(c) Ne and O
(d) Na  and K 
The num ber of electrons in one m olecule of CO 2 are
3
(a) Proton
The total number of neutrons in dipositive zinc ions with
mass number 70 is
[IIT 1979; Bih a r MEE 1997]
(a) 3 4
(b) 4 0
(c) 3 6
(d) 3 8
Which of the following are isoelectronic with one another
[NCERT 1983; EAMCET 1989]
[MP PET 1999]
37 .
(a) 2 5 protons and 3 0 neutrons
(b) 2 5 neutrons and 3 0 protons
(c) 55 protons
(d) 55 neutrons
If W is atom ic weight and N is the atom ic num ber of an
elem ent, then
[CPMT 1971, 80, 89]
The m ass of a m ol of proton and electron is
(a) 10 6 m
35.
(a) Atom ic weight
(b) Atom ic num ber
(c) Equiv alent weight
(d) Electron affinity
The nucleus of the element having atomic number 2 5 and
atom ic weight 55 will contain
(b) Nature of residual gas
(c) Nature of discharge tube (d)
32.
2.
49
12.
(a) 2 6
(b) 3 0
(c) 3 6
(d) 56
The m ost probable radius (in pm) for finding the electron
in He  is
[A IIMS 2005]
50 Structure
of atom
(a) 0.0
(c) 2 6 .5
(b) 52 .9
(d) 1 05.8
13.
The number of unpaired electrons in the Fe 2  ion is
14.
(a) 0
(b) 4
(c) 6
(d) 3
A sodium cation has different num ber of electrons from
[MP PET 1989; KCET 2000]
(c) Li 
(d) Al 3
An atom which has lost one electron would be
16.
17.
18.
An atom has the electronic configuration of 1s 2 ,2 s 2 2 p 6 ,
3 s 2 3 p 6 3d 10 ,4 s 2 4 p 5 . Its atom ic weight is 80. Its atom ic
num ber and the number of neutrons in its nucleus shall
be
19.
(a) C
(b) F
21.
CH 3
III
IV
[IIT 1993]
(b) I and IV
(d) II, III and IV
26.
(a) 2 2
(b) 2 4
(c) 2 0
(d) 2 8
The atomic number of an element having the valency shell
[A MU 1988]
electronic configuration 4 s 2 4 p 6 is
27 .
[MP PMT 1991]
(a) 3 5
(b) 3 6
(c) 3 7
(d) 3 8
The present atom ic weight scale is based on
[EA MCET 1988; MP PMT 2002]
(a) C
28.
(b) O 16
12
(c) H 1
Isoelectronic species are

(a) K , Cl

(c) Na, Ar
29.
(d) C 13
[EA MCET 1989]

(b) Na , Cl 
(d) Na  , Ar
If the atom ic weight of an element is 2 3 tim es that of the
lightest element and it has 1 1 protons, then it contains
[EA MCET 1986; AFMC 1989]
30.

2
20.
NH 3
II
Number of electrons in CONH 2 is
[MP PMT 1987]
(a) 3 5 and 4 5
(b) 4 5 and 3 5
(c) 4 0 and 4 0
(d) 3 0 and 50
Which of the following particles has m ore electrons than
neutrons
H3O
I
25.
[CPMT 1986]
(a) Negativ ely charged
(b) Positiv ely charged
(c) Electrically neutral
(d) Carry double positiv e charge
Num ber of electrons in the outermost orbit of the element
of atomic num ber 15 is
[CPMT 1988, 93]
(a) 1
(b) 3
(c) 5
(d) 7
The atom ic weight of an elem ent is double its atom ic
num ber. If there are four electrons in 2 p orbital, the
elem ent is
[A MU 1983]
(a) C
(b) N
(c) O
(d) Ca
CH 3
(a) I and II
(c) I and III
(b) F 
(a) O 2 
15.
24.
(b) The number of nucleons is double of the number of
electrons
(c) The number of protons is half of the number of neutrons
(d) The number of nucleons is double of the atomic num ber
Pick out the isoelectronic structures from the following
(a) 1 1 protons, 2 3 neutrons, 1 1 electrons
(b) 1 1 protons, 1 1 neutrons, 1 1 electrons
(c) 1 1 protons, 1 2 neutrons, 1 1 electrons
(d) 1 1 protons, 1 1 neutrons, 2 3 electrons
Which of the following oxides of nitrogen is isoelectronic
with CO 2
[CBSE PMT 1990]
(a) NO 2
(b) N 2 O
(c) NO
(d) N 2 O 2
(c) O
(d) Al 3
Com pared with an atom of atomic weight 1 2 and atom ic
num ber 6 , the atom of atom ic weight 1 3 and atom ic
num ber 6
[NCERT 1971]
(a) Contains m ore neutrons(b) Contains m ore electrons
(c) Contains m ore protons(d) Is a different elem ent
31.
In the nucleus of
32.
The ratio between the neutrons in C and Si with respect
to atom ic m asses 1 2 and 2 8 is
(a) 2 : 3
(b) 3 : 2
(c) 3 : 7
(d) 7 : 3
The atom ic num ber of an elem ent is alway s equal to
33.
(a) Atom ic weight div ided by 2
(b) Num ber of neutrons in the nucleus
(c) Weight of the nucleus
(d) Electrical charge of the nucleus
Which of the following is isoelectronic with carbon atom
20
Ca 40 there are
[CPMT 1990; EAMCET 1991]
(a)
(b)
(c)
(d)
22.
40
20
20
20
[MP PMT 1994]
electrons
electrons
neutrons
neutrons
Na  ion is isoelectronic with
(a) Li 
23.
protons and 2 0
protons and 4 0
protons and 2 0
protons and 4 0
[CPMT 1990]
[MP PMT 1994; UPSEAT 2000]
2
(b) Mg
(c) Ca 2
(d) Ba 2
Ca has atomic no. 2 0 and atomic weight 4 0. Which of the
following statem ents is not correct about Ca atom
(a) Na
34.

(c) O 2 
(d) N 
CO 2 is isostructural with
[MP PET 1993]
(a) The num ber of electrons is sam e as the num ber of
neutrons
(b) Al 3 
[IIT 1986; MP PMT 1986, 94, 95]
(a) SnCl 2
(b) SO 2
[
Structure of atom
(c) HgCl2
35.
(d) All the abov e
47 .
[A FMC 1995; Bihar MEE 1997]
(b) He 
(d) Be
36.
48.
[A FMC 1995]
37 .
38.
The nucleus of the elem ent
39.
(a) 4 5 protons and 2 1 neutrons
(b) 2 1 protons and 2 4 neutrons
(c) 2 1 protons and 4 5 neutrons
(d) 2 4 protons and 2 1 neutrons
Neutrons are found in atom s of all elem ents except in
21
49.
40.
50.
51.
I  CH 3 , II  NH 2 , III  NH 4 , IV  NH 3
42.
43.
44.
(a) I, II, III
(c) I, II, IV
The charge on the atom
neutrons and 18 electrons is
(a) 1
52.
53.
46.
55.






o
(b) e , e
(c) e , p
(d) p , n
Nuclei tend to have m ore neutrons than protons at high
mass numbers because
[Roor kee Qu a l ify in g 1998]
(a) Neutrons are neutral particles
(b) Neutrons hav e m ore m ass than protons
(c) More neutrons m inim ize the coulom b repulsion
(d) Neutrons decrease the binding energy
56.
[CBSE PMT 1994]
(b) F 
(c) Tl 
(d) Na 
(b)
and PO43 
(d) ClO4 and OCN 
Nitrogen atom has an atomic number of 7 and oxygen has
an atomic number 8. The total num ber of electrons in a
nitrate ion will be
[Pb. PMT 2000]
(a) 8
(b) 1 6
(c) 3 2
(d) 6 4
If m olecular mass and atom ic m ass of sulphur are 2 56
and 32 respectively, its atom icity is
[RPET 2000]
(a) 2
(b) 8
(c) 4
(d) 1 6
The nitride ion in lithium nitride is com posed of
(a) 7 protons + 1 0 electrons
(b) 1 0 protons + 1 0 electrons
(c) 7 protons + 7 protons
(d) 1 0 protons + 7 electrons
The atomic number of an elem ent is 1 7 . The num ber of
orbitals containing electron pairs in its v alence shell is
(a) Eight
(b) Six
(c) Three
(d) Two
The atomic number of an element is 3 5 and mass number
is 81 . The number of electrons in the outer m ost shell is
[UPSEAT 2001]
57 .
58.
(a) 7
(b) 6
(c) 5
(d) 3
Which of the following is not isoelectronic[MP PET 2002]
(a) Na 
(b) Mg 2 
(c) O 2 
(d) Cl 
The charge of an electron is  1.6  10 19 C. The v alue of

free charge on Li ion will be
[A FMC 2002; KCET (Engg.) 2002]
(a) 3.6  10
Which one of the following is not isoelectronic with O 2 
(a) N 3 
and O 3
[CPMT 2001]
[RPMT 1997]

[UPSEA T 1999]
NO 2
[KCET 2000]
[A IIMS 1996]
(b) 2
(c) 1
(d) Zero
Number of unpaired electrons in inert gas is[CPMT 1996]
(a) Zero
(b) 8
(c) 4
(d) 1 8
In neutral atom , which particles are equiv alent
(a) p , e
45.
54.

(d) H 
(c) CO 2 , N 2 O, NO 3
[CPMT 1999]
(b) II, III, IV
(d) I and II
containing 1 7 protons, 1 8
NO 2
[EA MCET 1998]
(b) Na
(c) Li
An isostere is
(a)
[MP PMT 1999]
41.
2

(b) Oxy gen
(d) Hy drogen
(a) 1 0
(b) 1 4
(c) 7
(d) 5
Which of the following are isoelectronic species
(b) 89, 89, 2 4 2
(d) 89, 7 1 , 89
Be 2  is isoelectronic with
(a) Mg
E 45 contains
The m ass number of an anion, X 3  , is 14. If there are ten
electrons in the anion, the num ber of neutron s in the
nucleus of atom , X 2 of the elem ent will be
(a) 1 7
(b) 3 7
(c) 2
(d) 3 8
Num ber of protons, neutrons and electrons in the
231
elem ent 89
[A FMC 1997]
Y is
(a) 89, 2 3 1 , 89
(c) 89, 1 4 2 , 89
[MP PMT 1997]
(a) Chlorine
(c) Argon
(a) 1 9
(b) 2 0
(c) 1 8
(d) 4 0
The number of electrons and neutrons of an element is 1 8
and 2 0 respectiv ely . Its m ass num ber is
[CPMT 1997; Pb. PMT 1999; MP PMT 1999]
The num ber of electrons in the nucleus of C 12 is
(a) 6
(b) 1 2
(c) 0
(d) 3
An elem ent has electronic configuration 2 , 8, 1 8, 1 . If its
atom ic weight is 6 3 , t hen how m any neutrons will be
present in its nucleus
(a) 3 0
(b) 3 2
(c) 3 4
(d) 3 3
40
The num ber of electrons in [19
K ]1 is
[CPMT 1997; AFMC 1999]
The hy dride ions (H  ) are isoelectronic with
(a) Li
(c) He
51
59.
19
C
(c) 1.6  10 19 C
Iso-electronic species is

(a) F , O
2
(b) 1  10 19 C
(d) 2.6  10 19 C
[RPMT 2002]

(b) F , O
52 Structure
of atom
(c) F  , O 
60.
An elem ent hav e atom ic weight 4 0 and it’s electronic
61.
configuration is 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 . Then its atom ic
number and number of neutrons will be [RPMT 2002]
(a) 1 8 and 2 2
(b) 2 2 and 1 8
(c) 2 6 and 2 0
(d) 4 0 and 1 8
The nucleus of tritium contains
[MP PMT 2002]
(a) 1 proton + 1 neutron (b) 1 proton + 3 neutron
(c) 1 proton + 0 neutron (d) 1 proton + 2 neutron
Which one of the following groupings represents a
collection of isoelectronic species
[A IEEE 2003]
62.
63.
(a) Na  , Ca 2  , Mg 2 
(b) N 3  , F  , Na 
(c) Be, Al 3  , Cl 
(d) Ca 2  , Cs  , Br
Which of the following are isoelectronic and isostructural
NO 3 , CO 32  , ClO3 , SO 3
64.
65.
66.
67 .
(b) 4 protons and 7 electrons
(c) 4 protons and 1 0 electrons
(d) 1 0 protons and 7 electrons
(d) F  , O 2
73.
[MP PMT 1986]
(a) 0
74.
(b) SO 3 , NO 3
(c) ClO3 , CO 32 
(d) CO 32  , SO 3
69.
1.
(a) 6
(b) 1 1
(c) 1 7
(d) 2 3
The nucleus of an element contain 9 protons. Its v alency
would be
(a) 1
(b) 3
(c) 2
(d) 5
The compound in which cation is isoelectronic with anion is
2.
3.
7 1.
(c) Nucleus
(d) Neutrons
Rutherford's scattering experiment is related to the size of
the
(a) Nucleus
(b) Atom
(c) Electron
(d) Neutron
Rutherford's alpha particle scattering experim ent
ev entually led to the conclusion that[IIT 1986; RPMT 2002]
(d) The point of im pact with m atter can be precisely
determ ined
4.
Bohr's m odel can explain
[MP PET 2004]
[IIT 1985]
(a) The spectrum of hy drogen atom only
(b) Spectrum of atom or ion containing one electron only
(c) The spectrum of hy drogen m olecule
5.
When atoms are bombarded with alpha particles, only a
few in m illion suffer deflection, others pass out
undeflected. This is because[MNR 1979; NCERT 1980; A FMC 19
(a) The force of repulsion on th e m oving alpha particle is
sm all
2
(b) The force of attraction on the alpha particle to the
oppositely charged electrons is v ery sm all
(a) Boron
(b) Lithium
(c) Carbon
(d) Helium
The nitrogen atom has 7 protons and 7 electrons, the
(a) 7 protons and 1 0 electrons
(b) Protons
(c) Neutrons are buried deep in the nucleus
(c) F 
(d) Ca 2 
Six protons are found in the nucleus of
3
nitride ion ( N ) will hav e
(a) Electrons
(b) Electrons occupy space around the nucleus
(c) There is only one nucleus and large num ber of
electrons
[CPMT 1977, 80, 81; NCERT 1975, 78]
72.
Rutherford's experiment on scattering of particles showed
for the first tim e that the atom has
(a) Mass and energy are related
[DCE 2004]
(b) O
(d) Atom ic num ber
[IIT 1983; MADT Bihar 1995; BHU 1995]
Which among the following species have the same number of
electrons in its outerm ost as well as penultim ate shell
(a) Mg
(c) Equiv alent weight
(d) The solar spectrum
(b) CsF
(d) K 2 S
2
(b) Atom ic radii
CPMT 1984; Kurukshetra CEE 1998]
[UPSEAT 2004]
70.
(a) Atom ic weight
[IIT 1981; NCERT 1981; CMC Vellore 1991;
The number of electrons in Cl  ion is [MP PMT 2003]
(a) 1 9
(b) 2 0
(c) 1 8
(d) 3 5
The number of neutron in tritium is
[CPMT 2003]
(a) 1
(b) 2
(c) 3
(d) 0
Tritium is the isotope of
[CPMT 2003]
(a) Hy drogen
(b) Oxy gen
(c) Carbon
(d) Sulpher
The atomic number of an element is 3 5. What is the total
num ber of electrons present in all the p-orbitals of the
ground state atom of that elem ent
(a) NaCl
(c) NaI
(c) 2
(d) 3
Which of the following is alway s a whole num ber
Atomic models and Planck's quantum theory
[EA MCET (Engg.) 2003]
68.
(b) 1
[CPMT 1976, 81, 86]
[IIT Scr een in g 2003]
(a) NO 3 , CO 32 
Num ber of neutrons in heav y hy drogen atom is
[NCERT 1977]
(d) The nucleus occupies m uch sm aller v olum e
com pared to the v olum e of the atom
6.
Positronium consists of an electron and a positron (a
particle which has the sam e m ass as an electron, but
opposite charge) orbiting round their com m on centre of
Structure of atom
m ass. Calculate the value of the Rydberg constant for this
sy stem.
7.
(a) R / 4
(b) R / 2
(c) 2 R
(d) R
When  -particles are sent through a thin m etal foil, most
of them go straight through the foil because (one or m ore
are correct)
[IIT 1984]
(a) Positiv e ray analy sis
(b)  -ray scattering experim ents
(c) X-ray analy sis
15.
(b) Alpha particles are positiv ely charged
16.
(d) Alpha particles m ov e with high v elocity
8.
[CPMT 1983]
(c) The difference in the energy of the energy lev els
inv olv ed in the transition
(d) The v elocity of the electron undergoing the transition
(a) Energy is absorbed
(b) Energy is released
17.
(d) Energy is neither absorbed nor released
9.
10.
(a) A beam of protons
(b)  -ray s
(c) A beam of neutrons
(d) X-ray s
Which one of the following is not the characteristic of
Planck's quantum theory of radiation
[A IIMS 1991]
(a) The energy is not absorbed or em itted in whole
num ber or m ultiple of quantum
(b) Radiation is associated with energy
(c) Radiation energy is not em itted or absorbed conti nuously but in the form of sm all packets called
quanta
(d) This m agnitude of energy associated with a quantum
is proportional to the frequency
11.
The spectrum of He is expected to be sim ilar to
[A IIMS 1980, 91; DPMT 1983; MP PMT 2002]
(a) H
12.
13.
(b) Li 
(c) Na
(d) He 
Energy of orbit
[DPMT 1984, 91]
(a) Increases as we m ov e away from nucleus
(b) Decreases as we m ov e away from nucleus
(c) Rem ains sam e as we m ov e away from nucleus
(d) None of these
Bohr m odel of an atom could not account for
(a) Em ission spectrum
(b) Absorption spectrum
(b) Energy is absorbed
(c) Atom ic num ber increases
(d) Atom ic num ber decreases
18.
Dav isson and Germ er's experim ent showed that
[MA DT Bihar 1983]
(a)  -particles are electrons
(b) Electrons com e from nucleu s
(c) Electrons show wav e nature
(d) None of the abov e
19. When an electron jum ps from lower to higher orbit, its
energy
[MA DT Bi h a r 1982]
(a) Increases
(b) Decreases
(c) Rem ains the sam e
(d) None of these
20. Experimental ev idence for the existence of the a tom ic
nucleus com es from
[CBSE PMT 1989]
(a) Millikan's oil drop experim ent
(b) Atom ic em ission spectroscopy
(c) The m agnetic bending of cathode ray s
(d) Alpha scattering by a thin m etal foil
21. Which of the following statem ents does not form part of
Bohr's m odel of the hydrogen atom
[CBSE PMT 1989]
(a) Energy of the electrons in the orbit is quantized
(b) The electron in the orbit nearest the nucleus has the
lowest energy
(c) Electrons rev olv e in different orbits around the
nucleus
(d) The position and velocity of the electrons in the orbit
cannot be determ ined sim ultaneously
22. When  -particles are sent through a tin m etal foil, m ost
of them go straight through the foil as [EA MCET 1983]
(a)  -particles are m uch heav ier than electrons
(c) Line spectrum of hy drogen
(d) Fine spectrum
14.
When an electron drops from a higher energy lev el to a
low energy level, then
[A MU 1985]
(a) Energy is em itted
When bery llium is bom barded with  -particles,
extrem ely penetrating radiations which cannot be
deflected by electrical or m agnetic field are giv en out.
These are
[CPMT 1983]
(c) Heisenberg's principle
(b) Prout's hy pothesis
The wav elength of a spectral line for an electronic
transition is inversely related to
[IIT 1988]
(a) The number of electrons undergoing the transition
(b) The nuclear charge of the atom
When an electron jumps from L to K shell
(c) Energy is som etim es absorbed and som etim es
released
(d) Discharge tube experim ents
Electron occupies the av ailable orbital singly before
pairing in any one orbital occurs, it is [CBSE PMT 1991]
(a) Pauli's exclusion principle
(b) Hund's Rule
(a) Alpha particles are m uch heav ier than electrons
(c) Most part of the atom is em pty space
53
Existence of positively charged nucleus was established by
[CBSE PMT 1991]
(b)  -particles are positiv ely charged
(c) Most part of the atom is em pty space
54 Structure
of atom
(d)  -particles m ov e with high v elocity
23.
(a) E n  
The energy of second Bohr orbit of the hydrogen atom is –
3 2 8 kJ mol–1, hence the energy of fourth Bohr orbit would be
(c) E n  
[CBSE PMT 2005]
24.
(a) – 4 1 kJ mol–1
(b) –1 3 1 2 kJ mol–1
(c) –1 6 4 kJ mol–1
(d) – 82 kJ mol–1
When an electron rev olv es in a stationary orbit then
33.
[MP PET 1994]
25.
26.
27 .
(a) It absorbs energy
(b) It gains kinetic energy
(c) It em its radiation
(d) Its energy rem ains constant
A m ov ing particle m ay hav e wav e m otion, if
(a) Its m ass is v ery high
(b) Its v elocity is negligible
(c) Its m ass is negligible
(d) Its m ass is v ery high and v elocity is negligible
The postulate of Bohr theory that electrons jum p from
one orbit to the other, rather than flow is according to
(a) The quantisation concept
(b) The wav e nature of electron
(c) The probability expression for electron
(d) Heisenberg uncertainty principle
The frequency of an electrom agnetic radiation is
34.
35.
36.
37 .
2  10 Hz . What is its wav elength in m etres
(a) 6.0  10
38.
(b) 1.5  10
14
4
(c) 1.5  10
(d) 0.66  10 2
What is the packet of energy called
[A FMC 2005]
(a) Electron
(b) Photon
(c) Positron
(d) Proton
2
28.
29.
The energy of an electron in n th orbit of hydrogen atom is
[MP PET 1999]
30.
31.
(a)
13 .6
(c)
13 .6
n4
n2
eV
(b)
13 .6
eV
(d)
13 .6
eV
n
n3
39.
eV
If wav elength of photon is 2.2  10 11 m, h  6.6  10 34 Jsec, then m om entum of photon is
[MP PET 1999]
(a) 3  10 23 kg ms 1
(b) 3.33  10 22 kg ms 1
(c) 1.452  10 44 kg ms 1
(d) 6.89  10 43 kg ms 1
40.
41.
2 me z
4
(a) r 
(c) r 
n2h2
4 me z
2
4
2
n2h2
4 2 me 2 z 2
(b) r 
(d) r 
n2h2
4 2 me 2 z
(d) E n  
n2h2
2m 2 e 2 z 4
(a) Absorption of energy
(b) Release of energy
(c) Both release or absorption of energy
(d) Unpredictable
In an element going away from nucleus, the energy of
particle
[RPMT 1997]
(a) Decreases
(b) Not changing
(c) Increases
(d) None of these
The  -particle scattering experim ent of Rutherford
concluded that
[Or issa JEE 1997]
(a) The nucleus is m ade up of protons and neutrons
(b) The number of electrons is exactly equal to number of
protons in atom
(c) The positive charge of the atom is concentrated in a
v ery sm all space
(d) Electrons occupy discrete energy lev els
Wavelength associated with electron m otion [BHU 1998]
(a) Increases with increase in speed of electron
(b) Rem ains sam e irrespectiv e of speed of electron
(c) Decreases with increase in speed of e 
(d) Is zero
The elem ent used by Rutherford in his famous scattering
experim ent was
[KCET 1998]
(a) Gold
(b) Tin
(c) Silv er
(d) Lead
If electron falls from n  3 to n  2 , then emitted energy
is
[A FMC 1997; MP PET 2003]
42.
n2h2
4 2 m 2 e 2 z 2
The energy of an electron revolving in n th Bohr's orbit of
an atom is given by the expression
[MP PMT 1999]
2
2 2 me 2 z 2
n2h2
n2h2
Who m odified Bohr's theory by introducing ellipt ical
orbits for electron path [CBSE PMT 1999; A FMC 2003]
(a) Hund
(b) Thom son
(c) Rutherford
(d) Som m erfield
Bohr's radius can have
[DPMT 1996]
(a) Discrete v alues
(b) ve v alues
(c) ve v alues
(d) Fractional v alu es
The first use of quantum theory to explain the structure of
atom was made by[IIT 1997; CPMT 2001; J&K CET 2005]
(a) Heisenberg
(b) Bohr
(c) Planck
(d) Einstein
An electronic transition from 1s orbital of an atom causes
The expression for Bohr's radius of an atom is
[MP PMT 1999]
32.
n2h2
2
(b) E n  
[JIPMER 1997]
6
(Velocity of light  3  10 8 ms 1 )
2 2 m 4 e 2 z 2
43.
(a) 10.2eV
(b) 12.09eV
(c) 1.9 eV
(d) 0.65eV
The radius of the nucleus is related to the m ass num ber
A by
(a) R  R o A
1/2
(b) R  R o A
(c) R  R o A
2
(d) R  R o A 1 / 3
The specific charge of proton is 9.6  10 6 C kg 1 then for an
 -particle it will be
[MH CET 1999]
[
Structure of atom
44.
45.
46.
47 .
(a) 38.4  10 7 C kg 1
(b) 19.2  10 7 C kg 1
(c) 2.4  10 7 C kg 1
(d) 4.8  10 7 C kg 1
In hy drogen spectrum the different lines of Ly m an series
are present is
[UPSEA T 1999]
(a) UV field
(b) I R field
(c) Visible field
(d) Far I R field
Which one of the following is considered as the m ain
postulate of Bohr’s m odel of atom
[A MU 2000]
(a) Protons are present in the nucleus
(b) Electrons are rev olv ing around the nucleus
(c) Centrifugal force produced due to the rev olv ing
electrons balances the force of attraction between the
electron and the protons
(d) Angular momentum of electron is an integral multiple of
h
2
The electronic energy levels of the hydrogen atom in the
Bohr’s theory are called
[A MU 2000]
(a) Ry dberg lev els
(b) Orbits
(c) Ground states
(d) Orbitals
The energy of a photon is calculated by [Pb. PMT 2000]
(a) E  h
(b) h  E
(c) h 
48.
49.
E
55.
56.
energy of a radiation of wavelength 16000 Å is E 2 . What
is the relation between these two
[Kerala CET 2005]
(a) E1  6 E 2
(b) E1  2E 2
(c) E1  4 E 2
(b) 0 .22 Å
(c) 0 .28 Å
(d) 0 .53 Å
In Balm er ser ies of hy drogen atom spectrum which
electronic transition causes third line [MP PMT 2000]
58.
59.
60.
61.
(c) Fourth Bohr orbit to second one
(d) Fourth Bohr orbit to first one
Energy of electron of hydrogen atom in second Bohr orbit
is
[MP PMT 2000]
52.
(a)  5.44  10 19 J
(b)  5.44  10 19 kJ
(c)  5.44  10 19 cal
(d)  5.44  10 19 eV
If change in energy (E)  3  10 8 J , h  6.64  10 34 J - s
and c  3  10 8 m/s, then wavelength of the light is
[CBSE PMT 2000]
(a) 6.36  10 Å
3
8
53.
The form ation of energy bonds in solids are in accordance
with
[DCE 2001]
(a) Heisenberg’s uncertainty principle
(b) Bohr’s theory
(c) Ohm ’s law
(d) Rutherford’s atom ic m odel
The frequency of y ellow light having wavelength 600 nm
is
[MP PET 2002]
(a) Fifth Bohr orbit to second one
(b) Fifth Bohr orbit to first one
51.
(b) 6.36  10 5 Å
(c) 6.64  10 Å
(d) 6.36  10 18 Å
The radius of first Bohr’s orbit for hydrogen is 0.53 Å. The
radius of third Bohr’s orbit would be
[MP PMT 2001]
(a) 0.7 9 Å
(b) 1 .59 Å
(c) 3 .1 8 Å
(d) 4 .7 7 Å
(d) E1  1 / 2 E2
(e) E1  E 2
57 .
h
(d) E 


Visible range of hy drogen spectrum will contain the
following series
[RPET 2000]
(a) Pfund
(b) Ly m an
(c) Balm er
(d) Brackett
Radius of the first Bohr’s orbit of hy drogen atom is
(a) 1 .06 Å
Rutherford’s  -particle scattering experiment proved that
atom has
[MP PMT 2001]
(a) Electrons
(b) Neutron
(c) Nucleus
(d) Orbitals
Wav elength of spectral line em itted is inv ersely
proportional to
(a) Radius
(b) Energy
(c) Velocity
(d) Quantum num ber
The energy of a radiation of wavelength 8000 Å is E1 and
(a) 5.0  10 Hz
(c) 5.0  10 7 Hz
The v alue of the energy
hy drogen atom will be
(a)  13.6 eV
(b) 2.5  10 Hz
(d) 2.5  10 14 Hz
for the first excited state of
(c)  1.51 eV
(d)  0.85 eV
14
[RPET 2000]
50.
54.
55
62.
7
[MP PET 2002]
(b)  3.40 eV
Bohr m odel of atom is contradicted by [MP PMT 2002]
(a) Pauli’s exclusion principle
(b) Planck quantum theory
(c) Heisenberg uncertainty principle
(d) All of these
Which of the following is not true in Rutherford’s nuclear
m odel of atom
[Or issa JEE 2002]
(a) Protons and neutrons are present inside nucleus
(b) Volum e of nucleus is v ery sm all as com pared to
v olum e of atom
(c) The num ber of protons and neutrons are alway s
equal
(d) The num ber of electrons and protons are alway s
equal
The em ission spectrum of hydrogen is found to satisfy the
expression for the energy change. E (in joules) such
 1
1 
that E  2 .18  10 2  2  J where n 1 = 1, 2 , 3 ….. and
 n1 n2 
n 2 = 2 , 3 , 4 ……. The spectral lines correspond to Paschen
series to
[UPSEA T 2002]
(a) n1  1 and n 2  2, 3, 4
(b) n1  3 and n 2  4, 5, 6
(c) n1  1 and n 2  3, 4, 5
(d) n1  2 and n 2  3, 3, 5
(e) n1  1 and n 2  infinity
[
56 Structure
63.
of atom
The ratio between kinetic energy and the total energy of
the electrons of hydrogen atom according to Bohr’s m odel
is
74.
state 1 , would be (Ry dberg constant  1.097 10 7 m 1 )
[Pb. PMT 2002]
64.
65.
[A IEEE 2004]
(a) 2 : 1
(b) 1 : 1
(c) 1 : – 1
(d) 1 : 2
Energy of the electron in Hy drogen atom is giv en by
[A MU (Engg.) 2002]
(a) En  
131.38
131.33
kJ mol 1 (b) En  
kJ mol 1
n
n2
(c) En  
1313.3
2
kJ mol 1 (d) En  
313.13
2
kJ mol 1
n
n
Ratio of radii of second and first Bohr orbits of H atom
75.
76.
[BHU 2003]
(a) 2
(c) 3
66.
(b) 4
(d) 5
The frequency corresponding to transition n  2 to n  1
in hydrogen atom is
[MP PET 2003]
(a) 15.66  10 10 Hz
67 .
(a) 0.8268  10 34 kg
68.
69.
(b) 1.2876  10 33 kg
(c) 1.4285  10 32 kg
(d) 1.8884  10 32 kg
Splitting of spectral lines under the influence of m agnetic
field is called
[MP PET 2004]
(a) Zeem an effect
(b) Stark effect
(c) Photoelectric effect
(d) None of these
The radius of electron in the first excited state of
hy drogen atom is
[MP PMT 2004]
(a) a0
(b) 4a0
(c) 2a0
70.
7 1.
7 8.
79.
80.
(d) 8a0
(c)
4 2 mr 2
nh
(b)
nh
81.
4 2 mr
h
(d)
2mr
nh
(a) 4 06 nm
(b) 1 92 nm
(c) 91 nm
(d) 9.110 8 nm
In Bohr’s m odel, atomic radius of the first orbit is  , the
radius of the 3 rd orbit, is [MP PET 1997; Pb. CET 2001]
(a)  / 3
(b) 
(c) 3
(d) 9
According to Bohr’s principle, the relation between
principle quantum number (n) and radiu s of orbit is [BHU 2004]
(a) r  n
(b) r  n 2
1
1
(c) r 
(d) r  2
n
n
The ionisation potential of a hydrogen atom is –1 3 .6 eV.
What will be the energy of the atom corresponding to
n2
[Pb. CET 2000]
The ratio of area cov ered by second orbital to the first
orbital is
[A FMC 2004]
(a) 1 : 2
(b) 1 : 1 6
(c) 8 : 1
(d) 1 6 : 1
Tim e taken for an electron to com plete one rev olution in
the Bohr orbit of hydrogen atom is [Ker a l a PMT 2004]
(a)
72.
77.
(b) 24.66  10 14 Hz
(c) 30.57  10 14 Hz
(d) 40.57  10 24 Hz
The m ass of a photon with a wav elength equal to
[Pb. PMT 2004]
1.54  10 8 cm is
The wav elength of the radiation em itted, when in a
hy drogen atom electron falls from infinity to stationary
4 2 mr 2
The radius of which of the following orbit is sam e as that
of the first Bohr's orbit of hy drogen atom
(a) –3 .4 eV
(b) –6.8 eV
(c) –1 .7 eV
(d) – 2 .7 eV
The energy of electron in hy drogen atom in its grounds
state is –13.6 eV. The energy of the level corresponding to
the quantum number equal to 5 is
[Pb. CET 2002]
(a) –0.54 eV
(b) – 0.85 eV
(c) – 0.64 eV
(d) – 0.4 0 eV
The positive charge of an atom is
[A FMC 2002]
(a) Spread all ov er the atom
(b) Distributed around the nucleus
(c) Concentrated at the nucleus
(d) All of these
A metal surface is exposed to solar radiations [DPMT 2005]
(a) The em itted electrons hav e energy less than a
m aximum value of energy depending upon frequency
of incident radiations
(b) The em itted electrons hav e energy less than
m aximum value of energy depending upon intensity
of incident radiation
(c) The em itted electrons hav e zero energy
(d) The em itted electrons have energy equal to energy of
photos of incident light
Which of the following transitions hav e m inim um
wav elength
[DPMT 2005]
(a) n4  n1
(b) n 2  n1
(c) n4  n 2
Dual nature of electron
[IIT Screening 2004]
(a) He  (n  2)
73.
(b) Li 2 (n  2)
2
3
(c) Li (n  3)
(d) Be (n  2)
The frequency of radiation em itted when the electron falls
from n  4 to n  1 in a hy drogen atom will be (Giv en
ionization
h  6.625  10
energy
34
Js )
(a) 3.08  1015 s 1
(c) 1.54  1015 s 1
of
H  2.18  10
18
J atom
1
and
[CBSE PMT 2004]
(b) 2.00  1015 s 1
(d) 1.03  1015 s 1
(d) n3  n1
1.
De broglie equation describes the relationship of wavelengt h
associated with the motion of an electron and its[MP PMT
1986]
2.
(a) Mass
(b) Energy
(c) Mom entum
(d) Charge
The wav e nature of an electron was first giv en by
[CMC V ellore 1991; Pb. PMT 1998; CPMT 2004]
(a) De-Broglie
(c) Mosley
(b) Heisenberg
(d) Som m erfield
57
Structure of atom
3.
4.
Am ong the following for which one m athem atical
h
expression  
stands
p
(a) De Broglie equation
(b) Einstein equation
(c) Uncertainty equation (d) Bohr equation
Which one of the following explains light both as a stream
of particles and as wav e m otion
[A IIMS 1983; IIT 1992; UPSEAT 2003]
(c) Interference
(d) Photoelectric effect
In which one of the following pairs of experim ental
observ ations and phenom enon does the experim ental
observation correctly account for phenomenon
[A IIMS 2001]
(a) 6.63  10
6.
m
hv
de-Broglie equation is
(d)  
(c)  
7.
14.
15.
16.
8.
17.
v
mh
(b) E  hv
h
(d)  
mv
18.
19.
A FMC 1999; AIIMS 2000]
(a) 6.63  10
9.
(b) 6.63  10 34 m
m
(c) 6.63  10 35 m
(d) 6.65  10 35 m
Minimum de-Broglie wavelength is associated with [RPMT
(b) Proton
(d) SO 2 m olecule
10.
The de-Broglie wav elength associated with a m aterial
particle is
[JIPMER 2000]
(a) Directly proportional to its energy
(b) Directly proportional to m om entum
(c) Inv ersely proportional to its energy
(d) Inv ersely proportional t o m om entum
11.
An electron has kinetic energy 2.8  10 23 J . de-Broglie
wav elength will be nearly
(m e  9.1  10 31 kg )
(a) 9.28  10
4
[MP PET 2000]
(b) 9.28  10
m
(c) 9.28  10 8 m
12.
20.
1999]
(a) Electron
(c) CO 2 m olecule
7
21.
(d) 6.63  10 34 m
(a) 9.96  10 10 cm
(b) 9.96  10 8 cm
(c) 9.96  10 4 cm
(d) 9.96  10 8 cm
If the v elocity of hy drogen m olecule is 5  10 4 cm sec 1 ,
[Ade-Broglie
IIMS 1983]
then its
wavelength is
[MP PMT 2003]
(a) 2 Å
(b) 4 Å
(c) 8 Å
(d) 1 00 Å
A 2 00g golf ball is m oving with a speed of 5 m per hour.
The associated wav e length is (h  6.625  10 34 J - sec)
(a) 10 10 m
(b) 10 20 m
(c) 10 30 m
(d) 10 40 m
A cricket ball of 0.5 kg is m ov ing with a v elocity of
100 m / sec . The wavelength associated with its m otion is
with a velocity of 1.2  10 ms
1
(a) 6.068  10 9
(b) 3.133  10 37
(c) 6.626  10 9
(d) 6.018  10 7
(c) 1.32  10 35 m
(d) 6.6  10 28 m
Dual nature of particles was proposed by
[DCE 2004]
(a) Heisenberg
(b) Lowry
(c) de-Broglie
(d) Schrodinger
Calculate de-Broglie wavelength of an electron trav elling
at 1% of the speed of light
[DPMT 2004]
(a) 2.73  10 24
(b) 2.42  10 10
(c) 242.2  10 10
(d) None of these
Which is the correct relationship between wavelength and
m omentum of particles
[Pb. PMT 2000]
h
h
(a)  
(b)  
P
P
h
P
(c) P 
(d) h 


The de-Broglie equation applies
[MP PMT 2004]
(a) To electrons only
(b) To neutrons only
(c) To protons only
(d) All the m aterial object in m otion
1.
The uncertainty principle was enunciated by
2.
(a) Einstein
(b) Heisenberg
(c) Rutherford
(d) Pauli
According to heisenberg uncertainty principle
[NCERT 1975; Bihar MEE 1997]
m
What will be de-Broglie wavelength of an electron m ov ing
(b) 6.6  10 34 m
Uncertainty principle and Schrodinger wave
equation
(d) 9.28  10 10 m
5
m
[DCE 2004]
The de-Broglie wavelength of a particle with m ass 1 gm
and velocity 100m / sec is[CBSE PMT 1999; EA MCET 1997;
33
(b) 6.63  10
[MP PET 2003]
[MP PMT 1999; CET Pune 1998]
(c) E  mc 2
m
29
What is the de-Broglie wav elength associated with the
hydrogen electron in its third orbit [A MU (En gg.) 2002]
(a) 1 / 100cm
(a) n  2d sin
22
(c) 6.63  10 31 m
Experimental observa t ion Ph en om en on
(a) X -ray spectra
Charge on the nucleus
(b)  -particle scattering Quantized electron orbit
(c) Em ission spectra
The quantization of energy
(d) The photoelectric effect The nuclear atom
Which of the following expressions giv es the de-Broglie
relationship[MP PMT 1996, 2004; MP PET /PMT 1998]
h

(a) h 
(b)  
mv
mv
The de-Broglie wavelength associated with a particle of
m ass 10 6 kg m ov ing with a v elocity of 10 ms 1 , is
(b)   h / p
(a) Diffraction
5.
13.
[A MU 1990; BCECE 2005]
[MP PET 2000]
(a) E  mc 2
(b) x  p 
h
4
58 Structure
of atom
h
h
(d) x  p 
6
p
“The position and velocity of a sm all particle like electron
cannot be simultaneously determined.” This statement is
(c)  
3.
(c) Heisenberg, Planck
12.
1  10 5 kg  m / s . The uncertainty in its position will be
( h  6.62  10 34 kg  m 2 / s )
[NCERT 1979; BHU 1981, 87]
(a)
(b)
(c)
(d)
4.
5.
Heisenber g uncertainty principle
Principle of de Broglie's wav e nature of electron
Pauli's exclusion principle
Aufbau's principle
h
In Heisenberg's uncertainty equation x  p 
; p
4
stands for
(a) Uncertainty in energy
(b) Uncertainty in v elocity
(c) Uncertainty in m om entum
(d) Uncertainty in m ass
Which one is not the correct relation in the following
(a) h 
6.
7.
E
v
[A FMC 1998; CBSE PMT 1999; JIPMER 2002]
13.
14.
(c) x  p 
h
h
(d)  
mv
4
The m aximum probability of finding an electron in the
[MP PET 1996]
d xy orbital is
15.
(a) Along the x-axis
(b) Along the y-axis
16.
(d) 5.25  10 28 m
The uncertainty in the position of a m oving bullet of m ass
1 0 gm is 10 5 m . Calculate the uncertainty in its v elocity
28
m / sec
(b) 3.0  10
28
m / sec
(d) 3  10 22 m / sec
h
shows
[MP PET 2000]
4
(a) de-Broglie relation
(b) Heisenberg’s uncertainty principle
(c) Aufbau principle
(d) Hund’s rule
Which quantum number is not related with Schrodinger
equation
[RPMT 2002]
(a) Principal
(b) Azim uthal
(c) Magnetic
(d) Spin
The equation x .p 
Uncertainty in position of a 0.2 5 g particle is 10 5 .
Uncertainty of velocity is (h  6.6  10 34 Js) [A IEEE 2002]
(b) 2.1  10 29
(c) At an angle of 45 o from the x and y-axes
(a) 1.2  10 34
(d) At an angle of 90 o from the x and y-axes
(c) 1.6  10
(d) 1.7  10 9
The uncertainty in m om entum of an electron is
1  10 5 kg m / s . The uncertainity in its position will be
Sim ultaneous determ ination of exact position and
m omentum of an electron is
[BHU 1979]
20
17.
(h  6.63  10 34 Js)
(a) 5.28  10
h
(c) 
2
30
m
18.
h
(b) 
2
19.
The possibility of finding an electron in an orbital was
conceiv ed by
[MP PMT 1994]
(a) Rutherford
(b) Bohr
(c) Heisenberg
(d) Schrodinger
Uncertainty principle gav e the concept of
(a) Probability
(b) An orbital
(c) Phy sical m eaning of  the  2
(d) All the abov e
The uncertainty principle and the concept of wave nature
of m atter was proposed by ...... and ...... respectiv ely
[MP PET 1997]
(a) Heisenberg, de Broglie (b) de-Broglie, Heisenberg
(b) 5.25  10 28 m
(c) 1.05  10 m
(d) 2.715  10 30 m
According to Heisenberg’s uncertainty principle, the
product of uncertainties in position and v elocities for an
electron of m ass 9.1  10 31 kg is
3
2 1
5
2 1
(a) 2.8  10 m s
(d) Infinite
[Pb. CET 2000]
26
If uncertainty in the position of an electron is zero, the
uncertainty in its m om entum would be
[CPMT 1988]
(a) Zero
11.
(c) 5.27  10 30 m
(c) 5.2  10 22 m / sec
(b) E  mc 2
(d) None of the abov e
10.
(b) 1.05  10 26 m
(a) 5.2  10
(c) Som etim es possible som etim es im possible
9.
(a) 1.05  10 28 m
[DCE 1999]
(a) Possible
(b) Im possible
8.
(d) Planck, Heisenberg
The uncertainty in m om entum of an electron is
2 1
(b) 3.8  10 m s
(c) 5.8  10 m s
(d) 6.8  10 6 m 2 s 1
For an electron if the uncertainty in v elocity is  , the
uncertainty in its position (x ) is given by [DPMT 2005]
hm
4
h
(c)
4m 
(a)
20.
5
4
hm 
4 m
(d)
h . 
(b)
Orbital is
[DPMT 2005]
(a) Circular path around the nucleus in which the
electron rev olv es
(b) Space around the nucleus where the probability of
finding the electron is m axim um
(c) Am plitude of electrons wav e
(d) None of these
Quantum number, Electronic configuration
[
59
Structure of atom
and Shape of orbitals
1.
Be's 4 th electron will hav e four quantum num bers
[MNR 1985]
2.
3.
l
n
m
s
(a) 1
0
0
+ 1 /2
(b) 1
1
+1
+ 1 /2
(c) 2
0
0
– 1 /2
(d) 2
1
0
+ 1 /2
The quantum number which specifies the location of an
electron as w ell as energy is
[DPMT 1983]
(a) Principal quantum num ber
(b) Azim uthal quantum num ber
(c) Spin quantum num ber
(d) Magnetic quantum num ber
The shape of an orbital is given by the quantum number
[NCERT 1984; MP PMT 1996]
4.
(a) n
(b) l
(c) m
(d) s
In a given atom no two electrons can have the same values
for all the four quantum num bers. This is called
[BHU 1979; A MU 1983; EAMCET 1980, 83;
MA DT Bihar 1980; CPMT 1986, 90, 92; NCERT 1978, 84;
RPMT 1997; CBSE PMT 1991; MP PET 1986, 99]
5.
1
1
(d) 6, 0, 0, 
2
2
The correct ground state electronic configuration of
chromium atom is[IIT 1989, 94; MP PMT 1993; EA MCET 1997;
(c) 5, 1, 1, 
(a) Hund's rule
(b) Aufbau's principle
(c) Uncertainty principle
(d) Pauli's exclusion principle
Nitrogen
has
the
electronic
configuration
1s 2 ,2 s 2 2 p 1x 2 p 1y 2 p 1z and not 1s 2 ,2 s 2 2 p x2 2 p 1y 2 p z0 which is
10.
ISM Dh anbad 1994; AFMC 1997; Bihar MEE 1996;
MP PET 1995, 97; CPMT 1999; Kerala PMT 2003]
11.
(a) [ Ar] 3d 5 4 s 1
(b) [ Ar] 3d 4 4 s 2
(c) [ AR]3d 6 4 s 0
(d) [ Ar]4 d 5 4 s 1
2 p orbitals have
(a) n  1, l  2
[NCERT 1981; MP PMT 1993, 97]
(b) n  1, l  0
(c) n  2, l  1
12.
13.
14.
(d) n  2, l  0
Electronic configuration of H  is
(a) 1s 0
(b) 1s 1
(c) 1s 2
(d) 1s 1 2 s 1
The quantum numbers for the outerm ost electron of an
1
elem ent are given below as n  2, l  0, m  0, s   . The
2
atom s is
[EA MCET 1978]
(a) Lithium
(b) Bery llium
(c) Hy drogen
(d) Boron
Principal quantum num ber of an atom represents
[EA MCET 1979; IIT 1983; MNR 1990;UPSEAT 2000, 02]
15.
(a)
(b)
(c)
(d)
An
Size of the orbital
Spin angular m om entum
Orbital angular m om entum
Space orientation of the orbital
elem ent has the electronic
determ ined by
(a) Aufbau's principle
(b) Pauli's
exclusion
principle
(c) Hund's rule
(d) Uncertainty principle
6.
Which one of the following configuration represents a
noble gas
[NCERT 1973]
16.
6
(a) 1s ,2 s 2 p ,3 s
7.
2
2
2
6
(b) 1s ,2 s 2 p ,3 s1
(c) 1s 2 ,2 s 2 2 p 6
(d) 1s 2 ,2 s 2 sp6 ,3 s 2 3 p 6 ,4 s 2
The electronic configuration of silver atom in ground state
is
17.
(a) [Kr]3d
4s
1
(c) [Kr] 4 d 5 s
10
8.
9.
1
(b) [ Xe] 4 f 5 d 6 s
14
10
1
(d) [Kr] 4 d 9 5 s 2
Principal, azimuthal and m agnetic quantum numbers are
respectively related to
[CPMT 1988; A IIMS 1999]
(a) Size, shape and orientation
(b) Shape, size and orientation
(c) Size, orientation and shape
(d) None of the abov e
Correct set of four quantum numbers for valence electron
of rubidium (Z = 3 7 ) is
(b) 5, 1, 0, 
1
2
l
m
ms
(a) 3
2
–2
(+ )
1
2
(b) 4
0
0
(–)
1
2
(c) 3
2
–3
(+ )
1
2
(d) 5
3
0
(–)
1
2
18.
If n  3 , then the v alue of ' l' which is incorrect
19.
(a) 0
(b) 1
(c) 2
(d) 3
Which orbital is dum b-bell shaped
[CPMT 1994]
[IIT 1984; JIPMER 1999; UPSEAT 2003]
1
(a) 5, 0, 0, 
2
(a) Size of orbitals
(b) Shape of orbitals
(c) Orientation of orbitals (d) Nuclear stability
Which of the following sets of quantum num bers
represent an impossible arrangement[IIT 1986; MP PET 1995]
n
[CPMT 1984, 93]
10
(a) 6
(b) 2
(c) 3
(d) 4
The m agnetic quantum num ber specifies
[MNR 1986; BHU 1982; CPMT 1989, 94;
[CPMT 1983,
89,
93;1999;
NCERT
1973;
MPAMU
PMT (Engg.)
1989; 1999]
MP
PET
AFMC
1999;
DPMT 1984]
2
configuration
1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 2 . Its v alency electrons are
[DPMT 1982, 83, 89; MP PMT/PET 1988; EAMCET 1988]
2
[CPMT 1985]
[MP PMT 1986; MP PET/PMT 1998]
60 Structure
20.
21.
of atom
(a) s -orbital
(b) p -orbital
(c) d -orbital
(d) f -orbital
[MNR 1988; UPSEAT 1999, 2000; Kerala PMT 2003]
The total number of unpaired electrons in d - orbitals of
atom s of element of atomic number 29 is [CPMT 1983]
(a) 1 0
(b) 1
(c) 0
(d) 5
(a)
(b)
(c)
(d)
32.
The shape of 2 p orbital is
[CPMT 1983; NCERT 1979]
22.
23.
(a) Spherical
(b) Ellipsoidal
(c) Dum b-bell
(d) Py ram idal
The m agnetic quantum number for a n electron when the
v alue of principal quantum num ber is 2 can hav e
33.
(a) 3 v alues
(b) 2 v alues
(c) 9 v alues
(d) 6 v alues
Which one is the correct outer configuration of chromium
34.


(b)  

(c)




  





[A IIMS 1997; J&K CET 2005]



(a) 2n
35.
The following has zero valency
[DPMT 1991]
(a) Sodium
(b) Bery llium
(c) Alum inium
(d) Kry pton
The number of electrons in the valence shell of calcium is
36.
[IIT 1975]
27 .
(a) 6
(b) 8
(c) 2
(d) 4
The valence electron in the carbon atom are [MNR 1982]
(a) 0
(b) 2
(c) 4
(d) 6
For the dum b-bell shaped orbital, the v alue of l is
[CPMT 1987, 2003]
(a) 3
(c) 0
28.
(b) 1
(d) 2
37 .
Chrom ium has the electronic configuration 4 s1 3d 5
rather than 4 s 2 3d 4 because
(a) 4 s and 3 d hav e the sam e energy
(b) 4 s has a higher energy than 3 d
(c) 4 s 1 is m ore stable than 4 s 2
(d) 4 s 1 3d 5 half-filled is m or e stable than 4 s 2 3d 4
29.
The electronic configuration of calcium ion (Ca 2  ) is
[CMC V ellore 1991]
(a) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2
38.
(b) 1s 2 ,2 s 2 sp 6 ,3 s 2 3 p 6 ,4 s 1
2
2
6
2
6
(c) 1s ,2 s 2 p ,3 s 3 p 3d
2
2
6
2
6
0
(e) 1s ,2 s 2 p ,3 s 3 p ,4 s
The structure of external m ost shell of inert gases is
[JIPMER 1991]
2 3
(a) s p
31.
(b) 2n 2
(c) n
(d) n 2
The number of orbitals in the fourth principal quantum
num ber will be
(a) 4
(b) 8
(c) 1 2
(d) 1 6
Which set of quantum numbers are not possible from the
following
1
(a) n  3, l  2, m  0, s  
2
1
(b) n  3, l  2, m  2, s  
2
1
(c) n  3, l  3, m  3, s  
2
1
(d) n  3, l  0, m  0, s  
2
The four quantum number for the valence shell electron
or last electron of sodium (Z = 11) is
[MP PMT 1999]
1
(a) n  2, l  1, m  1, s  
2
1
(b) n  3, l  0, m  0, s  
2
1
(c) n  3, l  2, m  2, s  
2
1
(d) n  3, l  2, m  2, s  
2
The explanation for the presence of three unpaired
electrons in the nitrogen atom can be giv en by
[NCERT 1979; RPMT 1999; DCE 1999, 2002;
CPMT 2001; MP PMT 2002; Pb. PMT / CET 2002]
2
2
2
6
2
6
5
(d) 1s ,2 s sp ,3 s 3 p 3d
30.
(a) Spherically symmetrical
(b) Has octahedral sy m m etry
(c) Has tetrahedral sy m m etry
(d) Depends on the atom
If m agnetic quantum number of a given atom represented
by –3 , then what will be its principal quantum num ber
[BHU 2005]
(d)
26.
[MNR 1987]
(a) 2
(b) 3
(c) 4
(d) 5
The total number of orbitals in an energy level designated
by principal quantum number n is equal to
[A IIMS 1980, 91; BHU 1995]
25.
A com pletely filled d -orbital (d 10 )
[CPMT 1984]
(a)
24.
Principal quantum num ber
Azim uthal quantum num ber
Magnetic quantum num ber
Spin quantum num ber
2 6
(b) s p
1 2
(c) s p
(d) d 10 s 2
The two electrons in K sub-shell will differ in
39.
(a) Pauli's exclusion principle
(b) Hund's rule
(c) Aufbau's principle
(d) Uncertainty principle
The m axim um energy is present in any electron at
(a) Nucleus
(b) Ground state
(c) First excited state
Structure of atom
(d) Infinite distance from the nucleus
40.
41.
42.
The electron density between 1s and 2 s orbital is
(a) High
(b) Low
(c) Zero
(d) None of these
For ns orbital, the magnetic quantum number has v alue
(a) 2
(b) 4
(c) – 1
(d) 0
The m axim um num ber of electrons that can be
(b) n  4 , l  0, m  0, s  
(c) n  3, l  1, m  1, s  
44.
45.
46.
47 .
48.
50.
53.
(a) 4 s
(b) 4 f
55.
The num ber of orbitals in 2 p sub-shell is
(c) 4 p
(d) 4 d
56.
(a) 6
(b) 2
(c) 3
(d) 4
The number of orbitals in d sub-shell is
(a) 1
(b) 3
(c) 5
(d) 7
If v alue of azimuthal quantum number l is 2 , then total
possible v alues of m agnetic quantum num ber will be
(a) 7
(b) 5
(c) 3
(d) 2
57 .
A sub-shell l  2 can take how m any electrons
The ty pe of orbitals present in Fe is
(a) s
(b) s and p
58.
(a) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 9 ,4 s 2
(b) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 10 ,4 s1
(c) 1s 2 .2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 4 p 6
(d) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 10
[NCERT 1973; MP PMT 1996]
Which electronic configuration for oxy gen is correct
according to Hund's rule of m ultiplicity
(a) 1s 2 ,2 s 2 2 p x2 2 p 1y 2 p 1z
(b) 1s 2 ,2 s 2 2 p x2 2 p y2 2 p z0
(c) 1s 2 ,2 s 2 2 p x3 2 p 1y 2 p z0
(d) None of these
Pauli's exclusion principle states that
[MNR 1983; A MU 1984]
(b) Two electrons in the sam e atom cannot hav e the
sam e spin
(a) Circular
(b) Dum b-bell
(c) Double dum b-bell
(d) Trigonal
In any atom which sub-shell will have the highest energy
in the following
(c) The electrons tend to occupy different orbitals as far
as possible
(d) Electrons tend to occupy lower energy orbitals
preferentially
(b) 3 d
(e) None of the abov e
59.
For d electrons, the azim uthal quantum num ber is
[MNR 1983; CPMT 1984]
2
2
6
2
6
7
2
(b) 1s ,2 s sp ,3 s 3 p 3d ,4 s
60.
(a) 0
(b) 1
(c) 2
(d) 3
For p -orbital, the m agnetic quantum number has v alue
(a) 2
(c) – 1 , 0, + 1
2
2
6
2
6
8
2
(d) 1s ,2 s 2 p ,3 s 3 p 3d ,4 s
The four quantum numbers of the outermost orbital of K
(atomic no. =19) are
[MP PET 1993, 94]
1
(a) n  2, l  0, m  0, s  
2
(b) 1 0
(d) 6
(a) Two electrons in the same atom can hav e the sam e
energy
The shape of d xy orbital will be
2
2
6
2
6
5
1
(c) 1s ,2 s 2 p ,3 s 3 p 3d ,4 s
[MNR 1981]
[NCERT 1973, 78]
(a) 3
(c) 5
(d) s, p, d and f
(c) 4 s
(d) 3 s
Which electronic configuration is not observ ing the
( n  l ) rule
The electronic configuration of copper ( 29 Cu ) is
[DPMT 1983; BHU 1980; AFMC 1981;
CBSE PMT 1991; MP PMT 1995]
(a) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 1 ,4 s 2
52.
1
2
The angular m omentum of an electron depends on
(a) Principal quantum num ber
(b) Azim uthal quantum num ber
(c) Magnetic quantum num ber
(d) All of these
54.
(a) 3 p
51.
1
2
For a giv en value of quantum num ber l , the num ber of
allowed v alues of m is giv en by
(a) l  2
(b) 2l  2
(c) 2l  1
(d) l  1
The num ber of radial nodes of 3 s and 2 p orbitals are
respectively.
[IIT -JEE 2005]
(a) 2 , 0
(b) 0, 2
(c) 1 , 2
(d) 2 , 1
Which of the sub-shell is circular
(c) s, p and d
49.
1
2
(d) n  4 , l  2, m  1, s  
accom m odated in the M th shell is
(a) 2
(b) 8
(c) 1 8
(d) 3 2
43.
61
61.
(b) 4 , – 4
(d) 0
For n  3 energy level, the number of possible orbitals (all
kinds) are
[BHU 1981; CPMT 1985; MP PMT 1995]
(a) 1
(b) 3
(c) 4
(d) 9
62 Structure
62.
63.
of atom
Which of the following ions is not hav ing the
configuration of neon
(b) Mg 2
(a) 4 s  3d
(b) 4 s  4 p
(c) Na 
(d) Cl 
(c) 4 s  3d
(d) 4 s  3 p
Elem ents upto atomic number 103 have been synthesized
and studied. If a newly discov ered elem ent is found to
have an atomic number 106, its electronic configuration
will be
[A IIMS 1980]
(a) [Rn]5 f ,6 d ,7 s
2
(b) [Rn]5 f ,6 d 1 ,7 s 2 7 p 3
(c) [Rn]5 f ,6 d ,7 s
0
(d) [Rn]5 f ,6 d ,7 s
4
14
6
73.
Fe (atom ic num ber = 2 6) atom has the electronic
arrangement
[NCERT 1974; MNR 1980]
(a) 2 , 8, 8, 8
(b) 2 , 8, 1 6
(c) 2 , 8,1 4 , 2
(d) 2 , 8, 1 2 , 4
74.
Cu 2  will hav e the following electronic configuration
14
14
5
[MP PMT 1985]
(a) 1s ,2 s 2 p ,3 s 3 p 3d
1
2
66.
68.
When the azim uthal quantum num ber has a v alue of
[MP PET 1995]
l  0 , the shape of the orbital is
(a) Rectangular
(b) Spherical
10
(d) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 10 ,4 s 1
75.
[MA DT Bihar 1982; AIIMS 1989]
(a) 1s ,2 s 2 p ,3 s 3 p 3d 6
(a) 1s 2 ,2 s 1 ,2 p x3
(b) 1s 2 ,2 s 2 2 p x2 2 p1y
(c) 1s 2 ,2 s 2 2 p 1x 2 p 1y 2 p 1z
(d) 1s 2 ,2 s 2 2 p 1x 2 p y2
In a m ulti-electron atom, which of the following orbitals
described by the three quantum m em bers will hav e the
sam e energy in the absence of m agnetic and electric fields
76.
[MP PMT 1986, 87; Orissa JEE 1997]
77.
(a) 2
(b) 8
(c) 1 8
(d) 3 2
Which element is represented by the following electronic
configuration
[MP PMT 1987]
2p
2s
(3 ) n  2, l  1, m  1
(4 ) n  3, l  2, m  0

(b) (2 ) and (3 )
7 8.
1
(a) 1s ,2 s 2 p ,3 s 3 p
6
2
2
(b) 1s ,2 s 2 p ,3 s 2 3 p 4 ,4 s1
(d) 1s 2 ,2 s 2 2 p 6 ,3 s 1 3 p 4 ,4 s 2
The shape of s -orbital is
(a) Py ram idal
(b) Spherical
(c) Tetrahedral
(d) Dum b-bell shaped
79.
(b) 4 f -orbital
(c) 4 s -orbital
(d) 4 d -orbital
(b) Oxy gen
(d) Neon
If the v alue of azim uthal quantum num ber is 3 , the
possible v alues of m agnetic quantum num ber would be
(a) 0, 1 , 2 , 3
(b) 0, – 1 , – 2 , – 3
(c) 0,  1 ,  2 ,  3
(d)  1 ,  2 ,  3
Kry pton (36 Kr) has the electronic configuration (18 Ar)
[CBSE PMT 1989; CPMT 1989; EAMCET 1991]
When 3 d orbital is com plete, the new electron will enter
the
(a) 4 p -orbital

4 s 2 ,3d 10 ,4 p 6 . The 37 th electron will go into which one of
the following sub-lev els
[NCERT 1978I]
[EA MCET 1980; MP PMT 1995]

[MP PMT 1987; RPMT 1999; AFMC 2002; KCET 2002]
6
(c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 5


(a) Nitrogen
(c) Fluorine
[A MU 1982]
6
6
com prise the 3 rd quantum shell n  3
1s
2
2
(d) None of these
How m any electrons can be fit into the orbitals that
(2 ) n  2, l  0, m  0
2
6
(c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 5 ,4 s 1
(1 ) n  1, l  0, m  0
(c) (3 ) and (4 )
(d) (4 ) and (5)
Which of the following represents the electronic
configuration of an elem ent with atom ic num ber 1 7
2
(b) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 4 ,4 s 2
(b) 2
(c) 1
(d) 0
The electronic configuration of an elem ent with atom ic
number 7 i.e. nitrogen atom is
[CPMT 1982, 84, 87]
Which one is the electronic con figuration of Fe 2
2
(c) Dum bbell
(d) Unsy m m etrical
The m agnetic quantum number for v alency electrons of
sodium is
[CPMT 1988; MH CET 1999]
(a) (1 ) and (2 )
7 1.
6
(c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 9
(b) Sodium and potassium
(5) n  3, l  2, m  0
70.
2
(a) Lithium and sodium
[A IEEE 2005]
69.
6
(b) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 9 ,4 s 1
(a) 3
67 .
2
Ions which have the sam e electronic configuration are
those of
(c) Potassium and calcium (d) Oxy gen and chlorine
65.
In a potassium atom, electronic energy lev els are in the
following order
[EA MCET 1979; DPMT 1991]
(a) F 
14
64.
72.
80.
(a) 4 f
(b) 4 d
(c) 3 p
(d) 5 s
1
and a
2
m agnetic quantum number of 1 , it cannot be presented
in an
[CBSE PMT 1989; UPSEA T 2001]
If an electron has spin quantum num ber of 
Structure of atom
81.
(a) d -orbital
(b) f -orbital
(c) p -orbital
(d) s -orbital
93.
The azim uthal quantum num ber is related to
[BHU 1987, 95]
(a) Size
82.
83.
84.
85.
(b) Shape
(c) Orientation
(d) Spin
The total number of electrons that can be accomm odated
in all the orbitals having principal quantum number 2 and
azimuthal quantum number 1 is
[CPMT 1971, 89, 91]
(a) 2
(b) 4
94.
(c) 6
(d) 8
Electronic configuration of C is
95.
(a) 1s 2 ,2 s 2 2 p 2
(b) 1s 2 ,2 s 2 2 p 3
(c) 1s 2 ,2 s 2
(d) 1s 2 ,2 s 2 2 p 6
There is no difference between a 2 p and a 3 p orbital
regarding
[BHU 1981]
(a) Shape
(b) Size
(c) Energy
(d) Value of n
The electronic configuration of chrom ium is
86.
(b) [ Ne ]3 s 2 3 p 6 3d 5 ,4 s 1
(c) [ Ne ]3 s 2 3 p 6 ,4 s 2 4 p 4
(d) [ Ne ]3 s 2 3 p 6 3d 1 ,4 s 2 4 p 3
The shape of p -orbital is
(a) Elliptical
(c) Dum b-bell
87 .
88.
89.
90.
96.
91.
EA MCET 1991; RPMT 2002; CBSE PMT 2002]
92.
(a) 2
(b) 6
(c) 0
(d) 1 4
An ion has 1 8 electrons in the outerm ost shell, it is
[CBSE PMT 1990]
(a) Cu

(b) Th
4
The quantum number which m ay be designated by s, p, d
BHU 1980]
(b) l
(d) m s
Which of the following represents the correct sets of the
four quantum num bers of a 4 d electron
1
2
(b) 4 , 2 , 1 , 0
1
1
(d) 4 , 2, 1, 
2
2
Which of the following statem ents is not correct for an
electron that has the quantum numbers n  4 and m  2
[MNR 1993]
(a) The electron m ay have the quantum num ber s  
1
2
(b) The electron m ay have the quantum num ber l  2
(c) The electron m ay have the quantum num ber l  3
(d) The electron m ay hav e the quantum num ber
l  0, 1, 2, 3
97 .
The set of quantum num bers not applicable for an
electron in an atom is
[MNR 1994]
(a) n  1, l  1, m l  1, m s  1 / 2
(b) n  1, l  0, m l  0, m s  1 / 2
(c) n  1, l  0, m l  0, m s  1 / 2
(b) 3 d 4 ,4 s 1
(a) 8
(b) 6
(c) 1 8
(d) 3 2
For azim uthal quantum num ber l  3 , the m axim um
num ber of electrons will be
[CBSE PMT 1991;
(d) 3d, 4 p, 4 s, 4 d, 5 s
(c) 4 , 3,  2, 
[MP PET 1993]
(c) 3 d 3 ,4 s 2
(d) 3d 2 ,4 s 2 4 p 2
The principal quantum number represents [CPMT 1991]
(a) Shape of an orbital
(b) Distance of electron from nucleus
(c) Num ber of electrons in an orbit
(d) Num ber of orbitals in an orbit
When the azim uthal quantum num ber has a v alue of
[MP PET 1993]
l  1 , the shape of the orbital is
(a) Unsy m m etrical
(b) Spherically symmetrical
(c) Dum b-bell
(d) Com plicated
How m any electrons can be accommodated in a sub-shell
for which n  3, l  1
[CBSE PMT 1990]
(c) 5 s, 4 p, 3d, 4 d, 5 s
(a) 4 , 3, 2,
The electronic configuration (outerm ost) of Mn 2 ion
(atom ic num ber of Mn  25 ) in its ground state is
(a) 3 d 5 ,4 s 0
(b) 4 s, 3d, 4 p, 5 s, 4 d
[MNR 1992; UPSEAT 2001; J&K CET 2005]
[MP PMT 1993]
(b) Spherical
(d) Com plex geom etrical
(a) 3d, 4 s, 4 p, 4 d, 5 s
(a) n
(c) m l
[MP PMT 1993; MP PET 1995; BHU 2001; BCECE 2005]
(a) [ Ne ]3 s 2 3 p 6 3d 4 ,4 s 2
(c) Cs 
(d) K 
The order of filling of electrons in the orbitals of an atom
will be
and f instead of number is
[CPMT 1975]
63
(d) n  2, l  0, m l  0, m s  1 / 2
98.
Correct configuration of Fe 3 [2 6] is
[CPMT 1994; BHU 1995; KCET 1992]
2
2
6
2
(a) 1s ,2 s 2 p ,3 s 3 p 6 3d 5
(b) 1s 2 ,2 s 2 sp 6 ,3 s 2 3 p 6 3d 3 ,4 s 2
(c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 6 ,4 s 2
(d) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 5 ,4 s 1
99.
Azim uthal quantum number for last electron of Na atom
is
[BHU 1995]
(a) 1
(b) 2
(c) 3
(d) 0
100. A 3 p orbital has
[IIT 1995]
(a) Two spherical nodes
(b) Two non-spherical nodes
(c) One spherical and one non -spherical nodes
(d) One spherical and two non -spherical nodes
101. All electrons on the 4 p sub-shell m ust be characterized
by the quantum number(s)
[MP PET 1996]
1
(a) n  4 , m  0, s  
(b) l  1
2
[
64 Structure
of atom
1
1
(d) s  
2
2
102. The electronic configuration of the elem ent of atom ic
num ber 2 7 is
(c) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 4
(c) l  0, s  
(d) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 1 3d 3
110. Which of the following configuration is correct for iron
[CBSE PMT 1999]
(a) 1s , 2 s 2 p , 3 s 3 p , 4 s () 4 p ()()() 5 s ()
2
2
6
2
6
(a) 1s 2 s 2 p 3 s 3 p 3d
2
(b) 1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 3d ()()(), 4 s () 4 p ()
2
6
2
6
2
6
2
6
103. When the value of the principal quantum number n is 3 ,
the permitted values of the azimuthal quantum num bers
l and the m agnetic quantum num bers m , are
l
0
(a) 1
2
 1, 0,  1
 2,1, 0,  1,2
1
(b) 2
3
1
 2, 1,  1
 3,2, 1,  2,3
0
(c) 1
2
0
1, 2, 3
 3,  2, 1,  2,3
m
0
1
0, 1
0, 1, 2
0, 1, 2, 3
(d) 2
3
104. The number of possible spatial orientations of an electron
in an atom is giv en by its
(a) Spin quantum num ber
(b) Spin angular m om entum
(c) Magnetic quantum num ber
(d) Orbital angular m om entum
105. Which of the following sets of orbitals m ay degenerate
(a) 2 s, 2 p x , 2 p y
(b) 3 s, 3 p x ,3d xy
(c) 1s, 2 s, 3 s
111.
1
2
(b) n  3, l  0, m  0, s  
1
2
(c) n  3, l  1, m  1, s  
1
2
112. Which quantum number will determine the shape of the
subshell
[CPMT 1999; Pb. PMT 1998]
(a)
(b)
(c)
(d)

(a)













(c)
[MP PET 1997]
(d)





2
108. What is the electronic configuration of Cu (Z  29) of
least position
[MP PET /PMT 1998; MP PET 2001]
1
8
(a) [ Ar] 4 s 3d
(b) [ Ar] 4 s 2 3d 10 4 p 1
(d) [ Ar] 3 d 9
109. The correct electronic configuration of Ti(Z  22) atom is
[MP PMT 1999]
2
2
6
2
6
4
(b) 1s 2 s 2 p 3 s 3 p 3d
Principal quantum num ber
Azim uthal quantum num ber
Magnetic quantum num ber
Spin quantum num ber
113. For the n  2 energy level, how many orbitals of all kinds
are possible
[Bih a r CEE 1995]
(a) 2
(b) 3
(c) 4
(d) 5
114. Which one is in the ground state
[DPMT 1996]
(c) Ne
(d) F
107 . An electron has principal quantum num ber 3 . The
num ber of its (i) sub-shells and (ii) orbitals would be
respectiv ely
2
2
6
2
6
2
2
(a) 1s 2 s 2 p 3 s 3 p 4 s 3d
1
2
(d) n  3, l  2, m  1, s  
(b)
(c) [ Ar] 4 s 1 3d 10
5
(a) n  4 , l  0, m  0, s  
(d) 2 p x , 2 p y , 2 p z
(b) 3 and 7
(d) 2 and 5
6
(d) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 6
Which of the following set of quantum numbers belong to
highest energy
[CPMT 1999]
106. The set of quantum num bers n  3, l  0, m  0, s  1 / 2
belongs to the elem ent
(a) Mg
(b) Na
(a) 3 and 5
(c) 3 and 9
2
(c) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 7
(d) 1s , 2 s 2 p , 3 s 3 p , 3d ()()()()() 4 s ()
2
6
(b) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 5
(c) 1s , 2 s 2 p , 3 s 3 p , 3d ()()()(), 4 s ()
2
2
115. When the principal quantum number (n  3) , the possible
v alues of azim uthal quantum num ber ( l ) is
[Bihar MEE 1996; KCET 2000]
(a) 0, 1 , 2 , 3
(b) 0, 1 , 2
(c) – 2 , – 1 , 0, 1 , 2
(d) 1 , 2 , 3
(e) 0, 1
116. Which statem ent is not correct for n  5 , m  3
[CPMT 1996]
(a) l  4
(b) l  0, 1, 3; s  
1
2
Structure of atom
(c) l  3
117 .
2
2
(d) All are correct
6
1s 2 s 2 p 3 s
3
(a) Al
1
shows configuration of
[CPMT 1996]
in ground state (b) Ne in excited state

(c) Mg in excited state
(d) None of these
118. Fiv e v alence electrons of p 15 are labelled as
AB
3s
X
Y
3p
Z
1
, the group of
2
electrons with three of the quantum num ber sam e are
If the spin quantum of B and Z is 
[JIPMER 1997]
119.
(a) AB, XYZ , BY
(b) AB
(c) XYZ, AZ
(d) AB, XYZ
Electronic configuration of Sc 21 is
(a) 1s 2 s 2 p 3 s 3 p 4 s 3d
2
2
6
2
6
2
1
2
2
6
2
6
1
2
(b) 1s 2 s 2 p 3 s 3 p 4 s 3d
[BHU 1997]
(d) 1s 2 2 s 2 2 p 6 3 s 2 3 p 2 4 s 2 3d 2
120. If n  l  6 , then total possible num ber of subshells
would be
[RPMT 1997]
(a) 3
(b) 4
(c) 2
(d) 5
electron
hav ing the quantum num bers
1
would be in the orbital
n  4, l  3, m  0 , s  
2
[Or issa JEE 1997]
(a) 3 s
(b) 3 p
(c) 4 d
(d) 4 f
122. Which of the following sets of quantum num bers is not
allowed
[Or issa JEE 1997]
(a) n  1, l  0, m  0, s  
(d) [ Xe]4 f 6 5 d 2 6 s 2
125. An e  has m agnetic quantum number as 3 , what is its
principal quantum number
[BHU 1998]
(a) 1
(b) 2
(c) 3
(d) 4
126. The number of quantum numbers required to describe an
electron in an atom com pletely is
[CET Pu n e 1998]
(a) 1
(b) 2
(c) 3
(d) 4
127 . The electronic configuration 1s 2 2 s 2 2 p 1x 2 p 1y 2 p 1z
[A FMC 1997; Pb. PMT 1999; CBSE PMT 2001; AIIMS 2001]
(a) Oxy gen
(b) Nitrogen
(c) Hy drogen
(d) Fluorine
128. Which one of the following set of quantum numbers is not
possible for 4 p electron
[EA MCET 1998]
(a) n  4 , l  1, m  1, s  
(c) 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 0 3d 3
121. An
(c) [ Xe]4 s 3 5 d 5 6 s 2
1
2
65
1
2
(b) n  4 , l  1, m  0, s  
1
2
(c) n  4 , l  1, m  2, s  
1
2
1
2
129. Which of the following orbital is not possible[RPMT 1999]
(a) 3 f
(b) 4 f
(c) 5 f
(d) 6 f
130. Which set of quantum numbers for an electron of an atom
is not possible
[RPMT ; DCE 1999]
(a) n  1, l  0, m  0, s  1 / 2
(d) n  4 , l  1, m  1, s  
(b) n  1, l  1, m  1, s  1 / 2
(c) n  1, l  0, m  0, s  1 / 2
(d) n  2, l  1, m  1, s  1 / 2
131. Electronic configuration of ferric ion is
(a) [ Ar] 3 d
[RPET 2000]
(b) [ Ar] 3 d
5
7
(c) [ Ar] 3 d
(d) [ Ar] 3 d 8
132. What is the m aximum number of electrons which can be
accom modated in an atom in which the highest principal
quantum num ber value is 4
[MP PMT 2000]
(a) 1 0
(b) 1 8
(c) 3 2
(d) 54
133. Which of the following electronic configurations is not
possible
3
(b) n  1, l  1, m  0, s  
1
2
(c) n  2, l  1, m  1, s  
1
2
(d) n  2, l  1, m  0, s  
1
2
123. For which of the following sets of four quantum numbers,
an electron will have the highest energy [CBSE PMT 1994]
n
(a) 3
(b) 4
l
2
2
(c) 4
(d) 5
1
0
m
1
1
0
0
s
+ 1 /2
+ 1 /2
–1 /2
–1 /2
124. The electronic configuration of gadolinium (atom ic no.
64) is
8
9
2
(a) [ Xe]4 s 5 d 6 s
7
1
2
(b) [ Xe]4 s 5 d 6 s
[CPMT 2000]
(a) 1s 2 s
2
(c) 3d
10
2
2
2
(b) 1s 2 s 2 p
2
6
2
(d) 1s 2 2 s 2 2 p 2 3 s 1
134. The electronic configuration of an elem ent is
1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 5 4 s1 . This represents its
4s 4 p
[IIT Screening 2000]
(a) Excited state
(b) Ground state
(c) Cationic form
(d) Anionic form
135. Which of the following set of quantum num bers is
[CBSE
PMT 1997]
possible
[A IIMS 2001]
66 Structure
of atom
1
2
1
(b) n  3; l  4 ; m  0 and s  
2
1
(c) n  4 ; l  0; m  2 and s  
2
1
(d) n  4 ; l  4 ; m  3 and s  
2
136. Which of the following set of quantum num ber is not
v alid
(a) n  3; l  2; m  2 and s  
[A IIMS 2001]
(a) n  1, l  2
(b) 3  2, m  1
(c) m  3, l  0
(d) 3  4, l  2
137 . Which one pair of atom s or ions will hav e sam e
configuration
[JIPMER 2001]
(a) F  and Ne
(b) Li  and He 

(c) Cl and Ar
(d) Na and K
138. Which of the following sets of quantum num ber is not
possible
[MP PET 2001]
1
(a) n  3; l  2; m  0; s  
2
(b) n  3; l  0; m  0; s  
1
2
(a) Hund’s rule
(b) Aufbau’s principle
(c) Pauli’s exclusion principle
(d) Heisenberg’s uncertainty principle
145. Which of the following has maximum energy
[A IIMS 2002]
3s
3p
3d
3s
3p
3d
3s
3p
3d
3s
3p
3d
(a)
(b)
(c)
(d)
146. The total m agnetic quantum num bers for d-orbital is
given by
(b) 0,  1 ,  2
(d) 5
(a) 2
(c) 0, 1 , 2
147 . The outer electronic structure 3 s 2 3 p 5 is possessed by
[Pb. PMT 2002; Pb. CET 2001]
1
(c) n  3; l  0; m  1; s  
2
1
2
139. Which of the following set of quantum numbers is correct
for the 19th electron of chromium
[DCE 2001]
n
l
m
s
(d) n  3; l  1; m  0; s  
(a)
3
0
0
1 /2
(b)
3
2
–2
1 /2
(c)
4
0
0
1 /2
(d)
4
1
–1
1 /2
(a) Cl
(b) O
(c) Ar
(d) Br
148. Which of the following set of quantum num ber is not
possible
[Pb. PMT 2002]
n
l
m1
m2
(a)
(b)
(c)
(d)
3
3
3
5
2
2
2
2
1
1
1
–1
+ 1 /2
– 1 /2
0
+ 1 /2
149. The configuration 1s 2 , 2 s 2 2 p 5 , 3 s 1 shows[Pb. PMT 2002]
(a) Excited state of O 2
140. When the v alue of azim uthal quantum num ber is 3 ,
(b) Excited state of neon
magnetic quantum number can have values[DPMT 2001]
(c) Excited state of fluorine
(a) + 1 , 0, – 1
(d) Ground state of fluorine atom
(b) + 2 , + 1 , 0, – 1 , – 2
150. The quantum num ber ‘m’ of a free gaseous atom is
(c) – 3 , – 2 , – 1 , – 0, + 1 , + 2 , + 3
associated with
[A IIMS 2003]
(d) + 1 , – 1
(a) The effectiv e v olum e of the orbital
141. The quantum num bers n  2, l  1 represent [A FMC 2002]
(b) The shape of the orbital
(a) 1 s orbital
(b) 2 s orbital
(c) The spatial orientation of the orbital
(c) 2 p orbital
(d) 3 d orbital
(d) The energy of the orbital in the absence of a m agnetic
142. The m agnetic quantum num ber of v alence electron of
field
sodium (Na) is
[RPMT 2002]
151. Correct statement is
[BHU 2003]
(a) 3
(b) 2
1
4
2
10
2
(a) K  4 s , Cr  3d 4 s , Cu  3d 4 s
(c) 1
(d) 0
143. Azimuthal quantum number defines
[A IIMS 2002]
(b) K  4 s 2 , Cr  3d 4 4 s 2 , Cu  3d 10 4 s 2
(a) e/m ratio of electron
(c) K  4 s 2 , Cr  3d 5 4 s 1 , Cu  3d 10 4 s 2
(b) Spin of electron
(c) Angular m om entu m of electron
(d) K  4 s 1 , Cr  3d 5 4 s 1 , Cu  3d 10 4 s 1
(d) Magnetic m om entum of electron
152. Number of orbitats in h sub-shell is
[BHU 2003]
144. Quantum numbers of an atom can be defined on the basis
of
[A IIMS 2002]
(a) 1 1
(b) 1 5
[
Structure of atom
(c) 1 7
153. Electronic configuration
(d) 1 9
1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 3d 5 , 4 s 1 represents
[CPMT 2003]
(a) Ground state
(b) Excited state
(c) Anionic state
(d) All of these
154. Which of the following sets is possible for quantum
num bers
(a) Heisenberg’s principle
(b) Hund’s rule
(c) Aufbau principle
(d) Pauli exclusion principle
162. The electronic configuration of elem ent with atom ic
number 24 is
[Pb. CET 2004]
(a) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 4 ,4 s 2
[RPET 2003]
(b) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 10
(a) n  4, l  3, m  2, s  0
(c) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 6
(b) n  4 , l  4 , m  2, s  
1
2
(c) n  4 , l  4 , m  2, s  
1
2
(d) 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 5 4 s 1
163. The m axim um num ber of electrons in p -orbital with
[Pb. CET 2003]
n  5, m  1 is
1
(d) n  4 , l  3, m  2, s  
2
155. For principle quantum number n  4 the total number of
orbitals having l  3
[A IIMS 2004]
(a) 3
(b) 7
(c) 5
(d) 9
156. The num ber of 2 p electrons hav ing spin quantum
num ber s  1 / 2 are
[KCET 2004]
(a) 6
(b) 0
(c) 2
(d) 3
157 . Which of the following sets of quantum num bers is
correct for an electron in 4 f orbital
[A IEEE 2004]
(a) 6
(b) 2
(c) 1 4
(d) 1 0
164. Num ber of two electron can have the sam e v alues of ……
quantum num bers
[UPSEA T 2004]
(a) One
(b) Two
(c) Three
(d) Four
165. The number of orbitals present in the shell with n  4 is
[UPSEAT 2004]
(a) 1 6
(b) 8
(c) 1 8
(d) 3 2
166. Which of the following electronic configuration is not
possible
[MHCET 2003]
(a) 1s 2 s
2
1
(a) n  4 , l  3, m  1, s  
2
(b) n  4 , l  4 , m  4 , s  
167 .
1
2
1
2
1
2
(b) 4 , 0, 1 and
1
2
1
1
(d) 4 , 1 , 1 and
2
2
160. Which of the following electronic configuration is not
possible according to Hund’s rule
(c) 4 , 0, 0 and 
(a) 1s 2 2 s 2
(b) 1s 2 2 s 1
(c) 1s 2 2 s 2 2 p 1x 2 p 1y 2 p 1x
(d) 1s 2 2 s 2 2 p x2
2
2
(b) 1s ,2 s 2 p 6
(d) 1s 2 ,2 s 2 2 p 2 ,3 s 1
p x orbital can accom m odate
[MNR 1990; IIT 1983; MADT Bihar 1995; BCECE 2005]
158. Consider the ground state of (Z  24) . The num bers of
electrons with the azimuthal quantum numbers, l  1 and
2 are, respectiv ely
[A IEEE 2004]
(a) 1 6 and 4
(b) 1 2 and 5
(c) 1 2 and 4
(d) 1 6 and 5
159. The four quantum num bers of the v alence electron of
potassium are
[DPMT 2004]
(a) 4 , 1 , 0 and
2
(c) [ Ar] 3d 10 ,4 s 2 4 p 2
1
(c) n  4 , l  3, m  4 , s  
2
(d) n  3, l  2, m  2, s  
67
(e) 1s 2 2 s 2 2 p x2 2 p 1y 2 p 1z
161. The ground state term sy m bol for an electronic state is
governed by
[UPSEA T 2004]
(a) 4 electrons
(b) 6 electrons
(c) 2 electrons with parallel spins
(d) 2 electrons with opposite spins
168. The m axim um num ber of electrons that can be
accom m odated in ' f ' sub shell is
[CPMT 1983, 84; MP PET/PMT 1988; BITS 1988]
(a) 2
(b) 8
(c) 3 2
(d) 1 4
169. The number of electrons which can be accom m odated in
an orbital is
[DPMT 1981; A FMC 1988]
(a) One
(b) Two
(c) Three
(d) Four
170. The num ber of electrons in the atom which has 2 0
protons in the nucleus[CPMT 1981, 93; CBSE PMT 1989]
(a) 2 0
(b) 1 0
(c)a l3a0PMT 2004]
(d) 4 0
[Ker
171. The m aximum number of electrons accommodated in 5 f
orbitals are
[MP PET 1996]
(a) 5
(c) 1 4
(b) 1 0
(d) 1 8
172. The m aximum number of electrons in an atom with l  2
and n  3 is
[MP PET /PMT 1998]
68 Structure
of atom
(a) 2
(c) 1 2
(a) 6
(c) 3
(b) 6
(d) 1 0
173. The configuration 1s 2 2 s 2 2 p 5 3 s 1 shows
[A IIMS 1997]
174. For sodium atom the number of electrons with m  0 will
be
[RPMT 1999]
(b) 7
(d) 8
175. The number of electrons that can be accom m odated in
[Ku r u ksh et r a
CEE
dz 2 orbital is
2002]
(b) 1
(c) 4
(d) 2
176. Num ber of unpaired electrons in 1s 2 2 s 2 2 p 3 is
[CPMT 1982; MP PMT 1987; BHU 1987;
CBSE PMT 1990; CET Pune 1998; AIIMS 2000]
(a) 2
(b) 0
(c) 3
(d) 1
177. Total number of unpaired electrons in an atom of atom ic
num ber 2 9 is
[CPMT 1984, 93]
(a) 1
(b) 3
(c) 4
(d) 2
17 8. The num ber of unpaired electrons in 1s 2 , 2 s 2 2 p 4 is
(a) Fe 2
(b) CO 2
(c) Ni 2
(d) Mn 2
187 . Which of the m etal ion will hav e highest num ber of
unpaired electrons
(a) Cu 
(b) Fe 2 
(c) Fe 3 
(d) Co 2 
188. The m axim um num ber of unpaired electron can be
present in d orbitals are
(a) 1
(b) 3
(c) 5
(d) 7
189. The m olecule hav ing one unpaired electron is
(a) NO
(b) CO
(c) CN 
(d) O 2
190. A filled or half-filled set of p or d -orbitals is spherically
sy m m etric. Point out the species which has spherical
sy m m etry
[NCERT 1983]
(a) Na
(b) C
191.
[NCERT 1984; CPMT 1991; MP PMT 1996, 2002]
(a) 4
(b) 2
(c) 0
(d) 1
179. The m axim um num ber of electrons that can be
accom m odated in a 3 d subshell is
(a) 2
(b) 1 0
(c) 6
(d) 1 4
180. The m aximum number of electrons which each sub-shell
can occupy is
[Pb. CET 1989]
(a) 2n 2
(c) 2(2l  1)
(b) Cu  
(c) Cd  
(d) Hg  
186. Which of the following m etal ions will hav e m axim um
number of unpaired electrons
[CPMT 1996]
(d) Excited state of ion O 2
(a) 1 0
3d 10 4 s 0 electronic configuration exhibits by
(a) Zn  
(a) Ground state of fluorine atom
(b) Excited state of fluorine atom
(c) Excited state of neon atom
(a) 2
(c) 9
185.
(b) 4
(d) 1
192.
193.
(b) 2n
(d) (2l  1)
(c) Cl 
(d) Fe
The atom of the element having atomic number 14 should
hav e
[A MU 1984]
(a) One unpaired electron (b) Two unpaired electrons
(c) Three unpaired electrons (d)Four unpaired electrons
An atom has 2 electrons in K shell, 8 electrons in L shell
and 6 electrons in M shell. The num ber of s -electrons
present in that elem ent is
[CPMT 1989]
(a) 6
(b) 5
(c) 7
(d) 1 0
The num ber of unpaired electrons in carbon atom in
excited state is
[MNR 1987]
(a) One
(b) Two
(c) Three
(d) Four
Maxim um num ber of electrons present in ' N ' shell is
181. Num ber of unpaired electrons in the ground state of
bery llium atom is
(a) 2
(b) 1
(c) 0
(d) All the abov e
194.
182. How m any unpaired electrons are present in Ni 2  cation
(atom ic num ber = 2 8)
[IIT 1981; MNR 1984;
195. The number of d electrons in Fe 2 (atom ic num ber of
[MNR 1993]
Fe  26 ) is not equal to that of the
(a) p -electrons in Ne (At. No.= 1 0)
MP PMT 1995; Kerala PMT 2003]
(a) 0
(c) 4
(b) 2
(d) 6
(b) 3 2
(d) 8
(b) s -electrons in Mg (At. No.= 1 2 )
183. The number of unpaired electrons in an O 2 m olecule is
[MNR 1983]
(a) 0
(c) 2
[EA MCET 1984]
(a) 1 8
(c) 2
(b) 1
(d) 3
184. The number of unpaired electrons in a chrom ic ion Cr 3 
(atomic number = 24) is
[MNR 1986; CPMT 1992]
(c) d -electrons in Fe
(d) p -electrons in Cl  (At. No. of Cl = 1 7 )
4
196. A transition m etal X has a configuration [ Ar]3 d in its
3 oxidation state. Its atomic number is[EA MCET 1990]
(a) 2 5
(b) 2 6
(c) 2 2
(d) 1 9
Structure of atom
197 . The total num ber of electrons present in all the p orbitals of brom ine are
[MP PET 1994]
(a) Fiv e
(b) Eighteen
(c) Sev enteen
(d) Thirty fiv e
198. Which of the following has the m axim um num ber of
unpaired electrons
[IIT 1996]
(a) Mg 2 
(b) Ti 3 
(c) V 3 
(d) Fe 2 
199. Which of the following has m ore unpaired d -electrons
[CBSE PMT 1999]
(a) Zn 
(b) Fe 2 
(c) N 3 
(d) Cu 
200. Maximum electrons in a d -orbital are
(a) 2
(b) 1 0
(c) 6
(d) 1 4
69
(c) Pauli's exclusion principle
(d) Uncertainty principle
208. According to Aufbau's principle, which of the three
4 d , 5 p and 5 s will be filled with electrons first[MA DT Bi h a r 19
(a) 4 d
(b) 5 p
(c) 5 s
(d) 4 d and 5 s will be filled sim ultaneously
209. The energy of an electron of 2 p y orbital is [A MU 1984]
(a) Greater than that of 2 p x orbital
(b) Less than that of 2 p x orbital
[CPMT 1999]
201. The num ber of unpaired electrons in Fe 3  (Z  26) are
[KCET 2000]
(a) 5
(b) 6
(c) 3
(d) 4
202. How m any unpaired electrons are present in cobalt [ Co]
m etal
[RPMT 2002]
(a) 2
(b) 3
(c) 4
(d) 7
203. The num ber of unpaired electrons in nitrogen is
[Pb. CET 2002]
(a) 1
(b) 3
(c) 2
(d) None of these
204. Which of the following has the least energy
(a) 2 p
(b) 3 p
(c) 2 s
(d) 4 d
205. Pauli's exclusion principle states that [CPMT 1983, 84]
(a) Nucleus of an atom contains no negativ e charge
(b) Electrons m ove in circular orbits around the nucleus
(c) Electrons occupy orbitals of lowest energy
(d) All the four quantum numbers of two electrons in an
atom cannot be equal
206. For the energy lev els in an atom , which one of the
following statements is correct
[A IIMS 1983]
(a) There are sev en principal electron energy lev els
(b) The second principal energy lev el can hav e four subenergy levels and contains a maximum of eight electrons
(c) The M energy lev el can hav e m axim um of 3 2
electrons
(d) The 4 s sub-energy level is at a higher energy than
the 3 d sub-energy lev el
207 . The statem ents
[A IIMS 1982]
(i) In filling a group of orbitals of equal energy , it is
energetically preferable to assign electrons to em pty
orbitals rather than pair them into a particular
orbital.
(ii) When two electron s are placed in two different
orbitals, energy is lower if the spins are parallel.
are v alid for
(a) Aufbau principle
(b) Hund's rule
(c) Equal to that of 2 s orbital
(d) Sam e as that of 2 p z orbital
210. Which of the following principles/rules lim its the
maximum number of electrons in an orbital to two[CBSE PMT 198
(a) Aufbau principle
(b) Pauli's exclusion principle
(c) Hund's rule of m axim um m ultiplicity
(d) Heisenberg's uncertainty principle
211. The electrons would go to lower energy lev els first and
then to higher energy lev els according to which of the
following
[BHU 1990; MP PMT 1993]
(a) Aufbau principle
(b) Pauli's exclusion principle
(c) Hund's rule of m axim um m ultiplicity
(d) Heisenberg's uncertainty principle
212. Energy of atomic orbitals in a particular shell is in the
order
[A FMC 1990]
(a) s  p  d  f
(b) s  p  d  f
(c) p  d  f  s
(d) f  d  s  p
213. Aufbau principle is not satisfied by
[MP PMT 1997]
(a) Cr and Cl
(b) Cu and Ag
(c) Cr and Mg
(d) Cu and Na
214. Which of the following explains the sequence of filling the
electrons in different shells
[A IIMS 1998; BHU 1999]
(a) Hund's rule
(b) Octet rule
(c) Aufbau principle
(d) All of these
215. Aufbau principle is obey ed in which of the following
electronic configurations
[A FMC 1999]
(a) 1s 2 2 s 2 2 p 6
(b) 1s 2 3 p 3 3 s 2
(c) 1s 2 3 s 2 3 p 6
(d) 1s 2 2 s 2 3 s 2
216. Following Hund’s rule which elem ent contains six
unpaired electron
[RPET 2000]
(a) Fe
(b) Co
(c) Ni
(d) Cr
217 . Electron enters the sub-shell for which (n  l) v alue is
m inim um . This is enunciated as
[RPMT 2000]
(a)
(b)
(c)
(d)
Hund’s rule
Aufbau principle
Heisenberg uncertainty principle
Pauli’s exclusion principle
70 Structure
of atom
218. The atom ic orbitals are progressiv ely filled in order of
increasing energy . This principle is called as
(a) F 
[MP PET 2001]
(a) Hund’s rule
(b) Aufbau principle
(c) Exclusion principle
(d) de-Broglie rule
219. The correct order of increasing energy of atom ic orbitals
is
2.
[MP PET 2002]
(a) 5 p  4 f  6 s  5 d
(b) 5 p  6 s  4 f  5 d
(c) 4 f  5 p  5 d  6 s
(d) 5 p  5d  4 f  6 s
220. The orbital with maximum energy is
[CPMT 2002]
(a) 3 d
(b) 5p
(c) 4 s
(d) 6 d
221. p-orbitals of an atom in presence of m agnetic field are
3.
[Pb. PMT 2002]
(a) Two fold degenerate (b) Non degenerate
(c) Three fold degenerate (d) None of these
4.
222. Orbital angular momentum for a d-electron is[MP PET 2003]
(a)
6h
2
(b)
6h
2
12 h
12h
(d)
2
2
223. Number of nodal centres for 2s orbital
[RPET 2003]
(a) 1
(b) 0
(c) 4
(d) 3
224. The orbital angular m om entum of an electron in 2 s orbital is
1 h
h
(a)
(b)
2 2
2
h
(d) Zero
2
225. The m aximum num ber of electrons present in an orbit
[Pb. PMT 2004]
l  3 , is
(a) 6
(b) 8
(c) 1 0
(d) 1 4
2
226. Number of unpaired electrons in Mn 4  is [DPMT 2005]
(a) 3
(b) 5
(c) 6
(d) 4
227 . Which of the following sequence is correct as per Aufbau
principle
[DPMT 2005]
(a) 3 s  3d  4 s  4 p
(b) 1s  2 p  4 s  3d
5.
(c) 2s 1
(d) 1s 2
(c) n, p, , e
(d) n, , p, e
The electronic configuration of a dipositive metal M 2  is
2 , 8, 14 and its atomic weight is 56 a.m.u. The num ber of
neutrons in its nuclei would be
(b) 3 2
(d) 4 2
The ratio of the energy of a photon of 2000 Å wavelength
radiation to that of 4000 Å radiation is
[IIT 1986; DCE 2000; JIPMER 2000]
(a) 1 /4
(b) 4
[MP
(c)PET
1 /22004]
(d) 2
6.
Discov ery of the nucleus of an atom was due to the
experiment carried out by [CPMT 1983; MP PET 1983]
(a) Bohr
(b) Mosley
(c) Rutherford
(d) Thom son
7.
In a Bohr's m odel of atom when an electron jum ps from
n  1 to n  3 , how m uch energy will be em itted or
absorbed
[CBSE PMT 1996]
(a) 2.15  10 11 erg
8.
9.
(b) 0.1911  10 10 erg
(c) 2.389  10 12 erg
(d) 0.239  10 10 erg
The nucleus of an atom can be assum ed to be spherical.
The radius of the nucleus of m ass number A is giv en by
1.25  10 13  A1 / 3 cm Radius of atom is one Å . If the
m ass num ber is 64 , then the fraction of the atom ic
volume that is occupied by the nucleus is [NCERT 1983]
(a) 1.0  10 3
[J&K CET 2005]
(b) 2s 2
(a) Will rem ain approxim ately the sam e
(b) Will becom e approxim ately two tim es
(c) Will rem ain approxim ately half
(d) Will be reduced by 2 5%
The increasing order (lowest first) for the v alues of e / m
(charge/m ass) for
[IIT 1984]
(a) e, p, n, 
(b) n, p, e, 
(a) 3 0
(c) 3 4
(c) 2 s  5 s  4 p  5 d
(d) 2 s  2 p  3d  3 p
228. Electronic configuration of deuterium atom is
(a) 1s 1
(c) Mg
(d) N 
Atom s consists of protons, neutrons and electrons. If the
m ass of neutrons and electrons were m ade half and two
tim es respectively to their actual masses, then the atomic
m ass of 6 C 12
[NCERT 1982]
[MNR 1984, 89; Kerala PMT 1999]
(c)
(c)
(b) Oxy gen atom
(b) 5.0  10 5
(c) 2.5  10 2
(d) 1.25  10 13
The energy of an electron in the first Bohr orbit of
H atom is 13.6 eV . The possible energy v alue(s) of the
excited state(s) for electrons in Bohr orbits to hy drogen
is(are)
[IIT 1998; Orissa JEE 2005]
(a) 3.4 eV
(c) 6.8 eV
10.
1.
Which of the following atom s and ions are isoelectronic
i.e. hav e the sam e num ber of electrons with the neon
atom
[NCERT 1978]
(b) 4.2eV
(d) 6.8 eV
The energy of the electron in the first orbit of He  is
 871.6  10 20 J . The energy of the electron in the first
orbit of hydrogen would be[Roor kee Qu a l ify i n g 1998]
(a)  871.6  10 20 J
(b)  435.8  10 20 J
Structure of atom
(c)  217.9  10 20 J
11.
12.
(d)  108.9  10 20 J
21.
The total number of v alence electrons in 4.2 gm of N 3
ion is ( N A is the Av ogadro's number) [CBSE PMT 1994]
(a) 1.6 N A
(b) 3.2 N A
(c) 2.1 N A
(d) 4.2 N A
(a) From n  1 to n  2
22.
approximately 0.530 Å . The radius for the first excited
state (n  2) orbit is
[CBSE PMT 1998; BHU 1999]
14.
(b) 1.06 Å
17.
(c) 4  1
(d) 2  5
frequency of 8  10 15 s 1
(d) 2.12 Å
(b) 4  10 8 cm 1
(a) 3  10
(c) 2  10 7 m 1
(d) 4  10 4 cm 1
The series lim it for Balm er series of H-spectra is
(a) 3 800
(b) 4 2 00
(c) 3 6 4 6
(d) 4 000
24.
The ionization energy of hydrogen atom is 13.6 eV . The
energy required to excite the electron in a hydrogen atom
from the ground state to the first excited state is
(Avogadro’s constant = 6.022 × 10 23)
[BHU 1999]
(a)  13.6 eV
(b)  13.6 eV
(c)  13.6 and  3.4 eV
(d)  3.4 eV
The num ber of nodal planes in a p x is
[IIT Screening 2000]
(c) 5  10 18
(d) 4  10 1
As electron m oves away from the nucleus, its potential
energy
[UPSEA T 2003]
(a) Increases
(b) Decreases
(c) Rem ains constant
(d) None of these
(b) Two
(d) Zero
Read the assertion and reason carefully to m ark the correct
option out of the options giv en below :
(a)
I f both assertion and reason are true and the reason is
the correct explanation of the assertion.
(b)
I f both assertion and reason are true but reason is not
the correct explanation of the assertion.
(c)
I f assertion is true but reason is false.
(d)
I f the assertion and reason both are false.
(e)
I f assertion is false but reason is true.
1.
The third line in Balm er series corresponds to an
electronic transition between which Bohr’s orbits in
hy drogen
Assertion :
Reason
[MP PMT 2001]
19.
20.
(a) 5  3
(b) 5  2
(c) 4  3
(d) 4  2
(a) Fe
(b) Fe (II)
(c) Fe (III)
(d) Fe (IV)
The frequency of one of the lines in Paschen series of
hy drogen atom is 2.340  10 11 Hz. The quantum num ber
[DPMT 2001]
n 2 which produces this transition is
(b) 5
(d) 3
The position of an electron can be
determined exactly with the help of an
electron m icroscope.
: The product of uncertainty in the
m easurement of its m om entum and the
uncertainty in the m easurem ent of the
position cannot be less than a finite limit.
[NDA 1999]
2.
Which of the following has m aximum number of unpaired
electron (atomic number of Fe 26)
[MP PMT 2001]
(a) 6
(c) 4
[CBSE PMT 2003]
(b) 2  10 25
7
(b) 1.69  10 23 J
(c) 1.69  10 23 J
(d) 1.69  10 25 J
The energy required to dislodge electron from excited
isolated H-atom, IE1  13.6 eV is
[DCE 2000]
(a) One
(c) Three
18.
(b) 5  2
(a) 5  10 7 m
(a) 1.69  10 20 J
16.
(a) 3  2
The frequency of a wave of light is 12  10 14 s 1 . The wav e
number associated with this light is
[Pb. PMT 1999]
23.
[A MU (Engg.) 1999]
15.
(b) From n  2 to n  3
(c) From n   to n  1
(d) From n  3 to n  5
In Bohr series of lines of hy drogen spectrum , the third
line from the red end corresponds to which one of the
following inter-orbit jumps of the electron for Bohr orbits
in an atom of hy drogen
[A IEEE 2003]
The v alue of Planck’s constant is 6.63  10 34 Js. The
v elocity of light is 3.0  10 8 ms 1 . Which value is closest to
the wavelength in nanometres of a quantum of lig ht with
(c) 4.77 Å
13.
Which of the following electron transition in a hy drogen
atom will require the largest am ount of energy
[UPSEAT 1999, 2000, 01]
The Bohr orbit radius for the hy drogen atom (n  1) is
(a) 0.13 Å
71
3.
Assertion :
A spectral line will be seen for a
2 p x  2 p y transition.
Reason
:
Energy is released in the form of wav e of
light when the electron drops from
2 p x  2 p y orbital.
[A IIMS 1996]
Assertion :
The cation energy of an electron is largely
determ ined by its principal quantum
num ber.
Reason
The principal quantum num ber n is a
m easure of the m ost probable distance of
finding the electron around the nucleus.
:
[A IIMS 1996]
72 Structure
4.
of atom
Assertion :
Nuclide
Reason
Nuclides having odd num ber of protons
and neutrons are generally unstable
:
30
Al13 is less stable than
40
Reason
Ca 20
17.
:
Num ber of orbitals in a shell equals to
2n .
Energy of the orbitals increases as
Assertion :
1s  2 s  2 p  3 s  3 p  3d  4 s  4 p
[IIT 1998]
5.
Assertion :
Reason
6.
Assertion :
Reason
7.
:
:
Assertion :
Reason
:
The atoms of different elem ents hav ing
sam e m ass number but different atom ic
num ber are known as isobars
The sum of protons and neutrons, in the
isobars is always different [A IIMS 2000]
Two electrons in an atom can hav e the
sam e values of four quantum num bers.
Two electrons in an atom can be present
in the sam e shell, sub-shell and orbital
and have the same spin
[A IIMS 2001]
The v alue of n for a line in Balm er series
of hy drogen spectrum having the highest
wav e length is 4 and 6 .
For Balm er series of hydrogen spectrum ,
the v alue n1  2 and n2  3 , 4 , 5.
 4 d  4 f  ......
Reason
18.
19.
Assertion :
Reason
:
Absorption spectrum conists of som e
bright lines separated by dark spaces.
Em ission spectrum consists of dark lines.
:
Assertion :
Reason
20.
Energy of the electron depends
com pletely on principal quantum
num ber.
Splitting of the spectral lines in the
presence of m agnetic field is known as
stark effect.
Line spectrum is sim plest for hy drogen
atom .
Thom son’s atom ic m odel is known as
‘raisin pudding’ m odel.
The atom is v isualized as a pudding of
positiv e charge with electrons (raisins)
em bedded in it.
Atom ic orbital in an atom is designated
by n, l, m l and m s .
Assertion :
Reason
[A IIMS 1992]
8.
:
:
Assertion :
Reason
Assertion :
These are helpful in designating electron
present in an orbital.
The transition of electrons n3  n 2 in H
atom will em it greater energy than
n4  n3 .
Reason
:
n 3 and n 2 are closer to nucleus tan n 4 .
22.
Assertion :
Reason
:
23.
Assertion :
Cathode rays are a stream of  -particles.
They are generated under high pressure
and high v oltage.
In case of isoelectronic ions the ionic size
increases with the increase in atom ic
num ber.
The greater the attraction of nucleus,
greater is the ionic radius.
21.
:
[A IIMS 2002]
9.
Assertion :
Reason
10.
Assertion :
Reason
11.
:
:
Assertion :
Reason
:
A resonance hybrid is always m ore stable
than any of its canonical structures.
This stability is due to delocalization of
electrons.
[A IIMS 1999]
Cathode ray s do not trav el in straight
lines.
Cathode ray s penetrate through thick
sheets
[A IIMS 1996]
Electrons revolving around the nucleus
do not fall into the nucleus because of
centrifugal force.
Rev olv ing electrons are planetary
electrons.
Reason
:
[A IIMS 1994]
12.
Assertion :
Reason
13.
Assertion :
Reason
14.
:
:
Assertion :
Reason
:
15.
Assertion :
16.
Reason
:
Assertion :
Threshold frequency is a character istic
for a m etal.
Threshold frequency is a m axim um
frequency required for the ejection of
electron from the m etal surface.
The radius of the first orbit of hy drogen
atom is 0.52 9 Å.
Radius for each circular orbit
(rn )  0.52 9 Å (n 2 / Z) , where n  1 ,2 ,3
Discovery and Properties of anode, cathode rays
neutron and Nuclear structure
1
d
2
a
3
c
4
c
5
b
6
a
7
b
8
a
9
d
10
c
and Z  atom ic num ber.
3 d z 2 orbital is spherically sy m m etrical.
11
b
12
d
13
b
14
a
15
b
16
b
17
c
18
c
19
c
20
b
3 d z 2 orbital is the only d -orbital which
21
a
22
d
23
c
24
b
25
d
26
c
27
b
28
d
29
c
30
a
31
b
32
d
33
b
34
c
35
c
36
a
37
b
38
a
39
d
40
c
41
c
is spherical in shape.
Spin quantum number can have the value
+ 1 /2 or –1 /2 .
(+ ) sign here signifies the wave function.
Total number of orbitals associated with
principal quantum num ber n  3 is 6.
Structure of atom
Atomic number, Mass number, Atomic species
1
b
2
a
3
b
4
b
5
a
6
a
7
c
8
b
9
c
10
b
11
b
12
c
13
b
14
c
15
c
16
c
17
c
18
a
19
c
20
a
21
c
22
b
23
c
24
d
25
b
26
b
27
a
28
a
29
c
30
b
31
c
32
d
33
d
34
c
35
c
36
c
37
c
38
b
39
d
40
c
41
b
42
c
43
a
44
c
45
b
46
c
47
d
48
a
49
c
50
c
51
a
52
c
53
b
54
a
55
c
56
a
57
d
58
c
59
a
60
a
61
d
62
b
63
a
64
c
65
b
66
a
67
c
68
a
69
d
70
d
71
c
72
a
73
b
74
d
Atomic models and Planck's quantum theory
73
1
b
2
b
3
a
4
c
5
c
6
c
7
b
8
d
9
d
10
a
11
a
12
c
13
a
14
b
15
d
16
b
17
a
18
c
19
c
20
b
Quantum number, Electronic configuration
and Shape of orbitals
1
c
2
a
3
b
4
d
5
c
6
c
7
c
8
a
9
a
10
a
11
c
12
c
13
a
14
a
15
d
16
c
17
c
18
d
19
b
20
c
21
c
22
a
23
c
24
d
25
c
26
c
27
b
28
d
29
e
30
b
31
d
32
a
33
c
34
d
35
d
36
c
37
b
38
b
39
d
40
c
41
d
42
c
43
c
44
a
45
a
46
a
47
b
48
c
49
c
50
b
51
c
52
b
53
b
54
b
55
c
56
c
57
b
58
e
59
c
60
c
1
c
2
a
3
b
4
b
5
d
6
b
7
c
8
b
9
c
10
a
61
d
62
d
63
d
64
c
65
b
d
67
c
68
d
69
c
70
b
11
b
12
a
13
d
14
b
15
b
66
16
c
17
a
18
c
19
a
20
d
71
a
72
c
73
c
74
c
75
a
21
d
22
c
23
d
24
d
25
c
76
c
77
c
78
c
79
d
80
d
26
a
27
c
28
b
29
c
30
a
81
b
82
c
83
a
84
a
85
b
31
b
32
c
33
d
34
b
35
b
36
a
37
c
38
c
39
c
40
a
86
c
87
a
88
b
89
c
90
b
41
c
42
d
43
d
44
a
45
d
91
d
92
a
93
b
94
b
95
d
46
b
47
a
48
c
49
d
50
a
96
d
97
a
98
a
99
d
100
c
51
a
52
c
53
d
54
c
55
b
101
b
102
d
103
a
104
c
105
d
a
107
c
108
d
109
a
110
d
56
b
57
b
58
a
59
b
60
c
106
61
c
62
b
63
c
64
c
65
b
111
d
112
b
113
c
114
b
115
b
66
b
67
c
68
a
69
b
70
d
116
a
117
c
118
b
119
a
120
a
71
a
72
d
73
a
74
c
75
d
76
b
77
a
78
a
79
c
80
a
121
d
122
b
123
b
124
b
125
d
81
a
126
d
127
b
128
c
129
a
130
b
131
a
132
c
133
d
134
b
135
a
136
a
137
c
138
c
139
c
140
c
141
c
142
d
143
c
144
c
145
b
146
d
147
a
148
c
149
b
150
c
151
d
152
a
153
a
154
d
155
b
156
d
157
a
158
b
159
c
160
d
161
c
162
d
163
b
164
c
165
a
166
d
167
d
168
d
169
b
170
a
171
c
172
d
173
c
174
b
175
d
176
c
177
a
178
b
179
b
180
c
Dual nature of electron
1
c
2
a
3
a
4
b
5
c
6
b
7
d
8
a
9
d
10
d
11
c
12
c
13
b
14
b
15
b
16
c
17
c
18
c
19
b
20
a
21
d
Uncertainty principle and Schrodinger wave
equation
74 Structure
of atom
181
c
182
b
183
c
184
c
185
a
186
d
187
c
188
c
189
a
190
c
191
b
192
a
193
d
194
b
195
d
196
a
197
c
198
d
199
b
200
b
201
a
202
b
203
b
204
c
205
d
206
b
207
b
208
c
209
d
210
b
211
a
212
a
213
b
214
c
215
a
216
d
217
b
218
b
219
b
220
d
221
b
222
b
223
a
224
d
225
d
226
a
227
b
228
a
15.
16.
19.
1
a
2
d
3
d
4
a
5
d
6
c
7
b
8
d
9
a
10
c
11
a
12
d
13
d
14
c
15
b
16
d
17
a
18
b
19
c
20
b
21
a
22
a
23
d
24
a
1
d
2
d
3
a
4
a
5
c
6
d
7
e
8
d
9
a
10
e
11
b
12
c
13
a
14
d
15
c
16
d
17
c
18
e
19
a
20
e
21
b
22
d
23
d
Discovery and Properties of anode, cathode rays
neutron and Nuclear structure
(d) Neutrons and protons in the nucleus and electrons in
the extranuclear region.
(a) It consists of proton and neutron and these are also
known as nucleones.
15
20.
26.
27 .
Assertion & Reason
2.
11.
13.
18.
Critical Thinking Questions
1.
10.
3.
(c) Radius of nucleus ~ 10
4.
(c) Positiv e ions are formed from the neutral atom by the
loss of electrons.
5.
(b) The  -ray particle constitute electrons.
6.
(a) Jam es Chadwick discov ered neutron (0 n1 ) .
7.
(b) Charge/m ass for
2
1
1
n  0,   , p  and e 
4
1
1 / 1837
9.
11
(d) The density of neutrons is of the order 10 kg / cc.
m.
(c) This is because chargeless particles do not undergo
any deflection in electric or m agnetic field.
(b) Neutron and proton found in nucleus.
(b) Cathode ray s are m ade up of negativ ely charged
particles (electrons) which are deflected by both the
electric and m agnetic fields.
(b) Mass of neutron is greater than that of proton, m eson
and electron.
Mass of neutron = mass of proton + m ass of electron
(b) Proton is 1 837 (approx 1800) times heav ier than an
1
electron. Penetration power 
mass
(c) Nucleus of helium is 2 He 4 m ean 2 neutrons and 2
protons.
(c) Proton is the nucleus of H  atom ( H  atom dev oid
of its electron).
(b) Cathode ray s are m ade up of negativ ely charged
particles (electrons, e  )
(c) Size of nucleus is m easured in Fermi (1 Ferm i
 10 15 m) .
(b) A m olecule of an elem ent is a incorrect statem ent.
The correct statement is “an element of a m olecule”.
Structure of atom
29.
31.
32.
(c) Proton is represented by p having charge +1 discovered in 1988
by Goldstein.
(b) The nature of anode rays depends upon the nature of residual
gas.
(d)
H  (proton) will have very large hydration energy due to its
very small ionic size
Hydration energy 
33.
(b)
1
Size
12.
(c) Most probable radius = a / Z
where a = 52.9 pm. For helium ion, Z = 2.
0
0
r =
mp
(b) Four unpaired electron are present in the
2
Fe 26
 [ Ar] 3d 6 ,4 s 0
14.
(c)
16.
 9.1  10 24  6.02  10 23  10.07  10 1  1.008 g
17.
(c)
8O
Mass of a electron  9.1  10 28 g
18.
(a)
35 Br
28
 6.02  10
23
 54.78  10
g  0.55 mg .
(c) One mole of electron = 6.023 × 10 electron
Mass of one electron = 9.1 × 10 gm
Mass of one mole of electrons
23
= 5.48  10 4  1000 mg = 0.548 gm  0.55 mg .
39.
40.
 1s 2 2 s 2 2 p 4
80
 1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 10 4 s 2 4 p 5
A  80 , Z  35 , N  ?
5
19.
= 6.023 × 10 23  9.1  10 28 gm = 5.48  10 4 gm
38.
ion
 Mass of one mole of proton
Mass of a proton  1.673  10 24 g
–28
37.
Fe 2 
Na  has 10 electron and Li  has 2 electron so these are
different number of electron from each other.
(c) P15  2, 8, 5
 9.1  10
36.
52 . 9
= 26.45 pm.
2
13.
 Mass of one mole of electron
35.
75
20.
N  A  Z  80  35  45
atomic number (Proton) is 35 and no. of neutron is 45.
16 O  
(c)
have
more
electrons
than
neutron
8
p  8, e  10, n  8 .
(a)
(a) Charge on proton  1 unit, charge on  particle = + 2 units, 2 :
1.
(b) m p / me ~ 1837 ~ 1.8  10 3 .
(a) Splitting of signals is caused by protons attached to adjacent
carbon provided these are not equivalent to the absorbing
proton.
(d) Nucleus consists of proton and neutron both are called as
nucleon.
(c) Positron (1e 0 ) has the same mass as that of an electron
(1e ) .
21.
(c)
A 12 and
neutrons.
6
6
X 13 both are isotopes but have different no. of
6
A12 , For A have p  6, e  6 and n  6 and
6
X 13 , For B have p  6, e  6 and n  7
P  20, mass no. (A) = 40
N  A–P = 40  20  20
P  N  20.
22.
(b) Electrons in Na   11  1  10
Electrons in Mg2   12  2  10
0
41.
1
(c) Electron
time lighter than proton so their mass ratio
1837
will be 1 : 1837
23.
24.
(c)
20 Ca
40
(d)
CH 3
 6  3  1  8e ,
H 3 O   3  8  1  10e  ,
NH 3  7  3  10e  , CH 3  6  3  1  10e 
Atomic number, Mass number, Atomic species
1.
3.
(b) The number of electrons in an atom is equal to its atomic
number i.e. number of protons.
(a) No. of protons = Atomic no. = 25 and no. of neutron = 55 –
25 = 30.
(b) No. of neutrons = mass number – no. of protons. = W – N.
4.
(b)
5.
6.
(a) Na  and Ne are isoelectronic which contain 10 electrons.
(a) One molecule of CO 2 have 22 electrons.
7.
(c)
2.
30
Zn 70 , Zn 2  has No. of Neutrons = 70 – 30 = 40.
(b)
26.
(b) Complete E.C.  [ Ar]18 3d 10 4 s 2 4 p 6 .
Hence no. of e   no. of protons  36  Z .
28.
9.
10.
CO  6  8  14 and CN   6  7  1  14 .
(c) Mass of an atom is due to nucleus (neutron + proton).
(b) Atomic number is defined as the number of protons in the
nucleus.
11.
(b)
26
X 56 A  P  N  Z  N  E  N
N  A  E  56  26  30
(a)
K   1s 2 2 s 2 2 p 6 3 s 2 3 p 6
Cl   1s 2 2 s 2 2 p 6 3 s 2 3 p 6 .
29.
(c) Mass no.  At. Wt.
Mass no. = no. of protons + no. of neutrons
At. no. = no. of protons.
30.
(b) N 2O  14  8  22
while Cl  has 18 e  .
(b) CO and CN  are isoelectronic.
 CONH 2  6  8  7  2  1 (from other atom to form
covalent bond) = 24.
25.
Cl and Cl  differs in number of electrons. Cl has 17 e 
8.
has 20 proton, 20 neutron.
CO 2  6  16  22.
31.
(c) Neutron in
12
6 C
 6, , Neutrons in
Ratio = 6 : 14 = 3 : 7.
33.
(d)
N 7  1s 2 2 s 2 2 p 3
N   1s 2 2 s 2 2 p 2
28
14
Si  14
76 Structure of atom
C  1s 2 2 s 2 2 p 2 .
34.
(c)
O  C  O, linear structure 180 o angle
Cl  Hg  Cl, linear structure 180 o angle.
35.
36.
37.
38.
(c) H   1s 2 and He   1s 2 .
(c) In the nucleus of an atom only proton and neutrons are
present.
(c)
63
Cu 29
Number of neutrons = atomic mass – atomic number = 63
– 29 = 34.
(b) 21 Protons and 24 Neutrons are present in nucleus and element
is Sc.
(c)
42.
43.
(c) Cl  have 17 proton, 18 neutron and 18 electron.
(a) Number of unpaired electrons in inert gas is zero because they
have full filled orbitals.
(c) Electrons and Protons are same in neutral atom.
48.
49.
51.
52.
(b) N 3  , F  and Na  (These three ions have e   10 , hence
they are isoelectronic)
63.
(a)
64.
(c) Atomic number of chlorine 17 and in Cl  ion total no. of
electron =18.
65.
(b) Tritium (H 13 ) has one proton and two neutron.
67.
(c)
68.
Total no. of e  is all p–orbitals  6  6  5  17 .
(a) Since its nucleus contain 9 proton so its. atomic number is 9
and its electronic configuration is 2, 7. So it require one more
electron to complete its octet. Hence its valency is 1.
69.
(d)
14
, n  14  7  7
7 X
40.
44.
62.
(d) No. of proton and no. of electron = 18 [ Ar1836 ] and No. of
neutron = 20
Mass number = P + N = 18  20  38.
(a)
and O 3 are isostere. The number of atoms in these
( 3) and number of electrons (24 ) are same.
(c) Number of electrons in nitrogen = 7 and number of electron is
oxygen = 8 we know that formula of nitrate ion is NO 3 we
also know that number of electron
= (1 × Number of electrons in nitrogen)
+ (3 × number of electrons in oxygen) + 1
= (1 × 7) + (3 × 8) + 1 = 32.
53.
Molecular mass 256
(b) Atomicity =

 8  S8 .
Atomic mass
32
54.
(a) In case of N 3  , p  7 and c  10
55.
(c) Chlorine Cl 17  [Ne ]
3s
2
3p
70.
71.
57.
(d)
Na   1s 2 2 s 2 2 p 6
Mg  1s 2 2 s 2 2 p 6
Hence, Ca 2  has 8 electrons each in outermost and
penultimate shell.
(c) Atomic no. of C = 6 so the number of protons in the nucleus =
6
(a) No. of P  Z  7; No. of electrons in N 3   7  3  10.
73.
(b) Heavy hydrogen is
74.
(d) Atomic number is always whole number.
2
1
D .Number of neutrons = 1
Atomic models and Planck's quantum theory
2.
3.
4.
5.
8.
9.
10.
(a) The central part consisting whole of the positive charge and
most of the mass caused by nucleus, is extremely small in size
compared to the size of the atom.
(b) Electrons in an atom occupy the extra nuclear region.
(b) According to the Bohr model atoms or ions contain one
electron.
(d) The nucleus occupies much smaller volume compared to the
volume of the atom.
(c) -particles pass through because most part of the atom is
empty.
(b) An electron jumps from L to K shell energy is released.
(c) Neutron is a chargeless particles, so it does not deflected by
electric or magnetic field.
(a) Energy is always absorbed or emitted in whole number or
multiples of quantum.
19.
Ar1840 = atomic number 18 and no. of Neutron in case of
20.
21.
23.
(d)
27.
(c) When c     than  
Ar22
Neutron = Atomic mass – Atomic number
 40  18  22
61.
 2, 8, 8, 2
Cl   1s 2 2 s 2 2 p 6 3 s 2 3 p 6
O 2   1s 2 2 s 2 2 p 6
(a)
20 Ca
(b) Both He and Li  contain 2 electrons each.
(c) During the experimental verification of de-Broglie equation,
Davisson and Germer confirmed wave nature of electron.
(a) Increases due to absorption of energy and it shows absorption
spectra.
(d) Rutherford -Scattering experiment.
(d) It represents Heisenberg’s uncertainty principle.
++
60.
(d)
72.
7.
Three electron pair
(a) Bromine consists of outer most electronic configuration
[ Ar] 3d 10 4 s 2 4 p 5 .
K 2 S formed by K  and S 2  ion. We know that atomic
Ca 2   2, 8, 8
5
56.
X 35  1s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 10 4 s 2 4 p 5
number of K is 19 and in K  ion its atomic number would be
18 similarly atomic number of S is 16 and in form S 2  ion its
atomic number would be 18 so the K  and S 2  are
isoelectronic with each other in K 2 S .
231
(c) In Xe 89
number of protons and electrons is 89 and No. of
neutrons = A – Z = 231 – 89 = 142.
NO 2
NO 3 and CO 32  consist of same electron and show same
isostructural.
(d) Nucleus of tritium contain [H 13 ]
p 1 , e 1 , n  2
11.
18.
E4
E
22
4
1
328
 2 
 ; E4  2 
 82 kJ / mol.
E2 4
16 4
4
4
c


3  10 8
 1 .5  10 2 m
2  10 6
Structure of atom
28.
30.
(b) According to quantum theory of radiation, a hot body emits
radiant energy not continuously but discontinuously in the
form of small packets of energy called quanta or photons.
(a)
p
h


40.
n 2h2
(b) Bohr’s radius 
. Which is a positive quantity.
4 2 me 2 z
(a) Gold used by Rutherford in scatting experiment.
41.
(c)
 1
1 
E  E3  E2  13.6  2  2   1 .9 eV
(3) 
 (2)
42.
(d)
R  R 0 ( 1.4  10 13 cm)  A 1 / 3
34.
43.
44.
45.
46.
47.
49.
 12.1567  10 8 m
1q 
1
q
(d)        9 .6  10 7  4 .8  10 7 C kg 1
 m  2  m  p 2
(a) According to Hydrogen spectrum series.
(d) The electron can move only in these circular orbits where the
h
angular momentum is a whole number multiple of
or it
2
is quantised.
(b) Generally electron moving in orbits according to Bohr‘s
principle.
(a) According to the planck’s law that energy of a photon is
directly proportional to its frequency i.e. E  h
(d) Bohr’s radius of the hydrogen atom
n 2  0.529Å
; where z = Atomic number,
z
n = Number of orbitals
3  10 8
 24.66  1014 Hz .
 12.567  10 8
h
h
(c) We know that  
;  m
m
mv
67.
52.
(a)
(c)
2 .172  10 18  2 .172  10 18

n2
22
19
 5.42  10 J .
hc
hc
or  
E 

E
m
68.
69.
70.
54.
55.
56.
(d) rn  n 2 : An  n 4
r3  0.53  3 2  0.53  9  4.77 Å
(c) According to Rutherford an atom consists of nucleus which is
small in size but carries the entire mass (P+ N ).
(b) Wavelength of spectral line emitted is inversely proportional to
1
energy  
.
E
1
1
1
(b) E  ; E1 
; E2 

8000
16000
E1 16000

 2  E1  2 E2
E2
8000
3  10 8 ms 1
 5 .0  1014 Hz .
 600  10  9 m
13.6
13.6 13.6
E
eV 

 3.40 eV
4
n2
22
59.
(b)
c
(a) It will take
72.
(d) rH  0 .529
n2
Å
z
For hydrogen ; n  1 and z  1 therefore
rH  0.529 Å
For Be 3  : Z  4 and n  2 Therefore
rBe 3  
73.
(a)
0 .529  2 2
 0 .529 Å .
4
Eionisation  E  En 
(b) v 
E  h 
74.
1
1 
1 1 
 R  2  2   109678     82258.5

1 4 
 n1 n 2 
  1.21567  10 5 cm or   12.1567  10 6 cm
n2
eV



13.6  1 2 13.6  1 2
; h  13.6  0.85

(1)2
(4 )2
 h  6.625  10 34
13.6  0 .85

 1.6  10 19 = 3.08  10 15 s 1 .
6 .625  10 34
1
1
1 
 R 2  2 
(c)

 n1 n 2 
1 
1
 1 .097  10 7 m 1  2  2 

 
1
1
   91  10 9 m
We know 10 9  1 nm So   91nm
75.
r2 (2)2
 2  4.
r1
(1)
1
2
13.6 Z eff
 13.6 Z 2 13.6 Z 2

= 
2
n12
 n 2

(b) Bohr radius 
4 2 mr 2
nh
71.
(d) rn  r1  n 2
(a)  
66.
1 .54  10 10 m  3  10 8 m sec 1
 1.4285  10 32 kg .
(a) The spliting of spectral line by the magnetic field is called
Zeeman effect.
(b) r  n 2 (excited state n  2 )
r  4a 0
E
58.
65.
6 .626  10 34 Js
A 2 n 24
2 4 16
 4  4 
 16 : 1
A1 n1
1
1
6 .64  10 34  3  10 8

 6 .64  10 8 Å
3  10 8
53.

The velocity of photon (v) = 3  10 8 m sec 1
  1.54  10 8 cm  1.54  10 10 meter
r 
51.
c
v
34
6 .6  10
 3  10 23 kgms 1
2 .2  10 11
77
(d) r  n 2
For I orbit   1
st
For III orbit =   3 2  9
So it will 9  .
(b) Bohr suggest a formulae to calculate the radius and energy of
each orbit and gave the following formulae
rd
76.
78 Structure of atom
rn 
n2h2
4 2 kme 4 Z
Where except n 2 , all other unit are constant so rn  n 2 .
77.
h
6 .63  10 34

 6 .63  10  29 m
mv
10 6  10
(b) According to de–Broglie

(a) Energy of an electron E 
15.
 E0

n2
h

mv
For energy level (n  2)
E
78.
79.
81.
13.6
(2)
2

13.6
 3 .4 eV .
4

(a) Energy of ground stage (E 0 ) = 13.6 eV and energy level =
5
13.6
13.6
13.6
=
E5 
eV 
 0 .54 eV .
25
n2
52
(c) Positive charge of an atom is present in nucleus.
(a) For n4  n1 , greater transition, greater the energy difference,
lesser will be the wavelength.
4.
5.
6.
h 

(d) According to de-Broglie   
.
mv 

8.
(a)
10.
11.

h

2meV
19.
 1 
6
1
8
1
v 
 (3 .00  10 ms ) = 3.00  10 ms
 100 
Momentum of the electron (p ) = m 
= (9.11  10 31 kg ) (3.00  10 6 ms 1 )
= 2.73  10 24 kg ms 1

2eV
m
20.
21.
1.
2.
(b) The uncertainty principle was enunciated by Heisenberg.
(b) According to uncertainty principle, the product of uncertainties
of the position and momentum, is x  p  h / 4 .
5.
(c) x  p 
13.
h
, p  mv
p
7.

6 .62  10 34
h

mv
9 .1  10  31  1 .2  10 5
8.
(b) Mass of the particle (m)  10 6 kg and velocity of the particle
(v)  10 ms 1
h
4
is not the correct relation. But correct
h
.
4
(b) According to the Heisenberg’s uncertainty principle momentum and
exact position of an electron can not be determined simultaneously.
h
(d) x. p 
, if  x  0 then p   .
4
h
(c) According to x  p 
4
Heisenberg’s uncertainty equation is x  p 

  6.626  10 9 m .
(a) We know that the correct relationship between wavelength and
h
momentum is   . Which is given by de-Broglie.
p
(d) De-broglie equation applies to all the material object in motion.
Uncertainty principle and Schrodinger wave
equation
6 .62  10 34
2  9 .1  10  31  2 .8  10  23
h
6 .626  10 34

p 2 .73  10  24 kgms 1
  2.424  10 10 m .
  9.28  10 meter .
(c)
h
mv
(b) One percent of the speed of light is

8
12.
6 .62  10 34
h
=
= 1.32  10 35 m .
mv
0 .5  100
(c) Dual nature of particle was proposed by de-broglie who gave
the following equation for the wavelength.

The de-broglie wavelength of this electron is
h
6 .63  10 34

 6 .63  10  33 m
mv
10  3  100
h
1
(d)  
. For same velocity   .
m
mv
SO 2 molecule has least wavelength because their molecular
mass is high.
h
(d) de-Broglie equation is   .
p
(c) Formula for de-Broglie wavelength is
h
h
1
or  
 eV  mv 2 or  
p
mv
2
cm  4  10 8 cm  4 Å .
6 .625  10 34
 10  30 m .
5
0 .2 kg 
60  60 ms 1
(c) From de Broglie equation


h

mv
17.
h
h
h
or
(b)   or
de-Broglie equation.
p
mv
mc
(c) Emission spectra of different  accounts for quantisation of
energy.
(b) According to de-Broglie equation
h
h
h

, p  mv ,   ,  
mv
mc
p
7.
9.
h
h
h
or
or
.
p
mv
mc

2  5  10 4
(c)
18.
(c) According to de-Broglie equation  
6 .62  10 27  6 .023  10 23
16.
Dual nature of electron
1.
6 .62  10 20 erg. sec
2
 5  10 4 cm / sec
6 .023  10 23
12.
x 
h
6 .62  10 34

 5.27  10 30 m .
p  4 1  10 5  4  3 .14
Structure of atom
13.
10.
(a) Uncertainty of moving bullet velocity
h
6 .625  10 34
v 

4  m  v 4  3 .14  .01  10 5
 5.2  10
14.
15.
16.
h
This equation shows Heisenberg’s uncertainty
4
principle. According to this principle the product of uncertainty
in position and momentum of particle is greater than equal to
h
.
4
(d) Spin quantum number does not related with Schrodinger
equation because they always show 1 / 2 , 1 / 2 value.
h
h
(b) According to x  m  v 
; v 
x  m  4
4
6 .6  10 34
10
5
 0 .25  3 .14  4
 2 .1  10  29 m / s
h
(a) Uncertainity in position x 
4  p

18.
m/sec .
(b) x .p 

17.
28
6 .63  10 34
 5.28  10 30 m .
4  3 .14  (1  10 5 )
(c) Given that mass of electron  9.1  10
Planck’s constant  6.63  10
34
31
kg m s
h
h
; x  v  m 
4
4
where : x = uncertainity in position
v = uncertainity in velocity
h
x  v 
4  m
By using x  p 

34
6 .63  10
4  3 .14  9 .1  10 31
 5.8  10 5 m 2 s 1 .
Quantum number, Electronic configuration
and Shape of orbitals
3.
5.
(b) The shape of an orbital is given by azimuthal quantum number
'l' .
(c) Hund’s rule states that pairing of electrons in the orbitals of a
subshell (orbitals of equal energy) starts when each of them is
singly filled.
2
2
11.
(c) In 2p – orbital, 2 denotes principal quantum number (n) and
p denotes azimuthal quantum number (l  1) .
12.
(c) Electronic configuration of H  is 1s 2 . It has 2 electrons in
extra nuclear space.
13.
(a) The electronic configuration must be 1s 2 2 s 1 . Hence, the
14.
15.
(a) Principal quantum no. tells about the size of the orbital.
(d) An
element
has
the
electronic
configuration
1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 2 , (Si) . It’s valency electrons are four.
16.
(c) The magnetic quantum number specifies orientation of orbitals.
17.
(c) If l  2, m  3. (e to e ) .
18.
(d) If n  3 then l  0, 1, 2 but not 3.
20.
(c) Atomic number of Cu is 29  ( Ar) 4 s 1 3d 10 .
21.
(c) The shape of 2 p orbital is dumb-bell.
22.
(a) When the value of n  2 , then l  1 and the value of
m  1, 0,  1 i.e. 3 values.
23.
(c) Cr24  ( Ar) 3d 5 4 s 1 electronic configuration because half filled
orbital are more stable than other orbitals.
(d) Kr has zero valency because it contains 8 electrons in
outermost shell.
element is lithium (z  3).
24.
25.
(c) 2 electron in the valence shell of calcium Ca 20  (2, 8, 8, 2) .
27.
(b) Value of l  1 means the orbital is p (dumb-bell shape).
31.
33.
(d) Cr has Ar4 s 1 3d 5 electronic configuration because half
filled orbital are more stable than other orbitals.
(d) The two electrons will have opposite spins.
(c) If m = –3, then l = 3, for this value n must be 4.
34.
(d) No. of electrons  2n 2 hence no. of orbital 
2n 2
 n2 .
2
35.
(d) No. of electrons  2n 2 hence no. of orbital 
2n 2
 n2 .
2
36.
(c) If n  3 then l  0 to n  1 & m  l to l
37.
(b)
28.
6
6.
(c) 1s , 2 s , 2 p represents a noble gas electronic configuration.
7.
(c) The electronic configuration of Ag in ground state is
[Kr] 4 d 10 5 s 1 .
38.
8.
(a) n, l and m are related to size, shape and orientation
respectively.
39.
9.
(a) Electronic configuration of Rb(37 ) is
n  5, l  0, m  0 , s  
1
2
Na 11  2, 8, 1  1s 2 , 2 s 2 2 p 6 , 3 s 1
n  3, l  0, m  0, s  1 / 2
1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 10 4 s 2 4 p 6 5 s 1
So for the valence shell electron (5 s 1 )
(a) 3d subshell filled with 5 electrons (half-filled) is more stable
than that filled with 4 electrons. 1, 4 s electrons jumps into
3 d subshell for more sability.
kg
2 1
79
40.
(b) Hund’s rule states that pairing of electrons in the orbitals of a
subshell (orbitals of equal energy) starts when each of them is
singly filled.
(d) As a result of attraction, some energy is released. So at infinite
distance from the nucleus energy of any electron will be
maximum. For bringing electrons from  to the orbital of
any atom some work has to be done be electrons hence it bill
loose its energy for doing that work.
(c) This space is called nodal space where there is no possibility of
oressene of electrons.
41.
(d) For s orbital l  0 m  0 .
42.
(c) For M th shell, n  3; so maximum no. of electrons in M th
shell  2n 2  2  3 2  18 .
80 Structure of atom
m  l to l including zero.
43.
(c)
44.
(a) Number of radial nodes = (n  l  1)
For 3s: n = 3, l = 0
(Number of radial node = 2)
For 2p: n =2, l = 1
(Number of radial node = 0)
(a) It consists only s orbital which is circular.
(a) Hund’s rule states that pairing of electrons in the orbitals of a
subshell (orbitals of equal energy) starts when each of them is
singly filled.
45.
46.
47.
48.
(b) If value of l is 2 then m  2,  1, 0,  1,  2 . m  l to l
including zero.
(5 values of magnetic quantum number)
(c) s, p, d orbitals present in Fe
(c)
68.
69.
70.
72.
(d) (4) and (5) belong to d -orbital which are of same energy.
(c) Atomic no. 17 is of chlorine.
(b) The s-orbital has spherical shape due to its non- directional
nature.
(a) According to the Aufbau’s principle the new electron will enter
in those orbital which have least energy. So here 4 p -orbital
has least energy then the others.
(c) According to Aufbau’s principle.
73.
(c)
74.
(c) Ground
71.
(b) According to Aufbau rule.
(c) 3d subshell filled with 5 electrons (half-filled) is more stable
than that filled with 4 electrons. 1, 4 s electrons jumps into
3 d subshell for more sability.
52.
(b)
1
.
2
(b) 3d subshell filled with 5 electrons (half-filled) is more stable
than that filled with 4 electrons. 1, 4 s electrons jumps into
3 d subshell for more sability.
55.
(c) It has 3 orbitals p x , p y , p z .
57.
(b) If l  2 then it must be d orbital which can have 10
electrons.
59.
(c) for d orbital l  2 .
60.
(c)
61.
(d) When n  3 shell, the orbitals are n 2  3 2  9 .
m  l to l including zero.
2n 2
 n2 .
2
(d) Configuration of Ne = 1s 2 2 s 2 2 p 6
F

2
2
= 1s 2 s 2 p
6
Na  = 1s 2 2 s 2 2 p 6
Cl
63.
2
2
6
(c)
78.
(c) When l  3 then
80.
(d) m  1 is not possible for s orbital (l  0).
84.
85.
(a) Both 2p and 3p-orbitals have dumb-bell shape.
(b) 3d subshell filled with 5 electrons (half-filled) is more stable
than that filled with 4 electrons. 1, 4 s electrons jumps into
3 d subshell for more sability.
86.
(c) The shape of 2 p orbital is dumb-bell.
87.
(a)
89.
(c) For p-orbital, l  1 means dumb-bell shape.
91.
(d) l  3 means f subshell maximum number of e in f subshell =
14.
(b) As per Aufbau principle.
25
Mn  [ Ar] 3d 5 4 s 2
-2 electrons
–
94.
(b) l  0 is s, l = 1is p and l  2 is d and so on hence s p d may
be used in state of no..
95.
(d) For 4 d, n  4, l  2, m  2,1,0,1,2, s  
96.
(d) m cannot be greater than l ( 0, 1).
97.
(a) For n  1, l  0.
99.
(d)
1
.
2
Na 11  1s 2 2 s 2 p 6 3 s 2
6
= 1s 2 s 2 p 3 s 3 p .
Mn 2   [ Ar] 3d 5 4 s 0
1
.
2
102.
(d) According to Aufbau’s rule.
105.
(d)
2 p x , 2 p y , 2 p z sets of orbital is degenerate.
106.
(a)
Mg12 have 1s 2 2 s 2 2 p 6 3 s 2 electronic configuration

64.
(c)
K and Ca
have the same electronic configuration
(1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 )
65.
(b) For s-orbital, l  0.
66.
(d)
3s 1
F9  1s 2 2 s 2 2 p 5
n  3, l  0, m  0 and s  
2
(d) Unh106  [Rn] 5 f 14 , 6 d 5 , 7 s 1

Cu 29  1s 2 2 s 2 2 p 6 3 s 2 3 p 6 3d 10 4 s1
77.
Mg   = 1s 2 2 s 2 2 p 6

of
(c) No. of electrons in 3 rd shell = 2n 2 = 2 (3) 2  18
93.
No. of electrons  2n 2
62.
state
m  3,  2,  1, 0,  1,  2,  3 . m  l to l including
zero.
n  4, l  0, m  0 and s  
Hence no. of orbital 
1s 2 2 s 2 2 p 6 ,3 s 2 3 p 6 ,4 s 2 3d 6  2, 8, 14, 2 .
76.
K19  1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 , 4 s 1
for 4 s1 electrons.
54.
7
Cu 2   1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 9 .
Fe26  1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 , 4 s 2 3d 6
50.
51.
N  1s 2 ,2 s 2 2 p1x , 2 p1y , 2 p1z . Hund’s rule states that pairing
of electrons in the orbitals of a subshell (orbitals of equal
energy) starts when each of them is singly filled.
67.
is
valency
electrons
1
n  3, l  0, m  0, s 
2
n  3, l  0, m  0, s  
of
Na
for
this
1
.
2
Structure of atom
107.
108.
(c) The principle quantum number n  3 . Then azimuthal
quantum number l  3 and number of orbitals
 n 2  3 2  9 . 3 and 9
(d)
29
Cu  [ Ar] 3d 10 4 s 1 , Cu 2  [ Ar] 3d 9 .4 s 0 .
Ground state of Cu
Cu
2
29
 1s 2 s 2 p 3 s 3 p 3 d 4 s
2
2
6
2
6
10
146.
(d) m  (2l  1) for d orbital l  2 m  (2  2  1)  5 .
147.
(a) The atomic number of chlorine is 17 its configuration is
1s 2 2 s 2 2 p 6 3 s 2 3 p 5
148.
(c)
1
2
6
2
6
9
2
110.
(d) 1s 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3d 6
configuration of Iron.
111.
(d) Orbitals are 4 s, 3 s, 3 p and 3 d . Out of these 3d has highest
energy.
113.
(c) For the n  2 energy level orbitals of all kinds are possible
2n , 2 2  4 .
it
shows
electronic
114.
(b) n  2 than no. of orbitals  n 2 , 2 2  4
118.
(b) For both A & B electrons s  1 / 2 & 1 / 2 respectively,
n  3, l  0, m  0
119.
120.
(a) According to Aufbau’s rule.
(a) Possible number of subshells would be (6s, 5p, 4d).
121.
(d) For f orbital l  3 .
123.
(b) 4d-orbital have highest energy in given data.
125.
(d) If m  3, l  3 and n  4.
127.
(b)
128.
(c) m can't be greater than l.
130.
(b) n  1 and m  1 not possible for s-orbitals.
131.
(a)
149.
152.
(d) When 2p orbital is completely filled then electron enter in the
3s. The capacity of 2p orbital containing e  is 6. So
1s 2 , 2 s 2 2 p 2 3 s 1 is a wrong electronic configuration the
 1s 2 s 2 p 3 s 3 p 3 d 4 s
2
6
2
6
5
(c) No. of electron are same (18) in Cl  and Ar .
138.
(c) For s-subshell l  0 then should be m  0.
139.
(c) 19 electron of chromium is 4 s1
1
n  4 , l  0, m  0, s  
2
140.
(c) The value of m is – l to l including zero so for l = 3, m would
be –3, –2, –1, 0, +1, +2, +3.
141.
(c)
142.
(d) Magnetic quantum number of sodium ( 3 s1 ) final electron is m
= 0.
(c) Generally azimuthal quantum number defines angular
momentum.
143.
g
4
h
5
156.
(d) 2p have contain maximum 6 electron out of which there are 3
are of + 1/2 spin and 3 are of – 1/2 spin
So l  (n  1)  4  1  3 which is f orbit contain 7 orbital.
  

+1/2

–1/2
(a) For 4f orbital electron, n  4
l  3 (Because 0, 1, 2, 3)
s, p, d, f
m = + 3, + 2, +1, 0, – 1, – 2, – 3
s = +1/2
(b)
24
Cr  1s 2 , 2 s 2 , 2 p 6 , 3 s 2 , 3 p 6 , 3 d 5 , 4 s1
l 1
l 1 l  2
(We know that for p the value of l  1 and for d , l  2)
For l  1 total number of electron = 12
159.
For l  2 total number of electron = 5.
(c) Atomic number of potassium is 19 and hence electronic
configuration will be 1s 2 , 2 s 2 , 2 p 6 , 3 s 2 , 3 p 6 , 4 s1
Hence for 4 s1 electron value of Quantum number are
Principal quantum number n  4
Azimuthal quantum number l  0
Magnetic quantum number m  0
Spin quantum number s  1 / 2
160.
th
l  1 is for p orbital.
f
3
(b) n  4  1s 2 , 2 s 2 , 2 p 6 , 3 s 2 , 3 p 6 , 3d 10 , 4 s 2 , 4 p 6 , 4 d 10 , 4 f 14
1
137.
d
2
155.
(b) This electronic configuration is Cr (chromium element) in the
ground state
2
p
1
Number of orbitals  5  2  1  11
(a) It is the ground state configuration of chromium.
write is 1s 2 2 s 2 2 p 3 .
134.
m2
153.
158.
133.
s
(a)
Fe 26  [ Ar] 3d 6 4 s 2
(c) Maximum number of electron
 2n 2 (where n  4 )  2  4 2  32 .
m1
(b) The ground state of neon is 1s 2 2 s 2 2 p 6 on excitation an
electron from 2 p jumps to 3 s orbital. The excited neon
l=0
157.
132.
l
configuration is 1s 2 2 s 2 2 p 5 3 s 1 .
N 714  1s 2 2 s 2 2 p 1x 2 p 1y 2 p 1z .
Fe 3   [ Ar] 3d 5 4 s 0 .
n
3
2
1
0
This set (c) is not possible because spin quantum number
1
values   .
2
 1s , 2 s 2 p ,3 s 3 p 3 d .
2
81
(d) According to Hund's rule electron first fill in unpaired form in
vacant orbital then fill in paired form to stabilized the molecule
by which 1s 2 ,2 s 2 ,2 p x2 is not possible. According to Hund's
rule. Because 2 p x , py , p z have the same energy level so
electron first fill in unpaired form not in paired form so it
should be 1s 2 , 2 s 2 , 2 p1x ,2 p1y .
161.
162.
(c) It is governed by Aufbau principle.
(d) The electronic configuration of atomic number
24  1s 2 , 2 s 2 , 2 p 6 , 3 s 2 ,3 p 6 ,3d 5 , 4 s1
163.
(b) The maximum number of electron in any orbital is 2.
82 Structure of atom
164.
(c) According to pauli principle 2 electron does not have the same
value of all four quantum number. They have maximum same
value are 3.
165.
166.
(a) Number of orbitals  n 2  4 2  16 .
(d) We know from the Aufbau principle, that 2p orbital will be
filled before 3s orbital. Therefore, the electronic configuration
1s 2 , 2 s 2 , 2 p 2 , 3 s1 is not possible.
167.
(d) Each orbital may have two electrons with opposite spin.
168.
(d) Maximum no. of electrons in a subshell  2(2l  1) for fsubshell, l = 3 so 14 electrons accommodated in f -subshell.
(b) Each orbital has atleast two electron.
(a) Nucleus of 20 protons atom having 20 electrons.
169.
170.
174.
176.
177.
(a) Shell = K, L, M
1s
2s
2p
(a) Atomic number of Cu is 29 so number of unpaired electrons is
1
4 s1
3d 10
1s 2
193.
(d) C 6  1s 2 , 2 s 2 2 p 2 (Ground state)
 1s 2 2 s1 2 Px1 2 p1y 2 p1z (Excited state)
In excited state no. of unpaired electron is 4.
194.
(b) Max. no. of electrons in N-shell n  4 
 2n2  2  4 2  32 .
195.
(d)
26 Fe
 [ Ar] 3d 6 , 4 s 2
Fe2   [ Ar] 3d 6 , 4 s0
Number of d-electrons = 6
17 Cl
 [ Ne ] 3 s 2 , 3 p 5
Cl   [ Ne ] 3 s 2 , 3 p 6
Number of p-electrons = 6.
196.
197.
2p4
2s 2
= 1s 2 2 s 2 2 p 6 3 s 2 3 p 4
Hence the number of s electron is 6 in that element.
(b) For m  0 , electron must be in s-orbital.
(c) In this type of electronic configuration the number of unpaired
electrons are 3.
=3
Cu  (Ar)
178.
192.
(b) O8 
(a) Electrons in the atom  18  4  3  25 i.e. Z  25 .
(c) The atomic number of bromine is 35 and the electronic
configuration of Br is
Br35  1s 2 , 2 s 2 , 2 p 6 , 3 s 2 , 3 p 6 , 3d 10 , 4 s 2 , 4 p 5
total electron present in p-orbitals of Br is –
2 p 6  3 p 6  4 p 5  17.
Unpaired electron
Be 4  1s , 2 s = (Ground state)
2
2
181.
(c)
182.
Number of unpaired electrons in the ground state of Beryllium
atom is zero.
(b) Two unpaired electrons are present in
198.
(d)
Fe 2  has 1s 2 ,2 s 2 2 p 6 ,3 s 2 3 p 6 3d 6 configuration with 4
unpaired electron.
199.
(b)
Fe 2  [ Ar] 3d 6 4 s 0
Ni   (z  28) cation

Ni28
=4
 [ Ar]
4s
3d
8
183.
(c)
Fe 2  consist of maximum 4 unpaired electrons.
0
201.
O 2  1s 2 2 s 2 2 p 6 3 s 2 3 p 4
3s
Fe 3  (z  26)
Fe 3   [ Ar] 3d 5 4 s 0
3p4
2
(a)
=5
3d
2 Unpaired electrons
3
4 s 1 but Cr24
184.
(c)
Cr24  ( Ar) 3d
185.
(a)
Zn 30  [ Ar] 3d 10 4 s 2
5
 ( Ar) 3d 4 s
3
0
Total no. of unpaired electron=5
202.
(b) Co 27  [ Ar] 3d 7 4 s 2
3d 7
Zn    [ Ar] 3d 10 4 s 0
186.
(d)
Mn 2 ion will have five (maximum) unpaired electrons
[Ar]
203.
4s
3d 5
3
187.
(c)
190.
191.
(c) Due to full filled d-orbital Cl  has spherical symmetry.
(b) Atomic number 14 leaving 2 unpaired electron
Fe
14 Si
1s
ion will have five (maximum) unpaired electrons.
 1s 2 s 2 p 3 s 3 p
2
2s
2
6
2
2p
4s
3s
3p
N 14  1 s 2 , 2 s 2 , 2 p x 2 p y 2 p z
1
7
1
1
Hence it has 3 unpaired electron in 2p-orbital.
204.
2
3 unpaired electron are present in cobalt metal.
(b) According to Hund's rule, the pairing of electrons will not
occur in any orbital of a subshell unit and unless, all the
available of it have one electron each.
Electronic configuration of
(c) 2 s orbital have minimum energy and generally electron filling
increases order of energy according to the Aufbau’s principle.
Structure of atom
205.
206.
207.
208.
210.
212.
214.
215.
216.
(d) According to Pauli’s exclusion principle no two electrons in the
same atom can have all the set of four quantum numbers
identical.
(b) The second principal shell contains four orbitals viz
2 s, 2 p x , 2 p y and 2 p z .
(b) Follow Hund’s multiplicity rules.
(c) According to the Aufbau’s principle, electron will be first enters
in those orbital which have least energy. So decreasing order of
energy is 5 p  4 d  5 s.
(b) No two electrons in an atom can have identical set of all the
four quantum numbers.
(a) In particular shell, the energy of atomic orbital increases with
the value of l .
(c) Aufbau principle explains the sequence of filling of orbitals in
increasing order of energy.
(a) According to Aufbau principle electron are filling increasing
order of energy. Therefore the electronic configuration
1s 2 2 s 2 2 p 6 obeys Aufbau principle.
(iii) Proton 
(a) Metal is
5.
(d)
220.
222.
(b) We know that for d-electron l  2.
219.
h
h
;   2 (2  1)
  l(l  1)
2
2
  2 (2  1)
 2 .179  10 11 

8.
(d) Radius of nucleus  1.25  10 13  A1 / 3 cm
Radius of atom = 1Å  10 8 cm.
Volume of nucleus (4 / 3) (5  10 13 )3

Volume of atom
(4 / 3) (10  8 )3
 1.25  10 13 .
9.
(a) Values of energy in the excited state

225.
11.
227.
228.
(a) Atomic number of deuterium = 1; 1 D2  1s1
12.
10.
(c)
(ii)   particle 
2
 0.5
4

13.6
eV
n2
13.6
 3 .4 eV in which n  2, 3, 4 etc.
4
E1 He   E1 H  z 2
 871.6  10 20  E1 H  4
E1 H  217.9  10 20 J
(a) 42g of N 3 ions have 16 N A valence electrons 4.2g of N 3
ion have 
16 N A
 4 .2  1 .6 N A .
42
(d) Ist excited state means n  2
r  r0  2 2  0.53  4  2.12 Å
Critical Thinking Questions
0
0
1
8
 2 .179  10 11  1.91  10 11  0.191  10 10 erg
9
 1.25  10 13  64 1 / 3  5  10 13 cm
h
h
Angular momentum = l (l  1) 
= 0
0 .
2
2
(d) Azimuthal quantum number (l) = 3 shows the presence of f
orbit, which contain seven orbitals and each orbital have 2
electrons. Hence 7  2  14 electrons.
(b) According to Aufbau principle.
(i) neutron 
2 .179  1011
9
Since electron is going from n  1 to n  3 hence energy is
absorbed.
(d) Since s-orbital have l  0
e
for
m
1

(b) According to Bohr’s model E  E1  E3
224.
(d)
i.e. E 
7.
(a) Number of nodal centre for 2 s orbitals ( n  1)  2  1  1 .
3.
c

(c) Rutherford discovered nucleus.
h
h
;  6
.
2
2
(a) F  have the same number of electrons with the neon atom.
(d) No change by doubling mass of electrons however by reducing
mass of neutron to half total atomic mass becomes 6  3
instead of 6  6 . Thus reduced by 25%.
E  hv  h
6.
223.
1.
2.
M 2  (2, 8, 14) than n  A  Z
E1

4000
 2 
 2.
E2
1 2000
(d) Electronic configuration of the Cr24 is [ Ar] 4 s1 3d 5 or
3d
4s
(b) According to the Aufbau principle electron filling minimum to
higher energy level.
(b) According to Aufbau principle electron are filled in various
atomic orbital in the increasing order of energy
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f <
5 d < 6p < 7 s .
(d) According to Aufbau’s rule.
56
1
 1837 .
1 / 1837
 56  26  30 .
=6
217.
1
1
1
(iv) electron 
4.
83
13.
  12  1014 s 1
(d) Frequency
and
velocity
of
light
c  3  10 cm s 1 . We know that the wave number
10
v 12  1014

 4  10 4 cm 1
c
3  1010
(c) The last line in any series is called series limit. Series limit for
Balmer series is 3646 Å.

14.
15.
(b)
E
13.6 13.6

 3.4 eV
4
n2
84 Structure of atom
We know that energy required for excitation E  E 2  E1
6.
 3.4  (13.6)  10.2 eV
Therefore energy required for excitation of electron per atom
10.2

 1.69  10  23 J
6.02  10 23
17.
7.
(a) The number of nodal plane are present in a p x is one or no.
of nodal place = l
for p x orbital l = 1
x
(d) We know from the Pauli exclusion principle, that two electrons
in the same atom can not have same value of all four quantum
numbers. This means each electron in an atom has only one set
of values for n, l, m and s . Therefore both the Assertion and
Reason are false.
(e) We know that the line in Balmer series of hydrogen spectrum the
highest wavelenght or lowest energy is between n1  2 and
n 2  3 . And for Balmer series of hydrogen spectrum, the value of
8.
(d)
9.
(a)
10.
(e)
11.
(b)
12.
(c)
13.
(a)
Nodal plane
18.
(b) In Balmer series of hydrogen atomic spectrum which electronic
transition causes third line O  L , n2  5  n1  2
20.
(b)  
1
1 
 RH  2  2 

 n1 n2 
1
1
1 
  R H  2  2   n2  3 for Paschen series.

n2 
 3
1
1
1 
E 2  2
 n2 n1 
21.
(a)
23.
(d)  
c
3  10 8

 3 .75  10 8
v 8  1015
 3.75  10 8  10 9 nm  4  101 nm .
Assertion & Reason
1.
(d) The assertion is false but the reason is true exact position and
exact momentum of an electron can never be determined as
according to Hesenberg’s uncertainity principle even with the
help of electron microscope because when e  beam of
electron microscope strikes the target e  of atom, the impact
causes the change in velocity of e  thus attempt to locate the
e  changes ultimately, the momentum & position of e  .
x .p 
2.
3.
4.
5.
h
 0.57 ergs sec/ gm.
4
(d) Both assertion and reason are false. 2 p x and 2 p y orbitals
are degenerate orbitals, i.e., they are of equal energy and hence
no possibility of transition of electron.
(a) We know that principal quantum number represent the main
energy level or energy shell. Since each energy level is
associated with a definite amount of energy, this quantum
number determines to a large extent te energy of an electron. It
also determines the average distance of an electron around the
nucleus. Therefore both Assertion and Reason are true and the
Reason is a correct explanation of the Assertion.
(a) It is observed that a nucleus which is made up of even number
of nucleons (No. of n & p ) is more stable than nuclie which
consist of odd number of nucleons. If number of neutron or
proton is equal to some numbers i.e., 2,8, 20, 50, 82 or 126
(which are called magic numbers), then these passes extra
stability.
(c) The assertion that the isobars are the atoms of different
elements having same mass number but different atomic
number, is correct but reason is false because atomic mass is
sum of number of neutron and protons which should be same
for isobars.
n1  2 and n 2  3,4,5 . Therefore the Assertion is false but the
Reason is true.
We know that Absorption spectrum is produced when white
light is passed through a substance and transmitted light is
analysed by a spectrograph. The dark spaces corresponds to the
light radiation absorbed by the substance. And emission
spectrum is produced by analysing the radiant energy emitted
by an excited substance by a spectrograph. Thus discontinuous
spectra consisting of a series of sharp lines and separated by
dark bands are obtained. Therefore both the Assertion and
Reason are false.
We know that a resonance hybrid or the actual molecule is
always more stable than any of its canonical structures which is
also called hypothetical or imaginary structures. This stability is
due to delocalization of electrons and is measured in terms of
resonance energy or delocalization energy, it is defined as the
difference in internal energy of the resonance hybrid and the
most stable canonical structure. Therefore both the Assertion
and Reason are true and the Reason is a correct explantion of
the Assertion.
We know that cathode rays cast shadows of solid objects placed
in their path. During experiment performed on these rays,
fluorescene (flash of light) is observed in the region, outside
the shadow. This shows that cathode rays travel in straight
lines. We also known that cathode rays penetrate through a
thin sheet of metals but are stopped by thick sheets. Therefore
both Assertion and Reason are false.
We know that electrons are revolving around the nucleus at
high speed in circular paths. The centrifugal force (which arises
due to rotation of electrons) acting outwards, balances the
electrostatic force of attraction (which arises due to attraction
between electrons and nucleus). This prevent the electron from
falling into the nucleus. We also know that Rutherford’s model
of atom is comparable to the “solar system”. The nucleeus
represent the sun whereas revolving electrons represent the
planets revolving around the sun. Thus revolving electron are
also called planetary electrons. Therefore both Assertion and
Reason are true but Reason is not a correct explanation of
Assertion.
Assertion is true but Reason is false. Threshold frequency is a
minimum frequency required for the emission of electrons
from the metal surface.
Both assertion and reason are true and reason is the correct
explanation of assertion.
n2h2
n2

 0 .529 Å. rn also increases
2
Z
4e mZ
indicating a greater separation between the orbit and the
nucleus.
(d) Both assertion and Reason are false. Only s -orbital is
spherically symmertrical. Shape of different d orbitals is as
below.
(c) Assertion is true but reason is false. Spin angular momentum
of the electron, a vector quantity, can have two orientations
(represented by + and – sign) relative to a chosen axis. These
two orientation are distinguished by the spin quantum number
1
1
m s equals to 
or  . These are called the two spin
2
2
Radius,
14.
15.
r
Structure of atom
states of the electron and are normaly represented by the two
arrows  (spin up) and  (spin down) respectively.
Both assertion and reason are false. Total number of orbitals
associated with Principal quantum number n  3 is 9. One
3 s orbital + three 3 p orbital + five 3 d orbitals. 
Therefore there are a total number of nine orbitals. Number of
orbitals in a shell equals to n 2 .
Assertion is true but reason is false. The order
1s  2s  2 p  3 s  3 p  3d  .... is true for the energy of
an electron in a hydrogen atom and is solely determined by
Principal quantum number. For multielectron system energy
also depends on azimuthal quantum number. The stability of
an electron in a multi electron atom is the net result of the
attraction between the electron and the uncleus and the
repulsion between the electron and the rest of the electron
present. Energies of different subshell (azimuthal quantum
number) present within the same principal shell are found to
be in order of s  p  d  f .
Assertion is false but reason is true. Splitting of the spectral
lines in the presence of a magnetic field is known as Zeeman
effect or in electric field it is known as stark effect. The
splitting of spectral lines is due to different orientations which
the orbitals can have in the presence of magnetic field.
Both assertion and reason are true and reason is the correct
explanation of assertion.
Assertion is false but reason is true. Atomic orbital is
designated by n, l and m l while state of an electron in an
16.
(d)
17.
(c)
18.
(e)
19.
(a)
20.
(e)
21.
(b) Both assertion and reason are true but reason is not the
correct explanation of assertion. The difference between the
energies of adjacent energy levels decreases as we move away
from the nucleus. Thus in H atom
E2  E1  E3  E2  E4  E3 ......
22.
(d) Both assertion and reason are false. Cathode rays are stream of
electrons. They are generated through gases at low pressure
and high voltage.
(d) Both assertion and reason are false. In case of isoelectronic, i.e.,
ions, having the same number of electrons and different
nuclear charge, the size decreases with increase in atomic
number.
atom is specified by four quantum numbrs n, l, m l and m s .
23.
Ion
Na+
Mg2+
Al3+
At. No.
11
12
13
No. of electrons
10
10
10
Ionic radii
0.95Å
0.65Å
0.50Å
85
86 Structure of atom
1.
The correct set of quantum numbers for the unpaired electron of
chlorine atom is
[IIT 1989; MP PET 2004]
n
2.
l
8.
1
0
(b) 2
1
1
(c) 3
1
1
(d) 3
0
0
The uncertainty in the position of an electron ( mass =
9.1  10 28 g) moving with a velocity of
accurate upto 0.001% will be
(Use
The orbital diagram in which the Aufbau's principle is violated is
2s
2px
2 py














(b) 
(c)
(d)
9.
10
kg
(b) 10 24 kg
(c)
10 26 kg
(d) 10 27 kg
10.
Which electronic level would allow the hydrogen atom to absorb a
photon but not to emit a photon
[IIT 1984; CPMT 1997]
5.
3s
(b)
(c)
2s
(d) 1s
6.
2p
11.
Which of the following is not correct for electron distribution in the
ground state
[AIIMS 1982]
4s
3d
(a)
Co (Ar)






(b)
Ni(Ar)






(c)
Cu (Ar)






(d)
Zn(Ar)






If electron, hydrogen, helium and neon nuclei are all moving with
the velocity of light, then the wavelengths associated with these
particles are in the order
[MP PET 1993]
(a) Electron > hydrogen > helium > neon
(b) Electron > helium > hydrogen > neon
(c) Electron < hydrogen < helium < neon
(d) Neon < hydrogen < helium < electron
7.
From the given sets of quantum numbers the one that is inconsistent
with the theory is
[IIT Screening 1994]
(a)
n  3; l  2; m  3; s  1 / 2
(b) n  4 ; l  3; m  3; s  1 / 2
where
[CBSE PMT 1995]
1.92cm
(b) 7.68 cm
(c)
5.76cm
(d)
3.84 cm
The orbital angular momentum of an electron in s orbital is
[IIT 1996; AIEEE 2003; MP PET 2004]
(a)
(a)
3.0  10 4 cm s 1
expression,
(a)
1 h
.
2 2
(a)

(c)
h
2
[MNR 1988; UPSEAT 1999, 2000, 02]
4.
h
in
the
uncertainty
4 [IIT 1988; AMU 1999]
h  6.626  10 27 erg  s )
2 pz
The mass of neutron is nearly
23
n  2; l  1; m  0l s  1 / 2
(d) n  4 ; l  3; m  2; s  1 / 2
m
(a) 2
(a)
3.
(c)
(b) Zero
(d)
2.
h
2
Values of the four quantum numbers for the last electron in the
atom are n  4, l  1, m  1 and s  1 / 2 . Atomic number of
the atom will be
(a) 22
(b) 32
(c) 33
(d) 36
The atomic weight of an element is 39. The number of neutrons in
its nucleus is one more than the number of protons. The number of
protons, neutrons and electrons respectively in its atom would be[MP PMT 1997
(a) 19, 20, 19
(b) 19, 19, 20
(c) 20, 19, 19
(d) 20, 19, 20
12.
The electrons identified by quantum numbers n and l (i)
n  4, l  1 (ii) n  4, l  0 (iii) n  3, l  2 (iv) n  3, l  1
can be placed in order of increasing energy from the lowest to
highest, as
[IIT 1999]
(a) (iv) < (ii) < (iii) <(i)
(b) (ii) < (iv) < (i) < (iii)
(c) (i) < (iii) < (ii) < (iv)
(d) (iii) < (i) < (iv) < (ii)
13.
Ground state electronic configuration of nitrogen atom can be
represented by
[IIT 1999]
(a)





(b)





(c)





Structure of atom 87
(d)





18.
14.
Which of the following statements (s) is (are) correct
[IIT 1998]
5
1
(a) The electronic configuration of Cr is [ Ar] 3d 4 s (Atomic
3+
19.
no. of Cr  24 )
(b) The magnetic quantum number may have a negative value
(c) In silver atom, 23 electrons have a spin of one type and 24 of
the opposite type (Atomic no. of Ag  47 )
(d) The oxidation state of nitrogen in HN 3 is 3
15.
The position of both an electron and a helium atom is known within
1.0nm and the momentum of the electron is known within
50  10 26 kg ms 1 . The minimum uncertainty
measurement of the momentum of the helium atom is
in
the
20.
[CBSE PMT 1998; AIIMS 2001]
16.
(a)
50kg ms 1
(b) 60 kg ms 1
(c)
80  10 26 kg ms 1
(d) 50  10 26 kg ms 1
(c)
17.
21.
p xy , d x 2 y 2
(b) d xy , d zx
p xy , d zx
(d) d z 2 , d x 2 y 2
22.
The number of atoms in 0.004 g of magnesium are
[AFMC 2000]
(a)
4  10
20
(b) 8  10
3+
(a)
10 33 metres
(b) 10 31 metres
(c)
10 16 metres
(d) 10 25 metres
Which of the following are isoelectronic and isostructural
NO 3 , CO 32  , ClO3 , SO 3
Which of the following pair of orbitals posses two nodal planes
(a)
(c) 10 20
(d) 6.02  10 20
Which of the following have the same number of unpaired electrons
in ‘d’ orbitals
[Roorkee 2000]
(a) Cr
(b) Mn
(c) Fe
(d) Co
The quantum numbers + 1/2 and – 1/2 for the electron spin
represent
[IIT Screening 2001]
(a) Rotation of the electron in clockwise and anticlockwise
direction respectively
(b) Rotation of the electron in anticlockwise and clockwise
direction respectively
(c) Magnetic moment of the electron pointing up and down
respectively
(d) Two quantum mechanical spin states which have no classical
analogue
The de-Broglie wavelength of a tennis ball of mass 60 g moving with
a velocity of 10 metres per second is approximately
[IIT Screening 2003]
[RPMT
2000]

2
SO 3 , NO 3
(a)
NO 3 , CO 3
(b)
(c)
ClO3 , CO 32 
(d) CO 32  , SO 3
The total number of electrons present in all the s-orbitals, all the porbitals and all the d-orbitals of cesium ion are respectively
(a) 8, 26, 10
(b) 10, 24, 20
(c) 8, 22, 24
(d) 12, 20, 22
20
(SET -2)
1.
(c) Electronic configuration of Cl is
3s 2
[ Ne ] 3 s 2 3 p 5 or [Ne ]
3p
So for the unpaired electron (3 p1z ) :
n  3, l  1, m  1, S  
3 p 2 x 3p 2 y 3p 1 z
1
2
88 Structure of atom
2.
(b) According to Aufbau principle the orbitals of lower energy (2s)
should be fully filled before the filling of orbital of higher
energy starts.
3.
4.
(d) Mass of neutron  1.675  10 27 kg .
(d) 1s-orbital is of lowest energy. Absorption of photon can raise
the electron in higher energy state but emission is not possible.
(c) The correct electronic configuration
5.
Cu 29  [ Ar], 4 s1
14.
HN 3 : 1  3 x  0  3 x  1 or x 
15.
3d 10
(a)
7.
(a) When l  2, m  3 .
8.
(a)
(d) The product of uncertainties in the position and the
momentum of a sub atomic particle  h/4 . Since x is
16.
(b) d xy and d zx has two modal planes.
17.
(c) No. of atoms in magnesium =
18.
(a,b,c) Cr , Mn and Fe3  have 5 unpaired electron in dorbitals.
0 .001
100
24 Cr
P  2.73  10 24
9.
 3d 5 4 s1  5
25
Mn  3d 5 4 s 2  5
26
Fe3   3d 5 4 s 0  5
(a,d) Both statement are correct.
x  1.92 cm.
20.
(a)
21.
(a) NO 3 and CO 32  consist of same electron and show same
isostructural.
22.
(b) (Cs 35 )  1s 2 , 2 s 2 , 2 p 6 ,3 s 2 ,3 p 6 ,3d 10 ,4 s 2
h
0.
2
10.
(d) Atomic number is 36 and element is Kr .
11.
(a)
h
6 .63  10 34

 10  33 m
mv 60  10  3  10
Cs   1s 2 , 2 s 2 ,2 p 6 , 3 s 2 ,3 p 6 ,3d 10 , 4 s 2 ,
39
K 19
, P  19 , E  19 , N  20
(i) 4 p (ii) 4 s (iii) 3d (iv) 3 p order

4 p 6 ,4 d 10 ,5 s 2 ,5 p 6 ,6 s1
4 p 6 ,4 d 10 , 5 s 2 ,5 p 6
of
increasing
energy
is
3 p  4 s  3d  4 p.
13.
20
19.
(b) For 2s orbital, l = 0; azimuthal quantum number is not show
angular momentum for the 2 s orbitals.
(a)
0 .004
 6 .023  10 23 =10
24
h
6 .626  10 27

Hence x 
p  4 2 .73  10  28  4  3 .14
Angular momentum  l(l  1)
12.
1
3
the particle i.e. 50  10 26 kg ms 1 (given).
p  m  v
p  9.1  10  28  3.0  10 4 
1
.
3
same for electron and helium so p must be same for both
1
  , m e  m H  m He  m Ne .
m
6.
(a,b,c) The oxidation state of nitrogen in HN 3 is 
Total no. of e  in s-orbitals  10
Total no. of e  in p-ortbitals  24
(a,d) According to Hund’s principle.
Total no. of e  in d-ortbitals  20 .
***
Chemical Bonding
89
Chapter
3
Chemical Bonding
Atoms of different elements excepting noble gases donot have
complete octet so they combine with other atoms to form chemical bond.
The force which holds the atoms or ions together within the molecule is
called a chemical bond and the process of their combination is called
Chemical Bonding. It depends on the valency of atoms.
Some other examples are: MgCl , CaCl , MgO, Na S, CaH , AlF , NaH,
KH, K 2O , KI, RbCl, NaBr, CaH etc.
2
2
2
2
3
2
(1) Conditions for formation of electrovalent bond
Cause and Modes of chemical combination
(i) The atom which changes into cation (+ ive ion) should possess 1,
2 or 3 valency electrons. The other atom which changes into anion (–ve
ion) should possess 5, 6 or 7 electrons in the valency shell.
Chemical bonding takes place due to acquire a state of minimum
energy and maximum stability and to convert atoms into molecule to
acquire stable configuration of the nearest noble gas. We divide atoms into
(ii) A high difference of electronegativity (about 2) of the two atoms
is necessary for the formation of an electrovalent bond. Electrovalent bond
is not possible between similar atoms.
three classes,
(1) Electropositive elements which give up one or more electrons
easily. They have low ionisation potentials.
(2) Electronegative elements, which can gain electrons. They have
higher value of electronegativity.
(3) Elements which have little tendency to lose or gain electrons.
Different types of bonds are formed from these types of atoms.
Atoms involved
Type
A+B
B+B
A+A
Electrovalent
Electrons deficient molecule or ion
(Lewis acid) and electrons rich
molecule or ion (Lewis base)
Coordinate
H and electronegative element (F,
N,O)
Hydrogen
Covalent
r   r  is internuclear distance.
The energy changes involved in the formation of ionic compounds
from their constituent elements can be studied with the help of a
thermochemical cycle called Born Haber cycle.
Na(s)
H


 Cl 

 
  Na 
 
+IE
1/2Hdiss.
sub
   
  Cl  
   

or
Cl (g)
Na  Cl 
– EA
– e–
Na  (g)

1
Cl 2 (g)
2
+
Na(g)
An electrovalent bond is formed when a metal atom transfers one or
more electrons to a non-metal atom.

(iv) Higher the lattice energy, greater will be the case of forming an ionic
compound. The amount of energy released when free ions combine together to
K
form one mole of a crystal is called lattice energy (U). Lattice energy    ;
r r
Metallic
Electrovalent bond
Na 
(iii) There must be overall decrease in energy i.e., energy must be
released. For this an atom should have low value of Ionisation potential and
the other atom should have high value of electron affinity.
+
+e
–
Cl  (g)
H
f
Na Cl (s)
+
–
(Crystal)
–U
(Lattice energy)
(Born Haber Cycle)
According to Hess's law of constant heat summation, heat of
formation of an ionic solid is net resultant of the above changes.
H f  H Subl. 
1
H diss.  IE  EA  U
2
90 Chemical Bonding
(2) Characteristics of electrovalent compounds
2
(i) Electrovalent compounds are generally crystalline is nature. The
constituent ions are arranged in a regular way in their lattice.
(ii) Electrovalent compounds possess high melting and boiling
points. Order of melting and boiling points in halides of sodium and oxides
of II group elements is as,
nd
NaF  NaCl  NaBr  NaI, MgO  CaO  BaO
(iii) Electrovalent compounds are hard and brittle in nature.
(iv) Electrovalent solids do not conduct electricity. While
electrovalent compounds in the molten state or in solution conduct
electricity.
(v) Electrovalent compounds are fairly soluble in polar solvents and
insoluble in non-polar solvents.
(vi) The electrovalent bonds are non-rigid and non-directional. Thus
these compound do not show space isomerism e.g. geometrical or optical
isomerism.
(vii) Electrovalent compounds furnish ions in solution. The chemical
reaction of these compounds are known as ionic reactions, which are fast.






K  Cl   Ag NO 3  Ag Cl   K NO 3
(Precipitate )
(viii) Electrovalent compounds show isomorphism.
(ix) Cooling curve of an ionic compound is not smooth, it has two
break points corresponding to time of solidification.
(x) Ionic compounds show variable electrovalency due to unstability
of core and inert pair effect.
Covalent bond
Covalent bond was first proposed by Lewis in 1916. The bond
formed between the two atoms by mutual sharing of electrons so as to
complete their octets or duplets (in case of elements having only one shell)
is called covalent bond or covalent linkage. A covalent bond between two
similar atoms is non-polar covalent bond while it is polar between two
different atom having different electronegativities. Covalent bond may be
single, double or a triple bond. We explain covalent bond formation by
Lewis octet rule.
Chlorine atom has seven electrons in the valency shell. In the
formation of chlorine molecule, each chlorine atom contributes one electron
and the pair of electrons is shared between two atoms. both the atoms
acquire stable configuration of argon.
**


*
 Cl   Cl *

**
( 2 , 8 , 7 ) ( 2, 8 , 7 )






 Cl * Cl 


( 2, 8 , 8 ) ( 2, 8 , 8 )
Some other examples are :
(ii) Diamond, Carborandum (SiC), Silica (SiO ), AlN etc. have giant
three dimensional network structures; therefore have exceptionally high
melting points otherwise these compounds have relatively low melting and
boiling points.
(iii) In general covalent substances are bad conductor of electricity.
Polar covalent compounds like HCl in solution conduct electricity. Graphite
can conduct electricity in solid state since electrons can pass from one layer
to the other.
(iv) These compounds are generally insoluble in polar solvent like
water but soluble in non-polar solvents like benzene etc. some covalent
compounds like alcohol, dissolve in water due to hydrogen bonding.
(v) The covalent bond is rigid and directional. These compounds,
thus show isomerism (structural and space).
(vi) Covalent substances show molecular reactions. The reaction
rates are usually low.
(vii) The number of electrons contributed by an atom of the element
for sharing with other atoms is called covalency of the element. Covalency
= 8 – [Number of the group to which element belongs]. The variable
or Cl  Cl
H 2 S , NH 3 , HCN , PCl3 , PH3,
C2 H 2 , H 2 , C2 H 4 , SnCl 4 , FeCl3 , BH3 , graphite, BeCl 2 etc.
(1) Conditions for formation of covalent bond
(i) The combining atoms should be short by 1, 2 or 3 electrons in
the valency shell in comparison to stable noble gas configuration.
(ii) Electronegativity difference between the two atoms should be
zero or very small.
(iii) The approach of the atoms towards one another should be
accompanied by decrease of energy.
(2) Characteristics of covalent compounds
(i) These exist as gases or liquids under the normal conditions of
temperature and pressure. Some covalent compounds exist as soft solids.
covalency of an element is equal to the total number of unpaired electrons
in s, p and d-orbitals of its valency shell.
The element such as P, S, Cl, Br, I have vacant d-orbitals in their
valency shell. These elements show variable covalency by increasing the
number of unpaired electrons under excited conditions. The electrons from
paired orbitals get excited to vacant d-orbitals of the same shell.
Four elements, H, N, O and F do not possess d-orbitals in their
valency shell. Thus, such an excitation is not possible and variable valency is
not shown by these elements. This is reason that NCl exists while NCl does
not.
3
5
(3) The Lewis theory : The tendency of atoms to achieve eight
electrons in their outermost shell is known as lewis octet rule.
Lewis symbol for the representative elements are given in the
following table,
Group
Lewis
symbol
1
IA
2
IIA
13
IIIA
X
X

X
14
IVA
15
VA
16
VIA
17
VIIA

X


 X


 X


X 
 
(4) Failure of octet rule : There are several stable molecules known
in which the octet rule is violated i.e., atoms in these molecules have
number of electrons in the valency shell either short of octet or more than
octet.
BeF2 , BF3 , AlH3 are electron- deficients (Octet incomplete) hence
are Lewis acid.
In PCl5 , P has 10 electrons in valency shell while in SF6 , S has 12
electrons in valence shell. Sugden introduced singlet linkage in which one
electron is donated (Instead of one pair of electrons) to the electron
deficient atom so that octet rule is not violated. This singlet is represented
as (⇁). Thus, PCl5 and SF6 have structures as,
Cl
Cl
P
Cl
Cl
Cl
F
F
S
F
F
F
F
(5) Construction of structures for molecules and poly atomic ions :
The following method is applicable to species in which the octet rule is not
violated.
(i) Determine the total number of valence electrons in all the atoms
present, including the net charge on the species (n ).
(ii) Determine n = [2 × (number of H atoms) + 8 × (number of
other atoms)].
1
2
Chemical Bonding
(iii) Determine the number of bonding electrons, n , which equals n
– n . No. of bonds equals n /2.
(iv) Determine the number of non-bonding electrons, n , which
equals n – n . No. of lone pairs equals n /2.
(v) Knowing the central atom (you’ll need to know some chemistry
here, math will not help!), arrange and distribute other atoms and n /2
bonds. Then complete octets using n /2 lone pairs.
(vi) Determine the ‘formal charge’ on each atom.
3
1
91
Examples : CO, N O, H O , N O , N O , N O , HNO , NO 3 , SO , SO ,
2
2
H SO ,
2
2
2
2
3
2
4
2
5
3
SO 42  , SO 22  ,
3
4
2
3
H 3 PO4 , H 4 P2O7 ,
4
1
3
4
3
4
(vii) Formal Charge = [valence electrons in atom) – (no. of bonds) –
(no. of unshared electrons)]
(viii) Other aspects like resonance etc. can now be incorporated.
Illustrative examples
(i) CO 32  ; n1  4  (6  3)  2  24 [2 added for net charge]
n 2 = (2 × 0) + (8 × 4) = 32 (no. H atom, 4 other atoms (1’C’ and 3
‘O’)
n 3 = 32 – 24 = 8, hence 8/2 = 4 bonds
n 4 = 24 – 8 = 16, hence 8 lone pairs.
Since carbon is the central atom, 3 oxygen atoms are to be arranged
around it, thus,
H 3 PO3 , Al2Cl6 (Anhydrous), O3 , SO 2Cl2 , SOCl 2 , HIO3 , HClO4 ,
HClO3 , CH 3 NC , N 2 H 5 , CH 3 NO 2 , NH 4 , [Cu(NH 3 )4 ]2 etc.
Characteristics of co-ordinate covalent compound
(1) Their melting and boiling points are higher than purely covalent
compounds and lower than purely ionic compounds.
(2) These are sparingly soluble in polar solvent like water but readily
soluble in non-polar solvents.
(3) Like covalent compounds, these are also bad conductors of
electricity. Their solutions or fused masses do not allow the passage to
electricity.
(4) The bond is rigid and directional. Thus, coordinate compounds
show isomerism.
Dipole moment
“The product of magnitude of negative or positive charge (q) and
the distance (d) between the centres of positive and negative charges is
called dipole moment”.
 = Electric charge  bond length
As q is in the order of 10 esu and d is in the order of 10 cm,  is in
the order of 10 esu cm. Dipole moment is measured in “Debye” (D) unit.
1D  10 18 esu cm = 3.33  10 30 coulomb metre (In S.I. unit).
–10
–8
–18
O
|
O  C  O , but total bonds are equal to 4.
O
|
Hence, we get O  C  O . Now, arrange lone pairs to complete
..
.O :
|
.
..
Dipole moment is indicated by an arrow having a symbol (
)
pointing towards the negative end. Dipole moment has both magnitude and
direction and therefore it is a vector quantity.
Symmetrical polyatomic molecules are not polar so they do not have
any value of dipole moment.
H
F
octet : O  C  O :
..
(ii)
O
CO 2 ; n = 4 + (6 × 2 ) = 16
C
O
C
B
1
n = (2 × 0) + (8 × 3) = 24
n = 24 – 16 = 8, hence 4 bonds
n = 16 – 8 = 8, hence 4 lone-pairs
Since C is the central atom, the two oxygen atoms are around to be
arranged it thus the structure would be; O – C – O, but total no. of bonds
F
H
F
H
H
2
3
4
 = 0 due to symmetry
Unsymmetrical polyatomic molecules always have net value of dipole
moment, thus such molecules are polar in nature. H O, CH Cl, NH , etc are
polar molecules as they have some positive values of dipole moments
.
Cl
2
3
3
=4
Thus, O = C = O. After arrangement of lone pairs to complete
..
..
..
octets, we get, : O  C  O : and thus final structure is : O  C  O :
This is a special type of covalent bond where the shared pair of
electrons are contributed by one species only but shared by both. The atom
which contributes the electrons is called the donor (Lewis base) while the
other which only shares the electron pair is known as acceptor (Lewis acid).
This bond is usually represented by an arrow (  ) pointing from donor to
the acceptor atom.
BF molecule, boron is short of two electrons. So to complete its
octet, it shares the lone pair of nitrogen in ammonia forming a dative bond.
3
H
 


F 
 
H * N **  B
 
H


 
F 

H



 F

 


 H * N


H


 
H
Water
F 
B


 
F
 

H
|
F
General
formula
|
|
H
F
H
H
Molecular geometry
AX
Linear
AX 2
AX 3
|
F   H  N  B  F



C
H
H
H
Ammonia
Methyl chloride
 = 1.46D
 = 1.86D
  0 due to unsymmetry
(1) Dipole moment is an important factor in determining the
geometry of molecules.
Table : 3.1 Molecular geometry and dipole moment

 
N
H
H
 = 1.84D
Co-ordinate covalent or Dative bond

O
..
Dipole
moment
Example
HF, HCl
Linear
Bent or V-shape
May
be
non zero
Zero
Non zero
Triangular planar
Pyramidal
T-shape
Zero
Non zero
Non zero
BF3
Tetrahedral
Zero
CO 2 ,CS 2
H 2O, NO 2
NH 3 , PCl3
ClF3
Formation of a co-ordinate
bond between NH3 and BF3
AX 4
CH 4 ,CCl 4
92 Chemical Bonding
Square planar
See saw
Zero
Non zero
Trigonal bipyramidal
Square pyramidal
Zero
Non zero
PCl5
AX 6
Octahedral
Distorted octahedral
Zero
Non zero
SF6
AX 7
Pentagonal bipyramidal
Zero
IF7
AX 5
XeF4
SF4 ,TeCl 4
BrCl5
XeF6
(2) Every ionic compound having some percentage of covalent
character according to Fajan's rule. The percentage of ionic character in
compound having some covalent character can be calculated by the
following equation.
The % ionic character 
Observed 
 100 .
Theoretica l 
(3) The trans isomer usually possesses either zero dipole moment or
very low value in comparison to cis–form
H  C  Cl
||
H  C  Cl
H  C  Cl
||
Cl  C  H
(8) Between two sub shells of same energy level, the sub shell more
directionally concentrated shows more overlapping. Bond energy :
2s  2s < 2s  2 p < 2 p  2 p
(9) s -orbitals are spherically symmetrical and thus show only head
on overlapping. On the other hand, p -orbitals are directionally
concentrated and thus show either head on overlapping or lateral
overlapping.Overlapping of different type gives sigma () and pi () bond.
Sigma () bond
Pi () bond
It results from the end to end
overlapping of two s-orbitals or
two p-orbitals or one s and one porbital.
Stronger
Bond energy 80 kcals
More stable
Less reactive
Can exist independently
It result from the sidewise (lateral)
overlapping of two p-orbitals.
The electron cloud is symmetrical
about the internuclear axis.
Fajan’s rule
The magnitude of polarization or increased covalent character
depends upon a number of factors. These factors are,
(1) Small size of cation : Smaller size of cation greater is its
polarizing power i.e. greater will be the covalent nature of the bond.
(2) Large size of anion : Larger the size of anion greater is its
polarizing power i.e. greater will be the covalent nature of the bond.
(3) Large charge on either of the two ions : As the charge on the
ion increases, the electrostatic attraction of the cation for the outer
electrons of the anion also increases with the result its ability for forming
the covalent bond increases.
(4) Electronic configuration of the cation : For the two ions of the
same size and charge, one with a pseudo noble gas configuration (i.e. 18
electrons in the outermost shell) will be more polarizing than a cation with
noble gas configuration (i.e., 8 electron in outer most shell).
Valence bond theory or VBT
It was developed by Heitler and London in 1927 and modified by
Pauling and Slater in 1931.
(1) To form a covalent bond, two atoms must come close to each
other so that orbitals of one overlaps with the other.
(2) Orbitals having unpaired electrons of anti spin overlaps with
each other.
(3) After overlapping a new localized bond orbital is formed which
has maximum probability of finding electrons.
(4) Covalent bond is formed due to electrostatic attraction between
radii and the accumulated electrons cloud and by attraction between spins
of anti spin electrons.
(5) Greater is the overlapping, lesser will be the bond length, more
will be attraction and more will be bond energy and the stability of bond
will also be high.
(6) The extent of overlapping depends upon: Nature of orbitals
involved in overlapping, and nature of overlapping.
(7) More closer the valence shells are to the nucleus, more will be
the overlapping and the bond energy will also be high.
Less strong
Bond energy 65 kcals
Less stable
More reactive
Always exist along with a -bond
The electron cloud is above and
below the plane of internuclear
axis.
Hybridization
The concept of hybridization was introduced by Pauling and Slater.
Hybridization is defined as the intermixing of dissimilar orbitals of the same
atom but having slightly different energies to form same number of new
orbitals of equal energies and identical shapes. The new orbitals so formed
are known as hybrid orbitals.
Characteristics of hybridization
(1) Only orbitals of almost similar energies and belonging to the
same atom or ion undergoes hybridization.
(2) Hybridization takes place only in orbitals, electrons are not
involved in it.
(3) The number of hybrid orbitals produced is equal to the number
of pure orbitals, mixed during hybridization.
(4) In the excited state, the number of unpaired electrons must
correspond to the oxidation state of the central atom in the molecule.
(5) Both half filled orbitals or fully filled orbitals of equivalent
energy can involve in hybridization.
(6) Hybrid orbitals form only sigma bonds.
(7) Orbitals involved in  bond formation do not participate in
hybridization.
(8) Hybridization never takes place in an isolated atom but it occurs
only at the time of bond formation.
(9) The hybrid orbitals are distributed in space as apart as possible
resulting in a definite geometry of molecule.
(10) Hybridized orbitals provide efficient overlapping than
overlapping by pure s, p and d-orbitals.
(11) Hybridized orbitals possess lower energy.
How to determine type of hybridization : The structure of any
molecule can be predicted on the basis of hybridization which in turn can
be known by the following general formulation,
H 
1
(V  M  C  A)
2
Chemical Bonding
Where H = Number of orbitals involved in hybridization viz. 2, 3, 4,
5, 6 and 7, hence nature of hybridization will be sp, sp , sp , sp d, sp d , sp d
respectively.
2
3
3
3
2
3
3
V = Number electrons in valence shell of the central atom,
M = Number of monovalent atom
C = Charge on cation,
Bond order 
A = Charge on anion
The phenomenon of resonance was put forward by Heisenberg to
explain the properties of certain molecules.
In case of certain molecules, a single Lewis structure cannot explain
all the properties of the molecule. The molecule is then supposed to have
many structures, each of which can explain most of the properties of the
molecule but none can explain all the properties of the molecule. The actual
structure is in between of all these contributing structures and is called
resonance hybrid and the different individual structures are called
resonating structures or canonical forms. This phenomenon is called
resonance.
To illustrate this, consider a molecule of ozone O3 . Its structure
can be written as
. .
..
..
O
O
.O.
. ..
..
.
.
.
. O.
.O
O.
O
.O
. .
.. .
.O. .
..
. . (c ) . . .
(a )
(b )
As a resonance hybrid of above two structures (a) and (b. For
simplicity, ozone may be represented by structure (c), which shows the
resonance hybrid having equal bonds between single and double.
Resonance is shown by benzene, toluene, O , allenes
(>C = C =
3
C<), CO, CO , CO 3 , SO , NO, NO while it is not shown by H O , H O, NH ,
CH , SiO .
4
O
||
C


/ \
O O
O
|

C
/ \\

O
O
2 11
 1.33
3
Bond characteristics
Resonance
2
O
|
C
// \
O
O
93
3
2
2
2
2
3
(1) Bond length
“The average distance between the centre of the nuclei of the two
bonded atoms is called bond length”.
It is expressed in terms of Angstrom (1 Å = 10 10 m) or picometer
(1pm = 10 12 m).
In an ionic compound, the bond length is the sum of their ionic
radii ( d  r  r ) and in a covalent compound, it is the sum of their
covalent radii (e.g., for HCl, d  rH  rCl ).
Factors affecting bond length
(i) The bond length increases with increase in the size of the atoms.
For example, bond length of H  X
are in the order,
HI  HBr  HCl  HF .
(ii) The bond length decreases with the multiplicity of the bond.
Thus, bond length of carbon–carbon bonds are in the order,
C CC CC –C.
(iii) As an s-orbital is smaller in size, greater the s-character shorter
is the hybrid orbital and hence shorter is the bond length.
For example, sp 3 C – H  sp 2 C – H  sp C – H
2
As a result of resonance, the bond lengths of single and double bond
in a molecule become equal e.g. O–O bond lengths in ozone or C–O bond
lengths in CO 32 – ion.
(2) Bond energy
The resonance hybrid has lower energy and hence greater stability
than any of the contributing structures.
Greater is the number of canonical forms especially with nearly
same energy, greater is the stability of the molecule.
Difference between the energy of resonance hybrid and that of the
most stable of the resonating structures (having least energy) is called
resonance energy. Thus,
Resonance energy = Energy of resonance hybrid – Energy of the
most stable of resonating structure.
In the case of molecules or ions having resonance, the bond order
changes and is calculated as follows,
Bond order 
Total no. of bonds between two atoms in all the structures
In benzene
Total no. of resonating structures


double bond  singlebond 2  1
Bond order 

 1.5
2
2
In carbonate ion
(iv) Polar bond length is usually smaller than the theoretical nonpolar bond length.
“The amount of energy required to break one mole of bonds of a
particular type so as to separate them into gaseous atoms is called bond
dissociation energy or simply bond energy”. Greater is the bond energy,
stronger is the bond. Bond energy is usually expressed in kJ mol –1 .
Factors affecting bond energy
(i) Greater the size of the atom, greater is the bond length and less
is the bond dissociation energy i.e. less is the bond strength.
(ii) For the bond between the two similar atoms, greater is the
multiplicity of the bond, greater is the bond dissociation energy.
(iii) Greater the number of lone pairs of electrons present on the
bonded atoms, greater is the repulsion between the atoms and hence less is
the bond dissociation energy.
(iv) The bond energy increases as the hybrid orbitals have greater
amount of s orbital contribution. Thus, bond energy decreases in the
following order, sp  sp 2  sp 3
(v) Greater the electronegativity difference, greater is the bond
polarity and hence greater will be the bond strength i.e., bond energy,
H  F  H  Cl  H  Br  H  I ,
94 Chemical Bonding
(vi) Among halogens Cl – Cl > F – F > Br – Br > I – I, (Decreasing
order of bond energy) Resonance increases bond energy.
In case the central atom remains the same, bond angle increases
with the decrease in electronegativity of the surrounding atom.
(3) Bond angle
In case of molecules made up of three or more atoms, the average
angle between the bonded orbitals (i.e., between the two covalent bonds) is
known as bond angle .
Factors affecting bond angle
(i) Repulsion between atoms or groups attached to the central atom
may increase or decrease the bond angle.
(ii) In hybridisation as the s character of the s hybrid bond
increases, the bond angle increases.
Bond type
sp
sp
Bond angle
109º28
120°
3
PCl3 PBr3 PI3 , AsCl3 AsBr3 AsI3
100 o 101.5 o 102 o 98.4 o 100.5 o 101o
Valence shell electron pair repulsion theory (VSEPR )
The basic concept of the theory was suggested by Sidgwick and
Powell (1940). It provides useful idea for predicting shapes and geometries
of molecules. The concept tells that, the arrangement of bonds around the
central atom depends upon the repulsion’s operating between electron
pairs(bonded or non bonded) around the central atom. Gillespie and
Nyholm developed this concept as VSEPR theory.
The main postulates of VSEPR theory are
sp
2
Bond angle
(1) For polyatomic molecules containing 3 or more atoms, one of the
180°
atoms is called the central atom to which other atoms are linked.
(iii) By increasing lone pair of electron, bond angle decreases
approximately by 2.5%.
CH
Bond angle
NH
4
109º
107
HO
3
2
105
o
o
(iv) If the electronegativity of the central atom decreases, bond angle
decreases.
Bond angle
H 2O
104.5 o
 H 2S
92.2 o
 H 2 Se
91.2 o
 H 2 Te
89.5 o
(2) The geometry of a molecule depends upon the total number of
valence shell electron pairs (bonded or not bonded) present around the
central atom and their repulsion due to relative sizes and shapes.
(3) If the central atom is surrounded by bond pairs only. It gives the
symmetrical shape to the molecule.
(4) If the central atom is surrounded by lone pairs (lp) as well as
bond pairs (bp) of e  then the molecule has a distorted geometry.
(5) The relative order of repulsion between electron pairs is as
follows : lp – lp > lp – bp > bp – bp.
A lone pair is concentrated around the central atom while a bond
pair is pulled out between two bonded atoms. As such repulsion becomes
greater when a lone pair is involved.
Table : 3.2 Geometry of Molecules/Ions having bond pair as well as lone pair of electrons
Type of
mole-cule
No. of bond
pairs of
electron
No. of lone
pairs of
electrons
Hybridization
Bond angle
Expected
geometry
AX 3
2
1
sp 2
< 120
Trigonal planar
AX 4
2
2
sp 3
< 109 28
Tetrahedral
AX 4
3
1
sp 3
< 109 28
AX 5
4
1
sp 3 d
< 109 28
AX 5
3
2
sp 3 d
90
AX 5
2
3
sp 3 d
180
AX 6
5
1
sp 3 d 2
< 90
2
3
2
3
3
AX 6
AX
7
4
6
1
sp d
sp d
Actual geometry
Examples
V-shape, Bent,
Angular
V-shape,
Angular
H O, H S, SCl , OF , NH , ClO
Tetrahedral
Pyramidal
NH , NF , PCl, PH, AsH, ClO , H O
Trigonal
bipyramidal
Trigonal
bipyramidal
Trigonal
bipyramidal
Irregular
tetrahedron
SF , SCl , TeCl
T-shaped
ICl , IF , ClF
Linear
XeF , I , ICl
Octahedral
Square
pyramidal
ICl , BrF , IF
–
Octahedral
Square planar
XeF , ICl
–
Pentagonal
pyramidal
Distorted
octahedral
XeF
o
o
o
o
o
o
o
Molecular orbital theory or MOT
Molecular orbital theory was given by Hund and Mulliken in 1932.
The main ideas of this theory are,
(1) When two atomic orbitals combine or overlap, they lose their
identity and form new orbitals. The new orbitals thus formed are called
molecular orbitals.
SO , SnCl , NO
2
2
–
2
–
2
2
2
2
2
–
2
–
3
3
3
3
4
3
4
3
+
3
4
3
3
–
2
3
3
–
2
5
5
4
5
–
4
6
(2) Molecular orbitals are the energy states of a molecule in which
the electrons of the molecule are filled just as atomic orbitals are the energy
states of an atom in which the electrons of the atom are filled.
(3) In terms of probability distribution, a molecular orbital gives
the electron probability distribution around a group of nuclei just as an
atomic orbital gives the electron probability distribution around the single
nucleus.
(4) Only those atomic orbitals can combine to form molecular
orbitals which have comparable energies and proper orientation.
Chemical Bonding
(5) The number of molecular orbitals formed is equal to the number
of combining atomic orbitals.
(6) When two atomic orbitals combine, they form two new orbitals
called bonding molecular orbital and antibonding molecular orbital.
(7) The bonding molecular orbital has lower energy and hence
greater stability than the corresponding antibonding molecular orbital.
(8) The bonding molecular orbitals are represented by  ,  etc,
whereas the corresponding antibonding molecular orbitals are represented
Increasing energy (for electrons > 14)
(b)  1s,  1s,  2 s,  * 2 s,  2 p y ,  2 p x ,
*
 * 2pz
Increasing energy (for electrons  14)
(12) Number of bonds between two atoms is called
order and is given by
(11) Electrons are filled in the increasing energy of the MO which is
in order
(a)  1s,  *1s,  2 s,  * 2 s,  2 p x ,  2 p y  * 2 py ,  * 2 p x
 2 pz ,  * 2 pz
where N B  number of electrons in bonding MO.
N A  number of electrons in antibonding MO.
For a stable molecule/ion, N B  N A
(13) Bond order  Stability of molecule  Dissociation energy 
1
.
Bond length
(14) If all the electrons in a molecule are paired then the substance
is a diamagnetic on the other hand if there are unpaired electrons in the
molecule, then the substance is paramagnetic. More the number of unpaired
electron in the molecule greater is the paramagnetism
of the substance.
*
 (2pz)
*(2pz)
2p
2p
*(2py)
2p
*(2px)
bond
 N  NA 
Bond order   B

2


by  * ,  * etc.
(9) The shapes of the molecular orbitals formed depend upon the
type of combining atomic orbitals.
(10) The filling of molecular orbitals in a molecule takes place in
accordance with Aufbau principle, Pauli's exclusion principle and Hund's
rule. The general order of increasing energy among the molecular orbitals
formed by the elements of second period and hydrogen and their general
electronic configurations are given below.
95
*(2py)
2p
*(2px)
 (2py)  (2px)
 (2pz)
2s
2s
*(2s)
Increasing energy
Increasing energy
 (2pz)
 (2py)  (2px)
2s
 (2s)
 (2s)
1s
1s
1s
*(1s)
Molecular
orbitals
1s
*(1s)
 (1s)
 (1s)
Atomic
orbitals
2s
*(2s)
Atomic
orbitals
Molecular orbital energy level diagram
(Applicable for elements with Z > 7)
Hydrogen bonding
In 1920, Latimer and Rodebush introduced the idea of “hydrogen
bond”.
For the formation of H-bonding the molecule should contain an
atom of high electronegativity such as F, O or N bonded to hydrogen atom
and the size of the electronegative atom should be quite small.
Types of hydrogen bonding
(1) Intermolecular hydrogen bond : Intermolecular hydrogen bond is
formed between two different molecules of the same or different
substances.
(i) Hydrogen bond between the molecules of hydrogen fluoride.
Atomic
orbitals
Molecular
orbitals
Atomic
orbitals
Molecular orbital energy level diagram obtained by the overlap of 2s and 2pz
atomic orbitals after mixing
(Applicable for elements with Z < 7)
(ii) Hydrogen bond in alcohol and water molecules
(2) Intramolecular hydrogen bond (Chelation)
Intramolecular hydrogen bond is formed between the hydrogen atom
and the highly electronegative atom (F, O or N) present in the same
molecule. Intramolecular hydrogen bond results in the cyclisation of the
molecules and prevents their association. Consequently, the effect of
intramolecular hydrogen bond on the physical properties is negligible.
For example : Intramolecular hydrogen bonds are present in
molecules such as o-nitrophenol, o-nitrobenzoic acid, etc.
O
H
|
|
C
C
O
O
H
O
H
O
O
H
N
N
O
|
|
Salicyldehyde
O
O
(o-hydroxy benzaldehyde)
Ortho nitrophenol
Ortho nitrobenzoic acid
96 Chemical Bonding
The extent of both intramolecular and intermolecular hydrogen
bonding depends on temperature.
Effects of hydrogen bonding
Hydrogen bond helps in explaining the abnormal physical properties
in several cases. Some of the properties affected by H-bond are given below,
(1) Dissociation : In aqueous solution, hydrogen fluoride dissociates
and gives the difluoride ion (HF2 ) instead of fluoride ion (F  ) . This is
due to H-bonding in HF. This explains the existence of KHF2 . H-bond
formed is usually longer than the covalent bond present in the molecule
(e.g. in H O, O–H
bond = 0.99 Å but H-bond = 1.77 Å).
2
(2) Association : The molecules of carboxylic acids exist as dimers
because of the hydrogen bonding. The molecular masses of such compounds
are found to be double than those calculated from their simple formulae.
For example, molecular mass of acetic acid is found to be 120.
(5) As the compounds involving hydrogen bonding between
different molecules (intermolecular hydrogen bonding) have higher boiling
points, so they are less volatile.
(6) The substances which contain hydrogen bonding have higher
viscosity and high surface tension.
(7) Explanation of lower density of ice than water and maximum
density of water at 277K : In case of solid ice, the hydrogen bonding gives
rise to a cage like structure of water molecules as shown in following figure.
As a matter of fact, each water molecule is linked tetrahedrally to four other
water molecules. Due to this structure ice has lower density than water at
273K. That is why ice floats on water. On heating, the hydrogen bonds
start collapsing, obviously the molecules are not so closely packed as they
are in the liquid state and thus the molecules start coming together
resulting in the decrease of volume and hence increase of density. This goes
on upto 277K. After 277 K, the increase in volume due to expansion of the
liquid water becomes much more than the decrease in volume due to
(3) High melting and boiling point : The compounds having
hydrogen bonding show abnormally high melting and boiling points.
breaking of
H-bonds. Thus, after 277 K , there is net increase of
volume on heating which means decrease in density. Hence density of water
The high melting points and boiling points of the compounds
(H 2 O, HF and NH 3 ) containing hydrogen bonds is due to the fact that
is maximum 277 K .
(4) Solubility : The compound which can form hydrogen bonds with
the covalent molecules are soluble in such solvents. For example, lower
alcohols are soluble in water because of the hydrogen bonding which can
take place between water and alcohol molecules as shown below,




H
0.90 Å
(99 pm)
some extra energy is needed to break these bonds.

H
H
H
The intermolecular hydrogen bonding increases solubility of the
compound in water while, the intramolecular hydrogen bonding decreases.
Vacant spaces
O
O

C2 H 5
H
H
H
H
H
O
H  O .......... ...... H  O .......... ..... H  O
C2 H 5
1.77 Å
(177 pm)
O
H
O
H
H
O
H
H
O
H
H
H
O
O
N
|
O
H – O …… H – O – H
H
O
o- Nitrophenol
Due to chelation, – OH group is not
available to form hydrogen bond with
water hence it is sparingly soluble in
water.
H
O
H
Cage like structure of H2O in the ice
ON=O
p- Nitrophenol
– OH group available to form
hydrogen bond with water, hence it
is completely soluble in water.
98 Chemical Bonding
Electrovalent bonding
1.
Which forms a crystal of NaCl
[CPMT 1972; NCERT 1976; DPMT 1996]
 A chemical bond is expected to be formed when the energy of the
2.
aggregate formed is about 40 kJ mole lower than the separate
particles.
–1
 Formation of a chemical bond is always an exothermic process.
 Lattice energies of bi-bivalent solids > bi-univalent solids > uniunivalent
solids.
For
example,
lattice
energy
3.
[CPMT 1976; BHU 1998]
of
(a)
Mg 2  O 2  (3932 kJ mole 1 )  Ca 2  (F  )2 (2581 kJ mole 1 ) 


1
4.
Li F (1034 kJ mole ) .
elements of the middle columns of the periodic table.
is more than that of Fe
5.
. Similarly SnCl 4 is more
covalent than SnCl 2 .
 Boron forms the maximum number of electron deficient
2
 
2 p , 3 s ; R 1s , 2 s
6
1
2
2
6.
[CPMT 1978, 81; MNR 1979]
 Roughly each lone pair decreases the bond angle by 2.5°.
 Greater the number of the lone pairs at the two bonding atoms,
7.
 The actual number of s- and p-electrons present in the outermost
shell of the element is called maximum covalency of that atom.
8.
 The hydrogen bonds are tetrahedral in their directions and not
planar.
 The hydrogen bond is stronger in HF and persists even in vapour
9.
state. Such bonds account for the fact that gaseous hydrogen
fluoride is largely polymerised into the molecular species
H 2 F2 , H 3 F3 , H 4 F4 , H 5 F5 and H 6 F6 .
 Hydrogen bonding is strongest when the bonded structure is
(a) Electrically charged molecules
(b) Neutral molecules
(c) Neutral atoms
(d) Electrically charged atoms or group of atoms
Electrovalent bond formation depends on
(a) Ionization energy
(b) Electron affinity
(c) Lattice energy
(d) All the three above
In the following which substance will have highest boiling point [NCERT 1973; M
(a) He
(b) CsF
(c) NH 3
(d) CHCl 3
An atom of sodium loses one electron and chlorine atom accepts one
electron. This result the formation of sodium chloride molecule. This
type of molecule will be
[MP PMT 1987]
(a) Coordinate
(c) Electrovalent
10.
stabilised by resonance.
 Critical temperature of water is higher than that of O 2 because
H 2 O molecule has dipole moment.
2
(c) P2 L, RL, PQ and RQ2 (d) LP , R2 L, P2Q and RQ
Electrovalent compound's
[MP PMT 1984]
(a) Melting points are low
(b) Boiling points are low
(c) Conduct current in fused state
(d) Insoluble in polar solvent
A electrovalent compound is made up of
compounds than any other elements in the periodic table.
greater is the repulsion between them and weaker is the bond.

2 p , 3s 
6
The formulae of ionic compounds that can be formed between these
elements are
[NCERT 1983]
(a) L2 P, RL, PQ and R2 Q (b) LP , RL, PQ and RQ
 As a general rule, atomic crystals are formed by the lighter
Fe
(c) BaCl2
(d) CaCl 2
The electronic configuration of four elements L, P, Q and R are
given in brackets
2
 Ionic solids have negative vapour pressure.
2
KCl
L 1s 2 , 2 s 2 2 p 4 ; Q 1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 5
attraction increases and hence stability increases.
3
(b)
AgCl

P 1s , 2 s
 When co-ordination number increases, the coulombic forces of
 FeCl3 is more covalent than FeCl2 because polarising power of
(a) NaCl molecules
(b) Na  and Cl  ions
(c) Na and Cl atoms
(d) None of the above
When sodium and chlorine reacts then
[NCERT 1973]
(a) Energy is released and ionic bond is formed
(b) Energy is released and a covalent bond is formed
(c) Energy is absorbed and ionic bond is formed
(d) Energy is absorbed and covalent bond is formed
Which one is least ionic in the following compounds
11.
12.
(b) Covalent
(d) Matallic bond
Formula of a metallic oxide is MO. The formula of its phosphate
will be
[CPMT 1986, 93]
(a)
M 2 PO4 2
(b)
M PO4 
(c)
M 2 PO4
(d)
M 3 PO4 2
From the following which group of elements easily forms cation
(a)
F, Cl, Br
(b)
Li, Na, K
(c)
O, S , Se
(d)
N , P, As
Which type of compounds show high melting and boiling points
(a) Electrovalent compounds
(b) Covalent compounds
Chemical Bonding
13.
(c) Coordinate compounds
(d) All the three types of compounds have equal melting and
boiling points
Lattice energy of an ionic compound depends upon
14.
(a) Charge on the ion only
(b) Size of the ion only
(c) Packing of ions only
(d) Charge on the ion and size of the ion
In the given bonds which one is most ionic
23.
[AIEEE 2005]
24.
15.
16.
17.
Cs  Cl
(b)
Al  Cl
(c) C  Cl
(d) H  Cl
Element x is strongly electropositive and y is strongly
electronegative. Both element are univalent, the compounds formed
from their combination will be
[IIT 1980]
(a)
x y 
(b)
x  y
(c)
x y
(d)
x y
In the transition of Zn atoms to Zn   ions there is a decrease in
the
[CPMT 1972]
(a) Number of valency electrons
(b) Atomic weight
(c) Atomic number
(d) Equivalent weight
Phosphate of a metal M has the formula M 3 PO4 2 . The formula
for its sulphate would be
[CPMT 1973; MP PMT 1996]
[EAMCET 1980]
(a)
25.
26.
In the formation of NaCl from Na and Cl
[CPMT 1985]
(a) Sodium and chlorine both give electrons
(b) Sodium and chlorine both accept electrons
(c) Sodium loses electron and chlorine accepts electron
(d) Sodium accepts electron and chlorine loses electron
Which of the following is an electrovalent linkage
(a)
MSO 4
(b)
M SO 4 2
(c)
M 2 SO 4 3
(d)
M 3 SO 4 2
The molecular formula of chloride of a metal M is MCl 3 . The
formula of its carbonate would be [CPMT 1987]
(a)
MCO 3
(b)
M 2 CO 3 3
(c)
M 2 CO 3
(d)
M CO 3 2
Sodium chloride easily dissolves in water. This is because
[NCERT 1972; BHU 1973]
(a)
(b)
(c)
(d)
27.
It is a covalent compound
Salt reacts with water
It is a white substance
Its ions are easily solvated
When NaCl is dissolved in water the sodium ion becomes
[CPMT 1974; DPMT 1984, 91; AFMC 1988]
18.
19.
(a)
CH 4
(b)
MgCl2
(c)
SiCl4
(d)
BF3
[NCERT 1974; CPMT 1989; MP PMT 1999]
(a) Oxidized
(c) Hydrolysed
Electrovalent compounds do not have [CPMT 1991]
(a) High M.P. and Low B.P.
(b) High dielectric constant
(c) High M.P. and High B.P.
(d) High polarity
Many ionic crystals dissolve in water because
28.
(a)
(b)
(c)
(d)
Water is an amphiprotic solvent
Water is a high boiling liquid
The process is accompanied by a positive heat of solution
Water decreases the interionic attraction in the crystal
lattice due to solvation
The electronic structure of four elements A, B, C, D are
(A) 1s 2
(B) 1s 2 , 2 s 2 2 p 2
(C) 1s 2 , 2 s 2 2 p 5
(D) 1s 2 , 2 s 2 2 p 6
Solid NaCl is a bad conductor of electricity since
(a) In solid NaCl there are no ions
(b) Solid NaCl is covalent
(c) In solid NaCl there is no motion of ions
29.
30.
(d) In solid NaCl there are no electrons
Favourable conditions for electrovalency are
(a) Low charge on ions, large cation, small anion
(b) High charge on ions, small cation, large anion
(c) High charge on ions, large cation, small anion
(d) Low charge on ions, small cation, large anion
The sulphate of a metal has the formula M 2 SO 4 3 . The formula
for its phosphate will be
The tendency to form electrovalent bond is largest in
[MNR 1987, 95]
(a) A
(c) C
21.
22.
(a)
(b) B
(d) D
Chloride of metal is MCl 2 . The formula of its phosphate will be
(a)
M 2 PO4
(b)
M 3 PO4 2
(c)
M 2 PO4 3
(d)
MPO4
The phosphate of a metal has the formula MPO4 . The formula of
its nitrate will be
[CPMT 1971; MP PMT 1996]
(a)
MNO 3
(b)
M 2 NO 3 2
(c)
M NO 3 2
(d)
M NO 3 3
(b) Reduced
(d) Hydrated
[AFMC 1980]
[NCERT 1982]
20.
99
(c)
31.
32.
M HPO4 2
M 2 PO4 3
[DPMT 1982; CPMT 1972; MP PMT 1995]
(b)
M 3 PO4 2
(d)
MPO4
Ionic bonds
are usually formed by combination of elements with [CBSE PMT 199
[CPMT 1979]
(a) High ionisation potential and low electron affinity
(b) Low ionisation potential and high electron affinity
(c) High ionisation potential and high electron affinity
(d) Low ionisation potential and low electron affinity
Molten sodium chloride conducts electricity due to the presence of
(a) Free electrons
(b) Free ions
(c) Free molecules
(d) Atoms of sodium and chlorine
100 Chemical Bonding
33.
The phosphate of a metal has the formula MHPO4 . The formula
of its chloride would be
45.
[NCERT 1974; CPMT 1977]
(a)
MCl
(b)
MCl 2
(c)
MCl 3
(d)
M 2 Cl 3
46.
35.
A number of ionic compounds e.g. AgCl, CaF2 , BaSO 4 are
insoluble in water. This is because
[NCERT 1984]
(a) Ionic compounds do not dissolve in water
(b) Water has a high dielectric constant
(c) Water is not a good ionizing solvent
(d) These molecules have exceptionally high alternative forces in
the lattice
What is the nature of chemical bonding between Cs and F
36.
(a) Covalent
(b) Ionic
(c) Coordinate
(d) Metallic
Which one of the following compound is ionic
34.
47.
48.
[MP PMT 1987; CPMT 1976]
[MNR 1985]
(a)
KCl
(c) Diamond
37.
(b) CH 4
(d)
H2
Which of the following compound has electrovalent linkage
[CPMT 1983, 84, 93]
38.
39.
40.
41.
(a)
CH 3 Cl
(b)
(c)
CH 4
(d) Cl 2
Chemical formula for calcium pyrophosphate is Ca 2 P2 O7 . The
formula for ferric pyrophosphate will be
[NCERT 1977]
(a)
Fe3 (P2 O7 )3
(b)
Fe4 P4 O14
(c)
Fe4 (P2 O7 )3
(d)
Fe3 PO4
Among the bonds formed by a chlorine atom with atoms of
hydrogen, chlorine, sodium and carbon, the strongest bond is
formed between
[EAMCET 1988; MP PMT 1993]
43.
(a)
BeF2
(b)
SrF2
(c)
CaF2
(d)
MgF2
NaCl
(b)
He
(b) CsCl
(c)
NH 3
(d) CHCl 3
What is the effect of more electronegative atom on the strength of
ionic bond
[AMU 1999]
(a) Decreases
(b) Increases
[AMU 1982]
(c) Decreases
slowly
(d) Remains the same
52.
An element X with the electronic configuration 1s 2 , 2 s 2 2 p 6 , 3 s 2
would be expected to form the chloride with the formula
53.
54.
55.
[CPMT 1989]
Which of the following halides has maximum melting point
(a)
51.
(b) Cl  Cl
(c) Na  Cl
(d) C  Cl
Which of the following is least soluble
(a)
44.
H  Cl
50.
NaCl
An ionic compound is generally a
[MADT Bihar 1981]
(a) Good electrolyte
(b) Weak electrolyte
(c) Non-electrolyte
(d) Neutral
What metals combine with non-metals, the metal atom tends to
(a) Lose electrons
(b) Gain electrons
(c) Remain electrically neutral
(d) None of these
(a)
42.
49.
Out of the following, which compound will have electrovalent
bonding
(a) Ammonia
(b) Water
(c) Calcium chloride
(d) Chloromethane
The force which holds atoms together in an electrovalent bond is
(a) Vander Waal's force
(b) Dipole attraction force
(c) Electrostatic force of attraction
(d) All the above
The main reaction during electrovalent bond formation is
(a) Redox reaction
(b) Substitution reaction
(c) Addition reaction
(d) Elimination reaction
Electrovalent compounds are
[CPMT 1996]
(a) Good conductor of electricity
(b) Polar in nature
(c) Low M.P. and low B.P.
(d) Easily available
Ionic compounds do not have
[RPMT 1997]
(a) Hard and brittle nature
(b) High melting and boiling point
(c) Directional properties
(d) Soluble in polar solvents
Highest melting point would be of
[RPMT 1999]
(a)
XCl 3
(b)
XCl 2
(c)
XCl
(d)
X 2 Cl
Two element have electronegativity
between them would be
(a) Ionic
(b)
(c) Co-ordinate
(d)
Which of the following is least ionic
of 1.2 and 3.0. Bond formed
[CPMT 1982; DCE 2000]
Polar covalent
Metallic
[MP PET 2002]
(a)
C 2 H 5 Cl
(b)
(c)
BaCl2
(d) C 6 H 5 N  H 3 Cl 
KCl
Which type of bonding exists in Li 2 O and CaF2 respectively
[RPET 2000]
56.
(a) Ionic, ionic
(b) Ionic, covalent
(c) Covalent, ionic
(d) Coordinate, ionic
An atom with atomic number 20 is most likely to combine
chemically with the atom whose atomic number is
NaBr
(c) NaI
(d) NaF
The high melting point and insolubility in organic solvents of
sulphanilic acid are due to its ...... structure.
[IIT 1994]
(a) Simple ionic
(b) Bipolar ionic
(c) Cubic
(d) Hexagonal
[BHU 2000]
57.
(a) 11
(b) 14
(c) 16
(d) 10
Bond formed in crystal by anion and cation is
[CBSE PMT 2000]
(a) Ionic
(b) Metallic
Chemical Bonding
58.
59.
(c) Covalent
(d) Dipole
Atoms or group of atoms which are electrically charged are known
(a) Anions
(b) Cations
(c) Ions
(d) Atoms
Which one is the strongest bond
[Pb. PMT 2001]
(a)
Br – F
(b)
3.
61.
[Kerala CET (Med.) 2002]
2
62.
(d) H 2
Zn 2  and Ni 2 
63.
64.
(a) Hydration energy
(b) Lattice energy
(c) Internal energy
(d) Bond energy
Which of the following has highest melting point
7.
[DPMT 1986; CPMT 1986]
[DPMT 1985, 86; NCERT 1975]
[CPMT 2002]
8.
65.
66.
67.
BeCl 2
(b)
MgCl2
(c)
CaCl 2
(d)
BaCl2
(c)
68.
CaH 2
(b)
SrH 2
(d)
9.
BeH 2
Which of the following conduct electricity in the fused state
13.
[Roorkee 2000]
BeCl 2
(b)
MgCl2
(c)
SrCl 2
(d)
BaCl2
14.
Covalent bonding
1.
2.
The valency of sulphur in sulphuric acid is
[NCERT 1974]
(a) 2
(b) 4
(c) 6
(d) 8
The number of electrons involved in the bond formation of
N 2 molecule
[IIT 1980; CPMT 1983, 84, 85; CBSE PMT 1992]
(a) 2
(b) 4
2
2
2
2
2
2
4
2
[CPMT 1987]
BaH 2
(a)
2
2
Which of the following statements is not true for ionic compounds [RPET 2003]
(a) High melting point
(b) Least lattice energy
10.
(c) Least solubility in organic compounds
(d) Soluble in water
Electrolytes are compound containing [MADT Bihar 1981]
11.
(a) Electrovalent bond
(b) Covalent bond
(c) Coordinate bond
(d) Hydrogen bond
Which of the following hydrides are ionic
[Roorkee 1999]
12.
(a)
(a) Iodine crystal
(b) Solid CO 2
(c) Silica
(d) White phosphorus
With which of the given pairs CO resembles
[BHU 2005]
(a) HgCl , C H
(b) HgCl , SnCl
(c) C H , NO
(d) N O and NO
The electron pair which forms a bond between two similar nonmetallic atoms will be
[IIT 1986]
(a) Dissimilar shared between the two
(b) By complete transfer from one atom to other
(c) In a similar spin condition
(d) Equally shared in between the two
For the formation of covalent bond, the difference in the value of
electronegativities should be
[EAMCET 1982]
(a) Equal to or less than 1.7
(b) More than 1.7
(c) 1.7 or more
(d) None of these
Which type of bond is formed between similar atoms
(a) Ionic
(b) Covalent
(c) Coordinate
(d) Metallic
Covalent compounds are generally ...... in water
2
[RPET 2003]
(a)

(a) Covalent
(b) Ionic
(c) Metallic
(d) Coordinate
Which of the following substances has giant covalent structure
(b) Co 3 and Ni 4 
(c) Co 2  and Ni 2 
(d) Ti 4  and V 3 
The energy that opposes dissolution of a solvent is
 
2 p 6 , 3 s1 ; R 1s 2 , 2 s 2 2 p 2
6.
4.
5.
Which of the following pairs of species has same electronic
configuration
[UPSEAT 2002]
(a)
2

(a) Q
(b) M
(c) R
(d) L
In covalency
[CPMT 1974, 76, 78, 81; AFMC 1982]
(a) Electrons are transferred
(b) Electrons are equally shared
(c) The electron of one atom are shared between two atoms
(d) None of the above
Which compound is highest covalent
(a) LiCl
(b) LiF
(c) LiBr
(d) LiI
The nature of bonding in graphite is
[UPSEAT 2002]
(c) Diamond
 
[NCERT 1983]
(a) Solute-Solute
(b) Solvent-Solvent
(c) The charges
(d) Molecular properties
Which of the following compounds is ionic
(b) CH 4

Q 1s , 2 s
The element that would most readily form a diatomic molecule is
60.
KI
(c) 6
(d) 10
The electronic
of four elements are given in brackets
[UPSEATconfiguration
2001]
L 1s 2 , 2 s 2 2 p 1 ; M 1s 2 , 2 s 2 2 p 5
F–F
(c) Cl – F
(d) Br – Cl
The interionic attraction depends on interaction of
(a)
101
15.
16.
(a) Soluble
(b) Insoluble
(c) Dissociated
(d) Hydrolysed
Which one is the electron deficient compound
(a) ICl
(b) NH 3
[AIIMS 1982]
(c) BCl3
(d) PCl3
Which among the following elements has the tendency to form
covalent compounds
(a) Ba
(b) Be
(c) Mg
(d) Ca
Silicon has 4 electrons in the outermost orbit. In forming the bonds
(a) It gains electrons
(b) It loses electrons
(c) It shares electrons
(d) None of these
Which of the following occurs when two hydrogen atoms bond with
each others
(a) Potential energy is lowered
(b) Kinetic energy is lowered
102 Chemical Bonding
18.
(c) Electronic motion ceases
(d) Energy is absorbed
A bond with maximum covalent character between non-metallic
elements is formed
[NCERT 1982]
(a) Between identical atoms
(b) Between chemically similar atoms
(c) Between atoms of widely different electronegativities
(d) Between atoms of the same size
Amongst the following covalent bonding is found in
19.
(a) Sodium chloride
(b) Magnesium chloride
(c) Water
(d) Brass
Indicate the nature of bonding in diamond
20.
(a) Covalent
(b) Ionic
(c) Coordinate
(d) Hydrogen
Octet rule is not valid for the molecule
17.
27.
21.
CO 2
(b)
(c)
CO
(d) O 2
22.
(b) CaO
(c)
KCl
(d)
Na 2 S
28.
29.
33.
24.
(c)
..
(d) : X .
..
35.
Which is the most covalent
(a)
(c)
25.
34.
C O
CS
[AFMC 1982]
(b) C  Br
[CPMT 1987]
(b) 3
(d) 5
Hydrogen chloride molecule contains a [CPMT 1984]
(a) Covalent bond
(b) Double bond
(c) Coordinate bond
(d) Electrovalent bond
As compared to covalent compounds, electrovalent compounds
generally have
[CPMT 1990, 94; MP PMT 1997]
(a) Low melting points and low boiling points
(d) C  F
(b) Low melting points and high boiling points
The covalent compound HCl has the ionic character as
(c) High melting points and low boiling points
[EAMCET 1980]
26.
(b) Sodium chloride
(d) Copper
The covalency of nitrogen in HNO 3 is
(a) 0
(c) 4
If the atomic number of element X is 7, the best electron dot
symbol for the element is
[NCERT 1973; CPMT 2003]
.
. X :
.
[DPMT 1984]
[AMU 1985]
(a) Germanium
(c) Solid neon
(d) Electrovalent in CCl 4 and covalent in CaH 2
(b) . X .
(a) Molecular solid
(b) Ionic solid
(c) Pseudo solid
(d) Does not exist
A covalent bond is likely to be formed between two elements which
(a) Have similar electronegativities
(b) Have low ionization energies
(c) Have low melting points
(d) Form ions with a small charge
The bond between two identical non-metal atoms has a pair of
electrons
[CPMT 1986]
(a) Unequally shared between the two
(b) Transferred fully from one atom to another
(c) With identical spins
(d) Equally shared between them
32.
(c) Covalent in both CCl 4 and CaH 2
X .
[DPMT 1983]
(a) 2
(b) 5
(c) 4
(d) 1
Which of the following substances has covalent bonding
(b) Electrovalent in both CCl 4 and CaH 2
(a)
Solid CH 4 is
The valency of phosphorus in H 3 PO4 is
[NCERT 1973]
23.
Bond energy of covalent O  H bond in water is
31.
Indicate the nature of bonding in CCl 4 and CaH 2
(a) Covalent in CCl 4 and electrovalent in CaH 2
BeCl 2  LiCl  NaCl
(c) Less than bond energy of H  bond
(d) None of these
[IIT 1980; MLNR 1982]
H2
(d)
(b) Equal to bond energy of H  bond
Which of the following compounds are covalent
(a)
NaCl  LiCl  BeCl 2
[EAMCET 1982]
30.
H 2O
BeCl 2  NaCl  LiCl
(a) Greater than bond energy of H  bond
[IIT 1979; MP PMT 1995]
(a)
(b)
(c)
[CPMT 1973]
[EAMCET 1980; BHU 1996; KCET 2000]
(a) LiCl  NaCl  BeCl 2
(a) The electronegativity of hydrogen is greater than that of
chlorine
(b) The electronegativity of hydrogen is equal to that of chlorine
(c) The electronegativity of chlorine is greater than that of
hydrogen
(d) Hydrogen and chlorine are gases
The correct sequence of increasing covalent character is represented
by [CBSE PMT 2005]
(d) High melting points and high boiling points
36.
The interatomic distances in H 2 and Cl 2 molecules are 74 and
198 pm respectively. The bond length of HCl is
[MP PET 1993]
37.
(a) 272 pm
(b) 136 pm
(c) 124 pm
(d) 248 pm
On analysis, a certain compound was found to contain iodine and
oxygen in the ratio of 254 gm of iodine and 80 gm of oxygen.
Chemical Bonding
The atomic mass of iodine is 127 and that of oxygen is 16. Which of
the following is the formula of the compound
38.
(a)
IO
(b)
I2 O
(c)
I5 O 2
(d)
I2 O5
(e)
47.
Ionic and covalent bonds are present in
[CBSE PMT 1990; MNR 1990; KCET 2000; UPSEAT 2001]
39.
40.
41.
42.
(a)
CCl 4
(b) CaCl 2
(c)
NH 4 Cl
(d)
H 2O
Highest covalent character is found in [EAMCET 1992]
(a)
CaF2
(b) CaCl 2
(c)
CaBr2
(d) CaI 2
Among the following which property is commonly exhibited by a
covalent compound
[MP PET 1994]
(a) High solubility in water
(b) High electrical conductance
(c) Low boiling point
(d) High melting point
Atoms in the water molecule are linked by
[MP PAT 1996]
(a) Electrovalent bond
(b) Covalent bond
(c) Coordinate covalent bond
(d) Odd electron bond
48.
49.
[MP PET 1996]
.. .. ..
N N O
..
..
(c)
43.
44.
..
(b) : N  N   O : 
..
51.
..
(d) : N  N  O :
..
52.
A covalent bond between two atoms is formed by which of the
following
[MP PMT 1996]
(a) Electron nuclear attraction
(b) Electron sharing
(c) Electron transfer
(d) Electrostatic attraction
The
2
electronic
2
6
configuration
of
a
metal
M
1s , 2 s 2 p , 3 s . The formula of its oxides will be
53.
45.
46.
MO
(b)
M 2O
(c)
M 2 O3
(d)
MO 2
54.
55.
(a)
M 2 A3
(b)
MA 2
(c)
M2 A
(d)
MA3
CB
240
328
C C
276
485
CD
(a) A
(b) B
(c) C
(d) D
If a molecule X 2 has a triple bond, then X will have the electronic
configuration
[CET Pune 1998]
(a)
1s 2 2 s 2 2 p 5
(b) 1s 2 2 s 2 2 p 3
(c)
1s 2 2 s 1
(d) 1s 2 2 s 2 2 p 1
Which of the following compounds does not follow the octet rule for
electron distribution
[CET Pune 1998]
(a) PCl5
(b) PCl 3
H 2O
(d)
PH 3
The valency of A  3 and B  2 , then the compound is
(a)
A2 B3
(b)
A3 B2
(c)
A3 B3
(d)
A2 B2
(e) None of these
The number of electrons shared by each outermost shell of N 2 is
(a) 2
(b) 3
(c) 4
(d) 5
Which of the following substances when dissolved in water will give
a solution that does not conduct electricity
(a) Hydrogen chloride
(b) Potassium hydroxide
(c) Sodium acetate
(d) Urea
Which of the following atoms has minimum covalent radius
(a) B
(b) C
(c) N
(d) Si
Boron form covalent compound due to [Pb. PMT 2000]
(a) Small size
(b) Higher ionization energy
(c) Lower ionization energy
(d) Both (a) and (b)
Two elements X and Y have following electron configurations
and Y  1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6
The compound formed by combination of X and Y is
[DPMT 2001]
56.
[Bihar MEE 1996]
CA
X  1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 , 4 s 2
Which of the following statements regarding covalent bond is not
true
[MP PET/PMT 1998]
(a) The electrons are shared between atoms
(b) The bond is non-directional
(c) The strength of the bond depends upon the extent of
overlapping
(d) The bond formed may or may not be polar
If the electronic configuration of M  2, 8, 3 and that of
A  2, 8, 7, the formula of the compound is
E diss (kJ mol 1 )
[DPMT 2000]
[MP PET/PMT 1998]
(a)
Bond
[JIPMER 1999]
is
1
[CPMT 1981]
[Bihar MEE 1997]
Which is the correct electron dot structure of N 2 O molecule
..
(a) : N  N  O
..
M3 A
The table shown below gives the bond dissociation energies (E diss )
for single covalent bonds of carbon (C) atoms with element
A, B, C and D. Which element has the smallest atoms[CBSE PMT 1994]
(c)
50.
103
(a)
XY5
(b)
X 2 Y5
(c)
X 5 Y3
(d)
XY 2
Covalent compounds have low melting point because
[KCET 2002]
(a) Covalent bond is less exothermic
(b) Covalent molecules have definite shape
(c) Covalent bond is weaker than ionic bond
104 Chemical Bonding
57.
(d) Covalent molecules are held by weak Vander Waal’s force of
attraction
p and n-type of semiconductors are formed due to
58.
(a) Covalent bonds
(b) Metallic bonds
(c) Ionic bonds
(d) Co-ordinate bond
Which of the following is Lewis acid
[RPET 2003]
72.
(a) 8
(b) 5
(c) 6
(d) 7
Hydrogen atoms are held together to form hydrogen molecules by
(a) Hydrogen bond
(b) Ionic bond
(c) Covalent bond
(d) Dative bond
Strongest bond is
[AFMC 1987]
(a) C  C
(b) C  H
(c) C  N
(d) C  O
The major binding force of diamond, silicon and quartz is
73.
(a) Electrostatic force
(b) Electrical attraction
(c) Co-valent bond force
(d) Non-covalent bond force
Multiple covalent bonds exist in a molecule of
[NCERT 1973]
70.
[UPSEAT 2002]
59.
(a)
BF3
(b)
NH 3
(c)
PH 3
(d)
SO 2
71.
[Kerala CET (Med.) 2002]
Among the species : CO 2 , CH 3 COO  , CO , CO 3 2  , HCHO
which has the weakest carbon- oxygen bond
[Kerala PMT 2004]
(a)
CO 2
(c) CO
(b) CH 3 COO 
74.
2
(d) CO 3
Valency of sulphur in Na 2 S 2 O 3 is
(a) Two
(c) Four
61.
[DPMT 1984]
(b) Three
(d) Six
75.
The acid having O  O bond is
[IIT JEE Screening 2004]
62.
(a)
H 2 S 2 O3
(b)
H 2 S 2 O6
(c)
H 2 S 2 O8
(d)
H 2 S 4 O6
76.
The following salt shows maximum covalent character
[UPSEAT 2004]
63.
(a)
AlCl3
(b)
MgCl2
(c)
CsCl
(d)
LaCl 3
65.
(a) O is more electronegative than S
(b) O  H bond is stronger than S  H bond
(c) O  H bond is weaker than S  H bond
(d) None of these
Which of the following has covalent bond
F2
(c)
C2 H 4
(d)
N2
Which of the following does not obey the octet rule
(a)
CO
(b)
NH 3
(c)
H 2O
(d)
PCl5
Which of the following statements is correct for covalent bond
(a) Electrons are shared between two atoms
(b) It may be polar or non-polar
(c) Direction is non-polar
(d) Valency electrons are attracted
Among CaH 2 , NH 3 , NaH and B2 H 6 , which are covalent
hydride
[Orissa JEE 2005]
(a) NH 3 and B2 H 6
(b) NaH and CaH 2
(c) NaH and NH 3
1.
[MHCET 2003; Pb CET 2001]
64.
(b)
Which species has the maximum number of lone pair of electrons
on the central atom?
[IIT 2005]
(a) [ClO ]
3
66.
67.
68.
69.
Na 2 S
(b)
(b) XeF
4
[BVP 2004]
2.
3.
valency in acetylene C 2 H 2  is
[NCERT 1971]
(a) 1
(b) 2
(c) 3
(d) 4
Number of electrons in the valence orbit of nitrogen in an ammonia
molecule are
[MH CET 2004]
3
(a)
C2 H 2
(b)
H 2 SO 4
(c)
NH 3
(d)
HCl
The bond that exists between NH 3 and BF3 is called
[AFMC 1982; MP PMT 1985; MNR 1994;
KCET 2000; MP PET 2001; UPSEAT 2001]
AlCl3
(c) NaH
(d) MgCl2
The following element forms a molecule with eight its own weight
atoms
[MHCET 2004]
(a) Si
(b) S
(c) Cl
(d) P
In H 2 O2 , the two oxygen atoms have
(a) Electrovalent bond
(b) Covalent bond
(c) Coordinate bond
(d) No bond
Carbon has a valency of 2 in CO and 4 in CO 2 and CH 4 . Its
4
(c) SF
(d) [I ]
A simple example of a coordinate covalent bond is exhibited by
[AFMC 1988; DCE 2004]
(a)
(d) CaH 2 and B2 H 6
Co-ordinate or Dative bonding
Which type of bond is present in H 2 S molecule
(a) Ionic bond
(b) Covalent bond
(c) Co-ordinate
(d) All of three
H 2 S is more acidic than H 2 O , due to
H2
[EAMCET 1993]
(e) HCHO
60.
(a)
4.
(a) Electrovalent
(b) Covalent
(c) Coordinate
(d) Hydrogen
Which of the following does not have a coordinate bond
[MADT Bihar 1984]
5.
(a)
SO 2
(b)
HNO 3
(c)
H 2 SO 3
(d)
HNO 2
Coordinate covalent compounds are formed by
[CPMT 1990, 94]
6.
(a) Transfer of electrons
(b) Sharing of electrons
(c) Donation of electrons
(d) None of these process
In the coordinate valency
[CPMT 1989]
(a) Electrons are equally shared by the atoms
(b) Electrons of one atom are shared with two atoms
(c) Hydrogen bond is formed
[
Chemical Bonding
7.
[DPMT 1985]
(d) None of the above
Which of the following contains a coordinate covalent bond
[MNR 1990; IIT 1986]
8.
9.
10.
(a)
N 2 O5
(b)
BaCl 2
(c)
HCl
(d)
H 2O
3.
SO 32 
(b) CH 4
(c)
CO 2
(d)
13.
NH 3
O3
(b)
(c)
H 2 SO 4
(d) All of these
Which molecule has the largest dipole moment
(b)
(a)
CH 3 NC
(b) CH 3 OH
(c)
CH 3 Cl
(d)
6.
(a) Water
(b) Ethanol
(c) Ethane
(d) Ether
Which of the following molecules will show dipole moment
[NCERT 1972, 74; DPMT 1985]
(c)
[KCET 2003]
|
O
O
|
|
H
H
(d)
O

H O P  O
9.
10.
|
(a) Covalent
(b) Co-ordinate covalent
(c) Ionic bond
(d) Banana shaped bond
Sulphuric acid provides a example of
11.
[Kerala CET (Med.) 2002]
(a)
(b)
(c)
(d)
Co-ordinate bonds
Non-covalent compound
Covalent and co-ordinate bond
Non-covalent ion
Dipole moment
12.
(a)
H 2O
(c)   150 o
(d)   180 o
Which of the following would have a permanent dipole moment[CBSE PMT 2005
(a) BF3
(b)
SiF4
(c) SF4
(d)
XeF4
Carbon tetrachloride has no net dipole moment because of
(a) Its planar structure
(b) Its regular tetrahedral structure
(c) Similar sizes of carbon and chlorine atoms
(d) Similar electron affinities of carbon and chlorine
The molecule which has the largest dipole moment amongst the
following
[MNR 1983]
(a)
CH 4
(b) CHCl 3
(c)
CCl 4
(d) CHI 3
Positive dipole moment is present in
13.
14.
(a)
CCl 4
(b) C 6 H 6
(c)
BF3
(d)
HF
The polarity of a covalent bond between two atoms depends upon
(a) Atomic size
(b) Electronegativity
(c) Ionic size
(d) None of the above
Pick out the molecule which has zero dipole moment
[CPMT 1989; EAMCET 1993; MP PMT 1999]
(b) CO 2
(c) HF
(d) HBr
In the following which one have zero dipole moment
(b)   120 o
[MNR 1986; MP PET 2000]
Which molecules has zero dipole moment
[AIIMS 1980, 82, 91; Roorkee 2000; MH CET 2001]
Which bond angle  would result in the maximum dipole moment
for the triatomic molecule YXY [AIIMS 1980]
[IIT 1982, 83; MP PMT 1985, 91; EAMCET 1988;
AMU 1999]
H
What is the nature of the bond between B and O in
(C 2 H 5 )2 OBH 3
[Orissa JEE 2003]
(a) Methane
(b) Carbon tetrachloride
(c) Chloroform
(d) Carbon dioxide
Which of the following compounds possesses the dipole moment[NCERT 1978; E
[RPET 2003]
(a) Water
(b) Boron trifluoride
(c) Benzene
(d) Carbon tetrachloride
(a)   90 o
H
(b) O  P  O  H
|
O  P OH
7.
8.
NH 3
The structure of orthophosphoric acid is
O

H O  P O H
HI
5.
SO 3
(a) 0
(b) 1
(c) 2
(d) 4
Which of the following compounds has coordinate (dative) bond
|
2.
(d) All of these
The number of dative bonds in sulphuric acid molecules is
H
1.
BeCl 2
[NCERT 1975; Kurukshetra CEE 1998]
(a)
|
15.
(c)
The compound containing co-ordinate bond is
(a)
14.
(b) CCl 4
(c) HBr
(d) HF
The unequal sharing of bonded pair of electrons between two atoms
in a molecule causes
[EAMCET 1986]
(a) Dipole
(b) Radical formation
(c) Covalent bond
(d) Decomposition of molecule
Which of the following will show least dipole character
4.
[MP PET 2002]
12.
BF3
[CBSE
1992]
(a)PMTHCl
[AFMC 1999; Pb. CET 2002]
11.
(a)
[CPMT 1991]
A coordinate bond is formed when an atom in a molecule has
(a) Electric charge on it
(b) All its valency electrons shared
(c) A single unshared electron
(d) One or more unshared electron pair
Which has a coordinate bond
[RPMT 1997]
(a)
105
(a)
NH 3
(b)
H 2O
(c)
BCl 3
(d)
SO 2
106 Chemical Bonding
15.
Zero dipole moment is present in [DPMT 1986; IIT 1987]
(a)
16.
(b)
NH 3
(c) Sulphur dioxide
H 2O
28.
(c) cis 1, 2-dichloroethene
(d) trans 1, 2-dichloroethene
Which of the following is the most polar
[AFMC 1988]
(a)
CCl 4
(b) CHCl 3
(c)
CH 3 OH
(d) CH 3 Cl
Which one has minimum (nearly zero) dipole moment
18.
(a) Butene-1
(b) cis butene-2
(c) trans butene-2
(d) 2-methyl-1-propene
Which one of the following is having zero dipole moment
[IIT Screening 1994; CBSE PMT 1996]
[RPMT 1997; EAMCET 1988; MNR 1991]
19.
(a)
CCl 4
(b) CH 3 Cl
(c)
CH 3 F
(d) CHCl 3
Which of the following molecules does not possess a permanent
dipole moment
[CBSE PMT 1994]
(a) H 2 S
(b) SO 2
(c)
20.
(d)
CS 2
21.
30.
31.
CH 2 Cl 2
(b) CH 4
(c)
NH 3
(d)
32.
PH 3
Fluorine is more electronegative than either boron or phosphorus.
What conclusion can be drawn from the fact that BF3 has no
33.
dipole moment but PF3 does
[Pb. PMT 1998]
(a)
BF3 is not spherically symmetrical but PF3 is
(b)
BF3 molecule must be linear
34.
(c) The atomic radius of P is larger than the atomic radius of B
(d) The BF3 molecule must be planar triangular
22.
[RPET 1997, 99]
23.
BF3
(b)
(c)
CCl 4
(d) CH 4
NH 3
35.
The dipole moment of HBr is 1.6  10 30 cm and interatomic
spacing is 1Å. The % ionic character of HBr is
[MP PMT 2000]
24.
25.
(a) 7
(b)
(c) 15
(d)
Non-polar solvent is
(a) Dimethyl sulphoxide
(b)
(c) Ammonia
(d)
Which shows the least dipole moment
10
27
Carbon tetrachloride
Ethyl alcohol
(a)
CCl 4
(b) CHCl 3
(c)
CH 3 CH 2 OH
(d) CH 3 COCH 3
Which molecule has zero dipole moment
(a) H 2 O
(b) AgI
(c) PbSO 4
27.
(b)
H 2 Se
(c)
CH 4
(d)
HI
Which of the following has no dipole moment
(a)
CO 2
(b)
SO 3
(c)
O3
(d)
H 2O
Which of the following is non-polar
[DCE 2002]
(a)
PCl5
(b)
PCl3
(c)
SF6
(d)
IF7
Identify the non-polar molecule in the set of compounds given :
[UPSEAT 2004]
HCl, HF, H 2 , HBr
(a)
H2
(b)
HCl
(c)
HF, HBr
(d)
HBr
Dipole moment is shown by
(a) 1, 4-dichlorobenzene
[IIT 1986]
(d) trans 1, 2-dichloro-2-pentene
If HCl molecule is completely polarized, so expected value of dipole
moment is 6.12D (deby), but experimental value of dipole moment is
1.03D. Calculate the percentage ionic character
[Kerala CET 2005]
(a) 17
(b) 83
(c) 50
(d) Zero
(e) 90
Polarisation and Fajan's rule
1.
BF3 and NF3 both molecules are covalent, but BF3 is non-polar
and NF3 is polar. Its reason is
[CPMT 1989; NCERT 1980]
(a) In uncombined state boron is metal and nitrogen is gas
B  F bond has no dipole moment whereas N  F bond has
dipole moment
(c) The size of boron atom is smaller than nitrogen
(b)
[UPSEAT 2001]
(d) HBr
(d)
The dipole moment is zero for the molecule
[IIT 1989; MP PMT 2002]
(a) Ammonia
HCl
[RPET 2000]
[UPSEAT 2001; DPMT 1982]
26.
(a)
(b) cis 1, 2-dichloroethene
(c) trans 1, 2-dichloroethene
Which molecule does not show zero dipole moment
(a)
In a polar molecule, the ionic charge is 4.8  10 10 e.s.u. If the
inter ionic distance is one Å unit, then the dipole moment is
(a) 41.8 debye
(b) 4.18 debye
(c) 4.8 debye
(d) 0.48 debye
Which of the following is a polar compound
[DCE 2002]
Which of the following has zero dipole moment
(a)
Presence of more electrons in orbitals
Absence of dipole moment
Difference in spin quantum no
None of these
[Pb. CET 2000]
SO 3
[CPMT 1997; AFMC 1998; CBSE PMT 2001]
N 2 is less reactive than CN due to [UPSEAT 2003]
(a)
(b)
(c)
(d)
29.
17.
(d) Water

(b) Boron trifluoride
2.
BF3 is planar whereas NF3 is pyramidal
Which one is polar molecule among the following
(a)
CO 2
(b) CCl 4
Chemical Bonding
(c)
3.
H 2O
If the electron pair forming a bond between two atoms A and B is
not in the centre, then the bond is
[AIIMS 1984]
(a) Single bond
(b) Polar bond
(c) Non-polar bond
4.
5.
6.
7.
8.
9.
(d) CH 4
(d)  bond
16.
Which of the following liquids is not deflected by a non-uniform
electrostatic field
[NCERT 1978]
(a) Water
(b) Chloroform
(c) Nitrobenzene
(d) Hexane
Which of the following is non-polar
[EAMCET 1983]
(a)
H2S
(b)
NaCl
(c)
Cl 2
(d)
H 2 SO 4
15.
Polarization is the distortion of the shape of an anion by an
adjacently placed cation. Which of the following statements is
correct
[NCERT 1982]
(a) Maximum polarization is brought about by a cation of high
charge
(b) Minimum polarization is brought about by a cation of
low radius
(c) A large cation is likely to bring about a large degree of
polarization
(d) A small anion is likely to undergo a large degree of polarization
107
(c) Na 
(d) Ca 2 
Maximum covalent character is associated with the compound [RPMT 1999]
(a) NaI
(b) MgI2
(c) AlCl3
(d) AlI3
Polarisibility of halide ions increases in the order
[DCE 1999]
17.
(a)
F  , I  , Br  , Cl 
(b) Cl  , Br  , I  , F 
(c)
I  , Br  , Cl  , F 
(d)
F  , Cl  , Br  , I 
According to Fajan’s rule, covalent bond is favoured by
[AIIMS 1999]
18.
(a) Large cation and small anion
(b) Large cation and large anion
(c) Small cation and large anion
(d) Small cation and small anion
Which of the following statements is correct
(a) SF4 is polar and non-reactive
[AMU 1999]
(b)
SF6 is non-polar and very reactive
(c)
SF6 is a strong fluorinating agent
(d)
SF4 is prepared by fluorinating SCl 2 with NaF
19.
Choose the correct statement
[RPMT 2000]
(a) Amino polarisation is more pronounced by highly charged
cation
The bonds between P atoms and Cl atoms in PCl5 are likely to
(b)
Small cation has minimum capacity to polarise an anion.
be
[MP PMT 1987]
(c)
Small anion has maximum polarizability
(a) Ionic with no covalent character
(d)
None of these
(b) Covalent with some ionic character
20.
The
[DPMT 2001]
ICl molecule is
(c) Covalent with no ionic character
(a)
Purely
electrovalent
(d) Ionic with some metallic character
(b) Purely covalent
Two electrons of one atom A and two electrons of another atom B
(c) Polar
with
negative end on iodine
are utilized to form a compound AB. This is an example of
[MNR
1981]
(d)
Polar
with
negative end on chlorine
(a) Polar covalent bond
(b) Non-polar covalent bond
21.
Which
of
the
following
is a polar compound
[AIIMS 2001]
(c) Polar bond
(d) Dative bond
(a) HF
(b) HCl
In which of the following molecule is the covalent bond most polar [AMU 1985; MP PET 2001]
(c) HNO 3
(d) H 2 SO 4
(a) HI
(b) HBr
22.
Which of the following has zero dipole moment
(c) HCl
(d) H 2
[MP PMT 2002]
10.
Amongst ClF3 , BF3 and NH 3 molecules the one with nonplanar geometry is
[MP PMT 1999]
(a)
ClF3
(b)
(c)
BF3
(d) None of these
NH 3
23.
Which of the following possesses highest melting point
12.
(a) Chlorobenzene
(b) o-dichlorobenzene
(c) m-dichlorobenzene
(d) p-dichlorobenzene
The polar molecule among the following is
[CPMT 1999]
(a)
CCl 4
(b) CO 2
(c)
CH 2 Cl 2
(d) CH 2  CH 2
Which of the following have both polar and non-polar bonds
(a) C 2 H 6
(b) NH 4 Cl
[CET Pune 1998]
(a)
Mg
(b)
Al
3
(d) CFCl 3
SiF4
Which of the following compounds has least dipole moment
(a)
PH 3
(b) CHCl 3
(c)
NH 3
(d)
BF3
Pauling’s electronegativity values for elements are useful in
predicting
[UPSEAT 2004]
(a) Polarity of bonds in molecules
(b) Position of elements in electrochemical series
(c) Co-ordination number
(d) Dipole moment of various molecules
25.
Amongst LiCl, RbCl, BeCl 2 and MgCl2 the compounds with the
[AIIMS 1997]
greatest and the least ionic character, respectively, are [UPSEAT 2002]
(c) HCl
(d) AlCl3
Which of the following has a high polarising power
2
(c)
24.
[Orissa JEE 1997]
14.
(b) PCl 3
[RPET 2003]
11.
13.
(a) ClF
26.
(a) LiCl and RbCl
(b) RbCl and BeCl 2
(c) RbCl and MgCl2
(d)
MgCl2 and BeCl 2
Bond polarity of diatomic molecule is because of
[UPSEAT 2002]
108 Chemical Bonding
(a)
(b)
(c)
(d)
Difference in electron affinities of the two atoms
Difference in electronegativities of the two atoms
Difference in ionisation potential
All of these
11.
(b) Sidewise overlap of orbitals takes place
(c) End to end overlap of orbitals takes place
(d) None of the above
The number of sigma and pi bonds in 1-butene-3-yne are
12.
(a) 5 sigma and 5 pi
(b) 7 sigma and 3 pi
(c) 8 sigma and 2 pi
(d) 6 sigma and 4 pi
The most acidic compound among the following is
[IIT 1989]
Overlaping-  and - bonds
1.
Triple bond in ethyne is formed from
[MP PET 1993]
[MP PMT 1990; NCERT 1979; EAMCET 1978; AMU 1985;
CPMT 1988; MADT Bihar 1982; MH CET 2000]
2.
(a) Three sigma bonds
(b) Three pi bonds
(c) One sigma and two pi bonds
(d) Two sigma and one pi bond
The bond in the formation of fluorine molecule will be
13.
[MP PMT 1987]
(a) Due to s  s overlapping
(b) Due to s  p overlapping
14.
(c) Due to p  p overlapping
3.
(d) Due to hybridization
Which type of overlapping results the formation of a  bond
15.
[DPMT 1981]
(a) Axial overlapping of s  s orbitals
(b) Lateral overlapping of p  p orbitals
16.
5.
(d) Axial overlapping of s  p orbitals
The number and type of bonds between two carbon atoms in
calcium carbide are
[AIEEE 2005]
(a) One sigma, one pi
(b) One sigma, two pi
(c) Two sigma, one pi
(d) Two singma, two pi
In a double bond connecting two atoms, there is a sharing of
[CPMT 1977, 80, 81; NCERT 1975;
Bihar MEE 1980; MP PET 1999]
6.
7.
8.
9.
10.
(a) 2 electrons
(b) 1 electron
(c) 4 electrons
(d) All electrons
Strongest bond is
[DPMT 1990]
(a) C  C
(b) C  C
(c) C  C
(d) All are equally strong
[JIPMER 2002]
 bond is formed
(a) By overlapping of atomic orbitals on the axis of nuclei
(b) By mutual sharing of pi electron
(c) By sidewise overlapping of half filled p-orbitals
(d) By overlapping of s-orbitals with p-orbitals
The double bond between the two carbon atoms in ethylene consists
of
[NCERT 1981; EAMCET 1979]
(a) Two sigma bonds at right angles to each other
(b) One sigma bond and one pi bond
(c) Two pi bonds at right angles to each other
(d) Two pi bonds at an angle of 60 o to each other
In the series ethane, ethylene and acetylene, the C  H bond
energy is
[NCERT 1977]
(a) The same in all the three compounds
(b) Greatest in ethane
(c) Greatest in ethylene
(d) Greatest in acetylene
In a sigma bond
(a) Sidewise as well as end to end overlap of orbitals take place
CH 3 CH 2 OH
(b) C 6 H 5 OH
(c)
CH 3 COOH
(d) CH 3 CH 2 CH 2 OH
Which of the following is not correct
[CBSE PMT 1990]
(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
Strongest bond formed, when atomic orbitals
(a) Maximum overlap
(b) Minimum overlap
(c) Overlapping not done
(d) None of them
The p  p orbital overlapping is present in the following molecule
(a) Hydrogen
(b) Hydrogen bromide
(c) Hydrogen chloride
(d) Chlorine
In N 2 molecule, the atoms are bonded by
[MP PET 1996; UPSEAT 2001]
(c) Axial overlapping of p  p orbitals
4.
(a)
17.
(a) One , Two 
(b) One , One 
(c) Two , One 
(d) Three  bonds
In which of following there exists a p  d bonding
[AFMC 2001]
(a) Diamond
(c) Dimethyl amine
18.
(b) Graphite
(d) Trisilylamine
Number of bonds in SO 2
[DCE 2001]
(a) Two  and two 
(b) Two  and one 
(c) Two , two  and one lone pair
(d) None of these
19.
20.
Which of the following has p  d bonding
(a)
NO 3
(b) CO 32
(c)
BO 33
(d)
[CBSE 2002]
SO 32
Number of sigma bonds in P4 O10 is [AIEEE 2002]
(a) 6
(c) 17
(b) 7
(d) 16
Hybridisation
1.
2.
Which molecule is not linear
[CPMT 1994]
(a)
BeF2
(b)
BeH 2
(c)
CO 2
(d)
H 2O
The bond angle in water molecule is nearly or Directed bonds in
water forms an angle of
[NCERT 1980; EAMCET 1981; MNR 1983, 85;
AIIMS 1982; CPMT 1989; MP PET 1994, 96;
MP PET/PMT 1998]
(a)
120 o
(b) 180 o
Chemical Bonding
(c)
3.
4.
109 o 28 '
12.
(d) 104 o 30'
2
The central atom in a molecule is in sp hybrid state. The shape of
molecule will be [MP PMT 1987; CBSE PMT 1989]
(a) Pyramidal
(b) Tetrahedral
(c) Octahedral
(d) Trigonal planar
Which molecule is linear
13.
[MP PMT 1984; IIT 1982, 88; EAMCET 1993;
CBSE PMT 1992; MP PET 1995; RPMT 1997]
5.
In which of the following the central atom does not use sp 3 hybrid
orbitals in its bonding
[MNR 1992]
(a)
BeF3 
(b) OH 3
NH 2 
(d)
(d)
XeF2 involves hybridisation
14.
Which of the following molecules has trigonal planer geometry [CBSE PMT 2005]
(b)
(d)
NH 3
(a)
PCl3
(c)
BF3
15.
A sp3 hybridized orbital contains
1
s  character
4
2
s  character
3
Structure of ammonia is
(b)
1
s  character
2
(d)
3
s  character
4
16.
(a) Trigonal
(c) Pyramidal
The bond angle in ethylene is
180
(b) 120
(c)
109 o
(d) 90 o
17.
18.
[BHU 1982; RPMT 1999]
[CBSE PMT 1990]
(b)
F
19.
20.
O
21.
ON
O
(d)
(d) None of these
(b) dsp 2
(c)
sp 2
(d)
sp
Octahedral molecular shape exists in .......... hybridisation
3
(b)
sp d
3
sp d
3
3
sp d
2
(d) None of these
The electronic structure of molecule OF2 is a hybrid of
(a)
sp
(b)
sp 2
(c)
sp 3
(d)
sd 3
H C C
OH
OH
Which of the following statement is not correct
Percentage of s-character in sp 3 hybrid orbital is
(b) 50
(d) 75
Shape of XeF4 molecule is
[BHU 1987; AFMC 1992; CET Pune 1998;
Roorkee Qualifying 1998; DCE 2002]
F
O
(c)
sp d
sp 3
(a) 25
(c) 66
(a) Planar
(b) Pyramidal
(c) Angular
(d) Trigonal bipyramidal
Which of the following formula does not correctly represent the
bonding capacity of the atom involved
(a)
2
o
Compound formed by sp3 d hybridization will have structure
H


|


H  P  H 
|




H
sp d
(b)
(a)
(c)
[CPMT 1987]
(a)
3
[DPMT 1990]
3
Which of the following hybridisation results in non-planar orbitals
(a)
(b) Tetrahedral
(d) Trigonal pyramidal
o
sp
3
NF3
[DPMT 1990]
[MP PMT 1987, 89, 91; CPMT 1975, 82;
RPMT 1999; JIPMER 2002]
11.
(d) None of these
CO 2
(c)
10.
sp 3
(c)
(a)
9.
(c)
(c)
[DPMT 1984; BHU 1985; CPMT 1976]
8.
(b)
(b) ClO2
(c)
7.
sp
sp
[CPMT 1991]
2
(a)
NO 2
(a) IF3
6.
The mode of hybridisation of carbon in CO 2 is
(a)
H 2S
109
22.
(a) Linear
(b) Pyramidal
(c) Tetrahedral
(d) Square planar
For which of the following hybridisation the bond angle is maximum
(a)
sp 2
(c)
3
sp
(b)
sp
(d) dsp 2
The C  H bond distance is the longest in
(a)
C2 H 2
(b) C 2 H 4
(c)
C 2 H 4 Br2
(d) C 6 H 6
The nature of hybridization in CH 2 Cl  CH 2 Cl for carbon is
[AIIMS 1983]
(a)
sp
(b)
sp 2
(a) Hybridization is the mixing of atomic orbitals prior to their
combining into molecular orbitals
(c)
sp 3
(d)
sp 2 d
(b)
sp 2 hybrid orbitals are formed from two p atomic orbitals
and one s atomic orbital
23.
(c)
d 2 sp 3 hybrid orbitals are directed towards the corners of a
regular octahedron
24.
(d) dsp 3 hybrid orbitals are all at 90 o to one another
[MNR 1990]
Shape of methane molecule is
[MNR 1983]
(a) Tetrahedral
(b) Pyramidal
(c) Octahedral
(d) Square planer
Which one amongst the following possesses an sp hybridized
carbon in its structure
[CBSE PMT 1989]
(a)
CH 2  C.Cl  CH  CH 2
110 Chemical Bonding
(b) C.Cl 2  C.Cl 2
(c)
35.
CH 2  C  CH 2
H 3 C  C *  C *  CH 3
(d) CH 2  CH  CH  CH 2
25.
Which of the following is the correct electronic formula of chlorine
molecule
[CPMT 1971]
..
..
.. ..
(a) : Cl : Cl :
(b) : Cl  : : Cl  :
..
..
.. ..
(c)
.. ..
: Cl : Cl :
36.
27.
28.
30.
31.
33.
(b)
sp 2 orbital
(c)
sp orbital
(d)
s orbital
The bond angle in carbon tetrachloride is approximately
90
o
(b) 109
o
(c) 120
(d) 180 o
When two pairs of electrons are shared, bond is
In XeF4 hybridization is
3
(a)
sp d
(c)
sp 3 d
2
[MNR 1979]
(a) Single covalent bond
(c) Dative bond
3
(b)
sp
(d)
sp 2 d
38.
In HCHO, ' C' has hybridization
(b) Double covalent bond
(d) Triple bond
The nature of hybridization in the NH 3 molecule is
[EAMCET 1982]
[AIIMS 1987]
sp 2
(a)
sp
(b)
(c)
sp 3
(d) All the above
39.
Which has the shortest C  C bond length
(a)
C 2 H 5 OH
(b) C 2 H 6
(c)
C2 H 2
(d) C 2 H 4
The hybridization of Ag in the linear complex Ag NH 3 2  is
40.

(a)
dsp 2
(b)
sp
(c)
sp 2
(d)
sp 3
2
(a)
sp
(b)
sp
(c)
sp 3
(d)
sp 3 d
Which one of the following compounds has bond angle as nearly
90 o
[MP PMT 1985]
(a)
NH 3
(b)
(c)
H 2O
(d) CH 4
H2S
In ethene, the bond angle(s) is/are
[CPMT 1985; BHU 1981]
o
[CPMT 1976; AMU 1984; MP PMT 1985]
(a)
109 28 '
(b) 120 o
(c)
180 o
(d) Different
41.
Structure formula of H 2 O 2 is
Experiment shows that H 2 O has a dipole moment while CO 2 has
H
not. Point out the structures which best illustrate these facts [DPMT 1984; NCERT 1983; CPMT 1984]
(a)
O O
O
(a)
(b) O  C  O ; H  O  H
H
O  C  O; H
H
(b) H  O  O  H (straight line)
H
(c)
C
; H – H – O (d) O
||
|
H'
C O ; O  H
O
O
(c)
OO
Which species do not have sp 3 hybridization
H
(a) Only sp hybridized carbon atoms
(d)
(c) Both sp and sp 2 hybridized carbon atoms
(d)
sp, sp2 and sp 3 hybridized carbon atoms
The number of unpaired electrons in O 2 molecule is
[MNR 1983; Kerala PET 2002]
(a) 0
(c) 2
(b) 1
(d) 3
H'
OO
H
Where
42.
(b) Only sp 2 hybridized carbon atoms
[CPMT 1993]
Where H  O  O  O  O  H '  101.5 o and all the four
atoms are in the same plane
(a) Ammonia
(b) Methane
(c) Water
(d) Carbon dioxide
As compared to pure atomic orbitals, hybrid orbitals have
(a) Low energy
(b) Same energy
(c) High energy
(d) None of these
The compound 1, 2-butadiene has
[IIT 1983; MP PMT 1996]
34.
sp 3 orbital
o
[DPMT 1985]
32.
(a)
(a)
.. ..
(d) : Cl : : Cl :
[NCERT 1982; CPMT 1989]
29.
[NCERT 1984]
[MNR 1981; MP PMT 1987]
37.
26.
In the following molecule, the two carbon atoms marked by asterisk
(*) possess the following type of hybridized orbitals
43.
H  O  O  O  O  H '  97 o
and
the angle
between H  O  O plane and O  O  H ' plane is 101 o
Number of shared electrons in between carbon-carbon atoms in
ethylene molecule is
[MADT Bihar 1983]
(a) 2
(b) 4
(c) 6
(d) 3
The
structural
formula
of
a
compound
is
CH 3  CH  C  CH 2 . The type of hybridization at the four
carbons from left to right are
[CBSE PMT 1989]
(a)
sp 2 , sp, sp 2 , sp 3
(b)
sp 2 , sp 3 , sp 2 , sp
Chemical Bonding
(c)
44.
sp 3 , sp 2 , sp, sp 2
(d)
sp 3 , sp 2 , sp 2 , sp 2
Acetate ion contains
55.
[AMU 1983]
(b) Two C, O single bonds
(a)
2
(d) None of the above
The two carbon atoms in acetylene are
sp 2 hybridized
(a)
sp hybridized
(b)
(c)
sp hybridized
(d) Unhybridized
Among the following compounds which is planar in shape
47.
(a) Methane
(c) Benzene
In methane the bond angle is
57.
(a) 2
(b) 4
(c) 6
(d) 10
s-character in sp hybridised orbitals are
(b) Acetylene
(d) Isobutene
[AMU 1983]
58.
(a)
1
3
(b)
1
2
(c)
1
4
(d)
2
3
180 o
(b) 90 o
(c)
120 o
(d) 109 o
(a) Three centre bond
(c) Two centre bond
The angle between sp 2 orbitals in ethylene is
59.
90 o
(b) 120 o
(c)
180 o
(d) 109.5 o
The species in which the central atom uses sp 2 hybrid orbitals in
its bonding is
[IIT 1988]
PH 3
(b)
NH 3
(c)
H3C 
(d)
SbH 3
Carbon atoms in diamond are bonded to each other in a
configuration
[CPMT 1981]
(a) Tetrahedral
(b) Planar
(c) Linear
(d) Octahedral
Which of the following molecules can central atom said to adopt
52.
[CBSE PMT 1989; MP PET 1994]
BeF2
(b)
BCl 3
(c)
C2 H 2
(d)
NH 3
In the compound CH 3 OCl, which type of orbitals have been
used by the circled carbon in bond formation
60.
(a)
sp
(c)
sp
In Cu NH 3 4  SO 4 , ; Cu has following hybridization
54.
(a)
dsp 2
(b)
sp 3
(c)
sp 2
(d)
sp 3 d 2
The hybridization of carbon atoms in C  C single bond of
HC  C  CH  CH 2 is
[IIT 1991; MP PET 1995]
(a)
sp 3  sp 3
(b)
sp 2  sp 3
(c)
sp  sp 2
(d)
sp 3  sp
The compound in which C
is
*

(c)
( NH 3 )3 COH HgCl2
(a)
O2  O3  H 2 O2
(b) O 3  H 2 O 2  O 2
(c)
H 2 O2  O3  O2
(d) O 2  H 2 O 2  O 3
62.
(a) Tetrahedral
(b) Trigonal bipyramidal
(c) Square planar
(d) Pentagonal bipyramidal
Which of the following hybridisation has maximum s-characters
[AFMC 1995; JIPMER 2001]
63.
(a)
sp 3
(b)
sp 2
(c)
sp
(d) None of these
The PCl5 molecule is a result of the hybridisation of
[MP PET 1995; DCE 2000; MP PMT 2002]
64.
65.
[IIT 1989]
66.
(b) ( NH 2 )2 CO
2
(a)
sp d
(c)
spd 3
2
sp 3 d
(d)
sp 2 d 3
The shapes of BCl 3 , PCl3 and ICl 3 molecules are all
(c)
67.
(b)
Hybridisation involves
[MP PMT 1996]
(a) Addition of an electron pair
(b) Mixing up of atomic orbitals
(c) Removal of an electron pair
(d) Separation of orbitals
The geometry of sulphur trioxide molecule is
(a) Tetrahedral
(b) Trigonal planar
(c) Pyramidal
(d) Square planar
(a) Triangular

(d) CH 3 CHO
[CBSE PMT 1995]
The structure of PF5 molecule is
uses sp hybrids for bond formation

HCOOH
sp
(d) p
61.
3

(a)
(b)
2
The correct order of the O  O bond length in O 2 , H 2 O 2 and
[AIIMS 1988; UPSEAT 2001]
53.
3
O 3 is
(a)
(a)
(b) Hydrogen bond
(d) None of the above
[MP PET 1994]
(a)
sp 2 hybridization
The two types of bonds present in B 2 H 6 are covalent and
[IIT 1994]
(a)
[BHU 1987, 95; AMU 1985]
51.
sp
56.
[AMU 1992]
50.
The
[BHU 1981]
46.
49.
(b)
sp
(d) dsp 2
sp 3
The number of shared pairs of electrons in propane is
[AMU 1984; MADT Bihar 1982]
48.
120 o .
is
(c)
(c) Two C, O double bonds
3
angle
[BHU 1981; CBSE PMT 1999]
(a) One C, O single bond and one C, O double bond
45.
In diborane, the H  B  H bond
hybridization of boron is likely to be
111
T  shaped
(b) Pyramidal
(d) All above are incorrect
In benzene molecule all C  C bond lengths are equal because
112 Chemical Bonding
(a) All carbon atoms are equivalent
(a)
SnCl 2
(b)
HCl
(b) All carbon atoms are sp 2 hybridised
(c)
CO 2
(d)
HgCl 2
(c) All C  C bonds in benzene, have same order
68.
79.
(d) All C  C bonds are single covalent bond
Which one is false in the following statements
[MP PET 1997]
(a) Each carbon in ethylene is in sp
2
hybridisation
(b) Each carbon in acetylene is in sp 3 hybridisation
(c) Each carbon in benzene is in sp
(d) Each carbon in ethane is in sp
69.
3
2
hybridisation
hybridisation
81.
[MP PMT 1997]
(a)
d 2 sp 3
(b)
sp 3
(c)
sp 2
(d)
sp
As the p  character increases, the bond angle in hybrid orbitals
formed by s and atomic orbitals [MP PMT 1997]
(a) Decreases
(b) Increases
(c) Doubles
(d) Remains unchanged
71.
sp 3 hybridization leads to which shape of the molecule
72.
(a) Tetrahedron
(b) Octahedron
(c) Linear
(d) Plane triangle
Which of the following will be octahedral
[MP PET 1999]
82.
[MP PET/PMT 1998]
73.
83.
SF6
(b)
BF4
(c)
PCl5
(d)
BO 33 
The hybrid orbitals used by central atoms in BeCl 2 , BCl 3 and
84.
CCl 4 molecules are respectively [MP PMT 1999]
(a)
(c)
74.
75.
(d)
2
sp , sp and sp
N3H
NO 2
(c)
sp 3 d
sp
85.
3
(b)
(d)
H atoms are situated at the corners of a tetrahedron
The bond angle in sp 2 hybridisation is
86.
[RPMT 1997]
(b) 120 o
(c) 90 o
(d) 109 o 2'
The correct order towards bond angle is
(a)
sp  sp  sp
(b)
sp 2  sp  sp 3
(c)
sp 3  sp 2  sp
2
(d) Bond angle does not depend on hybridisation
The geometry and the type of hybrid orbital present about the
central atom in BF3 is
[IIT 1998; BHU 2001]
(a) Linear, sp
(b) Trigonal planar, sp 2
(c) Tetrahedral, sp 3
(d) Pyramidal, sp 3
In graphite, electrons are
[CBSE PMT 1997]
(c) Localised on each C atom
(d) Spread out between the structure
The ammonium ion is
[CET Pune 1998]
(a) Tetrahedral
(b) Trigonal pyramidal
(c) Square planar
(d) Square pyramidal
In sp hybridisation, shape is
[Bihar MEE 1997]
(b) Tetrahedral
(d) Linear
When the hybridisation state of carbon atom changes from sp 3 to
[AIIMS 1998]
(d) CO 2
87.
(a) Decreases gradually
(b) Increases gradually
(c) Decreases considerably
(d) All of these
The structure and hybridisation of Si(CH 3 )4 is
[CBSE PMT 1996]
(b) dsp 2
(d)
sp 3
Compound having planar symmetry is [DPMT 1996]
(a)
H 2 SO 4
(b)
(c)
HNO 3
(d) CCl 4
[RPMT 1997]
3
sp 2 to sp, the angle between the hybridised orbitals
H 2O
[DPMT 1996]
180 o
(a) Angular
(c) Bipyramidal
(e) None of these
[CBSE PMT 1999; AFMC 2003]
CCl 4 has the hybridisation
(a)
78.
sp , sp and sp
2
sp , sp 2 and sp 3
(a) Planar
(b) Non-planar
(c) Spherical
(d) Linear
Which of the following is isoelectronic as well as has same structure
as that of N 2 O
[CPMT 1999]
(c)
77.
3
(b)
The structure of H 2 O 2 is
(a)
76.
sp 2 , sp 3 and sp
H atoms are situated at the corners of a square
(a) Localised on every third C atom
(b) Present in antibonding orbital

(a)
(c)
(a)
Out of the following hybrid orbitals, the one which forms the bond
at angle 120 o , is
70.
80.
Which one of the following statements is true for ammonium ion
(a) All bonds are ionic
(b) All bonds are coordinate covalent
88.
H 2O
2
(a) Bent, sp
(b) Trigonal, sp
(c) Octahedral, sp 3 d
(d) Tetrahedral, sp 3
The type of hybridisation of boron in diborane is
[BHU 1999]
(a) sp - hybridisation
Which of the following compounds is not linear
(c)
[CPMT 1996]
89.
3
sp - hybridisation
2
(b)
sp - hybridisation
(d)
sp 3 d 2 - hybridisation
Which compound does not possess linear geometry
[
Chemical Bonding
[RPET 1999]
(a)
(c)
90.
92.
(b)
(d) CO 2
BeCl 2
(d) CF4
SF4
[RPET 1999]
(a)
NO 3
(b)
H 2O
(c)
H3O
(d)
NH 4
96.
103.
2
3
2
(a)
sp , sp , d sp
3
3
3
2
3
(b)
sp , sp d , sp d
(c)
sp, sp 3 d, sp 3 d 2
(d)
sp, sp 2 , sp 3
H 2O
(d)
H 2S
The correct order of hybridization of the central atom in the
following species NH 3 [PtCl4 ]2  , PCl5 and BCl 3 is
2
3
2
3
(a)
dsp , dsp , sp and sp (b)
(c)
dsp 2 , sp 2 , sp 3 , dsp 3
sp3 , dsp2 , dsp3 , sp2
(d) dsp 2 , sp 3 , sp 2 , dsp 3
dsp
[DCE 2000]
(b)
3
sp 2
2
(d) d sp
Which of the following molecule is linear
(a)
SO 2
(b)
NO 2
(c)
NO 2
(d)
SCl 2
The
is
(a)
(c)
106.
[MP PMT 2000]
107.
geometry of the molecule with sp 3 d 2 hybridised central atom
[NCERT 1981; AFMC 1982; RPMT 2000]
Square planar
Octahedral
108.
(b) Trigonal bipyramidal
(d) Square pyramidal
The bond angle in PH 3 is
CO 3
(b)
(c)
K 4 [Fe(CN )6 ]
(d) None of these
(c) 1.10, 1.15 and 1.22 Å
(d) 1.15, 1.10 and 1.22 Å
Shape of BF3 molecule is
[CPMT 2000; Pb. CET 2002]
(a) Linear
(b) Planar
(c) Tetrahedral
(d) Square pyramidal
In the complex [SbF5 ] 2  , sp 3 d hybridization is present.
Geometry of the complex is
[Pb. PMT 2000]
(a) Square
(b) Square pyramidal
(c) Square bipyramidal
(d) Tetrahedral
I 3
(d) None of these
Which of the following is not tetrahedral
[MP PMT 2001]
(a)
SCl 4
(b)
SO 42 
(c)
Ni(CO )4
(d)
NiCl 42 
As the s-character of hybridisation orbital increases, the bond angle [BHU 2002;
(a) Increases
(b) Decreases
(c) Becomes zero
(d) Does not change
The shape of IF7 molecule is
(b) Pentagonal bipyramidal
(d) Tetrahedral
A completely filled d orbital (d 10 )
(a) Spherically symmetrical
(b) Has octahedral symmetry
(c) Has tetrahedral symmetry
(d) Depends on the atom
110.
Which has sp 3 hybridization of central atom
[UPSEAT 2002]
[UPSEAT 2002]
111.
(a)
PCl 3
(b)
SO 3
(c)
BF3
(d)
NO 3
In which of the following species is the interatomic bond angle is
109 o 28 
The single, double and triple bond lengths of carbon in carbon
dioxide are respectively
[AIIMS 2000]
(b) 1.22, 1.15 and 1.10 Å
BF3
109.
NH 4
(a) 1.15, 1.22 and 1.10 Å
NH 4 and SO 42 
[AFMC 2002; MHCET 2003]
[CPMT 2000]
(a)
(d)
BeF2
(a) Octahedral
(c) Trigonal bipyramidal
[RPMT 2000]
Which of the following has tetrahedral structure
(d)
Which of the following is not linear
[DCE 2001]
(a) CO 2
(b) ClO2
(c)
3
PCl5 and SF6
The smallest bond angle is found in [AIIMS 2001]
(a) IF7
(b) CH 4
(c)
105.
The hybridization in PF3 is
sp 3
104.
2
(d) Slightly greater than NH 3
100.
(c)
(c)
(c) Much greater than NH 3
99.
H 2 Se
following compounds : NO 2 , SF4 , PF6 
(b) Equal to that of NH 3
98.
(b)
Which of the following pairs has same structure
[BHU 2001]
(a) PH 3 and BCl 3
(b) SO 2 and NH 3
(a) Much less than NH 3
97.
H 2 Te
What is the correct mode of hybridization of the central atom in the
(c)
95.
(a)
[IIT Screening 2001; BHU 2005]
Pyramidal shape would be of
(a)
94.
The bond angle is minimum in
[Pb. PMT 2001; MP PET 2003; UPSEAT 2004]
[AMU 1999]
93.
101.
HC  CH
Which of the following molecule does not show tetrahedral shape [RPET 1999]
102.
(a) CCl 4
(b) SiCl 4
(c)
91.
CH 2  CH 2
113
112.
113.
(a)
NH 3 , (BF4 )
(c)
NH 3 , BF4
[AIEEE 2002]
1

(b) (NH 4 ) , BF3
(d) (NH 2 )1 , BF3
A square planar complex is formed by hybridisation of which atomic
orbitals
[AIEEE 2002]
(a)
s, p x , p y , d yz
(b)
s, p x , p y , d x 2 y 2
(c)
s, p x , p y , d z 2
(d)
s, p y , p z , d xy
In benzene, all the six C – C bonds have the same length because
of
[MP PET 2002]
(a) Tautomerism
(b)
sp 2 hybridisation
114 Chemical Bonding
(c) Isomerism
114.
(d) Inductive effect
The bond energies of H  H and Cl – Cl are 430 kJ mol
1
1
(c)
and
242 kJ mol
respectively, H t for HCl is 91 kJ mol . The bond
energy of HCl will be
[MP PET 2003]
(a) 427 kJ
(b) 766 kJ
(c) 285 kJ
(d) 245 kJ
115.
125.
-1
116.
(c)
NH 4
(b)
SCl 4
(d)
PtCl42 
117.
118.
119.
(a)
NH 3
(b)
H3O
(c)
BCl 3
(d)
PCl 3
Which one of the following is a correct set with respect to molecule,
hybridisation and shape
[EAMCET 2003]
(a)
BeCl 2 , sp 2 , linear
(b)
BeCl 2 , sp 2 , triangular planar
(c)
BCl 3 , sp 2 , triangular planar
(d)
BCl 3 , sp 3 , tetrahedral
(a)
CO 2
(b)
(c)
BeCl 2
(d) C 2 H 2
121.
122.
123.
127.
sp3 d 2 hybrid orbitals are
(a) Linear bipyramidal
(c) Trigonal bipyramidal
128.
In an octahedral structure, the pair of d orbitals involved in d 2 sp3
hybridization is
[CBSE PMT 2004]
(a) d x 2 , d xz
(b) d xy , d yz
(c)
129.
[MP PET 2004]
(b) Pentagonal
(d) Octahedral
d x 2 y 2 , dz 2
(d) d xz , d x 2  y 2
The correct order of bond
H 2 S , NH 3 , BF3 and SiH 4 is
(a)
H 2 S  NH 3  SiH4  BF3
(b)
NH 3  H 2 S  SiH4  BF3
(c)
H 2 S  SiH4  NH 3  BF3
angles
(smallest
first)
in
[AIEEE 2004]
H 2 S  NH 3  BF3  SiH4
Which of the following compounds doesn’t have linear structure [RPET 1997,
2003]
130. Which one of the following has the regular tetrahedral structure
SO 2
Which of the following bonds require the largest amount of bond
energy to dissociate the atom concerned
(a)
BF4
(b)
SF4
(c)
XeF4
(d)
Ni(CN )4 2 
(Atomic no. : B  5, S  16, Ni  28, Xe  54 )
(a)
H – H bond in H 2
(b) C – C bond in CH 4
The states of hybridazation of boron and oxygen atoms in boric acid
[AIEEE 2004]
(H 3 BO3 ) are respectively
(c)
N  N bond in N 2
(d) O  O bond in O 2
(a)
sp 3 and sp 2
(b)
sp 2 and sp 3
(c)
sp 2 and sp 2
(d)
sp 3 and sp 3
(e) C – C bond in ethane
The percentage s-character of the hybrid orbitals in methane,
ethene and ethyne are respectively
[KCET 2003]
(a) 25, 33, 50
(b) 25, 50, 75
(c) 50, 75, 100
(d) 10, 20, 40
Arrange the hydra-acids of halogens in increasing order of acidity
131.
132.
[Pb. PMT 2004]
(a)
sp
(b)
sp 2
(c)
sp 3
(d)
sp3 d
[Orissa JEE 2003]
HF  HCl  HBr  HI (b)
HI  HBr  HCl  HF
(c)
HF  HBr  HI  HCl (d)
HF  HI  HBr  HCl
Among the compounds, BF3 , NCl 3 , H 2 S , SF4 and BeCl 2 ,
identify the ones in which the central atom has the same type of
hybridisation
[Kerala PMT 2004]
[MP PMT 2004]
(a)
BF3 and NCl 3
(b)
H 2 S and BeCl 2
(c)
BF3 , NCl 3 and H 2 S
(d)
SF4 and BeCl 2
(e)
NCl 3 and H 2 S
Which one has sp  hybridisation
2
(a)
CO 2
(b)
(c)
SO 2
(d) CO
133.
The hybridisation in BF3 molecule is
(a)
N 2O
Among the following compounds the one that is polar and has
central atom with sp  hybridization is
2
134.
[MP PMT 2004; IIT 1997]
124.
(d) C2 H 4
(d)
[UPSEAT 2003]
120.
SO 2
[MP PMT 2004]
In a regular octahedral molecule, MX 6 , the number X  M  X
bonds at 180° is
[CBSE PMT 2004]
(a) Six
(b) Four
(c) Three
(d) Two
Which one of the following is a planar molecule
[EAMCET 2003]
NO 3
126.
[MP PET 2003]
(a)
(d)
Which of the following has a linear structure
(a) CCl 4
(b) C2 H 2
(c)
Which of the following has dsp 2 hybridization
NiCl 42 
SO 3
(a)
H 2 CO 3
(b)
BF3
(c)
SiF4
(d)
HClO2
The molecule which is pyramid shape is
[MP PMT 2004; EAMCET 1985; IIT 1989]
(a)
PCl3
(b) CO 32 
135.
The molecule of CO 2 has 180° bond angle. It can be explanid on
the basis of
[AFMC 2004]
(a)
sp 3 hybridisation
(b)
sp 2 hybridisation
(c)
sp hybridisation
(d) d 2 sp3 hybridisation
sp 3 hybridisation is found in
[Pb. CET 2003; Orissa JEE 2005]
Chemical Bonding
136.
(a)
CO 32 
(b)
BF3
(c)
NO 3
(d)
NH 3
Which set hydridisation is correct for the following compounds[Pb. CET 2003]
NO 2 ,
137.
146.
SF4
2
PF6
147.
139.
[MADT Bihar 1983]
(a)
sp ,
sp ,
sp
(b)
sp ,
sp 3 d ,
sp 3 d 2
(c)
sp 2 ,
sp 3 ,
d 2 sp 3
(d)
sp 3 ,
sp 3 d 2 ,
sp 3 d 2
(a)
(b)
(c)
(d)
148.
The state of hybridisation of B in BCl 3 is
(a)
sp
(b)
sp 2
(c)
sp 3
(d)
sp 2 d 2
sp 3 d
(b)
sp 3
(c)
sp 3 d 2
(d)
sp 2
(a)
149.
[DCE 2004]
150.
(a)
PCl3
(b)
SO 3
(c)
CO 32 
(d)
NO 3
(a)
(c)
141.
142.
sp d
d 2 sp 3
(b)
(d)
152.
SnCl 2
(b)
NCO
(c)
CS 2
(d)
NO 2
Hybridisation of central atom in NF3 is
sp3
(b)
[Orissa JEE 2005]
sp
(c) sp
(d) dsp2
The pair having similar geometry is
[J&K CET 2005]
(a) PCl3 , NH 3
(b) BeCl 2 , H 2O
(d)
CH 4 , CCl 4
IF5 , PF5
3
The d-orbital involved in sp d hybridisation is
[J&K CET 2005]
sp 3
(a)
HF
(b)
HI
(c)
CH 4
(d)
PH 3
sp 3 -hybridization and tetrahedral geometry
(b)
sp 3 -hybridization and distorted tetrahedral geometry
(c)
sp 2 -hybridization and triangular geometry
143.
Be in BeCl 2 undergoes
144.
(a) Diagonal hybridization
(b) Trigonal hybridization
(c) Tetrahedral hybridization
(d) No hybridization
Which of the following is non-linear molecule
[MH CET 2004]
(a)
CO 3
(b) CO 2
(c)
CS 2
(d)
[DCE 2003]
(c)
dz 2
(b) d xy
(d) d zx
Which one in the following is not the resonance structure of CO 2
(a)
2.
3.
4.
BeCl 2
The trigonal bipyramidal geometry results from the hybridisation[UPSEAT 2004]
5.
(b) dsp 2 or sp 2 d
d x 2 y 2
Resonance
1.
(a)
(a)
`
Geometry of ammonia molecule and the hybridization of nitrogen
involved in it are
[MH CET 2004]
dsp 3 or sp 3 d
[IIT 1991]

sp d
In which compound, the hydrogen bonding is the strongest in its
liquid phase
[Pb. CET 2001]
(a)
sp 2  hybrid
2
(d) None of these
145.
sp  hybrid
(a)
(c)
[Pb. CET 2001]
(b)
2
2
151.
The hybrdization of IF7 is
3
sp d  hybrid
3
(c) sp d  hybrid
(d)
The linear structure is assumed by
(a)
Which of the following molecules has pyramidal shape
3
If a molecule MX 3 has zero dipole moment, the sigma bonding
orbital used by M are
3
The hybrid state of sulphur in SO 3 molecule is
(a)
Orbitals of different energy levels
Orbitals of different energy content
Orbitals of same energy content
None of the above
[IIT 1981; MP PMT 1994; Kerala PMT 2004]
[DCE 2004; J&K CET 2005]
140.
(c) d 2 sp 3 or sp 3 d 2
(d) d 3 sp 2 or d 2 sp 3
The valency of carbon is four. On what principle it can be explained
in a better way
(a) Resonance
(b) Hybridization
(c) Electron transfer
(d) None of the above
Hybridization is due to the overlapping of
3
[Pb. CET 2000; BHU 2004]
138.
115
OCO
(b)

O  C  O
(c)  O  C  O 
(d) O  C  O
Which of the following molecule contains one pair of non-bonding
electrons
(a)
CH 4
(b)
NH 3
(c)
H 2O
(d)
HF
Resonance is due to
[NCERT 1981; Kurukshetra CEE 1998]
(a) Delocalization of sigma electrons
(b) Delocalization of pi electrons
(c) Migration of H atoms
(d) Migration of protons
Resonating structures have different [AMU 1983]
(a) Atomic arrangements
(b) Electronic arrangements
(c) Functional groups
(d) Alkyl groups
In the cyanide ion, the formal negative charge is on
[AMU 1984]
116 Chemical Bonding
7.
(a) C
(b) N
(c) Both C and N
(d) Resonate between C and N
Which does not show resonance
(a) Benzene
(b) Aniline
(c) Ethyl amine
(d) Toluene
The enolic form of acetone contains
8.
(a) 9 sigma bonds, 1 pi bond and 2 lone pairs
(b) 8 sigma bonds, 2 pi bonds and 2 lone pairs
(c) 10 sigma bonds, 1 pi bond and 1 lone pair
(d) 9 sigma bonds, 2 pi bonds and 1 lone pair
Point out incorrect statement about resonance
6.
(c)
4.
[CPMT 1990]
5.
[IIT 1990; Bihar MEE 1997]
6.
7.
(a) 2
(b) 3
(c) 6
(d) 9
Resonance hybrid of nitrate ion is
[RPET 2000]
O ........ N ........O 1/ 2 (b)
2 / 3

1/ 3
8.

11.

O ...... N
In compound X , all the bond angles are exactly 109 o 28 ' , X is
The shape of SO 42  ion is
(a) Square planar
(b) Tetrahedral
(c) Trigonal bipyramidal
(d) Hexagonal
Which of the following molecules has one lone pair of electrons on
the central atom
......

(a)
NH 2
(b) CH 4
(c)
C2 H 2
(d)
CHF3
(b) CHCl 3
(c)
11.
CHBr3
(d) All have maximum bond angle
Of the following species the one having a square planar structure is
[NCERT 1981; MP PMT 1994]
12.
(a)
NH 4
(b)
BF4
(c)
XeF4
(d)
SCl 4
In which of the following is the angle between the two covalent
bonds greatest
[NCERT 1975; AMU 1982; MNR 1987;
IIT 1981; CPMT 1988; MP PMT 1994]
3.
90 o
(b) CH 4
(c)
NH 3
(d)
H 2O
14.
XeF2 molecule is
15.
(b) 105 o
(c) 109 o
(d) 120 o
In which molecule are all atoms coplanar
(a) CH 4
(b) BF3
CO 2
As the s-character of hybridized orbital decreases, the bond angle
(a) Decreases
(b) Increases
(c) Does not change
(d) Becomes zero
[NCERT 1983; MP PMT 1983]
(a)
(a)
13.
VSEPR Theory
2.
H 2O
10.
(b) sp 2 hybridization of C atom
(c) Resonance stabilization
(d) Same bond angles
(a) Square planar
(b) Tetrahedral
(c) Distorted rectangle
(d) Octahedral
The bond angle in PH 3 would be expected to be close to
NH 3
(c) CH 4
(d) PCl5
Of the following compounds, the one having a linear structure is[NCERT 1981; CP
[Roorkee 1999]

The structure of Cu H 2 O 4  ion is
(b)
(a) Octahedral
(b) Distorted octahedral
(c) Planar
(d) Tetrahedral
Which has maximum bond angle
[CPMT 1993]
O2 / 3
CO 32  anion has which of the following characteristics
H 2O
(a) Bonds of unequal length
1.
(b) Iodoform
(d) Chloroform
XeF6 is

O2 / 3

O 1/3
(d) CH 4
9.

2 / 3
H 2O
BeF2
MP PMT 1985; AIIMS 1996]
O ........ N ........O 2 / 3

O 2 / 3
O ........ N ........O 1/3 (d)
(c)
(a)

O 1/ 2
(c)
(b)
[EAMCET 1980; AMU 1982; MNR 1989]
[MP PMT 2000]
1/ 2
NH 3
(a)
The number of possible resonance structures for CO 32  is
(a)
(a)
[CPMT 1982; DPMT 1983, 84, 96; Bihar MEE 1997]
(a) Resonance structures should have equal energy
(b) In resonance structures, the constituent atoms should be in the
same position
(c) In resonance structures, there should not be the same number
of electron pairs
(d) Resonance structures should differ only in the location of
electrons around the constituent atoms
10.
NH 3
[NCERT 1973;
DPMT 1990; CBSE PMT 1990; UPSEAT 2003]
(a) Chloromethane
(c) Carbon tetrachloride
[MP PET 1997]
9.
(d)
PF3
Which has the least bond angle
[MP PMT 1994]
16.
[BHU 1982]
(a) Linear
(b) Triangular planar
(c) Pyramidal
(d) Square planar
Of the following sets which one does NOT contain isoelectronic
species
[AIEEE 2005]
(a) PO43  , SO 42  , ClO4
(b) CN  , N 2 , C 22 
(c) SO 32  , CO 32  , NO 3
(d)
BO33  , CO 32  , NO 3
A molecule which contains unpaired electrons is
Chemical Bonding
(b) C 2 H 2  NH 3  H 2 O  CH 4
[NCERT 1982]
(a) Carbon monoxide
(c) Molecular oxygen
17.
(a)
(b)
(c)
(d)
18.
19.
(b) Molecular nitrogen
(d) Hydrogen peroxide
H 2 O is
[MADT Bihar 1983]
A linear triatomic molecule
A bent (angular) triatomic molecule
Both of these
None of these
28.
29.
(a) Between 20  21%
(b) Between 19  20%
(c) Between 21  22%
(d) Between 22  23%
30.
The bond angle between H  O  H in ice is closest to
[CPMT 1989; UPSEAT 2002]
20.
31.
BCl 3 is a planar molecule while NCl 3 is pyramidal, because
(a)
22.
23.
24.
BCl 3 has no lone pair of electrons but NCl 3 has a lone pair
of electrons
(b) B  Cl bond is more polar than N  Cl bond
(c) Nitrogen atom is smaller than boron atom
(d) N  Cl bond is more covalent than B  Cl bond
The isoelectronic pair is
[AIIMS 2005]
(a)
Cl 2 O, ICl2
(c)
IF2 , I3
(b)
ICl2 ,ClO2
(d)
ClO2 ,
32.
33.
CIF2
According to VSEPR theory, the most probable shape of the
molecule having 4 electron pairs in the outer shell of the central
atom is
[MP PET 1996, 2001]
(a) Linear
(b) Tetrahedral
(c) Hexahedral
(d) Octahedral
The molecular shapes of SF4 , CF4 and XeF4 are
34.
[AIEEE 2005]
25.
(a) The same with 2, 0 and 1 lone pairs of electrons on the central
atom, respectively
(b) The same with 1, 1 and 1 lone pair of electrons on the central
atoms, respectively
(c) Different with 0, 1 and 2 lone pairs of electrons on the central
atom, respectively
(d) Different with 1, 0 and 2 lone pairs of electrons on the central
atom, respectively
Which of the following species is planar
[JIPMER 1997]
(a)
CO 32 
(c)
PCl 3
(b)
35.
The shape of
species is
27.
(a) Tetrahedral
(b) Square planar
(c) Trigonal planar
(d) Linear
Which of the following is the correct reducing order of bond-angle
(a) NH 3  CH 4  C 2 H 2  H 2 O
H 2S
(c) NH 3
(d) CH 4
A lone pair of electrons in an atom implies
[KCET 2002]
(a) A pair of valence electrons not involved in bonding
(b) A pair of electrons involved in bonding
(c) A pair of electrons
(d) A pair of valence electrons
The bond angle of water is 104.5 o due to
(a) Repulsion between lone pair and bond pair
[CPMT 2002]
sp 3 hybridization of O
(d) Higher electronegativity of O
The correct sequence of decrease in the bond angle of the following
hybrides is
[MP PET 2002]
(a)
NH 3  PH 3  AsH 3  SbH 3
(b)
NH 3  AsH 3  PH 3  SbH 3
(c)
SbH 3  AsH 3  PH 3  NH 3
(d)
PH 3  NH 3  AsH 3  SbH 3
[CBSE PMT 1995]
Central atom of the following compound has one lone pair of electrons and
three bond pairs of electrons
[JIPMER 2002]
(a)
H 2S
(b)
AlCl3
(c)
NH 3
(d)
BF3
Among KO 2 , AlO2 , BaO2
present in
(a)
NO 2
(c)
KO 2 only
and BaO2
and
NO 2 unpaired electron is
[MP PET 2003]
AlO2
(b)
KO 2 and
(d)
BaO2 only
True order of bond angle is
[RPET 2003]
(a)
H 2 O  H 2 S  H 2 Se  H 2 Te
(b)
H 2 Te  H 2 Se  H 2 S  H 2 O
(c)
H 2 S  H 2 O  H 2 Se  H 2 Te
(d)
H 2 O  H 2 S  H 2 Te  H 2 Se
Which of the following has not a lone pair over the central atom
(a)
NH 3
(b)
PH 3
(c)
BF3
(d)
PCl 3
In BrF3 molecule,the lone pairs occupy equatorial positions to
minimize
[CBSE PMT 2004]
(a) Lone pair- lone pair repuilsion and lone pair-bond pair
repulsion
(b) Lone pair- lone pair repulsion only
(c) Lone pair- bond pair repulsion only
(d) Bond pair- bond pair repulsion only
37.
H 2 O is dipolar, whereas BeF2 is not. It is because
NH 2
26.
(b)
H 2O
36.
(d) None of these
CH 3
H 2 O  NH 3  CH 4  C 2 H 2
Which compound has bond angle nearly to 90
(c) Bonding of H 2 O
(d) CO 2
BeH 2
(d)
(b)
o
(c) 90 o
(d) 105 o
Which of the following molecules does not have a linear
arrangement of atoms
[CBSE PMT 1989]
(a) H 2 S
(b) C 2 H 2
(c)
21.
(b) 60
120 28 '
NH 3  H 2 O  CH 4  C 2 H 2
[Pb. PMT 2001]
Bond angle between two hybrid orbitals is 105 .% s-orbital
character of hybrid orbital is
[MP PMT 1986]
(a)
(c)
(a)
o
o
117
[RPET 1999]
[BHU 2000]
(a)
H 2 O is linear and BeF2 is angular
[CBSE PMT 1989; 2004]
118 Chemical Bonding
(b)
H 2 O is angular and BeF2 is linear
(c) The electornegativity of F is greater than that of O
(d)
38.
39.
40.
H 2 O involves hydrogen bonding whereas BeF2 is a discrete
molecule
Maximum bond angle is present in
[BVP 2004]
(a)
BCl 3
(b)
(c)
BF3
(d) Same for all
(a)
42.
(d)
H 2O
91 8 
o
6.
7.
[EAMCET 1980]
o
44.
45.
(a)
(c)
SiF4 and SF4
and
NH 4
(b)
PF6 and
2.
3.
4.
O2
(b) O21
(c)
O21
(d) O22
Which of the following compounds of boron does not exist in the
free form
(a)
BCl 3
(b)
BF3
(c)
BBr3
(d)
BH 3
Molecular orbital theory was developed mainly by
(d)
IO3 and XeO 3
[IIT 1989; CBSE PMT 1995]
(a)
SF6
The maximum number of 90° angles between bond pair-bond pair
of electrons is observed in
[AIEEE 2004]
(a)
dsp 2 hybridization
(b)
sp 3 d hybridization
(c)
dsp3 hybridization
(d)
sp 3 d 2 hybridization
O 2
(b)
(c) Both (a) and (b)
NO
(d) CN 
11.
The bond order in N 2 ion is
12.
(a) 1
(b) 2
(c) 2.5
(d) 3
Out of the following which has smallest bond length
Molecular orbital theory
1.
(a)
10.
NH 3
Among the following the pair in which the two species are not
isostructural is
[CBSE PMT 2004]
BH 4
The bond order is maximum in
[NCERT 1984; IIT 1984]
[CBSE PMT 1990]
(d) CO 2
H 2O
(d) O21
(a) Bonding electrons are more than antibonding electrons
(b) Contains unpaired electrons
(c) Bonding electrons are less than antibonding electrons
(d) Bonding electrons are equal to antibonding electrons
Which one is paramagnetic from the following
(c) 106 o 45'
(d) 109 o 28'
Which of the following gives correct arrangement of compounds
involved based on their bond strength
(c)
O21
9.
8.
[BHU 2005]
43.
(c)
[BHU 1987; Pb. CET 2003]
(b) 93 3'
(a) HF > HCl > HBr > HI
(b) HI > HBr > HCl > HF
(c) HF > HBr > HCl > HI
(d) HCl > HF > HBr > HI
Which one has a pyramidal structure
(a) CH 4
(b)
(b) O22
(a) Pauling
(b) Pauling and Slater
(c) Mulliken
(d) Thomson
The bond order of a molecule is given by
[NCERT 1984]
(a) The difference between the number of electrons in bonding
and antibonding orbitals
(b) Total number of electrons in bonding and antibonding orbitals
(c) Twice the difference between the number of electrons in
bonding and antibonding electrons
(d) Half the difference between the number of electrons in bonding
and antibonding electrons
Oxygen molecule is paramagnetic because
PH 3
The bond angle in ammonia molecule is
O2
[AIIMS 1983, 85; CBSE PMT 1994; MP PET 2002]
BBr3
The shape of a molecule of NH 3 , in which central atoms contains
lone pair of electron, is
[MHCET 2003]
(a) Tetrahedral
(b) Planar trigonal
(c) Square planar
(d) Pyramidal
The largest bond angle is in
[DCE 2002; MNR 1984]
(a) AsH3
(b) NH 3
(c)
41.
5.
(a)
Bond order is a concept in the molecular orbital theory. It depends
13.
on the number of electrons in the bonding and antibonding orbitals.
Which of the following statements is true about it ? The bond order[AIIMS 1980]
(a) Can have a negative quantity
(b) Has always an integral value
14.
(c) Can assume any positive or integral or fractional value
including zero
(d) Is a non zero quantity
The bond order of NO molecule is [MP PET 1996]
15.
(a) 1
(b) 2
(c) 2.5
(d) 3
When two atomic orbitals combine they form
(a) One molecular orbital
(b) Two molecular orbital
16.
(c) Three molecular orbital
(d) Four molecular orbital
Which of the following species is the least stable
[Pb. CET 2004]
[RPMT 1997]
O 2
(a)
O2
(b)
(c)
O 2
(d) O 22 
Which of the following molecule is paramagnetic
[CPMT 1980; RPET 1999;MP PMT 1999; RPMT 2000]
(a) Chlorine
(b) Nitrogen
(c) Oxygen
(d) Hydrogen
Which molecule has the highest bond order
(a) N 2
(b) Li 2
(c)
(d) O 2
He 2
The molecular electronic configuration of H 2 ion is
(a)
 1s 2
(b)


 1s 2  x 1s 2
2
3
(c)  1s   x 1s
(d)  1s 
The paramagnetic nature of oxygen molecule is best explained on
the basis of
[BHU 1996]
(a) Valence bond theory
(b) Resonance
1
Chemical Bonding
17.
18.
(c) Molecular orbital theory
(d) Hybridization
In which case the bond length is minimum between carbon and
nitrogen
(a) CH 3 NH 2
(b) C 6 H 5 CH  NOH
(c) Equal to that of 2 s orbital
29.
[AIEEE 2005]
(a)
(b) H
2
H 2
19.
(c)
(d) H 2
Which one of the following oxides is expected exhibit paramagnetic
behaviour
[CBSE PMT 2005]
(a) CO 2
(b) SO 2
(c)
20.
(d)
ClO2
30.
SiO2
The bond order in N 2 molecule is
31.
[CBSE 1995; Pb. PMT 1999; MP PET 1997]
21.
(a) 1
(b) 2
(c) 3
(d) 4
Which one is paramagnetic and has the bond order 1/2
[NCERT 1983]
32.
(c) F2
(d) H 2
When two atoms of chlorine combine to form one molecule of
chlorine gas, the energy of the molecule
[AMU 1982]
(a) Greater than that of separate atoms
(b) Equal to that of separate atoms
(c) Lower than that of separate atoms
(d) None of the above statement is correct
An atom of an element A has three electrons in its outermost shell
and that of B has six electrons in the outermost shell. The formula
of the compound between these two will be
33.
(a)
22.
23.
O2
(b)
N2
[CPMT 1974, 84; RPMT 1999]
24.
(a)
A3 B4
(b)
A2 B3
(c)
A3 B2
(d)
A2 B
25.
The bond order of individual carbon-carbon bonds in benzene is
(a) One
(b) Two
(c) Between 1 and 2
(d) One and two alternately
PCl5 exists but NCl 5 does not because
26.
(a) Nitrogen has no vacant d-orbitals
(b) NCl 5 is unstable
(c) Nitrogen atom is much smaller
(d) Nitrogen is highly inert
Paramagnetism is exhibited by molecules
27.
(a) Not attracted into a magnetic field
(b) Containing only paired electrons
(c) Carrying a positive charge
(d) Containing unpaired electrons
Which one of the following is paramagnetic
34.
35.
[NCERT 1979; MP PET 2002]
28.
H 2O
(b)
(c)
SO 2
(d) CO 2
36.
H 2
(b)
He 2
(c)
He 2
(d)
Li 2
In P4 O10 , the number of oxygen atoms attached to each
phosphorus atom is
[IIT 1995]
(a) 2
(b) 3
(c) 4
(d) 2.5
Of the following statements which one is correct
(a) Oxygen and nitric oxide molecules are both paramagnetic
because both contain unpaired electrons
(b) Oxygen and nitric oxide molecules are both diamagnetic
because both contain no unpaired electrons
(c) Oxygen is paramagnetic because it contains unpaired electrons,
while nitric oxide is diamagnetic because it contains no
unpaired electrons
(d) Oxygen is diamagnetic because it contains no unpaired
electrons, while nitric oxide is paramagnetic because it contains
an unpaired electron
According to the molecular orbital theory, the bond order in C 2
molecule is
(a) 0
(b) 1
(c) 2
(d) 3
[IIT 1980]orbital configuration of a diatomic molecule is
The molecular
Its bond order is
(a) 3
(b) 2.5
(c) 2
(d) 1
The difference in energy between the molecular orbital formed and
the combining atomic orbitals is called
(a) Bond energy
(b) Activation energy
(c) Stabilization energy
(d) Destabilization energy
According to molecular orbital theory, the paramagnetism of O 2
molecule is due to presence of
[MP PMT 1997]
(b) Unpaired electrons in the antibonding  molecular orbital
(c) Unpaired electron in the bonding  molecular orbital
(d) Unpaired electrons in the antibonding  molecular orbital
[AMU 1983]
(b) More than that of 2 s orbital
(a)
(a) Unpaired electrons in the bonding  molecular orbital
[DPMT 1985]
NO 2
The energy of a 2 p orbital except hydrogen atom is
(a) Less than that of 2 s orbital
(a) 16 shared and 8 unshared electrons
(b) 8 shared and 16 unshared electrons
(c) 12 shared and 12 unshared electrons
(d) 18 shared and 6 unshared electrons
Which of the following does not exist on the basis of molecular
orbital theory
[AFMC 1990; MP PMT 1996]
  2 py2
 1s 2  * 1s 2  2 s 2  * 2 s 2  2 p x2 
2
  2 pz
[EAMCET 1977; MP PET/PMT 1988]
(a)
(d) Double that of 2 s orbital
In the electronic structure of acetic acid, there are
[AMU 1983]
(c) CH 3 CONH 2
(d) CH 3 CN
Which one of the following species is diamagnetic in nature
He 2
119
37.
The bond order in O 2 is
(a) 2
(c) 1.5
[MP PET 1999; BHU 2001]
(b) 2.5
(d) 3
120 Chemical Bonding
38.
39.
Which of the following is paramagnetic [MP PET 1999]
(a)
O2
(b) CN
(c)
CO
(d)
50.

[Pb. PMT 2000]
NO 
41.
42.
44.
Nx  Ny
(a) 1
(c)
Nx  Ny
(d)
Nx  Ny
(c)
(a)
 2s
(b)  2 p y
(c)
 * 2py
(d)  * 2 p x
(a) Ferromagnetism
(b) Diamagnetism
(c) Paramagnetism
(d) None of these
What is correct sequence of bond order
(a)
O 2
(c)
O 2  O 2  O 2

O 2
 O2
(b)
O 2
[BHU 1997]
 O2 
(c)
S
O 2
55.
[RPMT 1997]
(b)
(b)
(d)
N2
47.
(b)
SO 2
(c)
NO
(d)
H 2O
(c)
NO 2
(d)
NO
In which of the following pairs the two molecules have identical
bond order
[MP PMT 2000]
(a)
N 2 , O 22 
(c)
N 2
(a)
N 2
(b) O 22 
(c)
N2
(d)
O2
The bond order is not three for
,
(c)
58.
[CBSE PMT 1997]
60.
(b) In O 2 , the O  O bond order increases
90 o
Cu

61.
(b) Decreases
(d) None of these
49.
(a) Remains unaltered
(c) Increases
Which is not paramagnetic
(a)
O2
(b) O 2
(c)
O 22 
(d) O 2
[CET Pune 1998]
[DCE 1999, 2000]
,
N2
[MP PMT 2001]
NO 
(b) 101 o
(b) Cu 
(d)
H2
(a) 3s
(b) 1p, 2s
(c) 2p, 1s
(d) 3p
In which of the following pairs molecules have bond order three and
are isoelectronics
[MP PET 2003]
(a)
CN  , CO
(b)
(c)
CN  , O 2
(d) CO, O 2
62.
NO  , CO 

Which of the following is paramagnetic [MP PET 2003]
O 2
(c) CO
(c) In O 2 , bond length increases
With increasing bond order, stability of a bond
,
Which of the following is correct for N 2 triple bond
(a)
48.
(d)
O 2
[CPMT 2002]
59.
N 2 becomes diamagnetic
N 2 O 2
In H 2 O 2 molecule, the angle between the two O – H planes is
(a) Copper crystals
(a) In N 2 , the N  N bond weakens
(b)
57.
N 2 and O 2 are converted into monoanions N 2 and O 2
respectively, which of the following statements is wrong
(d)
(b)
[AIIMS 2002]
[Pb. PMT 1998]
CO 2
ClO2
(a) F  F
(b) C – C
(c) N – N
(d) O – O
Which of the following species would be expected paramagnetic
NO
(a)
(a)
56.

Which one of the following molecules is paramagnetic
[DPMT 2000]
ClO2
(c) 103 o
(d) 105 o
Which of the following molecule has highest bond energy
Br  F
[AIIMS 1997]
(b) 2
1
1
(d)
2
4
Which one
does
not exhibit paramagnetism
[KCET
1996]
(a)
(d) O 2  O 2  O 2
(c) Cl  F
(d) I  F
Which of the following is not paramagnetic
2
54.
O 2
Which bond is strongest
FF
52.
53.
The number of nodal planes ' d ' orbital has
[KCET 1996]
(a) Zero
(b) One
(c) Two
(d) Three
Atomic number of an element is 26. The element shows
(a)
46.
[Pb. PMT 2000; Pb CET 2001]
(b)
Which of the following molecular orbitals has two nodal planes
(b) 3
(d) 5
The bond order of He 2 molecule ion is
Nx  Ny
(a)
45.
51.
(a)
[CPMT 1996]
43.
(a) 4
(c) 2
If N x is the number of bonding orbitals of an atom and N y is the
number of antibonding orbitals, then the molecule/atom will be
stable if
[DPMT 1996]
40.
The number of antibonding electron pairs in O 22  molecular ion on
the basis of molecular orbital theory is
(b) CN 
(d)
N2
How many bonding electron pairs are there in white phosphorous
(a) 6
(b) 12
(c) 4
(d) 8
ˆ P bond angle in the
The atomicity of phosphorus is X and the PP
molecule is Y. What are X and Y [EAMCET 2003]
(a) X = 4, Y = 90 o
(b) X = 4, Y = 60 o
(c) X = 3, Y = 120 o
(d) X = 2, Y = 180 o
Chemical Bonding
63.
From elementary molecular orbital theory we can give the electronic
configuration of the singly positive nitrogen molecular ion
(a)


N 2
2
2
2
4
as
73.
 (1s)2  (1s)2  (2 s)2  (2 p)2  (2 p)4


2
2
2
2
(a)
( 2 p x ) and ( 2 p x )
(b)
1
( 2 p x )1 and ( 2 p y )
(c)
( * 2 p y )1 and ( * 2 p z )1
*
74.
65.
75.
(d)
[DPMT 2004]
77.
0.75, 1.0
78.
[Pb. PMT 2004]
(a) Zero
(c) 1.33
67.
(c)
69.
(b) CN
(d)
CO

80.
[DPMT 2004]
(c)
81.
 O2 
O2
O2
O2
O2 

(b)
O2
O2

(d)
O2
 O2  O2
 O2
(a) Bond length in NO  is equal to that in NO
I2
O 22
(b) O 22 
O2
(c) O 22 
(d) O 2
Which of the following species have maximum number of unpaired
electrons
[AIIMS 1983]
(b) O 2
O2
(c) O 2
(d) O 22 
The correct order in which the O – O bond length increases in the
following is
[BHU 2000; CBSE PMT 2005]
(a) H 2 O2  O2  O3
(b) O2  H 2 O2  O3
(c) O2  O3  H 2 O2
PETof2004]
Correct[MP
order
bond length is
CO 32   CO 2  CO
(d) O3  H 2 O2  O2
[Orissa JEE 2005]
(b) CO 2  CO  CO 32 
84.
(d) Dimagnetic and bond order> O 2
[AIEEE 2004]
(d)
Br2
[DCE 2002]
(c) N 2
(d) O22 
Among the following molecules which one have smallest bond angle[Orissa JEE 2
(a) NH 3
(b) PH3
(c) Dimagnetic and bond order< O 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
F2
83.
(b) Paramagnetic and bond order> O 2
72.
(d)
(c) CO  CO 2  CO 32 
(d) None of these
Which of the following is paramagnetic
(a) N 2
(b) C 2
[IIT JEE Screening 2004]
(a) Paramagnetic and bond order< O 2
NO  NO  NO 
In
molecular species, the total number of
O 2 and
antibonding electrons respectively are
[DCE 2003]
(a) 7, 6, 8
(b) 1, 0, 2
(c) 6, 6, 6
(d) 8, 6, 8
Which of the following is not paramagnetic
[DCE 2002]
(a)
According to molecular orbital theory which of the following
statement about the magnetic character and bond order is correct
regarding O 2
82.
(b)


N2
O 2 ,
(a)
(b) 4
(d) 10
The bond length the species O2 , O2 and O 2 are in the order of
O2
NO  NO  NO

(c) N 2
(d) CO
Which has the highest bond energy
(a) F2
(b) Cl 2
(a)

(a) 2
(b) 1.5
(c) 3
(d) 3.5
The total number of electron that takes part in forming bonds in
[MP PET 2004]
N 2 is
(a)
71.
NO
[CPMT 2004]
Bond order of O 2 is
(a) 2
(c) 6
70.
79.
The bond order of O 2 is the same as in
(a)
68.
(b) 0.88
(d) 2
N 2

The paramagnetic molecule at ground state among the following is[UPSEAT 200
(a) H 2
(b) O 2
(c)
3, 1.25
The bond order in CO 32  ion between C  O is
and NO  are such as
(c) NO  NO  NO
(d) NO   NO   NO
Which of the following is paramagnetic [UPSEAT 2004]
(a) B 2
(b) C 2
(c)
76.
P  O bond order respectively are
(a) 0.75, 1.25
(b)
66.
(a)
( * 2 p z )1 and ( 2 p z )1
0.75, 0.6
Bond energies in NO, NO

In PO43  ion, the formal charge on each oxygen atom and
(c)
(d) None

[Pb. CET 2004]
1
(d) ( * 2 p y )1 and ( 2 p y )1
(e)
N 2
2
[Kerala PMT 2004]
1
(c) Bond length in NO  is greater than in NO
(d) Bond length is unpredictable
Which of the following is diamagnetic [BVP 2004]
(a) Oxygen molecule
(b) Boron molecule
(c)
(d)  (1s)  (1s)  (2 s)  (2 s)  (2 p)  (2 p)
The paramagnetic property of the oxygen molecule due to the
presence of unpaired electorns present in
2
64.
[UPSEAT 2003]
1
(b)  (1s)2  (1s)2  (2 s)2  (2 s)2  (2 p)1 (2 p)3
(c)
(b) Bond length in NO is greater than in NO 
 (1s)  (1s)  (2 s)  (2 s)  (2 p)  (2 p)
2
121
(c)
H 2O
(e)
H 2S
(d)
[DPMT 2005]
H 2 Sc
Hydrogen bonding
1.
In the following which bond will be responsible for maximun value
of hydrogen bond
(a) O  H
(b) N  H
122 Chemical Bonding
2.
3.
(c) S  H
(d) F  H
In which of the following hydrogen bond is present
(a) H 2
(b) Ice
(c) Sulphur
(d) Hydrocarbon
In the following which has highest boiling point
13.
[NCERT 1978; DPMT 1985; CBSE PMT 1992]
(a) Hydrogen bonds
(c) Phosphate groups
[MP PMT 1989; RPMT 1997]
(a)
4.
5.
HI
(b)
(c) HBr
Which contains hydrogen bond
(a) HF
(d)
7.
H  bonding
(b)
HCl
(d) None of the above
15.
[MP PMT 1989]
HCl
16.
NH 3
(b) CH 4
(c)
H 2O
(d) CO 2
As a result of sp hybridization, we get
(a)
SiH 4
(b)
LiH
(c)
HI
(d)
NH 3
17.
18.
Which one among the following does not have the hydrogen bond
In the following which species does not contain sp3 hybridization
(a)
In which of the following compounds does hydrogen bonding occur
Which among the following compounds does not show hydrogen
bonding
[MP PMT 1989]
(a) Chloroform
(b) Ethyl alcohol
(c) Acetic acid
(d) Ethyl ether
Acetic acid exists as dimer in benzene due to
[CPMT 1982]
(a) Condensation reaction
[DPMT 1985]
(b) Hydrogen
bonding
(c) Presence of carboxyl group
(d) Presence of hydrogen atom at   carbon
HF is a weak acid
[IIT 1984]
MP PMT 1993; AIIMS 1996; KCET 2001; CPMT 2003]
(a) Its high specific heat
(b) Its high dielectric constant
(c) Low ionization of water molecules
(d) Hydrogen bonding in the molecules of water
Which concept best explains that o-nitrophenol is more volatile than
p-nitrophenol
19.
20.
(b) Liquid NH 3
(c) Water
(d) Liquid HCl
The bond that determines the secondary structure of a protein is[NCERT 1984; M
(a) Coordinate bond
(b) Covalent bond
(c) Hydrogen bond
(d) Ionic bond
HCl is a gas but HF is a low boiling liquid. This is because
[NCERT 1984; MP PMT 2001]
(a)
H  F bond is strong
(b) H  F bond is weak
(c) Molecules aggregate because of hydrogen bonding
(d)
21.
HF is a weak acid
The relatively high boiling point of HF is due to
[AIIMS 1980, 82; Kurukshetra CEE 1998; MP PET 2002]
(a) Resonance
(c) Hydrogen bonding
10.
[NCERT 1984]
(b) Hyperconjugation
(d) Steric hindrence
(a) Hydrogen bonding
(b) Covalent bonding
Which contains strongest H  bond
[IIT 1986; MP PET 1997, 2003; UPSEAT 2001, 03]
(a)
11.
O  H..... S
(b)
S  H..... O
22.
(c) F  H..... F
(d) F  H..... O
Which of the following compound can form hydrogen bonds
[NCERT 1978; MP PMT 1997]
12.
(a)
CH 4
(b)
NaCl
(c)
CHCl 3
(d)
H 2O
23.
(c) Unshared electron pair on F
(d) Being a halogen acid
Water is liquid due to
[MADT Bihar 1983]
(a) Hydrogen bonding
(b) Covalent bond
(c) Ionic bond
(d) Vander Waals forces
The maximum possible number of hydrogen bonds in which an
H 2 O molecule can participate is
[MP PMT 1986; MNR 1991; IIT 1992;MP PET 1999]
Of the following hydrides which has the lowest boiling point
[CBSE PMT 1987]
(a)
NH 3
(b)
PH 3
(c)
SbH 3
(d)
AsH 3
[
UPSEAT 2001]
(a) Phenol
(b) Two orbitals at 180 o
(c) Four orbitals in tetrahedral directions
(d) Three orbitals in the same plane
The reason for exceptionally high boiling point of water is
[DPMT 1986; NCERT 1976; AMU 1984; EAMCET 1979;
9.
[AFMC 1982]
(c) Ionic bonding
(a) Two mutual perpendicular orbitals
8.
Water has high heat of vaporisation due to
(a) Covalent bonding
(c) HF molecule are hydrogen bonded
(d) Fluorine is highly reactive
6.
14.
(b) Ionic bonds
(d) Deoxyribose groups
HF
(b)
(c) HBr
(d) HI
Contrary to other hydrogen halides, hydrogen fluoride is a liquid
because
[MP PMT 1990; AMU 1983; EAMCET 1980]
(a) Size of F atom is small
(b)
The pairs of bases in DNA are held together by
24.
(a) 1
(b) 2
(c) 3
(d) 4
Hydrogen bonding is maximum in
[IIT 1987; MP PMT 1991; MP PET 1993, 2001;
MNR 1995; CPMT 1999; KCET (Med.) 2002]
(a) Ethanol
(b) Diethyl ether
Chemical Bonding
25.
(c) Ethyl chloride
(d) Triethyl amine
The hydrogen bond is strongest in
35.
[BHU 1987; CBSE PMT 1990, 92]
(a) Water
(c) Hydrogen fluoride
26.
(b) Ammonia
(d) Acetic acid
27.
36.
[MP PMT 1993]
(a) Hydrogen bonding
(b) Ionic bonding
(c) Coordinate covalent bonding
(d) Resonance
Methanol and ethanol are miscible in water due to
37.
Covalent character
Hydrogen bonding character
Oxygen bonding character
None of these
29.
(a) Vander Waal's forces
(b) Covalent bond
(c) Hydrogen bond
(d) Ionic bond
Strength of hydrogen bond is intermediate between
39.
33.
H 2O
(b)
(c)
NH 3
(d) C 6 H 6
NH 3
(b) P
[KCET 2001]
(a) Vander Waal and covalent
(b) Ionic and covalent
(c) Ionic and metallic
(d) Metallic and covalent
In which of the following compounds intramolecular hydrogen bond
is present
[MP PET 1994]
(a) Ethyl alcohol
(b) Water
(c) Salicylaldehyde
(d) Hydrogen sulphide
Hydrogen bonding is formed in compounds containing hydrogen
and
[MP PET 1995]
(a) Highly electronegative atoms
(b) Highly electropositive atoms
(c) Metal atoms with d-orbitals occupied
(d) Metalloids
Which of the following compounds in liquid state does not have
hydrogen bonding
[MP PMT 1996]
(a)
HF    HCl
(c) As
(d) Sb
The boiling point of a compound is raised by
[DPMT 2001]
(a) Intramolecular hydrogen bonding
(b) Intermolecular hydrogen bonding
(c) Covalent bonding
(d) Ionic covalent
The boiling point of water is exceptionally high because
(a)
(b)
(c)
(d)
[DPMT 1991]
32.
(b)
(c) HCl    HCl
(d) HF    HI
Which of the following shows hydrogen bonding
(a)
38.
B.P. of H 2 O (100 o C) and H 2 S (42 o C) explained by
31.
HF    HF
[CPMT 2000]
28.
30.
Which of the following hydrogen bonds are strongest in vapour
phase
[AMU 1999]
(a)
[CPMT 1989]
(a)
(b)
(c)
(d)
CH 3 group in ethanol
(d) CH 3 group in dimethyl ether
ether (23.6 o C) , though both having the same molecular formulae
C 6 H 6 O , is due to
Ethanol and dimethyl ether form a pair of functional isomers. The
boiling point of ethanol is higher than that of dimethyl ether due to
the presence of
[AIIMS 1998]
(a) Hydrogen bonding in ethanol
(b) Hydrogen bonding in dimethyl ether
(c)
The high boiling point of ethanol (78.2 o C) compared to dimethyl
123
HF
Compounds showing hydrogen bonding among HF, NH 3 , H 2 S
40.
Water molecule is linear
Water molecule is not linear
There is covalent bond between H and O
Water molecules associate due to hydrogen bonding
NH 3 has a much higher boiling point than PH 3 because
[UPSEAT 2002; MNR 1994]
(a)
NH 3 has a larger molecular weight
(b)
NH 3 undergoes umbrella inversion
(c)
NH 3 forms hydrogen bond
(d)
41.
42.
NH 3 contains ionic bonds whereas PH 3 contains covalent
bonds
Which one has the highest boiling point
[MP PET 2002]
(a) Acetone
(b) Ethyl alcohol
(c) Diethyl ether
(d) Chloroform
Which of the following compounds has the highest boiling point
(a) HCl
(b) HBr
(c)
H 2 SO 4
(d)
HNO 3
43.
Which of the following has minimum melting point
44.
(a) CsF
(b)
(c) HF
(d)
Hydrogen bond energy is equal to
(a) 3 – 7 cals
(b)
(c) 3 – 10 kcals
(d)
[UPSEAT 2003]
and PH 3 are
(a) Only HF, NH 3 and PH 3
(b) Only HF and NH 3
(c) Only NH 3 , H 2 S and PH 3
34.
(d) All the four
The high density of water compared to ice is due to
[CBSE PMT 1997; BHU 1999; AFMC 2001]
(a)
(b)
(c)
(d)
Hydrogen bonding interactions
Dipole-dipole interactions
Dipole-induced dipole interactions
Induced dipole-induced dipole interactions
45.
HCl
LiF
[UPSEAT 2003]
30 – 70 cals
30 – 70 kcals
H 2 O is a liquid while H 2 S is gas due to
(a) Covalent bonding
(b) Molecular attraction
(c) H – bonding
(d) H – bonding and molecular attraction
[BHU 2003]
124 Chemical Bonding
46.
47.
48.
H – bonding is maximum in
[BHU 2003]
(a)
C 6 H 5 OH
(b) C 6 H 5 COOH
(c)
CH 3 CH 2 OH
(d) CH 3 COCH 3
Select the compound from the following which dissolves in water
(a)
CCl 4
(b) CS 2
(c)
CHCl 3
(d) C2 H 5 OH
8.
9.
[CPMT 1994]
When two ice cubes are pressed over each other, they unit to form
one cube. Which of the following force is responsible for holding
them together
[NCERT 1978]
(a) Vander Waal's forces
(a)
10.
(b) Hydrogen bond formation
(c) Covalent attraction
(d) Dipole–dipole attraction
49.
Which is the weakest among the following types of bond
[NCERT 1979; MADT Bihar 1984]
50.
(a) Ionic bond
(b) Metallic bond
(c) Covalent bond
(d) Hydrogen bond
H-bond is not present in
11.
12.
[BCECE 2005]
(a) Water
(b) Glycerol
(c) Hydrogen fluoride
(d) Hydrogen Sulphide
(a) Overlapping valency orbitals
(b) Mobile valency electrons
(c) Delocalized electrons
(d) Highly directed bonds
In melting
[CPMT 1982]
[IIT lattice,
1980] structure of solid
(a) Remains unchanged
(b) Changes
(c) Becomes compact
(d) None of the above
Which of the following has the highest melting point
Pb
(b) Diamond
(c) Fe
(d) Na
In the formation of a molecule by an atom
(a) Attractive forces operate
(b) Repulsive forces operate
(c) Both attractive and repulsive forces operate
(d) None of these
Which has weakest bond
(a) Diamond
(b) Neon (Solid)
[AFMC 1995]
[RPMT 1997]
(c) KCl
(d) Ice
Which of the following exhibits the weakest intermolecular forces[AIIMS 1999; BH
(a)
He
(b)
HCl
(c)
NH 3
(d)
H 2O
13.
Glycerol has strong intermolecular bonding therefore it is
14.
(a) Sweet
(b) Reactive
(c) Explosive
(d) Viscous
Among the following the weakest one is
15.
(a) Metallic bond
(b) Ionic bond
(c) Van der Waal's force
(d) Covalent bond
Lattice energy of alkali metal chlorides follows the order
[RPET 2000]
Types of bonding and Forces in solid
1.
In a crystal cations and anions are held together by
[EAMCET 1982]
2.
3.
(a) Electrons
(b) Electrostatic forces
(c) Nuclear forces
(d) Covalent bonds
In the following metals which one has lowest probable interatomic
forces
[MP PMT 1990]
[Pb. PMT 2004; CPMT 2002]
[DPMT 2004]
(a) Copper
(b) Silver
(c) Zinc
(d) Mercury
(b) CsCl  NaCl  KCl  RbCl  LiCl
(c)
In solid argon, the atoms are held together by
[NCERT 1981; MP PET 1995]
4.
16.
(a) Ionic bonds
(b) Hydrogen bonds
(c) Vander Waals forces
(d) Hydrophobic forces
Which one is the highest melting halide
(a)
5.
NaCl
(b)
[AIIMS 1980]
NaBr
(c) NaF
(d) NaI
The enhanced force of cohesion in metals is due to
17.
[NCERT 1972]
6.
7.
LiCl  NaCl  KCl  RbCl  CsCl
(a)
(a) The covalent linkages between atoms
(b) The electrovalent linkages between atoms
(c) The lack of exchange of valency electrons
(d) The exchange energy of mobile electrons
Which one of the following substances consists of small discrete
molecules
[CPMT 1987]
(a) NaCl
(b) Graphite
(c) Copper
(d) Dry ice
Which of the following does not apply to metallic bond
[CBSE PMT 1989]
LiCl  CsCl  NaCl  KCl  RbCl
(d) NaCl  LiCl  KCl  RbCl  CsCl
In the following which molecule or ion possesses electrovalent,
covalent and co-ordinate bond at the same time
(a)
HCl
(b)
NH 4
(c)
Cl 
(d)
H 2 O2
Both ionic and covalent bond is present in the following
[MNR 1986; MP PMT 2004]
(a)
CH 4
(b)
KCl
(c)
SO 2
(d)
NaOH
18.
The formation of a chemical bond is accompanied by
19.
(a) Decrease in energy
(b) Increase in energy
(c) Neither increase nor decrease in energy
(d) None of these
Chemical bond implies
[MP PET 1995]
[KCET 2002]
Chemical Bonding
20.
21.
22.
23.
24.
(a) Attraction
(b) Repulsion
(c) Neither attraction nor repulsion
(d) Both (a) and (b)
Which of the following statements is true
[AIEEE 2002]
(a) HF is less polar than HBr
(b) Absolutely pure water does not contain any ions
(c) Chemical bond formation take place when forces of attraction
overcome the forces of repulsion
(d) In covalency transference of electron takes place
Which of the following statements is true about Cu NH 3 4 SO 4
(a) It has coordinate and covalent bonds
(b) It has only coordinate bonds
(c) It has only electrovalent bonds
(d) It has electrovalent, covalent and coordinate bonds
Blue vitriol has
(a) Ionic bond
(b) Coordinate bond
(c) Hydrogen bond
(d) All the above
125
[UPSEAT 2001]
(a)
N 2 H 5
(c) HCl
6.
7.
(b)
BaCl2
(d)
H 2O
Which combination is best explained by the co-ordinate covalent
bond [JIPMER 2001; CBSE PMT 1990]
(a)
H   H 2O
(c)
Mg 
(b) Cl + Cl
1
O2
2
H 2  I2
(d)
Arrange the following compounds in order of increasing dipole
[CPMT 1988]
moment.
(I) Toluene
(II) m  dichlorobenzene
(III) o  dichlorobenzene
(IV) p  dichlorobenzene
[IIT 1996]
(a)
8.
[MP PET 1995]
IV  I  II  III
(b)
(c) IV  I  III  II
(d) IV  II  I  III
The correct order of dipole moment is [Roorkee 1999]
(a)
The number of ionic, covalent and coordinate bonds in NH 4 Cl
are respectively
[MP PMT 1999]
(a) 1, 3 and 1
(b) 1, 3 and 2
(c) 1, 2 and 3
(d) 1, 1 and 3
Covalent molecules are usually held in a crystal structure by
I  IV  II  III
CH 4  NF3  NH 3  H 2 O
(b) NF3  CH 4  NH 3  H 2 O
9.
(c)
NH 3  NF3  CH 4  H 2 O
(d)
H 2 O  NH 3  NF3  CH 4
Which of the following has the highest dipole moment
[AIIMS 2002]
(a) Dipole-dipole attraction
H
(b) Electrostatic attraction
H
H
(a)
(c) Hydrogen bonds
|
CO
(b)
C
|
CH 3
(d) Vander Waal's attraction
CH 3 H
(c)
|
|
|
|
Cl
|
CC
(d)
C
|
CH 3
|
 C
|
H
CH 3
|
 C
|
CH 3 H
10.
1.
2.
3.
4.
The values of electronegativity of atoms A and B are 1.20 and 4.0
respectively. The percentage of ionic character of A – B bond is
(a) 50 %
(b) 43 %
(c) 55.3 %
(d) 72.24%
O 22  is the symbol of ….. ion
(b) Superoxide
(c) Peroxide
(d) Monoxide
(b) 8
(c) 14
(d) 16
The type of hybrid orbitals used by the chlorine atom in ClO 2 is
12.
(a) sp 3
(b) sp 2
(c) sp
(d) None of these
Among the following species, identify the isostructural pairs,
(a)
[IIT 1996]
(b)
Increase, decrease
(c) Increase, increase
(d) None of these
Which of the following contains a coordinate covalent bond
[IIT 1995]
[ NF3 ,
NO 3 ]
[IIT 1996]

and [BF3 , H 3 O ]
(b) [ NF3 , HN 3 ] and [ NO 3 , BF3 ]
(c)
When N 2 goes to N 2 , the N  N bond distance ..... and when
(a) Decrease, increase
NH 3  NF3  BF3
(d)
11.
The number of electrons that are paired in oxygen molecule is
(a) 7
[MP PET 2003]
NH 3  BF3  NF3
NF3 , NO 3 , BF3 , H 3 O  , HN 3
O 2 goes to O 2 , the O  O bond distance .......
5.
(c)
[EAMCET 2003]
(a) Oxide
CH 3
Cl
Which of the following arrangement of molecules is correct on the
basis of their dipole moments
[AIIMS 2002]
(a) BF3  NF3  NH 3
(b) NF3  BF3  NH 3
[ NF3 , H 3 O  ] and [ NO 3 , BF3 ]
(d) [ NF3 , H 3 O  ] and [HN 3 , BF3 ]
13.
In the compound CH 2  CH  CH 2  CH 2  C  CH , the
C2  C3 bond is of the type
(a)
sp  sp 2
[IIT 1999]
(b)
sp 3  sp 3
126 Chemical Bonding
(c)
14.
sp  sp 3
(d)
The correct order of increasing
sp 2  sp 3
C O
24.
bond length of
CO , CO 32  , CO 2 is
15.
[IIT 1999]
(a)
CO 32   CO 2  CO
(b) CO 2  CO 32   CO
(c)
CO  CO 32   CO 2
(d) CO  CO 2  CO 32 
In the dichromate dianion
(a)
25.
O
||
(b) CH 3  C  CH 3 and CHCl 3
[IIT 1999]
4 Cr  O bonds are equivalent
(b) 6 Cr  O bonds are equivalent
(c) All Cr  O bonds are equivalent
(d) All Cr  O bonds are non-equivalent
16.
17.
18.
Bond length of ethane (I), ethene (II), acetylene (III) and benzene
(IV) follows the order
[CPMT 1999]
(a)
I  II  III  IV
(b)
I  II  IV  III
(c)
I  IV  II  III
(d)
III  IV  II  I
Hybridisation state of chlorine in ClF3 is
(a)
sp 3
(b)
sp 3 d
(c)
sp 3 d 2
(d)
sp 3 d 3
26.
20.
21.
22.
(c)
sp3 d , Irregular tetrahedral
H 2 O and H 2 O 2
A compound contains atoms X , Y , Z. The oxidation number of
(a)
XYZ 2
(b)
X 2 YZ 3 2
(c)
X 3 YZ 4 2
(d)
X 3 Y 4 Z 2
Bonds present in CuSO 4 .5 H 2 O is
28.
(a) Electrovalent and covalent
(b) Electrovalent and coordinate
(c) Electrovalent, covalent and coordinate
(d) Covalent and coordinate
The ionization of hydrogen atom would give rise to
[IIT Screening 2000]
19.
(d)
27.
Molecular shapes of SF4 , CF4 and XeF4 are
The same with 2, 0 and 1 lone pairs of electrons respectively
The same, with 1, 1 and 1 lone pairs of electrons respectively
Different, with 0, 1 and 2 lone pairs of electrons respectively
Different, with 1, 0 and 2 lone pairs of electrons respectively
(c)
O
O
||
||
H  C  OH and CH 3  C  OH
X is  2, Y is + 5 and Z is  2 . Therefore, a possible formula
of the compound is
[CPMT 1988]
[RPET 1999]
(a)
(b)
(c)
(d)
(d) Isoelectronic and weak field ligands
The number of S – S bonds in sulphur trioxide trimer S 3 O 9 is
(a) Three
(b) Two
(c) One
(d) Zero
Strongest intermolecular hydrogen bond is present in the following
molecules pairs
[IIT 1981; DCE 2000]
(a) SiH 4 and SiF
[IIT 1983; DCE 2001]
[UPSEAT 2001]
(a) Hybrid ion
(b) Hydronium ion
Structure of IF4 and hybridization of iodine in this structure are[UPSEAT 2001]
(c) Proton
(d) Hydroxyl ion
(a) sp 3 d , Linear
29.
Which can be described as a molecule with residual bonding
capacity
[JIPMER 2000]
(b) sp 3 d 2 , T-shaped
(d) sp 3 d 2 , Octahedral
In which of the following the central atom does not use sp hybrid
orbitals in its bonding
[UPSEAT 2001, 02]
(a)
BeCl 2
(b)
NaCl
(c)
CH 4
(d)
N2
3
(a)
BeF3
(b) OH 3
(c)
NH 2
(d)
NF3
The magnetic moment of K 3 [Fe(CN )6 ] is found to be 1.7 B.M.
How many unpaired electron (s) is/are present per molecule
(a) 1
(b) 2
(c) 3
(d) 4
N 2 and O 2 are converted into monocations N 2 and O 2
respectively. Which is wrong
[CBSE PMT 1997]
(a) In N 2 , the N  N bond weakens
(b) In O 2 , the O  O bond order increases
Read the assertion
and
[Orissa
JEEreason
2003] carefully to mark the correct option out of
the options given below :
(a)
(c)
(d)
(e)
If both assertion and reason are true and the reason is the correct
explanation of the assertion.
If both assertion and reason are true but reason is not the correct
explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.
If assertion is false but reason is true.
1.
Assertion
:
Water is a good solvent for ionic compounds but
poor one for covalent compounds.
Reason
:
Hydration energy of ions releases sufficient
energy to overcome lattice energy and break
hydrogen bonds in water, while covalent bonded
compounds interact so weakly that even Vander
(b)
(c) In O 2 , paramagnetism decreases
(d)
23.
N 2 becomes diamagnetic
The common features among the species CN  , CO and NO  are
[IIT Screening 2001]
(a) Bond order three and isoelectronic
(b) Bond order three and weak field ligands
(c) Bond order two and -acceptors
[
Chemical Bonding
Wall's forces between molecules of covalent
compounds cannot be broken.
[AIIMS 1996]
2.
3.
4.
5.
Assertion
:
Reason
:
Assertion
Reason
:
:
Assertion
:
Reason
:
Assertion
:
Reason
:
16.
Assertion
:
Reason
:
17.
In a polar covalent molecule, the shared electrons
spend more time on the average near one of the
atoms.
[AIIMS 1996]
Diborane is electron deficient
18.
There are no enough valence electrons to form
the expected number of covalent bonds[AIIMS 2001]
A resonance hybrid is always more stable than
19.
any of its canonical structures
This stability is due to delocalization of electrons[AIIMS 1999]
Assertion
:
Reason
:
Assertion
Reason
:
:
Assertion
:
All F  S  F angle in SF4 greater than 90° but
less than 180°
The lone pair-bond pair repulsion is weaker than
bond pair-bond pair repulsion
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
22.
Assertion
:
BF3 has greater dipole moment than H 2 S .
23.
Reason
Assertion
:
:
Reason
:
Assertion
:
Fluorine is more electronegative than sulphur.
The bond between two identical nonmetal atoms
has a pair of electrons with identical spin.
Electrons are transferred fully from one atom to
another.
B molecule is diamagnetic.
Reason
:
Assertion
:
Reason
:
Assertion
Reason
:
:
The atoms in a covalent molecule are said to
share electrons, yet some covalent molecules are
polar.
20.
The crystal structure gets stabilized even though
the sum of electron gain enthalpy and ionization
enthalpy is positive.
Energy is absorbed during the formation of
crystal lattice.
Order of lattice energy for same halides are as
LiX  NaX  KX .
Size of alkaline – earth metal increases from Li
to K .
Born-Haber cycle is based on Hess’s law.
Lattice enthalpy can be calculated by BornHaber cycle.
Bond
energy
has
order
like
C C  C  C  C  C .
Bond energy increases with increase in bond
order.
Electron affinity refers to an isolated atom’s
attraction for an additional electron while
electronegativity is the ability of an element to
attract electrons towards itself in a shared pair of
electrons.
Electron affinity is a relative number and
electronegativity is experimentally measurable.
[AIIMS 2004]
6.
Assertion
Reason
:
:
The electronic structure of O 3 is
O
O
O
O
structure is not allowed because
O
O
octet around cannot be expanded.
21.
Geometry of SF4 molecule can be termed as
distorted tetrahedron, a folded square or see saw.
Four fluorine atoms surround or form bond with
sulphur molecule.
[IIT 1998]
7.
Assertion
:
Reason
:
Assertion
:
Bond order can assume any value number
including zero
Higher the bond order, shorter is bond length
and greater is bond energy
[AIIMS 1999]
8.
9.
Reason
:
Assertion
:
Ortho nitrophenol molecules are associated due to the
presence of intermolecular hydrogen bonding while
paranitrophenol involves intramolecular, hydrogen
bonding
Ortho nitrophenol is more volatile than the para
nitrophenol
[AIIMS 1999]
Nitrogen molecule diamagnetic.
Reason
:
N 2 molecule have unpaired electrons.
10.
Assertion
Reason
:
:
Ice is less dense than liquid water.
There are vacant spaces between hydrogen
bonded water molecules in ice.
11.
Assertion
:
Water is liquid but H 2 S is a gas.
12.
Reason
Assertion
:
:
13.
Reason
Assertion
:
:
Reason
:
Assertion
Reason
Assertion
Reason
:
:
:
:
Oxygen is paramagnetic.
Iodine is more soluble in water then in carbon
tetrachloride.
Iodine is a polar compound.
o and p -nitrophenols can be separated by
steam distillation.
o -nitrophenol have intramolecular hydrogen
bonding while
p -nitrophenol exists as
associated molecules.
The fluorine has lower reactivity.
F  F bond has low bond dissociation energy.
 is strong while  is a weak bond.
Atoms rotate freely about  bond.
14.
15.
127
24.
25.
26.
2
The highest occupied molecular orbital is of 
type.
[AIIMS 2005]
The nearly tetrahedral arrangement of the
orbitals about the oxygen atom allows each water
molecule to form hydrogen bonds with as many
as four neighbouring water molecules.
In ice each molecule forms four hydrogen bonds
as each molecule is fixed in the space.
The bond order of helium is always zero.
The number of electrons in bonding molecular
orbital and antibonding molecular orbital is
equal.
Electrovalent bonding
1
b
2
a
3
a
4
c
5
c
6
d
7
d
8
b
9
c
10
d
11
b
12
a
13
d
14
a
15
a
16
c
17
b
18
a
19
d
20
c
128 Chemical Bonding
21
b
22
d
23
a
24
a
25
b
26
d
27
d
28
c
29
a
30
d
31
b
32
b
33
b
34
d
35
b
36
a
37
b
38
a
39
a
40
c
41
c
42
b
43
d
44
b
45
c
46
c
47
a
48
b
49
c
50
b
51
b
52
b
53
a
54
a
55
a
56
c
57
a
58
c
59
a
60
c
61
a
62
b
63
d
64
d
65
b
66
a
67
abc
68
bd
Covalent bonding
Polarisation and Fajan’s rule
1
d
2
c
3
b
4
d
5
c
6
a
7
b
8
a
9
c
10
b
11
d
12
c
13
b
14
b
15
d
16
d
17
c
18
b
19
a
20
d
21
a
22
c
23
d
24
a
25
b
26
b
Overlaping -  and - bonds
1
c
2
c
3
b
4
b
5
c
6
c
7
c
8
b
9
d
10
c
1
c
2
c
3
B
4
b
5
d
11
b
12
c
13
a
14
a
15
d
6
a
7
c
8
a
9
d
10
a
16
a
17
d
18
c
19
d
20
d
11
b
12
b
13
c
14
b
15
c
16
a
17
a
18
c
19
a
20
b
21
a
22
a
23
c
24
c
25
c
26
c
27
a
28
a
29
a
30
d
31
b
32
a
33
d
34
a
35
d
36
b
37
d
38
c
39
d
40
c
41
b
42
b
43
b
44
b
45
b
46
d
47
d
48
b
49
a
50
a
51
b
52
d
53
c
54
d
55
d
56
d
57
a
58
a
59
d
60
a
61
c
62
a
63
b
64
b
65
b
66
b
67
b
68
d
69
b
70
c
71
c
72
c
73
cd
74
ad
75
ab
76
a
Co-ordinate or Dative bonding
1
d
2
b
3
c
4
d
5
c
6
b
7
a
8
d
9
a
10
d
11
c
12
a
13
a
14
b
15
c
Dipole moment
Hybridisation
1
d
2
d
3
d
4
c
5
d
6
a
7
c
8
b
9
d
10
d
11
d
12
a
13
a
14
b
15
a
16
b
17
c
18
a
19
d
20
b
21
c
22
c
23
a
24
c
25
a
26
a
27
b
28
c
29
b
30
a
31
d
32
a
33
d
34
c
35
c
36
b
37
b
38
c
39
b
40
b
41
d
42
b
43
c
44
a
45
c
46
c
47
d
48
b
49
c
50
a
51
b
52
a
53
c
54
c
55
c
56
d
57
b
58
a
59
b
60
c
61
b
62
c
63
b
64
b
65
b
66
a
67
c
68
b
69
c
70
a
71
a
72
a
73
b
74
b
75
d
76
d
77
c
78
a
79
d
80
b
81
c
82
b
83
d
84
a
85
d
86
b
87
d
88
c
89
a
90
c
91
c
92
c
93
a
94
b
95
c
1
b
2
d
3
d
4
a
5
c
96
a
97
b
98
b
99
b
100
b
6
c
7
a
8
a
9
c
10
b
101
a
102
b
103
d
104
a
105
b
11
b
12
d
13
b
14
c
15
d
106
a
107
a
108
b
109
b
110
a
16
c
17
c
18
a
19
c
20
b
111
a
112
b
113
b
114
d
115
d
21
d
22
b
23
b
24
b
25
a
116
c
117
c
118
b
119
c
120
a
26
b
27
b
28
b
29
c
30
a
121
a
122
c
123
a
124
a
125
b
31
a
32
c
33
a
34
bd
35
a
126
c
127
d
128
c
129
c
130
a
131
b
132
b
133
e
134
c
135
d
Chemical Bonding
129
136
b
137
b
138
d
139
a
140
a
6
d
7
b
8
d
9
c
10
c
141
a
142
b
143
a
144
a
145
a
11
d
12
b
13
a
14
b
15
d
146
b
147
c
148
d
149
bcd
150
a
16
d
17
b
18
d
19
c
20
c
151
ac
152
a
21
a
22
a
23
d
24
a
25
c
26
a
27
b
28
c
29
a
30
c
31
a
32
b
33
d
34
a
35
a
36
a
37
a
38
b
39
d
40
c
41
a
42
c
43
b
44
c
45
c
46
b
47
d
48
b
49
d
50
d
Resonance
1
d
2
b
3
b
4
b
5
b
6
c
7
a
8
c
9
b
10
c
11
abcd
Types of bonding and Forces in solid
VSEPR Theory
1
b
2
d
3
c
4
c
5
d
6
d
7
d
8
b
9
b
10
c
11
d
12
a
13
d
14
c
15
a
16
b
17
d
18
a
19
d
20
c
21
d
22
d
23
a
24
d
b
5
a
9
a
10
d
d
14
d
15
b
18
d
19
c
20
a
d
23
a
24
d
25
c
c
28
c
29
a
1
a
2
a
3
b
4
c
5
c
6
b
7
b
8
c
9
b
10
a
11
c
12
a
13
a
14
a
15
c
16
c
17
b
18
d
19
d
20
a
21
a
22
d
23
b
24
d
25
a
26
c
27
b
28
b
29
a
30
a
31
a
32
c
33
c
34
a
35
c
36
b
37
b
38
d
39
d
40
b
1
d
2
c
3
c
4
41
c
42
a
43
b
44
c
45
d
6
a
7
b
8
a
11
a
12
c
13
16
c
17
b
21
a
22
26
c
27
Molecular orbital theory
1
a
2
c
3
b
4
b
5
c
6
d
7
c
8
b
9
c
10
b
11
c
12
b
13
c
14
a
15
c
16
c
17
d
18
b
19
c
20
c
21
d
22
c
23
b
24
c
25
a
26
d
27
b
28
b
29
a
30
c
31
c
32
a
33
c
34
a
35
c
36
d
37
b
38
a
39
a
40
c
41
c
42
a
43
b
44
a
45
a
46
c
47
b
48
c
49
c
50
a
51
c
52
b
53
a
54
a
55
a
56
c
57
c
58
c
59
a
60
a
61
a
62
b
63
a
64
c
65
a
66
c
67
a
68
a
69
c
70
a
71
b
72
b
73
d
74
c
75
a
76
b
77
b
78
a
79
c
80
a
81
c
82
a
83
c
84
d
5
c
Hydrogen bonding
1
d
2
b
3
b
4
a
Critical Thinking Question
Assertion & Reason
1
a
2
a
3
a
4
a
5
c
6
b
7
b
8
e
9
c
10
a
11
b
12
d
13
a
14
e
15
c
16
c
17
c
18
b
19
a
20
c
21
b
22
e
23
d
24
d
25
a
26
a
Chemical Bonding
24.
(a)
133
M 3 (PO4 )2 means M is divalent so formula of its sulphate is
MSO 4 .
25.
(b) As the molecular formula of chloride of a metal M is MCl3 , it
is trivalent so formula of its carbonate will be M 2 (CO 3 )3 .
26.
Electrovalent bonding
27.


(d) Sodium chloride is electrovalent compound so it dissolves in
water which is a polar solvent.
(d) When sodium chloride is dissolved in water, the sodium ion is
hydrated.
1.
(b) NaCl is ionic crystal so it is formed by Na
2.
(a) Bond formation is always exothermic. Compounds of sodium
are ionic.
3.
(a) According to Fajan’s rule ionic character is less.
32.
4.
(c) Valencies of L, Q, P and R is – 2, – 1, + 1 and + 2 respectively so
they will form P2 L, RL, PQ and RQ 2 .
(b) Molten sodium chloride conducts electricity due to the
presence of free ions.
33.
(b) The phosphate of a metal has the formula MHPO4 it means
34.
35.
(d)
(b) Cs is highly electropositive while F is highly
electronegative so they will form ionic bond.
(b) Na is highly electropositive while Cl is highly
electronegative so they will form ionic bond.
(a) Ionic compounds are good conductors of heat and electricity so
they are good electrolyte.
(a) Metal tends to lose electrons due to low ionization energy.
(c) As the formula of calcium pyrophosphate is Ca 2 P2 O7 means
valency of pyrophosphate radical is – 4 so formula of ferric
pyrophosphate is Fe4 P2 O7 3 .
5.
and Cl
ions.
(c) Electrovalent compounds are good conductor of heat and
electricity in molten state or in aqueous solution.
30.
M 3  ion so formula of its phosphate would be MPO4 .
metal is divalent so its chloride would be MCl 2 .
7.
(d) Electrovalent bond formation depends on ionization energy of
cation, electron affinity of anion and on lattice energy.
8.
(b) Because CsF is electrovalent compound.
37.
9.
(c)
38.
10.
(d) Valency of metal is + 2 by formula MO so its phosphate would
be M 3 (PO4 )2 because valency of [PO4 ] is – 3.
NaCl is formed by electrovalent bonding.
11.
(b) Li, Na and K are alkali metals with low ionization energy and
one electron in their outermost shell so they will form cation
easily.
12.
(a) Melting point and boiling point of electrovalent compounds are
high due to strong electrostatic force of attraction between the
ions.
13.
(d) The value of lattice energy depends on the charges present on
the two ions and distance between them. It shell be high if
charges are high and ionic radii are small.
39.
40.
41.
42.
43.
14.
(a) Cs is more electropositive.
15.
(a) X loses electron, Y gains it.
44.
16.


(c) Formation of NaCl occurs by Na ion
and Cl ion
.
45.
17.
(b)
18.
(a) Electrovalent compounds generally have high m.pt and high
b.pt due to stronger coulombic forces of attractions.
19.
(d) Water is a polar solvent so it decreases the interionic attraction
in the crystal lattice due to solvation.
20.
21.
MgCl2 has electrovalent linkage because magnesium is
electropositive metal while chlorine is electronegative.
2
2
5
(c) Element C has electronic structure 1s , 2 s 2 p , it requires
only one electron to complete its octet and it will form anion
so it will form electrovalent bond.
(b) Since the chloride of a metal is MCl 2 therefore metal ‘M’
must be divalent i.e. M 2 . As a result the formula of its
phosphate is M 3 (PO4 )2 .
22.
48.
50.
51.
52.
53.
(a) Ion is formed by gaining or losing electrons. To form cation
electron are lost from the valency shell, so Zn atoms to Zn 
ions there is a decrease in the no. of valency electron.
61.
of nitrate is M ( NO 3 )3 .
23.
47.
54.
55.
57.
58.
59.
(d) In MPO4 the oxidation state of M is +3. Hence, the formula
(d) Yet the formula of sulphate of a metal (M) is M 2 (SO 4 )3 , it is
M  X bond is a strongest bond so between Na  Cl is a
strongest bond.
(b) The solubility order is :
BeF2  MgF2  CaF2  SrF2 so SrF2 is least soluble.
(c)
(d) NaF has maximum melting point, melting point decreases of
sodium halide with increase in size of halide their bond energy
get lower.
(b) Sulphanilic acids have bipolar structure so their melting point
is high and insoluble in organic solvents.
(c) CaCl 2 will have electrovalent bonding because calcium is
electropositive metal while chlorine is electronegative so they
will combined with electrovalent bond.
(a) Electrovalent bond is formed by losing electrons from one atom
and gaining electron by other atom i.e. redox reaction.
(b) Electrovalent compound are polar in nature because they are
formed by ions.
(b) CsCl has ionic bonding.
(b) As soon as the electronegativity increases, ionic bond strength
increases.
(b) This X element is a second group element so its chloride will
be XCl 2 .
(a) When electronegativity difference is from 1.7 to 3.0. This bond
is called as ionic bond.
(a) Ethyl chloride is an organic compound so it will be covalent.
(a) Lithium oxide and calcium fluoride show ionic characters.
(a) Generally cation and anion form ionic bond.
(c) Those atoms which contain +ve and –ve sign are known as ion.
(a) Generally Br-F contain maximum electronegativity difference
compare to other compound.
(a) Due to greater electronegativity difference.
3d
4s
4s
134 Chemical Bonding
31.
62.
(b) Valency of phosphorus in H 3 PO4 is supposed ‘x’ then
(b) Co 3   3d 6 4 s 0 ,
Ni 4   3d 6 4 s 0 ,
3  x 8  0, x 5  0 , x  5.
33.
(d) (1)  x  3(2)  0  1  x  6  0  x  6  1  5 .
34.
35.
(a) HCl molecule has covalent bond.
(d) Electrovalent compounds have high melting point and high
boiling point.
36.
(b)
64.
(d)
BaCl2 contain higher ionic character.
66.
(a) Electrolytes are compound which get dissociated into their ion
in water so it contains electrovalent bond.
67.
(abc) CaH 2 , BaH2 , SrH 2 are ionic hydride.
Length of H 
68.
(bcd) Generally MgCl2 , SrCl 2 , BaCl2 are ionic compounds so
they conduct electricity in fused state.
Middle length of Cl 2  198 pm
3.
(c) In N 2 molecule each Nitrogen atom contribute 3 e  so total
no. of electron’s are 6.
(b) Non-metals readily form diatomic molecules by sharing of
electrons. Element M (1s 2 2 s 2 2 p 5 ) has seven electrons in its
valence shell and thus needs one more electron to complete its
octet. Therefore, two atoms share one electron each to form a
diatomic molecule (M 2 )
198
 99 pm
2
Bond length of HCl = Length of H + Length of Cl
= 37 + 99 = 136 pm
37.
(d) Covalent character depend on the size of cation and anion.
6.
(a) In graphite all carbon atoms are sp 2 -hybridised and have
covalent bond.
(c) Silica has tendency to form long chain covalent structure such
as carbon so it has giant covalent structure.
7.
8.
(a) All have linear structure.
O = C = O, Cl – Hg – Cl, HC  CH
9.
10.
(d) Similar atoms form covalent bond.
(a) Covalent bond forms when electronegativity difference of two
atom is equal to 1.7 or less than 1.7
(b) Similar atoms form covalent bond.
(b) Water is a polar solvent while covalent compounds are nonpolar so they usually insoluble in water.
11.
12.
13.
14.
BCl 3 is electron deficient compound because it has only ‘6’
electrons after forming bond.
(b) Due to its small size and 2 electrons in s-orbital Be forms
covalent compound.
(c)
18.
(c)
21.
(a) Two identical atoms are joined with covalent bond so H 2 will
be covalent.
(c) Element ‘X’ has atomic no. 7 so its electronic configuration will
.
be 2, 5. So its electron dot symbol would be : X .
.
(c) C-S will be most covalent. Covalent character depend on the
size of cation and anion.
(c) HCl has ionic character yet it has covalent compound because
electronegativity of chlorine is greater than that of hydrogen.
23.
24.
25.
26.
H 2 O will formed by covalent bonding.
(c) Order of polarising power Be   Li   Na 
Hence order of covalent character BeCl 2  LiCl  NaCl .
(d) Compound has 254 gm of I 2 means
80 gm O 2 means
254
 2 mole, while
127
80
 5 mole so they will form compound
16
I2 O5 .
38.
..
..
.. ..
: M ..M :  : M : M :
..
.. ..
..
5.
74
 37 pm
2
Length of Cl 
Covalent bonding
2.
Middle length of H 2  74 pm
39.
41.
(c)
NH 4 Cl has covalent as well as ionic bond.
H




|

 H  N  H  Cl 


|


H


(d) Covalent character increases when we come down a group so
CaI2 will have highest covalent character.
(b) In water molecule three atom are linked by covalent bond.
O
H
H
..
(b) : N  N   O : or N  N  O.
..
(b) The
electronic
configuration
Structure is
42.
44.
of
Na (Z  11) is
1s 2 , 2 s 2 2 p 6 , 3 s1 . The oxide of Na is Na 2 O .
45.
47.
48.
(b) Covalent bond is directional.
(d) Bond dissociation energy decreases with increase in size. So D
is smallest.
(b) Molecule X is nitrogen because nitrogen molecule has triple
bond. It’s configuration will be 1s 2 , 2 s 2 2 p 3 .
49.
(a)
50.
(a) The compound will be A2 B3 (By criss cross rule).
51.
52.
(b) Each nitrogen share 3 electrons to form triple bond.
(d) Urea solution does not conduct electricity because it is a
covalent compound.
(d) Due to the small size and higher ionization energy, boron
forms covalent compound.
54.
PCl5 does not follow octet rule, it has 10 electrons in its
valence shell.
BF 3 contain 6 electron so it is lewis acid.
58.
(a)
59.
(d) Among the given species. The bond dissociation energy of
C  O bond is minimum in case of CO 32  by which
Chemical Bonding
135
C  O bond become more weaker in CO 32  or the bond
order of CO 32  (1.33) is minimum so the bond become weaker.
60.
(a) Valency of
Na 2 S 2 O3
is supposed to be
x, then
2  2 x  ( 6)  0 , 2 x  4  0 , x  2 .
O
O
||
61.
(c)
||
H  O  S  O  O  S  O  H (Marshall acid)
||
62.
63.
64.
65.
||
O
O
(a) Among the given choice Al is least electropositive therefore, the
bond between Al and Cl will be least ionic or most covalent or
the difference in electronegativeity of two atom is less than 1.8.
2.
(b)
O
(b) Electronic configuration of 16 S 32  1s 2 ,2 s 2 ,2 p 6 ,3 s 2 ,3 p 4 . In
the last orbit it has only 6 electron. So it require 2 electron to
complete its octet, therefore it share 2 electron with two
hydrogen atom and forms 2 covalent bond with it.
(b) The acidity of hydrides of VI group elements increase from top
to bottom as the bond strength X  H decrease from top to
bottom
H 2 O  H 2 S  H 2 Se  H 2 Te
3.
4.
(d)
(b) We know that Al 3 cation is smaller than Na  (because of
greater nuclear change) According to Fajan's rule, small cation
7.
H O  N  O .
(a) Structure of N 2 O5 is O  N  O  N  O .
3

H O  SO H

O
(c) NH 3 has lone pair of electron while BF3 is electron
deficient compound so they form a co-ordinate bond.
NF3  BF3
66.
between Al and Cl atoms.
(b) Sulphur has the second highest catenation property after
carbon. Its molecule has eight atom bonded together (i.e. S 8 )
67.
(b)
9.
10.
H 2 O 2 has open book structure.
(a)
H
O
H
7
electronic
configuration
of
nitrogen
72.
73.
74.
76.
(ad) CO has only 6 electrons while PCl5 has 10 electrons after
sharing so both don’t follow octet rule.
(a) Among these, NaH and CaH 2 are ionic hydrides and B2 H 6
(d)
F
–
Cl
F
F
F
Xe
O

(a)
13.
(a)
H 3 PO4 is orthophosphoric acid.
O
F
H O  PO  H
|
O
|
15.
H
(c) Sulphuric acid contain, covalent and co-ordinate bond.
Dipole moment
1.
2.
3.
5.
6.
8.
(b) CO 2 is a symmetrical molecule so its dipole moment is
zero.
(d) These all have zero dipole moment.
(d) HF has largest dipole moment because electronegativity
difference of both is high so it is highly polar.
(c) Due to its symmetrical structure.
(c) Chloroform has 3 chlorine atom and one hydrogen atom
attached to the carbon so it is polarised and it will show dipole
moment.
(a) The dipole moment of two dipoles inclined at an angle  is
given by the equation  
Co-ordinate or Dative bonding
O
SO 32  has one coordinate bond.  O  S  O 

..
..
H   N   H or H  N  H
|
.
H

H
(c) Multiple bonds have more bond energy so C  N will be the
strongest.
(c) Diamond, silicon and quartz molecule bounded by covalent
bond.
(cd) C 2 H 4 and N 2 has multiple bonds.
O

O
12.
N  1s 2 , 2 s 2 , 2 p 3
and NH 3 are covalent hydrides.
1.

O
H 2 SO 4 . Therefore all these contains coordinate bond.

CH 3 N  C contain dative bond.
is
It has 5 electrons in valency shell, hence in ammonia molecule
it complete its octet by sharing of three electron with three H
atom, therefore it has 8 electrons in its valence shell in
ammonia molecule
71.
does not have co-ordinate bond. Structure is
O
(d) Co-ordinate bond is a special type of covalent bond which is
formed by sharing of electrons between two atoms, where both
the electrons of the shared pair are contributed by one atom.
Since this type of sharing of electrons exits in O 3 , SO 3 and
O
(b) The
HNO 2

polarise anion upto greater extent. Hence Al polarise Cl
ion upto greater extent, therefore AlCl3 has covalent bond
69.
H 2 SO 4 has co-ordinate covalent bond.
F
X 2  Y 2  2 XY cos
cos 90  0 . Since the angle increases from 90  180 , the
value of cos becomes more and more – ve and hence
resultant decreases. Thus, dipole moment is maximum when
  90 .
136 Chemical Bonding
9.
(c) Due to distorted tetrahedral geometry SF4 has permanent
dipole moment

F
F
:
1.03
 100  16.83%  17%
6.12
Polarisation and Fajan's rule
S
F
F
10.
(b) CCl 4 has no net dipole moment because of its regular
tetrahedral structure.
12.
(d)
14.
(c)
16.
20.
1.
(d)
BF3 is planar while NF3 is pyramidal due to the presence of
lone pair of electron on nitrogen in NF3 .
2.
(c)
3.
(b) When electronegativity difference is more between two joined
atoms then covalent bond becomes polar and electron pair
forming a bond don’t remain in the centre.
(c) Dipole moment of CH 3 OH is maximum in it.
4.
(d) Hexane has symmetrical structure so does not have polarity.
(b) CH 4 have regular tetrahedron so its dipole moment is zero.
5.
(c) When two identical atoms form a bond, bond is non-polar.
22.
(b) Ammonia have some dipole moment.
6.
23.
(b) Charge of e   1.6  10 19
(a) According to Fajan’s rule, polarisation of anion is influenced by
charge and size of cation more is the charge on cation, more is
polarisation of anion.
8.
(a) When two atoms shares two electrons it is an example of
covalent bond. This covalent bond may be polar or may be
non-polar depends on the electronegativity difference. In given
example formula is AB. So it is polar.
9.
(c) HCl is most polar due to high electronegativity of Cl.
10.
(b)
11.
(d) p-dichloro benzene have highest melting point.
13.
(b)
H-F is polar due to difference of electronegativity of hydrogen
and fluorine so it shows positive dipole moment.
BCl 3 has zero dipole moment because of its trigonal planar
geometry.
Dipole moment of HBr  1.6  10 30
Inter atomic spacing  1Å  1  10 10 m
% of ionic character in
HBr 

dipolemoment of HBr  100
inter spacing distance q
1 .6  10 30
 100
1 .6  10 19  10 10
 10 30  10 29  100  10 1  100  0.1  100  10%
25.
(b)
29.
(c) Given ionic charge  4.8  10 10 e.s.u. and ionic distance
BF3 has zero dipole moment.
× ionic distance
30.
 4.8  10 10  10 8
 4.8  10 8 e.s.u. per cm  4.8 debye.
(a) Higher is the difference in electronegativity of two covalently
bonded atoms, higher is the polarity. In HCl there is high
difference in the electronegativity of H and Cl atom so it is a
polar compound.
31.
(a) Linear molecular has zero dipole moment CO 2 has linear
32.
(c)
structure so it does not have the dipole moment O  C  O .
33.
34.
35.
SF6 is symmetrical and hence non polar because its net dipole
moment is zero.
(a) Polarity create due to the difference in electronegativity of both
atom in a molecule except H 2 all other molecule have the
different atom so they will have the polarity while H 2 will be
non polar.
(bd) cis isomer shows dipole moment while that of trans is
zero or very low value. Trans 1, 2 di-chloro-2-pentene will
also show dipole moment due to unsymmetry.
(a) % of ionic character
Experiment al value of dipole moment
=
Expected value of dipole moment
NH 4 Cl has both types of bonds polar and non polar




H  Cl 


14.
(b) Greater the charge of cation more will be its polarising power
(according to Fajan’s rule).
15.
(d)
16.
(d) As the size of anion increases, polarity character increases.
20.
(d) Due to the electronegativity difference.
21.
(a) We know that greater the difference in electronegativity of two
atoms forming a covalent bond. More is its polar nature. In HF
there is a much difference in the electronegatives of hydrogen
and flourine. Therefore (HF) is a polar compound.
22.
(c) Silicon tetrafloride has a centre of symmetry.
23.
(d)
25.
(b) According to Fajan’s rule largest cation and smallest anion form
ionic bond.
(b) Polarity character is due to the difference in electronegativity of
two atoms or molecule.
8
 1 A  10 cm we know that dipole moment = ionic charge
NH 3 has sp 3 hybridised central atom so it is non planar.
H

|

 H  N|

H

(a) Carbon tetrachloride has a zero dipole moment because of its
regular tetrahedral structure.
27.
H 2 O is a polar molecule due to electronegativity difference of
hydrogen and oxygen.
26.
AlI3 Aluminiumtriiodide shows covalent character. According
to Fajan’s rule.
BF3 have zero dipole moment.
Overlaping-  and - bonds
1.

(c)
H–C
2.

C–H
 molecule formation p-p orbitals take part in bond
(c) In fluorine
formation.
Chemical Bonding
3.
(b) -bond is formed by lateral overlapping of unhybridised p-p
orbitals.
5.
(d) No. of e pair  3 
–
C
4.
1
[3  3] =0
2
No. of e pair  3 +0
1 and 2
F
C
5.
(c) In a double bond connecting two atom sharing of 4 electrons
take place as in H 2 C  CH 2 .
6.
9.
(c) C  C is a multiple bond so it is strongest.
(d) As the bond order increases, C  H bond energy also
increases so it will be greatest in acetylene because its B.O. is 3.
H H
11.
(b)
|
16.
17.
(a)
N
|
8.
(b) In ethylene both Carbon atoms are sp 2 - hybridised so 120 o .
|
9.
(d) Structure of
bipyramidal.
||

(c)
2, 2 bond and one lone pair.
..
..
(d) : O  S  O : 5 atoms has 12 electrons in its outermost
||
(d) In H  C  C  O  H the asterisked carbon has a valency of
*
5 and hence this formula is not correct.
11.
(d) dsp 3 hybrid orbitals have bond angles 120 o , 90 o .
13.
(a) In BeF3 , Be is not sp 3 –hybridised it is sp 2 hybridised.
17.
(c) In molecule OF2 oxygen is sp 3 hybridised.
18.
(a) In sp 3 hybrid orbitals s-character is 1/4 means 25%.
19.
(d)
20.
th
XeF4 molecule has ‘Xe’ sp 3 d 2 hybridised and its shape is
square planar.
(b) The bond angle is maximum for sp hybridisation because two
sp hybridised orbitals lies at angle of 180 o .
O:
..
shell. One (S  O) bond will be (p-p)  bond while two
(S  O) bond will be (p-d)  bond.
20.
hybridized compound is Trigonal
10.
(d) We know that trisilylamine is sp 2 –hybridized therefore
..
:O
sp 3 d
O
S
19.
No. of atom bonded to the central atom  3
In case of 3, 3 geometry is Trigonal planar.
N
..
:O
F
(a) In sp 3 –hybridisation each sp 3 hybridised orbital has 1/4 scharacter.
p  d bonding is possible due to the availability of vacant
d-orbitals with silicon.
18.
120°
120°
H

B
6.
H C  C  C  C

F
120°
–
(b) Ca
137
(d) Structure of P4 O10 is
C 2 H 4 Br2 has all single bonds so C  H bond distance is
the largest.
21.
(c)
23.
(a) In methane molecule C is sp 3 hybridised so its shape will be
tetrahedral.
24.
(c) In compound CH 2  C  CH 2 the second carbon sphybridised.
.. ..
(a) : Cl   Cl : is the correct electronic formula of Cl 2 molecule
.. ..
because each chlorine has 7 electrons in its valence shell.
O
3
P
O
O
25.
O
P
O
P
O
O
O
1
XeF4 has sp 3 d 2 hybridisation, its shape is square planar.
26.
(a)
27.
(b) In HCHO, carbon is sp 2 hybridized
O
P
2
H
O
|
H  C2  O
Each phosphorus is attached to 4 oxygen atoms.
sp
Hybridisation
1.
(d)
H 2 O is not linear because oxygen is sp 3 hybridised in
H 2O .
2.
(d)
O
95.7 pm
(104.5)
H
4.
(c)
28.
(c) Because of the triple bond, the carbon-carbon bond distance in
ethyne is shortest.
29.
(b) The hybridisation of Ag in complex [ Ag (NH 3 )2 ] will be sp
because it is a Linear complex.
30.
(a) Structure of CO 2 is linear O  C  O while that of H 2 O
O
o
is
H
CO 2 has sp – hybridization and is linear.
H
H
i.e. bent structure so in CO 2 resultant dipole
moment is zero while that of H 2 O has some value.
31.
(d) CO 2 is not sp 3 hybridised, it is sp hybridised.
138 Chemical Bonding
32.
(a) As compare to pure atomic orbitals, hybrid orbitals have low
energy.
sp 2
(a) As p-character increases the bond angle decreases.
sp 3
sp 2
sp
70.
33.
(d) CH 2  C  CH – CH 3 1, 2-butadiene.
36.
(b) CCl 4 is sp 3 hybridised so bond angle will be approximately
109 o .
40.
o
2
1
, bond angle - 180 o
2
In sp 2 - p-character
2
, bond angle - 120 o
3
In sp 3 - p-character
3
, bond angle - 109 o
4
(b) Ethene has sp hybridised carbon so bond angles are 120 .
O
44.
In sp - p-character
(a) Acetate ion is CH 3
(a)
72.
(a) S-atom in SF6 has sp 3 d 2 hybridisation. So, the structure of
i.e. one C  O single bond
C
O

sp 3 -hybridization called tetrahedral because it provides
tetrahedral shape to the molecule.
71.
and one C  O double bond.
SF6 will be octahedral.
46.
(c) Benzene has all carbons sp 2 hybridised and planar in shape.
74.
(b) Structure of H 2 O 2 is non-planar. It has open book structure.
47.
(d) In methane C is sp 3 hybridized and bond angle is 109 o .
75.
(d) Structure of N 2 O is similar to CO 2 both have linear
structure.
H H H
56.
(d)
|
|
|
78.
(a)
|
|
|
79.
(d) In NH 4 nitrogen is sp 3 hybridised so 4 hydrogen situated
at the corners of a tetrahedron.
81.
(c) Increasing order of bond angle is sp 3  sp 2  sp .
H  C C C H
H H H
58.
There are 10 shared pairs of electrons.
(a) The diborane molecule has two types of B – H bond :
(ii) B  H b – It is a three centred bond.
86.
(b) Bond angle increases with change in hybridisation in following
Ht
B
Ht
Hb
(b) PF5 involves sp d hybridization and hence has trigonal
bipyramidal structure.
3
(c) s-character in sp 
NH 4
3
has sp –hybridized nitrogen so its shape is tetrahedral.
order sp 3  sp 2  sp .
Hb
1
 100  50%
2
s-character in sp 2 
88.
(c) In Diborane boron shows sp 3 –hybridization.
89.
(a) Alkene does not show linear structure but it has planar
structure due to sp 2 –hybridisation.
90.
(c) Generally SF4 consist of 10 electrons, 4 bonding electron pair
92.
and one lone pair of electron, hence it shows
hybridization.
(c) Atom/Ion
Hybridisation
1
 100  33.3%
3
1
s-character in sp   100  25%
4
3
Hence, maximum s-character is found in sp-hybridisation.
64.
180 
(a)
Ht
63.
120 
84.
B
62.
109 
(i) B  H t – It is a normal covalent bond.
Ht
61.
SnCl 2 is V–shaped.
93.
(a)
sp
SF4
sp 3 d with one lone pair of electron
PF6
sp 3 d 2
PF3 consist of three bonding pair electrons and one lone pair
of electron hence it shows sp 3 – hybridization.
3
(b) The molecule of PCl5 has sp d hybridisation, structure is
trigonal bipyramidal.
(b) Merging (mixing) of dissimilar orbitals of different energies to
form new orbitals is known as hybridisation and the new
orbital formed are known as hybrid oribitals. They have similar
energy.
NO 2
sp 3 d
NO 2 shows sp–hybridization. So its shape is linear.
94.
(b)
95.
(c) Generally octahedral compound show sp 3 d 2 – hybridization.
96.
(a) In fifth group hydride bond angle decreases from top to
bottom
65.
(b) In SO 3 sulphur is sp 2 hybridized so its shape will be
trigonal planar.
66.
(a) These all are triangular with sp 2 hybridization.
97.
(b) Generally NH 4 shows sp 3 hybridization.
67.
(c) Bond length depends upon bond order and in benzene all
C  C bonds have same bond order.
98.
(b) We know that single, double and triple bond lengths of carbon
(b) In
99.
(b) It shows sp 2 –hybridization so it is planar.
101.
(a) Bond angle of hydrides decreases down the group.
68.
C2 H 2
H  C  C H
sp
sp
each
carbon
has
sp -hybridization
NH 3  PH3  AsH3  SbH 3  BiH3 .
in carbon dioxide are 1.22 Å,1.15 Å and 1.10Å respectively.
Chemical Bonding
102.
(b) Hybridization of N in NH 3 is sp 3 that of Pt in [PtCl4 ]2
2
125.
3
is dsp that P in PCl5 is sp d and that of B in BCl 3 is
126.
sp 2 .
NH 4 and SO 42 
tetrahedral structure.
(d)
104.
(a) It is shows sp 3 d 3 –hybridization. Hence the bond angle is
107.
both show
sp
33.3
120 o
M
sp3 d 2 hybrid orbital have octahedral shape
127.
(d)
128.
(c) In the formation of d 2 sp3 hybrid orbitals two (n  1)d orbitals
of e.g., set [i.e., (n  1) dz 2 and (n  1)dx 2  y 2 orbitals] one
25
109.28
3 1
20
90 o and 120 o
3
X
according to thisX geometry, the number
of X  M  X bond
X
at 180° must be three. X
3
sp d
X
X
about 72 o .
(a) s-character increases with increase in bond angle.
Hybridization
s%
Angle
sp
50
180 o
sp 2
(b) There is sp hybridization in C 2 H 2 so it has the linear
structure.
(c) In octahedral molecule six hybrid orbitals directed towards the
corner of a regular octahedron with a bond angle of 90°.
sp 3 –hybridization and
103.
o
ns and three np [ np x , npy and np z ]
2
together and form six d sp hybrid orbitals.
3
129.
(c) The correct order of bond angle (Smallest first) is
(b)
IF7 molecule show sp d –hybridization.
110.
(a)
PCl3 contain three bonding and one lone pair electron. Hence
H 2 S  NH 3  SiH4  BF3
shows sp 3 –hybridization.
92.6  107  10928'  120
112.
113.
92.6°
2
(b) For square planar geometry hybridization is dsp involving
y2
S
117.
(c)
(c)
BCl 3 molecule
structure.
show
sp 2 –hybridization
119.
(c) Due to multiple bonding in N 2 molecule.
120.
(a) % of s-character in
CH 4 
3
(sp )
and
122.
H
130.
(a)
( 120) due to 2 lone pair of electron over S atom. CO 2
F
CN
F
Square planar
Regular tetrahedral
F
F
F
F
S
131.
F
H
F
Square planar
See saw shaped
(b)
F
Xe
F
··
O
··
(a) In H 2 CO 3 and BF3 central atom are in sp 2 hybridization
but in H 2 CO 3 due to the ionic character of O  H bond it
will be polar (High electronegativity of oxygen).
124.
F
B
NC
and N 2 O have the sp hybridization.
123.
–
CN
Ni
100
100
 25 , C 2 H 4 
 33 ,
2
3
4
(sp )
SO 2 has sp 2 hybridization have the V shape structure
F
2–
NC
(a) Acidic character increases when we come down a group, so HI
is the strongest acid.
(c)
H
H
F
F
planar
100
C2 H 2 
 50
2
( sp )
121.
Si
B
SO 2 molecule shows sp 2 –hybridization and bent structure.
(b)
109° 28'
120°
BCl 3 Boron trichloride molecule show sp 2 –hybridization and trigonal
planar structure.
118.
H
F
(b) All carbon atoms of benzene consist of alternate single and
H
H
H
orbital.
double bond and show sp 2 hybridization.
116.
N
H 107°
H
(a) Ammonia and (BF4 )1 shows sp 3 –hybridization.
s, p x , p y and d x 2 
orbitals combine
3
108.
111.
139
H
sp
(a) Due to sp 3 hybridization and presence of lone pair of electron
on p atom PCl3 are of pyramidal shape like that of NH 3 .
132.
3
··
O
··
B
sp2
··
H
O sp3
··
(b) In the formation of BF3 molecule, one s and 2p orbital
hybridise. Therefore it is sp 2 hybridization.
140 Chemical Bonding
133.
(e) In NCl 3 and H 2 S the central atom of both (N and S) are in
I  [ Xe] 5 s 2 ,5 p 5 hence
sp 3 hybridization state
··
··
N
Cl
S
··
107°
Cl
Cl
92.5°
H
H
5d
5s
5p
5d
and I F7 in excited
state
sp 3 hybridization respectively. In H 2 S and BeCl 2 central
F
F
F
I
F
F
F
atom are in sp 3 and sp 2 hybridization In BF3 , NCl 3 &
IF shows sp d hybridization. So, sp
itsd structure is pentagonal
H 2 S central atom are in sp 2 , sp 3 & sp 3 hybridization
bipyramidal.
141. (a) Compound containing highly electronegative element (F, O, N)
attached to an electropositive element (H) show hydrogen
bonding. Fluorine (F) is highly electronegative and has smaller
size. So hydrogen fluoride shows the strongest hydrogen
bonding in the liquid phase.
and in the central atom are in sp 3 and sp hybridization.
(c)
5p
I in excited state
while in BF3 and NCl 3 central atoms are in sp 2 and
134.
5s
I in ground state
1 1 1
Cground state  2 s 2 ,2 p x 1 py1 ; Cexcited state  2 s1 ,2 p x py p z
Oground state  2 s 2 ,2 p x p y p z
2
1
1
In the formation of CO 2 molecule, hybridization of orbitals of
carbon occur only to a limited extent involving only one s and one
p orbitals there is thus sp hybridisation of valence shell orbitals of
the carbon atom resulting in the formation of two sp hybrid
orbitals.
3
2
3 3
7
142.
(b) In the ammonia molecule N atom is sp 3  hybridized but due
to the presence of one lone pair of e  (i.e. due to greater
L p  b p repulsion) it has distorted tetrahedral (or pyramidal)
geometry.
Oxygen atom in
ground state
sp – p
 bonded
N
p – p
Carbon atom in
excited state
H
p – p
sp – p
 bonded
143.
(a)
136.
137.
1s
3
(d) In NH 3 , N undergoes sp hybridization. Due to the presence
of one lone pair, it is pyramidal in shape.
(b) NO 2 SF4
PF6
sp sp 3 d sp 3 d 2
(b) The configuration of
5B
2s
2p
Be in excited state
BeCl2
Cl
 1s 2 , 2 s 2 2 p 1
Cl
sp hybridisation
B in ground state
144.
1s
H
Be in ground state
Oxygen atom in
ground state
135.
H
2
2
0
4 Be  1s ,2 s ,2 p
2s
2p
(Linear diagonal hybridization)
(a) Except CO 3 other choice CO 2 , CS 2 and BeCl 2 have
sp  hybridization and shows the linear structure while
B in excited state
CO 3 have
1s
2s
2p
1s
2s
2p
In BCl3 state
sp3 hybridization and show the non linear
structure because sp3 generate tetrahedral structure.
145.
(a)
dsp3 or sp3 d hybridization exhibit trigonal bipyramidal
geometry e.g., PCl
5
Cl
Cl
Cl
Cl
sp2hybridisation
138.
(d) In SO 3 molecule, S atom remains sp hybrid, hence it has
O
trigonal planar structure
Cl
O
O
(a) In PCl3 molecule, phosphorous is sp 3  hybridised but due
to presence of lone pair of electron, it has pyramidal structure
P
140.
(a) The electronic configuration of
Cl
Cl Cl
sp3d2 (Trigonal bipyramidal)
P
Cl
S
139.
Cl
2
Cl
146.
(b) Carbon has only two unpaired electrons by its configuration
but hybridization is a concept by which we can explain its
valency 4.
147.
(c) Hybridization is due to overlapping of orbitals of same energy
content.
Chemical Bonding
148.
(d) MX 3 show the sp 2 hybridization in which 3sp 2 hybridized
orbital of M bonded by 3 X from  bond and having the
zero dipole moment.
149.
(bcd) SnCl 2 has V–shaped geometry.
150.
(a)
XeF6 is distorted Octahedral. It has sp 3 d 3 hybridisation
with lone pair of electron on Xe, so its shape is distorted.
9.
(b)
10.
11.
(a)
(c) Xe ground state
Xe double excitation
NF3 is predominantly covalent in nature and has pyramidal
structure (the central atom is sp 3 hybridised) with a lone pair
of electrons in the fourth orbital.
151.
(a)
12.
13.
dsp 3 or sp 3 d : one s  three p  one d (d z 2 ) .
2.
(b) In NH 3 nitrogen has one lone pair of electron.

(b) In CN ion formal negative charge is on nitrogen atom due
to lone pair of electrons.
 
9.
(b) There are three resonance structure of CO 32  ion.
O
C
11.
O
4.
17.
18.
SO 42  has 42 electrons; CO 32  has 32 electrons; NO 3 has
32 electrons.
(c) Molecular oxygen contains unpaired electron so it is
paramagnetic (according to MOT).
(b) Structure of H 2 O is a bent structure due to repulsion of lone
pair of oxygen.
(d) Bond angle between two hybrid orbitals is 105 o it means
orbitals are sp 3 hybridised but to lone pair repulsion bond
angle get changed from 109 o to 105 o . So its % of scharacter is between 22-23%.
(I)
(II)
(abcd) It has all the characteristics.

O
O
(III)
22.
(d) Number of electrons in ClO2–
= 7 + 6 + 6 + 1 = 20
Number of electrons in ClF2+ = 7+7+7 – 1=20.
23.
(b) Central atom having four electron pairs will be of tetrahedral
shape.
(a) The bond angle in PH 3 would be expected to be close to
F
F
24.
6.
7.
8.
..
F
(d)
F
C
S. .
F
F
F
F
Xe
F
F
F
F
..
sp 2 –hybridization and show trigonal planar
(b) In BF3 molecule Boron is sp 2 hybridised so its all atoms are
co-planar.
(c) Due to lp  lp repulsions, bond angle in H 2 O is lower
26.
(c) It shows
structure.
(104 o .5 o ) than that in NH 3 (107 o ) and CH 4 (109 o 28 ) .
28.
(b)
BeF2 on the other hand, has sp-hybridization and hence has a
31.
(a) Bond angle of hydrides is decreases top to bottom in the group.
NH 3  PH3  AsH3  SbH 3
32.
(c)
o
5.
5d
(c)
C
O
90 o . (The bond angle H  P  H in PH 3 is 93 o )
3.
5p
15.
VSEPR Theory
2.
5s
(a) CO 2 has bond angle 180 o .
(a) As the s-character of hybridized orbitals decreases the bond
angle also decreases
O
C
O
5d
14.
|
(a) CH 3  C  CH 2 has 9, 1 and 2 lone pairs.
(c) In resonance structure there should be the same number of
electron pairs.
O
5p
In sp hybridisation: s-character 1/2, bond angle 180 o
(a) XeF2 molecule is Linear because Xe is sp hybridised.
16.
O H
7.
8.

5s
In sp 2 hybridisation: s-character 1/3, bond angle 120 o
(d) Choice (a), (b), (c) are the resonance structures of CO 2 .
O
5d
In sp 3 hybridisation: s-character 1/4, bond angle 109 o
1.


5p
sp 3 d 2 - hybridization
Resonance
5.
5s
XeF4
(ac) PCl3 , NH 3  Pyramidal.
CH 4 , CCl 4  Tetrahedral.
152.
141
bond angle of 180 .
(c) Compound is carbontetrachloride because CCl 4 has sp 3 –
hybridization 4 orbitals giving regular tetrahedron geometry. In
others the geometry is little distorted inspite of sp 3
hybridization due to different atoms on the vertices of
tetrahedron.
(b) SO 42  ion is tetrahedral since hybridization of S is sp 3 .
N Three bond pair and one lone pair of electron.
H
33.
(b) NH 3 molecule has one lone pair of electrons on the central
atom i.e. Nitrogen.
(c) C 2 H 2 has linear structure because carbons are sp-hybridised
o
and lies at 180 .
H 2 S show bond angle nearly 90 o .
H
H
(c) Unpaired electrons are present in KO 2 while others have
paired electron
NO 2 = 22 electrons ; BaO2 = 72 electrons
AlO2  30 electrons ; KO 2  35 electrons
34.
(a) Bond angle decreases from H 2 O to H 2 Te .
142 Chemical Bonding
35.
(c)
36.
(b)
BF3 does not contain lone pair of electron.
Bent T-shaped geometry in which
both lone pairs occupy the
equatorial position of the trigonal
bipyramidal
here
F (l p  l p ) repulsion = 0
F
..
Br
(lp  b p ) repulsion = 4 and
..
F
37.
(b p  b p ) repulsion =2
(b) The overall value of the dipole moment of a polar molecule
depends on its geometry and shape i.e., vectorial addition of
dipole moment of the constituent bonds water has angular
structure with bond angle 105° as it has dipole moment.
However BeF2 is a linear molecule since dipole moment
summation of all the bonds present in the molecule cancel each
other.
Molecular orbital theory
3.
No. of bonding e   No. of antibonding e 
2
8 3 5

  2 .5 .
2
2
(b) One bonding M.O. and one anti-bonding M.O.
4.
(b) O 22  is least stable.
5.
(c) B.O. of O 2 is 2, B.O. of O 21 is 1.5, B.O. of O 21 is 2.5 and of
2.
38.
(d)
Be
F
6.
40.
H
P
107°
(b) Bond order of O 2 is highest so its bond length is smallest.
(c) Oxygen is paramagnetic due to the presence of two unpaired
electron :
O 2   (1s)2   (1s)2  (2 s)2   (2 s)2
 (2 p x )2  (2 p y )2  (2 p x )2   (2 p y )1   (2 p z )1
107°
Less than
17.
H
H
H
H
12.
13.
H
. .H
As
..
H
11.
NO(7  8  15) has odd number of electrons and hence it is
paramagnetic.
No. of N b  No. of N a
5
(c) B.O. 
  2 .5 .
2
2
N
H
18.
H
(d) In CH 3 CN bond order between C and N is 3 so its bond
length is minimum.
(b)
As the electronegativity of central atom decreases bond angle is
decreases
 NH 3 has largest bond angle.
41.
 (1s)
*
 (1s)
(c) In NH 3 , sp 3 -hybridization is present but bond angle is
B.O.
106 o 45  because Nitrogen has lone pair of electron according
to VSEPR theory due to bp-lp repulsion bond angle decreases
from 109 o 45' to 106 o 45  .
42.
44.
(b)
(c)
NH 3 has pyramidal structure, yet nitrogen is sp 3 hybridised.
This is due to the presence of lone pair of electron.
geometry which arise due to sp3 d hybridization of central
sulphur atom and due to the presence of lone pair of electron
in one of the equatorial hybrid orbital.
45.
O
O
dsp2 hybridization
sp3d hybridization
sp3d2 hybridization
(Four 90° angles
between bond pair
and bond pair)
(Six 90° angle
between bond pair
and bond pair)
(Twelve 90° angle
between bond pair
and bond pair)
H 2

H 2

1
1
2

1
2
(c) Due to unpaired e  ClO2 is paramagnetic.
20.
(c) The Bond order in N 2 molecule is 3, N  N
Here,
N b  2  4  2  8 and N a  2
 B.O.  ( 8  2) / 2  3.
21.
22.
(d)
O

1
2
H2

19.
3
SiF4 has symmetrical tetrahedral shape which is due to sp
hybridization of the central sulphur atom in its excited state
configuration. SF4 has distorted tetrahedral or Sea- Saw
He 2

Magnetic
P
D
P
P
nature
(P = Paramagnetic, D = Diamagnetic)
(a) Bond strength decreases as the size of the halogen
increases from F to I.
43.
O 2 (2  8  1  17) has odd number of electrons and hence
it is paramagnetic. All the remaining molecules/ions, i.e.,
CN  (6  7  1  14) diamagnetic
trigonal planar ( sp hybridization) Hence bond angle is same
for all of them (i.e., equal to 120°)
(d) We know that molecule of ( NH 3 ) has maximum repulsion
due to lone pair of electron. Its shape is pyramidal and is
sp3 hybridization.
:O:
(b)
..
105°
(c)
F
BCl
H3 , BBr3 and
H BF3 , all of these have same structure i.e.
H
(d) Hydride of boron does not exist in BH 3 form. It is stable as its
dimer di borane (B2 H 6 ) .
2
39.
B.O. 
O 22  is 1.
10.
O
(c)
23.
24.
(d)
H 2 has the bond order
1
, it has only one electron so it will
2
be paramagnetic.
(c) When bond forms between two atom then their energy get
lower than that of separate atoms because bond formation is
an exothermic process.
(b) Valency of A is 3 while that of B is 2 so according to Criss
Cross rule the formula of the compound between these two
will be A2 B3 .
(c) Due to resonance bond order of C  C bonds in benzene is
between 1 and 2.
Chemical Bonding
26.
27.
(a) Nitrogen does not have vacant ‘d’-orbitals so it can’t have +5
oxidation state i.e. the reason PCl5 exists but NCl 5 does
not.
(d) Molecules having unpaired electrons show paramagnetism.
(b) NO 2 has unpaired electrons so it would be paramagnetic.
30.
(c) Helium molecule does not exist as bond order of He 2  0 .
31.
(c) Structure of P4 O10 is
25.
55.
(a)
56.
(c)
57.
(c)
59.
(a)
60.
(a)
61.
(a)
64.
(c)
O
P
O
O
O
O
P
P
O
O
O
O
P
33.
34.
37.
38.
143
H 2 O 2 contain bond angle between two O  H planes about
90 o .
Nitrogen molecule has highest bond energy due to presence of
triple bond.
Cu 2  [ Ar18 ] 3d 9 4 s 0 it has one unpaired electron so it is
paramagnetic.
CN   14 electrons ; CO =14 electrons
1
6
B.O. = [10  4 ]   3 .
2
2
1
5
B.O. = [10  5]   2 .5 , paramagnetic
2
2
P
P
P
P
The paramagnetic property in oxygen came through unpaired
electron which can be explained by molecular orbital theory.
Each phosphorus is attached to 4Ooxygen atoms.
N  Na 8  4
(c) B.O. of carbon  b

2.
2
2
N  N a 10  4
(a) B.O.  b

3.
2
2
N  Na 8  3 5
(b) B.O.  b

  2.5 .
2
2
2
(a) Electronic configuration of O2 is
Antibonding
2px*
2pz* 2py*
Px
O2   (1s)2   (1s)2  (2 s)2   (2 s)2  (2 p x )2  (2 p y )2
Py
Pz
Px
Py
Pz
 (2 p z )2   (2 p y )1   (2 p z )1
The molecule has two unpaired electrons So, it is paramagnetic
 2 p y has two nodal planes.
*
40.
(c)
42.
43.
(a) Element with atomic number 26 is Fe. It is a ferromagnetic.
(b) Correct Sequence of bond order is
65.
44.
O 2  O 2  O 22
B.O – 2.5 2 1.5
(a) Due to small bond length.
45.
46.
(a) S 2 have all paired electrons so it is diamagnetic.
(c) NO has 15 electrons.
66.
47.
(b) In the conversion of O 2 into O 2 bond order decreases.
49.
(c)
O 22 
50.
(a)
O 22  consist of four antibonding electron pair [1s and 2s have
51.
two antibonding and 2 p x 2 p y have two antibonding electron
pair].
(c) The electron’s distribution in molecular orbitals is 1s 2 , 2 s 1
53.
54.
(b)
Total number
of bonds between atoms
2Px bonding
Total number of resonating structure
5
  1 .25
4
(c) We know that carbonate ion has following resonating
structures
(a) Bond order 
–
O
2 1 1
  0 .5 .
2
2
–
(a)
C – O–
O
O
Total number of bonds between atoms
Total number of resonating structure

67.
O
C–O 
–
–
O
Bond order 
ClO2
–
O
C=O 
does not have any unpaired electron so it is diamagnetic.
B.O. 
52.
So 2 unpaired of electron present in  2 p y* and  2 p *z .
1 1  2 4
  1 .33 .
3
3
O 2 (15e  )  K : K * ( 2 s)2 ( * 2 s)2 ( 2 p x )2
( 2 p y )2 ( 2 p z )2 ( * 2 p y )1 ( * 2 p z )0
has all paired electrons hence it does not show
paramagnetism.
1
(a) B.O.  [ N b  N a ]
2
1
6
1
6
N 2  [10  4 ]   3 ; O 22   [10  4 ]   3 .
2
2
2
2
1
1
5

(a) B.O. for N 2 = [ N b  N a ] = [9  4 ]   2 .5 .
2
2
2
Hence, bond order 
1
(10  5)  2 .5
2
N 2 (13e  )  KK * ( 2 s)2 ( * 2 s)2 ( 2 p x )2
( 2 p y )2 ( 2 p z )1
1
(9  4 )  2 .5 .
2
(a) Electronic configuration of O 2 is
Hence, bond order 
68.
144 Chemical Bonding
O 2  ( 1s)2 ( * 1s)2 ( 2 s)2 ( * 2 s)2 ( * 2 s)2 ( 2 p z )2
( 2 p x2
69.
70.

 2 p y2 ) ( * 2 p 1x

*
1
1
Hence bond order  N b  N a   [10  6]  2 .
2
2
(c) Nitrogen form triple bond N  N
In which 6 electron take part.
(a) As bond order increase bond length decrease the bond order of
species are
number of bonding electron - Number of a.b. electron

2
10  6
For O 2 
2 ;
2
10  5
O 2 
 2 .5
2
10  7
O 2 
 1 .5
2
O 2  O 2  O 2
So, bond order
O 2
 O2 
O 2
So only B 2 exist unpaired electron and show the
paramagnetism.
2 p 1y )
 2 py 2  * 2 py1
76.
(b) O2   1s 2 ,  * 1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x 2
 2 pz 2  * 2 pz 2
77.
78.
So two unpaired electron found in O 2 at ground stage by
which it shows paramagnetism.
(b) Due to greater electron affinity Cl 2 has the highest bond
energy.
(a) Molecular orbital electronic configuration of these species are :
O2 (17 e  )   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,
 2 p z 2 ,  * 2 p y 2 * 2 p z 1
O2 (16 e )   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,
 2 p z 2 * 2 p y 1 * 2 p z 1
and bond length are
O22  (18 e )   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,
.
 2 p z 2 * 2 p y 2 * 2 p z 2
 2py 2  * 2py1
71.
Hence number of antibonding electrons are 7,6,and 8
respectively.
(b) O 2 :  1s 2 ,  * 1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x 2
 2pz 2  * 2pz1
10  6
 2.0
2
unpaired electrons
79.
(c) Species with unpaired electrons is paramagnetic O 2 has 2
Bond order 
(Two
orbital)
72.
73.
74.
75.
unpaired electrons, O 2 has one unpaired, O 22  has zero
in
antibonding
unpaired electrons, O 22  has one unpaired.
molecular
2
*
1
2  2 py  2 py
O 2 :  1s 2 ,  * 1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x 
2 *
0
 2 pz  2 pz
10  5
Bond order 
 2 .5
2
(One unpaired electron in antibonding molecular orbital so it is
paramagnetic)
(b) Higher the bond order, shorter will be the bond length, thus
NO  having the higher bond order that is 3 as compared to
NO having bond order 2 so NO  has shorter bond length.
(d) Oxygen molecule (O 2 ) boron molecule (B 2 ) and N 2 ion, all
of them have unpaired electron, hence they all are
paramagnetic.
80.
(a)
unpaired electrons while O 22  does not have any unpaired
electron.
81.
(c)
(c) Bond order of NO , NO and NO are 3, 2 .5 and 2
respectively, bond energy  bond order
(a) Paramagnetic property arise through unpaired electron. B 2
molecule have the unpaired electron so it show paramagnetism.
O
C 2   1s  1s ,  2 s  2 s ,  2 p x . 2 p y
2
2
*
2
2
O

(a) From valency bond theory, bond order in CO, i.e. : C  O : is
3, that of O  C  O is 2 while that of CO 32  ion is 1.33.
Since the bond length increases as the bond order decreases,
i.e. CO  CO 2  CO 32  .
83.
(c)
N 2 : KK (2 s)2  * (2 s)2  (2 p x )2  (2 py )2  (2 p z )2
(diamagnetic)
C2 : KK (2 s)  * (2 s)  (2 p x )  (2 py )
2
2
2
2
(diamagnetic)
N 2 : KK (2 s)2  * (2 s)2  (2 p x )2  (2 py )2  (2 p z )2
(No unpaired electron)
(paramagnetic)
N 2   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 2 ,  2 p y 2 2 p z 2
O22 
(No unpaired electron)
: KK (2 s)  * (2 s)  (2 p z )  (2 p x )  (2 py )2
2
2
2
2
 * (2 p x )2  * (2 py )2 (diamagnetic)
F2  s 2 ,  *1s 2 ,  2 s 2 ,  * 2 s 2 ,  2 p x 2 ,  2 p y 2 ,  2 p z 2 ,
84.
 * 2 py 2 , * 2 pz 2
O

82.
2
(No unpaired electron)
O
O  O and O  O .
(2 unpaired electron)
*
O
Due to resonance in O 3 O  O bond length will be in b/w
B2   1s 2 * 1s 2 ,  2 s 2 * 2 s 2 ,  2 p x 1   2 p y 1
2
H OOH ,O  O  O,O  O
O


O 2 has 2 unpaired electron while O 2 and O 2 has one each
(d)
NH 3  107, PH3  93, H 2O  104.5
H 2 Se  91, H 2 S  92.5
Chemical Bonding
Hydrogen bonding
1.
2.
3.
6.
7.
8.
9.
10.
(d) Hydrogen bonding will be maximum in F-H bond due to
greater electronegativity difference.
(b) Ice has hydrogen bonding.
(b) H – F has highest boiling point because it has hydrogen
bonding.
(d) CO 2 is sp-hybridised
48.
(b) sp-hybridization gives two orbitals at 180 o with Linear
structure.
(d) Hydrogen bonding increases the boiling point of compound.
(c) o-Nitrophenol has intramolecular hydrogen bonding but pNitrophenol has intermolecular hydrogen bonding so boiling
point of p-Nitrophenol is more than o-Nitrophenol.
(c) The strongest hydrogen bond is in hydrogen fluoride because
the power of hydrogen bond  electronegativity of atom and
1.
electronegativity 
11.
(d)
1
atomic size
So fluorine has maximum electronegativity and minimum
atomic size.
H 2 O can form hydrogen bonds rest CH 4 and CHCl 3 are
organic compound having no oxygen while NaCl has itself
intraionic attraction in the molecule.
PH 3 has the lowest boiling point because it does not form
Hydrogen bond.
Hydrogen bonding increases heat of vaporisation.
Only NH 3 forms H-bonds.
12.
(b)
14.
15.
(b)
(d)
22.
25.
(a) Water molecule has hydrogen bonding so molecules get
dissociated so it is liquid.
(d) In case of water, five water molecules are attached together
through four hydrogen bonding.
(c) Hydrogen bond is strongest in hydrogen fluoride.
28.
(c) Boiling point of H 2 O is more than that of H 2 S because
23.
H 2 O forms hydrogen bonding while H 2 S does not.
30.
(c)
O
C
31.
34.
35.
36.
37.
38.
39.
40.
41.
43.
45.
46.
47.
H

Interamolecular H-bonding.

O
(a) HydrogenHbond is formed when hydrogen is attached with the
atom which is highly electronegative and having small radius.
(a) Water is dense than ice because of hydrogen bonding
interaction and structure of ice.
(a) Ethanol have hydrogen bonding so its boiling point is higher
than its isomer dimethyl ether.
(a) A compound having maximum electronegative element will
form strong Hydrogen bond.
(a) Due to electronegativity difference of N 2 and H 2 , NH 3
form hydrogen bond.
(b) Intermolecular hydrogen bonding compound contain more b.p.
compare to intramolecular hydrogen bonding compound.
(d) Water molecule contain hydrogen bonding.
(c) It contain intermolecular hydrogen bonding.
(b) Ethyl alcohol has a intermolecular hydrogen bond.
(b) HCl contain weak covalent bond.
(c) Due to intermolecular hydrogen bonding water molecules come
close to each other and exist in liquid state.
(b) Due to greater resonance stabilization.
49.
145
(d) C 2 H 5 OH will dissolve in water because it forms hydrogen
bond with water molecule.
(b) In ice cube all molecules are held by inter molecular hydrogen
bond.
(d) Hydrogen bonding is developed due to inter atomic attraction
so it is the weakest.
Types of bonding and Forces in solid
13.
14.
(b) In electrovalent crystal has cation and anion are attached by
electrostatic forces.
(d) Mercury has very weak interatomic forces so it remains in
liquid state.
(c) The melting and boiling points of argon is low hence, in solid
argon atoms are held together by weak Vander Waal’s forces.
(c) NaF is the strongest ionic crystal so its melting point would be
highest.
(b) Diamond is the hardest substance it’s melting point would be
highest.
(c) Bond is formed by attractive and repulsive forces of both the
atoms.
(a) Generally zero group elements are linked by the Vander Waal’s
force. Hence these show weakest intermolecular forces.
(d) Glycerol has a three OH group hence it is viscous in nature.
(c) Vander waal's forces is the weakest force of attraction.
16.
(b)
2.
3.
4.
9.
10.
12.
NH 4 contain all three types of bond in its structure
H


|


H

N

H


|


H


17.
18.
22.
23.

(d) In NaOH covalent bond is present in O  H bond while
ionic bond is formed between OH  and Na  .
(a) Bond formation is an exothermic reaction so there is decrease
in energy of product.
(d) Blue vitriol is CuSO 4 . 5 H 2 O and it has all types of bonds.

H


|


(a)  H  N  H  Cl 
|


H


Ionic bond = 1, Covalent bond = 3
Co-ordinate bond = 1.
Critical Thinking Questions
1.
(d) We know that ionic characters
 16 [EA  EB ]  3.5  [EA  EB ]2
or ionic characters = 72.24%
3.
(c) Configuration of O 2 molecule is
[ (1s)2   (1s)2  (2 s)2  * (2 s)2  (2 p x )2  (2 p y )2
 (2 p z )2   (2 p x )1   (2 p y )1 ]
No. of pair are 7 so total no. of paired electrons are 14.


H  O :  H  H  O  H
6.
(a)
7.
H
H
(b) The correct order of increasing dipole moment is
|
|
146 Chemical Bonding
p-dichlorobenzene < Toluene < m-dichlorobenzene
< o-
27.
(c)
dichlorobenzene.
8.
(a) The
dipole
moment
of
CH 4  0 D,
and
NF3  0.2 D, NH 3  1.47 D
H 2 O  1.85 D .
Therefore the correct order of the dipole moment is
CH 4  NF3  NH 3  H 2O .
10.
(d) Ammonia molecule is more basic than nitrogen trifluoride and
Boron trifluoride because ammonia molecule easily gives lone
pair of electron.
11.
(a) Chlorine atom in ClO2 is sp 3 hybridised but its shape is
angular.
12.
13.

(c) [ NF3 and H 3 O ] are pyramidal while [
are planar. Hence answer (c) is correct.
(d) CH 2  CH  CH 2  CH 2  C  CH
sp 2
and BF3 ]
Cu
1.
2.
5.
(d) B.O. in CO i.e., : C  O : is 3, that of O  C  O is 2 while
the bond order decreases i.e. CO  CO 2  CO 32 . Thus
option (d) is correct.
(b) Dichromate dianion has following structure
2
9.
6, Cr  O bonds are equivalent.
(b) ClF3 is a [ AB 3 ] type of molecule because it consist of three
bonding pair and two lone pair of electrons hence this
21.
BeF3 does not show
compound is not formed.
(a) K 3 [Fe(CN )6 ]
(a)
sp 3 –hybridization because this
Fe26  4 s 2 3d 6
10.
=
22.
23.
24.
11.
12.
Fe 3   3d 5 4 s 0
 

  
Unpaired electron
d 2sp 3 –hybridization

(d) N 2 has one unpaired electron so it would be paramagnetic.
(a) Each of the species has 14 electron so isoelectronic and shows
bond order 3.
1
1
6
B.O.  [ N b  N a ]  [10  4 ]   3 .
2
2
2
(d)
O
O
S
O O
O
O
S
Trimer of SO 3 .
O
O

SiF4 have sp 3 hybridization & shape of regular tetrahedral
where the bond angle of
F
F  S  F are found
which
is
o
but
109.5o
S
less than 180 o .
F
Repulsion sequence are
F
Lp  Lp  Lp  Bp  Bp  Bp
F
so assertion are true
but the reason are false.
(c) N 2 molecule is diamagnetic. The diamagnetic character is due
to the presence of paired electron N 2 molecule does not
contain any unpaired electron. Thus, assertion is coorect but
the reason is false.
(a) It is correct that during formation of Ice from water there are
vacant spaces between hydrogen bonded molecules of Ice. Ice
has a cage like structure. Due to this reason Ice is less dense
than liquid water. hence both assertion & reason are true &
reason are the correct explanation of assertion.
(b) Water is liquid while H 2 S is gas because oxygen is of small
size & more electronegative in comparision to sulphur. Hence
water molecules exist as associated molecules to form liquid
state due to hydrogen bonding H 2 S does not have hydrogen
bonding & can’t associated hence it is gas.
(d) Iodine is more soluble in CCl 4 than in H 2 O because iodine
13.
(a)
14.
(e)
15.
(c)
16.
(c)
17.
(c)
18.
(b)
O
S


 O . 5 H 2O .


greater than 90
compound shows sp 3 d hybridization.
20.
(c)
109.5 o
O
O






O  Cr  O  Cr  O 




O
O


17.
O

  
O  S


O

(a) Solubility in water depends on hydration energy and lattice
energy.
(a) Polarity in covalent bond developed due to shifting of electrons
towards one of the bonded atoms.

that of CO 32  ion is 1.33. Since the bond length increases as
15.
2
Assertion & Reason
sp 3
hybridised

14.
NO 3
CuSO 4 .5 H 2 O has electrovalent, covalent and coordinate
bonds.
is non polar & thus it dissolve in CCl 4 because like dissolves
like.
o & p -nitrophenols can be separated by steam distillation
because o -nitrophenol is steam volatile. Here, both assertion &
reason are correct & reason is correct explanation of assertion.
Fluorine is highly reactive F  F bond has low bond
dissociation energy. Here assertion is false but reason is true.
It is true that sigma ( ) bond is stronger than pi ( ) bond but
the reason that there is free rotation of atoms is false.
Energy is released in the formation of the crystal lattice. It is
qualitative measure of the stability of an ionic compound so
assertion is true & reason are false.
Li, Na & K are alkali metals & not alkaline earth metal so,
size of alkali metal increases So. Assertion is true & reason are
false.
Hess’s law states that the enthalpy of a reaction is the same,
whether it takes place in a single step or in more than one
step. In born haber cycle the formation of an cycle ionic
compound may occur either by direct combination of the
element or by a stepwise process involving vaporization of
elements, conversion of the gaseous atoms into ions & the
combination of the gaseous ions to form the ionic solid.
Chemical Bonding
19.
20.
21.
(a) With increase in bond order, bond length decreases & hence bond
energy increases so both assertion & reason are true & reason are
the correct explanation of assertion.
(c) Electron affinity is experimentally measurable while
electronegativity is a relative number so assertion is true but
reason are false.
(b) Assertion & reason both are correct but reason is not the
correct explanation of assertion sulphur has five electrons pairs
whose arrangement should be trigonal bipyramidal according
to VSEPR theory. Two structure are possible
F
F
F
|
..
:S
FS
F
F
F
|
F
(b)
Lone pair in the equatorial
position (two L.p – b.p
repulsion)
(a)
Lone pair in the axial
position (three l.p – b.p
repulsion at 90o)
22.
(e)
BF3 has zero dipole moment because of its structure.
F
F  B    0
F
23.
24.
H 2 S has two lone pairs on sulphur atom & hence. It has
irregular shape.
Thus it possess dipole moment. So assertion is false but reason
are true.
(d) Both assertion & reason are false because pairs of electron will
have different spins. Electrons are equally shared between
them.
(d) In B2, total number of electrons = 10
B  (1s) *(1s ) (2s) *(2s) (2p ) (2p )
Presence of unpaired electron shows the paramagnetic nature.
The highest occupied molecular orbital is of -type.
(a) Both assertion & reason are true & reason is the correct
explanation of the assertion because. At any given instant, at
room temperature each water molecules forms hydrogen bonds
with other water molecules. The H 2 O molecules are in
continuous motion. So hydrogen bonds are constantly & rapidly
broken & formed. In Ice H 2 O molecules are however fixed in
the space lattice.
(a) Both assertion & reason are true & reason is the correct
explanation of assertion, because helium molecule is formed by
linking two helium atoms. both have 1s orbitals. These will
2
2
25.
26.
2
2
2
1
x
1
y
combine to form two molecular orbitals  ( 1s ) &  * ( 1s )
four available electrons are accommodated as  (1s)2 &
 * (1s)2 .
147
Chemical Bonding
1.
2.
Nature of the bond formed between two elements depends on the
(a) Oxidation potential
(b) Electronegativity
(c) Ionization potential
(d) Electron affinity
(c)
12.
Two elements X and Y have following electronic configurations
X  1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 6 , 4 s 2
Y  1s 2 , 2 s 2 2 p 6 , 3 s 2 3 p 5 . The expected compound formed by
[BHU 1990]
(a)
(b)
(c)
3.
4.
5.
XY 2
X 2 Y5
(d)
13.
[IIT 1988]
(a)
Electricity do not pass through ionic compounds
(a) In solution
(b) In solid state
(c) In melted state
(d) None of these
From the following which compound on heating readily sublimes
(b)
MgCl2
(c)
BaCl2
(d)
AlCl3
14.


C H 2  CH  CH  O
The order of dipole moments of the following molecules is
[Roorkee 2000]
(a)
15.
7.
(a) Free atoms
(b) Free molecules
(c) Free ions
(d) Free atoms and free molecules
In which of the following reactions, there is no change in the valency
[NCERT 1972; MP PET 2000]
[NCERT 1974; CPMT 1971, 78]
(a)
4 KClO3  3 KClO4  KCl
(b)
SO 2  2 H 2 S  2 H 2 O  3 S
(c)
BaO2  H 2 SO 4  BaSO 4  H 2 O 2
(d)
2 BaO  O 2  2 BaO2
The octet rule is not followed in
(a) F2
(b)
NaF
(c)
BF3
CaF2
(d)
NaCl
CHCl 3  CH 2 Cl 2  CH 3 Cl  CCl 4
BH 4
[DPMT 1982; CBSE PMT 1992; CPMT 1999]
17.
(b) C  Br
(d) C  F
C O
CS
The geometry of H 2 S and its dipole moment are
[IIT 1999]
(a) Angular and non-zero
(b) Angular and zero
(c) Linear and non-zero
(d) Linear and zero
How many  and  bonds are there in the molecule of
tetracyanoethylene
N C
CN
CC
N C
CN
[NCERT 1980; MP PMT 1986, 95;Orissa JEE 1997]
(d) Cl 2
(a) Nine  and nine 
(c) Nine  and seven 

(d) Five  and eight 
The shape of H 3 O
20.
(a) Linear
(b) Angular
(c) Trigonal planar
(d) Triangular pyramidal
The hybridization in sulphur dioxide is[IIT 1986; DPMT 1990]
(a)
ion is
(b) Five  and nine 
19.
21.
[RPMT 2002]
(b) CO 32
The electronegativity of C, H , O, N and S are 2.5, 2.1, 3.5, 3.0
and 2.5 respectively. Which of the following bond is most polar
(a) O  H
(b) S  H
(c) N  H
(d) C  H
Which of the following bond has the most polar character
(a)
(c)
[BHU 1981]
Co-ordinate bond is absent in
(a)
16.
18.
Sodium chloride is an ionic compound whereas hydrogen chloride is
a gas because
[KCET 2002]
(a) Sodium is reactive
(b) Covalent bond is weaker than ionic bond
(c) Hydrogen chloride is a gas
(d) Covalent bond is stronger than ionic bond
Which one of the following molecules has a coordinate bond
(a) NH 4 Cl
(b) AlCl3
(c)
11.

C H 2  CH  CH  O
2
6.
10.

C H 2  C H  CH  O (d)

(b) CH 2 Cl 2  CH 3 Cl  CHCl 3  CCl 4
(c) KCN
(d) KCl
The solution of sugar in water contains
9.
C H 2  CH  C H  O (b)

(c)

(c) CH 3 Cl  CH 2 Cl 2  CHCl 3  CCl 4
Which one in the following contains ionic as well as covalent bond [IIT 1979; CPMT 1983; DPMT
1983]
(d)
CH
Cl
2
2  CHCl 3  CH 3 Cl  CCl 4
(a) CH
(b) H
4
8.
(a) 3.46 D
(b) 0.00 D
(c) 1.73 D
(d) 1.00 D
Polarization of electrons in acrolein may be written as

XY 5
NaCl
NH 4 
The dipole moment of chlorobenzene is 1.73 D. The dipole moment
of p -dichlorobenzene is expected to be
X 5Y2
(a)
(d)
[CPMT 1991]
and
combination of X and Y is
H3O
149
sp[CPMT 1988, 94]
[EAMCET 1993; CPMT 2001]
(b)
sp 3
(c) sp 2
(d) dsp 2
The number and type of bonds between two carbon atoms in
CaC 2 are
[IIT 1996]
(a) One sigma ( ) and one pi ( ) bonds
150 Chemical Bonding
(b) One sigma ( ) and two pi ( ) bonds
(a) 3
(c) 1
(c) One sigma ( ) and one and a half pi ( ) bonds
28.
(d) One sigma ( ) bond
22.
23.
N  N O
and
NH 4
(a)
sp, sp 3 and sp 2 respectively
are
(b) sp, sp and sp
29.
NN O
The hybridization of atomic orbitals of nitrogen in
2
24.
(d)
NO 2
,
NO 3
,
30.
27.
(c)
(d)
2 p z orbital
(c)
sp 2 , sp and sp 3 respectively
(d)
sp 2 , sp 3 and sp respectively
(a) 4
(b) 3
(c) 2
(d) 1
Hydrogen bonding is not present in
[AIIMS 1998; MP PET/PMT 1998]
The molecule having one unpaired electron is
(a)
NO
(b) CO
(c)
CN 
(d) O 2
The geometry of ClO3 , according to valence shell electron pair
repulsion (VSEPR) theory will be
31.
32.
(a) Planar triangle
(b) Pyramidal
(c) Tetrahedral
(d) Square planar
Which of the following halogens has the highest bond energy
(a)
F2
(b) Cl 2
(c)
Br2
(d)
What bond order does O 22  have
2 p x orbital
The maximum number of hydrogen bonds formed by a water
molecule in ice is
(a)
(b)
(c)
(d)
[KCET 1996; MP PET 1997]
26.
(b) Antibonding -orbital
respectively
[IIT 1985; MP PMT 1989]
25.
(a) Bonding -orbital
[MP PET 1993; AFMC 2002;UPSEAT 1999, 2001, 02]
[IIT Screening 2000]
3
In the process, O 2  O 22  e the electron lost is from
[Orissa JEE 2002]
Which of the following resonating structures of N 2 O is the most
contributing
[Roorkee Qualifying 1998]
(a) N  N  O
(b) N  N  O
(c)
(b) 2
(d) 1/2
Glycerine
Water
Hydrogen sulphide
Hydrogen fluoride
The bonds in K 4 [Fe (CN )6 ] are
[EAMCET 1991]
(a) All ionic
(b) All covalent
(c) Ionic and covalent
(d) Ionic, covalent and coordinate covalent
In which of the following ionic, covalent and coordinate bonds are
present
[UPSEAT 2002]
(a) Water
(b) Ammonia
[CPMT 1988]
(c) Sodium cyanide
(d) Potassium bromide
I2
[Pb. PMT 2001]
(SET -3)
1.
2.
(b) If the two elements have similar electronegativities,the bond
between them will be covalent, while a large difference in
electronegativities leads to an ionic bond.
(a) From electronic configuration valencies of X and Y are + 2 and
–1 respectively so formula of compound is XY 2 .
3.
4.
(b) Ionic compounds can’t pass electricity in solid state because
they don’t have mobile ion in solid state.
(d) AlCl3 sublimes readily on heating.
5.
(c) Structure of KCN is [K  (C   N )] .
..
Chemical Bonding
6.
(b) Sugar is an organic compound which is covalently bonded so in
water it remains as free molecules.
7.
(c) In the reaction BaO2  H 2 SO 4  BaSO 4  H 2 O valency
is not changing.
(d) BF3 does not have octet, it has only six electrons so it is
electron deficient compound.
(b) NaCl is a ionic compound because it consists of more
elelctronegativity difference compare to HCl.
(a) NH 4 Cl has a coordinate bond besides covalent and ionic
2
8.
9.
10.
H

|

bonds  H  N
|

H

O
6
(c)
28.
29.
1
2
[10  8 ]   1 .
2
2
(b) Electron lost from antibonding  orbital.
(a) In ice each water molecule forms four hydrogen bond through
which each water molecule is tetrahedrally attached with other
water molecule.
2 6
B.O. 
H
O
O
has slightly higher dipole
electronegativity then CH 2 Cl 2 . But CH 2 Cl 2 has greater
16.
17.
dipole moment than CHCl 3 .
(a) More the difference in electronegativity of atoms. Bond
between them will be more polar.
(d) C  F bond has the most polar character due to difference of
their electronegativity.
(a) H 2 S has angular geometry and have some value of dipole
moment.

N
N   C
C 


18.
(a)



C 
C

N 
C




N
C 

20.
9 and 9 bonds.
(d) H 3 O  has sp 3 hybridization and its shape is triangular
pyramidal due to lone pair on oxygen.
(c) SO 2 molecule has sp 2 hybridisation.
21.
(b) In
22.
C
(a) In N 2 O molecule N  N  O structure is most contributed.
19.
C
23.
24.
Ca two carbons are joined with 1 and 2 bonds.
***
(b) The shape of NO 2 , NO 3 and NH 4 are linear trigonal
planar and tetrahedral respectively. Thus the hybridization of
atomic orbitals of nitrogen in these species are sp, sp 2 and
sp 3 respectively.
(a) NO has one unpaired electron with Nitrogen.
..
: N :: O :
. ..

..
O  Cl  O
25.
(b)
26.
O
(b) Bond energy of Cl 2 is highest among all halogen molecule.
|
Bond energies of F2 , Cl2 , Br2 , I2 are 37, 58, 46 and 36 Kcal
mol 1 respectively.
H
H
H
H
moment which is equal to 1.86D. Now CH 3 Cl has less
15.
O
O
O
H
H
H
(b)  O  C  O has covalent bonds only.
(b) Due to symmetry dipole moment of p-dichloro benzene is zero.
(d)
(d) CCl 4 has zero dipole moment because of symmetric
tetrahedral structure. CH 3 Cl
H
H
H


H  Cl 


|
11.
12.
13.
14.
O 22  have bond order one
27.


151
H
H
O
O
H
H
H
O
O
31.
(c) Hydrogen bonding
H is present in molecules
H which have F, O, or
N atoms.
(d) Structure of K 4 [Fe(CN )6 ] is
32.


CN


C  N

C

N


4K 
Fe



C

N
C

N


CN


(c) Sodium cyanide contain ionic, covalent and coordinate bond.
30.
4
152 Solution and Colligative properties
Chapter
4
Solution & Colligative properties
“A solution is a mixture in which substances are intermixed so
intimately that they can not be observed as separate components”. The
dispersed phase or the substance which is to be dissolved is called solute,
while the dispersion medium in which the solute is dispersed to get a
homogenous mixture is called the solvent.
Solubility
“Solubility of a substance may be defined as the amount of solute
dissolved in 100gms of a solvent to form a saturated solution at a given
temperature”. A saturated solution is a solution which contains at a given
temperature as much solute as it can hold in presence of dissolveding
solvent. Any solution may contain less solute than would be necessary to
saturate it. Such a solution is known as unsaturated solution. When the
solution contains more solute than would be necessary to saturate it then it
is termed as supersaturated solution.
Kinds of solutions
All the three states of matter (gas, liquid or solid) may behave either
as solvent or solute. Depending on the state of solute or solvent, mainly
there may be following nine types of binary solutions.
Solvent
Gas
Gas
Gas
Liquid
Liquid
Liquid
Solid
Solute
Gas
Liquid
Solid
Gas
Liquid
Solid
Gas
Solid
Solid
Liquid
Solid
Example
Mixture of gases, air.
Water vapours in air, mist.
Sublimation of a solid into a gas, smoke.
CO2 gas dissolved in water (aerated drinks).
Mixture of miscible liquids, e.g., alcohol in water.
Salt in water, sugar in water.
Adsorption of gases over metals; hydrogen over
palladium.
Mercury in zinc, mercury in gold, CuSO4.5H2O.
Homogeneous mixture of two or more metals
(alloys), e.g., copper in gold, zinc in copper.
Among these solutions the most significant type of solutions are
those which are in liquid phase and may be categorised as,
(1) Solid in
liquid solutions, (2) Liquid in liquid solutions and
(3) Gas in liquid
solutions.
Methods of expressing concentration of solution
Concentration of solution is the amount of solute dissolved in a
known amount of the solvent or solution. The concentration of solution can
be expressed in various ways as discussed below,
(1) Percentage : It refers to the amount of the solute per 100 parts of
the solution. It can also be called as parts per hundred (pph). It can be
expressed by any of following four methods,
(i) Weight to weight percent
Wt. of solute
 100
Wt. of solution
% w/w 
Example : 10% Na 2 CO 3 solution w/w means 10 g of Na 2 CO 3
is dissolved in 100 g of the solution. (It means 10 g
Na 2 CO 3 is
dissolved in 90 g of H 2 O )
(ii) Weight to volume percent
Wt. of solute
 100
Volume of solution
% w/v 
Example : 10% Na 2 CO 3 (w/v) means 10 g
Na 2 CO 3
is
dissolved in 100 cc of solution.
(iii) Volume to volume percent
% v/v 
Vol. of solute
 100
Vol. of solution
Example : 10% ethanol (v/v) means 10 cc of ethanol dissolved in
100 cc of solution.
(iv) Volume to weight percent
% v/w 
Vol. of solute
 100
Wt. of solution
Example : 10% ethanol (v/w) means 10 cc of ethanol dissolved in
100 g of solution.
(2) Parts per million (ppm) and parts per billion (ppb) : When a
solute is present in trace quantities, it is convenient to express the
concentration in parts per million and parts per billion. It is the number of
parts of solute per million (10 6 ) or per billion (10 9 ) parts of the
solution. It is independent of the temperature.
ppm 
mass of solutecomponent
 10 6
Total mass of solution
Solution and Colligative properties
ppb 
mass of solutecomponent
 10 9
Total mass of solution
(xi) If W g of an acid is completely neutralised by V ml of base of
normality N
(3) Strength : The strength of solution is defined as the amount of
Wt. of acid
VN

g eq. wt. of acid 1000
solute in grams present in one litre (or dm 3 ) of the solution. It is
expressed in g/litre or (g / dm 3 ) .
Strength 
Similarly,
Mass of solutein grams
Volume of solutionin litres
(4) Normality (N) : It is defined as the number of gram equivalents
(equivalent weight in grams) of a solute present per litre of the solution.
Unit of normality is gram equivalents litre . Normality changes with
temperature since it involves volume. When a solution is diluted x times,
its normality also decreases by x times. Solutions in term of normality
generally expressed as,
Wt. of base
Vol. of acid  N of acid

g eq. wt. of base
1000
(xii) When Va ml of acid of normality N a is mixed with Vb ml
of base of normality N b
(a) If Va N a  Vb N b (Solution neutral)
–1
(b) If Va N a  Vb N b (Solution is acidic)
(c) If Vb N b  Va N a (Solution is basic)
N  Normal solution; 5 N  Penta normal,
(xiii) Normality of the acidic mixture 
Va N a  Vb N b
(Va  Vb )
(xiv) Normality of the basic mixture 
Vb N b  Va N a
(Va  Vb )
10 N  Deca normal; N / 2  semi normal
N / 10  Deci normal; N / 5  Penti normal
N / 100 or 0.01 N  centinormal,
N / 1000 or
Mathematically normality can be calculated by following formulas,
(ii) N 
Number of g.eq. of solute
Volume of solution(l)
Weight of solutein g.
g. eq. weight of solute Volume of solution(l)
Wt. of soluteper litreof solution
g eq. wt. of solute
(iii) N 
(iv) N 
Wt. of solute
1000

g.eq. wt. of solute Vol. of solutionin ml
(v) N 
Percent of solute 10
,
g eq. wt. of solute
(vi) N 
Strength in g l -1 of solution
g eq. wt. of solute
(vii) N 
No. of meq * of solute
Vol. of solutionin ml
(* 1 equivalent = 1000 milliequivalent or meq.)
(5) Molarity (M) : Molarity of a solution is the number of moles of
the solute per litre of solution (or number of millimoles per ml. of solution).
Unit of molarity is mol/litre or mol/dm For example, a molar (1 M ) solution
of sugar means a solution containing 1 mole of sugar (i.e., 342 g or
(xv) N 
0.001= millinormal
(i) Normality ( N ) 
3
6.02  10 23 molecules of it) per litre of the solution. Solutions in term of
molarity generally expressed as,
1 M = Molar solution, 2 M = Molarity is two,
M
or 0.5 M = Semimolar solution,
2
M
or 0.1 M = Decimolar solution,
10
M
or 0.01 M = Centimolar solution
100
M
or 0.001 M = Millimolar solution
1000
 Molarity is most common way of representing the concentration
of solution.
Wt%  density  10
Eq. wt.
 Molarity is depend on temperature as, M 
(viii) If volume V1 and normality N 1 is so changed that new
normality and volume N 2 and V2 then,
N 1 V1  N 2 V2 (Normality equation)
N 1 V1  N 2 V2
V1  V2
(x) Vol. of water to be added i.e., (V2  V1 ) to get a solution of
normality N 2 from V1 ml of normality N 1
 N  N2
V2  V1   1
 N2
1
T
 When a solution is diluted (x times), its molarity also decreases
(ix) When two solutions of the same solute are mixed then normality
of mixture (N ) is
N
153

 V1


(by
x times)
Mathematically molarity can be calculated by following formulas,
(i) M 
No. of moles of solute( n)
,
Vol. of solutionin litres
(ii) M 
Wt. of solute(in gm) per litreof solution
Mol.wt. of solute
(iii) M 
Wt. of solute(in gm)
1000

Mol.wt. of solute
Vol. of solutionin ml.
(iv) M 
No. of millimoles of solute
Vol. of solutionin ml
(v) M 
Percent of solute 10
Mol.wt. of solute
154
Solution and Colligative properties
(vi) M 
Strength in gl-1 of solution
Mol.wt. of solute
(vi) m 
10  solubility
Mol.wt. of solute
(vii) M 
10  Sp. gr. of the solution Wt.% of the solute
Mol.wt. of the solute
(vii) m 
1000  wt.% of solute(x )
(100  x )  mol. wt. of solute
(viii) m 
1000  Molarity
(1000  sp. gravity) (Molarity Mol.wt. of solute)
(viii) If molarity and volume of solution are changed from M 1 , V1
to M 2 , V2 . Then,
M 1 V1  M 2 V2 (Molarity equation)
(ix) In balanced chemical equation, if n 1 moles of reactant one
react with n 2 moles of reactant two. Then,
M 1 V1
M V
 2 2
n1
n2
Relation between molarity (M) and molality (m)
Molality (m) =
Molarity (M) 
(x) If two solutions of the same solute are mixed then molarity (M)
of resulting solution.
M
M 1 V1  M 2 V2
(V1  V2 )
(xi) Volume of water added to get a solution of molarity M 2 from
V1 ml of molarity M 1 is
 M  M2 
 V1
V2  V1   1

 M2 
Relation between molarity and normality
Normality of solution = molarity 
Molecular mass
Equivalent mass
Normality  equivalent mass = molarity  molecular mass
For an acid,
Molecular mass
= Basicity
Equivalent mass
So, Normality of acid = molarity  basicity.
Molecular mass
For a base,
= Acidity
Equivalent mass
So, Normality of base = Molarity  Acidity.
(6) Molality (m) : It is the number of moles or gram molecules of
the solute per 1000 g of the solvent. Unit of molality is mol / kg . For
Molarity
Molarity molecular mass
Density 
1000
Molality density
Molality molecular mass
1
1000
(7) Formality (F) : Formality of a solution may be defined as the
number of gram formula masses of the ionic solute dissolved per litre of the
solution. It is represented by F . Commonly, the term formality is used to
express the concentration of the ionic solids which do not exist as molecules
but exist as network of ions. A solution containing one gram formula mass
of solute per litre of the solution has formality equal to one and is called
formal solution. It may be mentioned here that the formality of a solution
changes with change in temperature.
Formality (F)=
=
Number of gram formula masses of solute
Volume of solutionin litres
Mass of ionic solute(g)
(gm. formula mass of solute) (Volume of solution(l))
Thus, F 
W B (g)
GFM  V (l)
or
WB (g)  1000
GFM  V (ml )
(8) Mole fraction (X) : Mole fraction may be defined as the ratio of
number of moles of one component to the total number of moles of all the
components (solvent and solute) present in the solution. It is denoted by
the letter X . It may be noted that the mole fraction is independent of the
temperature. Mole fraction is dimensionless. Let us suppose that a solution
contains the components A and B and suppose that W A g of A and
WB g of B are present in it.
example, a 0 .2 molal (0.2m ) solution of glucose means a solution
obtained by dissolving 0.2 mole of glucose in 1000 gm of water. Molality
(m) does not depend on temperature since it involves measurement of
weight of liquids. Molal solutions are less concentrated than molar solution.
Mathematically molality can be calculated by following formulas,
(i) m 
Number of moles of the solute
 1000
Weight of the solvent in grams
(ii) m 
Strength per 1000 grams of solvent
Molecular mass of solute
(iii) m 
No. of gm moles of solute
Wt. of solvent in kg
Wt. of solute
1000

(iv) m 
Mol.wt. of solute Wt. of solvent in g
(v) m 
No. of millimoles of solute
Wt. of solvent in g
Number of moles of A is given by, n A 
WA
and
MA
the number of moles of B is given by, n B 
WB
MB
where M A and M B are molecular masses of A and B
respectively.
Total number of moles of A and B  n A  n B
Mole fraction of A , X A 
nA
n A  nB
Mole fraction of B , X B 
nB
n A  nB
The sum of mole fractions of all the components in the solution is
always one.
nA
nB
XA  XB 

1.
n A  nB n A  nB
Thus, if we know the mole fraction of one component of a binary
solution, the mole fraction of the other can be calculated.
Solution and Colligative properties
Relation between molality of solution (m) and mole fraction of the
solute (X ).
m
XA 
55.5  m
(9) Mass fraction : Mass fraction of a component in a solution is the
mass of that component divided by the total mass of the solution. For a
solution containing w A gm of A and w B gm of B
A
wA
wB
; Mas fraction of B 
w A  wB
w A  wB
It may be noted that molality, mole fraction, mass fraction etc. are
preferred to molarity, normality, etc. because the former involve the weights
of the solute and solvent where as later involve volumes of solutions.
Temperature has no effect on weights but it has significant effect on
volumes.
(10) Demal unit (D) : The concentrations are also expressed in
“Demal unit”. One demal unit represents one mole of solute present in one
litre of solution at 0 o C .
Mass fraction of A 
Colligative properties
Certain properties of dilute solutions containing non-volatile solute
do not depend upon the nature of the solute dissolved but depend only
upon the concentration i.e., the number of particles of the solute present in
the solution. Such properties are called colligative properties. The four well
known examples of the colligative properties are,
(1) Lowering of vapour pressure of the solvent.
(2) Osmotic pressure of the solution.
(3) Elevation in boiling point of the solvent.
(4) Depression in freezing point of the solvent.
Since colligative properties depend upon the number of solute
particles present in the solution, the simple case will be that when the
solute is a non-electrolyte. In case the solute is an electrolyte, it may split to
a number of ions each of which acts as a particle and thus will affect the
value of the colligative property.
Each colligative property is exactly related to other, Relative lowering
of vapour pressure, elevation in boiling point and depression in freezing
point are directly proportional to osmotic pressure.
155
(3) Purity of liquid : Pure liquid always has a vapour pressure
greater than its solution.
Raoult’s law : When a non-volatile substance is dissolved in a
liquid, the vapour pressure of the liquid (solvent) is lowered. According
to Raoult’s law (1887), at any given temperature the partial vapour
pressure (p ) of any component of a solution is equal to its mole
fraction (X ) multiplied by the vapour pressure of this component in the
A
A
pure state ( p 0A ) . That is,
p A  p 0A  X A
The vapour pressure of the solution (Ptotal) is the sum of the
partial pressures of the components, i.e., for the solution of two volatile
liquids with vapour pressures p A and p B .
Ptotal  p A  p B  (p 0A  X A )  (p B0  X B )
Alternatively, Raoult’s law may be stated as “the relative lowering of
vapour pressure of a solution containing a non-volatile solute is equal to the
mole fraction of the solute in the solution.”
Relative lowering of vapour pressure is defined as the ratio of
lowering of vapour pressure to the vapour pressure of the pure solvent. It is
determined by Ostwald-Walker method.
Thus according to Raoult’s law,
w
p0  p
n
m


w W
nN
p0

m M
where, p  Vapour pressure of the solution
p 0  Vapour pressure of the pure solvent
n  Number of moles of the solute
N  Number of moles of the solvent
w and m  weight and mol. wt. of solute
W and M  weight and mol. wt. of the solvent.
Limitations of Raoult’s law
 Raoult’s law is applicable only to very dilute solutions.
Lowering of vapour pressure
The pressure exerted by the vapours above the liquid surface in
equilibrium with the liquid at a given temperature is called vapour pressure
of the liquid. The vapour pressure of a liquid depends on,
(1) Nature of liquid : Liquids, which have weak intermolecular forces,
are volatile and have greater vapour pressure. For example, dimethyl ether
has greater vapour pressure than ethyl alcohol.
(2) Temperature : Vapour pressure increases with increase in
temperature . This is due to the reason that with increase in
temperature more molecules of the liquid can go into vapour phase.

Raoult’s law is applicable to solutions containing non-volatile
solute only.

Raoult’s law is not applicable to solutes which dissociate or
associate in the particular solution.
Ideal and Non-Ideal solution
Table: 4.1 Ideal and non-ideal solutions
Non-ideal solutions
Ideal solutions
Positive deviation from Raoult's law
1.
Obey Raoult's law at every range of
concentration.
1.
2.
H mix  0; neither heat is evolved nor
2.
absorbed during dissolution.
3.
of
volumes
Do not obey Raoult's law.
1.
H mix  0. Endothermic dissolution; heat
2.
is absorbed.
Vmix  0; total volume of solution is
equal to sum
components.
Negative deviation from Raoult's law
of
the
3.
Vmix  0. Volume
dissolution.
Do not obey Raoult's law.
H mix  0. Exothermic dissolution; heat is
evolved.
is
increased
after
3.
Vmix  0. Volume is decreased during
dissolution.
156 Solution and Colligative properties
4.
P  p A  p B  p 0A X A  p B0 X B i.e.,
4.
A  A, A  B, B  B interactions should be
Escaping tendency of 'A' and 'B' should be
same in pure liquids and in the solution.
p A  p 0A X A ; p B  p B0 X B
 p A  p B  p 0A X A  p B0 X B
5.
A  B attractive force should be weaker
than A  A and B  B attractive forces.
'A' and 'B' have different shape, size and
character.
5.
A  B attractive force should be greater
than A  A and B  B attractive forces.
'A' and 'B' have different shape, size and
character.
6.
'A' and 'B' escape easily showing higher
vapour pressure than the expected value.
6.
Escaping tendency of both components 'A'
and 'B' is lowered showing lower vapour
pressure than expected ideally.
same, i.e., 'A' and 'B' are identical in shape, size
and character.
6.
4.
 p A  p B  p 0A X A  p B0 X B
p A  p 0A X A : p B  p B0 X B
5.
p A  p 0A X A ; p B  p B0 X B
Examples:
Examples:
Examples:
Dilute solutions;
Acetone +ethanol
Acetone + aniline;
benzene + toluene:
acetone + CS 2 :
acetone + chloroform;
n-hexane + n-heptane;
water + methanol;
CH 3 OH  CH 3 COOH ;
water + ethanol;
H 2O  HNO3
CCl 4  toluene;
chloroform + diethyl ether;
CCl 4  CHCl 3 ;
water + HCl;
chlorobenzene + bromobenzene;
ethyl bromide + ethyl iodide;
n-butyl chloride + n-butyl bromide
acetic acid + pyridine;
acetone + benzene;
chloroform + benzene
CCl 4  CH 3 OH ;
cyclohexane + ethanol
Graphical representation of ideal and non-ideal solutions
Positive deviation
Ideal solution
Negative deviation
Total vapour pressure
P=pA + pB
p°A
p°B
PB=p0B XB
Total vapour pressure
p°B
p°B
Vapour pressure
PA=p0A XA
p°A
Vapour pressure
Vapour pressure
p°A
Ideal
Ideal
XA = 1
XB = 0
Mole fraction
XA = 0
XB = 1
XA = 1
XB = 0
Mole fraction
Azeotropic mixture
Azeotropes are defined as the mixtures of liquids which boil at
constant temperature like a pure liquid and possess same composition of
components in liquid as well as in vapour phase. Azeotropes are also called
constant boiling mixtures because whole of the azeotropes changes into
vapour state at constant temperature and their components can not be
separated by fractional distillation. Azeotropes are of two types as described
below,
(1) Minimum boiling azeotrope : For the solutions with positive
deviation there is an intermediate composition for which the vapour
pressure of the solution is maximum and hence, boiling point is minimum.
At this composition the solution distills at constant temperature without
change in composition. This type of solutions are called minimum boiling
azeotrope. e.g.,
H 2O  C2 H5 OH, H 2O  C2 H5 CH 2OH
CHCl 3  C 2 H 5 OH , (CH 3 )2 CO  CS 2
(2) Maximum boiling azeotrope : For the solutions with negative
deviations there is an intermediate composition for which the vapour
pressure of the solution is minimum and hence, boiling point is maximum.
At this composition the solution distill`s at constant temperature without
the change in composition. This type of solutions are called maximum
boiling azeotrope. e.g.,
XB = 1
XA= 0
XA = 1
XB = 0
Mole fraction
XA = 0
XB = 1
H 2 O  HCl, H 2 O  HNO 3 , H 2 O  HClO4
Osmosis and Osmotic pressure of the solution
(1) Osmosis : The flow of solvent from pure solvent or from solution
of lower concentration into solution of higher concentration through a
semi-permeable membrane is called Osmosis. Osmosis may be divided in
following types,
(i) Exo-Osmosis : The outward osmotic flow of water from a cell
containing an aqueous solution through a semi-permeable membrane is
called as Exo-osmosis. For example, egg (after removing hard shell) placed
in conc. NaCl solutions, will shrink due to exo-osmosis.
(ii) Endo-osmosis : The inward flow of water into the cell containing
an aqueous solution through a semi-permeable membrane is called as endoosmosis. e.g., an egg placed in water swells up due to endo-osmosis.
(iii) Reverse osmosis : If a pressure higher than osmotic pressure is
applied on the solution, the solvent will flow from the solution into the pure
solvent through the semi-permeable membrane. Since here the flow of
solvent is in the reverse direction to that observed in the usual osmosis, the
process is called reverse osmosis.
Differences between osmosis and diffusion
Osmosis
In osmosis movement of molecules
Diffusion
In diffusion there is no role of semi-
Solution and Colligative properties
takes place through a semi-permeable
membrane.
permeable membrane.
It involves movement of only solvent
molecules from one side to the other.
It involves passage of solvent as well as
solute molecules from one region to
the other.
Osmosis is limited to solutions only.
Diffusion can take place in liquids,
gases and solutions.
Osmosis can be stopped or reversed by
applying additional pressure on the
solution side.
Diffusion can neither be stopped nor
reversed
 C
.....(i)
According to Gaylussac Van't Hoff law (at conc. temp.)
 T
From equation (i) and (ii)
.....(ii)
(Van't Hoff equation)
w RT
n
w


 C   ;  
 n  
m
V
V
m




Here, C = concentration of solution in moles per litre
R = gas constant ; T = temperature
n = number of moles of solute ; V = volume of solution
m = molecular weight of solute ; w = weight of solute
(iii) Conditions for getting accurate value of molecular mass are,
(a) The solute must be non-volatile.
(b) The solution must be dilute.
(c) The solute should not undergo dissociation or association in the
solution.
(iv) Relation of osmotic pressure with different colligative properties
: Osmotic pressure is related to relative lowering of vapour pressure,
elevation of boiling point and depression of freezing point according to the
following relations,

n
RT
V

(a)   

PAo
 PA
PAo
(c)   Tf 
 dRT

 M

dRT
1000  K f
In the above relations,  = Osmotic pressure; d = Density of solution
at temperature T; R = Universal gas constant; M = Mol. Mass of solute;
K f = Molal depression
Kb = Molal elevation constant of solvent;
constant of solvent
(v) Isotonic, Hypertonic and Hypotonic solutions
(a) Isotonic or iso-osmotic solutions : Two solutions of different
substances having same osmotic pressure at same temperature are known
as isotonic solutions.
For isotonic solutions,  1   2 Primary Condition …..(i)
(2) Osmotic pressure ()
The osmotic pressure of a solution at a particular temperature may
be defined as the excess hydrostatic pressure that builds up when the
solution is separated from the solvent by a semi-permeable membrane. It is
denoted by .
or
Osmotic pressure may be defined as the excess pressure which must
be applied to a solution in order to prevent flow of solvent into the solution
through the semi-permeable membrane.
or
Osmotic pressure is the excess pressure which must be applied to a
given solution in order to increase its vapour pressure until it becomes
equal to that of the solution.
(i) Measurements of osmotic pressure : Following methods are used
for the measurement of osmotic pressure,
(a) Pfeffer’s method, (b) Morse and Frazer’s method,
(c)
Berkeley and Hartley’s method, (d) Townsend’s negative pressure method,
(e) De Vries plasmolytic method.
(ii) Determination of molecular mass of non-volatile solute from
osmotic pressure () : The osmotic pressure is a colligative property. For a
given solvent the osmotic pressure depends only upon the molar
concentration of solute but does not depend upon its nature. The following
relation relates osmotic pressure to the number of moles of the solute,
According to Boyle Van't Hoff law (at conc. temp.)
  CT ,   CRT
157
(b)   Tb 
dRT
1000  Kb
Also,
C1  C2
or
n1 n2

V1 V2
or
w1
w2

m1 V1 m 2 V2
Secondary Conditions
…..(ii)
Eq. (ii) holds good only for those solutes which neither possess the
tendency to get associate nor dissociate in solution, e.g.,
Urea and glucose are isotonic then,  1   2 and C1  C2
Urea and NaCl
(dissociate)
are isotonic then,  1   2 but C1  C2
Urea and Benzoic acid are isotonic then,  1   2 but C1  C2
(associate)
(b) Hypertonic and hypotonic solution : The solution which has
more osmotic pressure than the other solution is called as hypertonic
solution and the solution which has lesser osmotic pressure than the other
is called as hypotonic solution.
The flow of solvent is always from lower osmotic pressure to higher
osmotic pressure i.e. from hypotonic to hypertonic solution.
Elevation in b.pt. of the solvent (Ebullioscopy)
Boiling point of a liquid may be defined as the temperature at which
its vapour pressure becomes equal to atmospheric pressure, i.e., 760 mm.
Since the addition of a non-volatile solute lowers the vapour pressure of the
solvent, solution always has lower vapour pressure than the solvent and
hence it must be heated to a higher temperature to make its vapour
pressure equal to atmospheric pressure with the result the solution boils at
a higher temperature than the pure solvent. Thus sea water boils at a
higher temperature than distilled water. If T is the boiling point of the
solvent and T is the boiling point of the solution, the difference in the boiling
point (T or  T ) is called the elevation of boiling point.
b
b
T  Tb  Tb or T
Elevation in boiling point is determined by Landsberger’s method
and Cottrell’s method. Study of elevation in boiling point of a liquid in
which a non-volatile solute is dissolved is called as ebullioscopy.
Important relations concerning elevation in boiling point
(1) The elevation of boiling point is directly proportional to the
lowering of vapour pressure, i.e., Tb  p 0  p
(2) Tb  Kb  m
where K b  molal elevation constant or ebullioscopic constant of
the solvent; m  Molality of the solution, i.e., number of moles of solute
per 1000 g of the solvent; Tb  Elevation in boiling point
(3) Tb 
1000  K b  w
1000  K b  w
or m 
m W
Tb  W
158 Solution and Colligative properties
where, K b is molal elevation constant and defined as the elevation
in b.pt. produced when 1 mole of the solute is dissolved in 1 kg of the
solvent.
w and W are the weights of solute and solvent and
molecular weight of the solute.
(4) Kb 
m is the
0 .002(T0 )2
lV
where T 0  Normal boiling point of the pure solvent; lV  Latent
heat of evaporation in cal / g
of pure solvent; K b for water is
1
0.52 deg  kg mol .
Depression in f.pt. of the solvent (Cryoscopy)
Freezing point is the temperature at which the liquid and the solid
states of a substance are in equilibrium with each other or it may be
defined as the temperature at which the liquid and the solid states of a
substance have the same vapour pressure. It is observed that the freezing
point of a solution is always less than the freezing point of the pure solvent.
Thus the freezing point of sea water is low than that of pure water. The
(T or Tf ) of a solvent is the difference
depression in freezing point
in the freezing point of the pure solvent (Ts ) and the solution (Tsol . ) .
Ts  Tsol  Tf or T
NaCl or CaCl 2 (anhydrous) are used to clear snow on roads.
They depress the freezing point of water and thus reduce the temperature
of the formation of ice.
Depression in freezing point is determined by Beckmann’s method
and Rast’s camphor method. Study of depression in freezing point of a
liquid in which a non-volatile solute is dissolved in it is called as cryoscopy.
Important relations concerning depression in freezing point.
(1) Depression in freezing point is directly proportional to the
lowering of vapour pressure. Tf  p 0  p
(2) Tf  K f  m
where K f  molal depression constant or cryoscopic constant;
m  Molality of the solution (i.e., no. of moles of solute per 1000 g of the
solvent); Tf  Depression in freezing point
1000  K f  w
(3) Tf 
m W
where K f
or m 
1000  K f  w
Tf  W
is molal depression constant and defined as the
depression in freezing point produced when 1 mole of the solute is dissolved
in 1kg of the solvent. w and W are the weights of solute and solvent
and m is the molecular weight of the solute.
(4) K f 
R(T0 )2 0 .002(T0 )2

l f 1000
lf
where T0  Normal freezing point of the solvent; l f  Latent heat
of fusion/g of solvent; K f for water is 1.86 deg  kg mol 1
Colligative properties of electrolytes
The colligative properties of solutions, viz. lowering of vapour
pressure, osmotic pressure, elevation in b.p. and depression in freezing
point, depend on the total number of solute particles present in solution.
Since the electrolytes ionise and give more than one particle per formula
unit in solution, the colligative effect of an electrolyte solution is always
greater than that of a non-electrolyte of the same molar concentration. All
colligative properties are used for calculating molecular masses of non-
volatile solutes. However osmotic pressure is the best colligative property
for determining molecular mass of a non-volatile substance.
Colligative properties are depending on following factory
(1) Colligative properties  Number of particles
 Number of molecules
(in case of non-electrolytes)
 Number of ions
(In case of electrolytes)
 Number of moles of solute
 Mole fraction of solute
(2) For different solutes of same molar concentration, the magnitude
of the colligative properties is more for that solution which gives more
number of particles on ionisation.
(3) For different solutions of same molar concentration of different
non-electrolyte solutes, the magnitude of the colligative properties will be
same for all.
(4) For different molar concentrations of the same solute, the
magnitude of colligative properties is more for the more concentrated
solution.
(5) For solutions of different solutes but of same percent strength,
the magnitude of colligative property is more for the solute with least
molecular weight.
(6) For solutions of different solutes of the same percent strength,
the magnitude of colligative property is more for that solute which gives
more number of particles, which can be known by the knowledge of
molecular weight and its ionisation behaviour.
Abnormal molecular masses
Molecular masses can be calculated by measuring any of the
colligative properties. The relation between colligative properties and
molecular mass of the solute is based on following assumptions.
(1) The solution is dilute, so that Raoult’s law is obeyed.
(2) The solute neither undergoes dissociation nor association in
solution.
In case of solutions where above assumptions are not valid we find
discrepencies between observed and calculated values of colligative
properties. These anomalies are primarily due to
(i) Association of solute molecules.
(ii) Dissociation of solute molecules.
(i) Association of solute molecules : Certain solutes in solution are
found to associate. This eventually leads to a decrease in the number of
molecular particles in the solutions. Thus, it results in a decrease in the
values of colligative properties.
1
molecular mass of solute
therefore, higher values are obtained for molecular masses than
normal values for unassociated molecules.
(ii) Dissociation of solute molecules : A number of electrolytes
dissociate in solution to give two or more particles (ions). Therefore, the
number of solute particles, in solutions of such substances, is more than the
expected value. Accordingly, such solutions exhibit higher values of
colligative properties. Since colligative properties are inversely proportional
to molecular masses, therefore, molecular masses of such substances as
calculated from colligative properties will be less than their normal values.
Van’t Hoff’s factor (i) : In 1886, Van’t Hoff introduced a factor ‘i’
called Van’t Hoff’s factor, to express the extent of association or dissociation
of solutes in solution. It is ratio of the normal and observed molecular
masses of the solute, i.e.,
Colligative property 
Solution and Colligative properties
Normal molecular mass
Observed molecular mass
In case of association, observed molecular mass being more than the
normal, the factor i has a value less than 1. But in case of dissociation, the
Van’t Hoff’s factor is more than 1 because the observed molecular mass has
a lesser value than the normal molecular mass. In case there is no
dissociation the value of ‘i’ becomes equal to one.
Since colligative properties are inversely proportional to molecular
masses, the Van’t Hoff’s factor may also be written as,
i
i
Observed value of colligative property
Calculated value of colligative property
assuming no association or dissociation
i
No. of particles after association or dissociation
No. of particles before association or dissociation
Introduction of the Van’t Hoff factor modifies the equations for the
colligative properties as follows,
Relative lowering of vapour pressure 
PAo  PA
 iXB
PAo
Elevation of boiling point, Tb  ikb m
Depression in freezing point, Tf  ikf m
inRT
;   iCRT
V
From the value of ‘i’, it is possible to calculate degree of dissociation
or degree of association of substance.
Osmotic pressure,  
Degree of dissociation () : It is defined as the fraction of total
molecules which dissociate into simpler molecules or ions.

i1
; m= number of particles in solution
m 1
Degree of association () : It is defined as the fraction of the total
number of molecules which associate or combine together resulting in the
formation of a bigger molecules.

i 1
; m = number of particles in solution.
1/m 1
159
160
Solution and Colligative properties
Solubility
1.
2.
[Orissa JEE 2002]
 A supersaturated solution is metastable.
(a) Boiling point is higher than H 2 O
 Dissolution of gases in liquid is always exothermic because
(b)
S  ve (non favourable factor) and in order to have
G  ve (spontaneous process), H has to be –ve.
 1 M aqueous solution is more concentrated than 1 m aqueous
3.
 Substances having high V.P. (e.g., petrol) evaporate more quickly
than substances having low V.P. (e.g., motor oil).
by addition of an non-volatile solute is called as Babo’s law.
4.
 Konowaloff’s rule : In case of a binary solution, at a fixed
temperature, the vapour phase is richer in that component whose
addition causes increase in total vapour pressure of the solution i.e.,
vapour phase is always richer in the more volatile component.
 When a non-volatile solute is added to the solvent, V.P. decrease,
5.
B.P. increase, F.P. decrease.
 Ethylene glycol is commonly added to car radiators to depress the
freezing point of water. It is known as antifreeze.
 NaCl or CaCl2 (anhydrous) are used to clear snow on roads. It
6.
depresses the freezing point of water and reduce the temperature
at which ice is expected to be formed.
 Plasmolysis : When a plant cell is placed in a hypertonic solution,
the fluid from the plant cell comes out and the cell shrinks. This
phenomenon is called plasmolysis and is due to osmosis.
1.
25 ml
of
3.0 M HNO 3
are
mixed
with
75 ml
of
4.0 M HNO3 . If the volumes are additive, the molarity of the final
mixture would be
[DPMT 1986; MH CET 2001]
(a) 3.25 M
(b) 4.0 M
 Gelatinous Cu 2 [Fe(CN )6 ] and gelatinous Ca3 (PO4 )2 are
artificial semipermeable membranes.
(c)
2.
non aqueous solutions because it get dissolved in non aqueous
solvents.
 Osmotic coefficient (g) is the ratio of van’t Hoff factor (i) to the no.
(d) Solubility of NaCl in it is more than H 2 O
The statement “ The mass of a gas dissolved in a given mass of a
solvent at any temperature is proportional to the pressure of the gas
above the solvent” is
[AMU 2002]
(a) Dalton’s Law of Partial Pressures
(b) Law of Mass Action
(c) Henry’s Law
(d) None of these
Which is correct about Henry’s law
[KCET 2002]
(a) The gas in contact with the liquid should behave as an ideal
gas
(b) There should not be any chemical interaction between the gas
and liquid
(c) The pressure applied should be high
(d) All of these
The statement “If 0.003 moles of a gas are dissolved in 900 g of
water under a pressure of 1 atmosphere, 0.006 moles will be
dissolved under a pressure of 2 atmospheres”, illustrates [JIPMER 1999]
(a) Dalton’s law of partial pressure
(b) Graham’s law
(c) Raoult’s law
(d) Henry’s law
The solution of sugar in water contains
[BHU 1973]
(a) Free atoms
(b) Free ions
(c) Free molecules
(d) Free atom and molecules
Method of expressing concentration of solution
 Bursting of red blood cells when placed in water is due to osmosis.
 Semipermeable membrane of Cu 2 [Fe(CN )6 ] dose not work in
D2 O reacts slowly than H 2 O
(c) Viscosity is higher than H 2 O at 25 o
solution.
 Babo’s law : The lowering in vapour pressure of a solution caused
The solubility of a gas in water depends on
[MP PET 2002]
(a) Nature of the gas
(b) Temperature
(c) Pressure of the gas
(d) All of the above
Which of the following is not correct for D2 O
3.
of ions furnished by one molecule of the electrolyte (N).
i.e., g  i / N .
(d)
3.50 M
The amount of anhydrous Na 2 CO 3 present in 250 ml of 0.25 M
solution is
[DPMT 2001]
(a) 6.225 g
(b) 66.25 g
(c) 6.0 g
(d) 6.625 g
Dilute one litre 1 molar H 2 SO 4 solution by 5 litre water, the
normality of that solution is
[DPMT 1983]
(a) 0.2 N
(b) 5 N
(c)
4.
3.75 M
10 N
(d) 0.33 N
If 5.85 gms of NaCl are dissolved in 90 gms of water, the mole
fraction of NaCl is
[CMC Vellore 1991; MP PMT 1994; AFMC 1998]
5.
(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.01
(e) 0.0196
The molarity of 0.006 mole of NaCl in 100ml solution is
Solution and Colligative properties
[Bihar MEE 1996]
6.
7.
(a) 0.6
(b) 0.06
(c) 0.006
(d) 0.066
(e) None of these
9 .8 g of H 2 SO 4 is present in 2 litres of a solution. The molarity
of the solution is
[EAMCET 1991; MP PMT 2002]
(a) 0.1M
(b) 0.05 M
(c) 0.2 M
(d) 0.01M
What will be the molarity of a solution containing 5 g of sodium
hydroxide in 250 ml solution
[MP PET 1999; BHU 1999; KCET 1999;
AIIMS 2000; Pb. CET 2000]
8.
(a) 0.5
(b) 1.0
(c) 2.0
(d) 0.1
The normality of 0.3 M phosphorus acid (H 3 PO3 ) is
[IIT 1999; AIIMS 2000]
9.
20.
21.
22.
23.
(a) 0.1
(b) 0.9
(c) 0.3
(d) 0.6
Which of the following has maximum number of molecules
10.
11.
(b) 16 gm of NO 2
(c) 7 gm of N 2
(d) 2 gm of H 2
Molarity is expressed as
[JIPMER 1991; CBSE PMT 1991]
(a) Gram/litre
(b) Moles/litre
(c) Litre/mole
(d) Moles/1000 gms
20 ml of HCl solution requires 19.85 ml of 0.01 M NaOH
solution for complete neutralization. The molarity of HCl solution
is
[MP PMT 1999]
(a) 0.0099
(b) 0.099
(c) 0.99
(d) 9.9
12.
13.
(a) Normality
(b) Molarity
(c) Mole fraction
(d) Mass percentage
(e) Molality
The normality of 2.3 M H 2 SO 4 solution is
[KCET 2000]
(a) 2.3 N
(b) 4.6 N
(c) 0.46 N
(d) 0.23 N
The molarity of a solution made by mixing 50ml of conc. H 2 SO 4
(36N) with 50 ml of water is
[MP PMT 2001]
(a) 36 M
(b) 18 M
(c) 9 M
(d) 6 M
171 g of cane sugar (C12 H 22 O11 ) is dissolved in 1 litre of water.
The molarity of the solution is
[MP PMT 2001]
(a) 2.0 M
(b) 1.0 M
(c) 0.5 M
(d) 0.25 M
The volumes of 4 N HCl and 10 N HCl required to make 1 litre
of 6 N HCl are
[CBSE PMT 2002]
(a) 16 gm of O 2
161
24.
[Kerala PMT 2004]
(a) 0.75 litre of 10 N HCl and 0.25 litre of 4 N HCl
(b) 0.25 litre of 4 N HCl and 0.75 litre of 10 N HCl
(c) 0.67 litre of 4 N HCl and 0.33 litre of 10 N HCl
(d) 0.80 litre of 4 N HCl and 0.20 litre of 10 N HCl
(e) 0.50 litre of 4 N HCl and 0.50 litre of 10 N HCl
Which statement is true for solution of 0.020 M H 2 SO 4
[DPMT 2001]
(a) 2 litre of the solution contains 0.020 mole of
SO 42 
(b) 2 litre of the solution contains 0.080 mole of H 3 O 
(c) 1 litre of the solution contains 0.020 mole H 3 O 
(d) None of these
25.
10 litre solution of urea contains 240g urea. The active mass of urea
How much of NaOH is required to neutralise 1500 cm 3 of 0.1 N
will be
[KCET 2000]
HCl (At. wt. of Na =23)
[KCET 2001]
(a) 0.04
(b) 0.02
(a) 4 g
(b) 6 g
(c) 40 g
(d) 60 g
(c) 0.4
(d) 0.2
If 5.85 g of NaCl (molecular weight 58.5) is dissolved in water and
26.
5 ml of N HCl, 20 ml of N/2 H 2 SO 4 and 30 ml of N/3 HNO are
the solution is made up to 0.5 litre, the molarity of the solution will be[AMU 1999; Pb PMT 2000; AFMC 2001]
mixed together and volume made to one litre. The normally of the
(a) 0.2
(b) 0.4
resulting solution is
[Kerala CET (Med.) 2003]
(c) 1.0
(d) 0.1
N
N
A mixture has 18g water and 414g ethanol. The mole fraction of
(a)
(b)
water in mixture is (assume ideal behaviour of the mixture)
10
5[MP PMT 2000]
(a) 0.1
(b) 0.4
N
N
(c)
(d)
(c) 0.7
(d) 0.9
20
40
The number of molecules in 4.25 g of ammonia is approximately
[CBSE PMT 2002]
N
23
23
(e)
(a) 0.5  10
(b) 1.5  10
25
23
23
(c) 3.5  10
(d) 2.5  10
27.
The amount of K 2 Cr2 O7 (eq. wt. 49.04) required to prepare 100
The largest number of molecules is in [Kurukshetra CEE 1998]
ml of its 0.05 N solution is
[JIPMER 2002]
(a) 25 g of CO 2
(b) 46 g of C 2 H 5 OH
(a) 2.9424 g
(b) 0.4904 g
(c)
1.4712
g
(d) 0.2452 g
(c) 36 g of H 2 O
(d) 54 g of N 2 O5
28.
With increase of temperature, which of these changes
If 1 M and 2.5 litre NaOH solution is mixed with another 0.5 M and
[AIEEE 2002]
3 litre NaOH solution, then molarity of the resultant solution will be[CBSE PMT 2002]
(a) Molality
(a) 1.0 M
(b) 0.73 M
(b) Weight fraction of solute
(c) 0.80 M
(d) 0.50 M
(c) Fraction of solute present in water
When a solute is present in trace quantities the following expression
(d) Mole fraction
is used
[Kerala CET (Med.) 2002]
29.
25ml of a solution of barium hydroxide on titration with a 0.1molar
(a) Gram per million
(b) Milligram percent
solution of hydrochloric acid gave a litre value of 35 ml. The
(c) Microgram percent
(d) Nano gram percent
molarity of barium hydroxide solution was
(e) Parts per million
[AIEEE 2003]
When the concentration is expressed as the number of moles of a
(a) 0.07
(b) 0.14
solute per litre of solution it known as
(c) 0.28
(d) 0.35
3
14.
15.
16.
17.
18.
19.
[Kerala CET (Med.) 2002]
162
30.
31.
32.
Solution and Colligative properties
2.0 molar solution is obtained , when 0.5 mole solute is dissolved in
(a) 250 ml solvent
(b) 250 g solvent
(c) 250 ml solution
(d) 1000 ml solvent
How many gram of HCl will be present in 150 ml of its 0.52 M
solution
[RPET 1999]
(a) 2.84 gm
(b) 5.70 gm
(c) 8.50 gm
(d) 3.65 gm
The number of moles present in 2 litre of 0.5 M NaOH is
43.
[BCECE 2005]
44.
33.
34.
35.
36.
45.
46.
37.
(b) Number of moles of solute per 1000 gm of the solvent
47.
(a)
3.0  10
(b) 6.02  10
48.
49.
41.
42.
(d) 800 mg
2m
A solution of Al2 (SO 4 )3 {d  1.253 gm / ml} contain 22% salt by
weight. The molarity, normality and molality of the solution is
(a) 0.805 M, 4.83 N, 0.825 M
(b) 0.825 M, 48.3 N, 0.805 M
(c) 4.83 M, 4.83 N, 4.83 M
(d) None
Which of the following should be done in order to prepare
0.40 M NaCl starting with 100 ml of 0.30 M NaCl (mol.wt.
[BIT 1992]
(a) Add 0.585 g NaCl
(b) Add 20 ml water
(c) Add 0.010ml NaCl
(d) Evaporate 10ml water
Which of the following solutions has the highest normality
(b) N phosphoric acid
(d) 0.5 M H 2 SO 4
What volume of 0 .8 M solution contains 0.1 mole of the solute
(a)
100 ml
(b) 125 ml
(c)
500 ml
(d) 62.5 ml
Hydrochloric acid solution A and B have concentration of
0.5 N and 0.1 N respectively. The volumes of solutions A and
B required to make 2 litres of 0.2 N HCl are
0 .5 l of A  1.5 l of B
[MP PET/PMT 1998]
(b) 1 .5 l of A  0.5l of B
(c)
(a) 2.5
(b) 0.25
(c) 1
(d) 0.75
The normality of a solution of sodium hydroxide 100 ml of which
contains 4 grams of NaOH is
[CMC Vellore 1991]
(a) 0.1
(b) 40
(c) 1.0
(d) 0.4
Two solutions of a substance (non electrolyte) are mixed in the
following manner 480 ml of 1.5M first solution + 520 mL of 1.2M
second solution. What is the molarity of the final mixture [AIEEE 2005]
(a) 1.20 M
(b) 1.50 M
(c) 1.344 M
(d) 2.70 M
The normal amount of glucose in 100ml of blood (8–12 hours
after a meal) is
[BHU 1981]
(a) 8 mg
(b) 80 mg
200 mg
(d)
23
16
16
(d)
 10 23
 10 23
6 .02
3.0
The number of moles of a solute in its solution is 20 and total
number of moles are 80. The mole fraction of solute is
(c)
0 .2 m
[KCET 1993]
1 .0 l of A  1.0 l of B
(d) 0.75 l of A  1.25l of B
50.
[MP PMT 1997]
40.
(c)
(a)
(c)
39.
(b) 0 .5 m
[JIPMER 1991]
(d) Number of gram equivalents of solute per 1000 ml of the
solution
The number of molecules in 16 gm of methane is
23
1m
(a) 8 gm of KOH / litre
(c) 6 gm of NaOH / 100 ml
(c) Number of moles of solute per 1000 ml of the solution
38.
(a)
of NaCl  58.5 )
CBSE PMT 1992, 95; MP PMT 1992; AIIMS 1997, 2001]
(a) Molarity
(b) Molality
(c) Formality
(d) Normality
The molality of a solution is
[MP PMT 1996]
(a) Number of moles of solute per 1000 ml of the solvent
(a) 1000g of solvent
(b) 1 litre of solvent
(c) 1 litre of solution
(d) 1000g of solution
What will be the molality of a solution having 18 g of glucose (mol.
wt. = 180) dissolved in 500 g of water
[MP PET/PMT 1998; CBSE PMT 2000; JIPMER 2001]
[MH CET 2001]
(a) 0.5
(b) 0.1
(c) 1
(d) 2
36g water and 828g ethyl alcohol form an ideal solution. The mole
fraction of water in it, is
[MP PMT 2003]
(a) 1.0
(b) 0.7
(c) 0.4
(d) 0.1
What will be the normality of a solution containing 4.9 g. H 3 PO4
dissolved in 500 ml water
[MP PMT 2003]
(a) 0.3
(b) 1.0
(c) 3.0
(d) 0.1
3.0 molal NaOH solution has a density of 1.110 g/ml. The molarity of
the solution is
[BVP 2003]
(a) 3.0504
(b) 3.64
(c) 3.05
(d) 2.9732
Which of the following modes of expressing concentration is
independent of temperature
[IIT 1988; CPMT 1999;
Molar solution
[MP PMTmeans
2003] 1 mole of solute present in
51.
52.
53.
54.
Conc. H 2 SO 4 has a density of 1.98 gm/ml and is 98% H 2 SO 4
by weight. Its normality is
[MP PET 2002]
(a) 2 N
(b) 19.8 N
(c) 39.6 N
(d) 98 N
The mole fraction of the solute in one molal aqueous solution is [CBSE PMT 200
(a) 0.027
(b) 0.036
(c) 0.018
(d) 0.009
N
With 63 gm of oxalic acid how many litres of
solution can be
10
prepared
[RPET 1999]
(a) 100 litre
(b) 10 litre
(c) 1 litre
(d) 1000 litre
Molarity of 0.2 N H 2 SO 4 is
[KCET 2005]
(a) 0.2
(b) 0.4
(c) 0.6
(d) 0.1
10.6 grams of a substance of molecular weight 106 was dissolved in
100ml . 10ml of this solution was pipetted out into a 1000ml
flask and made up to the mark with distilled water. The molarity of the
resulting solution is
[EAMCET 1998]
(a)
1 . 0M
(b) 10 2 M
Solution and Colligative properties
55.
56.
[EAMCET 1987]
(c) 10 3 M
(d) 10 4 M
The mole fraction of water in 20% aqueous solution of H 2 O 2 is
77
68
(b)
68
77
20
80
(c)
(d)
80
20
Mole fraction (X ) of any solution is equal to
66.
(a)
(a)
(b)
(c)
67.
No. of moles of solute
Volume of solutionin litre
No. of gram equivalent of solute
Volume of solutionin litre
No. of moles of solute
Mass of solvent in kg
68.
No. of moles of any constituent
Total no. of moles of all constituents
When WB gm solute (molecular mass M B ) dissolves in WA gm
(d)
57.
solvent. The molality M of the solution is
WB
MB
WB 1000
(a)
(b)


W A 1000
MB
WA
(c)
58.
W A 1000

WB
MB
(d)
WA  M B
WB  1000
59.
Normality (N ) of a solution is equal to
No. of moles of solute
(a)
Volume of solutionin litre
No. of gram equivalent of solute
(b)
Volume of solutionin litre
No. of moles of solute
(c)
Mass of solvent in kg
(d) None of these
The volume strength of 1.5 N H 2 O 2 solution is
60.
(a) 4.8
(b) 5.2
(c) 8.8
(d) 8.4
How many gm of H 2 SO 4 is present in 0.25 gm mole of
69.
70.
61.
62.
63.
KCET 2000; CPMT 2001]
65.
On passing H 2 S gas through a solution of Cu  and Zn 2 ions,
CuS is precipitated first because
[AMU 2001]
(a) Solubility product of CuS is equal to the ionic product of ZnS
(b) Solubility product of CuS is equal to the solubility product of
71.
72.
73.
74.
(c) Solubility product of CuS is lower than the solubility product
of ZnS
(d) Solubility product of CuS is greater than the solubility product
of ZnS
The number of moles of solute per kg of a solvent is called its[DPMT 1983; IIT 19
(a) Molarity
(b) Normality
(c) Molar fraction
(d) Molality
1.0 gm of pure calcium carbonate was found to require 50 ml of
dilute HCl for complete reaction. The strength of the HCl
solution is given by
[CPMT 1986]
(a) 4 N
(b) 2 N
(c) 0.4 N
(d) 0.2 N
Molecular weight of glucose is 180. A solution of glucose which
contains 18 gms per litre is
[AFMC 1978]
(a) 2 molal
(b) 1 molal
(c) 0.1 molal
(d) 18 molal
0.5 M of H 2 SO 4 is diluted from 1 litre to 10 litre, normality of
resulting solution is
[AFMC 2005]
(a) 1 N
(b) 0.1 N
(c) 10 N
(d) 11 N
If one mole of a substance is present in 1 kg of solvent, then
75.
what amount of AgNO3 should be added in 60 ml of solution [AFMC 2005]
(a) 1.8
(b) 0.8
(a) It shows molar concentration
(c) 0.18
(d) None of these
(b) It shows molal concentration
How many grams of dibasic acid (mol. wt. 200) should be present in
(c) It shows normality
100ml of its aqueous solution to give decinormal strength[AIIMS 1992; CBSE PMT (d)
1999; ItAFMC
1999;
shows
strength gm / gm
(a)
(c)
64.
(a) 0.1M
(b) 0.2 M
[EAMCET 1993]
(c) 0.3 M
(d) 0.4 M
Which of the following concentration factor is affected by change in
temperature
[DCE 2002]
(a) Molarity
(b) Molality
(c) Mole fraction
(d) Weight fraction
The distribution law is applied for the distribution of basic acid
between
[UPSEAT 2001]
(a) Water and ethyl alcohol
(b) Water and amyl alcohol
(c) Water and sulphuric acid
(d) Water and liquor ammonia
Which is heaviest
[CBSE PMT 1991]
(a) 25 gm of mercury
(b) 2 moles of water
(c) 2 moles of carbon dioxide
(d) 4 gm atoms of oxygen
The molarity of a solution of Na 2 CO 3 having 10.6 g / 500ml of
solution is
[AFMC 1992; DCE 2000]
(a) 0.2 M
(b) 2 M
(c) 20 M
(d) 0.02 M
ZnS
[CBSE PMT 1997; BHU 2002]
[CPMT 1990]
H 2 SO 4
(a) 24.5
(b) 2.45
(c) 0.25
(d) 0.245
20 g of hydrogen is present in 5 litre vessel. The molar
concentration of hydrogen is
[DPMT 2000]
(a) 4
(b) 1
(c) 3
(d) 2
To prepare a solution of concentration of 0.03 g/ml of AgNO3 ,
1g
10 g
(b)
(d)
2g
20 g
The weight of pure NaOH required to prepare 250cm 3 of
0.1 N solution is
[KCET 1991; Kerala PMT 2004]
(a) 4 g
(b) 1 g
(c) 2 g
(d) 10 g
If 20ml of 0.4 N NaOH solution completely neutralises 40ml
of a dibasic acid. The molarity of the acid solution is
163
76.
77.
[CPMT 1996]
The molality of 90% H 2 SO 4 solution is
[density=1.8 gm/ml]
(a) 1.8
(b) 48.4
(c) 9.18
(d) 94.6
[MP PMT 2004]
The volume of water to be added to 100cm 3 of 0.5 N H 2 SO 4 to get
decinormal concentration is
[KCET (Engg.) 2001]
(a) 400 cm 3
(b) 500 cm 3
(c) 450 cm 3
(d) 100 cm 3
164
78.
Solution and Colligative properties
If 25 ml of 0.25 M NaCl solution is diluted with water to a volume
of 500ml the new concentration of the solution is
[UPSEAT 2000, 01]
79.
80.
81.
82.
83.
84.
(a) 0.167 M
(b) 0.0125 M
(c) 0.833 M
(d) 0.0167 M
10 grams of a solute is dissolved in 90 grams of a solvent. Its mass
percent in solution is
(a) 0.01
(b) 11.1
(c) 10
(d) 9
What is the molality of a solution which contains 18 g of glucose
[UPSEAT 2001]
(C6 H12 O6 ) in 250 g of water
(a) 4.0 m
(b) 0.4 m
(c) 4.2 m
(d) 0.8 m
Calculate the molality of 1 litre solution of 93%
1.84 g
H 2 SO 4 (weight/volume). The density of the solution is
/ml [UPSEAT 2000]
(a) 10.43
(b) 20.36
(c) 12.05
(d) 14.05
Volume of water needed to mix with 10 ml 10N HNO 3 to get 0.1 N
[UPSEAT 2003]
HNO 3
(a) 1000 ml
(b) 990 ml
(c) 1010 ml
(d) 10 ml
The sum of the mole fraction of the components of a solution is
(a) 0
(b) 1
(c) 2
(d) 4
Increasing the temperature of an aqueous solution will cause
92.
93.
85.
86.
87.
94.
88.
89.
90.
(c)
x CH 3 OH  0.01
(e)
0.01 N CH 3 OH
95.
96.
97.
98.
(Avogadro constant, N A  6.02  10 mol )
91.
The number of moles of SO 2 Cl 2 in 13.5 gm is
(a) 0.1
(b) 0.2
[CPMT 1994]
In a solution of 7.8 gm benzene C 6 H 6 and 46.0 gm toluene
(a)
1/6
(b) 1 / 5
(c)
1/2
(d) 1 / 3
A
solution
contains
25% H 2 O , 25%C 2 H 5 OH
and
(a) 0.25
(b) 2.5
(c) 0.503
(d) 5.03
A 5 molar solution of H 2 SO 4 is diluted from 1 litre to 10 litres.
What is the normality of the solution [AFMC 2005]
(a) 0.25 N
(b) 1 N
(c) 2 N
(d) 7 N
Molarity of a solution containing 1g NaOH in 250ml of solution
is
[EAMCET 1990]
(a) 0.1M
(b) 1 M
(c) 0.01 M
(d) 0.001 M
What is molarity of a solution of HCl which contains 49% by
weight of solute and whose specific gravity is 1.41
(a) 15.25
(c) 18.92
NaClO
solution
(b) 16.75
(d) 20.08
reacts
with
H 2 SO 3
as,
NaClO  H 2 SO 3  NaCl  H 2 SO 4 . A solution of NaClO
99.
100.
101.
1
(d) 6 .3 g
63 g
used in the above reaction contained
normality of the solution would be
(a) 0.8
(b)
(c) 0.2
(d)
When 1.80 gm glucose dissolve in 90 gm of H 2 O , the mole
fraction of glucose is
[AFMC 2000]
(a) 0.00399
(b) 0.00199
(c) 0.0199
(d) 0.998
23
(c)
[EAMCET 1991]
[CPMT 2001; CBSE PMT 2001]
(d) 0.99 M H 2O
6.02  10 20 molecules of urea are present in 100 ml of its solution.
The concentration of urea solution is [AIEEE 2004]
(a) 0.02 M
(b) 0.01 M
(c) 0.001 M
(d) 0.1 M
(b) 12.6 g
50% CH 3 COOH by mass. The mole fraction of H 2 O would be
[AIIMS 1992; Pb. CET 2004]
(a) 18 litre
(b) 9 litre
(c) 0.9 litre
(d) 1.8 litre
The concentration of an aqueous solution of 0.01M CH 3 OH
solution is very nearly equal to which of the following
(a) 0.01% CH 3 OH
(b) 0.01m CH 3 OH
0.2 N solution is
(a) 126 g
(C6 H 5 CH 3 ) , the mole fraction of benzene in this solution is
[IIT Screening 1993]
(a) Decrease in molality
(b) Decrease in molarity
(c) Decrease in mole fraction
(d) Decrease in % w/w
1000 gms aqueous solution of CaCO 3 contains 10 gms of
carbonate. Concentration of the solution is
[CPMT 1985]
(a) 10 ppm
(b) 100 ppm
(c) 1000 ppm
(d) 10000 ppm
3.65 gms of HCl is dissolved in 16.2 gms of water. The mole fraction
of HCl in the resulting solution is
[EAMCET 2003]
(a) 0.4
(b) 0.3
(c) 0.2
(d) 0.1
An aqueous solution of glucose is 10% in strength. The volume in
which 1 gm mole of it is dissolved will be
(c) 0.3
(d) 0.4
The weight of H 2C2O4 . 2 H 2O required to prepare 500ml of
15g of NaClO per litre. The
[AMU 1999]
0.6
0.33
A solution contains 1.2046  10 24 hydrochloric acid molecules in
one dm 3 of the solution. The strength of the solution is [KCET 2004]
(a) 6 N
(b) 2 N
(c) 4 N
(d) 8 N
1
10 N and
N solution is called
10
[BITS 1992]and decanormal solution
(a) Decinormal
(b) Normal and decinormal solution
(c) Normal and decanormal solution
(d) Decanormal and decinormal solution
When 7.1gm Na 2 SO 4 (molecular mass 142) dissolves in
100 ml H 2 O , the molarity of the solution is
[CBSE PMT 1991; MP PET 1993, 95]
102.
103.
(a) 2.0 M
(b) 1.0 M
(c) 0.5 M
(d) 0.05 M
Molarity of 4% NaOH solution is
[EAMCET 1987]
(a) 0.1M
(b) 0.5 M
(c) 0.01M
(d) 1.0 M
When 6 gm urea dissolve in 180 gm H 2 O . The mole fraction of
urea is
[CPMT 1988]
10
10 . 1
(a)
(b)
10 . 1
10
Solution and Colligative properties
10 . 1
0 .1
(d)
0 .1
10 . 1
The normality of 10% (weight/volume) acetic acid is
(c)
104.
117.
(c) 0.33
(d) None of these
A solution of CaCl 2 is 0.5 mol / litre, then the moles of chloride
118.
ion in 500ml will be
[MP PMT 1986]
(a) 0.25
(b) 0.50
(c) 0.75
(d) 1.00
What is the molarity of H 2 SO 4 solution, that has a density 1.84
[CPMT 1983]
106.
(a) 1 N
(b)
(c) 1.7 N
(d)
Unit of mole fraction is
(a) Moles/litre
(b)
(c) Moles–litre
(d)
Normality of 2 M sulphuric acid is
107.
(a) 2 N
(b) 4 N
(c) N / 2
(d) N / 4
Molar concentration (M ) of any solution =
105.
10 N
0.83 N
[BHU 1998, 2005]
Moles/litre
gm/cc at 35 o C and contains solute 98% by weight
2
Dimensionless
[AIIMS 2001]
119.
(a] 4.18 M
(b) 8.14 M
(c) 18.4 M
(d) 18 M
A certain aqueous solution of FeCl 3 (formula mass =162) has a
120.
density of 1.1 g / ml and contains 20.0% FeCl3 . Molar
concentration of this solution is [Pb. PMT 1998]
(a) 0.028
(b) 0.163
(c) 1.27
(d) 1.47
If 0.50 mol of CaCl 2 is mixed with 0.20 mol of Na 3 PO4 , the
[AIIMS 1991, 92; Pb. CET 2002]
(a)
(b)
(c)
(d)
108.
109.
110.
111.
112.
114.
115.
How many grams of NaOH will be required to neutralize 12.2
grams of benzoic acid
[MP PMT 1999]
(a) 40 gms
(b) 4 gms
16 gms
198 gm
121.
(d)
212 gm
In a mixture of 1 gm H 2 and 8 gm O 2 , the mole fraction of
hydrogen is
[Orissa JEE 2002]
(a) 0.667
(b) 0.5
maximum number of moles of Ca 3 (PO4 )2 which can be formed,
is
[Pb. PMT 1998]
(a) 0.70
(b) 0.50
(c) 0.20
(d) 0.10
An X molal solution of a compound in benzene has mole fraction
of solute equal to 0.2. The value of X is
[KCET 1996; DCE 2001]
122.
123.
(a) 14
(b) 3.2
(c) 4
(d) 2
Molecular weight of urea is 60. A solution of urea containing 6 g
urea in one litre is
[BHU 1996, 99]
(a) 1 molar
(b) 1.5 molar
(c) 0.1 molar
(d) 0.01 molar
The molar solution of sulphuric acid is equal to
[MP PET 1999]
124.
125.
(a) N solution
(b) 2 N solution
(c) N / 2 solution
(d) 3 N solution
The weight of sodium carbonate required to prepare 500 ml of a
semi- normal solution is
[JIPMER 1999]
(a) 13.25
(b) 26.5 g
[MPgPET 1993]
(c) 53 g
(d) 6.125 g
200ml of a solution contains 5.85 g dissolved sodium chloride.
127.
The concentration of the solution will be (Na  23; Cl  35.5) [MP PMT 1999
(a) 1 molar
(b) 2 molar
(c) 0.5 molar
(d) 0.25 molar
Molarity of a solution prepared by dissolving 75.5 g of pure KOH in
540 ml solution is
[BHU 1999]
(a) 3.05 M
(b) 1.35 M
(c) 2.50 M
(d) 4.50 M
Which one of the following is an extensive property
128.
(a) Molar volume
(b) Molarity
(c) Number of moles
(d) Mole fraction
Addition of conc. HCl to saturated BaCl2 solution precipitates
126.
(d) 12.2 gms
10ml of conc. H 2 SO 4 (18 molar) is diluted to 1 litre. The
approximate strength of dilute acid could be
[JIPMER 1991]
(a) 0.18 N
(b) 0.09 N
(c) 0.36 N
(d) 1800 N
The normality of 10 lit. volume hydrogen peroxide is
[Kerala CET (Med.) 2003]
(a) 0.176
(b) 3.52
(c) 1.78
(d) 0.88
(e) 17.8
Essential quantity of ammonium sulphate taken for preparation of 1
molar solution in 2 litres is
(a) 132 gm
(b) 264 gm
(c)
116.
No. of moles of any constituent
Total no. of moles of all constituents
If 5.0 gm of BaCl2 is present in 10 6 gm solution, the
concentration is
(a) 1 ppm
(b) 5 ppm
(c) 50 ppm
(d) 1000 ppm
1 Molar solution contains
[DPMT 2002]
(a) 1000g of solute
(b) 1000g of solvent
(c) 1 litre of solvent
(d) 1 litre of solution
To neutralise completely 20 mL of 0.1 M aqueous solution of
phosphorous acid (H 3 PO3 ), the volume of 0.1 M aqueous KOH
solution required is
[AIEEE 2004]
(a) 40 mL
(b) 20 mL
(c) 10 mL
(d) 60 mL
On dissolving 1 mole of each of the following acids in 1 litre water,
the acid which does not give a solution of strength 1 N is
(a) HCl
(b) Perchloric acid
(c) HNO 3
(d) Phosphoric acid
(c)
113.
No. of moles of solute
Volume of solutionin litre
No. of gram equivalent of solute
Volume of solutionin litre
No. of moles of solute
Mass of solvent in kg
165
[KCET 1998]
BaCl2 ; because
[AMU 2000]
(a) It follows from Le Chatelier’s principle
(b) Of common-ion effect
(c) Ionic product (Ba  ), (Cl  ) remains constant in a saturated
solution
(d) At constant temperature, the product (Ba 2 ), (Cl  )2 remains
constant in a saturated solution
166
129.
Solution and Colligative properties
How much water is needed to dilute 10 ml of 10 N hydrochloric acid
to make it exactly decinormal (0.1 N)
[EAMCET 1982]
130.
131.
(c) 1.96 gm
141.
If 18 gm of glucose (C 6 H 12 O6 ) is present in 1000 gm of an
aqueous solution of glucose, it is said to be
[CPMT 1986]
(a) 1 molal
(b) 1.1 molal
(c) 0.5 molal
(d) 0.1 molal
142.
The number of moles of KCl in 1000 ml of 3 molar solution is
(a) 990 ml
(b) 1000 ml
(c) 1010 ml
(d) 100 ml
The formula weight of H 2 SO 4 is 98. The weight of the acid in
400ml of 0.1M solution is
(a) 2.45 g
(b)
(c)
(d) 9 .8 g
[EAMCET 1987]
4.90 g
3.92 g
143.
The molarity of pure water is
[CPMT 1974, 88, 90; CMC Vellore 1991; RPET 1999;
NCERT 1974, 76; MP PMT 1999; AMU 2002]
132.
(a) 55.6
(b) 5.56
(c) 100
(d) 18
The molarity of a 0.2 N Na 2 CO 3 solution will be
133.
(a) 0.05 M
(b) 0.2 M
(c) 0.1 M
(d) 0.4 M
How many moles of water are present in 180 g of water
134.
(a) 1 mole
(b) 18 mole
(c) 10 mole
(d) 100 mole
If we take 44 g of CO 2 and 14 g of N 2 what will be mole
144.
[MP PMT 1987; Pb. CET 2004]
(a) 1
(b) 2
(c) 3
(d) 1.5
The unit of molality is
[Pb. CET 2003]
(a) Mole per litre
(b) Mole per kilogram
(c) Per mole per litre
(d) Mole litre
A solution contains 1 mole of water and 4 mole of ethanol. The mole
fraction of water and ethanol will be
(a) 0.2 water + 0.8 ethanol
(b) 0.4 water + 0.6 ethanol
(c) 0.6 water + 0.8 ethanol
(d) 0.8 water + 0.2 ethanol
[JIPMER 1991; DPMT 1982; Manipal MEE 1995]
fraction of CO 2 in the mixture
135.
Colligative properties
1.
[KCET 1990]
(a) 1/5
(b) 1/3
(c) 2/3
(d) 1/4
What is the volume of 0.1 N HCl required to react completely
2.
[KCET 1998]
(a)
150 cm 3
(b)
(c)
200 cm 3
(d) 100 cm 3
The
amount
NaOH
in
138.
139.
gms
in
250 cm 3
of
(d) None
Equimolar solutions in the same solvent have
[AIEEE 2005]
(a) 1000 gm of the solvent
(b) One litre of the solvent
(c) One litre of the solution
(d) 22.4 litres of the solution
What weight of ferrous ammonium sulphate is needed to prepare
100 ml of 0.1 normal solution (mol. wt. 392)
[CPMT 1983]
(b) 3.92 gm
3.
4.
(a) Osmotic pressure
(b) Boiling point
(c) Vapour pressure
(d) Freezing point
The colligative properties of a solution depend on
5.
(a) Nature of solute particles present in it
(b) Nature of solvent used
(c) Number of solute particles present in it
(d) Number of moles of solvent only
Which of the following is not a colligative property
[AFMC 1992; CBSE PMT 1992; MP PMT 1996, 2003]
[CPMT 1984; MP PMT 1993; UPSEAT 2001; Kerala PMT 2002]
4.0 gm of NaOH are contained in one decilitre of solution. Its
molarity would be
(a) 4 M
(b) 2 M
(c) 1 M
(d) 1.5 M
When 90 gm of water is mixed with 300 gm of acetic acid. The total
number of moles will be
(a) 5
(b) 10
(c) 15
(d) 20
A molal solution is one that contains one mole of a solute in
(a) 39.2 gm
(c) Both
(d) Different boiling and different freezing points
Which of the following is a colligative property
a
[NCERT 1983; DPMT 1983; CPMT 1985; IIT 1986;
MP PMT 1987; EAMCET 1990; MP PET 1994, 99]
140.
(b) Higher
(c) Same boiling and same freezing points
250 cm 3
0.100 M NaOH solution would be
(a) 4 gm
(b) 2 gm
(c) 1 gm
(d) 2.5 gm
137.
(a) Lower
(b) Same freezing point but different boiling point
(Ca  40, C  12 and O  16)
of
The magnitude of colligative properties in all colloidal dispersions is
….than solution
[AMU 1999]
(a) Same boiling point but different freezing point
with 1 .0 g of pure calcium carbonate
136.
(d) 19.6 gm
[BHU 1982; CPMT 1988; DPMT 1985; MP PET 1999]
6.
7.
(a) Osmotic pressure
(b) Elevation in B.P.
(c) Vapour pressure
(d) Depression in freezing point
Which of the following is not a colligative property
[MP PET 2001; CPMT 2001; Pb. CET 2001]
(a) Optical activity
(b) Elevation in boiling point
(c) Osmotic pressure
(d) Lowering of vapour pressure
Colligative properties of a solution depends upon
[MP PMT 1994, 2002]
(a) Nature of both solvent and solute
Solution and Colligative properties
8.
(b) The relative number of solute and solvent particles
(c) Nature of solute only
(d) Nature of solvent only
Which is not a colligative property
9.
(a) Refractive index
(b) Lowering of vapour pressure
(c) Depression of freezing point
(d) Elevation of boiling point
Which of the following is a colligative property
[CPMT 1984; BHU 1982; Manipal MEE 1995]
(a)
(b)
(c)
(d)
8.
(a) Surface tension
(b) Viscosity
(c) Osmotic pressure
(d) Optical rotation
Colligative properties are used for the determination of
(a) Molar Mass
(b) Equivalent weight
(c) Arrangement of molecules
(d) Melting point and boiling point
(d) Both (a) and (b)
What does not change on changing temperature
9.
[DCE 2001]
(a) Mole fraction
(c) Molality
10.
(b) Normality
(d) None of these
Lowering of vapour pressure
11.
1.
Vapour pressure of CCl 4 at 25 o C is 143mm of Hg 0.5 gm
of a non-volatile solute (mol. wt. = 65) is dissolved in
100 ml CCl 4 . Find the vapour pressure of the solution (Density of
CCl 4  1.58 g / cm 2 )
[CBSE PMT 1998]
(a)
(b) 94.39 mm
(c)
2.
3.
141.43 mm
199.34 mm
6.
(b)
P  Po N2
(c) P o  P N 2
(d) P  P o (N1 / N 2 )
An aqueous solution of methanol in water has vapour pressure
(a) Equal to that of water
(b) Equal to that of methanol
(c) More than that of water
(d) Less than that of water
The pressure under which liquid and vapour can coexist at
equilibrium is called the
(a) Limiting vapour pressure
(b) Real vapour pressure
(c) Normal vapour pressure
(d) Saturated vapour pressure
Which solution will show the maximum vapour pressure at 300 K
(a) 1 M C 12 H 22 O11
(b) 1 M CH 3 COOH
(d) 1 M NaCl
The relative lowering of the vapour pressure is equal to the ratio
between the number of
[EAMCET 1991; CBSE PMT 1991]
(a)
(b)
(c)
(d)
(d) 143.99 mm
For a solution of volatile liquids the partial vapour pressure of each
component in solution is directly proportional to
(a) Molarity
(b) Mole fraction
(c) Molality
(d) Normality
“The relative lowering of the vapour pressure is equal to the mole
fraction of the solute.” This law is called
13.
(a) Henry's law
(b) Raoult's law
(c) Ostwald's law
(d) Arrhenius's law
The relative lowering of vapour pressure produced by dissolving 71.5
g of a substance in 1000 g of water is 0.00713. The molecular weight
of the substance will be
14.
[DPMT 2001]
5.
P  P o N1
(c) 1 M NaCl 2
12.
[MP PET 1997, 2001]
4.
An increase in the b.p. of the solution
A decrease in the b.p. of the solvent
The solution having a higher freezing point than the solvent
The solution having a lower osmotic pressure than the solvent
If P o and P are the vapour pressure of a solvent and its solution
respectively and N 1 and N 2 are the mole fractions of the solvent
and solute respectively, then correct relation is
(a)
[Kerala CET (Engg.) 2002]
11.
(a) Directly proportional to the mole fraction of the solvent
(b) Inversely proportional to the mole fraction of the solute
(c) Inversely proportional to the mole fraction of the solvent
(d) Directly proportional to the mole fraction of the solute
When a substance is dissolved in a solvent the vapour pressure of
the solvent is decreased. This results in
[NCERT 1981]
[BHU 1990; NCERT 1983; MP PMT 1983; DPMT 1981, 83;
MP PET/PMT 1998; AIIMS 1999; Pb. CET 2000]
10.
7.
167
(a) 18.0
(b) 342
(c) 60
(d) 180
When mercuric iodide is added to the aqueous solution of potassium
iodide, the
[IIT 1987]
(a) Freezing point is raised
(b) Freezing point is lowered
(c) Freezing point does not change
(d) Boiling point does not change
Vapour pressure of a solution is
[EAMCET 1988; MP PET 1994]
15.
Solute moleules and solvent molecules
Solute molecules and the total molecules in the solution
Solvent molecules and the total molecules in the solution
Solvent molecules and the total number of ions of the solute
5cm 3 of acetone is added to 100 cm 3 of water, the vapour
pressure of water over the solution
(a) It will be equal to the vapour pressure of pure water
(b) It will be less than the vapour pressure of pure water
(c) It will be greater than the vapour pressure of pure water
(d) It will be very large
At 300 K, when a solute is added to a solvent its vapour pressure
over the mercury reduces from 50 mm to 45 mm. The value of
mole fraction of solute will be
(a) 0.005
(b) 0.010
(c) 0.100
(d) 0.900
A solution has a 1 : 4 mole ratio of pentane to hexane. The vapour
pressure of the pure hydrocarbons at 20°C are 440 mmHg for
pentane and 120 mmHg for hexane. The mole fraction of pentane in
the vapour phase would be
[CBSE PMT 2005]
16.
(a) 0.549
(b) 0.200
(c) 0.786
(d) 0.478
Benzene and toluene form nearly ideal solutions. At 20°C, the
vapour pressure of benzene is 75 torr and that of toluene is 22 torr.
The parial vapour pressure of benzene at 20°C for a solution
containing 78g of benzene and 46g of toluene in torr is [AIEEE 2005]
(a) 50
(b) 25
168
17.
Solution and Colligative properties
(c) 37.5
(d) 53.5
The vapour pressure lowering caused by the addition of 100 g of
sucrose(molecular mass = 342) to 1000 g of water if the vapour
pressure of pure water at 25 o C is 23.8 mm Hg
28.
[RPET 1999]
18.
19.
20.
(a) 1.25 mm Hg
(b) 0.125 mm Hg
(c) 1.15 mm Hg
(d) 00.12 mm Hg
Which of the following is incorrect
[J & K 2005]
(a) Relative lowering of vapour pressure is independent
(b) The vapour pressure is a colligative property
(c) Vapour pressure of a solution is lower than the vapour
pressure of the solvent
(d) The relative lowering of vapour pressure is directly
propertional to the original pressure
Among the following substances the lowest vapour pressure is
exerted by
(a) Water
(b) Mercury
(c) Kerosene
(d) Rectified spirit
According to Raoult's law the relative lowering of vapour pressure of
a solution of volatile substance is equal to
29.
[Pb. PMT 1998]
30.
21.
31.
22.
23.
24.
[Kerala CET (Med.) 2002]
25.
26.
27.
(a) Pressure of the biomolecules
(b) Vapour pressure of atmospheric constituents
(c) Vapour pressure of chemicals and vapour pressure of volatiles
(d) Pressure created on to atmospheric molecules
The vapour pressure of pure liquid A is 0.80 atm. On mixing a nonvolatile B to A, its vapour pressure becomes 0.6 atm. The mole
fraction of B in the solution is
[MP PET 2003]
(a) 0.150
(b) 0.25
(c) 0.50
(d) 0.75
Lowering of vapour pressure is highest for
[BHU 1997]
(a) Urea
(b) 0.1M glucose
(c) 0.1 M MgSO4
(d) 0.1 M BaCl2
An aqueous solution of glucose was prepared by dissolving 18 g of
glucose in 90 g of water. The relative lowering in vapour pressure is
[KCET 2002]
(a) 0.02
(b) 1
(a) 140 torr
(b) 20 torr
(c) 68 torr
(d) 72 torr
The vapour pressure of benzene at a certain temperature is
640 mm of Hg . A non-volatile and non-electrolyte solid weighing
2.175 g is added to 39.08 g of benzene. The vapour pressure of
the solution is 600mm of Hg . What is the molecular weight of
solid substance
[MP PMT 1983; NCERT 1981]
(a) An increase in the boiling point of the solution
(b) A decrease in the boiling point of solvent
(c) The solution having a higher freezing point than the solvent
(d) The solution having a lower osmotic pressure than the solvent
The vapour pressure of a liquid depends on
(a) Temperature but not on volume
(b) Volume but not on temperature
(c) Temperature and volume
(d) Neither on temperature nor on volume
Which one of the statements given below concerning properties of
solutions, describes a colligative effect
[AIIMS 2003]
(a) Boiling point of pure water decreases by the addition of ethanol
(b) Vapour pressure of pure water decreases by the addition of
nitric acid
(c) Vapour pressure of pure benzene decreases by the addition of
naphthalene
(d) Boiling point of pure benzene increases by the addition
of toluene
The atmospheric pressure is sum of the
(a) 0.635 methanol, 0.365 ethanol
(b) 0.365 methanol, 0.635 ethanol
(c) 0.574 methanol, 0.326 ethanol
(d) 0.173 methanol, 0.827 ethanol
The vapour pressure of two liquids P and Q are 80 and 600 torr,
respectively. The total vapour pressure of solution obtained by
mixing 3 mole of P and 2 mole of Q would be
[CBSE PMT 2005]
[CBSE PMT 1995; BHU 2001]
(a) Mole fraction of the solvent
(b) Mole fraction of the solute
(c) Weight percentage of a solute
(d) Weight percentage of a solvent
When a substance is dissolved in a solvent, the vapour pressure of
the solvent is decreased. This results in
(c) 20
(d) 180
“Relative lowering in vapour pressure of solution containing nonvolatile solute is directly proportional to mole fraction of solute”.
Above statement is
[AFMC 2004]
(a) Henry law
(b) Dulong and Petit law
(c) Raoult's law
(d) Le-Chatelier's principle
An ideal solution was obtained by mixing methanol and ethanol. If
the partial vapour pressure of methanol and ethanol are
2.619 kPa and 4.556 kPa respectively, the composition of the
vapour (in terms of mole fraction) will be
[CBSE PMT 1999; AFMC 1999]
32.
(a) 49.50
(b) 59.6
(c) 69.5
(d) 79.8
Which one of the following is the expression of Raoult's law
p  ps
ps  p
N
n


(a)
(b)
p
nN
p
N n
(c)
p  ps
N

ps
N n
(d)
ps  p N  n

ps
N
p  vapour pressure of pure solvent
p s  vapour pressure of the solution
33.
34.
n  number of moles of the solute
N  number of moles of the solvent
Which has maximum vapour pressure [DPMT 2001]
(a) HI
(b) HBr
(c) HCl
(d) HF
When a non-volatile solute is dissolved in a solvent, the relative
lowering of vapour pressure is equal to
[BHU 1979; IIT 1983]
(a) Mole fraction of solute
(b) Mole fraction of solvent
(c) Concentration of the solute in grams per litre
35.
(d) Concentration of the solute in grams 100 ml
60 gm of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If
the vapour pressure of pure water is Po , the vapour pressure of
solution is
[DCE 2000]
(a) 0.10 Po
(b) 1.10 Po
(c) 0.90 Po
36.
(d) 0.99 Po
The vapour pressure of water at 20 o C is 17.54 mm. When 20g of
a non-ionic, substance is dissolved in 100g of water, the vapour
Solution and Colligative properties
37.
pressure is lowered by 0.30 mm. What is the molecular weight of
the substances
[UPSEAT 2001]
(a) 210.2
(b) 206.88
(c) 215.2
(d) 200.8
In an experiment, 1 g of a non-volatile solute was dissolved in 100 g
of acetone (mol. mass = 58) at 298K. The vapour pressure of the
solution was found to be 192.5 mm Hg. The molecular weight of the
solute is (vapour pressure of acetone = 195 mm Hg)
1.
38.
2.
(b) 35.24
(d) 55.24
How many grams of CH 3 OH should be added to water to
prepare 150 ml solution of 2 M CH 3 OH
(a) 9.6
(c)
39.
9.6  10
(d)
3.
[CBSE PMT 1994]
2.4  10 3
The vapour pressure of a solvent decreased by 10mm of mercury,
when a non-volatile solute was added to the solvent. The mole
fraction of the solute in the solution is 0.2. What should be the mole
fraction of the solvent, if decrease in the vapour pressure is to be
20 mm of mercury
4.
5.
[CBSE PMT 1998]
40.
(a) 0.8
(b) 0.6
(c) 0.4
(d) 0.2
For a dilute solution, Raoult's law states that
[CPMT 1987; BHU 1979; IIT 1985; MP PMT 2004;
MNR 1988; AMU 2002]
41.
42.
43.
44.
(a) The lowering of vapour pressure is equal to mole fraction of
solute
(b) The relative lowering of vapour pressure is equal to mole
fraction of solute
(c) The relative lowering of vapour pressure is proportional to the
amount of solute in solution
(d) The vapour pressure of the solution is equal to the mole
fraction of solvent
The vapour pressure of a solvent A is 0.80 atm When a non-volatile
substance B is added to this solvent its vapour pressure drops to 0.6
atm. What is mole fraction of B in solution
(a) 0.25
(b) 0.50
(c) 0.75
(d) 0.90
Determination of correct molecular mass from Raoult's law is
applicable to
(a) An electrolyte in solution
(b) A non-electrolyte in a dilute solution
(c) A non-electrolyte in a concentrated solution
(d) An electrolyte in a liquid solvent
If two substances A and B have PA0 : PB0  1 : 2 and have mole
fraction in solution 1 : 2 then mole fraction of A in vapours
(a) 0.33
(b) 0.25
(c) 0.52
(d) 0.2
A dry air is passed through the solution, containing the 10 gm of
solute and 90 gm of water and then it pass through pure water.
There is the depression in weight of solution wt by 2.5 gm and in
weight of pure solvent by 0.05 gm. Calculate the molecular weight
of solute
[Kerala CET 2005]
(a) 50
(b) 180
(c) 100
(d) 25
(e) 51
Ideal and Non-ideal solution
(a) Water-nitric acid
(b) Benzene-methanol
(c) Water-hydrochloric acid
(d) Acetone-chloroform
Which one of the following is non-ideal solution
(a) Benzene + toluene
(b) n -hexane + n -heptane
(c) Ethyl bromide + ethyl iodide
(d) CCl 4  CHCl 3
(b) 2.4
3
Which of the following liquid pairs shows a positive deviation from
Raoult's law
[MP PET 1993; UPSEAT 2001; AIEEE 2004]
[CPMT 2001; CBSE PMT 2001; Pb CET 2002]
(a) 25.24
(c) 45.24
169
6.
7.
8.
A non ideal solution was prepared by mixing 30 ml chloroform and
50 ml acetone. The volume of mixture will be
[Pb. CET 2003]
(a) > 80 ml
(b) < 80 ml
(c) = 80 ml
(d)  80 ml
Which pair from the following will not form an ideal solution
(a)
CCl 4  SiCl4
(b)
H 2 O  C 4 H 9 OH
(c)
C 2 H 5 Br  C 2 H 5 I
(d) C6 H 14  C7 H 16
An ideal solution is that which
[MP PMT 1996]
(a) Shows positive deviation from Raoult's law
(b) Shows negative deviation from Raoult's law
(c) Has no connection with Raoult's law
(d) Obeys Raoult's law
Which one of the following mixtures can be separated into pure
components by fractional distillation [CPMT 1987]
(a) Benzene – toluene
(b) Water – ethyl alcohol
(c) Water – nitric acid
(d) Water – hydrochloric acid
All form ideal solutions except
[DPMT 1983; MP PET 1997]
(a)
C 2 H 5 Br and C 2 H 5 I
(b) C 6 H 5 Cl and C 6 H 5 Br
(c)
C 6 H 6 and C 6 H 5 CH 3
(d) C 2 H 5 I and C 2 H 5 OH
Which property is shown by an ideal solution
[MP PET 2002]
PMTRaoult's
2000, 01]law
(a) It [MP
follows
(c)
9.
10.
12.
13.
H mix  0
(d) All of these
When two liquid A and B are mixed then their boiling points
becomes greater than both of them. What is the nature of this
solution
(a) Ideal solution
(b) Positive deviation with non ideal solution
(c) Negative deviation with non ideal solution
(d) Normal solution
In mixture A and B components show –ve deviation as
(a)
11.
Vmix  0
(b)
[DPMT 2005]
[AIEEE 2002]
Vmix  0
(b) H mix  0
(c) A-B interaction is weaker than A-A and B-B interaction
(d) A-B interaction is strong than A-A and B-B interaction
In which case Raoult's law is not applicable
(a) 1M NaCl
(b) 1 M urea
(c) 1 M glucose
(d) 1 M sucrose
A solution that obeys Raoult's law is [EAMCET 1993]
(a) Normal
(b) Molar
(c) Ideal
(d) Saturated
An example of near ideal solution is
(a) n -heptane and n -hexane
170
Solution and Colligative properties
(b) CH 3 COOH  C5 H 5 N
(c)
14.
15.
16.
[JIPMER 1997]
(a)
CHCl 3  (C 2 H 5 )2 O
S mix  0
H mix  0
(b)
(c) It obeys Raoult's law
(d) Vmix  0
Which of the following does not show negative deviation from
Raoult’s law
[MP PMT 2001]
(a) Acetone-Chloroform
(b) Acetone-Benzene
(c) Chloroform-Ether
(d) Chloroform-Benzene
(c) (C 2 H 5 )2 O  CHCl 3
(d) (CH 3 )2 CO  C6 H 5 NH 2
27.
A mixture of benzene and toluene forms
[MP PMT 1993]
All form ideal solution except
[UPSEAT 2001]
(a) An ideal solution
(b) Non-ideal solution
(a) C2 H 5 Br and C 2 H 5 I
(b) C 2 H 5 Cl and C6 H 5 Br
(c) Suspension
(d) Emulsion
28.
Which of the following is an ideal solution
(c) C6 H 6 and C6 H 5 CH 3
(d) C 2 H 5 I and C2 H 5 OH
(a) Water
+ ethanol
Formation of a solution from two components can be considered as
[CBSE PMT 2003]
(b) Chloroform + carbon tetrachloride
(i) Pure solvent  separated solvent molecules H
(c) Benzene + toluene
(ii) Pure solute  separated solute molecules H
(d) Water + hydrochloric acid
(iii) Separated solvent and solute molecules  solution H
29.
When ethanol mixes in cyclohexane; cyclohexane reduces the
intermolecular forces between ethanol molecule. In this, liquid pair
Solution so formed will be ideal if
shows
(a) H soln  H 3  H1  H 2
(a) Positive deviation by Raoult's law
(b) H soln  H1  H 2  H 3
(b) Negative deviation by Raoult's law
(c) No deviation by Raoult's law
(c) H soln  H1  H 2  H 3
(d) Decrease in volume
(d) H soln  H1  H 2  H 3
30.
Liquids A and B form an ideal solution [AIEEE 2003]
Identify the mixture that shows positive deviation from Raoult’s law [Kerala CET (Engg.)
(a) 2002]
The enthalpy of mixing is zero
(b) The entropy of mixing is zero
(a) CHCl 3  (CH 3 )2 CO
(b) (CH 3 )2 CO  C6 H 5 NH 2
(c) The free energy of mixing is zero
(c) CHCl 3  C6 H 6
(d) (CH 3 )2 CO  CS 2
(d) The free energy as well as the entropy of mixing are each zero
(e) C6 H 5 N  CH 3 COOH
(d) H 2 O  HNO 3
A mixture of liquid showing positive deviation in Raoult's law is
(a) (CH 3 )2 CO  C 2 H 5 OH (b) (CH 3 )2 CO  CHCl 3
26.
1
2
3
17.
18.
19.
20.
21.
22.
23.
24.
When acetone is added to chloroform, then hydrogen bond is
formed between them.These liquids show
(a) Positive deviation from Raoult's law
(b) Negative deviation from Raoult's law
(c) No deviation from Raoult's law
(d) Volume is slightly increased
Which of the following is true when components forming an ideal
solution are mixed
[AMU 2000]
(a) H m  Vm  0
(b) H m  Vm
(c) H m  Vm
(d) H m  Vm  1
The liquid pair benzene-toluene shows [MP PET 1995]
(a) Irregular deviation from Raoult's law
(b) Negative deviation from Raoult's law
(c) Positive deviation from Raoult's law
(d) Practically no deviation from Raoult's law
The solution which shows negative or positive deviation by Raoult's
law, is called
(a) Ideal solution
(b) Real solution
(c) Non-ideal solution
(d) Colloidal solution
Which of the following does not show positive deviation from
Raoult’s law
[MP PMT 2000]
(a) Benzene-Chloroform
(b) Benzene-Acetone
(c) Benzene-Ethanol
(d) Benzene-Carbon tetrachloride
Which of the following mixture shows positive deviation by ideal
behaviour
(a) CHCl 3  (CH 3 )2 CO
(b) C6 H 6  C6 H 5 CH 3
(c) H 2 O  HCl
(d) CCl 4  CHCl 3
Which property is not found in ideal solution
(a)
25.
PA  PAo  X A
(b)
H mix  0
(c) Vmix  0
(d) All of these
Which of the following is not correct for ideal solution
Azeotropic mixture
1.
The
azeotropic
mixture
of
o
water
(b. p.100 o C)
and
o
HCl (b. p. 85 C) boils at 108.5 C . When this mixture is
distilled it is possible to obtain
[IIT 1981]
(a) Pure HCl
(b) Pure water
(c) Pure water as well as pure HCl
(d) Neither HCl nor H 2 O in their pure states
2.
3.
An azeotropic solution of two liquids has boiling point lower than
either when it
[NCERT 1978; IIT 1981]
(a) Shows a negative deviation from Raoult's law
(b) Shows no deviation from Raoult's law
(c) Shows positive deviation from Raoult's law
(d) Is saturated
A liquid mixture boils without changing constituent is called
[DPMT 1982; CPMT 1987]
4.
5.
(a) Stable structure complex
(b) Binary liquid mixture
(c) Zeotropic liquid mixture
(d) Azeotropic liquid mixture
Azeotropic mixture are
[CPMT 1982]
(a) Constant temperature boiling mixtures
(b) Those which boils at different temperatures
(c) Mixture of two solids
(d) None of the above
A mixture of two completely miscible non-ideal liquids which distil
as such without change in its composition at a constant temperature
as though it were a pure liquid. This mixture is known as
(a) Binary liquid mixture
(b) Azeotropic mixture
Solution and Colligative properties
(c) Eutectic mixture
(d) Ideal mixture
10.
Osmosis and Osmotic pressure of the solution
1.
2.
If 3 gm of glucose (mol. wt. 180) is dissolved in 60 gm of water at
15 o C . Then the osmotic pressure of this solution will be
(a) 0.34 atm
(b) 0.65 atm
(c) 6.57 atm
(d) 5.57 atm
The concentration in gms per litre of a solution of cane sugar
(M  342) which is isotonic with a solution containing 6 gms of
11.
12.
urea (M  60) per litre is
[Orissa PMT 1989]
(a) 3.42
(c) 5.7
3.
4.
5.
6.
7.
(b) 34.2
(d) 19
Osmotic pressure is 0.0821 atm at temperature of 300 K . Find
concentration in mole/litre
[Roorkee 1990]
(a) 0.033
(b) 0.066
13.
(c) 0.33  10 2
(d) 3
Osmotic pressure of a solution containing 0.1 mole of solute per litre
at 273 K is (in atm)
[CPMT 1988]
(a)
0 .1
 0 .08205  273
1
(b) 0.1  1  0.08205  273
(c)
1
 0 .08205  273
0 .1
(d)
0.1
273

1 0 .08205
A solution contains non-volatile solute of molecular mass M p .
Which of the following can be used to calculate molecular mass of
the solute in terms of osmotic pressure (m = Mass of solute, V =
Volume of solution and  = Osmotic pressure)
(a)
m 
Mp    VRT
 
(b)
 m  RT
Mp   
V 
(c)
m  
Mp   
 V  RT
(d)
m 
Mp    RT
V
The osmotic pressure of a 5% (wt/vol) solution of cane sugar at
150 o C is
[AMU 1999]
(a) 2.45 atm
(b) 5.078 atm
(c) 3.4 atm
(d) 4 atm
The relationship between osmotic pressure at 273 K when 10 g
15.
16.
8.
9.
P1  P2  P3
(b)
P3  P1  P2
(c)
P2  P1  P3
(d)
P2  P3  P1
0.60 mol / L
(d) 0.45 mol / L
A solution of sucrose(molar mass = 342 g/mol) is prepared by
dissolving 68.4 g of it per litre of the solution, what is its osmotic
pressure (R = 0.082 lit. atm. k 1 mol 1 ) at 273k
(a) 6.02 atm
(b) 4.92 atm
(c) 4.04 atm
(d) 5.32 atm
Blood has been found to be isotonic with
[CPMT 1994]
[CBSE saline
PMT 2002]
(a) Normal
solution
(b) Saturated NaCl solution
(c) Saturated KCl solution
(d) Saturated solution of a 1 : 1 mixture of NaCl and KCl
If 20 g of a solute was dissolved in 500 ml of water and osmotic
pressure of the solution was found to be 600 mm of Hg at 15 o C,
then molecular weight of the solute is
[BHU 2004]
17.
[CBSE PMT 1996]
(a)
The average osmotic pressure of human blood is 7.8 bar at 37 o C .
What is the concentration of an aqueous NaCl solution that could be
used in the blood stream
[AIIMS 2004]
(a) 0.16 mol / L
(b) 0.32 mol / L
[UPSEAT 2001]
glucose (P1 ),10 g urea (P2 ) and 10 g sucrose (P3 ) are dissolved
in 250ml of water is
Two solutions A and B are separated by semi- permeable membrane.
If liquid flows form A to B then
[MH CET 2000]
(a) A is less concentrated than B
(b) A is more concentrated than B
(c) Both have same concentration
(d) None
these
[MP of
PMT
1986]
A 5% solution of canesugar (mol. wt. =342) is isotonic with 1%
solution of a substance X . The molecular weight of X is
[
(a) 34.2
(b) 171.2
(c) 68.4
(d) 136.8
Which of the following colligative properties can provide molar mass
of proteins (or polymers or colloids) with greater precision[Kerala PMT 2004]
(a) Relative lowering of vapour pressure
(b) Elevation of boiling point
(c) Depression in freezing point
(d) Osmotic pressure
(e) Rast's method
(c)
14.
171
18.
(a) 1000
(b) 1200
(c) 1400
(d) 1800
The osmotic pressure of 0.4% urea solution is 1.66 atm and. that of
a solution of suger of 3.42 % is 2.46 atm. When both the solution
are mixed then the osmotic pressure of the resultant solution will be[MP PMT 19
(a) 1.64 atm
(b) 2.46 atm
(c) 2.06 atm
(d) 0.82 atm
Blood is isotonic with
[DCE 2000]
(a) 0.16 M NaCl
(b) Conc. NaCl
(c) 50 % NaCl
(d) 30 % NaCl
Which inorganic precipitate acts as semipermeable membrane or
The chemical composition of semipermeable membrane is[CPMT 1984, 90; MP PM
(a) Calcium sulphate
(b) Barium oxalate
(c) Nickel phosphate
(d) Copper ferrocyanide
In osmosis
[DPMT 1985]
19.
(a) Solvent molecules move from higher concentration to lower
concentration
(b) Solvent molecules move from lower to higher concentration
(c) Solute molecules move from higher to lower concentration
20.
The osmotic pressure of 1 m solution at 27 o C is
[CPMT 1999]
(d) Solute molecules move from lower to higher concentration
(a) 84,2.46
(b) 24.6 atm
Semipermeable membrane is that which permits the passage of[BHU 1979; CPMT 1977,
90; atm
MP PMT 1994]
(c) 1.21 atm
(d) 12.1 atm
(a) Solute molecules only
21.
Osmotic
pressure
of
a
solution
can be measured quickly and
(b) Solvent molecules only
accurately by
[JIPMER 1991; CPMT 1983]
(c) Solute and solvent molecules both
(a) Berkeley and Hartley's method
(d) Neither solute nor solvent molecules
172
22.
23.
Solution and Colligative properties
(b) Morse's method
(c) Pfeffer's method
(d) De Vries method
The solution in which the blood cells retain their normal form are
with regard to the blood
[CBSE PMT 1991]
(a) Isotonic
(b) Isomotic
(c) Hypertonic
(d) Equinormal
The osmotic pressure of a solution is given by the relation
33.
34.
CPMT 1990; MP PMT 1999; AFMC 1999, 2001]
[CPMT 1983, 84, 87, 93, 94]
(a)
25.
26.
RT
C
(b)
P
CT
R
35.
RC
P
(d)
 RT
T
C
The osmotic pressure of a solution is directly proportional to
(a) The molecular concentration of solute
(b) The absolute temperature at a given concentration
(c) The lowering of vapour pressure
(d) All of the above
What would happen if a thin slice of sugar beet is placed in a
concentrated solution of NaCl
[CMC Vellore 1986]
(a) Sugar beet will lose water from its cells
(b) Sugar beet will absorb water from solution
(c) Sugar beet will neither absorb nor lose water
(d) Sugar beet will dissolve in solution
The osmotic pressure of a dilute solution is given by
(c)
24.
P
P
36.
27.
P  Po x
(b) V  nRT
(c)
P  Po N 2
(d)
38.
(d) Anhydrous sodium carbonate
The molecular weight of NaCl determined by osmotic pressure
method will be
(a) Same as theoritical value
(b) Higher than theoritical value
(c) Lower than theoritical value
(d) None of these
The osmotic pressure of solution increases, if
39.
(a) Temperature is decreased
(b) Solution concentration is increased
(c) Number of solute molecules is increased
(d) Volume is increased
At the same temperature, following solution will be isotonic
37.
29.
30.
P
1
if T is constant
V
(b) P  T if V is constant
(c) P  V if T is constant
(d) PV is constant if T is constant
Isotonic solutions have
[DPMT 1984; MP PMT 1986]
(a) Equal temperature
(b) Equal osmotic pressure
(c) Equal volume
(d) Equal amount of solute
Which of the following associated with isotonic solutions is not
correct
[AMU 2002]
(a) They will have the same osmotic pressure
(b) They have the same weight concentrations
(c) Osmosis does not take place when the two solutions are
separated by a semipermeable membrane
(d) They will have the same vapour pressure
Isotonic solution have the same
[EAMCET 1979; JIPMER 1991, 2002;
AFMC 1995; MP PMT 2002]
31.
(a) Density
(b) Molar concentration
(c) Normality
(d) None of these
A 0.6% solution of urea (molecular weight = 60) would be isotonic
with [NCERT 1982; DCE 2002]
(a)
0.1M glucose
(c) 0.6% glucose solution
32.
[CPMT 1985, 87, 91]
Which statement is wrong regarding osmotic pressure (P), volume
(V) and temperature (T)
[MP PMT 1985]
(a)
28.
P Po  P

Po
Po
(b) 0.1M KCl
(d) 0.6% KCl solution
The value of osmotic pressure of a 0.2 M aqueous solution at 293K
is
[AMU 2002]
(a) Hypertonic
(b) Hypotonic
(c) Normal
(d) Isotonic
At low concentrations, the statement that equimolal solutions under a
given set of experimental conditions have equal osmotic pressure is
true for
[EAMCET 1979; BHU 1979]
(a) All solutions
(b) Solutions of non-electrolytes only
(c) Solutions of electrolytes only
(d) None of these
Which one of the following would lose weight on exposure to
atmosphere
[NCERT 1975]
(a) Concentrated H 2 SO 4
(b) Solid NaOH
(c) A saturated solution of CO 2
[MP PMT 1987]
(a)
(a) 8.4 atm
(b) 0.48atm
(c) 4.8 atm
(d) 4.0 atm
Diffusion of solvent through a semi permeable membrane is called
(a) Diffusion
(b) Osmosis
(c) Active absorption
(d) Plasmolysis
Solutions having the same osmotic pressure under a given set of
conditions are known as
[BHU 1979; EAMCET 1979;
[MP PMT 1985]
(a) 3.24 gm of sucrose per litre of water and 0.18 gm glucose per
litre of water
(b) 3.42 gm of sucrose per litre and 0.18 gm glucose in 0.1 litre
water
(c) 3.24 gm of sucrose per litre of water and 0.585 gm of sodium
chloride per litre of water
(d) 3.42 gm of sucrose per litre of water and 1.17 gm of sodium
chloride per litre of water
40.
41.
42.
The osmotic pressure of a decinormal solution of BaCl2 in water
is
(a) Inversely proportional to its celsius temperature
(b) Inversely proportional to its absolute temperature
(c) Directly proportional to its celsius temperature
(d) Directly proportional to its absolute temperature
Blood cells will remain as such in
[CPMT 2004]
(a) Hypertonic solution
(b) Hypotonic solution
(c) Isotonic solution
(d) None of these
The osmotic pressure of a dilute solution is directly proportional to
the
[MP PMT 1987]
(a) Diffusion rate of the solute
(b) Ionic concentration
(c) Elevation of B.P.
Solution and Colligative properties
43.
44.
45.
(d) Flow of solvent from a concentrated to a dilute solution
The osmotic pressure in atmospheres of 10% solution of canesugar
at 69 o C is
[AFMC 1991]
(a) 724
(b) 824
(c) 8.21
(d) 7.21
Which of the following molecules would diffuse through a cell
membrane
[NCERT 1978]
(a) Fructose
(b) Glycogen
(c) Haemoglobin
(d) Catalase
52.
Two solutions of KNO 3 and CH 3 COOH are prepared separately.
54.
[DCE 2001]
53.
[CPMT 1983, 84; Pb CET 2004]
46.
47.
48.
49.
P2  P1
(b)
P1  P2
(c)
P1  P2
(d)
P1
P2

P1  P2 P1  P2
The osmotic pressure of a dilute solution of a non-volatile solute is
(a) Directly proportional to its temperature on the centigrade scale
(b) Inversely proportional to its temperature on the Kelvin scale
(c) Directly proportional to its temperature on the Kelvin scale
(d) Inversely proportional to its temperature on the centigrade
scale
Osmotic pressure of a urea solution at 10 o C is 500 mm .
Osmotic pressure of the solution become 105.3 mm. When it is
diluted and temperature raised to 25 o C. The extent of dilution is
(a) 6 Times
(b) 5 Times
(c) 7 Times
(d) 4 Times
If a 0.1M solution of glucose (mol. wt. 180) and 0.1 molar
solution of urea (mol. wt. 60) are placed on the two sides of a
semipermeable membrane to equal heights, then it will be correct to
say [CBSE PMT 1992]
(a) There will be no net movement across the membrane
(b) Glucose will flow across the membrane into urea solution
(c) Urea will flow across the membrane into glucose solution
(d) Water will flow from urea solution into glucose solution
At constant temperature, the osmotic pressure of a solution
[CPMT 1986]
(a)
(b)
(c)
(d)
50.
(a)
51.
55.
56.
57.
58.
59.
60.
Directly proportional to the concentration
Inversely proportional to the concentration
Directly proportional to the square of the concentration
Directly proportional to the square root of the concentration
The solution containing 4.0 gm of a polyvinyl chloride polymer in
1 litre of dioxane was found to have an osmotic pressure
6.0  10 4 atmosphere at 300 K , the value of R used is 0.082
litre atmosphere mole 1k 1 . The molecular mass of the polymer
was found to be
[NCERT 1978]
3.0  10 2
61.
If osmotic pressure of a solution is 2 atm at 273 K , then at
1999] pressure is
546 K [JIPMER
, the osmotic
(a) 0.5 atm
(b) 1 atm
(c) 2 atm
(d) 4 atm
In osmosis reaction, the volume of solution
(a) Decreases slowly
(b) Increases slowly
(c) Suddenly increases
(d) No change
As a result of osmosis the volume of solution
[JIPMER 2000]
(a) Increases
(c)
Decreases
[MP PET 2004]
(c) Remains constant
(d) Increases or decreases
A solution of urea contain 8.6 gm/litre (mol. wt. 60.0). It is isotonic
with a 5% solution of a non-volatile solute. The molecular weight of
the solute will be
[MP PMT 1986]
(a) 348.9
(b) 34.89
(c) 3489
(d) 861.2
One mole each of urea, glucose and sodium chloride were dissolved
in one litre of water Equal osmotic pressure will be produced by
solutions of
[MH CET 1999]
(a) Glucose and sodium chloride
(b) Urea and glucose
(c) Sodium chloride and urea
(d) None of these
Which of the following aqueous solutions produce the same osmotic
pressure
[Roorkee 1999]
(a) 0.1 M NaCl solution
(b) 0.1 M glucose solution
(c) 0.6 g urea in 100 ml solution
(d) 1.0 g of a non-electrolyte solute (X) in 50 ml solution (Molar
mass of X = 200)
Which of the following aqueous solutions are isotonic (R  0.082
(a)
[Roorkee Qualifying 1998]
0.01 M glucose
(b) 0.01 M NaNO 3
(b) 1.6  10 5
(c)
500 ml solution containing 0 .3 g urea
(d) 0.04 N HCl
Elevation of boiling boint of the solvent
[CPMT 1983; MP PMT 1987; RPET 2000; DCE 2004]
(b) Electrophoresis
(d) Osmosis
If solubility of NaCl at 20 o C is 35 gm per 100 gm of water.
Then on adding 50 gm of NaCl to the same volume at same
temperature the salt remains undissolved is
(a) 15 gm
(b) 20 gm
(c) 50 gm
(d) 35 gm
Which of the following associated with isotonic solution is not
correct
(a) They will have the same osmotic pressure
(b) They have the same weight concentration
(c) Osmosis does not take place when the two solutions are
separated by a semipermeable membrane
(d) They will have the same vapour pressure
atm K 1 mol 1 )
(c) 5.6  10 4
(d) 6.4  10 2
Solvent molecules pass through the semipermeable membrane is
called
(a) Electrolysis
(c) Cataphoresis
If molecular weight of compound is increased then sensitivity is
decreased in which of the following methods
(a) Elevation in boiling point(b) Viscosity
(c) Osmosis
(d) Dialysis
Molarity of both is 0 .1 M and osmotic pressures are P1 and P2
respectively. The correct relationship between the osmotic pressures is
(a)
173
1.
The latent heat of vapourisation of water is 9700 Cal / mole and
if the b.p. is 100 o C , ebullioscopic constant of water is
174
Solution and Colligative properties
[CBSE PMT 1989]
(a)
(c)
2.
4.
(b) 1.026 C
o
(d) 1.832 o C
0.513 C
10.26 C
100.52 o C
If for a sucrose solution elevation in boiling point is 0.1°C then what
will be the boiling point of NaCl solution for same molal
concentration
[BHU 1998, 2005]
(a) 0.1°C
(b) 0.2°C
(c) 0.08°C
(d) 0.01°C
The molal elevation constant is the ratio of the elevation in B.P. to
(a) Molarity
(b) Molality
(c) Mole fraction of solute
(d) Mole fraction of solvent
(c) 99.48 o C
The rise in the boiling point of
glucose in 100 g of a solvent
constant of the liquid is
(a) 0.01 K / m
(d) 98.96 o C
a solution containing 1.8 gram of
in 0.1 o C . The molal elevation
(b) 0.1 K / m
(a)
(c)
(d) 10 K / m
(c) 100.256 o C
Value of gas constant R is
1K /m
12.
If 0.15 g of a solute dissolved in 15 g of solvent is boiled at a
The molal boiling point constant for water is 0.513 o C kg mol 1 .
When 0.1 mole of sugar is dissolved in 200ml of water, the
solution boils under a pressure of one atmosphere at
[CPMT 1999]
[CBSE PMT 1999; BHU 1997]
13.
100.513 o C
(b) 100.0513 o C
(d) 101.025 o C
[AIEEE 2002]
1
(b) 0.987 cal mol K 1
(a) 0.082 litre atm
14.
(a) 1.01
(b) 10
(c) 10.1
(d) 100
Pressure cooker reduces cooking time for food because
(c) 8.3 J mol 1 K 1
(d) 83 erg mol 1 K 1
The temperature, at which the vapour pressure of a liquid becomes
equal to the atmospheric pressure is known as
[Pb. PMT 2000]
[MP PMT 1987; NCERT 1975; CPMT 1991; AIEEE 2003]
(a)
(b)
(c)
(d)
6.
11.
(b) 101.04 o C
temperature higher by 0.216 o C than that of the pure solvent.
The molecular weight of the substance (molal elevation constant for
the solvent is 2.16 o C ) is
5.
10.
o
The molal elevation constant of water  0.52 o C . The boiling point
of 1.0 molal aqueous KCl solution (assuming complete dissociation of
[BHU 1987]
KCl ), therefore, should be
(a)
3.
o
Heat is more evenly distributed in the cooking space
Boiling point of water involved in cooking is increased
The higher pressure inside the cooker crushes the food material
Cooking involves chemical changes helped by a rise in
temperature
Which of the following statements is correct for the boiling point of
solvent containing a dissolved solid substance
15.
(a) Freezing point
(b) Boiling point
(c) Absolute temperature
(d) None of these
The elevation in boiling point of a solution of 13.44g of CuCl in 1kg of
water using the following information will be
2
(Molecular weight of CuCl = 134.4 and K = 0.52 K molal)
1
b
2
[IIT 2005]
16.
[NCERT 1972, 74]
(a) 0.16
(b) 0.05
(c) 0.1
(d) 0.2
When 10g of a non-volatile solute is dissolved in 100 g of benzene, it
raises boiling point by 1o C then molecular mass of the solute is
[BHU 2002]
(K b for benzene =2.53k-m )
–1
7.
8.
(a) Boiling point of the liquid is depressed
(b) Boiling point of the liquid is elevated
(c) There is no effect on the boiling point
(d) The change depends upon the polarity of liquid
When a substance is dissolved in a solvent, the vapour pressure of
solvent decreases. It brings
[BHU 2004]
(a) A decrease in boiling point of solution
(b) An increase in boiling point of the solution
(c) A decrease in freezing point of the solution
(d) An increase in freezing point of the solution
(a) 223 g
(c) 243 g
17.
(a)
18.
[CPMT 1989]
19.
If the solution boils at a temperature T1 and the solvent at a
temperature T2 the elevation of boiling point is given by
[MP PET 1996]
(a)
(c)
T1  T2
(b) T1  T2
T2  T1
(d) T1  T2
(b) 100.5 C
100.75 C
(c) 100.25 C
(d) 100 o C
When common salt is dissolved in water
[CBSE PMT 1988; MP PET 1995; DCE 2000]
of X is (K b for water is 0.52 per 1000 gm of water)
9.
o
o
o
compound X was dissolved in 100 gm of water. Molecular weight
(b) 60
(d) 600
An aqueous solution containing 1g of urea boils at 100.25 o C . The
aqueous solution containing 3 g of glucose in the same volume will
boil at (Molecular weight of urea and glucose are 60 and 180
respectively)
[CBSE PMT 2000]
Elevation in boiling point was 0.52 o C when 6 gm of a
(a) 120
(c) 180
(b) 233 g
(d) 253 g
20.
(a) Melting point of the solution increases
(b) Boiling point of the solution increases
(c) Boiling point of the solution decreases
(d) Both melting point and boiling point decreases
During the evaporation of liquid
[DCE 2003]
(a) The temperature of the liquid will rise
(b) The temperature of the liquid will fall
(c) May rise or fall depending on the nature
(d) The temperature remains unaffected
At higher altitudes the boiling point of water lowers because
[NCERT 1972; CPMT 1994; J & K 2005]
(a) Atmospheric pressure is low
(b) Temperature is low
Solution and Colligative properties
21.
22.
23.
(c) Atmospheric pressure is high
(d) None of these
The elevation in boiling point for one molal solution of a solute in a
solvent is called
[MH CET 2001]
(a) Boiling point constant
(b) Molal elevation constant
(c) Cryoscopic constant
(d) None of these
A solution of 1 molal concentration of a solute will have maximum
boiling point elevation when the solvent is
[MP PMT 2000]
(a) Ethyl alcohol
(b) Acetone
(c) Benzene
(d) Chloroform
Mark the correct relationship between the boiling points of very
dilute solutions of BaCl2 (t1 ) and KCl (t 2 ) , having the same
molarity
[CPMT 1984, 93]
(a)
t1  t 2
(b)
t1  t 2
(c)
t2  t1
(d)
t 2 is approximately equal to t1
7.
175
What is the freezing point of a solution containing 8.1 g HBr in
100 g water assuming the acid to be 90% ionised (K f for water
 1.86 K mole 1 )
[BHU 1981; Pb CET 2004]
8.
9.
(a)
0.85 C
(b)
 3.53 o C
(c)
0o C
(d)
 0.35 o C
o
If K f value of H 2 O is 1.86. The value of T f for 0.1m solution
of non-volatile solute is
(a) 18.6
(b) 0.186
(c) 1.86
(d) 0.0186
1% solution of Ca(NO 3 )2 has freezing point
[DPMT 1982, 83; CPMT 1977]
(a)
10.
o
(b) Less than 0 o C
0 C
(c) Greater than 0 o C
(d) None of the above
A solution of urea (mol. mass 56g mol ) boils at 100.18°C at the
atmospheric pressure. If K f and K b for water are 1.86 and 0.512K
kg mol respectively the above solution will freeze at [CBSE PMT 2005]
(a) – 6.54°C
(b) 6.54°C
(c) 0.654°C
(d) –0.654°C
–1
–1
Depression of freezing point of the solvent
1.
Molal depression constant for water is 1.86 o C . The freezing point
of a 0.05 molal solution of a non-electrolyte in water is
11.
342 gm of canesugar (C12 H 22 O11 ) are dissolved in 1000 gm of
water, the solution will freeze at
[MNR 1990; MP PET 1997]
(a)
2.
3.
4.
 1.86 o C
(b)
[NCERT 1977; CPMT 1989; Roorkee 2000; DCE 2004]
 0.93 o C
(c)  0.093 o C
(d) 0.93 o C
The amount of urea to be dissolved in 500 ml of water (K =18.6 K
mole 1 in 100g solvent) to produce a depression of 0.186 o C
in freezing point is
[MH CET 2000]
(a) 9 g
(b) 6 g
(c) 3 g
(d) 0.3 g
The maximum freezing point falls in
[MP PMT 1986]
(a) Camphor
(b) Naphthalene
(c) Benzene
(d) Water
Which one of the following statements is FALSE
12.
6.
 1.86 C
(b) 1.86 o C
(c)
 3.92 C
(d)
o
o
13.
An aqueous solution of a non-electrolyte boils at 100.52 o C . The
freezing point of the solution will be
0o C
(b)
(c) 1.86 C
(d) None of the above
The freezing point of one molal NaCl solution assuming NaCl
to be 100% dissociated in water is (molal depression constant = 1.86)
 1.86 C
(b)
 3.72o C
(c)  1.86 o C
Heavy water freezes at
(d)
 3.72o C
(b)
3.8 o C
(a)
(b) The osmotic pressure ( ) of a solution is given by the
equation   MRT where M is the molarity of the solution.
(c) Raoult's law states that the vapour pressure of a component
over a solution is proportional to its mole fraction.
(d) Two sucrose solutions of same molality prepared in different
solvents will have the same freezing point depression.
Solute when dissolved in water
[MADT Bihar 1981]
(a) Increases the vapour pressure of water
(b) Decreases the boiling point of water
(c) Decreases the freezing point of water
(d) All of the above
15.
(a)
o
0o C
[CPMT 1993]
(c) 38 o C
(d)  0.38 o C
After adding a solute freezing point of solution decreases to – 0.186.
Calculate Tb if K f  1.86 and Kb  0.521 .
[Orissa JEE 2002, 04; MP PET/PMT 1998; AIEEE 2000]
16.
(a) 0.521
(c) 1.86
Given that
a solution
 T f
lim
m 0  m

(b) 0.0521
(d) 0.0186
T f is the depression in freezing point of the solvent in
of a non-volatile solute of molality m , the quantity

 is equal to


[IIT 1994; UPSEAT 2001]
The freezing point of a solution prepared from 1.25 gm of a non17.
(a) Zero
(b) One
(c) Three
(d) None of the above
The freezing point of 1 percent solution of lead nitrate in water will
be
constant is 1.86 K mole 1 , then molar mass of the solute will be[AFMC 1998; CPMT 1999]
(b) 106.7
(d) 93.9
 1.86 o C
[CPMT 1985; BHU 1981; MP PMT 1997; UPSEAT 2001]
14.
(a) 105.7
(c) 115.3
2.42o C
o
(a) The correct order of osmotic pressure for 0.01 M aqueous
solution
of
each
compound
is
BaCl2  KCl  CH 3 COOH  sucrose.
electrolyte and 20 gm of water is 271.9 K . If molar depression
(a)
(a)
[AIEEE 2004]
5.
The molar freezing point constant for water is 1.86 o C mole 1 . If
[NCERT 1971, 72; CPMT 1972; JIPMER 1991]
o
(a) Below 0 C
(b) 0 o C
176
18.
19.
20.
Solution and Colligative properties
(c) 1o C
(d) 2 o C
What is the effect of the addition of sugar on the boiling and
freezing points of water
[Kerala CET (Med.) 2003]
(a) Both boiling point and freezing point increases
(b) Both boiling point and freezing point decreases
(c) Boiling point increases and freezing point decreases
(d) Boiling point decreases and freezing point increases
During depression of freezing point in a solution the following are
in equilibrium
[IIT Screening 2003]
(a) Liquid solvent, solid solvent
(b) Liquid solvent, solid solute
(c) Liquid solute, solid solute
(d) Liquid solute solid solvent
1.00 gm of a non-electrolyte solute dissolved in 50 gm of benzene
lowered the freezing point of benzene by 0.40 K. K f for benzene is
5.12 kg mol . Molecular mass of the solute will be
[DPMT 2004]
Colligative properties of electrolyte
1.
[NCERT 1975; CPMT 1977; JIPMER 2001]
2.
3.
–1
21.
23.
256 g mol 1
(b)
2.56 g mol 1
(c)
512  10 3 g mol 1
(d)
2.56  10 4 g mol 1
0.440 g of a substance dissolved in 22.2 g of benzene lowered the
freezing point of benzene by 0.567 o C . The molecular mass of the
[BHU 2001; CPMT 2001]
(a) 178.9
(b) 177.8
(c) 176.7
(d) 175.6
Which of the following aqueous molal solution have highest freezing
point
[UPSEAT 2000, 01, 02; MNR 1988]
(a) Urea
(b) Barium chloride
(c) Potassium bromide
(d) Aluminium sulphate
Which will show maximum depression in freezing point when
concentration is 0.1M
[IIT 1989; MNR 1990; UPSEAT 2000; 03; BCECE 2005]
24.
(a) NaCl
(b) Urea
(c) Glucose
(d)
25.
K 2 SO 4
The freezing point of a 0.01M aqueous glucose solution at 1
atmosphere is  0.18 o C . To it, an addition of equal volume of
0.002 M glucose solution will; produce a solution with freezing point
of nearly
[AMU 1999]
(a)
 0.036 o C
(b)
 0.108 o C
(c)  0.216 o C
(d)  0.422 o C
What should be the freezing point of aqueous solution containing
17 gm of C 2 H 5 OH in 1000 gm of water (water K f = 1.86
deg  kg mol 1
(a)
26.
27.
 0.69 C
o
[MP PMT 1986]
(b)
 0.34 C
o
(a)
AgNO3
(b)
MgCl2
(c)
(NH 4 )3 PO4
(d)
Na 2 SO 4
Which of the following solution in water possesses the lowest
vapour pressure
[BHU 1996]
(a)
0.1 (M) NaCl
(b) 0.1 (N ) BaCl2
(c)
0.1 (M) KCl
(d) None of these
Which of the following solutions in water will have the lowest
vapour pressure
[Roorkee 2000]
(a) 0.1 M, NaCl
(b) 0.1 M, Sucrose
(c) 0.1 M, BaCl2
(a)
substance (K f  5.12 o C mol 1 )
22.
If O.P. of 1 M of the following in water can be measured, which one
will show the maximum O.P.
4.
(d) 0.1 M Na 3 PO4
The vapour pressure will be lowest for
[CPMT 2004]
(a) 0.1 M sugar solution
(b) 0.1 M KCl solution
(c) 0.1 M Cu (NO 3 )2 solution
5.
(d) 0.1 M AgNO3 solution
Osmotic pressure of 0.1 M solution of NaCl and
be
[AFMC 1978]
(a) Same
(b) Osmotic pressure of NaCl
Na 2 SO 4 solution
Na 2 SO 4 will
solution will be more than
(c) Osmotic pressure of Na 2 SO 4 solution will be more than
NaCl
6.
(d) Osmotic pressure of NaSO 4 will be less than that of NaCl
solution
Which of the following solutions has highest osmotic pressure
(a)
7.
(b) 1 M urea
1 M NaCl
(c) 1 M sucrose
(d) 1 M glucose
Which one has the highest osmotic pressure
[CBSE PMT 1991; DPMT 1991; MP PET 1994]
8.
(a)
M / 10 HCl
(b)
M / 10 urea
(c)
M / 10 BaCl2
(d)
M / 10 glucose
In equimolar solution of glucose, NaCl and BaCl2 , the order of
osmotic pressure is as follow
[CPMT 1988, 93; MP PMT/PET 1988; MP PET 1997, 2003]
(a) Glucose  NaCl  BaCl2
(b)
NaCl  BaCl2  Glucose
(c)
BaCl2  NaCl  Glucose
(c) 0.0 o C
(d) 0.34 o C
(d) Glucose  BaCl2  NaCl
In the depression of freezing point experiment, it is found that the[IIT 1999]
9.
The osmotic pressure of which solution is maximum (consider that
(a) Vapour pressure of the solution is less than that of pure
deci-molar solution of each 90% dissociated)
solvent
[MP PMT 2003]
(b) Vapour pressure of the solution is more than that of pure
(a) Aluminium sulphate
solvent
(b) Barium chloride
(c) Only solute molecules solidify at the freezing point
(c) Sodium sulphate
(d) Only solvent molecules solidify at the freezing point
(d) A mixture of equal volumes of (b) and (c)
Calculate the molal depression constant of a solvent which has
10.
At 25 o C , the highest osmotic pressure is exhibited by 0.1M
freezing point 16.6 o C and latent heat of fusion 180.75 Jg 1 .[Orissa JEE 2005] solution of
[CBSE PMT 1994; AIIMS 2000]
(a) 2.68
(b) 3.86
(a) CaCl 2
(b) KCl
(c) 4.68
(d) 2.86t6
(c) Glucose
(d) Urea
Solution and Colligative properties
11.
12.
(a)
(c)
(c)
Glucose  AlCl3  KNO 3
(d)
AlCl3  Glucose  KNO 3
14.
Which one of the following would produce maximum elevation in
boiling point
(c)
16.
17.
18.
19.
23.
[BHU 1982; MP PMT 1987, MP PET/PMT 1988]
0.01 M Na 2 SO 4
(b) 0.01 M KNO 3
(c)
(d) 0.015 M glucose
25.
(d) 0.1M magnesium sulphate
Which of the following has the lowest freezing point
26.
(a) 0.1 m sucrose
(b) 0.1 m urea
(c) 0.1 m ethanol
(d) 0.1 m glucose
Which of the following has minimum freezing point
Which of the following aqueous solutions containing 10 gm of solute
in each case has highest B.P.
(a) NaCl solution
(b) KCl solution
(c) Sugar solution
(d) Glucose solution
0.01 molar solutions of glucose, phenol and potassium chloride were
prepared in water. The boiling points of
(a) Glucose solution = Phenol solution = Potassium chloride
solution
(b) Potassium chloride solution > Glucose solution > Phenol
solution
(c) Phenol solution > Potassium chloride solution > Glucose
solution
(d) Potassium chloride solution > Phenol solution > Glucose
solution
Which one has the highest boiling point
[CBSE PMT 1990]
(a) 0.1 N Na 2 SO 4
(b) 0.1 N MgSO4
[Pb. PMT 1999]
(a)
21.
(a) 1 : 1 : 1
(b) 1 : 2 : 1
(c) 1 : 2 : 3
(d) 2 : 2 : 3
Which has the minimum freezing point
(a) One molal NaCl solution
(b) One molal KCl solution
(c) One molal CaCl 2 solution
(d) One molal urea solution
Which of the following has lowest freezing point
27.
28.
29.
22.
(d) 0.1 M Al2 (SO 4 )3
(a)
Al2 (SO 4 )3
(b) C 5 H 10 O5
(c)
KI
(d) C12 H 22 O11
For 0.1 M solution, the colligative property will follow the order
(a)
NaCl  Na 2 SO 4  Na 3 PO4
(b)
NaCl
Na 2 SO 4  Na 3 PO4
[AMU2001]
(c)
NaCl  Na 2 SO 4  Na 3 PO4
(d)
NaCl  Na 2 SO 4  Na 3 PO4
Which of the following will have the lowest vapour pressure
0.1M KCl solution
(b) 0.1M urea solution
(c) 0.1M Na 2 SO 4 solution
[CPMT 1991]
(d) 0.1M K 4 Fe(CN )6 solution
Abnormal molecular mass
1.
0 .1 M aqueous solution of glucose
2.
(b) 0 .1 M aqueous solution of NaCl
(c)
(b) 0.1 M NH 4 Cl
Which of the following 0.10 m aqueous solution will have the
lowest freezing point
[CBSE PMT 1997]
(a)
[NCERT 1981]
(a)
0.1M K 2 Cr2 O7
(c) 0.1 M BaSO 4
(c) 0.1M Al2 (SO 4 )3
(d) 0.1M BaSO4
Which of the following solutions boils at the highest temperature
(a) 0.1 M glucose
(b) 0.1 M NaCl
(c) 0.1 M BaCl2
(d) 0.1 M Urea
0.01M solution each of urea, common salt and Na 2 SO 4 are
taken, the ratio of depression of freezing point is
0.1M barium chloride
[UPSEAT 2004]
[Roorkee 1990]
20.
Which of the following will have the highest F.P. at one atmosphere
(a) 0.1 M glucose
(a) 0.1M NaCl solution
(b) 0.1M sugar solution
(b) 0.2 M sucrose
(c) 0.1 M barium chloride
(c) 0.1M BaCl2 solution
(d) 0.1M FeCl3 solution
(d) 0.1 M magnesium sulphate
24. CPMT
Which
Which of the following solutions will have the highest boiling point[DPMT 1991;
1991] of the following will produce the maximum depression in
freezing point of its aqueous solution
(a) 1% glucose
(b) 1% sucrose
[MP PMT 1996]
(c) 1% NaCl
(d) 1% CaCl 2
(a) 0.1M glucose
Which one of the following aqueous solutions will exhibit highest
(b) 0.1M sodium chloride
boiling point
[AIEEE 2004]
(a) 0.015 M urea
15.
(b) Glucose  KNO 3  AlCl3
(d) 0.1M glucose
[MP PMT 1985; CPMT 1990; NCERT 1982]
13.
AlCl3  KNO 3  Glucose
Which of the following will have the highest boiling point at 1 atm
pressure
[MP PET/PMT 1998]
(a) 0.1 M NaCl
(b) 0.1M sucrose
0.1M BaCl2
177
The Van't Hoff factor will be highest for
(a) Sodium chloride
(b) Magnesium chloride
(c) Sodium phosphate
(d) Urea
Which of the following salt has the same value of Van't Hoff factor
i as that of K 3 [Fe(CN )6 ]
[CBSE PMT 1994; AIIMS 1998]
0 .1M aqueous solution of ZnSO 4
(d) 0 .1 M aqueous solution of urea
(a)
Al2 (SO 4 )3
(b)
NaCl
The freezing points of equimolar solutions of glucose, KNO 3 and
(c)
Na 2 SO 4
(d)
Al(NO 3 )3
AlCl3 are in the order of
[AMU 2000]
3.
When benzoic acid dissolve in benzene, the observed molecular mass
is
178
Solution and Colligative properties
(a) 244
(c) 366
(b) 61
(d) 122
14.
4.
The ratio of the value of any colligative property for KCl solution
to that for sugar solution is nearly [MP PMT 1985]
(a) 1
(b) 0.5
(c) 2.0
(d) 3
5.
Van't Hoff factor of Ca(NO 3 )2 is
(a) 1
(c) 3
6.
[CPMT 1997]
(b) 2
(d) 4
[CET Pune 1998]
15.
Dry air was passed successively through a solution of 5 gm of a
solute in 80 gm of water and then through pure water. The loss in
weight of solution was 2.50 gm and that of pure solvent
16.
0.04 gm . What is the molecular weight of the solute
[MP PMT 1986]
7.
8.
(a) 70.31
(b) 7.143
(c) 714.3
(d) 80
The Van’t Hoff factor calculated from association data is
always...than calculated from dissociation data
[JIPMER 2000]
(a) Less
(b) More
(c) Same
(d) More or less
(a) 1  
(c)
9.
1  2

18.
(b)
(d) 1  2
19.
Van't Hoff factor i
(a)
Normal molecular mass
Observed molecular mass
Observed molecular mass
Normal molecular mass
(c) Less than one in case of dissociation
(d) More than one in case of association
Which of the following compounds corresponds Van't Hoff factor
' i' to be equal to 2 for dilute solution [NCERT 1978]
(a) K 2 SO 4
(b) NaHSO 4
(b)
10.
17.
If  is the degree of dissociation of Na 2 SO 4 , the Vant Hoff's
factor (i) used for calculating the molecular mass is
[AIEEE 2005]

(c) Sugar
(d)
MgSO4
1.
11.
The Van't Hoff factor i for a 0.2 molal aqueous solution of urea is
(a) 0.2
(b) 0.1
(c) 1.2
(d) 1.0
12.
One mole of a solute A is dissolved in a given volume of a solvent.
The association of the solute take place according to nA ⇄ ( A)n .
2.
The Van't Hoff factor i is expressed as
[MP PMT 1997]
(a)
(c)
13.
x
(b) i  1 
n
i 1 x
i
1 x 
x
n
3.
(d) i  1
1
Acetic acid dissolved in benzene shows a molecular weight of
(a) 60
(b) 120
(c) 180
(d) 240
The observed osmotic pressure of a solution of benzoic acid in
benzene is less than its expected value because
(a) Benzene is a non-polar solvent
(b) Benzoic acid molecules are associated in benzene
(c) Benzoic acid molecules are dissociated in benzene
(d) Benzoic acid is an organic compound
The experimental molecular weight of an electrolyte will always be
less than its calculated value because the value of Van't Hoff factor
“i” is [MP PMT 1993]
(a) Less than 1
(b) Greater than 1
(c) Equivalent to one
(d) Zero
The molecular mass of acetic acid dissolved in water is 60 and when
dissolved in benzene it is 120.This difference in behaviour of
[AMU 2000]
CH 3 COOH is because
(a) Water prevents association of acetic acid
(b) Acetic acid does not fully dissolve in water
(c) Acetic acid fully dissolves in benzene
(d) Acetic acid does not ionize in benzene
The correct relationship between the boiling points of very dilute
solutions of AlCl3 (t1 ) and CaCl 2 (t 2 ) , having the same molar
concentration is
[CPMT 1983]
(a)
t1  t 2
(b)
t1  t 2
(c)
t 2  t1
(d)
t 2  t1
The Van't Hoff factor for sodium phosphate would be
(a) 1
(b) 2
(c) 3
(d) 4
The molecular weight of benzoic acid in benzene as determined by
depression in freezing point method corresponds to
(a) Ionization of benzoic acid
(b) Dimerization of benzoic acid
(c) Trimerization of benzoic acid
(d) Solvation of benzoic acid
On adding solute to a solvent having vapour pressure 0.80 atm,
vapour pressure reduces to 0.60 atm. Mole fraction of solute is
(a) 0.25
(b) 0.75
(c) 0.50
(d) 0.33
A solution containing 30 gms of non-volatile solute in exactly 90 gm
water has a vapour pressure of 21.85 mm Hg at 25 o C . Further 18
gms of water is then added to the solution. The resulting solution
has a vapour pressure of 22.15 mm Hg at 25 o C . Calculate the
molecular weight of the solute
[UPSEAT 2001]
(a) 74.2
(b) 75.6
(c) 67.83
(d) 78.7
Vapour pressure of a solution of 5 g of non- electrolyte in 100 g
of water at a particular temperature is 2985 N / m 2 . The vapour
pressure of pure water is 3000 N / m 2 . The molecular weight of
the solute is
[IIT Screening 1993]
(a) 60
(b) 120
(c) 180
(d) 380
[MP PET 1993, 02]
Solution and Colligative properties
4.
Azeotropic mixture of HCl and water has
14.
An aqueous solution of a weak monobasic acid containing 0.1 g in
21.7g of water freezes at 272.813 K. If the value of K f for water is
1.86 K/m, what is the molecular mass of the monobasic acid [AMU 2002]
(a) 50 g/mole
(b) 46 g/mole
(c) 55 g/mole
(d) 60 g/mole
15.
K f of 1,4-dioxane is 4.9 mol 1 for 1000 g. The depression in
freezing point for a 0.001 m solution in dioxane is
[AFMC 1997; JIPMER 2002]
(a) 84% HCl
(c) 63% HCl
5.
(b) 22.2% HCl
(d) 20.2% HCl
The osmotic pressure at 17 o C of an aqueous solution containing
1.75 g of sucrose per 150 ml solution is
[BHU 2001]
6.
(a) 0.8 atm
(b) 0.08 atm
(c) 8.1 atm
(d) 9.1 atm
A 1.2 of solution of NaCl is isotonic with 7.2 of solution of glucose.
Calculate the van’t Hoff’s factor of NaCl solution
[DPMT 2001]
16.
7.
[EAMCET 1998]
(b) 1.50
(d) 1.00
0 .6 g of a solute is dissolved in 0.1 litre of a solvent which
17.
develops an osmotic pressure of 1.23 atm at 27 o C . The molecular
mass of the substance is
[BHU 1990]
8.
9.
(a)
149.5 g mole 1
(b) 120 g mole 1
(c)
430 g mole 1
(d) None of these
The boiling point of a solution of 0.1050 gm of a substance in 15.84
gram of ether was found to be 100 o C higher than that of pure
ether. What is the molecular weight of the substance [Molecular
elevation constant of ether per 100 g = 21.6]
(a) 144.50
(b) 143.18
(c) 140.28
(d) 146.66
Boiling point of chloroform was raised by 0.323 K, when 0.5143 g of
anthracene was dissolved in 35 g of chloroform. Molecular mass of
anthracene is
( K b for CHCl 3 = 3.9 kg mol )
[Pb PMT 2000]
(a) 79.42 g/mol
(b) 132.32 g/mol
(c) 177.42 g/mol
(d) 242.32 g/mol
–1
10.
The boiling point of water ( 100 o C ) becomes 100.52 o C , if 3
grams of a nonvolatile solute is dissolved in 200ml of water. The
molecular weight of solute is
( K b for water is 0.6 K  m )
[AIIMS 1998]
(a)
12.2 g mol
1
12.
[CPMT 2001]
13.
 0.31o C
Same as that of 2 solution
Nearly one-fifth of the 2 solution
Double that of 2 solution
Nearly five times that of 2 solution
nd
nd
nd
nd
[UPSEAT 2001]
Read the assertion and reason carefully to mark the correct option out of
the options given below :
(a)
(c)
(d)
(e)
If both assertion and reason are true and the reason is the correct
explanation of the assertion.
If both assertion and reason are true but reason is not the correct
explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.
If assertion is false but reason is true.
1.
Assertion
:
One molal aqueous solution of urea contains
60 g of urea in 1kg (1000 g) water.
Reason
:
Solution containing one mole of solute in
1000 g solvent is called as one molal solution.
Assertion
:
If 100 cc of 0.1 N HCl is mixed with 100 cc of
(b)
(b) 15.4 g mol
(c) 17.3 g mol
(d) 20.4 g mol
Normal boiling point of water is 373 K (at 760mm). Vapour
pressure of water at 298 K is 23 mm. If the enthalpy of evaporation
is 40.656 kJ/mole, the boiling point of water at 23 mm pressure will
be
[CBSE PMT 1995]
(a) 250 K
(b) 294 K
(c) 51.6 K
(d) 12.5 K
A 0.2 molal aqueous solution of a weak acid (HX) is 20% ionised.
The freezing point of this solution is (Given K f  1.86 o C / m for
water)
[IIT 1995]
(a)
(a) 22.4
(b) 2.24
(c) 0.224
(d) 5.6
A solution is obtained by dissolving 12 g of urea (mol.wt.60) in a
litre of water. Another solution is obtained by dissolving 68.4 g of
cane sugar (mol.wt. 342) in a litre of water at are the same
temperature. The lowering of vapour pressure in the first solution is
(a)
(b)
(c)
(d)
1
11.
(a) 0.0049
(b) 4.9 + 0.001
(c) 4.9
(d) 0.49
How many litres of CO 2 at STP will be formed when 100ml of
0.1 M H 2 SO 4 reacts with excess of Na 2 SO 3
[UPSEAT 2001]
(a) 2.36
(c) 1.95
(b)
2.
0.2 N HCl , the normality of the final solution
will be 0.30.
3.
 0.45 o C
(c)  0.53 o C
(d)  0.90 o C
A 0.001 molal solution of [Pt(NH 3 )4 Cl 4 ] in water had a freezing
point depression of 0.0054 o C . If K f for water is 1.80, the
correct formulation for the above molecule is
4.
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
[Kerala CET (Med.) 2003]
(a)
[Pt(NH 3 )4 Cl 3 ] Cl
(b) [Pt(NH 3 )4 Cl] Cl2
(c)
[Pt(NH 3 )4 Cl 2 ]Cl 3
(d) [Pt(NH 3 )4 Cl 4 ]
179
Normalities of similar solutions like HCl can be
added.
If a liquid solute more volatile than the solvent is
added to the solvent, the vapour pressure of the
solution may increase i.e., p s  p o .
In the presence of a more volatile liquid solute,
only the solute will form the vapours and solvent
will not.
Azeotropic mixtures are formed only by nonideal solutions and they may have boiling points
either greater than both the components or less
than both the components.
The composition of the vapour phase is same as
that of the liquid phase of an azeotropic mixutre.
180 Solution and Colligative properties
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
Assertion
:
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
Reason
Assertion
:
:
:
Reason
Assertion
Reason
:
:
:
Molecular mass of polymers cannot be calculated
using boiling point or freezing point method.
Polymers solutions do not possess a constant
boiling point or freezing point.
The molecular weight of acetic acid determined
by depression in freezing point method in
benzene and water was found to be different.
Water is polar and benzene is non-polar.
20.
Assertion
:
Reason
:
Assertion
:
CCl 4 and H 2 O are immiscible.
Reason
:
CCl 4 is a polar solvent.
Assertion
:
Reason
:
Isotonic solution do not show the phenomenon
of osmosis.
Isotonic solutions have equal osmotic pressure.
[AIIMS 2002]
[AIIMS 2002]
Increasing pressure on pure water decreases its
freezing point.
Density of water is maximum at 273 K.
[AIIMS 2003]
Ca   and K  ions are responsible for
maintaining proper osmotic pressure balance in
the cells of organism.
Solutions having the same osmotic pressure are
called isotonic solutions.
Reverse osmosis is used in the desalination of sea
1
d
water.
6
c
When pressure more than osmotic pressure is
applied, pure water is squeezed out of the sea
water through the membrane.
Method
Camphor is used as solvent in the determination
of molecular masses of naphthalene, anthracene
1
c
etc.
6
b
Camphor has high molal elevation constant.
Elevation in boiling point and depression in
11
a
freezing point are colligative properties.
16
c
All colligative properties are used for the
calculaltion of molecular masses.
21
c
An increase in surface area increases the rate of
26
d
evaporation.
31
a
Stronger the inter-molecular attractive forces,
fast is the rate of evaporation at a given
36
b
temperature.
[AIIMS 2002]
41
c
The boiling and melting points of amides are
higher than corresponding acids.
46
ac
It is due to strong intermolecular hydrogen
51
c
bonding in their molecules.
[AIIMS 2002]
The freezing point is the temperature at which
56
d
solid crystallizers from solution.
61
d
The freezing point depression is the difference
between that temperature and freezing point of
66
a
pure solvent. [AIIMS 2000]
71
d
On adding NaCl to water its vapour pressure
76
c
increases.
Addition of non-volatile solute increases the
81
d
vapour pressure.
[AIIMS 1996]
86
d
Molar heat of vaporisation of water is greater
than benzene.
91
a
Molar heat of vaporisation is the amount of heat
96
a
required to vaporise one mole of liquid at constant
temperature. [AIIMS 1996]
101
c
Ice melts faster at high altitude.
106
b
At high altitude atmospheric pressure is high. [AIIMS 1997]
111
d
Molecular mass of benzoic acid when determined
by colligative properties is found high.
116
a
Dimerisation of benzoic acid.
[AIIMS 1998]
121
b
Use of pressure cooker reduces cooking time.
126
c
At higher pressue cooking occurs faster.
[AIIMS 2000]
19.
21.
Solubility
2
d
3
c
4
b
5
d
of expressing concentration of solution
2
d
3
d
4
e
5
b
7
a
8
d
9
d
10
b
12
b
13
a
14
a
15
b
17
b
18
e
19
b
20
b
22
c
23
c
24
b
25
c
27
d
28
c
29
a
30
c
32
c
33
d
34
a
35
d
37
b
38
b
39
b
40
c
42
b
43
c
44
c
45
a
47
c
48
b
49
a
50
c
52
b
53
d
54
b
55
b
57
b
58
b
59
c
60
a
62
a
63
a
64
b
65
a
67
c
68
c
69
a
70
d
72
c
73
c
74
b
75
b
77
a
78
b
79
c
80
b
82
b
83
b
84
b
85
d
87
d
88
e
89
b
90
b
92
d
93
a
94
c
95
a
97
c
98
d
99
b
100
d
102
d
103
d
104
c
105
d
107
a
108
b
109
d
110
a
112
b
113
c
114
c
115
b
117
b
118
c
119
c
120
d
122
c
123
b
124
a
125
c
127
c
128
c
129
a
130
b
131
a
132
c
133
c
134
c
135
c
136
c
137
c
138
b
139
a
140
b
141
d
142
c
143
b
144
a
Colligative properties
Solution and Colligative properties
181
1
a
2
c
3
a
4
c
5
c
11
b
12
c
13
c
14
b
15
a
6
a
7
b
8
a
9
c
10
a
16
d
17
c
18
b
19
b
20
a
21
b
22
c
23
b
11
ac
Depression of freezing point of the solvent
Lowering of vapour pressure
1
a
2
b
3
b
4
d
5
b
1
c
2
c
3
a
4
d
5
c
6
a
7
a
8
a
9
c
10
b
6
a
7
b
8
b
9
b
10
d
11
a
12
b
13
b
14
c
15
d
16
a
17
b
18
d
19
b
20
b
11
a
12
b
13
b
14
b
15
b
21
a
22
a
23
b
24
b
25
b
16
d
17
a
18
c
19
a
20
a
26
d
27
a
28
c
29
b
30
d
21
a
22
a
23
d
24
c
25
a
31
c
32
a
33
c
34
a
35
c
26
ad
27
b
36
b
37
c
38
a
39
b
40
b
41
a
42
b
43
d
44
c
Colligative properties of electrolyte
Ideal and Non-ideal solution
1
c
2
b
3
d
4
c
5
c
6
a
7
c
8
c
9
a
10
a
1
b
2
d
3
b
4
b
5
d
6
a
7
d
8
d
9
c
10
b
11
c
12
c
13
d
14
c
15
a
11
a
12
c
13
a
14
a
15
d
16
d
17
c
18
b
19
c
20
c
16
b
17
d
18
b
19
a
20
d
21
b
22
a
23
b
24
c
25
c
21
c
22
a
23
d
24
d
25
a
26
d
27
a
28
b
29
d
26
b
27
a
28
c
29
a
30
a
Abnormal molecular mass
Azeotropic mixture
1
d
2
c
3
d
4
a
5
b
Osmosis and Osmotic pressure of the solution
1
c
2
a
3
a
4
c
5
c
6
a
7
a
8
c
9
a
10
d
11
d
12
c
13
b
14
b
15
b
16
b
17
b
18
d
19
b
1
c
2
b
3
c
4
a
5
b
6
b
7
c
8
b
9
b
10
a
11
c
12
d
13
b
14
b
15
a
16
b
17
c
18
a
19
d
20
b
21
a
22
a
23
d
24
d
25
a
1
a
2
c
3
c
4
d
5
a
26
b
27
c
28
b
29
b
30
b
6
c
7
b
8
b
9
c
10
c
31
a
32
c
33
b
34
d
35
b
11
b
12
b
13
b
14
d
15
a
36
c
37
c
38
c
39
b
40
d
41
c
42
b
43
c
44
a
45
c
16
c
17
a
46
c
47
b
48
a
49
a
50
b
51
d
52
d
53
a
54
b
55
d
56
b
57
d
58
a
59
b
60
bcd
61
ac
Elevation of boiling point of the solvent
1
a
2
b
3
c
4
d
5
b
6
b
7
b
8
b
9
b
10
b
Critical Thinking Questions
Assertion & Reason
1
a
2
e
3
c
4
b
5
c
6
a
7
d
8
a
9
c
10
b
11
c
12
a
13
b
14
d
15
b
16
d
17
a
18
a
19
c
20
b
21
c
182
Solution and Colligative properties
Mole fraction of water 
Method of expressing concentration of solution
1.
(c)
M1 V1  M 2 V2  MV
2.
(d)
M
3.
(d)
N 1 V1  N 2 V2
w
w
; 0 .25 
; w  6.625 gm
m  V (l)
106  0 .25
2 1  N 2  6
N 2  0.33
4.
(e) 5.85 g NaCl =
90 g H 2 O 
5 .85
mole  0 .1 mol
58.5
90
moles  5 moles
18
mole fraction of NaCl =
M
0.1
 0 .0196 .
5  0 .1
n
0 .006

 0 .06
V (l)
0 .1
5.
(b)
6.
(b) M 
7.
(a)
8.
(d) Basicity of H 3 PO3 is 2.
W  1000
9.8  1000

 0.05 M
mol. mass  Volume in ml. 98  2000
M
W
1000
5  1000


 0.5 M
m.wt . Volume in ml. 40  250
Hence 0.3 M H 3 PO3  0.6 N .
9.
11.
(d) 2 gm. Hydrogen has maximum number of molecules than
others.
(a) M1V1  M 2 V2
0.01  19.85  M 2  20
M 2  0.009925 ; M  0.0099 .
12.
(b) 1500 cm 3 of 0.1 N HCl have number of gm equivalence
N  V1 1500  0 .1
 1

 0.15
1000
1000
 0.15 gm. equivalent of NaOH  0.15  40  6 gm.
M
w
5 .85

 0.2 M
m. wt .  volume in litre 58.5  0 .5
13.
(a)
14.
(a) Molecular weight of C 2 H 5 OH  24  5  16  1  46
Molecular mass of H 2 O  18
414g of C 2 H 5 OH has
18g of H 2O has 
414
 9 mole
46
18
 1 mole
18
n1
1
1


 0 .1
n1  n 2 1  9 10
Solution and Colligative properties
15.
(b) 17 gm NH 3 = 1 mole.
39.
(b) Mole fraction of solute 
40.
(c)
N
17.
6 .02  10 23  4 .25
 1 .5  10 23
17
(b) (2.5  1  3  0.5)  M 3  5.5
41.
(c)
M1 V1  M 2 V2  M 3 V3 ;
20.
4
 0 .73 M.
5 .5
(b) Normality of 2.3 M H 2 SO 4  M  Valency
21.
 2. 3  2  4. 6 N
N 1 V1  N 2 V2 , 36  50  N 2  100
Molecules of NH 3 
or 2.5  1.5  M 3  5.5 or M 3 
(c)
N2 
36  50
 18 ; 18 N H 2 SO 4  9 M H 2 SO 4 .
100
(c) Molarity 
23.
(c) N 1 V1  N 2 V2  NV
(b) [H 3 O  ]  2  0.02  0.04 M

25.
26.
 2 litre solution contains 0.08 mole of H 3 O .
(c)  10 litre of urea solution contains 240 gm of urea
240
 Active mass 
 0.4 .
60  10
(d) NV  N 1 V1  N 2 V2  N 3 V3
or, 1000 N  1  5 
27.
29.
1
1
1
.
 20   30 or N 
2
3
40
N  eq .wt .  V (ml ) 0.05  49.04  100
 0.2452.

1000
1000
(a) For HCl M  N  0.1
N 1 V1  N 2 V2 ; 25  N 1  0.1  35
 2.0 
(a) Molarity 
31.
(a)
M
w
w
; 0 .52 
; w  2.84 gm
m  V (l)
36.5  0 .15
32.
(c)
M
n
n
; 0.5 
; n 1
V (l)
2
33.
(d)
N
W 828
w 36

 18, n 

2
M
46
m 18
x H 2O
n
2
2



 0 .1
2  18 20
nN
34.
(a)
w  1000
98
N
, E
 32.6
E  volume in ml.
3
N 
4 .9  1000
 0.3 N .
32.6  500
%  10  d
22  10  1.253

 0 .805 M .
342
GMM
Normality 
Molality 
46.
%  10  d
22  10  1 .253

 4 .83 N
342 / 6
GEM
22  1000
 0 .825m
342(100  22)
(a) 100 ml. of 0.30M 
100  0 .3
 0 .03 mole of NaCl
1000
100 ml of 0.40M 
100  0 .4
 0 .04 mole of NaCl
1000
Moles of NaCl to be added  0.04  0.03  0.01 mole
= 0.585 gm
6  1000
 1 .5 N
40  100
It is show highest normality than others.
N
47.
(c)
48.
(b)
50.
(c) Strength of H 2 SO 4  98  19.8 g / litre
M
n
0.1
 V  125 ml .
 0 .8 
V (l)
V (l)
S  eq . wt .  N ; N 
51.
(c)
W 1000

 55.55
M
18
x Solute =
53.
n
1
= 0.018.

n  N 1  55.55
(d) Normality of acid = molarity  basicity
i.e., 0.2=molarity  2
 Molarity = 0.2/2 = 0.1
55.
(b) Mole fraction of H 2 O =
59.
(d) Volume strength 
60.
(a)
61.
S
98  19.8

 39.6
eq . wt .
49
W  1000 gm (H 2 O) ; n  1 mole
N
0.5
Volume of solutionin litre
 Volume of solution in litre
0 .5

 0.250litre 250 ml.
2 .0
18  1000
 0 .2 m
180  500
45.
N1 
30.
m
(c)
(d) W 
0 .1  35
0.1  35
; M 
 0 .07 .
25
25  2
(c) We know that
Number of moles of solute
Molarity 
Volume of solutionin litre
w  1000
4  1000

 1.0 N .
m.wt .  Volume in ml 40  100
44.
4 x  10(1  x )  6  1 ; 6 x  4 ; x  0.66
24.
20
 0 .25 .
80
1.5  480  1.2  520  M  1000
720  624
 1.344 M .
M
1000
w
171

 0.5 M .
m.wt.  volume in litre 342  1
22.
185
80
18
80 20

18 34
=
68
.
77
1 .5  100
 8 .82.
17
w
; w  n  m  0.25  98  24.5 gm
m
20 2
Mole

 2.
(d) Molar concentration [H 2 ] 
V in litre
5
n
186
62.
Solution and Colligative properties
(a) Amount
of
added
AgNO3
in
60
ml of solution
84.
(b) An increases in temperature increase the volume of the
solution and thus decreases its molarity.
85.
(d) 10 3 parts of CaCO 3 has number of parts = 10
 60  0.03  1.8 g
63.
(a)
w
w
N
 0 .1 
 w  1 gm
E  V (l)
100  0.1
64.
(b)
N
65.
(a)
20  0.4  40  N or N  0 .2 or M 
69.
(a)
M
72.
(c) M.eq. of HCl = M.eq. of CaCO 3
w
w
 0 .1 
 w  1 gm
E  V (l)
40  0 .25
0.2
 0.1 M .
2
1
1  1000
 1000 ; N 
 0.4 N
50
50  50
molality 
86.
(c)
74.
(b) Molarity of H 2 SO 4  0.5
87.
w 3.65
W 16.2

 0.1, N 

 0 .9
m 36.5
M
18
X
0.1
 0 .1 .
0.1  0.9
(d) 10% glucose solution means 10 g 
Hence 1 mole will be present in
88.
(e) For methyl alcohol N = M.
N1V1  N 2 V2
89.
(b) Mole fraction of glucose =
1
 0 .1 N .
10
=
(c) The density of solution  1.8 gm / ml
 Weight of H 2 SO 4 in the solution 
90.
1620 100

 9 .18
98
180
(a) Suppose the total volume of water = x
91.
92.
100cm  0.5 N  x  0.1 N
100  0.5
x 
 500cm 3
0 .1
Therefore the volume of water added
0 .25  25
 0.0125 .
500
78.
(b)
79.
wt . of the solute(g)
 100
(c) % by wt . 
wt . of the solutiong
80.
81.
(d) Molality (m ) 
82.
(b)
w  1000
 14.05 .
mW
N 1 V1  N 2 V2
93.
94.
95.

10 3
 1000  0 .01 M .
100
w 13.5

 0 .1 .
m 135
(d) 1000 ml of 1 N oxalic solution = 63 g
500 ml of 0.2 N oxalic acid solution
63
=
 500  0.2  6 .3 g .
1000
7 .8
1
78
 .
(a) Mole fraction at C 6 H 6 
7 .8 46 6

78 92
nH 2 O
(c) X H 2 O 
n H 2 O  nC 2 H 5 OH  nCH 3 COOH
(a)
M1V1  M 2 V2
i.e. 5  1  M 2  10  M 2  0.5
Normality of the solution 
0.5
 0 .25.
2
96.
(a)
M
1  1000
w  1000

 0 .1 M .
m  Volume in ml. 40  250
98.
(d)
N
w  1000
 0.33 N .
eq . wt .  volume in ml.
99.
(b) Mole of HCl 
10  10  0.1 (10  V )
10  10
 10  1000  10  990 ml.
0 .1
(b) Sum of mole fraction is always 1.
6 .02  10 20
 10  3 moles
6 .02  10 23
(a) Gram molecule of SO 2 Cl 2 = 135
10

 100  10
90  10
w
18  1000
 0.4 m
(b) Molality 
 1000 
m W
180  250
n
nN
n
3
 Total volume – 100cm 3  500  100  400cm 3 .
0 .1  180
=1.8 litre.
10
Conc. of solution (in molarity)
 Molality 
V
100 cc.
0 .01
 0 .00199
0 .01  5
(b) Mole of urea 
1800  90
=1620gm
100
 Weight of solvent  1800  1620  180 gm
M 1 V1  M 2 V2 , M 2 
10
mole in
180
i.e., 0.1 litre
Weight of one litre of solution  1800 gm
83.
n
Normality of H 2 SO 4 (N1 )  0.5  2  1
1  1  N 2  10 or N 2 
77.
(d)
18
 0.1 molal .
180
73.
10
 10 6  10,000 ppm .
10 3
n
X 
nN

w  1000
10.6  1000

 0.2 M .
m. wt .  Volume in ml.
106  500
N  50 
76.
10 6 parts of CaCO 3 has number of parts
1 .2046  10 24
 2 mole
6 .023  10 23
Normality = molarity  basidity or acicity  2  1  2 N
Solution and Colligative properties
100. (d) 10 N  Deca - normal ,
101.
102.
103.
104.
1
N = Deci-normal.
10
w  1000
7 .1  1000
=
 0 .5 M .
ml wt.  Volume ml. 142  100
4  10
(d) M 
1M .
40
6
0 .1
n
60
(d) Mole fraction X 
.


6 180 10 . 1
nN

60 18
w  1000
10  1000

 1.66 N .
(c) N 
Eq.wt .  Volume
60  100
N  M  bosicity ; N  2  2  4 .
106.
(b)
108.
(b) Concentration 
110.
(a)
5  10 6
= 5 ppm.
10 6
H 3 PO3 is a dibasic acid
(b)
122.
(c)
C
(a)
N
123.
124.
6
 0 .1 molar.
60
(b) Molar solution of sulphuric acid is equal to 2N because it is
show dibasic nature.
0 .5  53  500
 13.25 .
1000
5.85  1000
125. (c) Molar concentration =
 0.5 Molar .
58.5  200
w  1000
75.5  1000
126. (c) M 

 2 .50 M
m.wt .  V in ml
56  540
129.
(a)
0.1  2  20  0.1  1  V2
111.
(d)
0 .1  2  20
 40 ml
0 .1  1
H 3 PO4 ⇌ H   H 2 PO4
H 2 PO4 ⇌ H   HPO42
112.
113.
HPO42  ⇌ H   PO43 
Phosphoric acid does not give 1N strength.
(b) C6 H 5 COOH  NaOH  C6 H 5 COONa  H 2 O
(c)
w 12.2

 4 gms.
40 122
(H 2 SO 4 ) N 1 V1 = N 2 V2 (dilute acid)
N 2  (10  36) / 1000  0.36 N .
114.
(c)
1
O2
2
1 M H 2 O 2 solution  2 N  34 gm / litre 11.2
2  10
 1 .75
11.2
(b) Weight = molarity × m.wt.× v = 1  132  2  264 gm.
115.
w
1
n
m
2


 0 .667 .
116. (a) Mole fraction 
1
8
w W
nN


2 32
m M
118.
(c) 98% H 2 SO 4 means 98g H 2 SO 4 in 100g solution.
120.
100
cc  54.3 cc ; 98 g H 2 SO 4  1 mol
1 .84
1
Hence molarity 
 1000  18.4 M
54.3
(d) 3 CaCl 2  2 Na 3 PO4  Ca 3 (PO4 )2  6 NaCl
 Mole of
Na 3 PO4  3
mole of
CaCl 2  1 mole
Ca 3 (PO4 )2
 0.2 mole of Na 3 PO4  0.3 mole of CaCl 2 = 0.1 mole of
Ca 3 (PO4 )2 .
N 1 V1  N 2 V2
10  10  0.1 × Volume of new solution
Volume of water = 1000 – 10 = 990 ml.
M  m . w t.  V
0.1  98  400
W

 3.92 g .
1000
1000
1000
Molarity of pure water 
 55.6 M .
18
N 0.2
M

 0 .1 M
2
2
180
Moles of water 
 10 mole.
18
44
n CO 2
2
44
Mole fraction of CO 

 .
44 14
3
n CO 2  n N 2

44 28
w
w4
 0 .1 
M
 w  1gm
m  V (l)
40  1
130.
(b)
131.
(a)
132.
(c)
133.
(c)
134.
(c)
136.
(c)
137.
(c)
138.
(b) Number of moles =
H 2O2  H 2O 
So Normality 
w  1000
106
eq . wt . 
 53
eq . wt .  volume in ml .
2
w
N 1 V1 (acid)  N 2 V2 (base)
 V2 
X
 0.2
1000
X
78
121.
(c) Molarity 
187
139.
140.
141.
142.
143.
144.
2
M
w  1 litre
4 1

1M .
m. wt .  Volume litre 40  0 .1
w1 w 2
90 300



 10
m 1 m 2 18
60
(a) The number of moles of solute dissolved in 1000 gm of the
solvent is called molal solution.
0.1  100  392
(b) w 
 3.92 g
1000
18
1
(d)

 0 .1 molal.
180  1 10
n
n
 3   n  3 moles.
V (l)
1
(b) The unit of molality is mole per kilogram.
(a) 0.2 water + 0.8 ethanol; X A  mole fraction of water,
(c)
M
X B  mole fraction of ethanol
XA 
N2
N1
, XB 
N1  N 2
N 2  N1
 Mole fraction of water = 0.2 and ethanol = 0.8.
Colligative properties
188 Solution and Colligative properties
3.
5.
(a) Osmotic pressure is colligative property.
(c) Vapour pressure is not colligative property.
xB 
Lowering of vapour pressure
P  Ps
0
1.
(a)
P
0

4.
5.
(d)
(b)
6.
(a)
7.
(a)
9.
(c)
11.
12.
13.
15.
16.
17.
18.
21.
25.
(a)
P0

w
m
w W

m M
or 0 .00713 
71.5 m
71.5 1000

m
18
Ps  P 0  P is maximum.
(b) The relative lowering of the vapour pressure of dilute solution
is equal to the mole fraction of the solute molecule present in
the solution.
(b) Acetone solution has vapour pressure less than pure water.
1
4
(d) PT  Pp0 x p  Ph0 x h = 440   120 
5
5
88
= 88  96 = 184 ; Pp0 x p  y p PT ;
 yp
184
y p  0.478
(a) P = PBo X B ;  PB 
78
78
 75 ; 
78 46

78 92
(b) Given molecular mass of sucrose = 342
100
Moles of sucrose 
 0 .292 mole
342
1000
Moles of water N 
 55.5 moles and
18
B
(a)
30.
(d)
 x B (Mole fraction of solute)
31.
33.
35.
38.
 molality (1 -   x   y  )
P0
P 0  Ps is maximum for BaCl2 .
P 0  Ps
P0

the
value
of
18  18
 0 .02 .
180  90
PT  PP0 X P  PQ0 X Q ; PT  80 
3
2
 60 
5
5
P 0  Ps
w
m
W w
640  600



w
W
M
m
640
P

m M
2.175  78 640
w M
40
2.175  78
;m
 



m W
640 m  39.08
39.08
40
m  69.45 .
(c) The lower is boiling point more is vapour pressure; boiling
point order, HCl  HBr  HI  HF
(c)
0
P 0  Ps
n
1


 9.9 P 0  9 .9 Ps  P 0
N
9 .9
P
P0
8.9 0
8.9 P 0  9.9 Ps  Ps 
P  0.90 P 0
9.9
(a) 1000 ml of CH 3 OH requires methanol = 32 g.
(c)
P 0  Ps

0

150 ml of 2 M CH 3 OH requires methanol

39.
40.
43.
P = 50 torr
32
 150  2  9 .6 g .
1000
(b)  P 0  Ps  P 0  mole fraction solute
10  P 0  0.2 ; 20  P 0  n  n  0.4  N  0.6 .
(b) According to the Raoult’s law for the non-volatile solute the relative
lowering of vapour pressure of a solution containing a non-volatile is
equal to the mole fraction of the solute.
(d) Relationship between mole fraction of a component in the
vapour phase and total vapour pressure of an ideal solution.
B
Vapour pressure of pure water P 0  23.8 mm Hg
According to Raoult’s law
P
n
P
0.292



23.8 0.292  55.5
P0 n  N
23.8  0.292
P 
 0.125 mm Hg.
55.792
(d) According to Raoult's law, the relative lowering in vapour
pressure of a dilute solution is equal to mole fraction of the
solute present in the solution.
(a) When vapour pressure of solvent decreases, then the boiling
point of solvent increases.
(b) According to Raoult’s Law
P0
27.
P 0  Ps
PT  48  24  72 torr .
m  180
HgI2 although insoluble in water but shows complex
formation with KI and freezing point is decreases.
For solutions containing non-volatile solutes, the Raoult’s law
may be stated as at a given temperature, the vapour pressure
of a solution containing non-volatile solute is directly
proportional to the mole fraction of the solvent.
1
Vapour pressure 
Boiling point
When vapour pressure decreases then b.pt. increases.
Methanol has low boiling point than H 2 O
Lower is boiling point of solvent more is vapour pressure.
Sucrose will give minimum value of P .
P  P 0  Ps
P 0  Ps
(d)
wM
0 .5  154
 143 
 143
65  158
mW
 143  1.03  141.97 mm .
P 0  Ps
26.
0 .8  0 .6
 0 .25 .
0 .8
yA 
PA
x A .PA0

Ptotal x A .PA0  x B .PB0
1 1
1
1

  0 .2
1 1  2  2 1  4 5
(c) Lowering in weight of solution  solution pressure

44.
Lowering in weight of solvent  P 0  Ps
(  p 0  vapour pressure of pure solvent)
p 0  ps
Lowering in weight of solvent

ps
Lowering in weight of solution
p 0  ps w  M

ps
m W
0 .05 10  18
2  2.5 2  250
 m


 100
2.5
90  m
0 .05
5
Ideal and Non-ideal solution
1.
(b) In solution showing positive type of deviation the partial
pressure of each component of solution is greater than the
vapour pressure as expected according to Raoult’s law.
Solution and Colligative properties
In solution of methanol & benzene methanol molecules are held
together due to hydrogen bonding as shown below.
CH 3
CH 3
CH 3
|
|
|
   O — H    O — H    O — H    
On adding benzene, the benzene molecules get in between the
molecule of methanol thus breaking the hydrogen bonds. As
the resulting solution has weak intermolecular attraction, the
escaping tendency of alcohol & benzene molecule from the
solution increases. Consequently the vapour pressure of the
solution is greater than the vapour pressure as expected from
Raoult’s law.
(b) Chloroform & acetone form a non-ideal solution, in which
A..... B type interaction are more than A...... A & B...... B
type interaction due to H -bonding. Hence, the solution shows,
negative deviation from Raoult’s Law i.e.,
3.
Vmix  ve ;
P2  P1  P3 .
8.
10.
11.
12.
H mix  ve
or  80 ml
(b)
H 2 O and C 4 H 9 OH do not form ideal solution because
there is hydrogen bonding between H 2 O and C 4 H 9 OH .
6.
(a) Aromatic compound generally separated
distillation. e.g. Benzene + Toluene.
by
7.
(d)
19.
(a) For the ideal solution H mix and Vmix  0 .
25.
(a) For the ideal solution S mix is not equal to zero.
fractional
13.
14.
(d) Azeotropic mixture is constant boiling mixture, it is not
possible to separate the components of azeotropic mixture by
boiling.
(d) Azeotropic mixture is a mixture of two liquids which boils at
on particular temperature like a pure liquid and distils over in
the same composition.
3.
  CRT 
3  1000
 0.0821  288  6.56 atm .
180  60
1.
(c)
2.
(b) Isotonic solution =

w1
6
342  6


 34.2 .
342  1 60  1
60
3.
(c)
  CRT , C 
4.
(a)

5.
(b)  
7.
(b) C 
(c)
w1
w2

m 1 V1
m 2 V2

RT

17.
19.
20.
23.
24.
31.
32.
35.
38.
41.
0.0821
 0.33  10  2 .
0.821  300
w
0 .1
 RT 
 0 .0821  273
m
1
43.
45.
5
1
50
mol/l

 1000 
342 100
342
47.
50
 0.082  423  5 .07 atm
342
P
w
1
R.T since wvT are constant thus P 
m
mv
n
m / MRT
RT 
V
V
1 .66  2.46
 2 .06 atm
2
(d) Copper ferrocyanide ppt. acts as a semipermeable membrane.
(b) Osmotic pressure = CRT where C = 1 m
  CRT  1  0.0821  300  24.6 atm
(c)

(d)
P  CRT or
P
 RT
C
 P 0  Ps  dRT

(d)   CRT or   
 P0  M


(a) Isotonic solutions are those which have same concentration.
(c)   CRT  0.2  0.0821  293  4.81 atm.
(b) Equal osmotic pressure only applicable of non-electrolytes
solution at low concentration.
(c) As soon as the solute molecules increases the osmotic pressure
of solution increase.
(c) Living cells shrinks in hypertonic solution (plasmolysis) while
bursts in hypotonic solution (endosmosis). There is no. effect
when living cells are kept in isotonic solution.
(c) V  nRT

n
 m  RT
RT  M P   
V
V 

(b)  
w  R  T 68.4  0.0821  273
= 4.92 atm

mV
342
600 20  0.0821  288  1000
; M  1200

760
500  M
Osmosis and Osmotic pressure of the solution
6.
(b)   CRT

16.
1 1000 10


m 100
m
10
50
or m  68.4 .

m
342
(d) Osmotic pressure method is especially suitable for the
determination of molecular masses of macromolecules such as
protein & polymer because for these substances the value of
other colligative properties such as elevation in boiling point or
depression in freezing point are too small to be measured on
the other hand osmotic pressure of such substances are
measurable.

7.8
(b)   C R T ; C 

 0  31 mol / litre
R T  082  310
C 2 H 5 I and C 2 H 5 OH do not form ideal solution.
Azeotropic mixture
1.
(b) In the osmosis solvent molecule move from lower concentration
to higher concentration.
(a) Osmosis occur from dilute solution to concentrate solution.
Therefore solution A is less concentrated than B.
5
1000
50
(c) Molar concentration of cane sugar 


342 100
342
Molar concentration of X =
 total volume of solution = less than (30 + 50 ml)
4.
189
(c)
w RT
10 0.821  (273  69)
= 8.21 atm.


m V
342
0 .1
KNO 3 dissociates completely while CH 3 COOH dissociates
to a small extent. Hence, P1  P2 .
(b) V  n RT
500 V1
V
nR  283
1

; 1 
so V2  5V1
105.3 V2 nR  298
V2 5
190 Solution and Colligative properties
48.
50.
(a) There is no net movement of the solvent through the
semipermeable membrane between two solution of equal
concentration.
w
(b) V  RT
m
4
 0.0821  300 ; m  1.64  10 5 .
m
According to the dialysis process molecular weight increases
but sensitivity decreases.
  T ; if T is doubled  is also doubled.
Osmosis reaction are takes place in increases the volume.
For two non-electrolytic solution if isotonic, C1  C 2
b
16.
(d)
18.
19.
(b)
(b)
20.
(a)
23.
(b)
 6  10  4  1 
52.
(d)
55.
56.
58.
(d)
(b)
(a)
8 .6
5  1000
 m  348.9

60  1 m . wt .  100

59.
1.
(a)
M1 RT02
18  1 .987  (373)2
 0.513 o C
Kb 

1000 H V
1000  9700
2.
(b)
Tb  imkb  0.52  1  2  1.04 .
 Tb  100  1.04  101.04 o C .
3.
(c)
4.
(d)
5.
6.
(b)
(b)
7.
(b)
8.
10.
higher boiling point T1  T2 .
(b) Both urea and glucose are non-electrolytes but NaCl being
electrolyte ionises.
Elevation of boiling point of the solvent
(b)
(b)
Tb
0 .1  100

1K /m .
1
.8
m
 1000
180
K  w  1000 2 .16  0 .15  1000
= 100 .
m b

Tb  W
0 .216  15
Due to higher pressure inside the boiling point elevated.
Dissolution of a non-volatile solute raises the boiling pt. of a
liquid.
As we know that
1
Boiling point 
vapour pre ssure of liquid
Hence, on decreasing vapour pressure, boiling point will
increase.
100  K b  w
100  5 .2  6
Tb 
 0 .52 
m W
m  100
100  5 .2  6
m
 60 .
0.52  100
Elevation in a boiling point is a colligative property as it
depends upon the number of particles.
∆T  n
For sucrose, n = 1, ∆T = 0.1°C
For NaCl, n = 2, ∆T = 0.2°C
Tb  Kb  m or Kb  Tb / m
Kb 
b
Depression of freezing point of the solvent
1.
(c)
2.
(c)
12.
(c)
 0.1

Tb  K b  m  0.513  
 1000 
 200

 0.2565 C , Tb  100 . 256 C
o
15.
o
(a) T = i.K .m
b
2+
2
0.186 
100  18.6  w
60  500
(a) Camphor has the maximum value of K f ( 39.7) .
4.
(d) The extent of depression in freezing point varies with the
number of solute particles for a fixed solvent only and it is a
characteristics feature of the nature of solvent also. So for two
different solvents the extent of depression may vary even if
number of solute particles be dissolved.
6.
(a) Molar mass 
K f  1000  w
T f  W

1 .86  1000  1 .25
20  1 .1
 105.68  105.7 .
7.
(b)
HBr ⇌ H   Br 
(1  )


Total = 1 +   i  1    1  0.9  1.9
T f  i K f  m  1.9  1.86 
8.1 1000

 3.53 o C
81 100
T f  3.53 o C .
T f  K f  m  1.86  0.1  0.186 .
8.
(b)
9.
10.
(b) Freezing point is lowered on addition of solute in it..
(d) Tb  0.18 ; Tb  mK b
0 .18 mK b 0 .18  1 .86

;
 T f ; T f  0.653
T f
m Kf
0 .512
T 0  Ts  0.653 ; T 0  Ts  0.653 ; Ts  0  0.653 o C .
11.
(a)
12.
(b)
2Cl
1
0
0
(1)

2
i = 1 + 2
Assuming 100% ionization
100  K  w
m W
3.
b
CuCl  Cu +
T f 
w  3g
b
(b)
Tf  K f  molality  1.86  0.05  0.093 °C
Thus freezing point = 0 – 0.093 =  0.093 o C .
b
11.
So, i = 3
T = 3  0.52  0.1 = 0.156  0.16
K  w  1000
Tb  b
m W
K b  w  1000 2.53  10  1000
m

 253 g .
Tb  W
1  100
Common salt is non-volatile and rises the b.pt.
In the process of evaporation, high energy molecules leave the
surface of liquid, hence average kinetic energy and
consequently the temperature of liquid falls.
The boiling occurs at lowers temperature if atmospheric
pressure is lower than 76cm Hg.
BaCl2 furnishes more ions than KCl and thus shows
 342 
o
o
T f  1 .86  
  1 .86 ;  T f  1.86 C .
 342 
Tb  Kb  m i.e. 0.52  0.52  m
Tf  K f  m  1.86  1  1.86 ; T f  1.86 o C .
13.
(b) For NaCl i  2
T f  2 K f m  2  1.86  1  3.72
Solution and Colligative properties
Ts  T  T f  0  3.72  3.72C
15.
(b) T f  K f  m  0.186  1.86  m
6.
(a) NaCl gives maximum ion hence it will show highest osmotic
pressure.
8.
(c)
So m  0.1 , Put the value of m in Tb  Kb  m
20.
(a) Dissolution of a non-volatile solute lowers the freezing pt. of
the solution H 2 O.
(a) By using, m 
9.
K f  1000  w
T f  WSolvent
5 .12  1000  1

0 .40  50
(gm)
 256 gm / mol
Hence, molecular mass of the solute  256 gm mol 1
K f  w  1000

5.12  0.440  1000
 178.9
0.567  22.2
10.
11.
12.
21.
(a)
m
22.
(a)
KBr  K   Br   2 ions
13.
BaCl2  Ba 2   2Cl   3 inos
14.
T f  W
Al2 (SO 4 )3  2 Al 3   3 SO 42   5 ions
 urea is not ionise hence it is shows highest freezing point.
23.
(d)
K 2 SO 4 give maximum ion in solution so it shows maximum
depression in freezing point.
24.
(c)
T f 
25.
(a)
T f 
K f  1000  w
m W
1000  1.86  17
 0.69 o C
46  1000
(ad) The depression of freezing point is less than that of pure
solvent and only solvent molecules solidify at the freezing point.
27.
(b)
Kf 
RTf2
1000  L f
, R  8.314 JK 1mol 1
Tf  273  16.6  289.6 K ; L f  180.75 Jg 1
Kf 
8.314  289.6  289.6
1000  180.75
Colligative properties of electrolyte
3.
4.
5.
(c)
BaCl2 gives maximum ion. Hence, its shows highest boiling
point.
(c) BaCl2 gives maximum ion. Hence, its boiling point is
maximum.
(d) CaCl 2 gives maximum ion hence it shows highest boiling
point.
(c) Elevation in boiling point is a colligative property which
depends upon the number of solute particles. Greater the
number of solute particle in a solution higher the extent of
elevation in boiling point.
Na 2 SO 4  2 Na   SO 42
15.
(a)
NaCl contain highest boiling point than other’s compound.
16.
(d)
KCl  C6 H 5 OH  C6 H 12 O6
Boiling point decreasing order 
Potassium chloride is ionic compound and phenol is formed
phenoxide ion hence it is shows greater boiling point then
glucose.
 0 .216 C
26.
2.
Al2 (SO 4 )3 gives maximum osmotic pressure because it is
gives 5 ion.
(a) Highest osmotic pressure is given by solution which produce
maximum number of ions i.e. CaCl2 .
o
T f  0  0.69  0.69 o C
1.
(a)
NaCl  Na   Cl   2 ions
K 2 SO 4  2 K   SO 42   3 ions
(NH 4 )3 PO4 gives maximum ion. Hence, its osmotic pressure
is maximum.
(b) BaCl2 gives maximum ion hence it is shows lowest vapour
pressure.
(d) Na 3 PO4 consist of maximum ions hence it show lowest vapour
pressure.
17.
(c)
18.
(b)
19.
(c)
20.
(c)
21.
23.
(b)
(b)
24.
(c)
26.
(d)
27.
(a)
(c)
Na 3 PO4  3 Na   PO43   4 ion.
(c) Vapour pressure of a solvent is lowered by the presence of
solute in it. Lowering in vapour pressure is a colligative
property i.e., it depends on the no. of particles present in the
solution. Cu (NO 3 )2 give the maximum no. of ions. (i.e., 3) so
it causes the greatest lowering in vapour pressure of water.
(c) Na2 SO 4 have more osmotic pressure than NaCl solution
because Na2 SO 4 gives 3 ions.
BaCl2  Ba2   2Cl   3 ion
NaCl  Na   Cl   2 ion
Glucose  No ionisation
 BaCl2  NaCl  Glucose
Tb  0.521  (0.1)  0.0521
17.
191
28.
Al2 (SO 4 )3 gives maximum ion hence it will show highest
boiling point.
NaCl is a more ionic compare to BaCl2 , glucose and urea
solution.
Urea = 1 ; Common salt = 1 ; Na 2 SO 4  3
Ratio = 1 : 2 : 3
CaCl 2 gives maximum ion hence it has minimum freezing
point.
NaCl gives maximum ion hence it shows lowest freezing point
Lesser the number of particles in solution. Lesser the
depression in freezing point, i.e. higher the freezing point.
BaCl2 gives maximum ion hence it shows maximum
depression in freezing point.
We know that lowering of freezing point is a colligative
property which is directly proportional to the number of
particles formed by one mole of compound therefore 0.1M
Al2 (SO 4 )3 solution will have minimum freezing point.
Al2 (SO 4 )3 gives maximum ion hence its gives lowest
freezing point.
(b) Colligative property in decreasing order
Na 3 PO4  Na 2 SO 4  NaCl
Na 3 PO4  3 Na   PO43   4
Na 2 SO 4  2 Na   SO 42   3
29.
NaCl  Na   Cl   2
(d) K 4 [Fe(CN )6 ] gives maximum ion. Hence it have lowest
vapour pressure.
Abnormal Molecular Mass
192
Solution and Colligative properties
1.
(c)
Na 3 PO4 gives maximum four ion it is show highest Vant’s
haff factor.
2.
(a)
K 4 [Fe(CN )6 ] dissociates as 4 K   [Fe(CN )6 ]4  , thus 1
molecule dissociates into five particles in the similar way
Al2 (SO 4 )3 also gives five particles per molecule.
3.
4.
5.
6.
8.
(a) Benzoic acid in benzene undergoes association through
intermolecular hydrogen bonding.
experiment al C.P.
(c) vont’s Hoff factor (i ) 
Calculated C.P.
 1    x  y , for KCl it is = 2 and for sugar it is equal
to 1.
Ca(NO 3 )2  Ca 2   2 NO 3 it gives three ions hence the
Van’t Hoff factor = 3.
5  18  2 .5
(a) m 
 70.31
0 .04  80
(c)
⇌
(c)
Na 2 SO 4
Mol. before diss. 1
Mol. after diss 1  
i
2 Na  
0
2
SO 42
0
1
4.
5.
6.
7.
8.
9.
(d) It is known that azeotropic mixture of HCl and water 20.2%
HCl .
1 .75
n
(a)   CRT  RT  342  0 .0821  290
150
V
1000
 0.8095  0.81 atm .
(c) Vant hoff factor of NaCl about 1.95 because it will be ionise
into two ions.
NaCl ⇌ Na   Cl 
wRT 0.6  0.082  300
(b) m 

 120
PV
1.23  0.1
K  w  1000
(b) m  b
 143.18
Tb  W
(c) Here: Tb  0.323 K
w  0.5143 g weight of Anthracene.
W  35 g weight of chloroform
K b  Molal elevation constant (3.9 K  Kg / mol)
Exp.C.P.
 1    2    1  2
Normal C.P.
m
10.
11.
13.
(d) MgSO4 dissociates to give 2 ions.
(d) Urea does not give ion in the solution.
(b) Molecular weight of CH 3 COOH  60
17.
Hence the molecular weight of acetic acid in
benzene  2  60  120 .
(b) AlCl3 furnishes more ions than CaCl 2 and thus shows
 177.42 g / mol
10.
(c) First boiling point of water = 100 o C
Final boiling point of water = 100.52 o
w  3 g , W  200 g , Kb  0.6 kg 1
Tb  100.52  100  0.52 o C
higher boiling point i.e. t1  t 2 .

Na 3 PO4  3 Na 
m
PO33  .
18.
(d)
19.
(b) Benzoic acid dimerises due to strong hydrogen bonding.
Critical Thinking Questions
1.
2.
P o  Ps
n
; P o  0.80, Ps  0.60

o
nN
P
n
0 .2


 0 .25 .
n  N 0 .8
(c) We have,
p 0  22.15
30  18
, for II case

22.15
108  m
By eq. (1)
p m0
11.
log
760
40656

23
2 .303  8 .314
 373  T1 
 373T 


This gives T1  294.4 K .
12.
(b)
Tf  molality K f  (1   )
  0.2 , Molality = 0.2, K f  1.86
Tf  0.2  1.2  1.86  0.4464 o
.....(ii)
13.
Freezing point =  0.45 o C .
(b) T f  imk f ; 0.0054  i  1.8  0.001
i  3 so it is [Pt(NH 3 )4 Cl]Cl 2 .
 21.85m  21.85  6  131.1
0.30m = 20.35
20.35
m
 67.83
0 .30
(c)
0.6  3  1000 1800

 17.3 gmol 1 .
0.52  200
104
(b) Applying clausius clapeytron equation
P
H V  T2  T1 
log 2 


P1 2 .303 R  T1  T2 
.....(i)
By eq. (2) p m0  22.15m  22.15  5  110.75
3.
K b  w  1000
Tb  W

(a)
p 0  21.85 30  18
, for I case

21.85
90  m
wt. of solvent  90  18  108 gm
Kb  w  1000 3 .9  0 .5143  1000

W  Tb
0 .323  35
5
W2
M2
P o  Ps
M 2 = 3000  2985

or M 2  180

100
W1
3000
Po
18
M1
K f  w  1000
14.
(d) m 
15.
(a)
T  K f  Molality  4.9  0.001  0.0049 K
16.
(c)
Na 2 CO 3  H 2 SO 4  Na 2 SO 4  CO 2 HO
T f  W
 60 g / mole .
98 gm (2 mole )
1 mole
0 .02  22.4
0.02 
 0 .224 .
2
1 mole
Solution and Colligative properties
17.
(a) We know that in the first solution number of the moles of urea
Mass of urea 1
12 1



  0 .2 and
m.wt. of urea V
60 1
In second solution the number of moles of cane sugar
68.4 1
Mass of cane sugar


  0 .2 .
m.wt. of cane sugar
342 1
Assertion & Reason
1.
2.
0.3  100
 0 .15
200
(c) Both the solute and solvent will form the vapour but vapour
phase will become richer in the more volatile component.
(b) Non-ideal solutions with positive deviation i.e., having more
vapour pressure than expected, boil at lower temperature while
those with negative deviation boil at higher temperature than
those of the components.
 N 3  200 or N 3 
3.
4.
abnormally high. This is because dimerisation of benzoic takes
place in solution resulting high molecular mass. Therefore,
assertion and reason are true and reason is correct explanation.
18.
(a) It is fact that use of pressure cooker reduces cooking time
because at higher pressure over the liquid due to cooker lid,
the liquid boils at higher temperature and cooking occurs
faster.
19.
(c) The assertion that CCl 4 & H 2 O are immiscible is true
because CCl 4 is non-polar liquid while water is polar hence
assertion is true and reason is false.
(a) Molecular weight of urea (NH 2CONH 2 )
 14  2  12  16  14  2  60
Weight
60

1
Number of moles 
molecular weight 60
(e) If 100 cc of 0.1 N HCl is mixed with 100 cc of 0.2 N HCl , the
normality of the final solution will be 0.15.
N1V1  N 2 V2  N 3 V3 i.e., 0.1  100  0.2  100
5.
(c) The polymer solutions possess very little elevation in boiling
point or depression in freezing point.
6.
(a) Depression in freezing point is a colligative property which
depends upon the number of particles. The number of particles
are different in case of benzene and water that is why
molecular weight of acetic acid determined by depression in
freezing point method is also different.
7.
(d) Sodium ion, Na  and potassium ion, K  are responsible for
maintaining proper osmotic pressure balance inside and outside
of the cells of organisms.
8.
(a) If a pressure larger than the osmotic pressure is applied to the
solution side, the pure solvent flows out of the solution
through semi-permeable membrane and this phenomenon is
called as reverse osmosis.
9.
(c) Camphor has high molal depression constant.
10.
(b) Elevation in boiling point and depression in freezing point are
colligative properties because both depend only on the number
of particles (ions or molecules) of the solute in a definite
amount of the solvent but not on the nature of the solute.
12.
(a) The boiling point and melting point are higher due to presence
of the intermolecular hydrogen bonding.
14.
(d) If a non-volatile solute is added to water its vapour pressure
always decreases. Therefore, both assertion and reason are
false.
15.
(b) We know that heat of vaporisation of water at 100 o C is
40.6 kJ and that of benzene is 31kJ at 80 o C . The
amount of heat required to vaporise one mole of liquid at
constant temperautre is known as heat of vapourisation
therefore, both assertion and reason are true but reason is nat
the correct explanation of assertion.
16.
(d) See melts slowly at high altitude because melting is favoured at
a high pressure at high altitude the atmospheric pressure is
low and so ice melts slowly.
17.
(a) Colligative properties are the properties of solutions containing
non volatile solute. It is correct that malecular mass of benzoic
acid when determined by colligative properties is found
193
20.
(b) It is true that isotonic solution doesn’t show the phenomenon
of osmosis. Isotonic solution are those solution which have
same osmotic pressure. Here both assertion and reason are
true but reason is not correct explanation.
194 Solution and Colligative properties
1.
The 2 N aqueous solution of H 2 SO 4 contains
9.
(a) 49 gm of H 2 SO 4 per litre of solution
(b) 4.9 gm of H 2 SO 4 per litre of solution
(c) 98 gm of H 2 SO 4 per litre of solution
(d) 9.8 gm of H 2 SO 4 per litre of solution
2.
3.
The amount of KMnO4 required to prepare 100 ml of 0.1 N
solution in alkaline medium is
[CPMT 1986]
(a) 1.58 gm
(b) 3.16 gm
(c) 0.52 gm
(d) 0.31 gm
10.
0.45 g
(b) 0.90 g
(c)
1.08 g
(d) 1.26 g
(a)
AlCl3  BaCl2  Urea
(b)
BaCl2  AlCl3  Urea
(c)
Urea  BaCl2  AlCl3
(d)
BaCl2  Urea  AlCl3
The osmotic pressure of a 5% solution of cane sugar at 150 o C is
(mol. wt. of cane sugar = 342)
[CPMT 1986; Manipal MEE 1995]
What weight of hydrated oxalic acid should be added for complete
neutralisation of 100ml of 0.2 N  NaOH solution ?
(a)
The O.P. of equimolar solution of Urea, BaCl2 and AlCl3 , will
be in the order
[DCE 2000]
(a) 4 atm
[MP PMT 1997]
(c) 5.07 atm
11.
5.
A 500 g tooth paste sample has 0 .2 g fluoride concentration.
0.1M urea and 0.1M NaCl
What is the concentration of F in terms of ppm level
(b) 0.[AIIMS
1M urea
1992]and 0.2 M MgCl2
(a) 250
(b) 200
(c)
(c) 400
(d) 1000
(a) 0.1 Normal
(c) 0.1 Molar
0.1M NaCl and 0.1M Na 2 SO 4
(d) 0.1M Ca(NO 3 )2 and 0.1M Na 2 SO 4
To 5.85 gm of NaCl one kg of water is added to prepare of
solution. What is the strength of NaCl in this solution (mol. wt. of
NaCl  58.5 )
[CPMT 1990; DPMT 1987]
6.
(d) 2.45 atm
Which one of the following pairs of solutions can we expect to be
isotonic at the same temperature
[NCERT 1982]
(a)
4.
(b) 3.4 atm
12.
Which of the following would have the highest osmotic pressure
(assume that all salts are 90% dissociated)
[NCERT 1982]
(b) 0.1 Molal
(a) Decimolar aluminium sulphate
(d) 0.1 Formal
(b) Decimolar barium chloride
The degree of dissociation of Ca(NO 3 )2 in a dilute aqueous
(c) Decimolar sodium sulphate
solution containing 14g of the salt per 200g of water 100 o C is 70
percent. If the vapour pressure of water at 100 o C is 760 cm.
Calculate the vapour pressure of the solution
(d) A solution obtained by mixing equal of (b) and (c) and filtering
13.
Which solution will have the highest boiling point
[UPSEAT 2000]
7.
8.
[NCERT 1981]
(a) 746.3 mm of Hg
(b) 757.5 mm of Hg
(a) 1% solution of glucose in water
(c) 740.9 mm of Hg
(d) 750 mm of Hg
(b) 1% solution of sodium chloride in water
The vapour pressure of pure benzene at a certain temperature is 200
mm Hg. At the same temperature the vapour pressure of a solution
containing 2g of non-volatile non-electrolyte solid in 78g of benzene is
195 mm Hg. What is the molecular weight of solid
(a) 50
(b) 70
(c) 85
(d) 80
(c) 1% solution of zinc sulphate in water
(d) 1% solution of urea in water
14.
ether was found to be 0.1o C higher than that of the pure ether.
The molecular weight of the substance will be (Kb  2.16) [MP PET 2002]
Which one of the following non-ideal solutions shows the negative
deviation
(a)
CH 3 COCH 3  CS 2
(b) C6 H 6  CH 3 COCH 3
(c)
CCl 4  CHCl 3
(d) CH 3 COCH 3  CHCl 3
[UPSEAT
The boiling
point2001]
of a solution of 0.11 gm of a substance in 15 gm of
15.
(a) 148
(b) 158
(c) 168
(d) 178
The boiling point of benzene is 353.23 K. When 1.80 gm of a
nonvolatile solute was dissolved in 90 gm of benzene, the boiling
point is raised to 354.11 K. the molar mass of the solute is
Solution and Colligative properties
[Kb for benzene = 2.53 K mol ]
[DPMT 2004]
–1
(a)
18.
58 g mol 1
(d) 0.88 g mol
16.
1
19.
(a) 1.8
(b) 0.18
(c) 18
(d) 18.6
compound (K f  5.1 km
0.60 o
(b) 0.80 o
(c)
1.60 o
(d)
The freezing point of equimolal aqueous solution will be highest for
C 6 H 5 NH 3 Cl  (aniline hydrochloride)
(c)
) , (freezing point of benzene = 5.5 C )
La(NO 3 )3
(d) C 6 H 12 O6 (glucose)
20.
The Van't Hoff factor of the compound K 3 Fe(CN )6 is
o
(a) 100
(b) 200
(c) 300
(d) 400
2.40 o
(b) Ca(NO 3 )2
The freezing point of a solution containing 4.8 g of a compound in
60 g of benzene is 4.48. What is the molar mass of the
1
(a)
(a)
The boiling point of 0.1 molal aqueous solution of urea is
100.18 o C at 1 atm. The molal elevation constant of water is
17.
When 0.01 mole of sugar is dissolved in 100 g of a solvent, the
depression in freezing point is 0.40 o . When 0.03 mole of glucose
is dissolved in 50 g of the same solvent, the depression in freezing
point will be
5.8 g mol 1
(b) 0.58 g mol 1
(c)
195
(a) 1
(c) 3
[AFMC 2000]
(b) 2
(d) 4
196 Solution and Colligative properties
(SET -4)
 2  49 = 98g.
1.
(c) Wt. of H 2 SO 4 per litre  N  eq. mass
2.
(a) In alkaline medium KMnO4 act as oxidant as follows.
2 KMnO4  2 KOH  2 K 2 MnO4  H 2 O  (O)
13.
(b)
than ZnSO 4 .
14.
(b) m 
Hence its eq .wt . = m. wt .  158
3.
100
g  1.58 g.
1000
m
15.

 354.11K  353.23 K  0.88 K
F  ions in PPm =
M Solute 
5.
(b) 5.85 g NaCl = 0.1 mol as it present in 1 kg of
M solute 
(a)
7.
(d)
8.
(d) CH 3 COCH 3  CHCl 3 is non ideal solution which shown
negative deviation.
9.
(a) The particle come of AlCl3 solution will be maximum due to
P o  Ps
P o  Ps w  M
n

;

 80
o
nN
m W
P
Po
(a)
17.
(d) m 
18.
(d)
0 .18
 1 .8
0 .1
K f  1000  w
W  T f
0.40 


5  0.0821  1000  423
 5.07 atm .
342  100
11.
(d) Osmotic pressure is a coligative properties equimolar solution
of Ca(NO 3 )2 and Na 2 SO 4 will produce same number of
solute particles.
Al2 (SO 4 )3 Deci-molar gives maximum ion. Hence, its
osmotic pressure is maximum.
0 .03  1000
4
50
 2.4
19.
(d)
La(NO 3 )3 will furnish four ions and thus will develop more
lowering in freezing point whereas glucose gives only one
particle and thus minimum lowering in freezing point.
20.
(d)
K 3 [Fe(CN )6 ]  3 K   [Fe(CN )6 ]3  .
CaNO 3 ⇌ Ca 2   2NO 3
Na 2 (SO 4 ) ⇌ 2 Na   SO 42 
5.1  1000  4.8
 400 .
60  1 .02
0.01  1000
kf  kf  4
100
AlCl3  Al 3   3Cl   4
(c)

T f  mk f
again T f  mk f
10.
(a)
Kb 
16.
More the number of particles in solution more is the osmotic
pressure a colligative properties.
12.
2.53  1.8  1000
 58 gm mol 1
0.88  90
ionisation less in BaCl2 and minimum in urea
BaCl2  Ba 2   2Cl   3
W  weight of solvent
Hence, molar mass of the solute  58 gm mol 1
wt .
5 .85
=0.1molal

m wt .  l 58.5  1
6.
K b  1000  w
Tb  W
Where, w  weight of solute,
0.2
 10 6  400
500
(c)
water ; molality 
Substituting these values in expression
w
0.2  100

 w  1.26 gm .
63
1000
4.
2.16  0.11  1000
 158.40 ~ 158 .
0.1  15
(c) The elevation (Tb ) in the boiling point
(d) For complete neutralization equivalent of oxalic acid =
equivalent of NaOH =
w
NV

eq . wt 1000
Kb  w  1000
Tb  W
Kb  2.16, w  0.11, W  15 g, Tb  0.1
Mass
1
Now, Normality 

Eq. mass V(L)
mass  0.1  158 
NaCl and ZnSO 4 gives 2 ions but NaCl is more ionic
***
Solid State 197
Chapter
5
Solid State
The solids are the substances which have definite volume and
definite shape. In terms of kinetic molecular model, solids have regular
order of their constituent particles (atoms, molecules or ions). These
particles are held together by fairly strong forces, therefore, they are present
at fixed positions. The properties of the solids not only depend upon the
nature of the constituents but also on their arrangements.
Types and Classification of solids
(1) Types of solids
Solids can be broadly classified into following two types,
Silica occurs in crystalline as well as amorphous states. Quartz is a
typical example of crystalline silica. Quartz and the amorphous silica differ
considerably in their properties.
Quartz
Amorphous silica
It is crystalline in nature
It is light (fluffy) white powder
All four corners of
SiO 44 
tetrahedron are shared by others to
give a network solid
It has high and sharp melting point
(1710°C)
The SiO 44  tetrahedra are randomly
joined, giving rise to polymeric chains,
sheets or three-dimensional units
It does not have sharp melting point
(i) Crystalline solids/True solids,
(3) Diamond and graphite
(ii) Amorphous solids/Pseudo solids
Crystalline solids
Amorphous solids
They have long range order.
They have short range order.
They have definite melting point
Not have definite melting point
They have a definite heat of fusion
Not have definite heat of fusion
They are rigid and incompressible
Not be compressed to any appreciable
extent
They are given cleavage i.e. they
break into two pieces with plane
surfaces
They are given irregular cleavage i.e.
they break into two pieces with irregular
surface
They are anisotropic because of these
substances show different property in
different direction
They are isotropic because of these
substances show same property in all
directions
There is a sudden change in volume
when it melts.
Diamond and graphite are tow allotropes of carbon. Diamond and
graphite both are covalent crystals. But, they differ considerably in their
properties.
Diamond
Graphite
It occurs naturally in free state
It occurs naturally, as
manufactured artificially
It is the hardest natural substance
known.
It is soft and greasy to touch
It has high relative density (about 3.5)
Its relative density is 2.3
It is transparent and
refractive index (2.45)
has
high
well
as
It has black in colour and opaque
It is non-conductor of heat and
electricity.
Graphite is a good conductor of heat
and electricity
There is no sudden change in volume on
melting.
It burns in air at 900°C to give CO2
It burns in air at 700°C to give CO2
These possess symmetry
Not possess any symmetry.
It occurs as octahedral crystals
It occurs as hexagonal crystals
These possess interfacial angles.
Not possess interfacial angles.
(2) Crystalline and amorphous silica (SiO2 )
(4) Classification of crystalline solids
Table : 5.1 Some characteristics of different types of crystalline solids
Types of
Solid
Constituents
Bonding
Examples
Physical
Nature
M.P.
B.P.
Electrical
Conductivity
198 Solid State
Ionic
Covalent
Molecular
Metallic
Atomic
Positive and
negative ions
network
systematically
arranged
Atoms connected in
covalent bonds
Coulombic
NaCl, KCl, CaO, MgO,
LiF, ZnS, BaSO4 and
K2SO4 etc.
Hard but
brittle
High (≃1000K)
High (≃2000K)
Conductor
(in molten state
and in aqueous
solution)
Electron
sharing
SiO2 (Quartz),
SiC, C (diamond),
Very high (≃4000K)
Very high
Insulator except
graphite
Polar or non-polar
molecules
(i) Molecular
interactions
(intermolecular forces)
(ii) Hydrogen
bonding
Hard
Hard
Hard
Soft
Cations in a sea of
electrons
Metallic
Atoms
London
dispersion
force
C(graphite) etc.
I2,S , P , CO , CH ,
8
4
2
4
Low
(≃300K to 600K)
(≃ 450 to 800 K)
Low
Low
(≃400K)
(≃373K to 500K)
Ductile
malleable
High
High
(≃800K to 1000 K)
(≃1500K to 2000K)
Soft
Very low
Very low
CCl etc.
4
Starch, sucrose,
water, dry ice or
drycold (solid CO2)
etc.
Sodium , Au, Cu,
magnesium, metals
and alloys
Noble gases
Crystallography
“The branch of science that deals with the study of structure,
geometry and properties of crystals is called crystallography”.
(≃5000K)
Low
Soft
Insulator
Insulator
Conductor
Poor thermal and
electrical
conductors
(iii) Centre of symmetry : It is an imaginary point in the crystal that
any line drawn through it intersects the surface of the crystal at equal
Y
distance on either side.
(1) Symmetry in Crystal : A crystal possess following three types of
symmetry,
(i) Plane of symmetry : It is an imaginary plane which passes
through the centre of a crystal can divides it into two equal portions which
are exactly the mirror images of each other.
Centre of
symmetry of a
cubic crystal
X
Z
Fig. 5.3
Only simple cubic system have one centre of symmetry. Other
system do not have centre of symmetry.
Plane of symmetry
Rectangular plane of
symmetry
Diagonal plane
of symmetry
Fig. 5.1
(ii) Axis of symmetry : An axis of symmetry or axis of rotation is an
imaginary line, passing through the crystal such that when the crystal is
rotated about this line, it presents the same appearance more than once in
one complete revolution i.e., in a rotation through 360°. Suppose, the same
appearance of crystal is repeated, on rotating it through an angle of 360°/n,
around an imaginary axis, is called an n-fold axis where, n is known as the
order of axis. By order is meant the value of n in 2 / n so that rotation
through 2 / n, gives an equivalent configuration.
Axis of two fold symmetry
The total number of planes, axes and centre of symmetries possessed
by a crystal is termed as elements of symmetry.
A cubic crystal possesses total 23 elements of symmetry.
Plane of symmetry
( 3 + 6)
=9
Axes of symmetry
( 3 + 4 + 6)
= 13
Centre of symmetry
(1)
=1
Total symmetry = 23
(2) Laws of crystallography : Crystallography is based on three
fundamental laws.
(i) Law of constancy of interfacial angles : This law states that angle
between adjacent corresponding faces is inter facial angles of the crystal of a
particular substance is always constant inspite of different shapes and sizes
and mode of growth of crystal. The size and shape of crystal depend upon the
conditions of crystallisation. This law is also known as Steno's Law.
Axis of three f old symmetry
Fig. 5.4. Constancy of interfacial angles
(ii) Law of rational indices : This law states that the ratio of
intercepts of different faces of a crystal with the three axes are constant and
can be expressed by rational numbers that the intercepts of any face of a
crystal along the crystallographic axes are either equal to unit intercepts
(i.e., intercepts made by unit cell) a, b, c or some simple whole number
multiples of them e.g., na, n' b, n''c, where n, n' and n'' are simple whole
numbers. The whole numbers n, n' and n'' are called Weiss indices. This law
was given by Hauy.
Axis of four fold symmetry
Fig. 5.2
Axis of six fold symmetry
Solid state
(iii) Law of constancy of symmetry : According to this law, all
crystals of a substance have the same elements of symmetry is plane of
symmetry, axis of symmetry and centre of symmetry.
Miller indices : Planes in crystals are described by a set of integers
(h, k and l) known as Miller indices. Miller indices of a plane are the
reciprocals of the fractional intercepts of that plane on the various
crystallographic axes. For calculating Miller indices, a reference plane, known
as parametral plane, is selected having intercepts a, b and c along x, y and
z-axes, respectively. Then, the intercepts of the unknown plane are given
with respect to a, b and c of the parametral plane.
A three dimensional group of lattice points which when repeated in
space generates the crystal called unit cell.
The unit cell is described by the lengths of its edges, a, b, c (which
are related to the spacing between layers) and the angles between the edges,
Space
,  ,  .
Lattice
c

b
a
k
b
intercept of the plane along y - axis
l
c
intercept of the plane along z - axis
Unit cell
The distance between the parallel planes in crystals are designated
as d hkl . For different cubic lattices these interplanar spacing are given by
the general formula,
d(hkl ) 


Thus, the Miller indices are :
a
h
intercept of the plane along x - axis
a
h2  k 2  l2
199
Fig. 5.5
Unit
Cell
Space lattice & unit cell
Types of units cells
A units cell is obtained by joining the lattice points. The choice of
lattice points to draw a unit cell is made on the basis of the external
geometry of the crystal, and symmetry of the lattice. There are four
different types of unit cells. These are,
(1) Primitive or simple cubic (sc) : Atoms are arranged only at the
corners of the unit cell.
(2) Body centred cubic (bcc) : Atoms are arranged at the corners
and at the centre of the unit cell.
(3) Face centred cubic (fcc) : Atoms are arranged at the corners and
at the centre of each faces of the unit cell.
(4) Side centered : Atoms are arranged at the centre of only one set
of faces in addition to the atoms at the corner of the unit cell.
Formation of crystal and Crystal systems
Where a is the length of cube side while h, k and l are the Miller
indices of the plane.
When a plane is parallel to an axis, its intercept with that axis is
taken as infinite and the Miller will be zero.
Negative signs in the Miller indices is indicated by placing a bar on
the intercept. All parallel planes have same Miller indices.
The crystals of the substance are obtained by cooling the liquid (or
the melt) of the solution of that substance. The size of the crystal depends
upon the rate of cooling. If cooling is carried out slowly, crystals of large
size are obtained because the particles (ions, atoms or molecules) get
sufficient time to arrange themselves in proper positions.
Dissolved
dissolved
Atoms of molecules 

 cluster 

dissolved embryo  nucleus  crystal
(unstable)
(If loosing units dissolves as embryo and if gaining unit grow as a
crystals).
Bravais (1848) showed from geometrical considerations that there
can be only 14 different ways in which similar points can be arranged. Thus,
there can be only 14 different space lattices. These 14 types of lattices are
known as Bravais Lattices. But on the other hand Bravais showed that there
are only seven types of crystal systems.
Space lattice and Unit cell
Crystal is a homogeneous portion of a crystalline substance,
composed of a regular pattern of structural units (ions, atoms or molecules)
by plane surfaces making definite angles with each other giving a regular
geometric form.
A regular array of points (showing atoms/ions) in three dimensions
is commonly called as a space lattice, or lattice.
Each point in a space lattice represents an atom or a group of
atoms.
Each point in a space lattice has identical surroundings throughout.
Table : 5.2 Bravais lattices corresponding to different crystal systems
Crystal system
Cubic
ab c,
Simple : Lattice points at the eight
corners of the unit cells.
      90 o
Face centered : Points at the
eight corners and at the six
face centres.
Pb, Hg, Ag,
Au, Cu , ZnS ,
diamond, KCl , CsCl,
NaCl, Cu 2O, CaF2

and alums. etc.

b
a
ab c ,
Body centered : Points at the eight
corners and at the body centred.
c

Tetragonal
Examples
Space lattice
Simple : Points at the eight corners of
the unit cell.
Body centered : Points at the eight corners and at the body centre
SnO 2 , TiO 2 ,
ZnO 2 , NiSO 4
      90 o
c



a
b
ZrSiO 4 . PbWO4 ,
white Sn etc.
200 Solid State
Orthorhombic (Rhombic)
ab c,
Simple: Points at the
eight corners of the
unit cell.
      90 o
End centered : Also called side
centered or base centered.
Points at the eight corners and
at two face centres opposite to
each other.
Body
centered
:
Points at the eight
corners and at the
body centre
Face centered: Points at
the eight corners and at
the six face centres.
c

KNO 3 , K2 SO 4 ,
PbCO 3 , BaSO 4 ,
rhombic sulphur,
MgSO 4 . 7 H 2O etc.


b
a
Rhombohedral
Simple : Points at the eight corners of the unit cell
NaNO 3 , CaSO 4 ,
calcite, quartz,
As, Sb , Bi etc.
or Trigonal
ab c,
      90 
Hexagonal
ab c ,
Simple : Points at the twelve corners of the unit cell
out lined by thick line.
or Points at the twelve corners of the hexagonal prism
and at the centres of the two hexagonal faces.
ZnO , PbS , CdS ,
HgS , graphite, ice,
Mg, Zn, Cd etc.
Simple : Points at the eight corners of the unit cell
End centered : Point at the eight corners and at two
face centres opposite to the each other.
Na 2 SO 4 .10 H 2 O,
    90 o
  120 o
Monoclinic
ab c,
    90 o ,   90 o
Triclinic
Simple : Points at the eight corners of the unit cell.
CaSO 4 .5 H 2O,
ab c,
      90
Na 2 B4 O7 .10 H 2 O,
CaSO 4 .2 H 2 O,
monoclinic sulphur etc.
K 2 Cr2 O7 , H 3 BO3
o
etc.
Analysis of cubic system
(1) Number of atoms in per unit cell
The total number of atoms contained in the unit cell for a simple
cubic called the unit cell content.
n f ni
n

The simplest relation can determine for it is, c 
8
2
1
Where n c  Number of atoms at the corners of the cube=8
n f  Number of atoms at six faces of the cube = 6
nc
nf
ni
Simple cubic (sc)
8
0
0
1
body centered cubic (bcc)
8
0
1
2
Face centered cubic (fcc)
8
6
0
4
(3) Density of the unit cell ( ) : It is defined as the ratio of mass
per unit cell to the total volume of unit cell.

ZM
a3 N
0
Where Z = Number of particles per unit cell
M = Atomic mass or molecular mass
n i  Number of atoms inside the cube = 1
Cubic unit cell
(2) Co-ordination number (C.N.) : It is defined as the number of
nearest neighbours or touching particles with other particle present in a
crystal is called its co-ordination number. It depends upon structure of the
crystal.
For simple cubic system C.N. = 6.
For body centred cubic system C.N. = 8
For face centred cubic system C.N. = 12.
Total atom in
per unit cell
N 0  Avogadro number (6.023  10 23 mol 1 )
a  Edge length of the unit cell= a pm  a  10 10 cm
a 3  volume of the unit cell
i.e.  
ZM
g / cm 3
a  N 0  10 30
3
Solid state
1 1
The density of the substance is same as the density of the unit cell.
(4) Packing fraction (P.F.) : It is defined as ratio of the volume of the
unit cell that is occupied by spheres of the unit cell to the total volume of
the unit cell.
Let radius of the atom in the packing = r
Edge length of the cube = a
4
Fig. 5.6. Square close packing
Volume of the atom (spherical)  
(ii) Hexagonal close packing : In which the spheres in every second
row are seated in the depression between the spheres of first row. The
spheres in the third row are vertically aligned with spheres in first row. The
similar pattern is noticed throughout the crystal structure. In this
arrangement each sphere is in contact with six other spheres.
4 3
r
3
4 3
r Z
Packing density 
 3 3
V
a
Z
r related to
a
Simple cubic
r
Face-centred
cubic
r
4 a
 
3 2
a
2 2
Bodycentred
cubic
r
11
Volume of the
atom ()
a
2
3a
4
 
4
3

Packing density
6
 3a 

 4 


3
3
2
% of void
3
6

3
 a 


2 2 
4
3
2
3
Volume of the cube V = a 3
Structure
201
 0 .52
100–52 =
48%
2
 0 .74
6
100 – 74
= 26%
3
 0 .68
8
100 – 68
= 32%
X-ray study of crystal structure
Study of internal structure of crystal can be done with the help of Xrays. The distance of the constituent particles can be determined from
diffraction value by Bragg’s equation.
n  2d sin
5
Fig. 5.7. Hexagonal close packing
(2) Close packing in three dimensions : In order to develop three
dimensional close packing, let us retain the hexagonal close packing in the
first layer. For close packing, each spheres in the second layer rests in the
hollow at the centre of three touching spheres in the layer as shown in
figure. The spheres in the first layer are shown by solid lines while those in
second layer are shown by broken lines. It may be noted that only half of
the triangular voids in the first layer are occupied by spheres in the second
layer (i.e., either b or c). The unoccupied hollows or voids in the first layer
are indicated by (c) in figure.
n

 2 sin
aa
a
a
where,  = Wave length of X-rays, n = order of reflection,
  Angle of reflection, d = Distance between two parallel surfaces
The above equation is known as Bragg’s equation or Bragg’s law.
The reflection corresponding to n = 1 (for a given family of planes) is called
first order reflection; the reflection corresponding to n = 2 is the second
order reflection and so on. Thus by measuring n (the order of reflection of
the X-rays) and the incidence angle , we can know d/.
d
4
b
b
a
a
b
a
a
a
a
a
c
c
c
a
a
a
a
Fig. 5.8. Close packing in three dimensions
There are two alternative ways in which species in third layer can be
arranged over the second layer,
From this, d can be calculated if  is known and vice versa. In X-ray
reflections, n is generally set as equal to 1. Thus Bragg’s equation may
alternatively be written as
  2 d sin = 2 d sin
hkl
Where d denotes the perpendicular distance between adjacent planes
with the indices hkl.
hkl
Close packing in crystalline solids
In the formation of crystals, the constituent particles (atoms, ions or
molecules) get closely packed together. The closely packed arrangement is
that in which maximum available space is occupied. This corresponds to a
state of maximum density. The closer the packing, the greater is the stability
of the packed system.
(1) Close packing in two dimensions : The two possible arrangement
of close packing in two dimensions.
(i) Square close packing : In which the spheres in the adjacent row
lie just one over the other and show a horizontal as well as vertical
alignment and form square. In this arrangement each sphere is in contact
with four spheres.
(i) Hexagonal close packing : The third layer lies vertically above the
first and the spheres in third layer rest in one set of hollows on the top of
the second layer. This arrangement is called ABAB …. type and 74% of the
available space is occupied by spheres.
This arrangement is found in
Be, Mg, Zn, Cd, Sc, Y, Ti, Zr, Tc, Ru.
A
A
A
B
B
A
B
B
A
A
A
Fig. 5.9. Hexagonal close packing (hcp) in three dimensions
(ii) Cubic close packing : The third layer is different from the first
and the spheres in the third layer lie on the other set of hollows marked ‘C’
202 Solid State
in the first layer. This arrangement is called ABCABC….. type and in this also
74% of the available space is occupied by spheres. The cubic close packing
has face centred cubic (fcc) unit cell. This arrangement is found in Cu, Ag,
(2) Tetrahedral void : A tetrahedral void is developed when triangular
voids (made by three spheres in one layer touching each other) have contact
with one sphere either in the upper layer or in the lower layer.
Au, Ni, Pt, Pd, Co, Rh, Ca, Sr.
B
B
A
The number of tetrahedral
Fig. 5.13.
voids
Tetrahedral
is double
void
the number of spheres in
the crystal structure.
C
A
B
r
 0 .225
R
A
C
where, r is the radius of the tetrahedral void or atom
occupying tetrahedral void.
R is the radius of spheres forming tetrahedral void.
C
Fig. 5.10. Cubic close packing (ccp or fcc) in three dimensions
(iii) Body centred cubic : This arrangement of spheres (or atoms) is
not exactly close packed. This structure can be obtained if spheres in the
first layer (A) of close packing are slightly opened up. As a result none of
these spheres are in contact with each other. The second layer of spheres
(B) can be placed on top of the first layer so that each sphere of the second
layer is in contact with four spheres of the layer below it. Successive
building of the third will be exactly like the first layer. If this pattern of
building layers is repeated infinitely we get an arrangement as shown in
figure. This arrangement is found in Li, Na, K, Rb, Ba, Cs, V, Nb, Cr, Mo, Fe.
A
A
B
A
B
A
A
A
A
A
(3) Octahedral void : This type of void is surrounded by six closely
packed spheres, i.e. it is formed by six
spheres.
The number of octahedral voids is
equal to the number of spheres.
r
 0 .414
R
B
r
 0 .732
R
B
A
A
A
A
A
B
A
B
A
Fig. 5.14. Octahedral void
(4) Cubic void : This type of void
is formed between
8 closely packed
spheres which occupy all the eight corner of
cube.
B
A
A
A
A
A
Fig. 5.15. Cubic void
The decreasing order of the size of the various voids is,
Cubic > Octahedral > Tetrahedral > Trigonal
Fig. 5.11. Body centred cubic (bcc) close packing in three dimensions
Table : 5.3 Comparison of hcp, ccp and bcc
Property
Arrangement of
packing
Type
of
packing
Available space
occupied
Coordination
number
Malleability and
ductility
hcp
ccp
bcc
Close packed
Close packed
AB AB A...
ABC ABC A...
Not
close
packed
AB AB A...
74%
74%
68%
12
12
8
Less malleable,
hard, brittle
Malleable
ductile
Ionic radii and Radius ratio
(1) Ionic radii : X-ray diffraction or electron diffraction techniques
provides the necessary information regarding unit cell. From the dimensions
of the unit cell, it is possible to calculate ionic radii.
Let, cube of edge length 'a' having cations and anions say NaCl
structure.
Then, rc  ra  a / 2
where rc and ra are radius
of cation and anion.
Radius of Cl 
and
(a / 2)2  (a / 2)2
a


2
4
Interstitial sites in close packing
Even in the close packing of spheres, there is left some empty space
between the spheres. This empty space in the crystal lattice is called site or
void or hole. Voids are of following types,
(1) Trigonal void : This site is formed when three spheres lie at the
vertices of an equilateral triangle. Size of the trigonal site is given by the
following relation,
Triogonal
void
r  0.155 R
r = Radius of the spherical
trigonal void
R = Radius of closely packed spheres
Fig. 5.12
For bcc lattice say CsCl.
rc  ra 
Cl–
a/2
90°
Cl–
Na+
a/2
Fig. 5.16. Radii of chloride ion
3a
2
(2) Radius ratio : Ionic compounds occur in crystalline forms. Ionic
compounds are made of cations and anions. These ions are arranged in
three dimensional array to form an aggregate of the type (A B ) . Since, the
Coulombic forces are non-directional, hence the structures of such crystals
+
–
n
Solid state
203
are mainly governed by the ratio of the radius of cation (r ) to that of
0.155 – 0.225
3
Planar triangle
anion (r ). The ratio r to r (r / r ) is called as radius ratio.
0.225 – 0.414
4
Tetrahedral
0.414 – 0.732
6
Octahedral
0.732 – 0.999 or 1
8
Body-centered cubic
Radius ratio 
r
r
Effect of temperature and Pressure on C.N.
r+/r– < 0.414
r+/r– = 0.414
r+/r– > 0.732
On applying high pressure NaCl structure having 6 : 6 coordination changes to CsCl structure having 8 : 8 co-ordination. Thus,
increase in pressure increases the co-ordination number.
Coordination number
decreases from 6 to 4
Coordination number
increases from 6 to 8
Similarly, CsCl structure on heating to about 760 K, changes to
NaCl structure. In other words, increase of temperature decreases the coordination number.
Fig. 5.17. Effect of radius ratio on co-ordination number
Table : 5.4 Limiting Radius ratios and Structure
NaCl
(6 : 6 )
Limiting radius ratio (r )/(r )
C.N.
Shape
< 0.155
2
Linear
+
–
Pressure
Temp
CsCl
(8 : 8)
Structure of ionic crystals
Table : 5.5 Types of ionic crystal with description
Crystal structure
type
Type AB
Rock salt (NaCl)
type
Zinc blende (ZnS)
type
Brief description
Examples
Co-ordination
number
It has fcc arrangement in which Cl  ions
occupy the corners and face centres of a cube
Halides of Li, Na, K, Rb, AgF,
AgBr, NH Cl, NH Br, NH I etc.
Na   6
CuCl , CuBr, CuI, AgI, BeS
Zn 2   4
2
Fluorite (CaF ) type
2
4
4
while Na ions are present at the body and
edge of centres.
It has ccp arrangement in which S 2  ions
form fcc and each Zn 2  ion is surrounded
tetrahedrally by four S
Type AB
4

2
Number of formula
units per unit cell
4
Cl   6
4
S 2  4
ions and vice versa.
It has arrangement in which Ca 2  ions form
BaF2 , BaCl2 , SrF2
Ca 2   8
fcc with each Ca 2  ions surrounded by 8 F 
SrCl2 , CdF2 , PbF2
F  4
Here negative ions form the ccp arrangement
so that each positive ion is surrounded by 4
negative ions and each negative ion by 8
positive ions
Na 2 O
Na   4
It has the bcc arrangement with Cs  at the
CsCl, CsBr, CsI, CsCN ,
Cs   8
TlCl, TlBr, TlI and TlCN
Cl   8

4
ions and each F ions by 4Ca ions.
Antifluorite type
Caesium chloride
(CsCl) type
body centre and Cl
cube or vice versa.
2+

ions at the corners of a
4
O2  8
Defects or Imperfections in solids
Any deviation from the perfectly ordered arrangement constitutes a
defect or imperfection. These defects sometimes called thermodynamic
defects because the number of these defects depend on the temperature.
(1) Electronic imperfections : Generally, electrons are present in fully
occupied lowest energy states. But at high temperatures, some of the
electrons may occupy higher energy state depending upon the temperature.
For example, in the crystals of pure Si or Ge some electrons are released
thermally from the covalent bonds at temperature above 0 K. these
electrons are free to move in the crystal and are responsible for electrical
conductivity. This type of conduction is known as intrinsic conduction. The
1
electron deficient bond formed by the release of an electron is called a hole.
In the presence of electric field the positive holes move in a direction
opposite to that of the electrons and conduct electricity. The electrons and
holes in solids gives rise to electronic imperfections.
(2) Atomic imperfections/point defects : When deviations exist from
the regular or periodic arrangement around an atom or a group of atoms in
a crystalline substance, the defects are called point defects. Point defect in a
crystal may be classified into following three types.
(i) Stoichiometric defects : The compounds in which the number of
positive and negative ions are exactly in the ratios indicated by their
chemical formulae are called stoichiometric compounds. The defects do not
204 Solid State
disturb the stoichiometry (the ratio of numbers of positive and negative
ions) are called stoichiometric defects. These are of following types,
(a) Interstitial defect : This type of defect is caused due to the
presence of ions in the normally vacant interstitial sites in the crystals.
(b) Schottky defect : This type of defect when equal number of
cations and anions are missing from their lattice sites so that the electrical
neutrality is maintained. This type of defect occurs in highly ionic
compounds which have high co-ordination number and cations and anions
of similar sizes. e.g., NaCl, KCl, CsCl and KBr etc.
(c) Frenkel defect : This type of defect arises when an ion is missing
from its lattice site and occupies an interstitial position. The crystal as a
whole remains electrically neutral because the number of anions and cations
remain same. Since cations are usually smaller than anions, they occupy
interstitial sites. This type of defect occurs in the compounds which have
low co-ordination number and cations and anions of different sizes. e.g.,
ZnS, AgCl and AgI etc. Frenkel defect are not found in pure alkali metal
halides because the cations due to larger size cannot get into the interstitial
sites. In AgBr both Schottky and Frenkel defects occur simultaneously.
A+
B–
B+
A–
A+
B–
A+
B–
A+
B–
A+
B–
A+
B–
A+
B–
A+
B–
A+
B–
A+
Fig. 5.18. Schottky defect
B–
Fig. 5.19. Frenkel defect
Consequences of Schottky and Frenkel defects
Presence of large number of Schottky defect lowers the density of
the crystal. When Frenkel defect alone is present, there is no decrease in
density. The closeness of the charge brought about by Frenkel defect tends
to increase the dielectric constant of the crystal. Compounds having such
defect conduct electricity to a small extent. When electric field is applied, an
ion moves from its lattice site to occupy a hole, it creates a new hole. In this
way, a hole moves from one end to the other. Thus, it conducts electricity
across the crystal. Due to the presence of holes, stability (or the lattice
energy) of the crystal decreases.
(ii) Non-stoichiometric defects : The defects which disturb the
stoichiometry of the compounds are called non-stoichiometry defects. These
defects are either due to the presence of excess metal ions or deficiency of
metal ions.
(a) Metal excess defects due to anion vacancies : A compound may
have excess metal anion if a negative ion is absent from its lattice site,
leaving a ‘hole’, which is occupied by electron to maintain electrical
neutrality. This type of defects are found in crystals which are likely to
possess Schottky defects. Anion vacancies in alkali metal halides are reduced
by heating the alkali metal halides crystals in an atmosphere of alkali metal
vapours. The ‘holes’ occupy by electrons are called F-centres (or colour
centres).
(b) Metal excess defects due to interstitial cations : Another way in
which metal excess defects may occur is, if an extra positive ion is present
in an interstitial site. Electrical neutrality is maintained by the presence of
an electron in the interstitial site. This type of defects are exhibit by the
crystals which are likely to exhibit Frenkel defects e.g., when ZnO is heated,
it loses oxygen reversibly. The excess is accommodated in interstitial sites,
with electrons trapped in the neighborhood. The yellow colour and the
electrical conductivity of the non-stoichiometric ZnO is due to these
trapped electrons.
A+
B–
A+
B–
A+
B–
A+
B–
A+
B–
A+
B–
A+
B–
A+
B–
A+
e–
A+
B–
A+
B–
B–
A+
A+
Fig. 5.21. Metal excess defect
due to anion vacancy
Fig. 5.20. Metal excess defect
due to extra cation
Consequences of Metal excess defects
The crystals with metal excess defects are generally coloured due to
the presence of free electrons in them.
The crystals with metal excess defects conduct electricity due to the
presence of free electrons and are semiconductors. As the electric transport
is mainly by “excess” electrons, these are called n-type (n for negative)
semiconductor.
The crystals with metal excess defects are generally paramagnetic
due to the presence of unpaired electrons at lattice sites.
When the crystal is irradiated with white light, the trapped electron
absorbs some component of white light for excitation from ground state to
the excited state. This gives rise to colour. Such points are called F-centres.
(German word Farbe which means colour) such excess ions are
accompanied by positive ion vacancies. These vacancies serve to trap holes
in the same way as the anion vacancies trapped electrons. The colour
centres thus produced are called V-centres.
(c) Metal deficiency defect by cation vacancy : In this a cation is
missing from its lattice site. To maintain electrical neutrality, one of the
nearest metal ion acquires two positive charge. This type of defect occurs in
compounds where metal can exhibit variable valency. e.g., Transition metal
compounds like NiO, FeO, FeS etc.
(d) By having extra anion occupying interstitial site : In this, an extra
anion is present in the interstitial position. The extra negative charge is
balanced by one extra positive charge on the adjacent metal ion. Since
anions are usually larger it could not occupy an interstitial site. Thus, this
structure has only a theoretical possibility. No example is known so far.
A+
B–
B–
A+
B–
B–
A+
A+
B–
A2
B–
B–
A+
B–
A+
+
Cation vacancy
Metal having higher
charge
Fig. 5.22
Consequences of metal deficiency defects
Due to the movement of electron, an ion A changes to A ions. Thus,
the movement of an electron from A ion is an apparent of positive hole and
the substances are called p-type semiconductor
(iii) Impurity defect : These defects arise when foreign atoms are
present at the lattice site (in place of host atoms) or at the vacant
interstitial sites. In the former case, we get substitutional solid solutions
while in the latter case, we get interstitial solid solution. The formation of
the former depends upon the electronic structure of the impurity while that
of the later on the size of the impurity.
+
+2
+
Properties of solids
Some of the properties of solids which are useful in electronic and
magnetic devices such as, transistor, computers, and telephones etc., are
summarised below,
Solid state
(1) Electrical properties : Solids are classified into following classes
depending on the extent of conducting nature.
Examples, Nb 3 Ge alloy (Before 1986)
La1.25 Ba0.15 CuO 4 (1986)
(i) Conductors : The solids which allow the electric current to pass
through them are called conductors. These are further of two types; Metallic
conductors and electrolytic conductors. The electrical conductivity of these
solids is high in the range 10 4  10 6 ohm 1 cm 1 . Their conductance
decrease with increase in temperature.
(ii) Insulators : The solids which do not allow the current to pass
through them are called insulators. e.g., rubber, wood and plastic etc. the
electrical conductivity of these solids is very low i.e.,
10
12
 10
22
1
ohm cm
1
.
(iii) Semiconductors : The solids whose electrical conductivity lies
between those of conductors and insulators are called semiconductors. The
conductivity of these solid is due to the presence of impurities. e.g. Silicon
and Germanium. Their conductance increase with increase in temperature.
The electrical conductivity of these solids is increased by adding impurity.
This is called Doping. When silicon is doped with P (or As, group 5
elements), we get n-type semiconductor. This is because P has five valence
electrons. It forms 4 covalent bonds with silicon and the fifth electron
remains free and is loosely bound. This give rise to n-type semiconductor
because current is carried by electrons when silicon is doped with Ga (or in
In/Al, group 3 elements) we get p-type semiconductors.
th
205
YBa 2 Cu 3 O7 (1987)
Following are the important applications of superconductivity,
(a) Electronics,
(b) Building supermagnets,
(c) Aviation transportation,
(d) Power transmission
“The temperature at which a material enters the superconducting
state is called the superconducting transition temperature, (Tc ) ”.
Superconductivity was also observed in lead (Pb) at 7.2 K and in tin (Sn) at
3.7K. The phenomenon of superconductivity in other materials such as
polymers and organic crystals. Examples are
(SN) , polythiazyl, the subscript x indicates a large number of
variable size.
x
(TMTSF) PF , where TMTSF is tetra methyl tetra selena fulvalene.
2
6
(2) Magnetic properties : Based on the behavior of substances when
placed in the magnetic field, there are classified into five classes.
rd
Superconductivity : When any material loses its resistance for
electric current, then it is called superconductor, Kammerlingh Onnes (1913)
observed this phenomenon at 4K in mercury. The materials offering no
resistance to the flow of current at very low temperature (2-5K) are called
superconducting materials and phenomenon is called superconductivity.
Table : 5.6 Magnetic properties of solids
Properties
Diamagnetic
Paramagnetic
Ferromagnetic
Description
Feebly repelled by the magnetic fields. Non-metallic
elements (excepts O2, S) inert gases and species with
paired electrons are diamagnetic
Attracted by the magnetic field due to the presence
of permanent magnetic dipoles (unpaired electrons).
In magnetic field, these tend to orient themselves
parallel to the direction of the field and thus, produce
magnetism in the substances.
Alignment of Magnetic Dipoles
All paired electrons
Permanent magnetism even in the absence of
magnetic field, Above a temperature called Curie
temperature, there is no ferromagnetism.
Dipoles are aligned in the same
direction
Antiferromagnetic
This arises when the dipole alignment is zero due to
equal and opposite alignment.
Ferrimagnetic
This arises when there is net dipole moment
(3) Dielectric properties : A dielectric substance is that which does
not allow the electricity to pass through them but on applying the electric
field, induced charges are produced on its faces. In an insulator, the
electrons are strongly held by the individual atoms. When an electric field is
applied polarization takes place because nuclei are attracted to one side and
the electron cloud to the other side. As a result, dipoles are created. Such
type of crystals shows the following properties,
(i) Piezoelectricity : In some of the crystals, the dipoles may align
themselves is an ordered way so as to give some net dipole moment. When
mechanical stress is applied in such crystals so as to deform them,
electricity is produced due to the displacement of ions. The electricity thus
produced is called piezoelectricity and the crystals are called piezoelectric
Examples
TiO2, V2O5, NaCl, C6H6
Applications
Insulator
(benzene)
O2 , Cu 2  , Fe3  , TiO,
Electronic
appliances
Ti2O3 , VO, VO2 , CuO
At least one unpaired electron
Fe, Ni, Co, CrO2
CrO2 is used in
audio and video
tapes
MnO, MnO2, Mn2O, FeO,
Fe2O3; NiO, Cr2O3, CoO,
Co3O4,
Fe3O4, ferrites
–
–
crystals. Examples, Quartz, Rochelle’s salt ( sod. pot. tartarate). Piezoelectric
crystals act as mechanical-electric transducer. These crystals are used as
pick-ups in record players where they produce electric signals by application
of pressure.
(ii) Pyroelectricity : On heating, some polar crystals produce a small
electric current. The electricity thus produced is called pyroelectircity.
(iii) Ferroelectricity : In some of the piezoelectric crystals, a
permanent alignment of the dipoles is always there even in the absence of
the electric field, however, on applying field the direction of polarization
changes. This phenomenon is called ferroelectricity and the crystals as
206 Solid State
ferroelectric crystal. Example, Potassium hydrogen phosphate (KH 2 PO4 ) ,
5.
Barium titanate (BaTiO3 ) .
(iv) Antiferroelectricity : In some crystals, the dipoles in
alternate polyhedra point up and down so that the crystals does not
possess any net dipole moment. Such crystals are said to be
antiferroelectric. Example, Lead zirconate (PbZrO3 ) . Ferroelectrics
are used in the preparation of small sized capacitors of high
capacitance. Pyroelectric infrared detectors are based on such
substances. These can be used in transistors, telephone, computer
etc.
6.
7.
Value of heat of fusion of NaCl is
(a) Very low
(b) Very high
(c) Not very low and not very high
(d) None of the above
Piezoelectric crystals are used in
(a) TV
(b) Radio
(c) Record player
(d) Freeze
Which of the following is true for diamond
[AFMC 1997]
(a) Diamond is a good conductor of electricity
(b) Diamond is soft
(c) Diamond is a bad conductor of heat
(d) Diamond is made up of C, H and O
8.
 The reverse of crystallization is the melting of the solid.
 The slower the rate of formation of crystal, the bigger is the crystal.
 The hardness of metals increases with the number of electrons
available for metallic bonding. Thus Mg is harder than sodium.
 Isomorphism is applied to those substances which are not only
similar in their crystalline form, but also possess an equal number
of atoms united in the similar manner. The existence of a
substance in more than one crystalline form is known as
polymorphism.
Properties and Types of solid
1.
The three states of matter are solid, liquid and gas. Which of the
following statement is/are true about them
[AIIMS 1991]
2.
3.
(a) Gases and liquids have viscosity as a common property
(b) The molecules in all the three states possess random
translational motion
(c) Gases cannot be converted into solids without passing through
the liquid phase
(d) Solids and liquids have vapour pressure as a common property
A pure crystalline substance, on being heated gradually, first forms a
turbid looking liquid and then the turbidity completely disappears.
This behaviour is the characteristic of substances forming [BHU 2000]
(a) Isomeric crystals
(b) Liquid crystals
(c) Isomorphous crystals
(d) Allotropic crystals
Which of the following is ferroelectric compound
[AFMC 1997]
4.
(a)
BaTiO3
(b)
K 4 [Fe(CN )6 ]
(c)
Pb2 O 3
(d)
PbZrO3
Solid CO 2 is an example of
(a) Molecular crystal
(c) Covalent crystal
(b) Ionic crystal
(d) Metallic crystal
NaCl is an example of
(a) Covalent solid
(b) Ionic solid
(c) Molecular solid
(d) Metallic solid
208 Solid state
9.
10.
11.
Amorphous substances show
(A) Short and long range order
(B) Short range order
(C) Long range order
(D) Have no sharp M.P.
(a) A and C are correct
(b) B and C are correct
(c) C and D are correct
(d) B and D are correct
The characteristic features of solids are
[AMU 1994]
(a) Definite shape
(b) Definite size
(c) Definite shape and size
(d) Definite shape, size and rigidity
Which one of the following is a good conductor of electricity
22.
23.
24.
25.
[MP PMT 1994; AFMC 2002]
13.
(a) Diamond
(b) Graphite
(c) Silicon
(d) Amorphous carbon
A crystalline solid
[Kerala CET (Med.) 2003]
(a) Changes abruptly from solid to liquid when heated
(b) Has no definite melting point
(c) Undergoes deformation of its geometry easily
(d) Has an irregular 3-dimensional arrangements
(e) Softens slowly
Diamond is an example of
14.
(a) Solid with hydrogen bonding
(b) Electrovalent solid
(c) Covalent solid
(d) Glass
The solid NaCl is a bad conductor of electricity since
12.
[MP PET/PMT 1998; CET Pune 1998]
[AIIMS 1980]
15.
(a) In solid NaCl there are no ions
(b) Solid NaCl is covalent
(c) In solid NaCl there is no velocity of ions
(d) In solid NaCl there are no electrons
The existence of a substance in more than one solid modifications is
known as or Any compound having more than two crystal structures
is called
26.
27.
28.
16.
17.
18.
19.
20.
21.
(b) Ionic solids
(a)
MnO2
(b) TiO2
(c)
VO2
(d) CrO 2
In graphite, carbon atoms are joined together due to
30.
(a) Ionic bonding
(b) Vander Waal’s forces
(c) Metallic bonding
(d) Covalent bonding
Which of the following is not correct for ionic crystals
31.
(a) They possess high melting point and boiling point
(b) All are electrolyte
(c) Exhibit the property of isomorphism
(d) Exhibit directional properties of the bond
Which of the following is a molecular crystal
[AFMC 2002]
[Orissa JEE 2002]
32.
33.
(a) SiC
(b) NaCl
(c) Graphite
(d) Ice
Quartz is a crystalline variety of
[Pb. PMT 2000]
(a) Silica
(b) Sodium silicate
(c) Silicon carbide
(d) Silicon
Which type of solid crystals will conduct heat and electricity
[RPET 2000]
34.
35.
36.
[Kerala CET (Med.) 2002]
(a) Covalent solids
The lustre of a metal is due to
[AFMC 1998]
(a) Its high density
(b) Its high polishing
(c) Its chemical inertness
(d) Presence of free electrons
A crystalline solid have
[DCE 2001]
(a) Long range order
(b) Short range order
(c) Disordered arrangement
(d) None of these
Crystalline solids are
[Pb. PMT 1999]
(a) Glass
(b) Rubber
(c) Plastic
(d) Sugar
Davy and Faraday proved that
[Kerala CET (Med.) 2002]
(a) Diamond is a form of carbon
(b) The bond lengths of carbon containing compounds are always
equal
(c) The strength of graphite is minimum compared to platinum
(d) Graphite is very hard
Which one of the following metal oxides is antiferromagnetic in
nature
[MP PET 2002]
29.
[MP PMT 1993; MP PET 1999]
(a) Polymorphism
(b) Isomorphism
(c) Allotropy
(d) Enantiomorphism
Which is not a property of solids
[MP PET 1995]
(a) Solids are always crystalline in nature
(b) Solids have high density and low compressibility
(c) The diffusion of solids is very slow
(d) Solids have definite volume
Which solid will have the weakest intermolecular forces
(a) Ice
(b) Phosphorus
(c) Naphthalene
(d) Sodium fluoride
Dulong and Petit’s law is valid only for
[KCET 2004]
(a) Metals
(b) Non-metals
(c) Gaseous elements
(d) Solid elements
Which of the following is an example of metallic crystal solid
(a) C
(b) Si
(c) W
(d) AgCl
Under which category iodine crystals are placed among the following
(a) Ionic crystal
(b) Metallic crystal
(c) Molecular crystal
(d) Covalent crystal
Among solids the highest melting point is established by
(c) Pseudo solids
(d) Molecular solids
To get a n- type semiconductor, the impurity to be added to silicon
should have which of the following number of valence electrons [KCET (Engg.)
(a) 1
(b) 2
(c) 3
(d) 5
Which of the following is non-crystalline solid
(a) CsCl
(b) NaCl
(c) CaF2
(d) Glass
37.
(a) Ionic
(b) Covalent
(c) Metallic
(d) Molecular
Which of the following is an example of covalent crystal solid
(a) Si
(b) NaF
(c) Al
(d) Ar
Which of the following is an example of ionic crystal solid
(a) Diamond
(b) LiF
(c) Li
(d) Silicon
Which one is an example of amorphous solid
(a) Glass
(b) Salt
(c) Cesium chloride
(d) Calcium fluoride
Silicon is
[MHCET 2004]
Solid state
38.
39.
40.
(a) Semiconductor
(b) Insulator
(c) Conductor
(d) None of these
Which of the following statements about amorphous solids is
incorrect
[KCET 2004]
(a) They melt over a range of temperature
(b) They are anisotropic
(c) There is no orderly arrangement of particles
(d) They are rigid and incompressible
The ability of a given substance to assume two or more crystalline
structure is called
[DCE 2004]
(a) Amorphism
(b) Isomorphism
(c) Polymorphism
(d) Isomerism
Glass is
(a) Supercooled liquid
(b) Crystalline solid
(c) Amorphous solid
(d) Liquid crystal
8.
9.
10.
11.
Crystallography and Lattice
1.
7.
The correct statement in the following is
(a) The ionic crystal of AgBr has Schottky defect
[MP PET 1997]
(b) The unit cell having crystal parameters,
a  b  c,
12.

is

(d) The coordination number of Na  ion in NaCl is 4
Which of the following is correct
[DPMT 1997]
(a)
Crystal
system
Cubic
13.
Axial distance
Axial angles
Examples
a b=c
==
90
Cu, KCl
Monoclinic
===
90
PbCrO ,
PbCrO
CaCO , HgS
a b=c
o
(c)
(d)
Rhombohedra
l
a=b=c
Triclinic
a=b=c
==
90
o
(b) 0.414  0.732
(c) 0.225  0.414
(d) 0.155  0.225
What type of lattice is found in potassium chloride crystal
14.
15.
(a) Face centred cubic
(b) Body centred cubic
(c) Simple cubic
(d) Simple tetragonal
The three dimensional graph of lattice points which sets the pattern
for the whole lattice is called
(a) Space lattice
(b) Simple lattice
(c) Unit cell
(d) Crystal lattice
Crystals can be classified into ...... basic crystal habits
[MP PMT 1994]
2
4
3
16.
o
=
90
0.732  1.000
[MP PMT 1996]
o
(b)
(b) Has 3 Na  ions
(d) Is electrically charged
r
For tetrahedral coordination number, the radius ratio c
is[KCET 2000]
ra 
(a)
high
2.
How many space lattices are obtainable from the different crystal
systems
[MP PMT 1996; MP PET/PMT 1998]
(a) 7
(b) 14
(c) 32
(d) 230
Example of unit cell with crystallographic dimensions
[AFMC 1998]
a  b  c,     90 o ,   90 o is
(a) Calcite
(b) Graphite
(c) Rhombic sulphur
(d) Monoclinic sulphur
In a face-centered cubic lattice, a unit cell is shared equally by how
many unit cells
[CBSE PMT 2005]
(a) 8
(b) 4
(c) 2
(d) 6
The maximum radius of sphere that can be fitted in the octahedral
hole of cubical closed packing of sphere of radius r is
(a) 0.732 r
(b) 0.414 r
(c) 0.225 r
(d) 0.155 r
The unit cell of a NaCl lattice
(a) Is body centred cube
(c) Has 4 NaCl units
    90 o ,   120 o is hexagonal
(c) In ionic compounds having Frenkel defect the ratio
K Cr O ,
CuSO .
5H O
2
2
7
4
17.
(a) 3
(b) 7
(c) 14
(d) 4
How many molecules are there in the unit cell of sodium chloride [MP PMT 1996
(a) 2
(b) 4
(c) 6
(d) 8
In a crystal, the atoms are located at the position of
2
3.
Tetragonal crystal system has the following unit cell dimensions[MP PMT 1993]
(a)
a  b  c and       90 o
18.
(b) a  b  c and       90 o
(c)
a  b  c and       90 o
(d) a  b  c and     90 o ,   120 o
4.
5.
6.
Rhombic sulphur has the following structure
(a) Open chain
(b) Tetrahedral
(c) Puckered 6-membered ring
(d) Puckered 8-membered ring
Space lattice of CaF2 is
19.
0.000  0.225
[AMU 1985]
(a) Maximum P.E.
(b) Minimum P.E.
(c) Zero P.E.
(d) Infinite P.E.
The total number of lattice arrangements in different crystal systems
is
[KCET (Engg.) 2001]
(a) 3
(b) 7
(c) 8
(d) 14
Monoclinic crystal has dimension
[DCE 2000]
(a) a  b  c,     90,   90
(b) a  b  c,       90
(c)
a  b  c,       90
(d) a  b  c,       90
[MP PMT 1993]
(a) Face centred cubic
(b) Body centred cubic
(c) Simple cubic
(d) Hexagonal closed packing
For cubic coordination the value of radius ratio is
(a) 0.732  1.000
(b) 0.225  0.414
(c)
209
(d) 0.414  0.732
20.
The low solubility of BaSO 4 in water can be attributed to
21.
(a) High lattice energy
(c) Low lattice energy
Bravais lattices are of
(a) 8 types
(c) 14 types
[CBSE PMT 1991]
(b) Dissociation energy
(d) Ionic bond
[MP PMT 1997]
(b) 12 types
(d) 9 types
210 Solid state
22.
23.
24.
25.
26.
27.
28.
The structure of TlCl is similar to CsCl. What would be the
radius ratio in TlCl
(a) 0.155  0.225
(b) 0.225  0.414
(c) 0.414  0.732
(d) 0.732  1.000
Structure similar to zinc blende is found in
(a) AgCl
(b) NaCl
(c) 14 and 9
7.
8.
(c) CuCl
(d) TlCl
The structure of Na 2 O crystal is
(a) CsCl type
(b) NaCl type
(c) ZnS type
(d) Antifluorite
Structure of ZnS is
(a) Body centred cubic
(b) Face centred cubic
(c) Simple cube
(d) Fluorite structure
The crystal system of a compound with unit cell dimensions
a  0.387 , b  0.387 and c  0.504nm and     90 o
and   120 o is
(a) Cubic
(c) Orthorhombic
The number of tetrahedral voids
cubic lattice of similar atoms is
(a) 4
(c) 8
[AIIMS 2004]
9.
10.
(b) Hexagonal
(d) Rhombohedral
in the unit cell of a face centered
11.
An fcc unit cell of aluminium contains the equivalent of how many
atoms
[DCE 2003]
(a) 1
(b) 2
(c) 3
(d) 4
12.
13.
If ‘Z’ is the number of atoms in the unit cell that represents the
closest packing sequence    A B C A B C   , the number
of tetrahedral voids in the unit cell is equal to
14.
[AIIMS 2005]
2.
3.
(a) Z
(b) 2 Z
(c) Z/2
(d) Z/4
The close packing represents ABC ABC...... order of
(a) Body centred cubic packing
(b) Face centred cubic packing
(c) Simple cubic packing
(d) Hexagonal cubic closed packing
The arrangement ABC ABC ABC ….. is referred as
15.
5.
6.
(a) Octahedral close packing
(b) Hexagonal close packing
(c) Tetragonal close packing
(d) Cubic close packing
The number of close neighbour in a body-centred cubic lattice of
identical sphere is
[MP PET 2001]
(a) 8
(b) 6
(c) 4
(d) 2
The number of equidistant oppositely charged ions in a sodium
chloride crystal is
[MP PET 2001]
(a) 8
(b) 6
(c) 4
(d) 2
Na and Mg crystallize in BCC and FCC type crystals respectively,
then the number of atoms of Na and Mg present in the unit cell
of their respective crystal is
[AIEEE 2002]
(a) 4 and 2
(b) 9 and 14
(a)
NaCl
(b)
Al2 O 3
(c)
CaF2
(d)
N 2O
Potassium crystallizes with a
[MP PET/PMT 1998]
(a) Face-centred cubic lattice
(b) Body-centred cubic lattice
(c) Simple cubic lattice
(d) Orthorhombic lattice
If the number of atoms per unit in a crystal is 2, the structure of
crystal is
(a) Octahedral
(b) Body centred cubic bcc
(c) Face centred cubic fcc
(d) simple cubic
The intermetallic compound LiAg crystallizes in cubic lattice in
which both lithium and silver have coordination number of eight.
The crystal class is
(a) Simple cube
(b) Body-centred cube
(c) Face-centred cube
(d) None of these
The number of octahedral sites per sphere in a fcc structure is [MP PMT 2000,
(a) 8
(b) 4
(c) 2
(d) 1
Hexagonal close packed arrangement of ions is described as
(a) ABC ABA
(b)
(c) ABABA
(d)
An example of a body cube is
(a) Sodium
(b)
(c) Zinc
(d)
An example of fluorite structure is
(a) NaF
(b)
ABC ABC
ABBAB
(c)
SiF4
AlCl3
(d)
[AIIMS 1996]
Magnesium
Copper
SrF2
In which of the following crystals alternate tetrahedral voids are
occupied?
[IIT 2005]
(a) NaCl
(b) ZnS
(c) CaF
(d) Na O
Which of the following contains rock salt structure
(a) SrF2
(b) MgO
2
16.
[MP PET 2001]
4.
[AIIMS 2002]
[MP PMT 1994]
Crystal packing
1.
An AB 2 type structure is found in
[CBSE PMT 1997]
[Kerala PMT 2004]
(b) 6
(d) 10
(d) 2 and 4
(c)
17.
18.
19.
Al 2 O 3
2
(d) All
In the fluorite structure, the coordination number of Ca 2  ion is
(a) 4
(b) 6
(c) 8
(d) 3
The ratio of close-packed atoms to tetrahedral holes in cubic close
packing is
[Pb. PMT 1998]
(a) 1 : 1
(b) 1 : 2
(c) 1 : 3
(d) 2 : 1
A solid is made of two elements X and Z . The atoms Z are in
CCP arrangement while the atom X occupy all the tetrahedral
sites. What is the formula of the compound
[UPSEAT 2004]
(a)
XZ
(b)
XZ2
(c)
X2Z
(d)
X 2 Z3
Solid state
20.
An ionic compound has a unit cell consisting of A ions at the
corners of a cube and B ions on the centres of the faces of the cube.
The empirical formula for this compound would be [CBSE PMT 2004; AIEEE 6.
2005]
(a) AB
(b) A 2 B
21.
The vacant space in the bcc unit cell is
(a) 32%
(b) 23%
(c) 26%
(d) None of these
The number of octahedral voids in a unit cell of a cubical closest
packed structure is
(a) 1
(b) 2
(c) 4
(d) 8
In the closest packed structure of a metallic lattice, the number of
nearest neighbours of a metallic atom is
(c) AB 3
22.
23.
(d)
A3 B
7.
24.
25.
[MH CET 2002]
(a)
26.
Na
(b)
Mg
8.
Na 
(b)
Mg 2 
(c)
Al 3 
(d)
Si 4 
(c)
a M
g cm  3
N  No
4.
5.
M  No
g cm  3
a N
Potassium fluoride has NaCl type structure. What is the distance
between K  and F  ions if cell edge is a cm
3
(a)
1 .86 Å
(b)
(c)
4.3 Å
(d) 7 .44 Å
3 .72 Å
In octahedral holes (voids)
(a) A simple triangular void surrounded by four spheres
(b) A bi-triangular void surrounded by four spheres
(c) A bi-triangular void surrounded by six spheres
(d) A bi-triangular void surrounded by eight spheres
Bragg's law is given by the equation
[MP PMT 1995, 2002]
(a)
n  2 sin
(b) n  2d sin
(c)
2n  d sin
(d) n

2

d
sin
2
The number of atoms in 100 g of an fcc crystal with density
(a)
12.
4  10 25
(b)
3  10 25
(c) 2  10 25
(d) 1  10 25
In the crystals of which of the following ionic compounds would you
expect maximum distance between centres of cations and anions[CBSE PMT 1998
(a)
LiF
(b) CsF
(c)
CsI
(d)
LiI
The number of unit cells in 58.5 g of NaCl is nearly
[MP PMT 2000, 01]
(a)
6  10
20
(b)
3  10
22
(c) 1.5  10
(d) 0.5  10 24
How many unit cells are present in a cube-shaped ideal crystal of
NaCl of mass 1.00 g [Atomic masses: Na  23, Cl  35.5] [AIEEE 2003]
23
13.
(a)
2.57  10 21 unit cells
(b) 5.14  10 21 unit cells
(c) 1.28  10 21 unit cells
(d) 1.71  10 21 unit cells
14.
In the Bragg’s equation
for diffraction of X- rays, n
(c) 4 a cm
(d) a / 4 cm
represents for
[MP PMT 2000]
(a) Quantum number
(b) An integer
An element occurring in the bcc structure has 12.08  10 23 unit
cells. The total number of atoms of the element in these cells will be[MP PET 1994](c) Avogadro’s numbers
(d) Moles
15.
In a face centred cubic cell, an atom at the face contributes to the
(a) 24.16  10 23
(b) 36.18  10 23
unit cell
(c) 6.04  10 23
(d) 12.08  10 23
[Karnataka (Engg./Med.) 2000; AFMC 2001]
If an atom is present in the centre of the cube, the participation of
that atom per unit cell is
(a) 1/4 part
(b) 1/8 part
1
(c)
1
part
(d) 1/2 part
(a)
(b) 1
4
16.
The interionic distance for cesium chloride crystal will be
1
1
[MP PET 2002]
(c)
(d)
2
8
a
(a) a
(b)
For an ionic crystal of the general formula AX and coordination
2
number 6, the value of radius ratio will be
[MP PMT 1993]
2a
3a
(c)
(d)
(a) Greater than 0.73
2
3
(b) In between 0.73 and 0.41
(a)
3.
(d)
CsBr crystal has bcc structure. It has an edge length of 4 .3 Å.
d  10 g / cm 3 and cell edge equal to 100 pm, is equal to[CBSE PMT 1994; K
The formula for determination of density of unit cell is
a3  N o
NM
(a)
(b)
g cm  3
g cm  3
NM
a3  N o
3
2.
10.
11.
Mathematical analysis of cubic system and
Bragg’s equation
1.
9.
(c) Al
(d) None of these
Which ion has the largest radius from the following ions
(a)
(c) In between 0.41 and 0.22
(d) Less than 0.22
The number of spheres contained (i) in one body centred cubic unit
cell and (ii) in one face centred cubic unit cell, is
(a) In (i) 2 and in (ii) 4
(b) In (i) 3 and in (ii) 2
(c) In (i) 4 and in (ii) 2
(d) In (i) 2 and in (ii) 3
The shortest interionic distance between Cs  and Br  ions is[IIT 1995]
[JIPMER 2002]
(a) Twelve
(b) Four
(c) Eight
(d) Six
In the rock salt structure, the number of formula units per unit cell
is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Hexagonal close packing is found in crystal lattice of
211
2a cm
(b) a / 2 cm
212 Solid state
17.
Sodium metal crystallizes as a body centred cubic lattice with the
cell edge 4.29 Å. What is the radius of sodium atom
[AIIMS 1999]
18.
19.
20.
21.
8
7
(a)
1.857  10 cm
(b)
(c)
3.817  10 8 cm
(d) 9.312  10 7 cm
Crystal structure and Coordination number
1.
2.371  10 cm
For an ionic crystal of the type AB, the value of (limiting) radius
ratio is 0.40. The value suggests that the crystal structure should be
(c) Na 2 W O3
(d) NaW O4
(a) Octahedral
(b) Tetrahedral
(c) Square planar
(d) Plane triangle
2.
Potassium crystallizes in a bcc lattice, hence the coordination
number of potassium in potassium metal is
Potassium has a bcc structure with nearest neighbour distance
[KCEE 1993]
3
4 .52 Å. Its atomic weight is 39. Its density (in kg m ) will be [AIIMS 1991]
(a) 0
(b) 4
(a) 454
(b) 804
(c) 6
(d) 8
(c) 852
(d) 908
3.
Body centered cubic lattice has a coordination number of
[AIIMS 1996; MP PMT 2002]
r 
If the value of ionic radius ratio  c  is 0.52 in an ionic
(a)
4
(b)
8
 ra 
(c) 12
(d) 6
compound, the geometrical arrangement of ions in crystal is
A and B. This crystallizes in
4.
A
compound
is
formed
by
elements
(a) Tetrahedral
(b) Planar
A
the
cubic
structure
when
atoms
are
the corners of the cube and
(c) Octahedral
(d) Pyramidal
B
atoms
are
at
the
centre
of
the
body.
The simplest formula of the
The number of atoms/molecules contained in one face centred
compounds
is
cubic unit cell of a monoatomic substance is
[KCET 1993; CBSE PMT 2000; Kerala PMT 2002]
[CPMT 1989, 94; CBSE PMT 1989, 96; NCERT 1990;
MP PET 1993; KCET 1999]
22.
(a) 1
(b) 2
(c) 4
(d) 6
The number of atoms/molecules contained in one body centered
cubic unit cell is
(a) 1
(b) 2
(c) 4
(d) 6
24.
It the distance between Na  and Cl  ions in sodium chloride
crystal is X pm, the length of the edge of the unit cell is [KCET 2004]
(a) 4X pm
(b) X/4 pm
(c) X/2 pm
(d) 2X pm
The edge of unit cell of FCC Xe crystal is 620 pm . The radius of
25.
[MP PET 2004]
Xe atom is
(a) 219.25 Pm
(b) 235.16 Pm
(c) 189.37 Pm
(d) 209.87 Pm
In orthorhombic, the value of a, b and c are respectively
23.
26.
5.
6.
7.
(a)
AB
(b)
AB 2
(c)
A2 B
(d)
AB 4
Coordination number for Cu is
[AMU 1982]
(a) 1
(b) 6
(c) 8
(d) 12
In the crystal of CsCl, the nearest neighbours of each Cs ion are[MP PET 199
(a) Six chloride ions
(b) Eight chloride ions
(c) Six Cs ions
(d) Eight Cs ions
In a cubic structure of compound which is made from X and Y,
where X atoms are at the corners of the cube and Y at the face
centres of the cube. The molecular formula of the compound is [AIIMS 2000]
(a) X 2 Y
(b) X 3 Y
(c)
8.
(d)
XY 2
XY 3
Ferrous oxide has a cubic structure and each edge of the unit cell is
5.0 Å. Assuming density of the oxide as 4.0 g  cm 3 , then the
4.2 Å, 8.6 Å and 8.3 Å . given the molecular mass of the solute is
number of Fe2  and O 2  ions present in each unit cell will be [MP PET 2000
155 gm mol 1 and that of density is 3.3 gm / cc , the number of
formula units per unit cell is
(a) Four Fe2  and four O 2 
[Orrisa JEE 2005]
(c) Four Fe2  and two O 2 
(a) 2
(b) 3
(c) 4
(d) 6
A metal has bcc structure and the edge length of its unit cell is
3.04 Å . The volume of the unit cell in cm 3 will be
(b) Two Fe 2  and four O 2 
(d) Three Fe 2  and three O 2 
9.
[Orrisa JEE 2005]
(a)
27.
A solid has a structure in which 'W ' atoms are located at the
corners of a cubic lattice ' O ' atoms at the centre of edges and
' Na ' atoms at the centre of the cube. The formula for the
compound is
[KCET 1996]
(a) NaW O2
(b) NaW O3
1.6  10 cm
21
3
(b)
2.81  10
23
cm
4
r
(b)
3
(c)
2r
4
Cl  ions are in fcc arrangement
(b)
Na  ions has coordination number 4
(c)
Cl  ions has coordination number 6
(d) Each unit cell contains 4 NaCl molecules
r
2
(d)
(a)
3
(c) 6.02  10 23 cm 3
(d) 6.6  10 24 cm 3
In face centred cubic unit cell edge length is
[DPMT 2005]
(a)
Which of the following statements is not true about NaCl
structure
[DCE 2001]
3
r
2
10.
In CsCl structure, the coordination number of Cs  is
[MP PMT 2001]

(a) Equal to that of Cl , that is 6
(b) Equal to that of Cl  , that is 8
(c) Not equal to that of Cl  , that is 6
Solid state
11.
(d) Not equal to that of Cl  , that is 8
In a solid ‘AB’ having the NaCl structure, ‘A’ atoms occupy the
corners of the cubic unit cell. If all the face-centered atoms along
one of the axes are removed, then the resultant stoichiometry of the
solid is
[IIT Screening 2001]
(a) AB 2
(b) A 2 B
(c)
12.
A4 B3
(d)
In CsCl lattice the coordination number of Cs  ion is
(a) 2
(b) 4
(c) 8
(d) 12
25.
Crystal structure of NaCl is
(a)
In solid CsCl each Cl is closely packed with how many Cs [MP PET 2003]
(a) 8
(b) 6
26.
(c) 10
(d) 2
In A  B  ionic compound, radii of A  and B  ions are 180 pm
14.
and 187 pm respectively. The crystal structure of this compound
will be
(a) NaCl type
(b) CsCl type
(c) ZnS type
(d) Similar to diamond
In which of the following substances the carbon atom is arranged in
a regular tetrahedral structure
[NCERT 1978]
(a) Diamond
(b) Benzene
(c) Graphite
(d) Carbon black
The coordination number of a metal crystallizing in a hexagonal
close packed structure is
16.
17.
18.
19.
20.
21.
22.
Al 2 O 3
(b)
Fe3 O 4
(c)
NiO2
(d)
PbO
What is the coordination number of sodium in Na 2 O
[BVP 2004]
(a) 12
(c) 8
29.
30.
31.
32.
23.
The number of Cl  ions around one Na  in NaCl crystal
lattice is
[MP PET 1996; BVP 2004]
(a) 12
(b) 4
(c) 8
(d) 6
The number of atoms present in unit cell of a monoatomic
substance of simple cubic lattice is
[Pb. PMT 2004]
(a) 6
(b) 3
(c) 2
(d) 1
The coordination number of a metal crystallizing in a hexagonal
close packed chep structure is
[MP PMT 2004]
(a) 12
(b) 8
(c) 4
(d) 6
Which of the following statement(s) is(are) correct
[IIT 1998]
The co-ordination number of Na  in NaCl is
[Orrisa JEE 2005]
34.
(a) 6
(b)
(c) 4
(d)
In the calcium fluoride structure the
cation and anions are respectively
(a) 6, 6
(b)
(c) 4, 4
(d)
8
1
co-ordination number of the
[J & K 2005]
8, 4
4, 8
Defects in crystal
1.
[KCET 2003]
(b) 4
(d) 6
(a) The coordination number of each type of ion in CsCl crystal
is 8
(b) A metal that crystallizes in bcc structure has a coordination
number of 12
(c) A unit cell of an ionic crystal shares some of its ions with other
unit cells
(d) The length of the unit cell in NaCl is 552 pm
(rNa   95 pm; rCl   181 pm)
[AIIMS 2003]
(a) 6
(b) 4
(c) 8
(d) 2
The ratio of cationic radius to anionic radius in an ionic crystal is
greater than 0.732. Its coordination number is
In NaCl lattice the coordination number of Cl  ion is
(a) 2
(b) 4
(c) 6
(d) 8
Coordination number of Na  ion in rock salt is
33.
If the radius ratio is in the range of 0.731  1, then the
coordination number will be
(a) 2
(b) 4
(c) 6
(d) 8
If the radius ratio is in the range of 0.414  0.732, then the
coordination number will be
(a) 2
(b) 4
(c) 6
(d) 8
(d) None
28.
[NCERT 1983]
(a)
(b) bcc
fcc
(c) Both (a) and (b)
In zinc blende structure the coordination number of Zn 2  ion is
(a) 2
(b) 4
(c) 6
(d) 8
BHU 1982, 87; MP PET 1995, 99]
(a) 3
(b) 8
(c) 4
(d) 6
Most crystals show good cleavage because their atoms, ions or
molecules are
[CBSE PMT 1991]
(a) Weakly bonded together
(b) Strongly bonded together
(c) Spherically symmetrical
(d) Arranged in planes
An example of a non-stoichiometric compound is
[NCERT 1982; BHU 1995]
27.
[NCERT 1978; IIT 1999]
(a) 4
(b) 12
(c) 8
(d) 6
The structure of MgO is similar to NaCl. What would be the
coordination number of magnesium
(a) 2
(b) 4
(c) 6
(d) 8
How many chloride ions are there around sodium ion in sodium
chloride crystal
[NCERT 1979, 80; CPMT 1988;
(b) 8
(d) 4
24.
A3 B4
13.
15.
(a) 6
(c) 1
213
2.
Certain crystals produce electric signals on application of pressure.
This phenomenon is called
[BHU 2005]
(a) Pyroelectricity
(b) Ferroelectricity
(c) Peizoelectricity
(d) Ferrielectricity
Which defect causes decrease in the density of crystal
[KCET 2000, 05]
(a) Frenkel
(b) Schottky
214 Solid state
(c) Interstitial
3.
F  centre
The correct statement regarding F  centre is
(a) Electron are held in the voids of crystals
(b)
4.
(d)
14.
F  centre produces colour to the crystals
(c) Conductivity of the crystal increases due to F  centre
(d) All
Doping of silicon (Si) with boron (B) leads to
15.
[UPSEAT 2004]
(a)
n -type semiconductor
(c) Metal
5.
p -type semiconductor
16.
(d) Insulator
If NaCl is doped with 10 3 mol % SrCl2 ,
concentration of cation vacancies will be
(a)
6.
(b)
1  10 3 mol%
(b)
then the
2  10 3 mol%
(c) 3  10 3 mol%
(d) 4  10 3 mol%
In the laboratory, sodium chloride is made by burning the sodium in
the atmosphere of chlorine which is yellow in colour. The cause of
yellow colour is
17.
18.

(a) Presence of Na ions in the crystal lattice
7.
8.
(b) Presence of Cl  ions in the crystal lattice
(c) Presence of electron in the crystal lattice
(d) Presence of face centered cubic crystal lattice
Frenkel defect is caused due to
[MP PET 1994]
(a) An ion missing from the normal lattice site creating a vacancy
(b) An extra positive ion occupying an interstitial position in the
lattice
(c) An extra negative ion occupying an interstitial position in the
lattice
(d) The shift of a positive ion from its normal lattice site to an
interstitial site
Which one of the following has Frenkel defect
[MP PMT 2000]
10.
(a) Sodium chloride
(b) Graphite
(c) Silver bromide
(d) Diamond
Schottky defect generally appears in
(a) NaCl
(b) KCl
(c) CsCl
(d) All of these
Schottky defect in crystals is observed when
11.
(a) Density of crystal is increased
(b) Unequal number of cations and anions are missing from the
lattice
(c) An ion leaves its normal site and occupies an interstitial site
(d) Equal number of cations and anions are missing from the
lattice
Ionic solids, with Schottky defects, contain in their structure
9.
19.
20.
21.
(a) Increases
(b) Decreases
(c) Does not change
(d) Changes
Point defects are present in
[MP PMT 1997]
(a) Ionic solids
(b) Molecular solids
(c) Amorphous solids
(d) Liquids
If a non-metal is added to the interstitial sites of a metal then the
metal becomes
[DCE 2001]
(a) Softer
(b) Less tensile
(c) Less malleable
(d) More ductile
In AgBr crystal, the ion size lies in the order Ag   Br  . The
AgBr crystal should have the following characteristics
(a) Defectless (perfect) crystal
(b) Schottky defect only
(c) Frenkel defect only
(d) Both Schottky and Frenkel defects
Frenkel and Schottky defects are
[BHU 2003]
(a) Nucleus defects
(b) Non-crystal defects
(c) Crystal defects
(d) None of these
Which one of the following is the most correct statement
(a) Brass is an interstitial alloy, while steel is a substitutional alloy
(b) Brass is a substitutional alloy, while steel is an interstitial alloy
(c) Brass and steel are both substitutional alloys
(d) Brass and steel are both interstitial alloys
The flame colours of metal ions are due to
[KCET 2003]
(a) Frenkel defect
(b) Schottky defect
(c) Metal deficiency defect
(d) Metal excess defect
Which one of the following crystals does not exhibit Frenkel defect [MP PET 200
(a) AgBr
(b) AgCl
(c) KBr
(d) ZnS
In a solid lattice the cation has left a lattice site and is located at an
interstitial position, the lattice defect is
[AIIMS 1982, 1991; DCE 2002; J & K 2005]
22.
[DCE 2004]
23.
(a) Interstitial defect
(b) Valency defect
(c) Frenkel defect
(d) Schottky defect
When electrons are trapped into the crystal in anion vacancy, the
defect is known as
[BHU 2005]
(a) Schotky defect
(b) Frenkel defect
(c) Stoichiometric defect
(d) F-centres
Schottky defect defines imperfection in the lattice structure of a [AIIMS 2002]
(a) Solid
(b) Liquid
(c) Gas
(d) Plasma
[CBSE PMT 1998; KCET 2002]
1.
[CBSE PMT 1994]
12.
13.
(a) Equal number of cation and anion vacancies
(b) Anion vacancies and interstitial anions
2.
(c) Cation vacancies only
(d) Cation vacancies and interstitial cations
The following is not a function of an impurity present in a crystal[MP PET 1995]
(a) Establishing thermal equilibrium
(b) Having tendency to diffuse
3.
(c) Contributing to scattering
(d) Introducing new electronic energy levels
Due to Frenkel defect, the density of ionic solids
4.
[MP PET 1996; MP PMT 2002]
Amorphous solids are
(a) Solid substance in real sense
(b) Liquid in real sense
(c) Supercooled liquid
(d) Substance with definite melting point
Silicon is found in nature in the form of
[MH CET 2002]
(a) Body centered cubic structure
(b) Hexagonal close-packed structure
(c) Network solid
(d) Face centered cubic structure
A match box exhibits
[MP PET 1993, 95]
(a) Cubic geometry
(b) Monoclinic geometry
(c) Orthorhombic geometry
(d) Tetragonal geometry
Which has no rotation of symmetry
[Orrisa JEE 2004]
(a) Hexagonal
(b) Orthorhombic
Solid state
5.
(c) Cubic
(d) Triclinic
Which of the following molecules has three-fold axis of symmetry[UPSEAT 2004]
(a) NH 3
(b) C2 H 4
6.
(c) CO 2
(d) SO 2
Which one possess a antifluorite structure
(a) Na 2 O
(b) MgO
7.
8.
3
(a)
Al
(c)
Mg 2
[MP PET 1993]
18.
2
(b)
Ba
(d)
Na 
Na  Cl  Na  Cl  Na  Cl 
Cl   Cl 
Na   Na 
Na  Cl   Cl 

(c) Fe 2 O 3
(d) Al 2 O 3
Which one of the following is the biggest ion
215

Na  Cl 

Cl
Na Cl
Na   Na 
(a) Interstitial defect
(b) Schottky defect
(c) Frenkel defect
(d) Frenkel and Schottky defects
Which of the following is a three dimensional silicate
[MHCET 2003]
(a) Mica
(c) Zeolite
(e) 12
The edge length of face centred unit cubic cell is 508 pm. If the
radius of the cation is 110 pm, the radius of the anion is
(b) Spodumene
(d) None of these
[CBSE PMT 1998]
9.
(a)
285 pm
(b)
(c)
144 pm
(d) 618 pm
398 pm
An element (atomic mass 100 g / mol) having bcc structure has
unit cell edge 400 pm. Then density of the element is
[CBSE PMT 1996; AIIMS 2002]
(a)
(c)
10.
11.
10.376 g / cm 3
7.289 g / cm
3
(a)
(b) 5.188 g / cm 3
(d)
2.144 g / cm
3
(b)
If the pressure on a NaCl structure is increased, then its
coordination number will
[AFMC 2000]
(a) Increase
(b) Decrease
(c) Remain the same
(d) Either (b) or (c)
The pyknometric density of sodium chloride crystal is
2.165  10 3 kg m 3
while
its
X-rays
density
is
2.178  10 3 kg m 3 . The fraction of unoccupied sites in sodium
chloride crystal is
[CBSE PMT 2003]
(a)
12.
5.96  10
3
Read the assertion and reason carefully to mark the correct option out of
the options given below :
(c)
(d)
(e)
1.
Assertion
Reason
:
Assertion
:
Reason
:
3.
Assertion
Reason
:
:
4.
Assertion
:
Reason
:
5.
Assertion
:
Space or crystal lattice differ in symmetry of the
arrangement of points.
6.
Reason
Assertion
:
:
Reason
:
Assertion
:
n  2d sin , is known as Bragg’s equation.
In close packing of spheres, a tetrahedral void is
surrounded by four spheres whereas an
octahedral void is surrounded by six spheres.
A tetrahedral void has a tetrahedral shape
whereas an octahedral void has an octahedral
shape.
Cyclic silicates and chain silicates have the same
general molecular formula.
Reason
:
Assertion
:
2.
(b) 5.96
(c) 5.96  10 2
(d) 5.96  10 1
Which of the following statements is correct for CsBr3
If both assertion and reason are true and the reason is the correct
explanation of the assertion.
If both assertion and reason are true but reason is not the correct
explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.
If assertion is false but reason is true.
[IIT 1996]
:
(a) It is a covalent compound
[AIIMS 1999]
(b) It contains Cs 3  and Br  ions
(c) It contains Cs  and Br3 ions
13.
(d) It contains Cs  , Br  and lattice Br2 molecule
In which compound 8 : 8 coordination is found
(a)
CsCl
(b)
(c)
Al 2 O 3
(d) All of these
MgO
14.
If the coordination of Ca 2  in CaF2 is 8, then the coordination
15.
number of F  ion would be
(a) 3
(b) 4
(c) 6
(d) 8
For some crystals, the radius ratio for cation and anion is 0.525, its
coordination number will be
(a) 2
(b) 4
(c) 6
(d) 8
The basic building unit of all silicates is
[UPSEAT 2002]
(a)
17.
SiO 4 square planar
In any ionic solid (MX) with Schottky defects, the
number of positive and negative ions are same.
Equal number of cation and anion vacancies are
present.
[IIT Screening 2001]
[EAMCET 1984]
16.
Diamond is a precious stone.
Carbon atoms are tetrahedrally arranged in
diamond.
[AIIMS 1994 ]
In crystal lattice, the size of the cation is larger in
a tetrahedral hole than in an octahedral hole.
The cations occupy more space than anions in
crystal packing.
[AIIMS 1996]
Crystalline solids have short range order.
Amorphous solids have long range order.
7.
(b) [SiO 4 ]4  tetrahedron
(c) SiO 4 octahedron
(d) SiO 4 linear
What type of crystal defect is indicated in the diagram below
[AIEEE 2004]
8.
In cyclic silicates, three corners of each SiO4
tetrahedron are shared while in chain silicates
only two are shared with other tetrahedra.
The presence of a large number of Schottky
defects in NaCl lowers its density.
216 Solid state
9.
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
11.
Assertion
:
12.
Reason
Assertion
Reason
:
:
:
10.
In NaCl , there are approximately 10 6
Schottky pairs per cm 3 at room temperature.
Anion vacancies in alkali halides are produced by
heating the alkali halide crystals with alkali metal
vapour.
Electrons trapped in anion vacancies are referred
to as F -centres.
Electrical conductivity of semiconductors
increases with increasing temperature.
With increase in temperature, large number of
electrons from the valence band can jump to the
conduction band.
On heating ferromagnetic or ferrimagnetic
substances, they become paramagnetic.
The electrons change their spin on heating.
Lead zirconate is a piezoelectric crystal.
Lead zirconate crystals have no dipole moment.
Type of solid and Their properties
1
a
2
b
3
a
4
a
5
b
6
c
7
c
8
b
9
d
10
d
11
b
12
a
13
c
14
c
15
a
16
a
17
a
18
d
19
c
20
c
21
b
22
d
23
d
24
d
25
a
26
d
27
a
28
a
29
d
30
31
d
32
a
33
c
34
a
36
a
37
a
38
b
39
c
Mathematical analysis of cubic system and
Bragg’s equation
1
b
2
b
3
a
4
b
5
b
6
a
7
b
8
c
9
b
10
a
11
c
12
c
13
a
14
b
15
d
16
c
17
a
18
b
19
d
20
c
21
c
22
b
23
d
24
a
25
c
26
b
27
b
Crystal structure and Coordination number
1
b
2
d
3
b
4
a
5
d
6
b
7
d
8
a
9
b
10
b
11
d
12
a
13
b
14
a
15
b
16
c
17
d
18
d
19
b
20
d
21
c
22
b
23
b
24
c
25
a
26
c
27
b
28
d
29
d
30
d
31
a
32
acd
33
a
34
b
Defects in crystal
1
c
2
b
3
d
4
d
5
a
d
6
c
7
d
8
c
9
d
10
d
35
b
11
a
12
a
13
c
14
a
15
b
40
ac
16
c
17
c
18
c
19
d
20
c
21
c
22
d
23
a
Crystallography and Lattice
1
b
2
c
3
b
4
d
5
a
6
a
7
b
8
d
9
d
10
b
11
c
12
c
13
a
14
c
15
b
16
b
17
b
18
b
19
a
20
a
21
c
22
d
23
c
24
d
25
b
26
b
27
c
28
d
Critical Thinking Questions
1
c
2
c
3
c
4
d
5
a
6
a
7
b
8
c
9
b
10
a
11
a
12
c
13
a
14
b
15
c
16
b
17
b
18
c
Assertion & Reason
Crystal packing
1
b
2
b
3
d
4
a
5
b
1
b
2
d
3
d
4
a
5
b
6
d
7
c
8
b
9
b
10
b
6
c
7
c
8
b
9
b
10
a
11
d
12
c
13
a
14
b
15
b
11
a
12
c
16
b
17
c
18
b
19
c
20
c
21
a
22
c
23
a
24
d
25
b
26
a
218 Solid state
35.
36.
Properties and Types of solids
37.
38.
2.
(a) Both gases and liquids possess fluidity and hence viscosity
molecules in the solid state do not have translational motion.
(b) It is a characteristic of liquid crystal.
3.
(a)
40.
5.
6.
(b) The value of heat of fusion of NaCl is very high due to fcc
arrangement of its ions.
(c) Piezoelectric crystals are used in record player.
8.
(b)
1.
9.
10.
12.
13.
BaTiO3 is a ferroelectric compound.
39.
NaCl is a ionic solid in which constituent particles are
positive ( Na  ) and negative (Cl  ) ions.
(d) Amorphous solids have short range order but no sharp in
melting point.
(d) Solids have definite shape, size and rigidity.
(a) In crystalline solid there is perfect arrangement of the constituent
particles only at 0K. As the temperature increases the chance that a
lattice site may be unoccupied by an ion increases. As the number
of defects increases with temperature solid change in liquid.
(c) Diamond is a covalent solid in which constituent particles are
atoms.
(b) LiF is an example of ionic crystal solid, in which constituent
particles are positive (Li  ) and negative (F  ) ions.
(a) Amorphous solids neither have ordered arrangement (i.e. no
definite shape) nor have sharp melting point like crystals, but
when heated, they become pliable until they assume the
properties usually related to liquids. It is therefore they are
regarded as super-cooled liquids.
(a) Silicon is a semiconductor because it is a thermal active and its
conductivity increased with increasing temperature.
(b) Amorphous solids are isotropic, because of these substances
show same properties in all directions.
(c) Polymorphism is a ability of a substances which show two or
more crystalline structure
(ac) Amorphous solids neither have ordered arrangement (i.e. no
definite shape) nor have sharp melting point like crystals, but
when heated, they become pliable until they assume the
properties usually related to liquids. It is therefore they are
regarded as super-cooled liquids.
Crystallography and Lattice
1.
2.
(b) A crystal system is hexagonal if its unit cell having a  b  c
axial ratio and     90,   120 axial angles.
(c) Rhombohedral crystal system
a  b  c ,       90 o
ex – NaNO 3 , CaSO 4 , calcite CaCO 3 , HgS
3.
(b) Tetragonal system has the
a  b  c,      90 .
(c) Solid NaCl is a bad conductor of electricity because ions are
not free to move.
(a) The existence of a substance in more than one crystalline form
is known as polymorphism.
(a) Solids are also non-crystalline in nature.
(a) Ice has the lowest melting point out of the given solids, hence
it has the weakest intermolecular forces.
(c) All metals and some alloys are metallic crystal.
(c) Iodine crystals are molecular crystals, in which constituent
particles are molecules having interparticle forces are Vander
Waal’s forces.
(b) Ionic solids have highest melting point due to strong
electrostatic forces of attraction.
(d) For n-type, impurity added to silicon should have more than 4
valence electrons.
(d) Glass is an amorphous solid.
(a) Crystalline solids have regular arrangement of constituent
particles, sharp melting points and are anisotropic.
(d) Sugar is a crystalline solid while glass, rubber and plastic are
amorphous solids.
5.
(a) Space lattice of CaF2 is face centred cubic.
6.
(a) For body centred cubic arrangement co-ordination number is 8
and radius ratio (r / r ) is 0.732  1.000 .
7.
8.
10.
(b) There are 14 Bravais lattices (space lattices).
(d) Monoclinic sulphur is an example of Monoclinic crystal system.
(b) r  0.414 r .
11.
(c) Each unit cell of NaCl contains 4 NaCl units.
12.
(c) For tetrahedral arrangement co-ordination number is 4 and
radius ratio (r / r ) is 0.225  0.414 .
13.
14.
(a) Face-centred cubic lattice found in KCl and NaCl .
(c) Definition of unit cell.
16.
28.
(a)
17.
18.
29.
(d) Graphite is sp 2 hybridised and a covalent crystal.
30.
31.
(d) Ionic crystals exhibit non-directional properties of the bond.
(d) Ice is a molecular crystal in which the constituent units are
molecules and the interparticle forces are hydrogen bonds.
(a) Quartz is a covalent crystal having a framework of silicates or
silica, i.e. a three dimensional network when all the four oxygen
atoms of each of SiO4 tetrahedron are shared.
(b) In NaCl (rock salt) : Number of Na  ions  12 (at edge
1
centers)   1 (at body centre) 1  4 . Number of
4
1
1
Cl  ions  8 (at corners)   6 (at face centre)   4 .
8
2
Thus 4 formula units per unit cell.
(b) Lowest potential energy level provides stable arrangement.
(b) The seven basic crystal lattice arrangements are :- Cubic,
Tetragonal,
Orthorhombic,
Monoclinic,
Hexagonal,
Rhombohedral and Triclinic.
(a) The conditions for monoclinic crystal system.
(a) High lattice energy of BaSO 4 causes low solubility of
14.
15.
16.
17.
19.
20.
21.
22.
23.
25.
26.
32.
33.
34.
MnO2 is antiferromagnetic.
(c) Metallic crystals are good conductor of heat and current due to
free electrons in them.
(a) Silicon is a covalent crystal in which constituent particles are
atoms.
19.
20.
21.
22.
23.
unit
cell
dimension
BaSO 4 in water.
(c) 14 kinds of Bravais lattices (space lattices) are possible in a
crystal.
(d) Radius ratio in TlCl is 0.732 – 1.000 and co-ordination number
is 8 and arrangement is body centred cubic.
(c) Zinc blende (ZnS ) has fcc structure and is an ionic crystal
having 4 : 4 co-ordination number.
Solid state
21.
24.
(d)
25.
(b) Zinc blende (ZnS ) has fcc structure and is an ionic crystal
having 4 : 4 co-ordination number.
1
(d)
 8 (at corners)  1
8
1
 6 (at face centre)  3
2
Z  1  3  4 (total no. of atoms)
28.
Na 2 O has antifluorite ( A2 B) type structure.
22.
23.
24.
Crystal packing
1.
2.
3.
4.
5.
6.
7.
(b) Number of tetrahedral voids in the unit cell
= 2  number of atoms = 2Z.
(b) The system ABC ABC…… is also referred to as face-centred
cubic or fcc.
(d) It represents ccp arrangement.
(a) BCC has a coordination number of 8.
(b) In rock salt structure the co-ordination number of Na  : Cl 
is 6 : 6 .
(d) The bcc cell consists of 8 atoms at the corners and one atom at
centre.
1

 n  8    1  2 .
8

The fcc cell consists of 8 atoms at the eight corners and one
atom at each of the six faces. This atom at the face is shared by
two unit cells.
1 
1
 n  8   6    4 .
8 
2
(c) AB2 type of structure is present in CaF2
2
8.
9.
10.
11.
12.
13.
14.
15.

2

 AB2 ⇌ A  2 B ; CaF2 ⇌ Ca  2 F
(b) Potassium (K) has bcc lattice.
(b) Number of atoms per unit cell in bcc system = 2.
(b) In body centred cubic, each atom/ion has a coordination
number of 8.
(d) Number of octahedral sites = Number of sphere in the packing.
 Number of octahedral sites per sphere  1 .
(c) ABAB …… is hexagonal close packing.
(a) Sodium (Na ) is a body cube.
(b) SrF2 has fluorite (CaF2 ) type structure.
(b) In ZnS structure, sulphide ions occupy all FCC lattice points
while Zn ions are present in alternate tetrahedral voids.
(b) MgO contains rock salt (NaCl) structure.
(c) CaF2 (fluorite) has fcc structure with 8 : 4 coordination
number.
(b) Every constituent has two tetrahedral voids. In ccp lattice
atoms
1
1
8 6  4
8
2
 Tetrahedral void  4  2  8 ,
Thus ratio  4 : 8 :: 1 : 2 .
(c) Tetrahedral sites one double comparable to octahedral sites
then ratio of X and Z respectively 2 : 1 since formula of the
compound X 2 Z .
(c) A atoms are at eight corners of the cube. Therefore, the no. of
8
 1 . B atoms are at the face
A atoms in the unit cell =
8
centre of six faces. Therefore, its share in the unit cell =
6
 3 . The formula is AB .
2
25.
26.
(a) In bcc structure 68% of the available volume is occupied by
spheres. Thus vacant space is 32%.
(c) Number of atoms in the cubic close packed structure = 8.
1
Number of octahedral voids   8  4 .
2
(a) Co-ordination number in HCP and CCP arrangement is 12
while in bcc arrangement is 8.
(d) In NaCl (rock salt) : Number of Na  ions  12 (at edge
1
centers)   1 (at body centre) 1  4 . Number of Cl  ions
4
1
1
 8 (at corners)   6 (at face centre)   4 . Thus 4
8
2
formula units per unit cell.
(b) Co-ordination number in HCP =12
Co-ordination number in Mg is also = 12
(a) All are the iso-electronic species but Na  has low positive
charge so have largest radius.
Mathematical analysis of cubic system and
Bragg’s equation
18.
19.
20.
3
N  mol . wt.( M )
g cm  3
V ( a 3 )  avogadro no. (N o )
1.
(b) Density of unit cell 
2.
(b) Distance between K  and F  
3.
1
 length of the edge
2
(a) There are two atoms in a bcc unit cell.
4.
5.
(b)
(b)
6.
(a)
7.
(b)
So, number of atoms in 12.08  10 23 unit cells
 2  12.08  10 23  24.16  10 23 atom .
bcc structure has one atom shared by 1 unit cell.
The structural arrangement of co-ordination number ‘6’ is
octahedral and its radius ratio is 0.414  0.732 . The
example of octahedral is KCl and NaCl .
The number of spheres in one body centred cubic and in one
face centred cubic unit cell is 2 and 4 respectively.
Closest approach in bcc lattice

10.
(a)
1
1
of body diagonal   3 a
2
2
M
2+
16.
17.
219


3
 4 .3  3 .72 Å .
2
  a 3  N 0  10 30
z
10  (100)  (6 .02  10 23 )  10 30
 15.05
4
3
No. of atoms in 100 g 
6.02  1023
 100  4  1025 .
15.05
11.
(c)
Cs  and I  have largest sizes.
12.
(c)
58.5 g NaCl  1 mole  6.02  10 23 Na  Cl units.
One unit cell contains 4 Na  Cl  units. Hence number of unit
cell present

13.
(a)
6 .02  10 23
 1.5  10 23 .
4
1
 6.023  10 23  1.029  10 22
58.5
A unit cell contains 4 Na  ion and 4 Cl  ions
Unit cell 
1 .029  10 22
 2.57  10 21 unit cell.
4
220 Solid state
(b) Bragg’s equation is n  2d sin
where n is an integer i.e. 1, 2, 3, 4 etc.
(d) Face centred cubic structure contribute of 1/8 by each atom
present on the corner and 1/2 by each atom present on the
face.
4.
16.
(c) As CsCl is body-centred, d  3a / 2 .
6.
17.
(a) Radius of Na (if bcc lattice) 
18.
 1.8574 Å  1.8574  10 8 cm
(b) The crystals in which radius ratio value is found between
0.225  0.414 shows tetrahedral crystal structure.
19.
(d) For bcc, d 
14.
15.

3a

4
3  4 .29
4
3
2d
2  4.52
a or a 

 5.219 Å  522 pm
2
1.732
3
5.
7.
8.
ZM
2  39

a 3  N 0  10 30
(522)3  (6 .023  10 23 )  10 30
21.
22.
23.
x
24.
(a)
25.
(c)
 (5 Å)3  125  10 24 cm 3
Density 
n
3079.2  10 1
 42.7  10 1  4.27  4
72
(b) In NaCl crystal Na  ions has coordination number 6.
10.
11.
(b) Cl  ions in CsCl adopt BCC type of packing.
(d) There were 6 A atoms on the face-centres removing facecentred atoms along one of the axes means removal of 2 A
atoms.
Now, number of A atoms per unit cell
1
1
 8  4 3
8
2
a = 2x
a
620
 219.25 Pm
; r
2 2
2 2
V  N0 d
Z
M
(corners )
(b) Volume of unit cell  a 3
27.
 (3.04  10 8 cm)3  2.81  10 23 cm 3
(b) In FCC
(edge centred)
15.
16.
(a) In Cs  Cl  crystal co-ordination number of each ion is 8.
180
 0 .962 which lies in the range of
(b) r / r 
187
0.732  1.000 , hence co-ordination number = 8 i.e. the
structure is CsCl type.
(a) In diamond, C-atoms are arranged in a regular tetrahedral
structure.
(b) In hcp, co-ordination no. is 12.
(c) Mg has 6 co-ordination number (fcc structure).
17.
(d) In NaCl crystal every Na  ion is surrounded by 6 Cl  ion and
12.
14.
Crystal structure and Coordination number
1
8 1
8
1.
(b) In a unit cell, W atoms at the corner 
2.
3.
1
 12  3
4
Na atoms at the centre of the cube = 1
W : O : Na  1 : 3 : 1, hence formula  NaW O3
(d) For bcc lattice, co-ordination number is 8.
(b) Body centered cubic lattice has a co-ordination number 8.
( body centred )
Hence the resultant stoichiometry is A3 B4
13.
4 r  2a
4r
a
2
(face  centred)
Number of B atoms per unit cell
1
+ 14
 12 
4
4 .2  8 .6  8 .3  10 24  6 .023  10 23  3.3
 3.84  4
155
26.
wt.of cell
72  n
, 4.09 
volume
6.023  10 23  125  10 24
9.
r

72  n
6 .023  10 23
Volume of one unit  (lengthof corner )3
(c) The value of ionic radius ratio is 0.52 which is between
0.414  0.732, then the geometrical arrangement of ions in
crystal is octahedral.
(c) The number of atoms present in sc, fcc and bcc unit cell are 1,
4, 2 respectively.
(b) The number of atoms present in sc, fcc and bcc unit cell are 1,
4, 2 respectively.
(d) Cl  Na  Cl 
a
(b) Each Cs  in CsCl is surrounded by eight Cl  and each
Cl  in CsCl is surrounded by eight Cs  .
(d) X atoms are at eight corners of the cube. Therefore, the
8
number of X atoms in the unit cell   1 .
8
Y atoms are at the face centre of six faces. Therefore, its share
6
in the unit cell   3 . The formula is XY 3 .
2
(a) Let the units of ferrous oxide in a unit cell  n , molecular weight
of ferrous oxide (FeO)  56  16  72 g mol 1 ,
weight of n units 
 0.91 g / cm 3  910 kg m 3
20.
(a) A atoms are at eight corners of the cube. Therefore, the
8
number of A atoms in the unit cell   1 , atoms B per unit
8
cell = 1. Hence the formula is AB.
(d) Co-ordination number for Cu is 12.
O atoms at the centre of edges 
18.
19.
20.
every chloride ion is surrounded by 6 Na  ion.
(d) Crystals show good cleavage because their constituent particles
are arranged in planes.
(b) Fe3 O4 is a non-stoichiometric compound because in it the
ratio of the cations to the anions becomes different from that
indicated by the chemical formula.
(d) The radius ratio for co-ordination number 4, 6 and 8 lies in
between the ranges [0.225  0.414], [0.414  0.732] and
[0.732  1] respectively.
Solid state
21.
22.
23.
24.
25.
26.
27.
30.
31.
32.
33.
34.
(c) The radius ratio for co-ordination number 4, 6 and 8 lies in
between the ranges [0.225  0.414], [0.414  0.732] and
[0.732  1] respectively.
2.
3.
5.
6.
7.
8.
(c)
17.
18.
(c)
(c)
ions and each Na  ion to 4 oxide ions. Hence it has 4 : 8 coordination.
(b) When radius ratio between 0.732  1, then co-ordination
number is 8 and structural arrangement is body-centred cubic.
19.
(d)
(c) Each Cs  is surrounded by eight Cl  ions in CsCl crystal
lattice because its co-ordination number is 8 : 8.
(a) NaCl has fcc arrangement of ions.
20.
21.
(c)
(c)
22.
(d)
(b) In Na 2 O, each oxide ions (O 2 ) is co-ordinated to 8 Na 
(c) Each Na  is surrounded by six Cl  ions in NaCl crystal
lattice because its co-ordination number is 6 : 6.
(b) Zinc blende (ZnS ) has fcc structure and is an ionic crystal
having 4 : 4 co-ordination number.
(d) In a simple cubic structure
1
z   8 (atoms one at a corners)
8
z 1
(a) Co-ordination number in hcp structure is 12.
(acd) A metal that crystallizes in bcc structure has a co-ordination
number of 8.

(a) In sodium chloride, each Na ion is surrounded by six Cl
ions and each Cl  ion is surrounded by six Na  ions. Thus,
both the ions have coordination number six.
(b) The Ca 2  ions are arranged in (ccp) arrangement, i.e. Ca 2 
ions are present at all corners and tat the centre of each face of
the cube. the fluoride ions occupy all the tetrahedral sites. This
is 8 : 4 arrangement i.e., each Ca 2  ion is surrounded by
8 F  ions and each F  ion by four Ca 2  ions.
(c) When polar crystal is subjected to a mechanical stress,
electricity is produced – a case of piezoelectricity. Reversely, if
electric field is applied, mechanical stress is developed.
Piezoelectric crystal acts as a mechanical electrical transductor.
(b) More is the Schottky defect in crystal more is the decrease in
density.
(d) All the given statements are correct about F-centres.
1.
2.
(c) Amorphous solids neither have ordered arrangement (i.e. no
definite shape) nor have sharp melting point like crystals, but
when heated, they become pliable until they assume the
properties usually related to liquids. It is therefore they are
regarded as super-cooled liquids.
(c) Silicon due to its catenation property form network solid.
4.
(c) Orthorhombic
geometry
has
and
abc
      90 . The shape of match box obey this
geometry.
(d) In a triclinic crystal has no notation of symmetry.
5.
(a) In NH 3 molecule, the original appearance is repeated as a
3.
result of rotation through 120 o . Such as axis is said to be an
axis of three-fold symmetry or a triad axis.
6.
(a)
7.
(b) Cationic radius increases down the group and decreases along
the period.
(c) Distance between centres of cation and anion
d 508
 
 254 pm
2
2
8.
9.
(d)
10.
(d)
11.
(a)
12.
(a)
13.
(c)
15.
(b)
size of Ag  and Br  ions.
Schottky defects occurs in highly ionic compounds which have
high co-ordination number ex. NaCl, KCl, CsCl .
Schottky defect is due to missing of equal number of cations
and anions.
Schottky defect is due to missing of equal number of cations
and anions.
Impurity present in a crystal does not establish thermal
equilibrium.
Since no ions are missing from the crystal as a whole, there is
no effect on density.
On adding non-metal in metal the metal becomes less tensile.
Na 2 O has antifluorite ( A2 B) type structure.
rc  ra  254 pm or 110  ra  254 or ra  144 pm
9.
(b)
(a) As each Sr 2 ion introduces one cation vacancy, therefore
concentration of cation vacancies = mol % of SrCl 2 added.
(c) Yellow colour on heating NaCl in presence of Na is due to
presence of electrons in anion vacancies (F-centres).
(d) Frenkel’s defect is due to shift of an ion from the normal lattice
site (Creating a vacancy) and occupy interstitial spaces.
(c) AgBr exhibits Frenkel defect due to large difference in the
AgBr exhibits Frenkel defect due to large difference in the
size of Ag  and Br  ions.
Both are stoichiometric crystalline defects.
Brass, Cu  80%, Zn  20% substitutional alloy.
Steel is an interstitial alloy because it is an alloy of Fe with C, C
atoms occupy the interstitial voids of Fe crystal.
In metal excess defect when holes created by missing of anions
are occupied by electrons, there sites are called F-centres and
are responsible for colour in the crystal.
KBr exhibits Schottky defect and not Frenkel defect.
When cation shifts from lattice to interstitial site, the defect is
called Frenkel defect.
F-centres are the sites where anions are missing and instead electrons
are present. they are responsible for colour.
Critical Thinking Questions

Defects in crystal
1.
16.
221


10.
(a)
nM
a 3  N 0  10  30
2  100
 5.188 g/cm 3
(400)  (6 .02  10 23 )  10 30
3
NaCl structure
High pressure
(6 : 6 co.  ord. )
11.
760 K
CsCl structure
(8 : 8 co.  ord. )
(a) Difference  2.178  10  2.165  10 3  0.013  10 3
3
Fraction unoccupied 
0 .013  10 3
 5.96  10 3
2 .178  10 3
CsBr3 consist of Cs  and Br3 ions.
12.
(c)
13.
(a) Each Cs  is surrounded by eight Cl  ions in CsCl crystal
lattice because its co-ordination number is 8 : 8.
14.
(b) In each CaF2 each calcium cation is surrounded by eight
fluoride anions in a body centred cubic arrangement. Each
fluoride ion is in contact with four calcium ions. Thus CaF2
has 8 : 4 co-ordination number.
(c) The radius ratio for co-ordination number 4, 6 and 8 lies in
between the ranges [0.225  0.414], [0.414  0.732] and
[0.732  1] respectively.
15.
O
O
O
Si
O
222 Solid state
16.
(b)
17.
(b) In this diagram, equal number of cations ( Na  ) and anions
(Cl  ) are missing, so it, shows schottky defect.
18.
(c) Zeolite is a three dimensional silicate because of in the silicates
all the four oxygen atoms at (SiO4 )4 tetrahedra are shared
with other tetrahedra, vesulting in a three dimensional
network.
Assertion & Reason
1.
(b) It is true that in the dimond structure, carbon atoms are
arranged in tetrahedrally ( sp 3 hybridized) but it is not the
correct explanation of assertion.
2.
(d) Tetrahedral holes are smaller in size than octahedral
holes. Cations usually occupy less space than anions.
3.
(d) Crystalline solids have regular arrangement of constituent
particles and are anisotropic whereas amorphous solids have no
regular arrangement and are isotropic.
4.
(a) Schottky defect is due to missing of equal number of cations and
anions.
5.
(b) Space or crystal lattice is a regular repeating arrangement of
points in space and forms the basis of classification of all
structures.
6.
(c) Tetrahedral void is so called because it is surrounded by four
spheres tetrahedrally while octahedral void is so called because
it is surrounded by six spheres octahedrally.
7.
(c) Two corners per tetrahedron one shared in both the cases.
8.
(b) When an atom or an ion is missing from its normal lattice site,
a lattice vacancy or defect is created, which is called schottky
defect. Due to missing density of crystal will be lowered.
9.
(b) On heating, the metal atoms deposit on the surface and finally
they deffuse into the crystal and after ionisation the alkali metal
ion occupies cationic vacancy where as electron occupies
anionic vacancy.
10.
(a) In case of semiconductors, the gap between valence band and
the conduction band is small and there fore some of the
electrons may jump from valence band to conduction band and
thus on increasing temperature conductivity is also increased.
11.
(a) All magnetically ordered solids (ferromagnetic and
antiferromagnetic solids) transform to the paramagnetic state
at high temperature due to the randomisation of spins.
12.
(c) In piezoelectric crystals, the dipoles may align them selves in an
ordered manner such that there is a net dipole moment in the
crystal.
Solid State
1.
Particles of quartz are packed by
7.
(a) Electrical attraction forces
223
The second order Bragg's diffraction of X  rays with   1 Å
from a set of parallel planes in a metal occurs at an angle of 60 o .
The distance between the scattering planes in the crystal is[CBSE PMT 1998; AFM
(b) Vander Waal's forces
(c) Covalent bond forces
(a)
0.575 Å
(b) 1 .00 Å
(c)
2 .00 Å
(d) 1 .15 Å
(d) Strong electrostatic attraction forces
2.
Crystals of covalent compounds always have
[BHU 1984]
(a) Atoms as their structural units
3.
4.
5.
8.
(b) Molecules as structural units
The edge length of the unit cell of NaCl crystal lattice is
552 pm. If ionic radius of sodium ion is 95 pm, what is the
(c) Ions held together by electrostatic forces
ionic radius of chloride ion
(d) High melting points
(a)
190 pm
(b)
368 pm
(c)
181 pm
(d)
276 pm
Wax is an example of
(a) Ionic crystal
(b) Covalent crystal
(c) Metallic crystal
(d) Molecular crystal
9.
Among the following which crystal will be soft and have low melting
point
The ionic radii of Rb  and I  are 1.46 Å and 2.16Å. the most
probable type of structure exhibited by it is
[UPSEAT 2004]
(a) Covalent
(b) Ionic
(a)
CsCl type
(b)
(c) Metallic
(d) Molecular
(c)
NaCl type
(d) CaF2 type
In zinc blende structure, zinc atom fill up
10.
(a) All octahedral holes
(b) All tetrahedral holes
(c) Half number of octahedral holes
6.
[KCET 1998]
11.
ZnS type
The coordination number of a cation occupying a tetrahedral hole is
(a) 6
(b) 8
(c) 12
(d) 4
(d) Half number of tetrahedral holes
If a electron is present in place of anion in a crystal lattice, then it is
called
Which ion has the lowest radius from the following ions
(a) Frenkel defect
[Kurukshetra CEE 1998]
(a)
Na 
(b)
Mg 2 
(c)
Al 3 
(d)
Si 4 
(b) Schottky defect
(c) Interstitial defects
(d)
F  centre
(SET -5)
224 Solid State
1.
(c) Quartz is a covalent solid in which constituent particles are
atoms which are held together by covalent bond forces.
2.
(a) Constituent particles of covalent compounds are atoms.
3.
(d) Iodine crystals are molecular crystals, in which constituent
particles are molecules having interparticle forces are Vander
Waal’s forces.
4.
(d) Molecular crystals are soft and have low melting point.
5.
(d) In zinc blende (ZnS ) half number of tetrahedral holes are
filled by zinc atoms.
6.
(d) All are the iso-electronic species but Si 4  has high positive
charge so have lowest radius.
7.
(d)
2d sin  n or 2  d  sin 60  2  1 Å
or 2  d  0.8660  2
or d  1.15 Å (sin 60  3 / 2 or 0.8660) .
***
8.
(c) Distance between centres of Na  and Cl 
rNa   rCl   276 pm or 95  rCl   276 pm
or rCl   276  95  181 pm
9.
(c)
rc 
ra 

1 .46
 0 .676
2 .16
It permits co-ordination number 6 and octahedral structure of
type NaCl .
10.
(d) The co-ordination number of a cation occupying a tetrahedral
hole is 4.
11.
(d) When electrons are trapped in anion vacancies, these are called
F-centres.
Gaseous State
225
Chapter
6
Gaseous state
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
1 m 3  10 3 dm 3  10 6 cm 3  10 6 mL  10 3 L
(3) 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.
forces.
(3) Gases have infinite expansibility and high compressibility.
(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 with perfectly elastic collisions.
(9) Gases can be liquified, if subjected to low temperatures (below
critical) or 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.
Measurable properties of gases
(1) 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).
(2) 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 (cm 3 ) or cubic metres (m 3 ) .
(iii) 1L  1000 mL ; 1 mL  10 3 L ; 1 L  1 dm 3  10 3 m 3
moles of gas (n) 
Mass in grams
m

Molar mass
M
(4) Temperature : (i) Gases expand on increasing the temperature. If
temperature is increased twice, the square of the velocity (v 2 ) also
increases two times.
(ii) Temperature is measured in centigrade degree ( o C) or celsius
degree with the help of thermometers. Temperature is also measured in
Fahrenheit (F ).
o
(iii) S.I. unit of temperature is kelvin (K) or absolute degree.
K  o C  273
C F o  32

5
9
(5) Pressure : (i) Pressure of the gas is the force exerted by the gas
per unit area of the walls of the container in all directions. Thus, Pressure
Force( F) Mass(m )  Acceleration(a)

(P) 
Area( A)
Area(a)
o
(iv) Relation between F and o C is
(ii) Pressure exerted by a gas is due to kinetic energy
1
(KE  mv 2 ) of the molecules. Kinetic energy of the gas molecules
2
increases, as the temperature is increased. Thus, Pressure of a gas 
Temperature (T).
(iii) Pressure of a pure gas is measured by manometer while that of
a mixture of gases by barometer.
(iv) Commonly two types of manometers are used,
(a) Open end manometer; (b) Closed end manometer
226 Gaseous State
(v) The S.I. unit of pressure, the pascal (Pa), is defined as
newton per metre square. It is very small unit.
1
1Pa  1 Nm 2  1 kg m 1 s 2
(vi) C.G.S. unit of pressure is dynes cm 2 .
(vii) M.K.S. unit of pressure is kgf / m 2 . The unit kgf / cm 2
sometime called ata (atmosphere technical absolute).
(viii)
Higher
unit
of
pressure
bar,
is
KPa
T1< T2< T3
PV
T3
T2
T1
MPa.
or
1 bar  10 5 Pa  10 5 Nm 2  100 KNm 2  100 KPa
log P
(ix) Several other units used for pressure are,
Name
Symbol
Value
O
O
(4) At constantP mass and temperature densitylogof1/Va gas is directly
bar
bar
1bar  10 Pa
atmosphere
atm
1 atm  1.01325  10 5 Pa
Torr
Torr
101325
1 Torr 
Pa  133.322 Pa
760
millimetre of
mercury
mm Hg
1 mm Hg  133.322 Pa
5
proportional to its pressure and inversely proportional to its volume.
Thus, d  P 
or
(x) The pressure relative to the atmosphere is called gauge pressure.
The pressure relative to the perfect vacuum is called absolute pressure.
Absolute pressure = Gauge pressure + Atmosphere pressure.
(xi) When the pressure in a system is less than atmospheric
pressure, the gauge pressure becomes negative, but is frequently designated
and called vacuum. For example, 16 cm vacuum will be
76  16
 1.013  0 .80 bar .
76
(xii) If ‘h’ is the height of the fluid in a column or the difference in
the heights of the fluid columns in the two limbs of the manometer, d is the
density of the fluid
d1
P
V
 1  2  .......  K
d 2 P2
V1
Charle's law
(1) French chemist, Jacques Charles first studied variation of volume
with temperature, in 1787.
(2) It states that, “The volume of a given mass of a gas is directly
proportional to the absolute temperature ( o C  273) at constant
pressure”.
Thus, V  T at constant pressure and mass
or V  KT  K(t( o C)  273.15) , (where k is constant),
K
pressure is given by, Pgas  Patm  h dg
Condition
S.T.P./N.T.P.
S.A.T.P .
*
T
273.15 K
298.15 K
V (Molar volume)
22.414 L
24.800 L
P
mass 

 V  d 


(5) At altitudes, as P is low d of air is less. That is why
mountaineers carry oxygen cylinders.
(Hg  13.6  10 3 Kg / m 3  13.6 g / cm 3 ) and g is the gravity, then
(xiii) Two sets of conditions are widely used as 'standard' values for
reporting data.
1
V
V
V
V
or 1  2  K (For two or more gases)
T
T1 T2
(3) If t  0 o C , then V  V0
V0  K  273.15
hence,
m
1 atm
1 bar
* Standard ambient temperature and pressure.
 K
V
Boyle's law
(1) In 1662, Robert Boyle discovered the first of several relationships
among gas variables (P, T, V).
(2) It states that, “For a fixed amount of a gas at constant
temperature, the gas volume is inversely proportional to the gas pressure.”
Thus, P  1 / V at constant temperature and mass
or P  K / V (where K is constant)
or PV  K or P1 V1  P2 V2  K (For two or more gases)
(3) Graphical representation of Boyle's law : Graph between P and V
at constant temperature is called isotherm and is an equilateral (or
rectangular) hyperbola. By plotting P versus 1 / V , this hyperbola can be
converted to a straight line. Other types of isotherms are also shown below,
V0
273.15
V0
t


[t  273.15]  V0 1 
  V0 [1   v t]
273.15
273
.
15


where  v is the volume coefficient,
v 
V  V0
1

 3 .661  10  3 o C 1
tV0
273.15
Thus, for every 1 o change in temperature, the volume of a gas
changes by
1
1 


 of the volume at 0 o C .
273.15  273 
(4) Graphical representation of Charle's law : Graph between V and
T at constant pressure is called isobar or isoplestics and is always a straight
line. A plot of V versus t( o C) at constant pressure is a straight line cutting
T3
P
O
T1
P
T3
T2
T1
V or 1/d
the temperature axis at
temperature.
T2
T1< T2< T3
T1< T2< T3
O
1/V or d
 273.15 o C . It is the lowest possible
Gaseous State
227
1/d or V
1/d
or V
22.4 L mol–1 = V0
O
–273.15o
T(k)
C
C
0o
t C)
(o
(5) At constant mass and pressure density of a gas is inversely
proportional to it absolute temperature.
Thus, d 
1
1

T
V
mass 

 V  d 


(1) According to this law, “Equal volumes of any two gases at the
same temperature and pressure contain the same number of molecules.”
Thus, V  n (at constant T and P)
or V  Kn (where K is constant)
or
V1 V2

 .......  K
n1 n 2
Example, 2 H 2 (g) O 2 (g)  2 H 2 O(g)
d1 T2 V2


 ......  K
d 2 T1
V1
or
Avogadro's law
2 moles
2 volumes
2 litres
1 litre
1n litre
(6) Use of hot air balloons in sports and meteorological observations
is an application of Charle's law.
(1) In 1802, French chemist Joseph Gay-Lussac studied the variation
of pressure with temperature and extende the Charle’s law so, this law is
also called Charle’s-Gay Lussac’s law.
(2) It states that, “The pressure of a given mass of a gas is directly
proportional to the absolute temperature ( o C  273) at constant
volume.”
(Avogadro's number  6.02  10 23 ) and by this law must occupy the
same volume at a given temperature and pressure. The volume of one mole
of a gas is called molar volume, V which is 22.4 L mol 1 at S.T.P. or
N.T.P.
m
(3) This law can also express as, “The molar gas volume at a given
temperature and pressure is a specific constant independent of the nature
of the gas”.
Thus, Vm  specific constant  22.4 L mol 1 at S.T.P. or N.T.P.
Thus, P  T at constant volume and mass
K
(where K is constant)
Ideal gas equation
(1) The simple gas laws relating gas volume to pressure, temperature
and amount of gas, respectively, are stated below :
P
P
P
or 1  2  K (For two or more gases)
T
T1 T2
Boyle's law :
(3) If t  0 C , then P  P0
P
o
K

P0
273.15
where  P is the pressure coefficient,
P  P0
1
P 

 3.661  10  3 o C 1
tP0
273.15
Thus, for every 1 o change in temperature, the pressure of a gas
1
1 

changes by

 of the pressure at 0 o C .
273.15  273 
(4) This law fails at low temperatures, because the volume of the
gas molecules be come significant.
(5) Graphical representation of Gay-Lussac's law : A graph between
P and T at constant V is called isochore.
V1< V2< V3< V4
V1
T(k)
O
T
(n and P constant)
V
or
V
nT
P
nRT
( R  Ideal gas constant)
P
PV  nRT
or
This is called ideal gas equation. R is called ideal gas constant. This
equation is obeyed by isothermal and adiabatic processes.
(2) Nature and values of R : From the ideal gas equation,
PV
Pressure  Volume
R

nT mole  Temperatur e
Force
 Volume
Force  Length
Area


mole  Temperatur e mole  Temperatur e

Work or energy
.
mole  Temperatur e
R is expressed in the unit of work or energy mol 1 K 1 .
V2
V3
V4
P
(n and T constant)
(T and P constant)
If all the above law's combines, then
P0
t


P
[t  273.15]  P0 1 
 P0 [1  t]
273.15
273.15 

O
1
1
or V 
V
P
Charle's law :
VT
Avogadro's law : V  n
Hence, P0  K  273.15
P
2 moles
2 volumes
2 litres
1 litre
1n litre
(2) One mole of any gas contains the same number of molecules
Gay-Lussac's law (Amonton's law)
or P  KT  K(t(o C)  273.15)
1 mole
1 volume
1 litre
1 / 2 litre
1 / 2 n litre
Since different values of R are summarised below :
R  0.0821 L atm mol 1 K 1
228 Gaseous State
Thus, Ptotal  P1  P2  P3  .........
 8.3143 joulemol 1 K 1 (S.I. unit)
Where P1 , P2 , P3 ,...... are partial pressures of gas number 1, 2, 3
 8.3143 Nm mol 1 K 1
.........
 8.3143 KPa dm 3 mol 1 K 1
 8.3143 MPacm mol
3
1
K
(2) Partial pressure is the pressure exerted by a gas when it is
present alone in the same container and at the same temperature.
Partial pressure of a gas
1
 5.189  10 19 eV mol 1 K 1
 1.99 cal mol
1
K
(P1 ) 
1
(3) Gas constant, R for a single molecule is called Boltzmann
constant (k)
k
R
8.314  10 7

ergs mole 1 degree 1
N 6 .023  10 23
 1.38  10 16 ergsmol 1 degree 1
or 1.38  10
23
joulemol
1
degree
m
RT
M
M
d
mRT
PV
PM
RT
m

 d  
V

dT M M

 Constant
,
P
R R
(  M and R are constant for a particular gas)
d T
d T
dT
Thus,
or 1 1  2 2 = Constant
P1
T2
P
(For two or more different temperature and pressure)
(5) Gas densities differ from those of solids and liquids as,
(i) Gas densities are generally stated in g/L instead of g / cm 3 .
(ii) Gas densities are strongly dependent on pressure and
temperature as, d  P  1 / T
Densities of liquids and solids, do depend somewhat on temperature,
but they are far less dependent on pressure.
(iii) The density of a gas is directly proportional to its molar mass.
No simple relationship exists between the density and molar mass for liquid
and solids.
d ( N 2 ) at STP 
PTotal 
P1 V1  P2 V2  P3 V3 .....
V
or  (n1  n 2  n 3 .....)
RT
V
molar mass
22.4
28
 1 .25 g L1 ,
22.4
32
 1 .43 g L1
d (O 2 ) at STP 
22.4
Dalton's law of partial pressures
(1) According to this law, “When two or more gases, which do not
react chemically are kept in a closed vessel, the total pressure exerted by the
mixture is equal to the sum of the partial pressures of individual gases.”
( PV  nRT )
RT
( n  n1  n 2  n3 .....)
V
(4) Applications : This law is used in the calculation of following
relationships,
or  n

(X 1 )
in a mixture of gas
Partial pressure of a gas (P1 )
PTotal
(ii) % of a gas in mixture 
or
(iv) Density of a gas at STP 
P1 , P2 , P3 ........ are mixed together in container of volume V, then,
(i) Mole fraction of a gas


mass of the gas (m )
 n 

Molecular weight of the gas (M ) 


(3) If a number of gases having volume V1 , V2 , V3 ...... at pressure
1
(4) Calculation of mass, molecular weight and density of the gas by
gas equation
PV  nRT 
Number of moles of the gas (n1 )  PTotal
 Mole fraction ( X 1 )  PTotal
Total number of moles (n) in the mixture
Partial pressure of a gas (P1 )
 100
PTotal
(iii) Pressure of dry gas collected over water : When a gas is
collected over water, it becomes moist due to water vapour which exerts its
own partial pressure at the same temperature of the gas. This partial
perssure of water vapours is called aqueous tension. Thus,
Pdry gas  Pmoist gas or PTotal  Pwater vapour
or Pdry gas  Pmoist gas  Aqueous tension (Aqueous tension is
directly proportional to absolute temperature)
(iv) Relative humidity (RH) at a given temperature is given by,
RH 
Partial pressure of water in air
.
Vapour pre ssure of water
(5) Limitations : This law is applicable only when the component
gases in the mixture do not react with each other. For example, N 2 and
O 2 , CO and CO 2 , N 2 and Cl 2 , CO and N 2 etc. But this law is not
applicable to gases which combine chemically. For example, H 2 and Cl 2 ,
CO and Cl 2 , NH 3 , HBr and HCl, NO and O 2 etc.
(6) Another law, which is really equivalent to the law of partial
pressures and related to the partial volumes of gases is known as Law of
partial volumes given by Amagat. According to this law, “When two or more
gases, which do not react chemically are kept in a closed vessel, the total
volume exerted by the mixture is equal to the sum of the partial volumes of
individual gases.”
Thus, VTotal  V1  V2  V3  ......
Where V1 , V2 , V3 ,...... are partial volumes of gas number 1, 2, 3.....
Graham's law of diffusion and Effusion
(1) Diffusion is the process of spontaneous spreading and
intermixing of gases to form homogenous mixture irrespective of force of
gravity. While Effusion is the escape of gas molecules through a tiny hole
such as pinhole in a balloon.
Gaseous State
 All gases spontaneously diffuse into one another when they are
brought into contact.
 Diffusion into a vacuum will take place much more rapidly than
diffusion into another place.
 Both the rate of diffusion of a gas and its rate of effusion depend
on its molar mass. Lighter gases diffuses faster than heavier gases. The gas
with highest rate of diffusion is hydrogen.
(2) According to this law, “At constant pressure and temperature,
the rate of diffusion or effusion of a gas is inversely proportional to the
square root of its vapour density.”
1
Thus, rate of diffusion (r) 
(T and P constant)
d
For two or more gases at constant pressure and temperature,
r1

r2
d2
d1
(3) Graham's law can be modified in a number of ways as,
(i) Since, 2  vapour density (V.D.) = Molecular weight
r
then, 1 
r2
d2  2

d1  2
d2

d1
M2
M1
where, M 1 and M 2 are the molecular weights of the two gases.
(ii) Since, rate of diffusion
then,
r1
V /t
w /t
 1 1  1 1 
r2 V2 / t 2 w 2 / t 2
Volume of a gas diffused
(r) 
Time taken for diffusion
then,
r1 t 2


r2 t1
d2
d1
(b) When volumes of the two gases diffuse in the same time, i.e.
t1  t 2
then,
r1
V
 1 
r2 V2
d2
d1
(iii) Since, r  p
(when p is not constant)

P 
 r 


M 

(4) Rate of diffusion and effusion can be determined as,
(i) Rate of diffusion is equal to distance travelled by gas per unit
time through a tube of uniform cross-section.
(ii) Number of moles effusing per unit time is also called rate of
diffusion.
(iii) Decrease in pressure of a cylinder per unit time is called rate of
effusion of gas.
(iv) The volume of gas effused through a given surface per unit time
is also called rate of effusion.
(5) Applications : Graham's law has been used as follows,
(i) To determine vapour densities and molecular weights of gases.
(ii) To prepare Ausell’s marsh gas indicator, used in mines.
(iii) Atmolysis : The process of separation of two gases on the basis
of their different rates of diffusion due to difference in their densities is
called atmolysis. It has been applied with success for the separation of
isotopes and other gaseous mixtures.
then,
r1
P
 1
r2 P2
M2
M1
Kinetic theory of gases
(1) Kinetic theory was developed by Bernoulli, Joule, Clausius,
Maxwell and Boltzmann etc. and represents dynamic particle or microscopic
model for different gases since it throws light on the behaviour of the
particles (atoms and molecules) which constitute the gases and cannot be
seen. Properties of gases which we studied earlier are part of macroscopic
model.
(2) Postulates
(i) Every gas consists of a large number of small particles called
molecules moving with very high velocities in all possible directions.
(ii) The volume of the individual molecule is negligible as compared
to the total volume of the gas.
(iii) Gaseous molecules are perfectly elastic so that there is no net loss of
kinetic energy due to their collisions.
(iv) The effect of gravity on the motion of the molecules is
negligible.
(v) Gaseous molecules are considered as point masses because they
do not posses potential energy. So the attractive and repulsive forces
between the gas molecules are negligible.
(vi) The pressure of a gas is due to the continuous bombardment on
the walls of the containing vessel.
(vii) At constant temperature the average K.E. of all gases is same.
(viii) The average K.E. of the gas molecules is directly proportional
to the absolute temperature.
(3) Kinetic gas equation : On the basis of above postulates, the
following gas equation was derived,
d2
d1
(a) When equal volume of the two gases diffuse, i.e. V1  V2
229
PV 
1
2
mnu rms
3
where, P = pressure exerted by the gas
V = volume of the gas
m = average mass of each molecule
n = number of molecules
u = root mean square (RMS) velocity of the gas.
(4) Calculation of kinetic energy
We know that,
K.E. of one molecule 
1
mu 2
2
K.E. of n molecules 
1
3
1
mnu 2  PV ( PV  mnu 2 )
2
2
3
n = 1, Then K.E. of 1 mole gas 
3
RT
2
( PV  RT )

3
 8.314  T  12.47 T Joules .
2

Average K.E.per mole 3 RT 3

 KT
N (Avogadro number ) 2 N
2
R


 Boltzmann constant 
K 
N


This equation shows that K.E. of translation of a gas depends only
on the absolute temperature. This is known as Maxwell generalisation. Thus
average K.E.  T.
If T  0 K (i.e.,  273.15 o C) then, average K.E. = 0. Thus,
absolute zero (0K) is the temperature at which molecular motion ceases.
Molecular collisions
230 Gaseous State
(1) The closest distance between the centres of two molecules taking
part in a collision is called molecular or collision diameter (). The
molecular diameter of all the gases is nearly same lying in the order of
10 8 cm .
(2) According to Maxwell, at a particular temperature the
distribution of speeds remains constant and this distribution is referred to
as the Maxwell-Boltzmann distribution and given by the following
expression,
dn0
 M 
 4 

n
 2RT 
molecules per unit time are given by, Z A  2 2 u av. n
where n is the number of molecules per unit molar volume,
Avogadro number( N 0 ) 6 .02  10 23 3

m
Vm
0 .0224
(ii) The total number of bimolecular collision per unit time are given
1
by, Z AA 
 2 u av. n 2
2
(iii) If the collisions involve two unlike molecules, the number of
bimolecular collision are given by,
n
where,  AB 
(b) At particular pressure; Z  T 3 / 2
(c) At particular volume; Z  T 1 / 2
(3) During molecular collisions a molecule covers a small distance
before it gets deflected. The average distance travelled by the gas molecules
between two successive collision is called mean free path ().
Average distance travelled per unit time( u av )
.

No. of collisionsmade by singlemolecule per unit time (Z A )

2 2 u avr.n

.u 2 dc
2
Ump
300 K (T1) T1<T2<T3
 A  B
(iv) (a) At particular temperature; Z  p 2

/ 2 RT
temperature. The exponential factor e  Mu / 2 RT is called Boltzmann factor.
(3) Maxwell gave distribution curves of molecular speeds for CO 2
at different temperatures. Special features of the curve are :
(i) Fraction of molecules with two high or two low speeds is very small.
(ii) No molecules has zero velocity.
(iii) Initially the fraction of molecules increases in velocity till the
peak of the curve which pertains to most probable velocity and thereafter it
falls with increase in velocity.
1/2
2
M A , M B are molecular weights (M  mN 0 )
u av
2
molecules n, having velocities between c and c  dc , dn0 / n  Fraction of
the total number of molecules, M = molecular weight, T = absolute
Fraction of molecules
(2) The number of collisions taking place in unit time per unit
volume, called collision frequency (z).
(i) The number of collision made by a single molecule with other
Z AB
.e  Mu
where, dn0  Number of molecules out of total number of

Molecular diameter
(M A  M B ) 
2 
  AB
8RT

MAMB 

3/2
1
2n  2
T
(4) Based on kinetic theory of gases mean free path,   . Thus,
P
(i) Larger the size of the molecules, smaller the mean free path, i.e.,
1
(radius)2
(ii) Greater the number of molecules per unit volume, smaller the
mean free path.
(iii) Larger the temperature, larger the mean free path.
(iv) Larger the pressure, smaller the mean free path.
(5) Relation between collision frequency (Z) and mean free path ()
u
is given by, Z  rms

Molecular speeds or velocities
(1) At any particular time, in the given sample of gas all the
molecules do not possess same speed, due to the frequent molecular
collisions with the walls of the container and also with one another, the
molecules move with ever changing speeds and also with ever changing
direction of motion.
Ump
Uav
Urms U
mp
1500 K (T2)
1800 K (T3)
Molecular speed
(4) Types of molecular speeds or Velocities
(i) Root mean square velocity (u ) : It is the square root of the mean
of the squares of the velocity of a large number of molecules of the same
gas.
rms
u rms 
u12  u 22  u 32  ..... u n2
n
u rms 
3 RT
3kT
3P
3 PV
3 RT




M
m
d
(mN 0 )  M
(mN 0 )  M
where k = Boltzmann constant 
R
N0
(a) For the same gas at two different temperatures, the ratio of
RMS velocities will be,
u1

u2
T1
T2
(b) For two different gases at the same temperature, the ratio of
RMS velocities will be,
u1

u2
M2
M1
(c) RMS velocity at any temperature t o C may be related to its
value at S.T.P. as, u t 
3 P(273  t)
.
273d
(ii) Average velocity (v av ) : It is the average of the various velocities
possessed by the molecules.
v av 
v1  v 2  v 3  ...... v n
; v av 
n
8 RT

M
8 kT
m
(iii) Most probable velocity ( mp ) : It is the velocity possessed by
maximum number of molecules of a gas at a given temperature.
Gaseous State
(iii) Z  1 , the gas is more compressible than expected from ideal
behaviour and shows negative deviation, usually at low P i.e. PV  RT .
2 RT
2 PV
2P


M
M
d
(5) Relation between molecular speeds or velocities,
 mp 
u rms and v av : v av  0.9213  urms
(i) Relation between
or urms  1.085  v av
(ii) Relation between  mp and u rms :  mp  0.816  urms
or urms  1.224   mp
(iii) Relation between  mp and v av : v av  1.128   mp
(iv) Relation between  mp , v av and u rms
 mp
:
:
v av
2 RT
M
:
8 RT
M
2
:

1.414
1
:
:
1.595
1.128
8
u rms
:
3 RT
M
:
3
:
Real and Ideal gases
(1) Gases which obey gas laws or ideal gas equation (PV  nRT ) at
all temperatures and pressures are called ideal or perfect gases. Almost all
gases deviate from the ideal behaviour i.e., no gas is perfect and the concept
of perfect gas is only theoretical.
(2) Gases tend to show ideal behaviour more and more as the
temperature rises above the boiling point of their liquefied forms and the
pressure is lowered. Such gases are known as real or non ideal gases. Thus,
a “real gas is that which obeys the gas laws under low pressure or high
temperature”.
(3) The deviations can be displayed, by plotting the P-V isotherms of
real gas and ideal gas.
H2
He
Z 
(v) The most easily liquefiable and highly soluble gases (NH 3 , SO 2 )
show larger deviations from ideal behaviour i.e. Z  1 .
(vi) Some gases like CO 2 show both negative and positive
deviation.
(6) Causes of deviations of real gases from ideal behaviour : The
ideal gas laws can be derived from the kinetic theory of gases which is
based on the following two important assumptions,
(i) The volume occupied by the molecules is negligible in
comparison to the total volume of gas.
(ii) The molecules exert no forces of attraction upon one another. It
is because neither of these assumptions can be regarded as applicable to
real gases that the latter show departure from the ideal behaviour.
(1) To rectify the errors caused by ignoring the intermolecular forces
of attraction and the volume occupied by molecules, Vander Waal (in 1873)
modified the ideal gas equation by introducing two corrections,
(i) Volume correction (ii) Pressure correction
(2) Vander Waal's equation is obeyed by the real gases over wide
range of temperatures and pressures, hence it is called equation of state for
the real gases.
(3) The Vander Waal's equation for n moles of the gas is,
2 

P  n a 

V 2 

Pressure correction
for molecular attraction
[V  nb]
 nRT
Volume correction for
finite size of molecules
a and b are Vander Waal's constants whose values depend on the
nature of the gas. Normally for a gas a  b .
(i) Constant a : It is a indirect measure of magnitude of attractive
forces between the molecules. Greater is the value of a, more easily the gas
can be liquefied. Thus the easily liquefiable gases (like
SO 2  NH 3  H 2 S  CO 2 ) have high values than the permanent gases
(like N 2  O 2  H 2  He) .
PV
nRT
(iv) Z  1 for H 2 and He at all pressure i.e., always shows
positive deviation.
Vander Waal's equation
1.732
1.224 i.e.,  mp  v av  urms
:
231
Units of 'a' are : atm. L2 mol 2
O2
N2
CH4
4
CO2
P
(4) It is difficult to determine quantitatively the deviation of a real
gas from ideal gas behaviour from the P-V isotherm curve as shown above.
Compressibility factor Z defined by the equation,
PV  ZnRT or Z  PV / nRT  PVm / RT
is more suitable for a quantitative description of the deviation from
ideal gas behaviour.
(5) Greater is the departure of Z from unity, more is the deviation
from ideal behaviour. Thus, when
(i) Z  1 , the gas is ideal at all temperatures and pressures. In case
o
of N 2 , the value of Z is close to 1 at 50 C . This temperature at which a
real gas exhibits ideal behaviour, for considerable range of pressure, is
known as Boyle's temperature or Boyle's point (TB ) .
(ii) Z  1 , the gas is less compressible than expected from ideal
behaviour and shows positive deviation, usual at high P i.e. PV  RT .
or atm. m 6 mol 2
or
2
N m mol (S.I. unit).
(ii) Constant b : Also called co-volume or excluded volume,
4

b  4 N 0 v  4 N 0  r 3 
3

It's value gives an idea about the effective size of gas molecules.
Greater is the value of b, larger is the size and smaller is the compressible
volume. As b is the effective volume of the gas molecules, the constant value
of b for any gas over a wide range of temperature and pressure indicates
that the gas molecules are incompressible.
Units of 'b' are : L mol 1 or m 3 mol 1 (S.I. unit)
(iii) The two Vander Waal's constants and Boyle's temperature (TB )
are related as,
a
TB 
bR
(4) Vander Waal's equation at different temperature and pressures
(i) When pressure is extremely low : For one mole of gas,
232 Gaseous State
a
ab
a 

 P  2  (V  b)  RT or PV  RT   Pb  2
V
V
V 

(ii) When pressure is extremely high : For one mole of gas,
PV
Pb
Pb
or Z  1 
1
PV  RT  Pb ;
RT
RT
RT
where Z is compressibility factor.
(iii) When temperature is extremely high : For one mole of gas,
PV  RT .
(iv) When pressure is low : For one mole of gas,
a ab
a 

 P  2  (V  b)  RT or PV  RT  Pb   2
V V
V 

PV
a
a
or Z  1 
1
RT
VRT
VRT
(v) For hydrogen : Molecular mass of hydrogen is small hence value
of 'a' will be small owing to smaller intermolecular force. Thus the terms
ab
a
and 2 may be ignored. Then Vander Waal's equation becomes,
V
V
or
PV  RT  Pb or
PV
Pb
1
RT
RT
Pb
RT
In case of hydrogen, compressibility factor is always greater than
or Z  1 
one.
(5) Merits of Vander Waal's equation
(i) The Vander Waal's equation holds good for real gases upto
moderately high pressures.
(ii) The equation represents the trend of the isotherms representing
the variation of PV with P for various gases.
(iii) From the Vander Waal's equation it is possible to obtain
expressions of Boyle's temperature, critical constants and inversion
temperature in terms of the Vander Waal's constants 'a' and 'b'.
(iv) Vander Waal's equation is useful in obtaining a 'reduced
equation of state' which being a general equation of state has the advantage
that a single curve can be obtained for all gases when the equation if
graphically represented by plotting the variables.
(6) Limitations of Vander Waal's equation
(i) This equation shows appreciable deviations at too low
temperatures or too high pressures.
(ii) The values of Vander Waal's constants a and b do not remain
constant over the entire ranges of T and P, hence this equation is valid only
over specific range of T and P.
(7) Other equations of state : In addition to Vander Waal's equation,
there are also equations of state which have been used to explain real
behaviour of gases are,


a
(V  b)  RT . Here 'c' is
(i) Clausius equation :  P 
2 
T
(
V

c
)


another constant besides a, b and R.
a 

(ii) Berthelot equation :  P 
 (V  b)  RT .
TV 2 

RT
a
c


(iii) Wohl equation : P 
(V  b) V (V  b) V 2
RT
.e a / RTV .
(iv) Dieterici equation : P 
V b
The expression is derived on the basis of the concept that molecules
near the wall will have higher potential energy than those in the bulk.
(v) Kammerlingh Onnes equation : It is the most general or
satisfactory expression as equation expresses PV as a power series of P at a
given temperature.
PV  A  BP  CP 2  DP3  ......
Here coefficients A, B, C etc. are known as first, second and third
etc. virial coefficients.
(a) Virial coefficients are different for different gases.
(b) At very low pressure, first virial coefficient, A = RT.
(c) At high pressure, other virial coefficients also become important
and must be considered.
The critical state
(1) A state for every substance at which the vapour and liquid states
are indistinguishable is known as critical state. It is defined by critical
temperature and critical pressure.
(2) Critical temperature (T ) of a gas is that temperature above
which the gas cannot be liquified however large pressure is applied. It is
8a
given by, Tc 
27 Rb
(3) Critical pressure (P ) is the minimum pressure which must be
applied to a gas to liquify it at its critical temperature. It is given by,
a
Pc 
27b 2
(4) Critical volume (V ) is the volume occupied by one mole of the
substance at its critical temperature and critical pressure. It is given by,
Vc  3b
c
c
c
(5) Critical compressibility
Pc Vc 3
Zc 
  0 .375
RTc
8
factor
(Z )
c
is
given
by,
A gas behaves as a Vander Waal’s gas if its critical compressibility
factor (Z c ) is equal to 0.375. A substance is the gaseous state below Tc is
called vapour and above Tc is called gas.
Degrees of freedom of a gaseous molecule
(1) The motion of atoms and molecules is generally described in
terms of the degree of freedom which they possess.
(2) The degrees of freedom of a molecule are defined as the
independent number of parameters required to describe the state of the
molecule completely.
(3) When a gaseous molecule is heated, the energy supplied to it
may bring about three kinds of motion in it, these are,
(i) The translational motion
(ii) The rotational motion
(iii) The vibrational motion.
This is expressed by saying that the molecule possesses translational,
rotational and vibrational degrees of freedom.
(4) For a molecule made up of N atoms, total degrees of freedom =
3N. Further split up of these is as follows,
Translational
Rotational
Vibrational
For linear molecule :
3
2
3N – 5
For non-linear molecule : 3
3
3N – 6
Specific and Molar heat capacity of gases
(1) Specific heat (or specific heat capacity) of a substance is the
quantity of heat (in calories, joules, kcal, or kilo joules) required to raise the
temperature of 1g of that substance through 1 o C . It can be measured at
constant pressure (c p ) and at constant volume (c v ) .
(2) Molar heat capacity of a substance is the quantity of heat
required to raise the temperature of 1 mole of the substance by 1 o C .
Gaseous State
 Molar heat capacity = Specific heat capacity  Molecular weight,
i.e.,
Cv  cv  M and C p  c p  M .
(3) Since gases upon heating show considerable tendency towards
expansion if heated under constant pressure conditions, an additional energy
has to be supplied for raising its temperature by 1 o C relative to that
required under constant volume conditions, i.e.,
C p  Cv or C p  Cv  Work done on expansion, PV( R)
where, C p  molar heat capacity at constant pressure; Cv 
molar heat capacity at constant volume.
(4) Some useful relations of C and C
(i) C p  Cv  R  2 calories 8.314 J
p
v
3
3
R (for monoatomic gas) and C v   x (for di and
2
2
polyatomic gas), where x varies from gas to gas.
Cp
(iii)
  (Ratio of molar capacities)
Cv
(ii) C v 
(iv)
For
monoatomic
gas
Cv  3 calories
whereas,
C p  Cv  R  5 calories
5
R
Cp
 2  1 .66 .
(v) For monoatomic gas, ( ) 
3
Cv
R
2
7
R
Cp
 2  1 .40
(vi) For diatomic gas ( ) 
5
Cv
R
2
Cp 8R
(vii) For triatomic gas ( ) 

 1 .33
Cv
6R
233
(iii) Liquid chlorine is used for bleaching and disinfectant purposes.
(iv) Liquid air is an important source of oxygen in rockets and jetpropelled planes and bombs.
(v) Compressed oxygen is used for welding purposes.
(vi) Compressed helium is used in airships.
(5) Joule-Thomson effect : When a real gas is allowed to expand
adiabatically through a porous plug or a fine hole into a region of low
pressure, it is accompanied by cooling (except for hydrogen and helium
which get warmed up).
Cooling takes place because some work is done to overcome the
intermolecular forces of attraction. As a result, the internal energy decreases
and so does the temperature.
Ideal gases do not show any cooling or heating because there are no
intermolecular forces of attraction i.e., they do not show Joule-Thomson
effect.
During Joule-Thomson effect, enthalpy of the system remains
constant.
Joule-Thomson coefficient.   (T / P)H .
For cooling,   ve (because dT and dP will be ve )
For heating   ve (because dT  ve , dP  ve ) .
For no heating or cooling   0 (because dT  0) .
(6) Inversion temperature : It is the temperature at which gas shows
neither cooling effect nor heating effect i.e., Joule-Thomson coefficient
  0 . Below this temperature, it shows cooling effect and above this
temperature, it shows heating effect.
Any gas like H 2 , He etc, whose inversion temperature is low
would show heating effect at room temperature. However, if these gases are
just cooled below inversion temperature and then subjected to JouleThomson effect, they will also undergo cooling.
Liquefaction of gases
(1) A gas may be liquefied by cooling or by the application of high
pressure or by the combined effect of both. The first successful attempt for
liquefying gases was made by Faraday.
(2) Gases for which the intermolecular forces of attraction are small
such as H 2 , N 2 , Ar and O 2 , have low values of Tc and cannot be
liquefied by the application of pressure are known as “permanent gases”
while the gases for which the intermolecular forces of attraction are large,
such as polar molecules NH 3 , SO 2 and H 2 O have high values of Tc
and can be liquefied easily.
(3) Methods of liquefaction of gases : The modern methods of cooling
the gas to or below their Tc and hence of liquefaction of gases are done by
Linde's method and Claude's method.
(i) Linde's method : This process is based upon Joule-Thomson
effect which states that “When a gas is allowed to expand adiabatically from
a region of high pressure to a region of extremely low pressure, it is
accompained by cooling.”
(ii) Claude's method : This process is based upon the principle that
when a gas expands adiabatically against an external pressure (as a piston in
an engine), it does some external work. Since work is done by the molecules
at the cost of their kinetic energy, the temperature of the gas falls causing
cooling.
(iii) By adiabatic demagnetisation.
(4) Uses of liquefied gases : Liquefied and gases compressed under a high
pressure are of great importance in industries.
(i) Liquid ammonia and liquid sulphur dioxide are used as
refrigerants.
(ii) Liquid carbon dioxide finds use in soda fountains.
 If the number of molecules present in 1 c.c. of the gas or vapour
at S.T.P., then that is called loschmidt number. Its value is 2.687
 10 per c.c.
19
 CO > SO > SO > PCl is order of rate of diffusion.
 Vapour density is independent of temperature and has no unit
2
2
3
3
while absolute density is dependent of temperature and has unit
of gm
–1
 The isotherms of CO were first studied by Andrews.
 1 Cal = 4.2 Joule, 1 Kcal = 4200 Joule
2
 The gas which has least mean free path has maximum value of a, is
easily liquefied and has maximum value of T .
b
 T<T <T
c
b
i
 For critical constants, compression factor Z is < 1.
 Boyle’s law and Avogadro’s law are applicable under limiting
condition. This limiting condition is P  0.
 T = 0.296 T ; T = 6.75 T
c
b
i
c
234 Gaseous State
 Mean free path increases if H is replaced by He.
2
Gaseous State
Characteristics and Measurable properties of gases
235
12.
(a) 1.5 lit.
(b) 2.8 lit.
(c) 11.2 lit.
(d) 22.4 lit.
Pressure of a gas in a vessel can be measured by
(a) Barometer
(b) Manometer
(c) Stalgometer
(d) All the baove
13.
Volume occupied by a gas at one atmospheric pressure and 0 o C is
V mL. Its volume at 273 K will be
[Bihar MADT 1982]
1.
2.
3.
4.
5.
6.
7.
8.
Which one of the following statements is not correct about the three
states of matter i.e. solid, liquid and gaseous
(a) Molecules of a solid possess least energy whereas those of a
gas possess highest energy
(b) The density of solid is highest whereas that of gases is lowest
(c) Gases like liquids possess definite volumes
(d) Molecules of a solid possess vibratory motion
The temperature and pressure at which ice, liquid water and water
vapour can exist together are
(a)
0 o C, 1 atm
(b)
2o C, 4.7 atm
(c)
0 o C, 4.7 mm
(d)
 2o C, 4.7 mm
Which of the following is true about gaseous state
(a) Thermal energy = Molecular attraction
(b) Thermal energy >> Molecular attraction
(c) Thermal energy << Molecular attraction
(d) Molecular forces >> Those in liquids
Kinetic energy of molecules is highest in
(a) Gases
(b) Solids
(c) Liquids
(d) Solutions
Which of the following statement is correct
(a) In all the three states the molecules possess random
translational motion
(b) Gases cannot be converted into solids without passing through
liquid state
(c) One of the common property of liquids and gases is viscosity
(d) According to Boyle's law V/P is constant at constant T
A volume of 1 m
(a)
1000 cm
(c)
10 dm 3
3
11.
15.
16.
100 cm 3
(d)
10 6 cm 3
(b)
9 / 5o F
H 2O
N 2 is found in a litre flask under 100kPa pressure and O 2 is
[Kerala PMT 2004]
(a) 310 kPa
(c) 420 kPa
(e) 265 kPa
(b) 210 kPa
(d) 365 kPa
Ideal gas equation and Related gas laws
1.
If P, V, T represent pressure, volume and temperature of the gas,
the correct representation of Boyle's law is
[BIT Ranchi 1988]
(a)
(b)
(d)
He
found in another 3 litre flask under 320 kPa pressure. If the two
flasks are connected, the resultant pressures is
1
V
(at constant P)
T
(b)
PV  RT
2.
(c) V  1 / P (at constant T) (d) PV  nRT
At constant temperature, in a given mass of an ideal gas
3.
(a) The ratio of pressure and volume always remains constant
(b) Volume always remains constant
(c) Pressure always remains constant
(d) The product of pressure and volume always remains constant
Air at sea level is dense. This is a practical application of
[CBSE PMT 1991]
1o C rise in temperature is equal to a rise of
1o F
(a) Gases do not have a definite shape and volume
(b) Volume of the gas is equal to the volume of the container
confining the gas
(c) Confined gas exerts uniform pressure on the walls of its
container in all directions
(d) Mass of the gas cannot be determined by weighing a container
in which it is enclosed
Which of the following exhibits the weakest intermolecular forces [AIIMS 2000]
(a) NH 3
(b) HCl
(c)
is equal to
3
o
10.
[CBSE PMT 1999]
Which one of the following is not a unit of pressure
(a) Newton
(b) Torr
(c) Pascal
(d) Bar
(a)
9.
14.
(a) V ml
(b) V/2 ml
(c) 2 V
(d) None of these
Which one of the following statements is wrong for gases
[Kerala CEE 2000]
o
(c) 5 / 9 F
(d) 33 F
Which of the following relations for expressing volume of a sample
is not correct
(a)
1L  10 3 ml
(b)
1 dm 3  1 L
(c)
1 L  10 3 m 3
(d)
1L  10 3 cm 3
(a) Boyle's law
(c) Avogadro's law
4.
(a)
10 6 dynes cm 2
(b)
10 2 dynes cm 2
(c)
10 4 dynes cm 2
(d)
10 8 dynes cm 2
o
If 20 cm 3 gas at 1 atm. is expanded to 50 cm 3 at constant T, then
what is the final pressure
[CPMT 1988]
(a)
One atmosphere is numerically equal to approximately
2gm of O 2 at 27 C and 760mm of Hg pressure has volume[BCECE 2005]
20 
1
50
(b) 50 
1
20
1
 50
(d) None of these
20
Which of the following statement is false
[BHU 1994]
(a) The product of pressure and volume of fixed amount of a gas
is independent of temperature
(c)
5.
(b) Charle's law
(d) Dalton's law
1
236 Gaseous state
6.
(b) Molecules of different gases have the same K.E. at a given
temperature
(c) The gas equation is not valid at high pressure and low
temperature
(d) The gas constant per molecule is known as Boltzmann constant
Which of the following graphs represent Boyle's law
(a)
(b)
P
PV
8.
9.
17.
(d)
V
P
Densities of two gases are in the ratio 1 : 2 and their
temperatures
are in the ratio 2 : 1, then the ratio of their respective pressures is[BHU 2000]
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 4 : 1
18.
At constant pressure, the volume of fixed mass of an ideal gas is
directly proportional to
[EAMCET 1985]
(a) Absolute temperature
(b) Degree centigrade
(c) Degree Fahrenheit
(d) None
Which of the following expression at constant pressure represents
Charle's law
[AFMC 1990]
(a)
V
1
T
(b)
V
19.
[Manipal PMT 1991]
1
T
(a) Avogadro number  6.02  10
(b) The relationship between average velocity (v ) and root mean
square velocity (u) is v  0.9213 u
2
21
(d) V  d
V T
Use of hot air balloons in sports and meteorological obsevations is
an application of
[Kerala MEE 2002]
(a) Boyle's law
(b) Newtonic law
(c) Kelvin's law
(d) Charle's law
(c)
10.
11.
12.
A 10 g of a gas at atmospheric pressure is cooled from 273o C to
0 o C keeping the volume constant, its pressure would become
(a) 1/2 atm
(b) 1/273 atm
(c) 2 atm
(d) 273 atm
Pressure remaining the same, the volume of a given mass of an ideal
gas increases for every degree centigrade rise in temperature by
definite fraction of its volume at
(c) The mean kinetic energy of an ideal gas is independent of the
pressure of the gas
(d) The root mean square velocity of the gas can be calculated by
20.
(a) 0 o C
(b) Its critical temperature
(c) Absolute zero
(d) Its Boyle temperature
A certain sample of gas has a volume of 0.2 litre measured at 1 atm.
pressure and 0 o C . At the same pressure but at 273o C , its
volume will be
[EAMCET 1992, 93; BHU 2005]
(a) 0.4 litres
(b) 0.8 litres
(c) 27.8 litres
(d) 55.6 litres
14.
15.
the formula (3 RT / M )1 / 2
The compressibility of a gas is less than unity at STP. Therefore [IIT 2000]
(a) Vm  22.4 litres
(b) Vm  22.4 litres
(c)
Vm  22.4 litres
(d)
In the equation of sate of an ideal gas PV  nRT , the value of the
universal gas constant would depend only on
22.
(a) The nature of the gas
(b) The pressure of the gas
(c) The units of the measurement
(d) None of these
In the ideal gas equation, the gas constant R has the dimensions of[NCERT 1982
(a) mole-atm K
(b) litre mole
(c) litre-atm K mole
(d) erg K
In the equation PV  nRT , which one cannot be the numerical
value of R
[BIT 1987]
[KCET 2005]
–1
–1
23.
–1
400 cm 3 of oxygen at 27 o C were cooled to  3 o C without
change in pressure. The contraction in volume will be
(a)
(a) 40 cm 3
(b) 8.31  107 dyne cm K 1mol 1
(b) 30 cm 3
(c) 44.4 cm 3
(d) 360 cm 3
The pressure p of a gas is plotted against its absolute temperature T
for two different constant volumes, V1 and V2 . When V1  V2 ,
the
(a) Curves have the same slope and do not intersect
(b) Curves must intersect at some point other than T  0
(c) Curve for V2 has a greater slope than that for V1
Vm  44.8 litres
21.
[CBSE PMT 1989]
13.
2 P1T2
2 P1
(d)
T1  T2
T1  T2
“One gram molecule of a gas at N.T.P. occupies 22.4 litres.” This fact
was derived from
[CPMT 1981, 1995]
(a) Dalton's theory
(b) Avogadro's hypothesis
(c) Berzelius hypothesis
(d) Law of gaseous volume
In a closed flask of 5 litres, 1.0 g of H 2 is heated from 300 to 600
K. which statement is not correct
[CBSE PMT 1991]
(a) Pressure of the gas increases
(b) The rate of collision increases
(c) The number of moles of gas increases
(d) The energy of gaseous molecules increases
Which one of the following statements is false
(c)
PV
(c)
Two closed vessels of equal volume containing air at pressure P1
and temperature T1 are connected to each other through a narrow
tube. If the temperature in one of the vessels is now maintained at
T1 and that in the other at T2 , what will be the pressure in the
vessels
2 P1T1
T1
(a)
(b)
T1  T2
2 P1T2
PV
P
V
7.
(d) Curve for V1 has a greater slope than that for V2
16.
(c)
24.
8.31  107 erg K 1mol 1
8.31 JK 1mol 1
(d) 8.31 atm. K 1mol 1
Which one of the following indicates the value of the gas constant R[EAMCET 19
(a) 1.987 cal K mol
(b) 8.3 cal K mol
(c) 0.0821 lit K mol
(d) 1.987 Joules K mol
The constant R is
[Orissa 1990]
(a) Work done per molecule
–1
–1
25.
–1
–1
–1
–1
–1
–1
–1
Gaseous State
26.
27.
(b) Work done per degree absolute
37.
(c) Work done per degree per mole
(d) Work done per mole
Select one correct statement. In the gas equation, PV  nRT [CBSE PMT 1992]
(a) n is the number of molecules of a gas
38.
(b) V denotes volume of one mole of the gas
(c) n moles of the gas have a volume V
(d) P is the pressure of the gas when only one mole of gas is
present
39.
The correct value of the gas constant R is close to
[CBSE PMT 1992]
(c) 0.082 litre- atmosphere1 K mole 1
40.
(d) 0.082 litre1 atmosphere1 K mol
–1
29.
30.
–1
–1
[CPMT 1994]
–1
41.
Gas equation PV  nRT is obeyed by
[BHU 2000]
(a) Only isothermal process
(b) Only adiabatic process
(c) Both (a) and (b)
(d) None of these
For an ideal gas number of moles per litre in terms of its pressure P,
gas constant R and temperature T is
(a) PT/R
(b) PRT
(c) P/RT
(d) RT/P
If two moles of an ideal gas at 546 K occupy a volume of 44.8 litres,
the pressure must be
[NCERT 1981; JIPMER 1991]
(a) 2 atm
(c) 4 atm
32.
33.
[CBSE PMT 1991]
42.
43.
35.
(b) 88.9 cc
(d) 100 cc
44.
45.
P1V1
T
 1
P2 V2 T2
(c)
36.
[BHU 1994]
(a) 2
(c) 4
(b) 1/2
(d) 1/4
o
21.6 C
(b)
o
240 C
(c)  33 C
(d) 89.5 o C
The total pressure exerted by a number of non-reacting gases is
equal to the sum of the partial pressures of the gases under the
same conditions is known as
[CPMT 1986]
(a) Boyle's law
(b) Charle's law
(c) Avogadro's law
(d) Dalton's law
“Equal volumes of all gases at the same temperature and pressure
contain equal number of particles.” This statement is a direct
consequence of
[Kerala MEE 2002]
(a) Avogadro’s law
(b) Charle's law
(c) Ideal gas equation
(d) Law of partial pressure
If three unreactive gases having partial pressures PA , PB and PC
and their moles are 1, 2 and 3 respectively then their total pressure
will be
[CPMT 1994]
PA  PB  PC
(a) P  PA  PB  PC
(b) P 
6
o
46.
P1T2 P2 V2
V1V2

 P1 P2
(d)
V1
T2
T1T2
Two separate bulbs contain ideal gases A and B. The density of gas
A is twice that of gas B. The molecular mass of A is half that of gas
B. The two gases are at the same temperature. The ratio of the
pressure of A to that of gas B is
A sample of gas occupies 100 ml at 27 o C and 740 mm pressure.
When its volume is changed to 80 ml at 740 mm pressure, the
temperature of the gas will be
(a)
[CBSE PMT 1989; CPMT 1991]
(b)
(b) 30 o C
(d) 300 K
[Vellore CMC 1991]
[DCE 2000]
(a) 80 cc
(c) 66.7 cc
Correct gas equation is
V1T2 V2 T1

(a)
P1
P2
(b) 20000 litres
(d) 12000 litres
The density of a gas at 27 o C and 1 atm is d. Pressure remaining
constant at which of the following temperatures will its density
become 0.75 d
[CBSE PMT 1992]
(a) 20 o C
(c) 400 K
(b) 11.2 litres
(d) 5.6 litres
At 0 o C and one atm pressure, a gas occupies 100 cc. If the
pressure is increased to one and a half-time and temperature is
increased by one-third of absolute temperature, then final volume of
the gas will be
A wheather balloon filled with hydrogen at 1 atm and 27 o C has
volume equal to 12000 litres. On ascending it reaches a place where
(a) 24000 litres
(c) 10000 litres
[EAMCET 1992]
34.
(b) 600 K
(d) 900 K
the temperature is  23o C and pressure is 0.5 atm. The volume of
the balloon is
(b) 3 atm
(d) 1 atm
How many moles of He gas occupy 22.4 litres at 30 o C and one
atmospheric pressure
[KCET 1992]
(a) 0.90
(b) 1.11
(c) 0.11
(d) 1.0
Volume of 0.5 mole of a gas at 1 atm. pressure and 273 K is
(a) 22.4 litres
(c) 44.8 litres
(a) 391.8 mL
(b) 380 ml
(c) 691.6 ml
(d) 750 ml
One litre of a gas weighs 2 g at 300 K and 1 atm pressure. If the
pressure is made 0.75 atm, at which of the following temperatures
will one litre of the same gas weigh one gram
(a) 450 K
(c) 800 K
[AIEEE 2002]
31.
20 o C and 2 atm pressure. The mass of the gas will be [CPMT 1989]
(a) 34 g
(b) 340 g
(c) 282.4 g
(d) 28.24 g
At N.T.P. the volume of a gas is found to be 273 ml. What will be
[CBSE PMT 1992]
S.I. unit of gas constant R is
(a) 0.0821 litre atm K mole
(b) 2 calories K mole
(c) 8.31 joule K mole
(d) None
–1
pressure and 0 o C . The total volume occupied by the mixture will
be nearly
[Vellore CMC 1991]
(a) 22.4 litres
(b) 33.6 litres
(c) 448 litres
(d) 44800 ml
Pure hydrogen sulphide is stored in a tank of 100 litre capacity at
[CPMT 1992]
(b) 0.082 litre-atmosphere K 1 mol 1
–1
16 g of oxygen and 3 g of hydrogen are mixed and kept at 760 mm
the volume of this gas at 600 mm Hg and 273o C
(a) 0.082 litre-atmopshere K
28.
237
47.
PA  PB  PC
(c)
P
(c)
H 2 and CO 2
(d) None
3
Dalton's law of partial pressure will not apply to which of the
following mixture of gases
[Bihar MADT 1981]
(a) H 2 and SO 2
(b) H 2 and Cl 2
(d) CO 2 and Cl 2
238 Gaseous state
48.
Which of the following mixtures of gases does not obey Dalton's law
of partial pressure
[CBSE PMT 1996: Kerala PMT 2000]
(a)
49.
50.
O 2 and CO 2
(b)
60.
N 2 and O 2
(c) Cl 2 and O 2
(d) NH 3 and HCl
To which of the following gaseous mixtures is Dalton's law not
applicable
(a) Ne  He  SO 2
(b) NH 3  HCl  HBr
(c) O2  N 2  CO 2
(d) N 2  H 2  O2
Equal amounts of two gases of molecular weight 4 and 40 are
mixed. The pressure of the mixture is 1.1 atm. The partial pressure of
the light gas in this mixture is
61.
62.
(a) 4
(b) 1/4
(c) 16
(d) 1/16
If rate of diffusion of A is 5 times that of B, what will be the density
ratio of A and B
[AFMC 1994]
(a) 1/25
(b) 1/5
(c) 25
(d) 4
The densities of two gases are in the ratio of 1 : 16. The ratio of their
rates of diffusion is
[CPMT 1995]
(a) 16 : 1
(b) 4 : 1
(c) 1 : 4
(d) 1 : 16
At constant volume and temperature conditions, the rate of diffusion
D A and DB of gases A and B having densities  A and  B are
related by the expression
[IIT 1993]
[CBSE PMT 1991]
51.
52.
(a) 0.55 atm
(b) 0.11 atm
(c) 1 atm
(d) 0.12 atm
Rate of diffusion of a gas is
[IIT 1985; CPMT 1987]
(a) Directly proportional to its density
(b) Directly proportional to its molecular mass
(c) Directly proportional to the square root of its molecular mass
(d) Inversely proportional to the square root of its molecular mass
Which of the following gas will have highest rate of diffusion
1/2
(a)
1/2
53.
NH 3
(b)
N2
(c)
CO 2
(d)
O2
63.
54.
r  1/d
(b) r  d
(c) r  d
(d) r  d
According to Grahman's law at a given temperature, the ratio of the
rates of diffusion rA / rB of gases A and B is given by
[IIT 1998]
(a)
56.
57.
58.
59.
65.
(PA / PB )(M A / M B )1 / 2

 
DA   DB  B 

A

1/2
 
 
DA  DB  A 
(d) DA  DB  B 
 B 
 A 
Atmolysis is a process of
(a) Atomising gas molecules
(b) The breaking of atoms to sub-atomic particles
(c) Separation of gases from their gaseous mixture
(d) Changing of liquids to their vapour state
A bottle of ammonia and a bottle of dry hydrogen chloride
connected through a long tube are opened simultaneously at both
ends, the white ammonium chloride ring first formed will be[IIT 1988]
(a) At the centre of the tube
(b) Near the hydrogen chloride bottle
(c) Near the ammonia bottle
(d) Throughout the length of the tube
Which of the following pairs will diffuse at the same rate through a
porous plug
[EAMCET 1990]
(a) CO, NO 2
(b) NO 2 , CO 2
(d) NO C2 H 6
NH 3 , PH3
If 4 g of oxygen diffuse through a very narrow hole, how much
hydrogen would have diffused under identical conditions [CPMT 1971]
(c) (PA / PB )(M B / M A )1 / 2
(a) 16 g
(b) 1 g
(c) 1/4 g
(d) 64 g
1/ 2
(d) (M A / M B )(PB / PA )
67.
A gas diffuse at a rate which is twice that of another gas B. The
(where P and M are the pressures and molecular weights of gases A
ratio of molecular weights of A to B is
[EAMCET 1986]
and B respectively)
(a) 1.0
(b) 0.75
The ratio of the rate of diffusion of a given element to that of
(c) 0.50
(d) 0.25
helium is 1.4. The molecular weight of the element is
68.
Two grams of hydrogen diffuse from a container in 10 minutes. How
[Kerala PMT 1990]
many grams of oxygen would diffuse through the same container in
the same time under similar conditions
[MNR 1980]
(a) 2
(b) 4
(a) 0.5 g
(b) 4 g
(c) 8
(d) 16
g
(d) 8 g
A gas diffuse 1/5 times as fast as hydrogen. Its molecular weight is[CPMT 1992; Bihar(c)CEE61982]
69.
The
rate
of
diffusion
of
methane
at
a given temperature is twice that
(a) 50
(b) 25
of X. The molecular weight of X is
[MNR 1995; Kerala CEE 2001]
(c) 25 2
(d) 50 2
(a) 64.0
(b) 32.0
The molecular weight of a gas which diffuses through a porous plug
(c) 40.0
(d) 80.0
at 1/6th of the speed of hydrogen under identical conditions is[EAMCET 1990]
70.
X ml of H 2 gas effuses through a hole in a container in 5 seconds.
(a) 27
(b) 72
The time taken for the effusion of the same volume of the gas
(c) 36
(d) 48
specified below under identical condition is
Molecular weight of a gas that diffuses twice as rapidly as the gas
[IIT 1996]
with molecular weight 64 is
[EAMCET 1994]
(a)
10
seconds
:
He
(b)
20
seconds
:
O
2
(a) 16
(b) 8
(c) 64
(d) 6.4
(c) 25 seconds : CO
(d) 55 seconds : CO 2
(b) (M A / M B )(PA / PB )1 / 2
55.
64.
Which of the following relationship is correct, where r is the rate of
diffusion of a gas and d is its density
[CPMT 1994]
(a)
1/2
(b)
(c)
[Pb. CET Sample paper 1993; CPMT 1990]
(a)

 
DA   DB  A 
B 

The densities of hydrogen and oxygen are 0.09 and 1.44 g L1 . If
the rate of diffusion of hydrogen is 1 then that of oxygen in the same
units will be
[RPMT 1994]
(c)
66.
71.
At what temperature, the rate of effusion of N 2 would be 1.625
times that of SO 2 at 50 o C
[CBSE PMT 1996]
Gaseous State
72.
73.
(a) 110 K
(b) 173 K
83.
A closed vessel contains equal number of nitrogen and oxygen
molecules at a pressure of P mm. If nitrogen is removed from the
(c) 373 K
(d) 273 K
system then the pressure will be
Given the reaction C(s)  H 2O(l)  CO(g)  H 2 (g) calculate the
[MP PMT 1985]
volume of the gases produced at STP from 48.0 g of carbon
(a) P
(b) 2P
(a) 179.2 L
(b) 89.6 L
(c) P / 2
(d) P 2
(c) 44.8 L
(d) 22.4 L
4.4 g of a gas at STP occupies a volume of 2.24 L, the gas can be[Haryana CEET
84. 2000]
If the four tubes of a car are filled to the same pressure with
N 2 , O2 , H 2 and Ne separately, then which one will be filled first[Manipal PMT
(a) O 2
(b) CO
(c)
74.
NO 2
(d)
CO 2
Under what conditions will a pure sample of an ideal gas not only
exhibit a pressure of 1 atm but also a concentration of 1 mole
litre 1
(R  0.082 litreatm mol 1 deg 1 )
(a) At STP
(b) When V  22.4 litres
76.
77.
78.
85.
[CBSE PMT 1993]
86.
There are 6.02  10 22 molecules each of N 2 , O2 and H 2 which
are mixed together at 760 mm and 273 K. The mass of the mixture
in grams is
[Pb. PMT 1997]
(a) 6.2
(b) 4.12
(c) 3.09
(d) 7
Volume of 4.4 g of CO 2 at NTP is
[Pb. CET 1997]
(a) 22.4 L
(b) 44.8 L
(c) 2.24 L
(d) 4.48 L
The energy of an ideal gas depends only on its
(a) Pressure
(b) Volume
(c) Number of moles
(d) Temperature
A bottle of cold drink contains 200 ml liquid in which CO 2 is 0.1
79.
80.
H2
(d) Ne
Which of the following gas mixture is not applicable for Dalton’s law
of partial pressure
[Pb. CET 2002]
(a) SO 2 and Cl 2
(b) CO 2 and N 2
(d) One fourth that of SO 2
Five grams each of the following gases at 87 o C and 750 mm
pressure are taken. Which of them will have the least volume[MNR 1991]
(a) HF
(b) HCl
(c) HBr
(d) HI
Who among the following scientists has not done any important
work on gases
[Bihar MADT 1980]
(a) Boyle
(b) Charles
(c) Avogadro
(d) Faraday
(d) CO and N 2
At what pressure a quantity of gas will occupy a volume of 60 ml ,
100 mm
(d) 1200 mm
At constant temperature and pressure which gas will diffuse first
[Pb. CET 2000]
H 2 or O 2 ?
89.
and H 2 is placed in a solution of sodium hydroxide, the solution
level will
[Pb. CET 2001]
(a) Rise
(b) Fall
(c) Remain constant
(d) Become zero
At S.T.P. 1g CaCO 3 on decomposition gives CO 2
[Pb. CET 2000]
[MNR 1982; CBSE PMT 1991]
90.
(a) 1 L
(b) 11.2 L
(c) 22.4 L
(d) 20 L
A pre-weighed vessel was filled with oxygen at N.T.P. and weighted.
91.
It was then evacuated, filled with SO 2 at the same temperature
and pressure, and again weighted. The weight of oxygen will be[NCERT 1989]
1
that of SO 2
2
CO and CO 2
88.
92.
(a) 22.4 litre
(b) 2.24 litre
(c) 0.224 litre
(d) 11.2 litre
At NTP, the density of a gas, whose molecular weight is 45 is
[Pb. CET 2001, 03]
(a) 44.8 gm/litre
(b) 11.4 gm/litre
(c) 2 gm/ litre
(d) 3 gm/litre
What is the ratio of diffusion rate of oxygen and hydrogen
[Pb. CET 2003]
(a) 1 : 4
(b) 4 : 1
(c) 1 : 8
(d) 8 : 1
The maximum number of molecules is present in
[CBSE PMT 2004]
(a) 0.5 g of H 2 gas
(c) Twice that of SO 2
82.
(c)
(a) Hydrogen
(b) Oxygen
(c) Both will diffuse in same time
(d) None of the above
When a jar containing gaseous mixture of equal volumes of CO 2
(a) The same as that of SO 2
81.
(b) O 2
(c)
87.
[CBSE PMT 1991]
(a) 0.224 litre
(b) 0.448 litre
(c) 22.4 litre
(d) 2.24 litre
The vapour density of a gas is 11.2. The volume occupied by 11.2 g of
this gas at N.T.P. is
(b)
N2
if it occupies a volume of 100ml at a pressure of 720mm ?
(while temperature is constant) :
[Pb. CET 2000]
(a) 700 mm
(b) 800 mm
molar. Suppose CO 2 behaves like an ideal gas, the volume of the
dissolved CO 2 at STP is
(a)
(c)
(c) When T  12 K
(d) Impossible under any conditions
75.
239
93.
94.
95.
(b) 10 g of O 2 gas
(c) 15 L of H 2 gas at STP
(d) 5 L of N 2 gas at STP
One litre oxygen gas at STP will weigh
[Pb. CET 2004]
(a) 1.43 g
(b) 2.24 g
(c) 11.2 g
(d) 22.4 g
How will you separate mixture of two gases
[AFMC 2004]
(a) Fractional distillation technique
(b) Grahams law of diffusion technique
(c) Osmosis
(d) Chromatography
The rate of diffusion of hydrogen gas is
[MH CET 2003; Pb. CET 2000]
(a) 1.4 times to He gas
(b) Same as He gas
240 Gaseous state
96.
97.
98.
99.
100.
101.
102.
(c) 5 times to He gas
(d) 2 times to He gas
Hydrogen diffuses six times faster than gas A . The molar mass of
gas A is
[KCET 2004]
(a) 72
(b) 6
(c) 24
(d) 36
At what pressure will a quantity of gas, which occupies 100 ml at
a pressue of 720 mm , occupy a volume of 84 ml
[DPMT 2004]
(a) 736.18 mm
(b) 820.20 mm
(c) 784.15 mm
(d) 857.14 mm
Containers A and B have same gases. Pressure, volume and
temperature of A are all twice that of B, then the ratio of
number of molecules of A and B are
[AFMC 2004]
(a) 1 : 2
(b) 2
(c) 1 : 4
(d) 4
A mixture of NO 2 and N 2 O4 has a vapour density of 38.3 at
300 K . What is the number of moles of NO 2 in 100 g of the
mixture
[Kerala PMT 2004]
(a) 0.043
(b) 4.4
(c) 3.4
(d) 3.86
(e) 0.437
A cylinder of 5 litres capacity, filled with air at NTP is connected
with another evacuated cylinder of 30 litres of capacity. The
resultant air pressure in both the cylinders will be
[BHU 2004]
(a) 10.8 cm of Hg
(b) 14.9cm of Hg
(c) 21.8 cm of Hg
(d) 38.8 cm of Hg
A certain mass of gas occupies a volume of 300 c.c. at 27 C and 620
mm pressure. The volume of this gas at 47 o C and 640 mm
pressure will be
[MH CET 2003]
(a) 400 c.c.
(b) 510 c.c.
(c) 310 c.c.
(d) 350 c.c.
What will be the volume of the mixture after the reaction?
NH 3  HCl  NH 4 Cl
[BVP 2004]
4 litre
1.5 litre
(solid)
103.
(a) 0.5 litre
(b) 1 litre
(c) 2.5 litre
(d) 0.1 litre
The pressure and temperature of 4dm 3 of carbon dioxide gas are
doubled. Then the volume of carbon dioxide gas would be[KCET 2004]
(a) 2 dm 3
(b) 3dm 3
104.
(c) 4 dm 3
(d) 8 dm 3
If the absolute temperature of an ideal gas become double and
pressure become half, the volume of gas would be
1.
2.
3.
[Bihar MADT 1980]
(a)
(b)
(c)
(d)
4.
5.
6.
7.
8.
9.
[Kerala CET 2005]
105.
(a) Remain unchange
(b) Will be double
(c) Will be four time
(d) will be half
(e) Will be one fourth
At what temperature, the sample of neon gas would be heated to
double of its pressure, if the initial volume of gas is/are reduced to
15% at 75 o C
[Kerala CET 2005]
(a)
(c)
106.
o
(b)
592 C
o
(d)
60 o C
319 C
128 C
(e) 90 o C
Equation of Boyle's law is
dP
dV
(a)

p
V
(c)
d2P
dV

P
dT
10.
[DPMT 2005]
dP
dV

P
V
(d)
d2P
d 2V

P
dT
Kinetic molecular theory of gases and
Molecular collisions
The average velocity of the molecules
The most probable velocity of the molecules
The square root of the average square velocity of the molecules
The most accurate form in which velocity can be used in these
calculations
Kinetic energy of a gas depends upon its[Bihar MADT 1982]
(a) Molecular mass
(b) Atomic mass
(c) Equivalent mass
(d) None of these
The kinetic theory of gases perdicts that total kinetic energy of a
gaseous assembly depends on
[NCERT 1984]
(a) Pressure of the gas
(b) Temperature of the gas
(c) Volume of the gas
(d) Pressure, volume and temperature of the gas
According to kinetic theory of gases, the energy per mole of a gas is
equal to
[EAMCET 1985]
(a) 1.5 RT
(b) RT
(c) 0.5 RT
(d) 2.5 RT
Internal energy and pressure of a gas per unit volume are related as[CBSE PMT
2
3
(a) P  E
(b) P  E
3
2
1
(c) P  E
(d) P  2 E
2
The translational kinetic energy of an ideal gas depends only on its
(a) Pressure
(b) Force
(c) Temperature
(d) Molar mass
Helium atom is two times heavier than a hydrogen molecule at 298
K, the average kinetic energy of helium is
[IIT 1982]
(a) Two times that of a hydrogen molecule
(b) Same as that of a hydrogen molecule
(c) Four times that of a hydrogen molecule
(d) Half that of a hydrogen molecule
Which of the following is valid at absolute zero
[Pb. CET 1985]
o
(b)
Postulate of kinetic theory is
[EAMCET 1980]
(a) Atom is indivisible
(b) Gases combine in a simple ratio
(c) There is no influence of gravity on the molecules of a gas
(d) None of the above
According to kinetic theory of gases,
[EAMCET 1980]
(a) There are intermolecular attractions
(b) Molecules have considerable volume
(c) No intermolecular attractions
(d) The velocity of molecules decreases after each collision
In deriving the kinetic gas equation, use is made of the root mean
square velocity of the molecules because it is
11.
12.
(a) Kinetic energy of the gas becomes zero but the molecular
motion does not become zero
(b) Kinetic energy of the gas becomes zero and molecular motion
also becomes zero
(c) Kinetic energy of the gas decreases but does not become zero
(d) None of the above
The average K.E. of an ideal gas in calories per mole is
approximately equal to
[EAMCET 1989]
(a) Three times the absolute temperature
(b) Absolute temperature
(c) Two times the absolute temperature
(d) 1.5 times the absolute temperature
According to kinetic theory of gases, for a diatomic molecule
[MNR 1991]
Gaseous State
13.
(a) The pressure exerted by the gas is proportional to the mean
velocity of the molecules
(b) The pressure exerted by the gas is proportional to the root
mean square velocity of the molecules
(c) The root mean square velocity is inversely proportional to the
temperature
(d) The mean translational kinetic energy of the molecules is
proportional to the absolute temperature
At STP, 0.50 mol H 2 gas and 1.0 mol He gas
[CBSE PMT 1993, 2000]
14.
(a) Have equal average kinetic energies
(b) Have equal molecular speeds
(c) Occupy equal volumes
(d) Have equal effusion rates
Which of the following expressions correctly represents the
21.
(a)
15.
(b)
KE CO  KE N 2
(c)
KE CO  KE N 2
22.
23.
25.
CO 2 (g) at 298 K and 1 atm pressure
(c)
26.
27.
With increase of pressure, the mean free path
[Pb. CET 1985]
19.
20.
(b)
P  2 atm, T  150 K
(c)
P  1 atm, T  300 K
(d)
P  1.0 atm, T  500 K
If the inversion temperature of a gas is  80 o C , then it will
produce cooling under Joule-Thomson effect at
[AFMC 1990]
[CBSE PMT 1990]
(b) 0 o C, 2 atm
273o C, 1 atm
(d)
273o C, 2 atm
Absolute zero is defined as the temperature
[CBSE PMT 1990]
(c) The density of N 2 is less than that of CO 2
(a) Decreases
(b) Increases
(c) Does not change
(d) Becomes zero
Which one of the following statements is NOT true about the effect
of an increase in temperature on the distribution of molecular
speeds in a gas
[AIEEE 2005]
(a) The most probable speed increases
(b) The fraction of the molecules with the most probable speed
increases
(c) The distribution becomes broader
(d) The area under the distribution curve remains the same as
under the lower temperature
If P, V, M, T and R are pressure, volume, molar mass, temperature
and gas constant respectively, then for an ideal gas, the density is
given by
[CBSE PMT 1989, 91]
RT
P
(a)
(b)
PM
RT
M
PM
(c)
(d)
V
RT
An ideal gas will have maximum density when
[CPMT 2000]
(a) P  0.5 atm, T  600 K
(b) 60 cm 3
(c) 0.6 cm 3
(d) 0.06 cm 3
The ratio  for inert gases is
(a) 1.33
(b) 1.66
(c) 2.13
(d) 1.99
The density of neon will be highest at
(a) S.T.P.
(b) The rms speed remains constant for both N 2 and CO 2
18.
At 100 o C and 1 atm, if the density of liquid water is 1.0 g cm 3
(a) 6 cm 3
and CO 2
17.
(c) 3.01  10
(d) 2.01  10 23
The density of air is 0.00130 g/ml. The vapour density of air will be[DCE 2000]
(a) 0.00065
(b) 0.65
(c) 14.4816
(d) 14.56
(d) Cannot be predicted unless the volumes of the gases are given
Indicate the correct statement for a 1-L sample of N 2 (g) and
(d) The total translational KE of both N 2 and CO 2 is the same
(b) 1.2  10
24
and that of water vapour is 0.0006 g m 3 , then the volume
occupied by water molecules in 1 litre of steam at that temperature
is
[IIT 2000]
24.
(a) The average translational KE per molecule is the same in N 2
16.
6.02  10
23
23
[CBSE PMT 2000]
KE CO  KE N 2
(a) 298 K
(b) 273 K
(c) 193 K
(d) 173 K
Ratio of C p and Cv of a gas 'X' is 1.4. The number of atoms of the
gas 'X' present in 11.2 litres of it at N.T.P. is
[CBSE PMT 1989]
relationship between the average molar kinetic energy, K.E. , of CO
and N 2 molecules at the same temperature
(a)
241
28.
29.
(a) At which all molecular motion ceases
(b) At which liquid helium boils
(c) At which ether boils
(d) All of the above
Consider the following statements :
(1) Joule-Thomson experiment is isoenthalpic as well as adiabatic.
(2) A negative value of  JT (Joule Thomson coefficient
corresponds to warming of a gas on expansion.
(3) The temperature at which neither cooling nor heating effect is
observed is known as inversion temperature.
Which of the above statements are correct
(a) 1 and 2
(b) 1 and 3
(c) 2 and 3
(d) 1, 2 and 3
Vibrational energy is
[Pb. CET 1985]
(a) Partially potential and partially kinetic
(b) Only potential
(c) Only kinetic
(d) None of the above
At the same temperature and pressure, which of the following gases
will have the highest kinetic energy per mole
[MNR 1991]
30.
(a) Hydrogen
(b) Oxygen
(c) Methane
(d) All the same
Dimensions of pressure are the same as that of
[CBSE PMT 1995]
31.
(a) Energy
(b) Force
(c) Energy per unit volume
(d) Force per unit volume
The density of a gas An is three times that of a gas B. if the
molecular mass of A is M, the molecular mass of B is
[CPMT 1987]
(a) 3 M
(b)
3M
(c)
(d)
M/ 3
M /3
Molecular speeds
242 Gaseous state
1.
2.
3.
The ratio of root mean square velocity to average velocity of gas
molecules at a particular temperature is
[IIT 1981]
(a) 1.086 : 1
(b) 1 : 1.086
(c) 2 : 1.086
(d) 1.086 : 2
Which is not true in case of an ideal gas
[CBSE PMT 1991]
(a) It cannot be converted into a liquid
(b) There is no interaction between the molecules
(c) All molecules of the gas move with same speed
(d) At a given temperature, PV is proportional to the amount of
the gas
The ratio among most probable velocity, mean velocity and root
mean square velocity is given by
[CBSE PMT 1993]
(a) 1 : 2 : 3
4.
5.
6.
(d)
O2
12.
(a)
13.
14.
15.
[EAMCET 1992]
7.
(b)
3/5
4/3
2/3
(c)
(d) 0.25
The average kinetic energy of an ideal gas per molecule in SI units at
25 o C will be
8.
(a)
6.17  10
(c)
6.17  10 20 J
(b)
kJ
6.17  10
10.
21
17.
J
(d) 7.16  10 20 J
[KCET 2001]
18.
(a) 273 K
(b) 606 K
(c) 303 K
(d) 403 K
Among the following gases which one has the lowest root mean
square velocity at 25 o C
(a) SO 2
(b)
N2
(c)
(d)
Cl 2
O2
[EAMCET 1983]
19.
The root mean square velocity of an ideal gas in a closed container
of
fixed
volume
is
increased
from
5  10 4 cm s 1
to
20.
10  10 4 cm s 1 . Which of the following statement correctly
explains how the change is accomplished
[Pb. CET 1986]
11.
(a) By heating the gas, the temperature is doubled
(b) By heating the gas, the pressure is quadrupled (i.e. made four
times)
(c) By heating the gas, the temperature is quadrupled
(d) By heating the gas, the pressure is doubled
The rms velocity at NTP of the species can be calculated from the
expression
[EAMCET 1990]
m1 / 2
(b) m 0
(c) m 1 / 2
(d) m
At constant volume, for a fixed number of moles of a gas, the
pressure of the gas increases with increase in temperature due to[IIT 1992]
(a) Increase in the average molecular speed
(b) Increased rate of collision amongst molecules
(c) Increase in molecular attraction
(d) Decrease in mean free path
Molecular velocities of the two gases at the same temperature are
u1 and u 2 . Their masses are m1 and m 2 respectively. Which of
the following expressions is correct
(a)
m1
(c)
m1 m 2

u1
u2
u12

m2
u 22
(b) m1u1  m 2u2
(d) m1u12  m 2u22
The temperature of the gas is raised from 27 o C to 927 o C , the
root mean square velocity is
[CBSE PMT 1994]
927 / 27 times the earlier value
(a)
(b) Same as before
(c) Halved
(d) Doubled
The ratio between the root mean square velocity of H at 50 K and
that of O 2 at 800 K is
[IIT 1996]
(a) 4
(b) 2
(c) 1
(d) 1/4
The root mean square velocity of an ideal gas at constant pressure
varies density (d) as
[IIT 2000]
2
(a)
At what temperature the RMS velocity of SO 2 be same as that of
O 2 at 303 K
9.
16.
[CBSE PMT 1996]
21
3 PV
M
[BHU 1994]
At 27 o C , the ratio of rms velocities of ozone to oxygen is
(a)
(b)
3 RT
(d) All the above
M
Root mean square velocity of a gas molecule is proportional to[CBSE PMT 1990]
H2
The temperature at which RMS velocity of SO 2 molecules is half
that of He molecules at 300 K is
[NTSE 1991]
(a) 150 K
(b) 600 K
(c) 900 K
(d) 1200 K
3P
d
(c)
(b) 1 : 2 : 3
2 : 3 : 8 /
2 : 8 / : 3
(c)
(d)
Which of the following has maximum root mean square velocity at
the same temperature
[Manipal PMT 2002]
(a) SO 2
(b) CO 2
(c)
(a)
d2
(b) d
(c)
(d) 1 / d
d
Consider a mixture of SO 2 and O 2 kept at room temperature.
Compared to the oxygen molecule, the SO 2 molecule will hit the
wall with
(a) Smaller average speed
(b) Greater average speed
(c) Greater kinetic energy
(d) Greater mass
The rms speed of N 2 molecules in a gas is u. If the temperature is
doubled and the nitrogen molecules dissociate into nitrogen atoms,
the rms speed becomes
(a) u / 2
(b) 2u
(c) 4u
(d) 14u
Choose the correct arrangement, where the symbols have their usual
meanings
(a) u  u p  urms
(b) urms  u  u p
21.
(c) u p  u  urms
(d) u p  urms  u
The ratio of most probable velocity to that of average velocity is [JEE Orissa 200
(a)  / 2
(b) 2 / 
22.
 /2
(c)
(d) 2 / 
The r.m.s. velocity of a certain gas is v at 300 K . The
temperature, at which the r.m.s. velocity becomes double
[Pb. CET 2002]
(a)
23.
1200 K
(b) 900 K
(c) 600 K
(d) 150 K
The r.m.s. velocity of a gas depends upon
(a) Temperature only
[DCE 2002]
Gaseous State
24.
25.
(b) Molecular mass only
(c) Temperature and molecular mass of gas
(d) None of these
What is the pressure of 2 mole of NH 3 at 27 o C when its
volume is 5 litre in vander Waal’s equation (a = 4.17, b = 0.03711)[JEE Orissa7.2004]
(a) 10.33 atm
(b) 9.33 atm
(c) 9.74 atm
(d) 9.2 atm
The root mean square velocity of one mole of a monoatomic having
molar mass M is U rms . The relation between the average kinetic
energy (E) of the U rms is
[IIT-JEE Screening 2004]
(a)
(b) U rms 
2E
(d) U rms 
M
Ratio of average to most probable velocity is
(c)
26.
U rms 
3E
2M
U rms 
(b) 100 o C and 2 atmospheric pressure
(c)
 100 o C and 5 atmospheric pressure
(d) 500 o C and 1 atmospheric pressure
The temperature at which the second virial coefficient of real gas is
zero is called
[AFMC 1993]
(a) Critical temperature
(b) Eutetic point
(c) Boiling point
(d) Boyle's temperature
When is deviation more in the behaviour of a gas from the ideal gas
equation PV  nRT
[DPMT 1981; NCERT 1982; CBSE PMT 1993]
E
3M
[Orissa JEE 2005]
27.
8.
2E
3M
243
(a) 1.128
(b) 1.224
(c) 1.0
(d) 1.112
If the vrms is 30R1 / 2 at 27 o C then calculate the molar mass of
gas in kilogram.
[DPMT 2005]
(a) 1
(b) 2
(c) 4
(d) 0.001
9.
10.
(a) At high temperature and low pressure
(b) At low temperature and high pressure
(c) At high temperature and high pressure
(d) At low temperature and low high pressure
Vander Waal's constants 'a' and 'b' are related with..... respectively[RPMT 1994]
(a) Attractive force and bond energy of molecules
(b) Volume and repulsive force of molecules
(c) Shape and repulsive forces of molecules
(d) Attractive force and volume of the molecules
Gas deviates from ideal gas nature because molecules
[CPMT 1996]
Real gases and Vander waal’s equation
1.
The Vander Waal's equation explains the behaviour of
[DPMT 1981]
2.
3.
(a) Ideal gases
(b) Real gases
11.
(c) Vapour
(d) Non-real gases
Gases deviate from the ideal gas behaviour because their molecules[NCERT 1981]
(a) Possess negligible volume
(b) Have forces of attraction between them
(c) Are polyatomic
(d) Are not attracted to one another
The compressibility factor of a gas is defined as Z  PV / RT . The
12.
compressibility factor of ideal gas is
[Pb. CET 1986]
4.
(a) 0
(b) Infinity
(c) 1
(d) –1
In Vander Waal's equation of state for a non-ideal gas, the term that
accounts for intermolecular forces is
13.
[CBSE PMT 1990; IIT 1988]
(a)
(b) (RT )1
(V  b)
a 

(d) RT
P  2 
V 

Vander Waal's equation of state is obeyed by real gases. For n moles
of a real gas, the expression will be
(c)
5.
14.
(a) Are colourless
(b) Attract each other
(c) Contain covalent bond
(d) Show Brownian movement
The Vander Waal's equation reduces itself to the ideal gas equation
at
[Kerala MEE 2001; CBSE PMT 2002]
(a) High pressure and low temperature
(b) Low pressure and low temperature
(c) Low pressure and high temperature
(d) High pressure and high temperature
The compressibility factor for an ideal gas is
[IIT 1997]
(a) 1.5
(b) 1.0
(c) 2.0
(d) 
When an ideal gas undergoes unrestrained expansion, no cooling
occurs because the molecules
[IIT 1984, 89]
(a) Are above the inversion temperature
(b) Exert no attractive force on each other
(c) Do work equal to loss in kinetic energy
(d) Collide without loss of energy
A gas is said to behave like an ideal gas when the relation
PV / T  constant . When do you expect a real gas to behave like
an ideal gas
[IIT 1999; CBSE PMT 1990; CPMT 1991]
[IIT 1992; Pb. CET 1986; DPMT 1986]
(a)
 P na 
  2
n V 
 V 

  RT
n b 
a 

(b)  P  2  (V  b)  nRT
V 

(c)
6.
15.
na 

 P  2  (nV  b)  nRT
V 


n 2a 
(d)  P  2  (V  nb)  nRT

V 

Any gas shows maximum deviation from ideal gas at
16.
[AFMC 1993; IIT 1981, 94]
[CPMT 1991]
(a)
0 o C and 1 atmospheric pressure
(a) When the temperature is low
(b) When both the temperature and pressure are low
(c) When both the temperature and pressure are high
(d) When the temperature is high and pressure is low
A real gas most closely approaches the behaviour of an ideal gas at[KCET 1992]
(a) 15 atm and 200 K
(b) 1 atm and 273 K
(c) 0.5 atm and 500 K
(d) 15 atm and 500 K
The temperature at which real gases obey the ideal gas laws over a
wide range of pressure is called
(a) Critical temperature
(b) Boyle temperature
(c) Inversion temperature
244 Gaseous state
17.
(d) Reduced temperature
At low pressure, the Vander Waal's equation is reduced to
(a)
pVm
ap
1
RT
RT
(b)
Z
pVm
b
1
p
RT
RT
pVm
a
1
RT
RT
At high temperature and low pressure, the Vander Waal's equation
is reduced to
(c)
18.
Z
pVm  RT
(a)


 p  a  (Vm )  RT
2

Vm 

(b)
pVm  RT
(c)
p(Vm  b)  RT
(d)
4.
Z
5.
(b) Its pressure is more than critical pressure Pc
(c) Its pressure is more than Pc at a temperature less than Tc
6.
20.
21.
When helium is allowed to expand into vacuum, heating effect is
observed. Its reason is that
[CPMT 1987]
(a) Helium is an ideal gas
(b) Helium is an inert gas
(c) The inversion temperature of helium is very low
(d) The boiling point of helium is the lowest among the elements
In van der Waal’s equation of state of the gas law, the constant ‘ b ’
is a measure of
[AIEEE 2004]
(a) Volume occupied by the molecules
(b) Intermolecular attraction
(c) Intermolecular repulsions
(d) Intermolecular collisions per unit volume
In which molecule the vander Waal’s force is likely to be the most
important in determining the m.pt. and b.pt.
[DPMT 2000]
(a)
22.
H2S
(b)
7.
8.
9.
[Orissa JEE 2005]
10.
11.
2.
Which set of conditions represents easiest way to liquefy a gas[NCERT 1983]
(a) Low temperature and high pressure
(b) High temperature and low pressure
(c) Low temperature and low pressure
(d) High temperature and high pressure
Adiabatic demagnetisation is a technique used for
3.
(a) Adiabatic expansion of a gas
(b) Production of low temperature
(c) Production of high temperature
(d) None
An ideal gas can't be liquefied because
O2
(b)
N2
(c) NH 3
(d) CH 4
A gas can be liquefied
[AFMC 2005]
(a) Above its critical temperature
(b) At its critical temperature
(c) Below its critical temperature
(d) At any temperature
Which of the following is correct for critical temperature
(a) It is the highest temperature at which liquid and vapour can
coexist
(b) Beyond the critical temperature, there is no distinction between
the two phases and a gas cannot be liquefied by compression
(c) At critical temperature (Tc ) the surface tension of the system
is zero
(d) At critical temperature the gas and the liquid phases have
different critical densities
A gas has a density of 2.68 g / L at stp. Identify the gas
(a)
Critical state and Liquefaction of gases
1.
–1–11–1–14545454
[IIT 1989]
(a)
Br2
(b) 152.51 atm
(d) 70.52 atm
–2
CH 4 are 1.3, 1.390, 4.170 and 2.253 L2 atm mol 2 respectively.
The gas which can be most easily liquefied is
(c) HCl
(d) CO
Pressure exerted by 1 mole of methane in a 0.25 litre container at
300K
using
vander
Waal's
equation
(given
1  2.253 atml 2 mol 2 , b  0.0428 litmol 1 ) is
(a) 82.82 atm
(c) 190.52 atm
(d) It cannot be liquefied at any value of P and T
The Vander Waal's parameters for gases W, X, Y and Z are
Gas
a (atm L mol )
b (L mol )
W
4.0
0.027
X
8.0
0.030
Y
6.0
0.032
Z
12.0
0.027
Which one of these gases has the highest critical temperature
(a) W
(b) X
(c) Y
(d) Z
The Vander Waal's constant 'a' for the gases O2 , N 2 , NH 3 and
2

a 
(d)  p  2  (Vm  b)  RT

Vm 

19.
(b) Its molecules are relatively smaller in size
(c) It solidifies before becoming a liquid
(d) Forces operative between its molecules are negligible
However great the pressure, a gas cannot be liquefied above its
(a) Boyle temperature
(b) Inversion temperature
(c) Critical temperature
(d) Room temperature
An ideal gas obeying kinetic theory of gases can be liquefied if[CBSE PMT 1995]
(a) Its temperature is more than critical temperature Tc
NO 2
(b) Kr
(c) COS
(d) SO 2
Weight of 112 ml of oxygen at NTP on liquefaction would be[DPMT 1984]
(a) 0.32 g
(b) 0.64 g
(c) 0.16 g
(d) 0.96 g
[BHU 1984]
1.
[CBSE PMT 1992]
o
(a) Its critical temperature is always above 0 C
As the temperature is raised from 20 o C to 40 o C the average
kinetic energy of neon atoms changes by a factor of which of the
following
[AIEEE 2004]
(a) 313/293
(b)
(313 / 293)
Gaseous State
2.
3.
4.
(c) 1/2
(d) 2
A gas is found to have a formula [CO] x . If its vapour density is 70,
the value of x is
[DCE 2004]
(a) 2.5
(b) 3.0
(c) 5.0
(d) 6.0
Which of the given sets of temperature and pressure will cause a gas
to exhibit the greatest deviation from ideal gas behavior [DCE 2003]
(a)
100 o C and 4 atm
(b) 100 o C and 2 atm
(c)
 100 o C and 4 atm
(d)
11.
0 o C and 2 atm
(a)
N /2
(c) 2N
(a)
7.
8.
9.
(n1C12  n2C22  n3 C32  .....) 1 / 2
n1  n2  n3  .....
(c)
(n1C12 )1 / 2 (n2C22 )1 / 2 (n3 C32 )1 / 2


 ......
n1
n2
n3
13.
At what temperature will the average speed of CH 4 molecules have
(d) 4N
the same value as O 2 has at 300 K
 : v : u :: 1 : 1.128 : 1.224
[CBSE PMT 1989]
(a) 1200 K
(c) 600 K
 : v : u :: 1.128 : 1.224 : 1
(d)  : v : u :: 1.124 : 1.228 : 1
Consider the following statements : For diatomic gases, the ratio
C p / Cv is equal to
(1) 1.40 (lower temperature)
(2) 1.66 (moderate temperature)
(3) 1.29 (higher temperature)
which of the above statements are correct
(a) 1, 2 and 3
(b) 1 and 2
(c) 2 and 3
(d) 1 and 3
The compressibility factor for an ideal gas is
[MP PET 2004]
(a) 1.5
(b) 1.0
(c) 2.0
(d) 
The compressibility factor of a gas is less than 1 at STP. Its molar
volume Vm will be
[MP PET 2004]
(a)
Vm  22.42
(b)
(c)
Vm  22.42
(d) None
(c)
45 2
18 2
 32
18 2
45  32
2
(b)
(d)
14.
18 2
45 2
A sample of O 2 gas is collected over water at 23 o C at a
barometric pressure of 751 mm Hg (vapour pressure of water at
15.
In an experiment during the analysis of a carbon compound, 145 l
of H 2 was collected at 760 mm Hg pressure and 27 o C
temperature. The mass of H 2 is nearly
[MNR 1987]
(a) 10 g
(c) 24 g
16.
(b) 12 g
(d) 6 g
The volume of 1 g each of methane (CH 4 ) , ethane (C2 H 6 ) ,
propane (C3 H 8 ) and butane (C4 H10 ) was measured at 350 K
and 1 atm. What is the volume of butane
[NCERT 1981]
(a) 495 cm 3
17.
 32
18.
45 2
18  32
The ratio of rates of diffusion of SO 2 , O2 and CH 4 is
(b) 1 : 2 : 4
(c)
2 : 2 :1
(d) 1 : 2 : 2
gas at 27 o C will be doubled keeping the pressure
19.
[Orissa 1993]
o
o
(a)
54 C
(b)
(c)
427 o C
(d) 527 o C
[BHU 1992]
1: 2 :2
(b) 600 cm 3
(c) 900 cm 3
(d) 1700 cm 3
The ratio of the rate of diffusion of helium and methane under
identical condition of pressure and temperature will be
[IIT 2005]
(a) 4
(b) 2
(c) 1
(d) 0.5
At what temperature in the celsius scale, V (volume) of a certain
mass of
constant
2
(a)
(b) 150 K
(d) 300 K
23 o C is 21 mm Hg). The partial pressure of O 2 gas in the
sample collected is
[CBSE PMT 1993]
(a) 21 mm Hg
(b) 751 mm Hg
(c) 0.96 atm
(d) 1.02 atm
Vm  22.42
If some moles of O 2 diffuse in 18 sec and same moles of other gas
diffuse in 45 sec then what is the molecular weight of the unknown
gas
[CPMT 1988]
(a)
10.
(b)
12.
(b)  : v : u :: 1.128 : 1 : 1.224
(c)




1/2
[AFMC 1994]
6.
 n1C12  n2 C 22  n3 C32  .....


n1  n2  n3  .....

(d)
(b) N
What is the relationship between the average velocity (v), root mean
square velocity (u) and most probable velocity (a)
1/2
(a)
 (n1C1  n2 C2  n3 C3  ....) 2 


(n1  n2  n3  ....)


50 ml of hydrogen diffuses out through a small hole from a vessel in
20 minutes. The time needed for 40 ml of oxygen to diffuse out is[CBSE PMT 19
(a) 12 min
(b) 64 min
(c) 8 min
(d) 32 min
If one litre of O 2 at 15 o C and 750 mm pressure contains 'N'
5.
If C1, C2 , C3 ...... represent the speeds of n1, n2 , n3 ..... molecules,
then the root mean square speed is
[IIT 1993]
The molecular weight of O 2 and SO 2 are 32 and 64 respectively.
molecules, the number of molecules in two litres of SO 2 under the
same conditions of temperature and pressure will be [CBSE 1990; MNR 1991]
245
327 C
Pressure of a mixture of 4 g of O 2 and 2 g of H 2 confined in a
bulb of 1 litre at 0 o C is
(a) 25.215 atm
[AIIMS 2000]
(b) 31.205 atm
246 Gaseous state
20.
(c) 45.215 atm
(d) 15.210 atm
If pressure becomes double at the same absolute temperature on 2
L CO 2 , then the volume of CO 2 becomes
8.
Assertion
:
1/4 of the gas is expelled if air present in an
open vessel is heated from 27 o C to 127 o C .
Rate of diffusion of a gas is inversely
proportional to the square root of its molecular
mass.
Compressibility factor for hydrogen varies with
pressure with positive slope at all pressures.
Even at low pressures, repulsive forces dominate
hydrogen gas.
[AIIMS 2005]
vander Waal’s equation is applicable only to nonideal gases.
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Ideal gases obey the equation PV  nRT .
Assertion
:
Pressure exerted by gas in a container with
increasing temperature of the gas.
Reason
:
With the rise in temperature, the average speed
of gas molecules increases.
Assertion
:
Gases do not settle to the bottom of container.
Reason
:
Gases have high kinetic energy.
th
[AIIMS 1992]
(a) 2 L
(c) 25 L
21.
(b) 4 L
(d) 1 L
9.
Volume of the air that will be expelled from a vessel of 300 cm 3
when it is heated from 27 o C to 37 o C at the same pressure will
be
(a) 310 cm 3
(c) 10 cm
22.
3
10.
(b) 290 cm 3
(d) 37 cm
3
11.
300 ml of a gas at 27 o C is cooled to  3 o C at constant
pressure, the final volume is
[NCERT 1981, MP PMT 1992]
(a) 540 ml
(c) 270 ml
(b) 135 ml
(d) 350 ml
[AIIMS 1995]
12.
[AIIMS 1997]
13.
Read the assertion and reason carefully to mark the correct option out of
the options given below :
(a)
(b)
(c)
(d)
(e)
1.
If both assertion and reason are true and the reason is the correct
explanation of the assertion.
If both assertion and reason are true but reason is not the correct
explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.
If assertion is false but reason is true.
2.
Assertion
Reason
Assertion
:
:
:
3.
Reason
Assertion
:
:
4.
5.
6.
7.
14.
15.
16.
Assertion
:
A mixture of He and O 2
respiration for deep sea divers.
is used for
Reason
:
He is soluble in blood.
Assertion
:
Wet air is heavier than dry air.
Reason
:
The density of dry air is more than density of
water.
[AIIMS 1999]
Assertion
:
All molecules in a gas have some speed.
Reason
:
Gas contains molecules of different size and
shape.
[AIIMS 2001]
Assertion
:
Effusion rate of oxygen is smaller than nitrogen.
Reason
:
Molecular size of nitrogen is smaller than oxygen.[AIIMS 200
[AIIMS 1998]
Plot of P Vs. 1 / V (volume) is a straight line.
Pressure is directly proportional to volume.
Jet aeroplane flying at high altitude need
pressurization of the cabin.
Oxygen is not present at higher altitude.
1 mol of H 2 and O 2 each occupy 22.4 L of
volume at 0 o C and 1 bar pressure.
Molar volume for all gases at the same
temperautre and pressure has the same volume.
Pressure exerted by a mixture of reacting gases is
equal to the sum of their partial pressures.
Reacting gases react to form a new gas having
pressure equal to the sum of both.
Greater the value of Vander Waal’s constant ' a'
greater is the liquefaction of gas.
' a' indirectly measures the magnitude of
attractive forces between the molecules.
Carbondioxide has greater value of root mean
square velocity  rms than carbon monoxide.
Characteristics and Measurable properties of gases
1
c
2
d
3
a
4
a
5
a
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
Assertion
:
Reason
:
 rms is directly proportional to molar mass.
6
bc
7
a
8
a
9
c
10
d
Assertion
:
11
a
12
a
13
a
14
a
15
c
Reason
:
4.58 mm and 0.0098 o C is known to be triple
point of water.
At this pressure and temperature all the three
states i.e., water, ice and vapour exist
simultaneously.
16
c
17
b
18
c
19
a
20
b
21
c
22
c
23
d
24
a
25
c
26
c
27
b
28
c
29
c
30
c
1
c
2
c
3
b
4
a
5
c
6
d
7
a
8
b
9
c
10
a
11
a
12
b
13
a
14
d
15
c
16
e
Ideal gas equation and Related gas laws
Gaseous State
31
a
32
a
33
b
34
b
35
b
36
c
37
d
38
c
39
c
40
a
41
b
42
c
43
c
44
d
45
a
46
a
47
b
48
d
49
b
50
c
51
d
52
a
53
a
54
c
55
a
56
a
57
b
58
a
59
b
60
a
61
b
62
d
63
c
64
b
65
d
66
b
67
d
68
a
69
a
70
b
71
c
72
a
73
d
74
c
75
a
76
c
77
d
78
b
79
b
80
b
81
d
82
d
83
c
84
c
85
a
86
d
87
a
88
a
89
c
90
c
91
a
92
c
93
a
94
b
95
a
96
a
97
d
98
b
99
e
100
a
101
c
102
c
103
c
104
c
105
a
106
a
Kinetic molecular theory of gases and
Molecular collisions
1
d
2
c
3
d
4
d
5
b
6
a
7
a
8
c
9
b
10
b
11
a
12
d
13
a
14
a
15
acd
16
a
17
b
18
d
19
b
20
d
21
a
22
d
23
c
24
b
25
b
26
a
27
d
28
a
29
d
30
c
31
c
Molecular speeds
1
a
2
c
3
d
4
d
5
d
6
c
7
b
8
b
9
d
10
b
11
d
12
c
13
a
14
d
15
d
16
c
17
d
18
d
19
b
20
b
21
c
22
a
23
c
24
b
25
c
26
a
27
d
Real gases and Vander waal’s equation
1
b
2
b
3
c
4
c
5
d
6
c
7
d
8
b
9
d
10
b
11
c
12
b
13
b
14
d
15
c
16
b
17
a
18
b
19
c
20
a
21
b
22
a
247
Gaseous State
P = 100 kPa , P1  ? , V = 1 litre ,
Critical state and Liquefaction of gases
a
2
b
3
d
4
c
5
d
6
d
7
c
8
c
9
abc
10
c
11
c
100  1 = P1  4 ; P1  25
If P2 is the pressure of O 2 gas in the mixture of O 2 and
N 2 then, 320  3 = P2  4 ; P2  240
Hence, Total pressure P  P1  P2  25  240
 265 kPa
Critical Thinking Questions
1
a
2
c
3
c
4
c
5
a
6
d
7
b
8
b
9
a
10
a
11
a
12
b
13
b
14
c
15
b
16
a
17
b
18
b
19
a
20
d
21
c
22
c
Ideal gas equation and Related gas laws
1.
2.
3.
Assertion & Reason
1
c
2
c
3
a
4
d
5
a
6
d
7
a
8
b
9
a
10
b
11
a
12
a
13
c
14
e
15
d
16
c
4.
5.
7.
Characteristics and Measurable properties of gases
3.
4.
5.
7.
(c) Gases do not have definite shape and volume. Their volume is
equal to the volume of the container.
(c) All the three phases of water can coexist at 0 o C & 4.7 mm
pressure.
(b) It is characteristic of gases i.e. Thermal energy >> molecular
attraction
(a) In gases, molecular attraction is very less and intermolecular
spaces are large hence kinetic energy of gases is highest.
(c) Gases and liquids, both can flow and posses viscosity.
(a) Newton is unit of force.
C o F o  32

5
9
(b)
9.
10.
12.
(c) 1 L  10 3 m 3  10 3 cm 3  1dm 3  10 3 ml .
(a) 1 atm = 10 6 dynes cm
(b) Barometer is used to measure atmospheric pressure of mixture
of gases. Staglometer is used to measure surface tension. Only
manometer is used to measure pressure of pure gas in a vessel.
(a) 0 o C is equivalent to 273o K i.e. conditions are same so
volume will be V ml.
(d) The mass of gas can be determined by weighing the container,
filled with gas and again weighing this container after removing
the gas. The difference between the two weights gives the mass
of the gas.
(c) Nobel gases has no intermolecular forces due to inertness.
(e) Total volume of two flasks = 1+ 3 = 4
If P1 the pressure of gas N 2 in the mixture of N 2 and
14.
15.
16.
20
1
50
(a) P.V = constant at constant temperature. As temperature
changes, the value of constant also changes.
(b,c)According to Boyle's Law PV = constant, at constant
temperature either P increases or V increases both (b) & (c)
may be correct.
T .d
d1 1 T1 2 P1 V2 T1
(a)
 ,
 


 1 1
d 2 2 T2 1 P2
V1 T2 T2 .d 2
P1 2 1 1
 . 
P2 1 2 1
8.
11.
(a) Absolute temperature is temperature measured in
expressed by T
(a) T1  273o C  273  273o K  546o K
Kelvin ,
T2  0 o C  273  0 o C  273o K
P1  1 ; P2  ?
According to Gay-Lussac’s law
PT
1  273 o K
1
P1 P2

atm;
atm.
 P2  1 2 
2
T1 T2
T1
546 o K
12.
(a)
Vt  Vo (1   v t)
if t2  t1  1o then V  Vo
For every 1o C increase in temperature, the volume of a given
mass of an ideal gas increases by a definite fraction
1
of Vo . Here Vo is volume at 0 o C temperature.
273.15
-2
O 2 then
o
 (V2  V1 )  V  Vo(t 2  t1 )
8.
13.
1
at constant T
P
1
(d) According to Boyle's law V 
P
Constant
V
; VP = Constant.
P
(a) At sea level, because of compression by air above the proximal
layer of air, pressure increases hence volume decreases i.e.
density increases. It is Boyle’s law.
(a) At constant T , P1V1  P2 V2
(c) Boyle's law is V 
1  20  P2  50 ; P2 
6.
2.
V1  4 litre
applying Boyle's law PV  P1V1
1
1.
249
13.
(a)
V1 T1
T
546 o K
 0 .2 L  0 .4 L.

 V2  2 V1 
V2 T2
T1
273o K
14.
(a)
V2 
T2
270 o K
.V1 
.400cm 3  360cm 3
T1
300 o K
contraction  V1  V2  400  360  40cm 3
15.
(c) At constant volumes P  T
P = constant T;
PV = nRT  P 
nR
T
V
250 Gaseous State
slope = m 
nR
 V2  V1
V
m1 V2

 m1  m 2 is curve for V has a greater slope than
m 2 V1
for V
P1 P1
P
P



T1 T1 T1 T2
35.
(b)
P1 V1 P2 V2
PV
T

 1 1  1
T1
T
P2 V2 T2
36.
(c)
d a  2d b ; 2 M a  M b
2
m
m RT dRT
RT ; P  .

M
V M
M
Pa d a M b 2d b 2 M a



4
Pb d b M a
db
Ma
PV  nRT 
1
16.
18.
19.
20.
(c)
 T  T2 
2 P1 (T1 T2 )
2 P1 T2
2 P1
; P 
 P  1


T
(
T

T
)
T
T1
T
T
1 1
2
1  T2
 1 2 
(c) At constant V of a definite mass
P1 P2 P1 300 1
i.e. pressure increases




T1 T2 P2 600 2
and on increasing temperature energy of molecules increases so
the rate of collisions also increases and number of moles
remains constant because there is neither addition nor
removal of gas in the occurring.
(a) Avogadro number = 6.0224  10
PV
 1 at STP (as given)
(b) Compressibility =
nRT
R = 0.821
nRT  PV
T = 273 K
N  0.0821  273  1Vm
37.
16 1

32 2
3
n of H 2 
2
(d) n of O2 
23
38.
(c)
39.
(c)
40.
(a)
o
P=1
n=1
22.41 litres Vm
21.
22.
(c) The value of universal gas constant can be expressed in
different units and its value would depend only on the units of
the measurement.
(c) PV = nRT
PV
R
 litre . atm. K mole
nT
(d) (atm. K mol ) is not a unit of R
(a) 8.31 J.K mol
1 cal = 4.2 J.
8 .31

cal.K 1mol 1 = 1.987 cal K mol
4 .2
n
P
(c) PV = nRT  
V RT
nRT 2  0 .0821  546

 2 atm.
(a) P 
V
44.8 l
–1
23.
24.
-1
–1
–1
–1
30.
31.
32.
(a)
41.
at STP n1  one mole.
at T  273o  30 o  303o K
P = 1 atm.
V = 22.4 lt
T = 273 K
P2  1 atm
1
1
42.
–1
P1 V1 P2 V2
PVT

 n 2  2 2 1 n1
n1 T1 n 2 T2
P1 V1 T2
V2  22.4
nRT 0 .5  0.082  273 K

 11.2lit
P
1
P1 V1T2
V2 
 P1  P ; T1  273o K
P2 T1
33.
(b) V 
34.
(b)
43.
(c)
47.
 240o K  33o C
(b) Because H 2 & Cl 2 gases may react with each other to
48.
49.
produce HCl gas hence Dalton's law is not applicable.
(d) Because HCl & NH 3 gases may react to produce NH 4 Cl
gas. Dalton's Law is applicable for non reacting gas mixtures.
(b) NH 3 and HCl & HBr is a reacting gas mixture to
produce NH 4 Cl & NH 4 Br so Dalton's law is not applicable.
o
T
4
3
P ; T2  T1  1   273 o K
2
3
3
2 P 4 273
800
V2 
 
 100cc 
cc  88.888 cc
3 P 3 273
9
= 88.9 cc
P2 
300
 400 o K
0 .75
P1 V1 P2 V2
PV
740 80

 T2  2 2  T1 

 300
T1
T2
P1 V1
740 100
T2 
o
1 22.4 273


 1  0.9 moles
1 22.4 303
0 .75 1 2
   300 o K  450 o K
1
1 1
P1 T2
1
250
.V1 

 12000lit.  20000lit.
(b) V2 
P2 T1
0.5 300
(c) At constant pressure
m
V  nT 
T
M
V1 m1 T1
T
V
m
d
300 o K 0 .75 d


 1  1  2  2 
T2
d
V2 m 2 T2 T2 m1 V2
d1
1
n2 
P1V1 P2 V2
P V
m

 T2  2 2 T1 1
n1T1
n2 T2
P1 V1
m2

–1
-1
3 1
 2
2 2
nRT 2  .082  273
V

 44.8 lit = 44800 ml
P
1
PV m
n

RT M
MPV 34  2  100
m

 282.4 gm
RT
0.082  293
PV T
760 546
V2  1 1 1 

 273  691.6 ml.
T1 P2 600 273
Total no. of moles =
50.
m
4
m
No. of moles of heavier gas 
40
m m 11m

Total no. of moles  
4 40
40
m
10
Mole fraction of lighter gas  4 
11m 11
40
(c) No. of moles of lighter gas 
Gaseous State
Partial pressure due to lighter gas  Po 
10
11
10
 1atm.
11
(a) m. wt. of NH 3  17 ; m.wt. of N 2  28
69.
 1 .1 
52.
beacuse NH 3 is lightest gas out of these gases
55.
(a)
rg
rHe
70.

1

MolecularWeight 
 rCH 
M g  M CH 4  4   16  2 2  64
 rg 


Volume effused V
1
(b) r 
r 

time taken
t
M
V
1
for same volumes (V constant)

t
M
M He
r 2 He
4
4
 M g  M He . 2 

2
2
Mg
1 .96
r g
(1 .4 )

t M
Note : 1.4  2 
56.
(a)
tCO  5
2
57.
58.
(b) rg 
(a)
 rH 
1
rH ; M g  M H 2 . 2   2  6 2  2  36  72
6 2
 rg 
71.
(c)
M1  64 ; r2  2r1
rN 2
rSO 2

59.
(b) rO  rH
2
60.
61.
(a)
(b)
r
d1
1
; 1 

d 2 16 r2
63.
64.
72.
2
(a)
1
2

(d)


(c) Gases may be separated by this process because of different rates of
diffusion due to difference in their densities.
(b) NH 4 Cl ring will first formed near the HCl bottle because
73.
HCl bottle & will react there with HCl to form NH 4 Cl
ring
(d) Because both NO and C2 H 6 have same molecular weights
NO
 M C 2 H 6  30

(a) rH 
rO  rH
M H2
M O2

2
10
2
32
64
28
TN 2 16
.
323 7
(1 .625)2  323  7
 373o K
16
C  H 2O  COg   H 2g 
(a)
6.02  10 22 molecules of each N 2 , O2 and H 2
6 .02  10 22
moles of each
6 .02  10 23
Weight of mixture = weight of 0.1 mole N 2  weight of 0.1

and rate of diffusion  molecular
xgm
2gm
if rO 
10 min
10 min
323

75.
2
68.
TN 2
PV  nRT
n
n
P  RT   C  P  CRT
V
V
P
1
T

 12o K
CR 1  .821
r
1
1
M A  rB 
1
    rA  2rB  B 
 2   .25
M B  rA 
4
rA 2
(2)
(d)
M N2

(c)
weight.
67.
M So 2
74.
M NH 3 : M HCl  17 : 36.5) . SO NH 3 will reach first to the
M
TSO 2
.
12 gm  1mol  1mol
12 gm C produces 2mole of gases (1mole CO & 1 mole of H )
48
 2  4  2  8 mole
48 gm C may produce
12
 22.4  8 L gases  179.2 L gas.
mRT 4 .4  .082  273

(d) Molecular weight =
= 44
PV
1  2 .24
So the gas should be CO 2
rate of diffusion of NH 3 is more than that of HCl because
65.
TN 2
2
d2
4
 16 
d1
1
 B
B  B  2

 ;  D A  DB 
A  A 
 A
DA

DB
28
44
 5 14 s ; tCO 2  5
 5 22 s
2
2
TN 2 
r 
d
1
1
ra  5rb ; a   b     
d b  ra 
25
5
1
62.
1 .625 
dH
0 09
1
1
1


dO
1 .44
16 4
32
 20 s
2
Vrms N 2

Vrms SO 2
2
r 
1
M 2  M 1  1   64   16
r
4
 2
M1
M2
M He
4
5
 5 2 s.
M H2
2
tO 2  t  5
2
 rH 
  2   (5)2  25 ; M g  2  25  50
 rg 
M H2
t1

t2
t He  t H 2
1
rg  .rH 2
5
Mg
x
2
1

 gm.  .5 gm
10 10  4 2
rCH 4  2rg
2
m.wt. of CO 2  44 ; m.wt. of O2  32

r 

(a)
251
mole H 2 + weight of 0.1 mole of O2
76.
 (28  0.1)  (2  0.1)  (32  0.1)  6.2 gm
(c) M.wt of CO 2 = 12+16+16 = 44
Volume of 44 gm of CO 2 at NTP = 22.4 litre
1 gm of CO 2 at NTP =
22 .4
44
252 Gaseous State
4.4 gm of CO 2 at N.T.P
78.
In
80.
(b) Molecular weight = V.d.  2 = 11.2  2 = 22.4
Volume of 22.4 gm Substance of NTP = 22.4 litre
22.4
1 gm substance at NTP =
litre
22.4
11.2 gm substance of NTP = 11.2 litre
M . wt . of O2
M
32 1
(b)
 1 

M . wt . of SO 2
M2
64 2
85.
(a) Mixture of SO 2 and Cl 2 are reacted chemically and forms
86.
SO 2 Cl 2 . That is why mixture of these gases is not applicable
for Dalton's law.
(d) According to Boyle's law
P1V1  P2 V2  P1  60  720  100
(a) Solution level will rise, due to absorption of CO 2 by sodium
hydroxide.
2 NaOH  CO 2  Na 2 CO 3  H 2 O
89.
(c)
92.
(c) In
22.4 l
r1

r2
of
 6.023  10
23
H2
98.
99.

r

6r
2
MA
P1 V1 720  100

 857.142
V2
84
Hence, P2  857.14 mm
(b) According to gas law
PV
PV  nRT , n 
RT
P1 V1
T
nA
RT1 n A P1 V1

 2
;

P
V
nB
T1
P2 V2
nB
2 2
RT2
(e) No. of molecules  2  V.d
2  38.3  76.3
wt. of NO 2  x
So that wt. of N 2O4  100  x
x 100  x 100 2 x  100  x 100




46
92
76.6
92
76.6
20.10
 0 .437
x = 20.10, no. of mole. of NO 2 
46
(a) Given that
P1  76cm of Hg (Initial pressure at N.T.P.)
Hence,
100.
P2  ? , V1  5litre, V2  30  5  35litres
According to Boyle's law
P1V1  P2 V2 ; 76  5  P2  35
2
1
1


32
16 4
number
MH
MA
4
 2  1 .4 ; r1  1.4 m
2
n A 2 P  2V T
n
2


; A 
nB
2T
PV n B 1
22.4
 .224 litre of
produce =
100
maximum
rA

rH
P2 
M1  32 g for O 2 , M 2  2 g for H 2
M2
;
M1
M2

M1
By using P1V1  P2 V2 [According to the Boyle's law]
22 .4 litre
Molecularwt . Of Metal
45

Volume
22 .4
r1

r2
97.
= 2 gmlitre
r1

r2
6 .023  10
 15  4.03  10 23 molecules.
22.4
22.4 l O2 at S.T.P.  32 gm of O 2
M A  6  6  2  72 g
(d) Given that:
V1  100ml, P1  720mm , V2  84 ml, P2  ?
1
(a)
molecules
(a)
CO 2
91.
of
96.
ˆ CaO  CO 2 

(c) The density of gas 
number
95
 At S.T.P. 100 g CaCO 3 produce= 22.4 litreof CO 2
90.
maximum
23
32
 1 .43 gm of O 2
22.4
(a) We know that molecular mass of hydrogen M1  2 and that
that is why H 2 gas diffuse first
88.
 At S.T.P. 1 g CaCO 3
molecules
(a)
720  100
 1200mm
60
1
(a) Rate of diffusion 
Molecular Mass
(40 12 16 3 ) 100 gm
of
93.
P1 
CaCo 3
number
of helium M 2  4 , we also know that Graham's law of
diffusion
84.
87.
maximum
1l O 2 at S.T.P. 
1
that of SO 2
2
(b) For HI has the least volume because of greater molecular
1
weight V 
M
(c) Since no. of molecules is halved so pressure should also be
halved.
(c) H 2 will be filled first because of lower molecular weight
83.
H2
23

The weight of oxygen will be
81.
of

200
= molarity  Volume (in lt.)  0 .1 
 .02
1000
Volume of 0.02 mole of CO 2  22.4  .02lt.  0.448lit.
79.
1l
6 .023  10
22.4
In 15l of
H2
22.4

 4 .4 litre  2.24 litre
44
(b) No. of moles of CO 2 present in 200 ml solution
of
76  5
 P2  10.8 cm of Hg
35
(c) Given initial volume V1   300cc , initial temperature
P2 
molecules
101.
T1   27o C  300 K , initial pressure P1   620mm , final
Gaseous State
temperature
T2  47 o C  320 K and
final
pressure
12.
P2   640mm .We know from the general gas equation
P1 V1 P2 V2
620  300 640  V2
 V2  310cc



300
320
T1
T2
102.
(c)
NH 3  HCl  NH 4 Cl
4 litre 1.5litre
14.
HCl is a limiting compound. That's why 1.5litre of HCl
reacts with 1.5litre of NH 3 and forms NH 4 Cl . Thus (4 -
15.
1.5) 2.5litre NH 3 remains after the reaction.
103.
(c)
13.
P1 V1 P2 V2 P1  4 2 P1  V2
;


T1
T2
T1
2T1
8  2  V2 so V2  4dm 3
104.
(c)
P1  P, V1  V, T1  T
P
, V2  ? , T2  T
2
According to gas equation
P1 V1 P2 V2
PV PV2
or


T
2T
T1
T2
16.

18.
P2
105.
(a)
V2  4 V
17.
85
, T2  ?
100
P  V 2 P  85 V
348  2  85
 T2 

100
398
T2  100
 T2  591.6 K  318.6 o C
106.
(d) All molecules of an ideal gas show random motion. They collide
with each other and walls of container during which they lose
or gain energy so they may not have same kinetic energy
always.
(a) For same temperature kinetic energies of H 2 & He molecules
will be same because kinetic energy depends only on
temperature.
(a) For same temp. kinetic energies would be equal for all
molecules, what ever their molecular weights will be, it doesn't
matter.
(a,c,d)Kinetic energies per molecule will be same because it is
proportional to absolute temperature only.
dN2
MN2
28


i.e. dN 2  dCO 2
d CO 2
M CO 2
44
Total translational kinetic energy will also be same because at
same temperature & pressure number of molecules present in
same volume would be same (according to Avogadro's Law)
(a) On increasing pressure, the volume decreases and density
increases. So molecules get closer to each other hence mean
free path also decreases.
(b) Most probable velocity increase and fraction of molecule
possessing most probable velocity decrease.
(d) PV  nRT 
P1  P, V1  V, T1  273  75  348 K

P2  2 P, V2 
(a) Boyle's law  PV = constant
On differentiating the equation,
d(PV)  d(C)  PdV  VdP  0
dP
dV

 VdP  PdV 
.
P
V
m
RT
M
m PM

 density
V
RT
P
P
the value of
is maximum for (b)
T
T
19.
(b) d 
20.
(d) If inversion temperature is 80 o C  193o K then the
temperature, at which it will produce cooling under Joule
Thomson’s effect, would be below inversion temperature except
173o K all other values given as
21.
(a) Since
CP
 1 .4 , the gas should be diatomic.
CV
If volume is 11.2 lt then, no. of moles =
Kinetic molecular theory of gases
and Molecular collisions
 no. of molecules =
4.
(d) Kinetic energy 
5.
(b) K.E =
3
RT
2
7.
8.
(c) Tr. K. E. 
9.
(b)
2
22.
3 RT
it means that the Translational Kinetic
2
energy of Ideal gas depends upon temperature only.
THe
so energies will be same for
TH 2
He & H 2 at same temperature.
11.
(a)
K. E. 
3
3
.RT  .2 .T
2
2
K. E.  3T
1
2
1
 Avagadro’s No.
2
no. of atoms = 2  no. of molecules
3
RT it means that K.E depends upon T (absolute
2
temperature) only.
3
3
(a) KE  RT  PV
2
2
2 E
2
P 
for unit volume (V = 1)  P  E
3 V
3
E He

EH2
253
 R  2calK1mol 1
23.
1
 Avagadro’s No.
2
 6.0223  10 23
M
(d) Density =
V
v.d  2
M  V.d  2
d
V
d V
V .d 
2
0.00130  22400
V .d 
 14.56 gm 1
2
(c) Volume of steam = 1lt = 10 3 cm 3
 m  d.V
mass of 10 3 cm 3 steam = density  Volume
0 .0006 gm
 10 3 cm 3  0.6 gm
=
cm 3
254 Gaseous State
Actual volume occupied by H 2 O molecules is equal to volume
of water of same mass
 Actual volume of H 2 O molecules in 6 gm steam
= mass of steam/density of water
= 0.6 gm /1 gm/cm  0.6 cm 3
1
12.
13.
3
CP 5
  1 .66 (For Monoatomic as He, Ne , Ar )
CV 3
24.
(b) r 
25.
(b) The density of neon will be highest at 0 o C 2 atm according
P
to d 
T
3
(d) K.E. per mole  RT
2
so all will have same K.E. at same temperature.
(c)  W  P.dV  E
 Energy per unit volume = P
d
M
3d
M
M
(c) d  M  1  1 ;
; M2 
.

3
d2 M 2
d
M2
29.
30.
31.
Molecular speeds
1.
(a)

3.
(d) most probable velocity : mean velocity : V
(d)
15.
(d)
2 RT
8 RT
:
:
M
M
3 RT
M
8
 2:

U SO 2

U He
1

2
M He TSO 2

M SO 2 THe

M O2

32

48
4 TSO 2
.
64 300
9.
10.
(b)
UO2
M O 2 TSO 2

M SO 2 TO 2

U1

U2
T1
T2

(d) U 
19.
(b)
 m1U1 2  m 2U2 2
T2
1200
 U1
 U1  2
T1
300
TH 2

UO 2
MH2
.
3 RT

M
U1

U2
n1T1

n2 T2
M O2
TH 2

3 PV

M
50 32
.
1
2 800
3P
1
U 
d
d
nT

2n  2 T
1 1

4
2
U2  2U1  2U
21.
(c)
22.
(a)
Vmp
Vav
2 RT
M  
2
8 RT
M

3 RT
 Vrms  T
M
Given V1  V , T1  300 K , V2  2V , T2  ?
Vrms 
V1

V2
24.
26.
32  TSO 2
1
64  303
T1  5  10 4

T2  10  10 4
UH 2
17.
2
T1  V 
300
; 
 T2  300  4  1200 K
 
T2  2 V 
T2

n 2a 
(b)  P  2 V  nb   nRT
V 

2  .0821  300 4 .7  2 2

 10  0.66  9.33atm
5  2  .03711
52
(a) Average speed : most probable speed
8 RT
:
M
27.
8

: 2  1.128 : 1.
3 RT
M
3 R  300
3 RT
 30  30 R 
M
M
3  300
1 gm  0.001 kg .
 M
30  30
Real gases and Vander waal’s equation
2

 1

4

(d) v rms 
2 RT

M
30 2 R 
32  TSO 2
1
 TSO 2  606
64 303
(d) Among these Cl 2 has the highest molecular weight so it will
posses lowest root mean square velocity.
(b)
m2
m1

P
2
3
(b) Average kinetic energy per molecule
3
3
 KT   1.38  10  23  300 J = 6.17  10 21 J
2
2
U SO 2
2
2


 P  (2)  4 .17 5  2  .03711  2  .0821  300
2


(
5
)


7.
M O3
U2
: 3
(c)
UO 2
2
(c)
4 TSO 2
1
 ; TSO 2  1200o K
64 300 4
UO 3
U1
m 2 T1
.
 T1  T2
m1 T2
(d) U2  U1
rms
6.
8.
U1

U2
16.
3 RT
1
 Vrms 
at same T
M
M
Vrms 

(d)
r.m.s. velocity will be doubled.
because H 2 has least molecular weight so its r.m.s. velocity
should be maximum.
5.
Vrms 
66
1 .086

56
1

4.
3
8
8 RT Vrms

;
Vav
M
3 RT
, Vav 
M
Vrms 
14.

3 KT
1
i.e. Vrms 
 (m ) 2
Molecularweight
m
(a) When average speed of molecule is increased due to increase in
temperature then the change in momentum during collision
between wall of container and molecules of gas also increases.
(c)
2.
(b) Because molecules of real gases have intermolecular forces of
attraction so the effective impact on the wall of container is
Gaseous State
a
factor hence
v2
behaviour of real gas deviate from ideal behaviour.
PV
Z
 for ideal gas PV = RT so Z = 1
RT
PV
; for ideal gas PV = RT; so Z = 1
Z
RT
Ideal gas has no attractive force between the particles
PV = nRT is a ideal gas equation it is allowed when the
temperature is high and pressure is low.
At Boyle temperature real gas is changed into ideal gas
When pressure is low
a 

 P  2 V  b   RT
V 

a ab
PV
a
1
or PV  RT  Pb   2 or
V V
RT
VRT
a
 PV

Z
 Z

VRT  RT

At high temperature and low pressure, Vander Waal’s equation
is reduced to ideal gas equation.
 0.16 gm of O 2
diminished. Pressure of real gas is reduced by
3.
(c)
12.
(b)
13.
14.
(b)
(d)
16.
17.
(b)
(a)
18.
(b)
Critical Thinking Questions
1.
2.
3.
4.
5.
(a) Average kinetic energy  (T Kelvin)
K. E2 T2 40  273
313
(Factor)



293
K. E1 T1 20  273
(c) M. wt. = V.d.  2
m.wt .
40
= 70  2 = 140  x 

5
wt .of [CO ] 12  16 
(c) Gas deviate from ideal gas behaviour to real gas (according to
Vander Waal's at low temperature and high pressure)
(c) At same temperature and pressure, equal volumes have equal
number of molecules. If 1lit. of oxygen consists N molecules
then at same temperature and pressure 1 lit of SO 2 will
consists N molecules. So 2 lit. of SO 2 will contain 2 N
molecules.
(a) Vav : Vrms : V
= V : U :
most probable
8 RT
:
M
PV = nRT
PV = RT (For 1 mole of gas)
20.
22.
(a)
or (P  36.048)(0.2072)  24.63
 P  36.048  118.87  P  82.82 atm.
6.
7.
8.
9.
Critical state and Liquefaction of gases
2.
3.
4.
5.
6.
7.
8.
10.
11.
(b) A diabatic demagnetisation is a technique of liquefaction of
gases in which temperature is reduced.
(d) An ideal gas can’t be liquefied because molecules of ideal gas
have not force of attraction between them.
(c) At above critical temperature, substances are existing in
gaseous state, since gas cannot be liquefied above it.
(d) Absence of inter molecular attraction ideal gas cannot be
liquefied at any volume of P and T.
(d) For Z gas of given gases, critical temperature is highest
8a
8  12
Tc 
 1603.98 K
 Tc 
27 Rb
27  .0821  .027
(c) Value of constant a is greater than other for NH 3 that’s why
NH 3 can be most easily liquefied.
(c) The temperature below which the gas can be liquefied by the
application of pressure alone is called critical temperature.
M
 M  d V
(c) d 
V
M  2.68  22.4 at N.T.P.  V  22.41
M  60.03 gm
m. wt of COS  12  16  32  60
(c) 22400 ml is the volume of O 2 at N.T.P =32gm of O 2
32
1ml is the volume of O 2 at NTP =
22400
112 ml is the volume of O 2 at NTP =
32
 112
22400
3 RT
:
M
(d)

: 3  1 : 1.128 : 1.224
Cp
ratio for diatomic gases is 1.40 at lower temperature &
Cv
1.29 at higher temperature so the answer is 'd'.
(b) PV  nRT (For ideal gas)
PV
Z
 1 (For ideal gas)
nRT
(b) If Z  1 then molar volume is less than 22.4 L
x
x
mole / sec  rg 
mol / sec
(a) rO 2 
18
45
 rO
M g  MO 2  2
 rg

10.
2 RT
M
8
 :V :U  2 :
(a) Vander waal's constant for volume correction b is the measure
of the effective volume occupied by the gas molecule.
2 

 P  n a  (V  nb)  nRT

V 2 

2.253 

P 
 (0 .25  0 .0428)  0 .0821  300
0.25  0.25 

255
(a) rSO 2 : rO2 : rCH 4
2
2
2

  32 x  45   32  45
2

18
 18 x 

1
1
1

:
:
M SO 2
O2
CH 4
1
1

64 : 32 : 16
4 : 2 :1
1 1 1 2 2 2
:
: ;
:
: ; 1: 2 :2
2
2 1 2
2 1

11.
 n c 2  n2 c 22  n3 c32  .... 
(a) Root mean square speed   1 1

 n1  n2  n3  ...... 
12.
(b) If 40 ml O 2 will diffuse in t min. then. rO2 
rH 
M H2
50
50

 rO  rH 2
20
M O2
20
(b)
Vav CH 4

Vab O2
TCH 4
14.
TCH 4
TO 2
.
M O2
M CH 4
40
t
2
50 1

.
32 20 4
40 50
40  80

t 
 60 min.
t
80
50
13.
1/2
1
32
.
 1 ; TCH 4  150 o K
300 16
(c) Pressure of O 2 (dry) = 751-21 = 730 mm Hg
.
256 Gaseous State
730
 0 .96 atm
760
PV
1  145
(b) PV  nRT , n 

 5.8  6 mole .
RT 0.082  3
nRT m RT
1 0.82  350
(a) V 

.


 0.495lit.
P
M P
58
1
= 495 cm

15.
16.
8.
(b)
17.
18.
M CH 4
(a)
10.
(b)
11.
(a)
4
 0 .125
32
12.
(a)
2
1
2
13.
(c)
14.
(e)
15.
(d)
16.
(c)
(b)
V1 T1
TV
2V

 T2  1 2  300 o K,
 600 o K
V2 T2
V1
V
M He
16
2
4
=
T2  600o K  (600  273)o C  327o C
19.
(a) no. of moles of O2 
no. of moles of H 2 
total no. of moles = 1  0.125  1.125
P
20.
(d)
nRT 1.125  0.082  273

 25.184 atm.
V
1
P1 
V
P
2
 , 1  2 
P2 2
V2 P1 1
2L 2

;
V2
1
21.
(c)
22.
(c)
V2  1L
V2 T2

V1 T1
 V2 
V2 
T2
310 o K
V1 
 300 cm 3  310 cm 3
T1
300 o K
T2
270 o K
.V1 
 300ml  270ml
T1
300 o K
Assertion & Reason
1.
2.
3.
4.
5.
(c) Pressure is inversly proportional to volume (Boyle’s law).
1
p
(n, T constant).
V
(c) The air pressure decreases with increase in altitude. So the
partial pressure of Oxygen is not sufficient for breathing at
higher altitude and thus pressurization is needed.
(a) At a given temperature and pressure the volume of a gas is
directly proportional to the amount of gas Vn (P and T
constant).
(d) According to Dalton’s law of partial pressure, the pressure
exerted by a mixture of non – interacting gases is equal to the
sum of their partial pressures (pressure exerted by individual
gases in mixture) PTotal  P1  P2  P3 … (T and V constant).
Both the gases if non-interacting would spread uniformly to
occupy the whole volume of the vessel.
(a) Considering the attractive force pressure in ideal gas equation
(PV = nRT) is correct by introducing a factor of
an2
where a
V2
is a vander waal’s constant.
6.
(d)
3 RT
is inversly related
M
Therefore, rms (CO )  rms (CO 2 ).
rms 
to
molecular
V1
 1 when temperature is
V2
mass.
V1 300 3



V2 400 4
3 1

4 4
In case of H , compressibility factor increases with the pressure.
At 273 K, Z > 1 which shows that it is difficult to compress the
gas as compared to ideal gas. In this case repulsive forces
dominate.
In real gases, the intermolecular forces of attraction and the
volume occupied by the gas molecules cannot be neglected.
When the temperature increase, the average speed of gas
molecules increases and by this increase the pressure of gas is
also increases.
It is correct that gases do not settle to the bottom of container
and the reason for this is that due to higher kinetic energy of
gaseous molecules they diffuse.
The assertion, that a mixture of helium and oxygen is used for
deep sea divers, is correct. The He is not soluble in blood.
Therefore, this mixture is used.
Dry air is heavier than wet air because the density of dry air is
more than water.
All molecule of a gas have different speed. Therefore, they move
by its own speed.
Assertion is true but reason is false because of effusion rate
1

(Molecular weight) but it does not depend on
M
molecular size.
air expelled  1 
9.
(b)
(Initial fraction
27 o C . At 127 o C the new fraction is
3
rHe

rCH 4
V1 V2

T1 T2
2
Gaseous State
1.
Same mass of CH 4 and H 2 is taken in container. The partial
pressure caused by H 2 is
2.
7.
[IIT 1989; CPMT 1996]
(a) 8 / 9
(b) 1 / 9
(c) 1 / 2
(d) 1
Which of the following volume (V) – temperature (T) plots
represents the behaviour of one mole of an ideal gas at one
atmospheric pressure
[IIT Screening 2002]
V(L)
(22.4 L
273 K)
The following graph illustrates
[JIPMER 2000]
(b) Charle's law
(c) Boyle's law
(d) Gay-Lussac's law
8.
(a) 2 litres
(b) 4 litres
(c) 5 litres
(d) 7 litres
9.
What is kinetic energy of 1 g of O 2 at 47 o C
[Orissa JEE 2004]
5.
(a)
1.24  10 J
(b)
2.24  10 J
(c)
1.24  10 3 J
(d)
3.24  10 2 J
2
10.
6.
HBr  O2  N 2  H 2
(c)
H 2  N 2  O2  HBr
(d)
HBr  O2  H 2  N 2
T(K)
By what ratio the average velocity of the molecule in gas change
1200 K
12.
The rate of diffusion of SO 2 and O 2 are in the ratio
[Assam JET 1991; EAMCET 1980]
[DCE 2003]
(c) 1.14 / 1
(d) 4 / 1
(b) 600 K
A gas of volume 100 cc is kept in a vessel at pressure 10 4 Pa
when the temperature is raised from 50 to 200 C
(b) 1.46 / 1
[Pb. CET 2001]
(c) 400 K
(d) 1800 K
Equal volumes of two gases which do not react together are
enclosed in separate vessels. Their pressures at 100 mm and 400
mm respectively. If the two vessels are joined together, then what
will be the pressure of the resulting mixture (temperature remaining
constant)
[CBSE PMT 1981]
(a) 125 mm
(b) 500 mm
(c) 1000 mm
(d) 250 mm
o
(a) 1.21 / 1
T(K)
11.
[Pb. CET 1994; CBSE PMT 1991]
(b)
(14.2 L
373 K)
(d)
to 10 5 Pa, keeping the temperature constant, then the volume of
the gas becomes
[AFMC 1992]
(a) 10 cc
(b) 100 cc
(c) 1 cc
(d) 1000 cc
If a gas is expanded at constant temperature
[IIT 1986]
(a) The pressure increases
(b) The kinetic energy of the molecules remains the same
(c) The kinetic energy of the molecules decreases
(d) The number of molecules of the gas increases
and HBr are in the order
H 2  N 2  O2  HBr
(22.4 L
273 K)
maintained at temperature 24 o C . If now the pressure is increased
The root mean square speeds at STP for the gases H 2 , N 2 , O2
(a)
(30.6 L
373 K)
If the average velocity of N 2 molecule is 0.3 m / s at 27 o C ,
(a)
doubled, the volume of CO 2 would become
[CBSE PMT 1991]
T(K)
V(L)
then the velocity will be 0.6 m / s at
If the pressure and absolute temperature of 2 litres of CO 2 are
2
T(K)
(c)
(a) Dalton's law
(28.6 L
373 K)
(22.4 L
273 K)
(b)
(22.4 L
273 K)
4.
(36.8 L
373 K)
V(L)
V
3.
V(L)
(a)
Temp. (oC)
257
(a)
1: 2
(c) 1 : 2
(b) 1 : 32
(d) 1 : 4
(SET -6)
Gaseous State
1.
(a)
N CH 4 = number of moles of CH 4 
m
16
m
2
N H 2 = number of moles of H 2 
7.
(c)
V1 T1
T
22.4  373

 V2  V1 . 2 
 30.6 L
V2 T2
T1
273
8.
(a)
Vrms 
fraction partial pressure of H 2 is
2.
3.
4.
m
8
H2 

 2 
9m
m m
n H 2  n CH 4
9

16
2 16
(b) According to Charle's Law V  T
Vt  Vo  Vot
compare it with Y = C + mx
PV T
P
2T
(a) V 2  1 1 . 2 
 2lt 
 2lt
T1 P2 2 P
T
5.
(a) K.E. 
(b)

9.
10.
2
300
 V 

 
T2
 2V 
(a)
P1  P2 100  400

 250mm
2
2
P1 V1  P2 V2 at constant T
10 4.100  10 5  V2
1
2
:
1
20
:
1
32
:
1
V2  10 cc
is
81
T1  150  273  423 K ; T2  50  273  323 K
(Vav )1

(Vav )2
T1
T2
(d) When two vessels are joined together, the volume will be
doubled hence effective pressure will be halved
P
 1.24  10 2 J
1
Vrms 
m
Hence,
V1

V2
T2  300  4  1200 K
11.
(b) Kinetic energy will also remain constant if Temperature is
constant.
12.
(a)
UHB r  UO2  UN 2  UH 2
(c)
T1  300 K
V2  2V , T2  ?
3
3 1
nRT  
 8.314  320 J .
2
2 32
U H 2 : U N 2 : U O2 : U HBr 
6.
3 RT
; Vrms  T
M
Given, V1  V ,
m
2
nH2
259
T1

T2
423 1 .14

323
1
***
rSO 2
rO2

M O2
M SO 2

32
1

64
2
Nuclear Chemistry
259
Chapter
7
Nuclear Chemistry
“The branch of chemistry which deals with the study of composition
of atomic nucleus and the nuclear transformations is known as nuclear
chemistry”.
The common examples of nuclear processes are radioactivity,
artificial transmutations, nuclear fission and nuclear fusion. The nuclear is
also an important aspect of chemistry because the energies involved in some
of these are million times greater than those in ordinary chemical reactions.
Radioactivity
“Radioactivity is a process in which nuclei of certain elements
undergo spontaneous disintegration without excitation by any external
means.’’ and the elements whose atoms disintegrate and emit radiations are
called radioactive elements.
Henry Becquerel (1891) observed the spontaneous emission of
invisible, penetrating rays from potassium uranyl sulphate
K 2 UO2 (SO4 )2 , which influenced photographic plate in dark and were
able to produce luminosity in substances like ZnS.
Later on, M.M. Curie and her husband P. Curie named this
phenomenon of spontaneous emission of penetrating rays as, Radioactivity.
Curies also discovered a new radioactive element Radium from
pitchblende (an ore of U i.e. U3 O8 ) which is about 3 million times more
radioactive than uranium. Now a days about 42 radioactive elements are
known.
The radioactivity may be broadly classified into two types,
(1) If a substance emits radiations by itself, it is said to possess
natural radioactivity.
(2) If a substance starts emitting radiations on exposure to rays
from some natural radioactive substance, the phenomenon is called induced
or artificial radioactivity.
Radioactivity can be detected and measured by a number of devices
like ionisation chamber, Geiger Muller counter, proportional counter, flow
counter, end window counter, scintillation counter, Wilson cloud chamber,
electroscope, etc.
Nature and characteristics of radioactive emissions
The phenomenon of radioactivity arises because of the decay of
unstable nuclei or certain element.
Photographic plate
The nature of the radiations emitted
from a radioactive substance was

investigated by Rutherford (1904) by


Lead block
Fig. 7.1
Radioactive
substance
applying electric and magnetic fields. When these radiation were subjected
to electric or magnetic field, these were split into three types ,  and  –
rays.
Characteristics of radioactive rays
-Ray
Charge and mass : It carries +2
Electron  1e 0
-Ray
It has no charge
and negligible
mass.
High energy
raditons.
Action of magnetic field :
Deflected to anode.
Not deflected.
Deflected towards the cathode.
Velocity : 1/10th to that of light.
Same as that of light.
Ionizing power : Very high
nearly 100 times to that of -
Low nearly 100 times to
that of -rays.
Same as that of
light.
Very low.
charge and 4 unit mass.
Identity : Helium nuclei or
helium ion 2 He 4 or He2+.
-Ray
It carries -1 charge and
no mass.
rays.
Effect on ZnS plate : They
cause luminescence.
Penetrating power : Low
Range : Very small.
Nature of product : Product
obtained by the loss of 1 particle has atomic number
less by 2 units and mass
number less by 4 units.
Very little effect.
Very little effect.
100 times that of
particles.
More than -particles.
Product obtained by the
loss of 1 -particle has
atomic number more by
1 unit, without any
change in mass number.
10 times that of particles.
More
There is no change
in the atomic
number as well as
in mass number.
Theory of radioactivity disintegration
Rutherford and Soddy, in 1903, postulated that radioactivity is a
nuclear phenomenon and all the radioactive changes are taking place in the
nucleus of the atom. They presented an interpretation of the radioactive
processes and the origin of radiations in the form of a theory known as
theory of radioactive disintegration. The main points of this theory are,
(1) The atomic nuclei of the radioactive elements are unstable and
liable to disintegrate any moment.
(2) The disintegration is spontaneous, i.e., constantly breaking. The
rate of breaking is not affected by external factors like temperature,
pressure, chemical combination etc.
260 Nuclear Chemistry
(3) During disintegration, atoms of new elements called daughter
elements having different physical and chemical properties than the parent
elements come into existence.
(4) During disintegration, either alpha or beta particles are emitted
from the nucleus.
The disintegration process may proceed in one of the following two
ways,
(i) -particle emission : When an -particle (2 He ) is emitted
from the nucleus of an atom of the parent element, the nucleus of the new
element, called daughter element possesses atomic mass or atomic mass
number less by four units and nuclear charge or atomic number less by 2
units because -particle has mass of 4 units and nuclear charge of two
units.
4
-α Daughter element

Parent element
W 4
Z 2
Atomic mass : W
Atomic number : Z
(ii) -particle emission : -particle is merely an electron which has
negligible mass. Whenever a beta particle is emitted from the nucleus of a
radioactive atom, the nucleus of the new element formed possesses the
same atomic mass but nuclear charge or atomic number is increased by 1
unit than the parent element. Beta particle emission is due to the result of
decay of neutron into proton and electron. 0 n1  1 p1  1e 0
The electron produced escapes as a beta-particle-leaving proton in
the nucleus.
-β
 Daughter element
Parent element
Atomic mass : W
Atomic number : Z
W
Z 1
(iii) -ray emission : -rays are emitted due to secondary effects. The
excess of energy is released in the form of -rays. Thus -rays arise from
energy re-arrangements in the nucleus. As -rays are short wavelength
electromagnetic radiations with no charge and no mass, their emission from
a radioactive element does not produce new element.
Special case : If in a radioactive transformation 1 alpha and 2 betaparticles are emitted, the resulting nucleus possesses the same atomic
number but atomic mass is less by 4 units. A radioactive transformation of
this type always produces an isotope of the parent element.
Z
A
W
α


Z 2 B
W 4
β


Z 1 C
W 4
β


W 4
ZD
A and D are isotopes.
Group displacement law
Soddy, Fajans and Russell (1911-1913) observed that when an particle is lost, a new element with atomic number less by 2 and mass
number less by 4 is formed. Similarly, when -particle is lost, new element
with atomic number greater by 1 is obtained. The element emitting then 
or -particle is called parent element and the new element formed is called
daughter element. The above results have been summarized as,
(1) When an -particle is emitted, the new element formed is
displaced two positions to the left in the periodic table than that of the
parent element (because the atomic number decreases by 2).
(2) When a -particle is emitted, the new element formed is
displaced one position to the right in the periodic table than that of the
parent element (because atomic number increased by 1).
(3) When a positron is emitted, the daughter element occupies its
position one group to the left of the parent element in periodic table.
Group displacement law should be applied with great care especially
in the case of elements of lanthanide series (57 to 71), actinide series (89 to
103), VIII group (26 to 28; 44 to 46; 76 to 78), IA and IIA groups.
nuclear
a
c
To determine the number of - and - particles emitted during the
transformation. It can be done in following manner,
X  bd Y  x 42 He  y 1 e 0
a  b  4 x or x 
ab
4
.....(i)
c  d  2x  y
.....(ii)
where x = no. of -emitted, y = no. of -emitted
substituting the value of x from eq. (i) in eq. (ii) we get
ab
a  b 
c d 
2  y ; y  d  
c
4


 2 
Radioactive disintegration series
The phenomenon of natural radioactivity continues till stable nuclei
are formed. All the nuclei from the initial element to the final stable
element constitute a series known as disintegration series. Further we know
that mass numbers change only when -particles are emitted (and not
when -particles are emitted) causing the change in mass of 4 units at each
step. Hence the mass numbers of all elements in a series will fit into one of
the formulae. 4n, 4 n  1 , 4 n  2 and 4 n  3 , hence there can be only
four disintegration series.
4n
58
n
Parent element
90 Th
4n + 1
59
232
94
Half life (yrs)
1.39  10 10
Half life (yrs)
1.39  10 10
Name of series
Thorium
(Natural)
End product
n
Number of lost
particles
82 Pb
4n + 2
59
Pu 241
10
92 U
238
4n + 3
58
92 U
235
4.5  10 9
7.07  10 8
2.2  10 6
4.5  10 9
13.5
Neptunium
(Artificial)
Uranium
(Natural)
Actinium
(Natural)
208
83
Bi 209
82 Pb
206
82 Pb
207
52
52
51
51
 6
 4
 8
 5
 8
 6
 7
 4
Nuclear structure and Nuclear forces
According to an earlier hypothesis, the nucleus is considered as
being composed of two building blocks, proton's and neutron's, which are
collectively called nucleons. The forces, which hold the nucleons together
means stronger proton – proton, neutron – neutron and even proton –
neutron attractive forces, exist in the nucleus. These attractive forces are
called nuclear forces. Nuclear forces operate only within small distance of
about 1  10 15 m or 1 fermi (1 fermi = 10 13 cm ) and drops rapidly to
zero at a distance of 1  10 13 cm. These are referred to as short range
forces. Nuclear forces are nearly 10 21 times stronger than electrostatic
forces.
Yukawa in 1935, put forward a postulate that neutrons and protons
are held together by very rapid exchange of nuclear particles called Pimesons (-mesons have mass equal to 275 times of the mass of an electron
and a charge equal to +1, 0 or –1. These are designated as   and 
respectively). The nuclear force which is used in rapid exchange of Pimesons between nucleons are also called exchange forces.
The binding forces between unlike nucleons (p and n) are explained
+
0
–
by the oscillation of a charged -meson ( or   )
+
(a) p1  n 2
(b) p1  n 2
n1     n 2

n1    p 2
n1  p 2
n1  p 2
Binding forces between like nucleons (p - p or n - n) result from the
exchange of neutral mesons ( ) as represented below.
0
Nuclear Chemistry
p 2   0 or p1   0
p2
(b) n1
n 2   0 or n1   0
n2
nucleons required for completion of the energy levels of the nucleus.
Nucleons are arranged in shells as two protons or two neutrons (with
paired spins) just like electrons arranged in the extra-nuclear part. Thus the
Nuclear stability
Nuclides can be grouped on the basis of nuclear stability, i.e. stable
and unstable nucleus. The most acceptable theory about the atomic nuclear
stability is based upon the fact that the observed atomic mass of all known
isotopes (except hydrogen) is always less from the sum of the weights of
protons and neutrons present in it. Electron (- particle) from a radioactive
nucleus may be regarded as derived from a neutron in the following way,
Neutron  Proton  Electron
Similarly, photons are produced from internal stresses within the
nucleus.
The stability of nucleus may be discussed in terms of any one of the
following,
(1) Nuclear Binding Energy and Mass defect : It is observed that
atomic mass of all nuclei (except hydrogen) is different from the sum of
masses of protons and neutrons. The difference is termed mass defect.
Mass defect = Total mass of nucleons – obs. atomic mass
The mass defect is converted into energy. This energy is
called the binding energy. This is the energy required to break the
nucleus into is constituents (p and n).
following nuclei 2 He 4 , 8 O16 ,20 Ca 40 and 82 Pb208 containing protons 2,
8, 20 and 82 respectively (all magic numbers) and neutrons 2, 8, 20 and 126
respectively (all magic numbers) are the most stable.
Magic numbers for protons : 2, 8, 20, 28, 50, 82,114
Magic numbers for neutrons : 2, 8, 20, 28, 50, 126, 184, 196
When both the number of protons and number of neutrons are
magic numbers, the nucleus is very stable. That is why most of the
radioactive disintegration series terminate into stable isotope of lead (magic
number for proton = 82, magic number for neutron = 126). Nuclei with
nucleons just above the magic numbers are less stable and hence these may
emit some particles to attain magic numbers.
(4) Neutron-proton ratio or causes of radioactivity
It has been
found that the stability of nucleus depends upon the neutron to proton
ratio (n/p). If we plot the number of neutrons against number of protons
for nuclei of various elements, it has been observed that most of the stable
(non-radioactive) nuclei lie in a belt shown by shaded region in figure this is
called stability belt or stability zone. The nuclei whose n/p ratio does not lie
in the belt are unstable and undergo spontaneous radioactive disintegration.
Neutron number
(n)
(a) p1
Binding energy = Mass defect 931MeV
The stability of the nucleus is explained on the value of binding
energy per nucleon and not on the basis of total binding energy . Binding
energy per nucleon is maximum (8.7 MeV) in the case of iron (56). The
value of binding energy per nucleon can be increased either by fusion of
lighter nuclei or fission of heavier nuclei.
Value of binding energy predicts the relative stability of the different
isotopes of an element. If the value of binding energy is negative, the
product nucleus or nuclei will be less stable than the reactant nucleus. Thus
the relative stability of the different isotopes of an element can be predicted
by the values of binding energy for each successive addition of one neutron
to the nucleus.
2 He
3
2 He
4
 0 n1 2 He 4  20.5 MeV
 0 n1  2 He 5  0.8 MeV
Therefore,
2 He
4
is more stable than
2 He
3
and
2 He
5
.
(2) Packing fraction : The difference of actual isotopic mass and the
mass number in terms of packing fraction is defined as,
Packing fraction 
261
Actual isotopicmass  Mass number
 10 4
Mass number
The value of packing fraction depends upon the manner of packing
of the nucleons with in the nucleus. Its value can be negative, positive or
even zero. A negative packing fraction generally indicates stability of the
nucleus.
In general, lower the packing fraction, greater is the binding energy
per nucleon and hence greater is the stability. The relatively low packing
It has been observed that,
Stability belt
Neutron rich
nuclei
n/p=1
Proton rich
nuclei
20 40 60 80 100 120
Atomic number (p)
Fig. 7.2
(i) n / p ratio for stable nuclei lies quite close to unity for elements
with low atomic numbers (20 or less) but it is more than one for nuclei
having higher atomic numbers. Nuclei having n / p ratio either very high or
low undergo nuclear transformation.
(ii) When n / p ratio is higher than required for stability, the nuclei
have the tendency to emit   rays i.e., a neutron is converted into a
proton.
(iii) When n / p ratio is lower than required for stability, the nuclei
increase the ratio, either by emitting   particle or by emitting a position
or by K-electron capture.
Rate of radioactive decay
“According to the law of radioactive decay, the quantity of a radioelement which disappears in unit time (rate of disintegration) is directly
proportional to the amount present.”
The law of radioactive decay may also be expressed mathematically.
Suppose N be the number of atoms of the radioactive element
present at the commencement of observation, t  0 and after time t, the
number of atoms remaining unchanged is N t . The rate of disintegration
0
dN t
 dN t 
  dt  at any time t is directly proportional to N. Then,  dt = N


fraction of He, C and O implies their exceptional stability, packing fraction
is least for Fe (negative) and highest for H (+78).
where  is a radioactive constant or decay constant.
Various forms of equation for radioactive decay are,
(3) Magic number : Nucleus of atom, like extra-nuclear electrons,
also has definite energy levels (shells).
N t  N 0 e t ; log N 0  log N t  0.4343 t
Nuclei with 2, 8, 20, 28, 50, 82 or 126 protons or neutrons have
been found to be particularly stable with a large number of isotopes. These
numbers, commonly known as Magic numbers are defined as the number of
log
N0
t
N
2 .303

log 0
; 
t
Nt
Nt
2 .303
262 Nuclear Chemistry
This equation is similar to that of first order reaction, hence we can
say that radioactive disintegration are examples of first order reactions.
However, unlike first order rate constant (K), the decay constant () is
independent of temperature.
Rate of decay of nuclide is independent of temperature, so its energy
of activation is zero.
(1) Half-life period (T or t ) : The half-life period of a radioelement
is defined, as the time required by a given amount of the element to decay
to one-half of its initial value.
1/2
1/2
t1 / 2 
0 .693

Now since  is a constant, we can conclude that half-life period of a
particular radioelement is independent of the amount of the radioelement.
In other words, whatever might be the amount of the radioactive element
present at a time, it will always decompose to its half at the end of one halflife period.
It is important to note that the term equilibrium is used for
reversible reactions but the radioactive reactions are irreversible, hence it is
preferred to say that B is in a steady state rather than in equilibrium state.
At a steady state,
N A  B T A t 1 / 2  A



N B  A T B t1 / 2 B
Thus at a steady state (at radioactive equilibrium), the amounts
(number of atoms) of the different radioelements present in the reaction
series are inversely proportional to their radioactive constants or directly
proportional to their half-life and also average life periods.
(5) Units of radioactivity : The standard unit in radioactivity is curie
(c) which is defined as that amount of any radioactive material which gives
3.7  1010 disintegration’s per second (dps), i.e.,1c = Activity of 1g of
Ra226  3.7  1010 dps
The millicurie (mc) and microcurie (c) are equal to 10 3 and
10 6 curies i.e. 3.7  10 7 and 3.7  10 4 dps respectively.
Let the initial amount of a radioactive substance be N 0
1c  10 3 mc  10 6 c ; 1c  3.7  1010 dps
Amount of radioactive substance left after n half-life periods
1mc  3.7  107 dps ; 1c  3.7  10 4 dps
n
1
N    N0
2
Total time T  n  t1 / 2 where n is a whole number.
(2) Average-life period (T) : Since total decay period of any element
is infinity, it is meaningless to use the term total decay period (total life
period) for radioelements. Thus the term average life is used.
Average life (T) 
Sum of livesof the nuclei
Total number of nuclei
Average life (T) of an element is the inverse of its decay constant,
1
i.e., T  , Substituting the value of  in the above equation,

10 6 dps. i.e., 1 rd  10 6 dps .
correspondingly rutherford units
microrutherford (rd) respectively.
The millicurie and microcurie
are millirutherford (mrd) and
1 c  3.7  1010 dps  37  10 3 rd
1 mc  3.7  107 dps  37 rd
1 c  3.7  10 4 dps  37 mrd
However, in SI system the unit of radioactivity is Becquerel (Bq)
1 Bq = 1 disintegration per second = 1 dps = 1rd,
10 6 Bq  1 rd , 3.7  1010 Bq  1 c
T
t1 / 2
0 .693
 1 .44 t1 / 2
Thus, Average life (T)  1.44  Half life(T1 / 2 )  2  t1 / 2
Thus, the average life period of a radioisotope is approximately
under-root two times of its half life period.
(3) Activity of population or specific activity : It is the measure of
radioactivity of a radioactive substance. It is defined as ' the number of
radioactive nuclei, which decay per second per gram of radioactive isotope.'
Mathematically, if 'm' is the mass of radioactive isotope, then
Specific activity 
But now a day, the unit curie is replaced by rutherford (rd) which is
defined as the amount of a radioactive substance which undergoes
Rate of decay
N
Avogadro number

 
m
m
Atomic mass in g
where N is the number of radioactive nuclei which undergoes
disintegration.
(4) Radioactive equilibrium : Suppose a radioactive element A
disintegrates to form another radioactive element B which in turn
disintegrates to still another element C.
A  B  C
B is said to be in radioactive equilibrium with A if its rate of
formation from A is equal to its rate of decay into C.
(6) The Geiger-Nuttal relationship : It gives the relationship between
decay constant of an - radioactive substance and the range of the particle emitted.
log  A  B log R
Where R is the range or the distance which an -particle travels
from source before it ceases to have ionizing power. A is a constant which
varies from one series to another and B is a constant for all series. It is
obvious that the greater the value of  the greater the range of the particle.
Artificial transmutation of elements
The conversion of one element into another by artificial means, i.e.,
by means of bombarding with some fundamental particles, is known as
artificial transmutation. The phenomenon was first applied on nitrogen
whose nucleus was bombarded with -particles to produce oxygen.
14
 2 He 4  8 O17  1 H 1
7N
Oxygen isotope Proton
Nitrogen isotope Alpha particle
The element, which is produced, shows radioactivity, the
phenomenon is known as Induced radioactivity. The fundamental particles
which have been used in the bombardment of different elements are,
-particle : 2 He 4 ; Proton : 1 H 1
Deutron : 1 H 2 or 1 D 2 ; Neutron : 0 n1
Nuclear Chemistry
Since -particles, protons and deutrons carry positive charge, they
are repelled by the positively charged nucleus and hence these are not good
projectiles. On the other hand, neutrons, which carry no charge at all, are
the best projectiles. Cyclotron is the most commonly used instrument for
accelerating these particles. The particles leave the instrument with a
velocity of about 25,000 miles per second. A more recent accelerating
instrument is called the synchrotron or bevatron. It is important to note
that this instrument cannot accelerate the neutrons, being neutral.
When a target element is bombarded with neutrons, product
depends upon the speed of neutrons. Slow neutrons penetrate the nucleus
while a high-speed neutron passes through the nucleus.
92 U

238
1
0n
slow speed
 92 U 239 ; 92 U 238 
1
0n
high speed
 92 U 237  20 n1
Thus slow neutrons, also called thermal neutrons are more effective
in producing nuclear reactions than high-speed neutrons.
Alchemy : The process of transforming one element into other is
known as alchemy and the person involved in such experiments is called
alchemist. Although, gold can be prepared from lead by alchemy, the gold
obtained is radioactive and costs very high than natural gold.
(i) Transmutation by -particles
(a) , n type
4
Be 9 ( , n) 6 C12 i.e.
94
4 Be
9
Pu239 (, n) 96 Cm 242 i.e.
 2 He 4  6 C12  0 n1
Pu239  2 He 4  94 Cm 242  0 n1
94
(, p) 10 Ne 22 ie. 9 F19  2 He 4  10 Ne 22  1 H 1
14
N
7
(, p)
8
O 17 i.e.,
N 14  2 He 4  8 O17  1 H 1
7
(c) ,  type
26
Fe59 (,  ) 29 Cu 63 i.e.,
26
Fe59  2 He 4  29 Cu 63  1e 0
(ii) Transmutation by protons
(a) p, n type
15
P 31 (p, n) 16 S 31 i.e.,
12
Be 9 (p, d ) 4 Be 8 i.e.,
4 Be
9
 1 H 1  4 Be 8  1 H 2
(d) p,  type
i.e., 8 O  1 H  7 N
8 O ( p,  ) 7 N
(iii) Transmutation by neutrons
(a) n,p type
16
13
31
16
Al 27 (n, p) 12 Mg 27 i.e.,
13
1
13
 2 He
4
Al 27  0 n1  12 Mg 27  1 H 1
(b) n, type
8O
Be 9    4 Be 8  0 n1
Synthetic elements : Elements with atomic number greater than 92
nature like other elements. All these elements are prepared by artificial
transmutation technique and are therefore known as transuranic elements
or synthetic elements.
Nuclear fission and Nuclear fusion
(1) Nuclear fission : The splitting of a heavier atom like that of
uranium – 235 into a number of fragments of much smaller mass, by
suitable bombardment with sub-atomic particles with liberation of huge
amount of energy is called Nuclear fission. Hahn and Startsman discovered
that when uranium-235 is bombarded with neutrons, it splits up into two
relatively lighter elements.
92 U
235
 0 n 1  56 Ba140  36 Kr 93  3 0 n 1 + Huge amount of energy
Spallation reactions are similar to nuclear fission. However, they
differ by the fact that they are brought by high energy bombarding particles
or photons.
Elements capable of undergoing nuclear fission and their fission
products. Among elements capable of undergoing nuclear fission, uranium is
the most common. The natural uranium consists of three isotopes, namely
U 234 (0.006%) , U 235 (0.7%) and U 238 (99.3%) . Of the three isomers
Uranium-238 undergoes fission by fast moving neutrons while U 235
undergoes fission by slow moving neutrons; of these two, U 235 fission is of
much significance. Other examples are Pu 239 and U 233 .
Uranium-238, the most abundant (99.3%) isotope of uranium,
although itself does not undergo nuclear fission, is converted into
plutonium-239.
93
(p,  ) 7 N 13 i.e., 6 C12  1 H 1  N 13  
(c) p, d type
4
4
i.e. the elements beyond uranium in the periodic table are not found in
92 U
P 31  1 H 1  16 S 31  0 n1
15
(b) p,  type
6C
Be 9 (, n) 4 Be 8 i.e.,
of uranium, nuclear fission of U 235 and U 238 are more important.
(b) , p type
19
9F
4
238
 0 n1  92 U 239 ;
92 U
238
(n, ) 92 U 239 i.e.,
92 U
238
 92 NP 239  1e 0
Which when bombarded with neutrons, undergo fission to emit
three neutrons per plutonium nucleus. Such material like U-238 which
themselves are non-fissible but can be converted into fissible material (Pu239) are known as fertile materials.
Nuclear chain reaction : With a small lump of U 235 , most of the
neutrons emitted during fission escape but if the amount of U 235 exceeds a
few kilograms (critical mass), neutrons emitted during fission are absorbed by
adjacent nuclei causing further fission and so producing more neutrons. Now
since each fission releases a considerable amount of energy, vast quantities
of energy will be released during the chain reaction caused by U 235
fission.
Uranium
E
(c) n,  type
92 U
239
Np 238  94 Pu239  1e 0
(n, ) 12 Mg 27 i.e., 8 O16  0 n1  6 C13  2 He 4
16
263
 0 n1  92 U 238  
Neutron
3E
(d) n, type
8O
18
(n,  ) 9 F19 i.e.,
8O
18
 0 n1  9 F19  1e 0
9E
(iv) Transmutation by deutrons
(a) d,p type
3 Li
6
(d, p) 3 Li7 i.e., 3 Li 6  1 H 2  3 Li7  1 H 1
75
76
i.e.,
32 As (d , p) 32 As
75
 1 H 2  32 As76  1 H 1
32 As
(v) Transmutation by -radiations
(a) , n type
Fig. 7.3
Atomic bomb : An atomic bomb is based upon the process of that
nuclear fission in which no secondary neutron escapes the lump of a fissile
material for which the size of the fissile material should not be less than a
minimum size called the critical size. There is accordingly a sudden release
264 Nuclear Chemistry
of a tremendous amount of energy, which represents an explosive force
much greater than that of the most powerful TNT bomb. In the world war
II in 1945 two atom bombs were used against the Japanese cities of
Hiroshima and Nagasaki, the former contained U-235 and the latter
contained Pu-239.
Atomic pile or Nuclear reactor : It is a device to obtain the nuclear
energy in a controlled way to be used for peaceful purposes. The most
common reactor consists of a large assembly of graphite (an allotropic form
of carbon) blocks having rods of uranium metal (fuel). Many of the
neutrons formed by the fission of nuclei of 92 U 235 escape into the
graphite, where they are very much slow down (from a speed of about
6000 or more miles/sec to a mile/sec) and now when these low speed
neutrons come back into the uranium metal they are more likely to cause
additional fissions. Such a substance like graphite, which slow down the
neutrons without absorbing them is known as a moderator. Heavy water,
D O is another important moderator where the nuclear reactor consists of
rods of uranium metal suspended in a big tank of heavy water (swimming
pool type reactor). Cadmium or boron are used as control rods for
absorbing excess neutrons.
Plutonium from a nuclear reactor : For such purposes the fissile
material used in nuclear reactors is the natural uranium which consists
mainly (99.3%) of U-238. In a nuclear reactor some of the neutrons
produced in U-235 (present in natural uranium) fission converts U-238 to a
long-lived plutonium isotope, Pu-239 (another fissionable material).
Plutonium is an important nuclear fuel. Such reactors in which neutrons
produced from fission are partly used to carry out further fission and partly
used to produce some other fissionable material are called Breeder reactors.
Nuclear reactors in India : India is equipped with the five nuclear
reactors, namely
(i) Apsara (1952)
(ii) Cirus (1960)
(iii) Zerlina (1961)
(iv) Purnima (1972) and R-5
Purnima uses plutonium fuel while the others utilize uranium as fuel.
(2) Nuclear fusion : “Oposite to nuclear fission, nuclear fusion is
2
defined as a process in which lighter nuclei fuse together to form a heavier
nuclei. However, such processes can take place at reasonable rates only at
very high temperatures of the order of several million degrees, which exist
only in the interior of stars. Such processes are, therefore, called
Thermonuclear reactions (temperature dependent reactions). Once a fusion
reaction initiates, the energy released in the process is sufficient to maintain
the temperature and to keep the process going on.
4 1H 1
Hydrogen

4
2 He
Helium
 2 1e 0  Energy
Positron
This is not a simple reaction but involves a set of the thermonuclear
reactions, which take place in stars including sun. In other words, energy of
sun is derived due to nuclear fission.
Controlled nuclear fusion : Unlike the fission process, the fusion
process could not be controlled. Since there are estimated to be some 1017
pounds of deuterium ( 1 H 2 ) in the water of the earth, and since each
pound is equivalent in energy to 2500 tonnes of coal, a controlled fusion
reactor would provide a virtually inexhaustible supply of energy.
Hydrogen bomb : Hydrogen bomb is based on the fusion of
hydrogen nuclei into heavier ones by the thermonuclear reactions with
release of enormous energy.
As mentioned earlier the above nuclear reactions can take place only
at very high temperatures. Therefore, it is necessary to have an external
source of energy to provide the required high temperature. For this
purpose, the atom bomb, (i.e., fission bomb) is used as a primer, which by
exploding provides the high temperature necessary for successful working of
hydrogen bomb (i.e., fusion bomb). In the preparation of a hydrogen bomb,
a suitable quantity of deuterium or tritium or a mixture of both is enclosed
in a space surrounding an ordinary atomic bomb. The first hydrogen bomb
was exploded in November 1952 in Marshall Islands; in 1953 Russia exploded
a powerful hydrogen bomb having power of 1 million tonnes of TNT
A hydrogen bomb is far more powerful than an atom bomb. Thus if it
is possible to have sufficiently high temperatures required for nuclear fusion,
the deuterium present in sea (as D O) is sufficient to provide all energy
requirements of the world for millions of years. The first nuclear reactor was
assembled by Fermi in 1942.
2
Difference between Nuclear fission and fusion
Nuclear fission
Nuclear fusion
The process occurs only in the nuclei of The process occurs only in the nuclei
heavy elements.
of light elements.
The process involves the fission of the The process involves the fission of the
heavy nucleus to the lighter nuclei.
lighter nuclei to heavy nucleus.
The process can take place at ordinary The process takes place at higher
temperature.
temperature ( 10 8 o C) .
The energy liberated during this process The energy liberated during the
is high
process is comparatively low
(200 MeV per fission)
(3 to 24 MeV per fusion)
Percentage efficiency of the energy Percentage efficiency of the energy
conversion is comparatively less.
conversion is high (four times to that
of the fission process).
The process can be controlled for useful The process cannot be controlled.
purposes.
Application of radioactivity
Radioisotopes find numerous applications in a variety of areas such as
medicine, agriculture, biology, chemistry, archeology, engineering and industry.
(1) Age determination : The age of earth has been determined by
uranium dating technique as follows. Samples of uranium ores are found to
contain Pb 206 as a result of long series of - and -decays. Now if it is
assumed that the ore sample contained no lead at the moment of its
formation, and if none of the lead formed from U 238 decay has been lost
then the measurement of the Pb 206 / U 238 ratio will give the value of time
t of the mineral.
No. of atoms of Pb 206
No. of atoms of U 238 left
 e  t 1
where  is the decay constant of uranium-238
Alternatively,
t
2 .303

log
Initial amount of U 238
Amount of U 238 in the mineral present till date
Similarly, the less abundant isotope of uranium, U 235 eventually decays
to Pb 207 ; Th 232 decays to Pb 208 and thus the ratios of Pb 207 / U 235 and
Pb 208 / Th 232 can be used to determine the age of rocks and minerals.
14
(half-life 5760 years) was used by Willard Libby (Nobel
lauret) in determining the age of carbon-bearing materials (e.g. wood,
animal fossils, etc.) Carbon-14 is produced by the bombardment of nitrogen
atoms present in the upper atmosphere with neutrons (from cosmic rays).
6C
7
N 14  0 n1  6 C14  1 H 1
Thus carbon-14 is oxidised to CO 2 and eventually ingested by
plants and animals. The death of plants or animals puts an end to the intake
of C 14 from the atmosphere. After this the amount of C 14 in the dead
tissues starts decreasing due to its disintegration.
6C
14
 7 N 14  1e 0
It has been observed that on an average, one gram of radioactive
carbon emits about 12 -particles per minute. Thus by knowing either the
amount of C-14 or the number of -particles emitted per minute per gram
of carbon at the initial and final (present) stages, the age of carbon material
can be determined by using the following formulae.
Nuclear Chemistry

N
N
2 .303
2 .303
log 0 or t 
log 0
t
Nt

Nt
where t = Age of the fossil,  = Decay constant, N 0 = Initial radioactivity
(in the fresh wood), N t = Radioactivity in the fossil
The above formula can be modified as,
t
2 .303

Similarly,
log
Initial ratio of C 14 / C 12 (in fresh wood)
C 14 / C 12 ratio in the old wood
tritium
1
H 3 has been used for dating purposes.
(2) Radioactive tracers (use of radio–isotopes) : A radioactive isotope
can be easily identified by its radioactivity. The radioactivity can, therefore
act as a tag or label that allows studying the behaviour of the element or
compounding which contains this isotope. An isotope added for this
purpose is known as isotopic tracer. The radioactive tracer is also known as
an isotopic tracer. The radioactive tracer is also known as an indicator
because it indicates the reaction. Radioisotopes of moderate half-life periods
are used for tracer work. The activity of radioisotopes can be detected by
means of electroscope, the electrometer or the Geiger-Muller counter.
Tracers have been used in the following fields,
(i) To diagnose many diseases : For example, Arsenic – 74 tracer is
used to detect the presence of tumours, Sodium – 24 tracer is used to
detect the presence of blood clots and Iodine –131 tracer is used to study
the activity of the thyroid gland. It should be noted that the radioactive
isotopes used in medicine have very short half-life periods.
265
The increased pace of synthesis and use of radio isotopes has led to
increased concern about the effect of radiations on matter, particularly in
biological systems. The accident of Chernobyl occurred in 1986 in USSR is no
older when radioisotopes caused a hazard there. The nuclear radiations (alpha,
beta, gamma as well as X-rays) possess energies far in excess of ordinary bond
energies and ionisation energies. Consequently, these radiations are able to
break up and ionise the molecules present in living organisms if they are
exposed to such radiations. This disrupts the normal functions of living
organisms. The damage caused by the radiations, however, depends upon the
radiations received. The resultant radiation damage to living system can be
classified as,
(1)Somatic or pathological damage : This affects the organism during
its own life time. It is a permanent damage to living civilization produced in
body. Larger dose of radiations cause immediate death whereas smaller
doses can cause the development of many diseases such as paralysis, cancer,
leukaemia, burns, fatigue, nausea, diarrhoea, gastrointestinal problems etc.
some of these diseases are fatal. Many scientists presently believe that the
effect of radiations is proportional to exposure, even down to low exposures.
This means that any amount of radiation causes some finite risk to living
civilization.
(2) Genetic damage : As the term implies, radiations may develop
genetic effect. This type of damage is developed when radiations affect genes
and chromosomes, the body's reproductive material. Genetic effects are more
difficult to study than somatic ones because they may not become apparent
for several generations.
(ii) In agriculture : The use of radioactive phosphorus 32 P in
fertilizers has revealed how phosphorus is absorbed by plants. This study
has led to an improvement in the preparation of fertilizers. 14 C is used to
study the kinetics of photo synthesis.
(iii) In industry : Radioisotopes are used in industry to detect the
leakage in underground oil pipelines, gas pipelines and water pipes.
Radioactive carbon has been used as a tracer in studying mechanisms
involved in many reactions of industrial importance such as alkylation,
polymerization, catalytic synthesis etc.
(iv) In analysis : Several analytical procedures can be used employing
radioisotopes as tracers.
(a) A small amount of radioactive isotope is mixed with the inactive
substance and the activity is studied before and after adsorption. Fall in
activity gives the amount of substance adsorbed.
(b) The solubility of lead sulphate in water may be estimated by
mixing a known amount of radioactive lead with ordinary lead.
(c) Ion exchange process of separation is readily followed by
measuring activity of successive fractions eluted from the column.
(d) By labelling oxygen of the water, mechanism of ester hydrolysis
has been studied.
(e) The efficiency of analytical procedures may be measured by
adding a known amount of radio-isotopes to the sample before analysis
begins. After the completion, the activity is again determined. The
comparison of activity tells about the efficiency of separation.
(3) Use of  rays :  rays are used for disinfecting food grains and
for preserving food stuffs. Onions, potatoes, fruits and fish etc., when
irradiated with  rays, can be preserved for long periods. High yielding
disease resistant varieties of wheat, rice, groundnut, jute etc., can be
developed by the application of nuclear radiations. The  rays radiations are
used in the treatment of cancer. The  radiations emitted by cobalt –60
can burn cancerous cells. The  radiations are used to sterilize medical
instruments like syringes, blood transfusion sets. etc. These radiations make
the rubber and plastics objects heat resistant.
Hazards of radiations
 The particle like mesons, positron, neutrino, etc, about 20 in
number are created by stresses in nucleus but do not exist as
component of nucleus.
 Highest degree of radioactivity is shown by radium.
 The nuclear forces are not governed by inverse square law.
 About 42 radioactive nuclides (Z > 82) occur in nature. Each of these
gives stable end product of an isotope of lead.
 The half life is independent of physical or chemical state of a
radioactive element.
 The average life of the natural radioactive element vary from 10 s 10
–6
10
years or more.
 It has been observed that fusion of 45 mg of hydrogen produce as
much energy as obtained from one ton of coal.
 Beryllium has been found to be the best moderator as it occupies small
space and have low absorption cross-section.
 The total life span of a radioactive element is infinite.
 The  - radiation of total energy 1.02 MeV, emitted when a positron
and an electron interact are known as annilation radiation.
Nuclear Chemistry
10.
267
Which can be used for carrying out nuclear reaction
[AFMC 2003]
(a) Uranium – 238
(c) Thorium – 232
11.
Nucleus (Stability and Reaction)
1.
2.
3.
4.
Nucleons are
[CPMT 1982]
(a) Protons and electrons
(b) Protons and neutrons
(c) Electrons and neutrons
(d) Electrons, protons and neutrons
A deutron contains
[NCERT 1982; CPMT 1994]
(a) A neutron and a positron
(b) A neutron and a proton
(c) A neutron and two protons
(d) A proton and two neutrons
The nucleus of radioactive element possesses
(a) Low binding energy
(b) High binding energy
(c) Zero binding energy
(d) High potential energy
6.
12.
13.
14.
15.
17
(b)
9F
F 17
(d)
8
(a)
8O
(c)
9
18
O 18
Nuclear energy is based on the conversion of
(a) Protons into neutrons
(b) Mass into energy
(c) Neutrons into protons
(d) Uranium into radium
Positron has nearly the same weight as that of
[NCERT 1975; JIPMER 1991; BHU 1995]
8.
9.
(b) C 14 is more reactive
(d) Both are equally active
234
The radionucleide 90
Th undergoes two successive  -decays
followed by one  -decay. The atomic number and the mass
number respectively of the resulting radionucleide are
(a) 92 and 234
(b) 94 and 230
(c) 90 and 230
(d) 92 and 230
Hydrogen and deuterium differ in
[CPMT 1980]
(a) Reactivity with oxygen
(b) Reactivity with chlorine
(c) Melting point
(d) Reducing action
A nuclear reaction must be balanced in terms of
(a) Only energy
(b) Only mass
(c) Mass and energy
(d) None of these
In the following nuclear reaction, the other product is
130
[MP PET 1991]
1 H 2 53 I131  ?
52 Te
[
(a) Positron
(b) Alpha particle
On bombarding 7 N 14 with  -particles, the nuclei of the product
(c) One neutron
(d) Proton
formed after the release of a proton will be or In nuclear reaction
14
4
A
1
A
8
8
0
MNR 1995;
 2 He  Z X  1 H , the term Z X represents[NCERT 1979;16.MP PMT
The1989;
reaction
is due to
[MP PMT 1991]
7N
5 B 4 Be 1 e
(a)  -particle
(c) Neutron
7.
On comparing chemical reactivity of C 12 and C 14 , it is revealed
that
(a) C 12 is more reactive
(c) Both are inactive
MP PET 1996; BHU 1996]
5.
(b) Neptunium – 239
(d) Plutonium – 239
In the reaction
is
(a) Electron
(c) Proton
(b) Proton
(d) Electron
3
17.
18.
19.
Li 6  (?)  2 He 4  1 H 3 . The missing particle
[CPMT 1983, 84]
(b) Neutron
(d) Deutron
20.
The 6 C 14 in upper atmosphere is generated by the nuclear
reaction
[MP PET 1993]
(a)
7
N 14  1 H 1  6 C 14 
(b)
7
N 14  6 C 14 
(c)
7
N 14  0 n 1  6 C 14  1 H 1
(d)
7
N 14  1 H 3  0 n 1  6 C 14  2 He 4
1 e
1 e
0
 1H1
21.
0
22.
17 Cl
35
(c)
17 Cl
37
(b)
19
K 27
(d)
19
K 39
In the nuclear reaction
nucleus is
27
(a)
13
Al
(c)
13
Al 28
12
Mg 24  2 He 4  0 n1  ? The product
[BHU 1987]
27
(b)
14
Si
(d)
12
Mg 25
is formed from 7 N 14 in the upper atmosphere by the
action of the fundamental particle
[Orissa JEE 2002]
(a) Positron
(b) Neutron
(c) Electron
(d) Proton
In the nuclear reaction
6C
14
92 U
238
82 Pb 206  x 2 He 4  y
1 
0
the value of x and y are respectively ……..
Deuterons when bombarded on a nuclide produce 18 Ar 38 and
neutrons. The target is
[CPMT 1982, 87]
(a)
(a) Loss of  -particles
(b) Loss of  -particles
(c) Loss of positron
(d) Electron loss
Positronium is the name given to an atom-like combination formed
between
[NCERT 1980; JIPMER 1991]
(a) A positron and a proton
(b) A positron and a neutron
(c) A positron and  -particle
(d) A positron and an electron
An electrically charged atom or a group of atoms is known as
(a) A meson
(b) A proton
(c) An ion
(d) A cyclotron
The charge on positron is equal to the charge on which one of the
following
[NCERT 1977]
(a) Proton
(b) Electron
(c)  -particle
(d) Neutron
[Orissa JEE 2002]
23.
(a) 8, 6
(b) 6, 4
(c) 6, 8
(d) 8, 10
If an isotope of hydrogen has two neutrons in its atom, its atomic
number and atomic mass number will respectively be
[
268 Nuclear Chemistry
24.
25.
[CBSE 1992]
The value of n will be
[MP PMT 1999]
(a) 2 and 1
(b) 3 and 1
(a) 3
(b) 4
(c) 1 and 1
(d) 1 and 3
(c) 5
(d) 6
Which one of the following nuclear transformation is (n, p ) type [AIIMS 1980, 83]
34.
The introduction of a neutron into the nuclear composition of an
atom would lead to a change in
[MNR 1995]
(a) 3 Li7  1 H 1  4 Be 7  0 n1
(a) The number of the electrons also
75
4
78
1
(b) 33 As  2 He  35 Br  0 n
(b) The chemical nature of the atom
(c) 83 Bi209  1 H 2  84 Po 210  0 n1
(c) Its atomic number
(d) Its atomic weight
45
1
45
1
(d) 21 Sc  0 n  20 Ca  1 H
35.
The composition of tritium ( 1 H 3 ) is
What is X in the following nuclear reaction
7
26.
27.
N 14  1 H 1 8 O 15  X
(a)
1 e
(c)

0
[Manipal MEE 1995; DPMT 1982,96]
[AIIMS 1983; MP PET 1997]
1
(b)
0n
(d)
1 e
0
In the reaction 93 Np 239 94 Pu 239 + (?), the missing particle
is [MNR 1987]
(a) Proton
(b) Positron
(c) Electron
(d) Neutron
According to the nuclear reaction
(a)
(b)
(c)
(d)
4 Be
 2 He 4 6 C 12  0 n 1 ,
mass number of (Be ) atom is
36.
37.
28.
29.
30.
31.
32.
(a)
50
Sn 115
(b)
82
Pb 206
(c)
82
Pb 208
(d)
50
Sn 118
In the carbon cycle, from which hot stars obtain their energy, the
14
nucleus is
6C
(a) Completely converted into energy
(b) Regenerated at the end of the cycle
(c) Combined with oxygen to form carbon monoxide
(d) Broken up into its constituents protons and neutrons
The atomic mass of lead is 208 and atomic number is 82. The
atomic mass of bismuth is 209 and atomic number is 83. The ratio
of n/p in the atom is
[EAMCET 1982]
(a) Higher of lead
(b) Higher of bismuth
(c) Same
(d) None of these
Which of the following is an n, p reaction
[BHU 1995]
13
 1 H 1  6 C 14
5C
(b)
7
(c)
13
(d)
235
 0 n1  54 Xe 140  38 Sr 93  3 0 n1
92 U
38.
39.
1H
(c)
0n
S 32  X  15 P 30  2 He 4
1
(b)
1D
2
(d) e 
1
In terms of energy 1 a. m.u. is equal to
40.
[AIIMS 1997]



X 
 Y 
 Z 
 W
In the above sequence of reaction, the elements which are isotopes
of each other are
[JIPMER 1997]
X and W
(b) Y and Z
(c) X and Z
(d) None of these
Stable nuclides are those whose n/p ratio is
[MP PMT 1993]
(a)
41.
(b) 931.1 MeV
(c) 931.1 kcal
(d) 10 7 erg
Positron is
(a) Electron with +ve charge
(b) A helium nucleus
(c) A nucleus with two protons
(d) A nucleus with one neutron and one proton
(a)
n/ p 1
(b) n / p  2
(c) n / p  1
(d) n / p  1
Neutrino has
[NCERT 1981]
(a) Charge +1, mass 1
(b) Charge 0, mass 0
(c) Charge – 1, mass 1
(d) Charge 0 , mass 1
Which one of the following nuclear reaction is correct
[CPMT 1997]
N 14  1 H 1  8 O 15
Al 27  0 n1  12 Mg 27  1 H 1
Which one of the following statements is incorrect
[MP PET 1997]
(a) Mass defect is related with binding energy
(b) ‘Meson’ was discovered by Yukawa
33.
(a)
16
(a) 100 J
42.
(a)
Identify ' X ' in
[MP PET/PMT 1998]
[AFMC 2002]
(a) 4
(b) 9
(c) 7
(d) 6
Which of the following nuclides has the magic number of both
protons and neutrons
[EAMCET 1989]
1 electron, 1 proton, 1 neutron
1 electron, 2 protons, 1 neutron
1 electron, 1 proton, 2 neutrons
1 electron, 1 proton, 3 neutrons
(c) The size of the nucleus is of the order of 10 12  10 13 cm
(d) Magnetic quantum number is a measure of ‘orbital angular
momentum’ of the electron
In
the
sequence
of
following
nuclear
reactions



n 
238
218

 Y 
 Z 
 L 84 M
92 X
43.
13
1 H 7 N
1
13

  v

(a)
6C
(b)
11
Na 23  1 H 1  10 Ne 20  2 He 4
(c)
13
Al 23  0 n 1  11 Na 23  e 0
(d) None of these
Formation of nucleus from its nucleons is accompanied by
[NCERT 1975; RPET 2000]
44.
(a) Decrease in mass
(b) Increase in mass
(c) No change of mass
(d) None of them
A particle having the same charge and 200 times greater mass than
that of electron is
(a) Positron
(b) Proton
(c) Neutrino
(d) Meson
Nuclear Chemistry
45.
46.
The positron is
(a)
1 e
0
(c)
1H
1
[AFMC 1997]
(b)
1 e
(d)
0n
58.
0
1
Which of the following is the most stable atom
[AFMC 1997]
48.
(a)
49.
50.
How many neutrons are present in the nucleus of Ra
51.
(a) 88
(b) 226
(c) 140
(d) 138
In a nuclear explosion, the energy is released in the form of
[CPMT 1980]
[CPMT 1994]
(a) Kinetic energy
(c) Potential energy
52.
In equation
11 Na
23
60.
61.
62.
54.
55.
56.
1
(a)
0n
63.
64.
1 H 1 12 Mg23  x , x represents
(c)
1H
1
(b)
1 e
0
(d)
1H
2
Which of the following sub-atomic particles is not present in an
atom
[JIPMER 1999]
(a) Neutron
(b) Proton
(c) Electron
(d) Positron
Electromagnetic radiation with maximum wave length is
65.
66.
67.
68.
[DCE 2000; UPSEAT 2000]
57.
(a) Ultraviolet ray
(b) Radiowave
(c) X-ray
(d) Infrared
Neutrons are obtained by
(a) Bombardment of Ra with -particles
(b) Bombardment of Be with -particles
(c) Radioactive disintegration of uranium
(d) None of these
4
2
6
3
(b)
He
Li
8
4
(c)
(d)
Li
Be
Which of the following does not contain number of neutrons equal to
40
that of 18
[MP PMT 2000]
Ar
(a)
41
19
(c)
40
21
K
Sc
(b)
43
21
Sc
(d)
42
20
Ca

Nuclear reactivity of Na and Na is same because both have
(a) Same electron and proton
(b) Same proton and same neutron
(c) Different electron and proton
(d) Different proton and neutron
Which of the following is the heaviest metal
[MH CET 2001]
[MP PMT 1990; MP PET 1999]
53.
[DCE 2000]
Be( p,  ) X , the X is
(a) Hg
(c) Ra
(b) Electrical energy
(d) None of these
(a) Neutron
(b) Deutron
(c)  -particle
(d) Positron
Which of the following atomic mass of uranium is the most
radioactive
[AFMC 1997]
(a) 238
(b) 235
(c) 226
(d) 248
Which of the following particle is emitted in the reaction
27
[DCE 1999]
 2 He 4 14 P 30  .....
13 Al
9
4
7
3
[NCERT 1980]
(a) X and Y are both even
(b) X and Y are both odd
(c) X is even and Y is odd
(d) X is odd and Y is even
Atom A possesses higher values of packing fraction than atom B.
The relative stabilities of A and B are
(a) A is more stable than B
(b) B is more stable than A
(c) A and B both are equally stable
(d) Stability does not depend on packing fraction
In the nuclear reaction
if Bi, belongs to
[MP PMT 2000]
(a)
47.
(b) Al
Bi
(c) U
(d) Pb
The positron is discovered by
[RPMT 1997]
(a) Pauling
(b) Anderson
(c) Yukawa
(d) Segar
The nucleus of an atom is made up of X protons and Y neutrons.
For the most stable and abundant nuclei
59.


In the reaction, Po 
 Pb 
 Bi,
group 15, to which Po belongs
(a) 14
(b) 15
(c) 13
(d) 16
269
(b) Pb
(d) U
In the following reaction, x will be
29 Cu
64
28 Ni64  x
(a) A proton
(b) An electron
(c) A neutron
(d) A positron
Which one out of the following statements is not correct for ortho
and para hydrogen
[Orissa JEE 2002]
(a) They have different boiling point
(b) Ortho form is more stable than para form
(c) They differ in the spin of their protons
(d) The ratio of ortho to para hydrogen increases with increase in
temperature and finally pure ortho form is obtained
For the nuclear reaction,
nucleide is
(a)
22
11 Na
(c)
23
12 Mg
ZX
M
24
2
12 Mg  1 D
   ? , the missing
[Kurukshetra CEE 2002]
(b)
23
11
(d)
26
12 Mg
Na
 2 He 4 15 P 30  0 n 1 . Then
[KCET 2002]
(a)
Z  12, M  27
(b)
Z  13, M  27
(c)
Z  12, M  17
(d)
Z  13, M  28
An element 96 X
emits 4 and 5  particles to form new
element Y. Then atomic number and mass number of Y are
(a) 93; 211
(b) 211; 93
(c) 212; 88
(d) 88; 211
Meson was discovered
[MH CET 2004]
(a) Yukawa
(b) Austin
(c) Moseley
(d) Einstein
227
Radioactivity and ,  and - rays
[JIPMER 1999]
1.
Which of the following does not contain material particles
2.
(a) Alpha rays
(b) Beta rays
(c) Gamma rays
(d) Canal rays
Radioactive substances emit -rays, which are
[BHU 2002]
270 Nuclear Chemistry
[Orissa JEE 2002]
3.
(a) + ve charged particle
(b) – ve charged particle
(c) Massive particle
(d) Packet of energy
Which statement is incorrect
[CPMT 1982]
(a)  -rays have more penetrating power than  -rays
(b)  -rays have less penetrating power than  -rays
(c)
4.
5.
6.
(d)  -rays have more penetrating power than  -rays
The velocity of  -rays is approximately
[CPMT 1982]
17.
(a) Equal to that of the velocity of light
(b) 1/10 of the velocity of light
(c) 10 times more than the velocity of light
(d) Uncomparable to the velocity of light
The radiations having high penetrating power and not affected by
electrical and magnetic field are
[Kerala CET 1992]
18.
(a) Alpha rays
(b) Beta rays
(c) Gamma rays
(d) Neutrons
Alpha particles are ...... times heavier (approximately) than neutrons [CPMT 1971]
(a) 2
(b) 4
8.
9.
10.
11.
Uranium
(d)
92 U
235
2
1
2
19.
on bombardment with slow neutrons produces[CPMT 1982]
20.
(b) Fusion reaction
(a) Deutrons
(c) Fission reaction
(d) Endothermic reaction
-particles can be detected using
[AIIMS 2005]
(a) Thin aluminum sheet
(b) Barium sulphate
(c) Zinc sulphide screen
(d) Gold foil
Alpha rays consist of a stream of
[BHU 1979]
(a) H 
(b) He 2
(c) Only electrons
(d) Only neutrons
Which is the correct statement
(a) Isotopes are always radioactive
(b)  -rays are always negatively charged particles
(c)  -rays are always negatively charged particles
(d)  -rays can be deflected in magnetic field
The  -particle is identical with
[DPMT 1986]
13.
 -rays
(b) Neutrons
(c)  -particles
Radioactivity was discovered by
15.
Penetrating power of -particle is
(d) -particle is not emitted
[CPMT 1973, 78; NCERT 1977]
 -rays have
(a) Positive charge
(b) Negative charge
(c) No charge
(d) Sometimes positive charge and sometimes negative charge
X-rays are produced due to
[JIPMER 2002]
(a) Bombarding of electrons on solids
(b) Bombarding of -particle on solids
(c) Bombarding of -rays on solids
(d) Bombarding of neutron on solids
Choose the element which is not radioactive
[CPMT 1988]
(a) Cm
(b) No
(c) Mo
(d) Md
A magnet will cause the greatest deflection of
[MP PMT 1991]
23.
(a) Radium
(b) Carbon
(c) Lead
(d) Neptunium
Which leaves no track on Wilson cloud chamber
[MP PET 1995]
24.
(a) Electrons
(b) Protons
(c)  -particles
(d) Neutrons
Which has the least penetrating power
(a)  -rays
(b)  -rays
[CPMT 1994]
25.
(c)  -rays
There exists on  -rays
26.
(a) Positive charge
(b) Negative charge
(c) No charge
(d) Sometimes positive charge, sometimes negative charge
Which is not emitted by radioactive substance
(d)
X -rays
[MP PMT 1996; Pb. PMT 2004; EAMCET 2004]
[AIIMS 1997]
[CPMT 1983, 88; DPMT 1982; AMU 1983;
MADT Bihar 1982]
14.
(a) Conversion of proton to neutron
(b) Form outermost orbit
(c) Conversion of neutron to proton
22.
21.
(d)  -particles
(a) Henry Becquerel
(b) Rutherford
(c) J. J. Thomson
(d) Madam Curie
Which of the following is radioactive element
(a) Sulphur
(b) Polonium
(c) Tellurium
(d) Selenium
(d) None of these
-particle is emitted in radioactivity by
[AFMC 1988]
(a) Helium nucleus
(b) Hydrogen nucleus
(c) Electron
(d) Proton
If by mistake some radioactive substance gets inside the human
body, then from the point of view of radiation damage, the most
harmful will be the one which emits
(a)
(c) Less than -rays
(a)  -rays
(b)  -rays
(c)  -rays
(d) Neutrons
Of the following radiations, the one most easily stopped by air is [MP PMT 1991
(a)  -rays
(b)  -rays
(c)  -rays
(d) X-rays
Uranium ultimately decays into a stable isotope of
[CPMT 1971]
[CPMT 1972, 82, 86; BHU 1984;
MP PMT 1990, 91, 93; MP PET 1999]
12.
(b) More than -rays
[AIEEE 2002; MP PMT 2004]
 -rays have less penetrating power than  -rays
(c) 3
7.
16.
(a) More than -rays
27.
(a)  -rays
(b)  -rays
(c) Positron
(d) Proton
The radiations from a naturally occurring radio element, as seen
after deflection in a magnetic field in one direction, are
[IIT 1984; MP PMT 1986;
MP PET/PMT 1988 JIPMER 1999]
(a) Definitely -rays
(c) Both  and -rays
[CPMT 1988]
28.
[MP PMT 2002]
The
88 Ra
226
is
(b) Definitely -rays
(d) Either  or -rays
[AIIMS 2001]
Nuclear Chemistry
(a) n-mesons
(c) Radioactive
(b) u-mesons
(d) Non-radioactive
(c) 0.0018 amu
(e) 18 amu
30.
During -decay
[UPSEAT 2001]
(a) An atomic electron is ejected
(b) An electron which is already present with in the nucleus is
ejected
(c) A neutron in the nucleus decays emitting an electron
(d) A part of binding of the nucleus is converted into an electron
The element californium belongs to the family of
31.
(a) Actinide series
(b) Alkali metal family
(c) Alkaline earth family
(d) Lantanide series
Which of the following is not deflected by magnetic field
29.
1.
32.
Which of the following can be used to convert
(a) Deuteron
33.
14
7
N into
17
8 O
34.
What happens when -particle is emitted
35.
(a) Mass number decreases by 12 unit, atomic number decreases by
4 unit
(b) Mass number decreases by 4 unit, atomic number decreases by
2 unit
(c) Only mass number decreases
(d) Only atomic number decreases
The charge on gamma rays is
[CBSE PMT 1989; JIPMER 2002]
4.
[Pb. PMT 2004; EAMCET 2004]
(a) Zero
(b) +1
(c) –1
(d) +2
A nuclear reaction is accompanied by loss of mass equivalent to
0.01864 amu . Energy liberated is
5.
[DCE 2002]
37.
(c)
17.36 MeV
(d)
6.
460 MeV
Nuclear theory of the atom was put forward by
39.
(a) Rutherford
(b) Aston
(c) Neils Bohr
(d) J.J. Thomson
Decrease in atomic number is observed during
(a) Alpha emission
(b) Beta emission
(c) Positron emission
(d) Electron capture
Calculate mass defect in the following reaction
1H
2
7.
[IIT 1998]
1 H 3 1 He 4 0 n1
(Given
:
mass
(a) 0.018 amu

particles
H 2  2.014, H 3  3.016, He  4.004,

particles
2 groups to the right
1 group to the right
2 groups to the right
1 group to the left
2 groups to the left
1 group to the left
2 groups to the left
1 group to the right
nuclides (A nuclide is the general name for any nuclear species)
12
56
and 92 U 238 have 12, 56 and 238 nucleons
6 C , 26 Fe
respectively in the nuclei. The total number of nucleons in a nucleus
is equal to
[NCERT 1975]
(a) The total number of neutrons in the nucleus
(b) The total number of neutrons in the atom
(c) The total number of protons in the nucleus
(d) The total number of protons and neutrons in the nucleus
Radioactivity is due to
[DPMT 1983, 89; AIIMS 1988]
(a) Stable electronic configuration
(b) Unstable electronic configuration
(c) Stable nucleus
(d) Unstable nucleus
Radioactive disintegration differs from a chemical change in being [MNR 1991]
(a) An exothermic change
(b) A spontaneous process
(c) A nuclear process
(d) A unimolecular first order reaction
U emits 8 -particles and 6 -particles. The neutron/proton
ratio in the product nucleus is
238
92
(a) 60/41
(b) 61/40
(c) 62/41
(d) 61/42
The element with atomic number 84 and mass number 218 change
to other element with atomic number 84 and mass number 214. The
number of  and  -particles emitted are respectively [CPMT 1989]
(b) 1, 4
(d) 1, 5
A radium 88 Ra 224 isotope, on emission of an  -particle gives
rise to a new element whose mass number and atomic number will
be
[CPMT 1980; EAMCET 1985; MP PMT 1993]
[Kerala CET 2005]
(b) 0.18 amu
On emission of
(a)
(b)
(c)
(d)
The
(a) 1, 3
(c) 1, 2
8.
n  1.008 amu )
4 n  1 and 4 n radioactive disintegration series
[AIIMS 2005]
[KCET 2004]
38.
(d)
2.
3.
(b) Lattice energy
(d) None of these
(b) 186.6 MeV
belong respectively to
(c)
(b)
(c) -particle
(d) Neutron
The amount of energy, which is required to separate the nucleons
from a nucleus. The energy is called
931 MeV
234
4 n and 4 n  1 radioactive disintegration series
4 n  1 and 4 n  2 radioactive disintegration series
4 n  1 and 4 n  3 radioactive disintegration series
On emission of
(a)
90 Th
[MP PMT 1999]
[UPSEAT 2001]
36.
Am 241 and
(a)
(b) Proton
(a) Binding energy
(c) Kinetic energy
95
Group displacement law states that the emission of  or 
particles results in the daughter element occupying a position, in the
periodic table, either to the left or right of that of the parent
element. Which one of the following alternatives gives the correct
[MP PMT 2001] position of the daughter element
[MP PMT 2001]
(b) Positron
(d) Photon
(d) 1.8 amu
Causes of radioactivity
and Group displacement law
[UPSEAT 2002]
(a) Deuteron
(c) Proton
271
(a) 220 and 86
(c) 228 and 88
(b) 225 and 87
(d) 224 and 86
272 Nuclear Chemistry
9.
89
Ac 231 gives
82
Pb 207
after emission of some  and  -
19.
particles. The number of such  and  -particles are respectively[MP PMT 1993;
(a) 5, 6
(c) 7, 5
10.
(b) 6, 5
(d) 5, 7
The number of  and  - particles emitted in the nuclear reaction
20.
228
83 Bi212 are respectively
90 Th
11.
The number of neutrons in the parent nucleus which gives N 14 on
 -emission and the parent nucleus is
(a)
8, C
(c)
4 , C 13
(b)
6, C
14.
15.
22.
(d) None of these
92
X 238 , the
(a) 138
(b) 140
(c) 144
(d) 150
When a radioactive element emits an electron the daughter element
formed will have
[EAMCET 1988; MP PET 1994]
(a) Mass number one unit less
(b) Atomic number one unit less
(c) Mass number one unit more
(d) Atomic number one unit more
If the amount of radioactive substance is increased three times, the
number of atoms disintegrated per unit time would [MP PMT 1994]
(a) Be double
(b) Be triple
(c) Remain one third
(d) Not change
(b) Group - 4
(d) Group - 2
Number of neutrons in a parent nucleus X , which gives
7
of
A X undergoes a series of m
Y
alpha and n beta
. The values of
[MP PET 1995]
24.
(a)
242
96
(c)
14
7
243
Cm ( , 2n) 97
Bk
N (n, p) 14
6 C
(b)
10
5
B ( , n) 13
7 N
(d)
28
14
29
Si (d, n) 15
P
25.
An atom has mass number 232 and atomic number 90. How many
 -particles should it emit after emission of two  -particles, so
that the new element's atom has mass number 212 and atomic
number 82
(a) 4
(b) 5
(c) 6
(d) 3
26.
After the emission of one  -particle followed by one   particle
27.
from the atom of 92 X 238 , the number of neutrons in the atom will
be [CBSE PMT 1995]
(a) 142
(b) 146
(c) 144
(d) 143
A nuclide of an alkaline earth metal undergoes radioactive decay by
emission of the   particles in succession. The group of the
periodic table to which the resulting daughter element would belong
is [CBSE PMT 2005]
(a) Gr.14
(b) Gr.16
(c) Gr.4
(d) Gr.6
28.
Which one of the following is not correct
N 14
[CBSE PMT 1998; MP PMT 2003]
(a) 9
(b) 8
(c) 7
(d) 6
The
disintegration
of
an
isotope
24
24
0
12 Mg  1 e shown is due to
11 Na
An isotope
(b)  -emission
(d) Proton emission
[MP PET 1996]
nucleus after two successive  emissions would be
sodium.
29.
(a)
7
1
7
1
3 Li  1 H  4 Be  0 n
(b)
21 Sc
(c)
33
As 75  2 He 4  35 Br 78  0 n1
(d)
83
Bi209  1 H 2  84 Po 210  0 n1
45
[MP PMT 1997]
 0 n1  20 Ca 45  0 n1
The end product of (4 n  2) radioactive disintegration series is
[MP PET 1997; Pb. PMT 1998; BHU 2000]
[AMU (Engg.) 2000]
The emission of -radiation
The formation of a stable nuclide
The fall in the neutron : proton ratio
None of these
Al is expected to disintegrate by
(a) Decreases by one unit
(b) Increases by one unit
(c) Decreases by two units
(d) Remains unaffected
Which one of the following notations shows the product incorrectly [MP PET/PM
Nd (Z  60) is a member of group -3 in periodic table. An isotope
(a)
(b)
(c)
(d)
29
13
During a  -decay the mass of the atomic nucleus
of it is  -active. The daughter nuclei will be a member of
18.
Al is a stable isotope.
23.
 -particles are emitted from the atom
(c) Due to removal of electron from K shell
(d) Due to removal of electron from outermost orbit
17.
Bi
(d) C
disintegration to form a stable isotope Y 10 B
m and n are respectively
(a) 6 and 8
(b) 8 and 10
(c) 5 and 8
(d) 8 and 6
12
After the emission of  -particle from the atom
number of neutrons in the atom will be
(a) Group -3
(c) Group -1
27
13
(b)
Sn
Pb
X 32
(a) Due to disintegration of neutron
(b) Due to disintegration of proton
16.
is changed to
total number of -particles lost in this process is[UPSEAT 1999, 2000
(a)  -emission
(c) Positron emission
[MNR 1993; UPSEAT 1999, 2001, 02]
13.
238
[IIT 1996; UPSEAT 2001]
[EAMCET 1985; MNR 1992;
Kurukshetra CEE 1998; UPSEAT 2000, 01]
12.
21.
92 U
(a) 10
(b) 5
(c) 8
(d) 32
Which element is the end product of each natural radioactive series[MP PMT 199
(c)
(b) 3, 7
(d) 4, 7
14
206
UPSEAT
2001]
. The
82 Pb
(a)
[MNR 1992; MP PMT 1993; AFMC 1998, 2001;
MH CET 1999; UPSEAT 2000, 01; AMU 2001; CPMT 2002]
(a) 4, 1
(c) 8, 1
After losing a number of  and -particles.
(a)
82
(c)
82 Pb
Pb
208
207
Pb 206
(b)
82
(d)
210
83 Bi
Nuclear Chemistry
30.
The element Th belongs to thorium series. Which of the following
will act as the end product of the series
232
90
41.
Whenever the parent nucleus emits a -particle, the daughter element
is shifted in the periodic table
[NCERT 1984]
(a) One place to the right
(b) One place to the left
(c) Two places to the right
(d) Two places to the left
42.
In the nuclear reaction 92 U 238  82 Pb 206 , the number of alpha
and beta particles decayed are
[BHU 2005]
(a)
(c)
31.
32.
Pb
Pb
(b) Bi
(d) Pb
208
82
209
82
206
207
82
82
On bombarding
formed will be
8
O 16 with deutrons, the nuclei of the product
[NCERT 1978]
18
(a)
9
F
(c)
8
O 17
17
(b)
9
F
(d)
7
N 14
[DPMT 1983; MNR 1985; Roorkee Qualifying 1998]
An element with atomic number 84 and mass number 218 loses one
 -particle and two  -particles in three successive stages, the
resulting element will have
43.
[NCERT 1979; CPMT 1990]
33.
34.
35.
(a) At. no. 84 and mass number 214
(b) At. no. 82 and mass number 214
(c) At. no. 84 and mass number 218
(d) At. no. 82 and mass number 218
Group displacement law was given by
[DPMT 1984]
(a) Becquerel
(b) Rutherford
(c) Soddy and Fajan
(d) Madam Curie
How many alpha particles are emitted in the nuclear transformation
215
[CPMT 1993]
82 Pb 211
84 Po
44.
45.
37.
(a) 234, 90
(b) 236, 92
(c) 238, 90
(d) 236, 90
Initial mass of a radioactive element is 40 g. How many grams of it
would be left after 24 years, if its half-life period is 8 years [MP PMT 1985]
46.
(a) 2
(b) 5
(c) 10
(d) 20
What is the symbol for the nucleus remaining after
undergoes  -emission
38.
39.
(b)
20 Sc
42
42
(d)
21 Sc
41
21 Ca
(c)
21 Sc
When a radioactive nucleus emits an  -particle, the mass of the
atom
[NCERT 1973, 82]
(a) Increases and its at. number decreases
(b) Decreases and its at. number decreases
(c) Decreases and its at. number increases
(d) Remains same and its at. number decreases
24
A photon of hard gamma radiation knocks a proton out of 12
Mg
nucleus to form
[AIEEE 2005]
(a) The isotope of parent nucleus
(b) The isobar of parent nucleus
(c) The nuclide
23
11
(d) The isobar of
40.
42
[MNR 1987; UPSEAT 2000, 02]
42
(a)
20 Ca
47.
Na
Pb 210  82 Pb 206  2 He 4 . From the above equation,
deduce the position of polonium in the periodic table (lead belongs
to group IV A)
[AIIMS 1980]
(a) II A
(b) IV B
(c) VI B
(d) VI A
4 , 3 
(b) 8 , 6 
(c)
6 , 4 
(d) 7 , 5 
Atomic number after a  -emission from a nucleus having atomic
number 40, will be
[BHU 1981]
(a) 36
(b) 39
(c) 41
(d) 44
A certain nuclide has a half-life period of 30 minutes. If a sample
containing 600 atoms is allowed to decay for 90 minutes, how many
atoms will remain
[NCERT 1978]
(a) 200 atoms
(b) 450 atoms
(c) 75 atoms
(d) 500 atoms
The reaction which disintegrates neutron is or neutron is emitted
(which completes first)
240
 2 He  97 Bk
4
(a)
96
Am
(b)
15
P 30  14 Si 30  1e 0
(c)
6C
(d)
13
12
244
 1e 0
 1 H 1  7 N 13
Al27  2 He 4  15 P 30
If 92 U 236 nucleus emits one  -particle, the remaining nucleus
will have
[MP PMT 1976, 80; BHU 1985; CPMT 1980]
(a) 119 neutrons and 119 protons
(b) 142 neutrons and 90 protons
(c) 144 neutrons and 92 protons
(d) 146 neutrons and 90 protons
 -rays have high ionization power because they possess
[CPMT 1982]
48.
49.
Na
23
11
(a)
[IIT 1988; MP PMT 1991; KCET 2005]
(a) 0
(b) 1
(c) 2
(d) 3
If uranium (mass no. 238 and atomic no. 92) emits  -particle, the
product has mass number and atomic number
[CPMT 1984, 90, 93, 94; MNR 1991; IIT 1981]
36.
273
50.
84
51.
(a) Lesser kinetic energy
(b) Higher kinetic energy
(c) Lesser penetrating power
(d) Higher penetrating power
When radium atom which is placed in II group, loses an  particle, a new element is formed which should be placed in group[CPMT 1979,
(a) Second
(b) First
(c) Fourth
(d) Zero
Starting from radium, the radioactive disintegration process
terminates when the following is obtained
[CPMT 1979]
(a) Lead
(b) Radon
(c) Radium A
(d) Radium B
The appreciable radioactivity of uranium minerals is mainly due to[NCERT 1980]
(a) An uranium isotope of mass number 235
(b) A thorium isotope of mass number 232
(c) Actinium
(d) Radium
After losing a number of  and  -particles,
82 Pb
206
92 U
238
changed to
. The total number of particles lost in this process is[MNR 1985]
274 Nuclear Chemistry
52.
(a) 14
(b) 5
(c) 8
(d) 32
When an radioactive element emits an alpha particle, the daughter
element is placed in the periodic table
[MP PET 1991; MADT Bihar 1981]
53.
(a) Two positions to the left of the parent element
(b) Two positions to the right of the parent element
(c) One position to the right of the parent element
(d) In the same position as the parent element
If the quantity of a radioactive element is doubled, then its rate of
disintegration per unit time will be
54.
(c) Increased by 2 times
(d) Doubled
The number of  and  -particles emitted during the transformation of
232
to 82 P 208 are respectively
90 Th
55.
56.
57.
(a)
83
Bi
(c)
82
Pb 208
84
Po
(d)
82
Pb 207
[MP PET 1990]
58.
(a) 2
(c) 6
59.
6.93  10 2 s 1
radioactive
change
66.
82 Pb
206
(a)
343.81 MeV
(b) 0.369096 MeV
(c)
931 MeV
(d) None of these
The number   and  - particles emitted respectively during the
transformation of
232
90 Th
to
208
82 Pb is
[Kerala PMT 2004]
(a) 3, 6
(c) 4, [EAMCET
6
1997]
(e) 6, 8
67.
68.
69.
[KCET 2000]
70.
(b) 6, 3
(d) 6, 4
Consider the following nuclear reactions,
x
y
238
92
M  yx N  2 42 He
N  BA L  2  
[AIEEE 2004]
The number of   and   particles emitted when a radioactive
90
E 232 changes into
86 G
220
will be
(a) 5 and 4
(b) 2 and 3
(c) 3 and 2
(d) 4 and 1
The disintegration constant of radium with half-life 1600 years is [MHCET 2004]
(a)
2.12  10 4 year 1
(b)
4.33  10 4 year 1
(c)
3.26  10 3 year 1
(d)
4.33  10 12 year 1
The number of  and   particles emitted in the nuclear reaction
92 U
238
90 Th 234 91 Pa234 are respectively
[Pb.CET 2001]
6.93  10 4 s
(c) 6.93  10 4 s 1
(d) 6.93  10 3 s
Radioactivity of naptunium stops when it is converted to
71.
(a) Bi
(b) Rn
(c) Th
(d) Pb
The highest binding energy per nucleon will be for
(a) 1 and 1
(b) 1 and 2
(c) 2 and 1
(d) 2 and 2
In which radiation mass number and atomic number will not change [JEE Orissa
(a) 
(c)

(b)

(d)  and 2 
Disintegration constant for a radioactive substance is 0.58 hr 1 . Its
half-life period
[BHU 2004]
(a)
8.2
hr
(b)
5.2
hr
(a) Fe
(b) H 2
(c) 1.2 hr
(d) 2.4 hr
(c) O 2
(d) U
73.
A radioactive nucleus will not emit
[DPMT 2005]
In the Thorium series, 90 Th 232 loses total of 6 -particles and 4
(a) Alpha and beta rays simultaneously
-particles in ten stages. The final isotope produced in the series is[MP PET 2001] (b) Beta and gamma rays simultaneously
(c) Gamma and alpha rays
72.
[AIIMS 2001]
62.
(d)
The nuclear binding energy for Ar (39.962384 amu) is: (given mass
of proton and neutron are 1.007825 amu and 1.008665 amu
respectively)
[Pb.CET 2002]
[JIPMER 2001]
61.
208
[MP PET 2004]
(b) 4
(d) 10
(b)
82 Pb
element
If half-life of a certain radioactive nucleus is 1000 s, the
disintegration constant is
[MP PET 2001]
(a)
60.
in
(c)
65.
Atomic number increases by two units
Atomic number increases by three units
Atomic number decreases by one unit
Atomic number increases by one unit
The number of -particles emitted
238
82 Pb 206  2 He 4 is
92 U
209
The number of neutrons in the element L is
(a) 140
(b) 144
(c) 142
(d) 146
When a  -particle emits from the atom of an element, then
(a)
(b)
(c)
(d)
83 Bi
The number of neutrons in the parent nucleus which gives N 14 on 
– emission is
[Pb.CET 2004]
(a) 7
(b) 14
(c) 6
(d) 8
210
(b)
(b)
64.
The end product of (4 n  1) radioactive disintegration series is[MP PMT 1999]
209
209
All the nuclei from the initial element to the final element constitute a
series which is called
[Kerala (Med.) 2002]
(a) g-series
(b) b-series
(c) b-g series
(d) Disintegration series
[MNR 1978; NCERT 1984;CPMT 1989;
RPET 1999; MP PMT 2001; KCET 2003]
(a) 4, 2
(b) 2, 2
(c) 8, 6
(d) 6, 4
The atomic number of a radioactive element increases by one unit in
(a) Alpha emission
(b) Beta emission
(c) Gamma emission
(d) Electron capture
82 Pb
63.
[NCERT 1972, 92; MP PET 1989]
(a) Unchanged
(b) Reduced to half
(a)
Nuclear Chemistry
(d) Gamma rays only
74.
75.
180
72
2


X 
  
A
Z X
. Z and A are
[DPMT 2005]
10.
[J & K 2005]
11.
12.
2.
(d)
4 .0 g
13.
3.
4.
14.
92
X 232  89 Y 220 , how many  and  -
15.
16.
5.
(b)
5 and 3 
(c) 3 and 5 
(d) 5 and 5 
Which of the following does not take place by  - decay
17.
[MP PMT 1996]
(a)
(c)
6.
7.
8.
9.
92 U
88
238
 90 Th
234
(b)
Ra 226  86 Rn 222 (d)
90
83
Th
232
1 .0 g of a radioactive isotope was found to reduce to 125 mg after
24 hours. The half-life of the isotope is [MP PET 1996]
(a) 8 hours
(b) 24 hours
(c) 6 hours
(d) 4 hours
A radioactive element decays at such a rate that after 15 minutes
only 1/10 of the original amount is left. How many more minutes will
be needed when only 1/100 of the original amount will be left
(a) 1.5 minutes
(b) 15.0 mintues
(c) 16.5 minutes
(d) 30 minutes
X 88
(b)
35
(c)
34
Z 88
(d)
36 W
7
N 14
5B
14
(b)
 - active,
(d)
 - active, C 14
10
1
of the original
10
radioactivity after 2.303 seconds. The half-life period is
(a) 2.303
(b) 0.2303
(c) 0.693
(d) 0.0693
A radioactive substance has t1 / 2 60 minutes. After 3 hrs, what
percentage of radioactive substance will remain
(a) 50%
(b) 75%
(c) 25%
(d) 12.5%
A freshly prepared radioactive source of half-life 2 hours emits
radiations of intensity which is 64 times the permissible safe level.
The minimum time after which it would be possible to work safely
with this source is
[IIT 1988]
(b) 12 hours
(d) 128 hours
18.
During a negative  -decay
[MNR 1990; IIT 1985]
(a) An atomic electron is ejected
(b) An electron which is already present within the nucleus is
ejected
(c) A neutron in the nucleus decays emitting an electron
(d) A part of the binding energy of the nucleus is converted into
an electron
19.
The decay constant of a radioactive sample is '  ' . The half-life and
mean life of the sample are respectively
[MNR 1990; IIT 1989]
87
20.
1 ln 2
,
(a)

(c)
 ln 2,
Y 89
What is the half-life of a radioactive substance if 75% of a given
amount of the substance disintegrates in 30 minutes
(a) 7.5 minutes
(b) 25 minutes
7 Be
Radioactivity of a radioactive element remains
(a) 6 hours
(c) 24 hours
The radioactive decay of 35 X 88 by a beta emission produces an
unstable nucleus which spontaneously emits a neutron. The final
product is
[MNR 1995; CBSE 2001]
37
 -active,
 88 Ra 228
Bi 213  84 Po 213
(a)
C 14 is radioactive. The activity and the disintegration product are
[BHU 1995]
[CBSE 1999]
3 and 3 
(b) 24 gm
(d) 48 gm
(c) Positron active,
particles are ejected from X to form Y
(a)
(d) 0.693  69.3
10 gm of a radioactive substance is reduced to 1.25 gm after 15
days. Its 1kg mass will reduce (in how many days) to 500 gm in
(a) 500 days
(b) 125 days
(c) 25 days
(d) 5 days
A radioactive isotope having a half-life of 3 days was received after
12 days. It was found that there were 3 gm of the isotope in the
container. The initial weight of the isotope when packed was
(a)
(b) 76.4 MeV
(d) 0.764 MeV
( e  =0.00055 amu, p =1.00814 amu, n =1.00893 amu)
Half-life period of a metal is 20 days. What fraction of metal does
remain after 80 days
[BHU 1996]
(a) 1
(b) 1/16
(c) 1/4
(d) 1/8
In the radioactive decay
(69.3)1
(a) 12 gm
(c) 36 gm
[MP PMT 1999]
(a) 7.64 MeV
(c) 764 MeV
(b) 10 2
[NCERT 1980; CPMT 1999; KCET 2000; Pb.CET 2001]
2.0 g
The atomic mass of an element is 12.00710 amu. If there are 6
neutrons in the nucleus of the atom of the element, the binding
energy per nucleon of the nucleus will be
 -particle
(c)  -rays
(d) Positron
The half-life of a radionuclide is 69.3 minutes. What is its average
life (in minutes)
(c)
The half-life period of a radioactive substance is 8 years. After 16
years, the mass of the substance will reduce from starting 16.0 g
to [MP PMT 1999]
(a) 8 .0 g
(b) 6 .0 g
(c)
(b)
(a) 100
Rate of decay and Half-life
1.
(c) 20 minutes
(d) 15 minutes
In radioactive decay which one of the following moves the fastest [MP PET/PMT
(a)  -particle
(a) 69, 172
(b) 172, 69
(c) 180, 70
(d) 182, 68
Loss of a beta particle is equivalent to
(a) Increase of one neutron only
(b) Decrease of one neutron only
(c) Both (a) and (b)
(d) None of these
275

1

(b)
(d)
ln 2 1
,



,
1
ln 2 
The half-life of a radio isotope is 20 hours. After 60 hours, how
much amount will be left behind
[MP PMT 1991]
276 Nuclear Chemistry
21.
(a) 1/8
(b) 1/4
(c) 1/3
(d) 1/2
Half-life period of a zero order reaction is
33.
If the half-life period of a first order reaction is 138.6 minutes, then the
value of decay constant for the reaction will be
[MH CET 1999]
(a) 5 minute
(b) 0.5 minute
(c) 0.05 minute
(d) 0.005 minute
Half-life of 10 gm of radioactive substance is 10 days. The half-life
of 20 gm is
[CPMT 1996]
(a) 10 days
(b) 20 days
(c) 25 days
(d) Infinite
8 gm of the radioactive isotope, cesium-137 were collected on
February 1 and kept in a sealed tube. On July 1, it was found that
only 0.25 gm of it remained. So the half-life period of the isotope
is [KCET 1989]
(a) 37.5 days
(b) 30 days
(c) 25 days
(d) 50 days
The half-life of radium (226) is 1620 years. The time taken to
convert 10 grams of radium to 1.25 grams is
–1
[AMU (Engg.) 1999]
–1
–1
22.
(a) Inversely proportional to the concentration
(b) Independent of the concentration
(c) Directly proportional to the initial concentration
(d) Directly proportional to the final concentration
If 12 g of sample is taken, and 6 g of a sample decays in 1 hr. The
amount of sample showing decay in next hour is
34.
35.
[AMU (Engg.) 1999]
23.
24.
25.
(a) 3 g
(b) 1 g
(c) 2 g
(d) 6 g
What will be half-life period of a nucleus if at the end of 4.2 days,
[MP PET 2000]
N  0.798 N 0
(a) 15 days
(b) 10 days
(c) 12.83 days
(d) 20 days
If 2.0 g of a radioactive substance has half-life of 7 days. The half-life
of 1 g sample is
[MP PET 2000]
(a) 7 days
(b) 14 days
(c) 28 days
(d) 35 days
The half-life of 90
38 Sr is 20 years. If its sample having initial activity
of 8000 dis/min is taken, what would be its activity after 80 years
(a) 500 dis/min
(b) 800 dis/min
(c) 1000 dis/min
(d) 1600 dis/min
36.
–1
[MP PET 1994; UPSEAT 2001]
37.
38.
(a) 810 years
(b) 1620 years
(c) 3240 years
(d) 4860 years
Half-life of a radioactive substance is 120 days. After 480 days, 4 gm
will be reduced to
[EAMCET 1993]
(a) 2
(b) 1
(c) 0.5
(d) 0.25
[MP PMT 2000]
The half-life of Co 60 is 7 years. If one gm of it decays, the
amount of the substance remaining after 28 years is
[EAMCET 1992]
26.
24
half-life is 15 hours. On heating it will
11 Na
(a)
0.25 gm
(b) 0.125 gm
(c)
0.0625 gm
(d) 0.50 gm
27.
(a) Reduce
(b) Remain unchanged
(c) Depend on temperature
(d) Become double
In a radioactive decay, an emitted electron comes from
39.
[CBSE 1994; Pb. PET 1999]
28.
(a) Nucleus of the atom
(b) Inner orbital of the atom
(c) Outermost orbit of the atom
(d) Orbit having principal quantum number one
What is the value of decay constant of a compound having half-life
time T1 / 2  2.95 days
[AFMC 1997]
(a)
29.
30.
2.7  10 5 s 1
(b)
32.
(c) 2.7  10 6 s 1
(d) 3  10 5 s 1
What kind of radioactive decay does not lead to the formation of a
daughter nucleus that is an isobar of the parent nucleus
(a) -rays
(b) -rays
(c) Positron
(d) Electron capture
The half-life of 6 C 14 if its K or  is 2.31  10 4 is
(a)
2  10 2 yrs
(b)
3  10 3 yrs
(c)
3.5  10 4 yrs
(d)
4  10 3 yrs
A radioactive isotope has a half-life of 10 days. If today 125 mg is
left over, what was its original weight 40 days earlier [KCET 2005]
(a) 2g
(b) 600 mg
(c) 1 g
(d) 1.5 g
The binding energy of
neutron is
(a) 0.794 MeV
(c) 7.94 MeV
8O
16
[Bihar CEE 1992; AMU 2002; MHCET 1999]
2.7  10 6 s 1
[BHU 1999]
31.
40.
41.
[MH CET 1999]
(a) 11520 years
(b) 2880 years
(c) 1440 years
(d) 17280 years
Half-life of a radioactive element is 100 yrs. The time in which it
disintegrates
to1999]
50% of its mass, will be
[JIPMER
[MP PMT 1995]
42.
43.
is 127 MeV. Its binding energy per
(b) 1.5875 MeV
(d) 15.875 MeV
A radioactive isotope decays at such a rate that after 96 minutes
1
th of the original amount remains. The half-life of this
only
8
nuclide in minutes is
[KCET 1992]
(a) 12
(b) 24
(c) 32
(d) 48
C  14 has a half-life of 5760 years. 100mg of a sample
containing C  14 is reduced to 25mg in
44.
(a) 50 yrs
(b) 200 yrs
(c) 100 yrs
(d) 25 yrs
The average life period of a radioactive element is the reciprocal of
its
[MP PET 1995]
(a) Half-life period
(b) Disintegration constant
(c) Number of atoms present at any time
(d) Number of neutrons
The half-life period of a radioactive element is 30 minutes. One
sixteenth of the original quantity of the element will remain
unchanged after
[CPMT 1983; MP PMT 1994]
(a) 60 minutes
(b) 120 minutes
(c) 70 minutes
(d) 75 minutes
For a radioactive substance with half-life period 500 years, the time
for complete decay of 100 milligram of it would be
[MADT Bihar 1984]
(a) 1000 years
(b) 100  500 years
Nuclear Chemistry
45.
(c) 500 years
(d) Infinite time
A substance of which one gram is taken, after half-life time what
fraction of it is left ?
[MADT Bihar 1983]
1
1
(a)
(b)
4
8
1
1
(c)
(d)
2
32
46.
The half-life of the radio element 83 Bi210 is 5 days. Starting with
20 g of this isotope, the amount remaining after 15 days is
(a) 10 g
(b) 5 g
(c) 2.5 g
(d) 6.66 g
47.
In radioactive decay of X into Y below,
6
Z
(a)
6
Y
(d) 1/16 of the original amount
56.
57.
58.
15
(b)
7
Y
17
14
48.
(c) 9 Y
(d) 8 Y 14
75% of the first order reaction was completed in 32 minutes. When
was 50% of the reaction completed
59.
[MNR 1983; MP PET 1997; EAMCET 1998]
49.
(a) 24 minutes
(b) 16 minutes
(c) 8 minutes
(d) 4 minutes
If 2 .0 g of a radioactive isotope has a half-life of 20 hr, the half-life
60.
[MP PMT 1990; MNR 1992]
50.
51.
(b) 80 hr
(d) 10 hr
61.
(a)
92 U
238
will be reduced to half of its present
[CPMT 1990; MP PET 1999]
9.0  10 9 years
(b) 13.5  10 9 years
(c) 4.5  10 9 years
(d) 4.5  10 4.5 years
Radium has atomic weight 226 and a half-life of 1600 years. The
number of disintegrations produced per second from 1 gm are[BHU 1990]
(a)
4.8  10 10
(b)
9.2  10 6
63.
64.
(c) 3.7  10
(d) Zero
The half-life of a radioactive element is 6 months. The time taken to
reduce its original concentration to its 1/16 value is
10
53.
54.
55.
(a)  -particles
(b)  -particles
(c) Neutrons
(d) None of these
Given that a radioactive species decays according to exponential law
N  N 0 e t . The half-life of the species is
Radioactive lead 82 Pb 201 has a half-life of 8 hours. Starting from
[Kerala (Med.) 2003]
one milligram of this isotope, how much will remain after 24 hours [MP PMT 1990]
(a) 
(b) No
(a) 1 / 2 mg
(b) 1 / 3 mg
(c) /ln 2
(d) ln 2/
(c) 1 / 8 mg
(d) 1 / 4 mg
62.
Half-life of a radioactive disintegration (A  B) having rate
The half-life of 92 U 238 is 4.5  10 9 years. After how many years,
constant 231 sec 1 is
[CPMT 1988]
the amount of
amount
52.
(a) 3000 years old
(b) 6000 years old
(c) 9000 years old
(d) 1200 years old
The decay of a radioactive element follows first order kinetics, as a
result
[NCERT 1982]
[BHU 1987]
(a) Half-life
period = constant / k , where k is the decay constant
(b) Rate of decay is independent of temperature
(c) Rate can be changed by changing chemical conditions
(d) The element will be completely transformed into a new element
after expiry of two half-life period
Half-life of a radioactive substance which disintegrates by 75 % in 60
minutes, will be
[MP PMT 2002]
(a) 120 min
(b) 30 min
(c) 45 min
(d) 20 min
87.5% decomposition of a radioactive substance complete in 3 hours.
What is the half-life of that substance
[MP PMT 2003]
(a) 2 hours
(b) 3 hours
(c) 90 minutes
(d) 1 hours
Tritium undergoes radioactive decay giving
[CPMT 1976; NCERT 1978]
of 0 .5 g of the same substance is
(a) 20 hr
(c) 5 hr
The radioactivity due to C 14 isotope (half-life 6000 years) of a
sample of wood from an ancient tomb was found to be nearly half
that of fresh wood, the tomb is therefore about
[NCERT 1980, 81; MP PET 1989]
Y m is
3 
X 14 
 Z Y m
277
[MP PET 1991]
65.
(a) 1 year
(b) 16 years
(c) 2 years
(d) 8 years
In the case of a radio isotope the value of T1 / 2 and  are identical
in magnitude. The value is
[KCET 2002]
66.
(a)
0.693
(b) (0.693)1 / 2
(c)
1/0.693
(d) (0.693)2
A radioactive element has half-life of one day. After three days, the
amount of the element left will be
3.0  10 2 sec
(b)
3.0  10 3 sec
(c)
3.3  10 2 sec
(d)
3.3  10 3 sec
The amount of 53 I 128 t1 / 2  25 minutes) left after 50 minutes
will be
[AIIMS 1982; DPMT 1982, 83]
(a) One – half
(b) One – third
(c) One – fourth
(d) Nothing
If 3/4 quantity of a radioactive element disintegrates in two hours,
its half-life would be
[MP PMT 1989; CPMT 1984]
(a) 1 hour
(b) 45 minutes
(c) 30 minutes
(d) 15 minutes
Radioactive decay is a
[MP PMT 1989, 97]
(a) Second order reaction
(b) First order reation
(c) Zero order reaction
(d) Third order reaction
The half-life of a radioactive element depends upon
[EAMCET 1980]
67.
[MNR 1985; UPSEAT 2000, 01; MH CET 2002]
(a) 1/2 of the original amount
(b) 1/4 of the original amount
(c) 1/8 of the original amount
(a)
68.
(a) The amount of the element
(b) The temperature
(c) The pressure
(d) None of these
The activity of radio isotope changes with
[MNR 1986]
(a) Temperature
(b) Pressure
(c) Chemical environment
(d) None of these
A certain nuclide has a half-life of 25 minutes. If one starts with 100
g of it, how much of it will remain at the end of 100 minutes
278 Nuclear Chemistry
(a) 1.0 g
(c) 6.25 g
69.
(b) 4.0 g
(d) 12.50 g
(a) 0.30 sec
(c) 3.3 sec
If U 235 is bombarded with neutrons, atom will split into
81.
[CPMT 1981]
70.
71.
(a) Sr + Pb
(b) Cs + Rb
(c) Kr + Cd
(d) Ba + Kr
If 8.0 of a radioactive isotope has a half-life of 10 hrs. The half-life of 2.0
g of the same substance is
[UPSEAT 2001]
(a) 2.5 hrs.
(b) 5 hrs.
(c) 10 hrs.
(d) 40 hrs.
82.
If the disintegration constant is 6.93  10 6 , then half-life of
83.
6C
72.
14
will be
[KCET 2001]
(a)
10 2 yrs
(b) 10 3 yrs
(c)
10 4 yrs
(d) 10 5 yrs
The decay constant of Ra 226 is 1.37  10 11 sec 1 . A sample of
84.
Ra 226 having an activity of 1.5 millicurie will contain ...... atoms
73.
(a)
4.1  1018
(b)
3.7  1017
(c)
2.05  1015
(d)
4.7  1010
Amount of
128
(t1 / 2
53 I
(b) 0.60 sec
(d) Data is insufficient
T1 / 2 of C 14 isotope is 5770 years. time after which 72% of
isotope left is
[Orissa JEE 2005]
(a) 2740 years
(b) 274 years
(c) 2780 years
(d) 278 years
A radioactive substance takes 20 min to decay 25%. How much time
will be taken to decay 75%
[Orissa JEE 2005]
(a) 96.4 min
(b) 68 min
(c) 964 min
(d) 680 min
A radioactive sample is emitting 64 times radiations than non
hazardous limit. if its half life is 2 hours, after what time it becomes
non-hazardous
[DPMT 2005]
(a) 16 hr
(b) 12 hr
(c) 8 hr
(d) 4 hr
If 8.0 g of a radioactive substance has a half-life of 10 hrs., the half
life of 2.0 g of the same substance is [J & K 2005]
(a) 2.6 hr
(b) 5 hr
(c) 10 hr
(d) 40 hr
Artificial transmutation
 25 min) left after 75 minutes is
1.
The age of most ancient geological formation is estimated by
[DCE 2002]
(a)
74.
75.
[NCERT 1981; MP PET/PMT 1988; CBSE 1989;
MP PET 1997; MP PMT 2002]
(b) 1 / 4
1/6
(c) 1 / 8
(d) 1 / 9
The half-life of a radioisotope is four hours. If the initial mass of the
isotope was 200 g, the mass remaining after 24 hours undecayed is
(a) 3.125 g
(b) 2.084 g
(c) 1.042 g
(d) 4.167 g
An artificial radioactive isotope gave
14
7
(a)
(b)
(c)
(d)
76.
The equation
3.
(a) Synthesis of helium
(b) Transmutation of element
(c) Fusion reaction
(d) Nuclear fission
The phenomenon of radioactivity arises from the
N after two successive
77.
(a)
6.932 yr
(c)
0.06932 yr 1
1
(b)
0.6932 yr
(d)
0.006932 yr 1
[Kerala CET 2004]
78.
4.
A radioactive isotope decays at such a rate that after 192 minutes
only 1 / 16 of the original amount remains. The half-life of the
radioactive isotope is
(a) 32 min
(c) 12 min
In the given reaction,
isotope are
(b) 48 min
(d) 24 min
92 U
235
5.




( A) 
(B) 
(C)
79.
80.
(b)
92 U
235
and C
6.
7.
(a) W.F.
Libby
[Orissa
JEE 2004]
(c) J. Willard Gibbs
The nuclear reaction
[MP PET 2001]
(c) A and B
(d) A, B and C
Rate constant for a reaction is  . Average life is representative by
(a)
1/
(b)
(c)

2
(d)
In2 / 
0 .693

For a reaction, the rate constant is 2.34 sec 1 . The half-life period
for the reaction is
(a) Binary fission
(b) Nuclear fusion
(c) Stable nuclei
(d) Decay of unstable nuclei
The first artificial disintegration of an atomic nucleus was achieved
by
[Kerala (Engg.) 2002]
(a) Geiger
(b) Wilson
(c) Madame curie
(d) Rutherford
(e) Soddy
Artificial elements have been prepared by bombardment reactions in
high energy accelerators. What is the mass number of the element X
produced
in
the
following
nuclear
reaction
249
15
1
[AMU (Engg.) 2002]
95 Cf  7 N 105 X  4 0 n
(a) 261
(b) 264
(c) 260
(d) 257
Radioactive carbon dating was discovered by
[Pb. CET 2000]
(a) A and C
3
[Kerala (Med.) 2002]
[DCE 2004]
1
Li 6  1 H 2  2 2 He 4  energy represents
2.
  particle emissions. The number of neutrons in the parent
nucleus must be
[KCET 2004]
(a) 9
(b) 14
(c) 5
(d) 7
If the half-life of an isotope X is 10 years, its decay constant is
Potassium – Argon method
Carbon – 14 dating method
Radium
Silicon method
[AIEEE –2004]
Uranium – Lead method
63
29 Cu
(b) G.N. Lewis
(d) W. Nernst
37
 42 He 17
Cl  14 11 H  16 10 n is referred to as
[MP PET 2002]
(a) Spallation reaction
(c) Fission reaction
(b) Fusion reaction
(d) Chain reaction
Nuclear Chemistry
8.
9.
The carbon dating is based on
[MP PMT 2001]
(a)
15
6
C
(b)
14
6
(c)
13
6
C
(d)
11
6 C
C
A possible material for use in the nuclear reactors as a fuel is
22.
[DPMT 1986]
10.
11.
(a) Thorium
(c) Beryllium
Heavy water freezes at
(b) Zirconium
(d) Plutonium
[UPSEAT 2001]
(a)
0C
(b)
3.8C
(c)
38C
(d)
 0.38C
23.
24.
To determine the masses of the isotopes of an element which of the
following techniques is useful
12.
13.
14.
The radioisotope, tritium (13 H ) has a half-life of 12.3 years. If the
initial amount of tritium is 32 mg. How many milligrams of it would
remain after 49.2 years
[CBSE 2003]
(a) 8 mg
(b) 1 mg
(c) 2 mg
(d) 4 mg
Neutron is used as a
[CPMT 1988]
(a) Reducing agent
(b) Moderator
(c) Tracer
(d) In biological programme
Hydrogen bomb is based on the phenomenon of
[EAMCET 1980; CPMT 1984, 96;
MP PMT 1993, 95, 2002; RPET 1999]
15.
16.
(a) Nuclear fission
(b) Nuclear fusion
(c) Nuclear explosion
(d) Disintegration
In the nuclear reactors the speed of the neutrons is slowed down by
(a) Heavy water
(b) Ordinary water
(c) Zinc rods
(d) Molten caustic soda
By which law, energy produced in nuclear reaction is given
25.
26.
17.
18.
A wood piece is 11460 years old. What is the fraction of
in the piece? (Half-life period of
14
14
19.
20.
(a) Long rods of Cd
(b) Heavy water
(c) Cubical blocks of steel
(d) Both (a) and (c)
The proper rays for radiocarbon dating are
[MP PET 2002]
(a) UV-rays
(b) IR-rays
(c) Cosmic rays
(d) X-rays
21.
1H
[MH CET 2001]
2
 1 H 2  2 He 3  0 n1 . Above nuclear reaction is called
[UPSEAT 2001]
(b)
N 2O
(c)
H 2O
(d)
NaOH
The fuel of atomic pile is
[NCERT 1973; AFMC 1989]
(a) Thorium
(b) Sodium
(c) Uranium
(d) Petroleum
Atom bomb is based on the principle of
(a) Nuclear fusion
(b) Nuclear fission
(c) Radioactivity
(d) Fusion and fission both
Who observed that when the nucleus of uranium atom was
bombarded with fast moving neutrons, it becomes so very unstable
that it is immediately broken into two nuclei of nearly equal mass
besides other fragments
(a) J.J. Thomson
(b) Chadwick
(c) Einstein
(d) Hahn and Strassmann
When a radioactive substance is subjected to vacuum, the rate of
disintegration per second
27.
28.
What is the packing fraction of
(a)
29.
30.
31.
14.167
[KCET 2002]
56
26 Fe
(Isotopic mass = 55.92066)
C is 5730 years)
[MP PMT 2000]
D2 O
[DPMT 1985; NCERT 1972]
C activity left
(a) 0.12
(b) 0.25
(c) 0.50
(d) 0.75
When nuclear energy is intended to be harnessed for generation of
electricity, potentially destructive neutron released in a nuclear
reactor are absorbed by
(a)
(a) Increases considerably
(b) Increases only if the products are gaseous
(c) Is not affected
(d) Suffers a slight decrease
A radio isotope will not emit
(a) Gamma and alpha rays simultaneously
(b) Gamma rays only
[CPMT
(c) Alpha
and1983,
beta84]
rays simultaneously
(d) Beta and gamma rays simultaneously
[MP PET 2000]
(a) Graham’s law
(b) Charle’s law
(c) Gas Lussac’s Law
(d) Einstein’s law
If two light nuclei are fused together in nuclear reaction the average
energy per nucleon
[Pb. PMT 2001]
(a) Increases
(b) Cannot be determined
(c) Remains same
(d) Decreases
(a) Nuclear fission
(b) Nuclear fusion
(c) Artificial transmutation
(d) Spontaneous disintegration
Which of the following is used as a moderator in a nuclear reactor
[CPMT 1982; BHU 1985]
[NCERT 1978; MNR 1979]
(a) The acceleration of charged atoms by an electric field and their
subsequent deflection by a variable magnetic field
(b) The spectroscopic examination of the light emitted by
vaporised elements subjected to electric discharge
(c) The photographing of the diffraction patterns which arise when
X-rays are passed through crystals
(d) The bombardment of metal foil with alpha particles
279
[JIPMER 2002]
(b) 173.90
(c) 14.187
(d) 73.90
The energy released in an atom bomb explosion is mainly due to
(a) Release of neutrons
(b) Release of electrons
(c) Greater mass of products than initial material
(d) Lesser mass of products than initial material
[KCET 2002]
C 14 is
(a) A natural radioactive isotope
(b) A natural non-radioactive isotope
(c) An artificial radioactive isotope
(d) An artificial non-radioactive isotope
A radioactive isotope has a half-life of 10 years. What percentage of the
original amount of it remain after 20 years
[KCET 2001]
32.
(a) 0
(b) 12.5
(c) 8
(d) 25
In a chain reaction, uranium atom gets fissioned forming two
different materials. The total weight of these put together is
(a) More than the weight of parent uranium atom
280 Nuclear Chemistry
33.
(b) Less than the weight of parent uranium atom
(c) More or less depends upon experimental conditions
(d) Neither more nor less
A substance used as a moderator in nuclear reactors is
45.
[MP PET 1995]
(a) Cadmium
(c) Lead
34.
35.
(b) Uranium-235
(d) Heavy water
46.
[DPMT 2000]
[MP PMT 1989]
 1 H  18 Ar  0 n is
17 Cl
(a) Nuclear fission
(b) Nuclear fusion
(c) Transformation of chlorine
(d) Synthesis of argon
1.0 gm radioactive sodium on decay becomes 0.25 gm in 16
Equation
37
2
38
(a) Deuterons
(c) -particles
1
need to become 3.0 gm
(a) 48 hours
(b) 32 hours
(c) 20 hours
(d) 16 hours
Large energy released in an atomic bomb explosion is mainly due to
(a) Products having a lesser mass than initial substance
(b) Conversion of heavier to lighter atoms
(c) Release of neutrons
(d) Release of electrons
2
A piece of wood was found to have C 14 /C 12 ratio 0.7 times that in
a living plant. The time period when the plant died is (Half-life of
[Pb. PMT 1999]
C 14  5760 yrs)
(a) 2770 yrs
(b) 2966 yrs
(c) 2980 yrs
(d) 3070 yrs
48.
When a slow neutron goes sufficiently close to a U 235 nucleus,
then the process which takes place is [AFMC 2000]
The reaction
38.
(a) Nuclear fission
(b) Nuclear fusion
(c) Artificial disintegration
(d) Transmutation of element
Carbon-14 dating method is based on the fact that
1H
(a) Fusion of U 235
[CPMT
73, 81,
(c) 1972,
Fusion
of 90]
neutron
49.
[MP PMT 1990; CPMT 1990; KCET 1992]
50.
[CBSE 1997]
39.
(a) Carbon-14 fraction is the same in all objects
(b) Carbon-14 is highly insoluble
(c) Ratio of carbon-14 and carbon-12 is constant
(d) All of these
Half-life period of a radioactive element is 10.6 yrs. How much time
will it take in its 99% decomposition
51.
40.
41.
42.
43.
(a)
H 2 O18
(b)
H 2 O16
(c)
H 2 O3
(d)
D2 O
15
P 31
and
(a) Proton
(b) Neutron
(c) Alpha particle
(d) Deuteron
Which of the following statements about radioactivity of an element
is incorrect
(a) It is a nuclear property
(b) It does not involve any rearrangement of electrons
(c) Its rate is affected by change in temperature and/or pressure
(d) It remains unaffected by the presence of other element or
elements chemically combined with it
Radioactive iodine is being used to diagnose the disease of
[MP PET 1996]
52.
(b) Kidneys
(d) Thyroid
C  14 is used in carbon dating of dead objects because
[DPMT 1996]
(a) Its half-life is 10 3 years
53.
(b) Its half-life is 10 4 years
(c) It [JIPMER
is found2001]
in nature abundantly and in definite ratio
(d) It is found in dead animals abundantly
A radioactive element resembling iodine in properties is
54.
(a) Astatine
(b) Lead
(c) Radium
(d) Thorium
For artificial transmutation of nuclei, the most effective one is
[Kurukeshetra CEE 1998]
[MP PMT 1996]
[CPMT 1997]
D2 O is used in
(a) Industry
(b) Nuclear reactor
(c) Medicine
(d) Insecticide
India conducted an underground nuclear test at
[KCET 1998]
44.
Al 28 when radiated by suitable projectile gives
neutron. The projectile used is
13
(a) Bones
(c) Blood cancer
[RPET 1999]
(a) 7046 yrs
(b) 7.046 yrs
(c) 704.6 yrs
(d) 70.4 yrs
Deuterium resembles hydrogen in chemical properties but reacts
(a) More vigorously than hydrogen
(b) Faster than hydrogen
(c) Slower than hydrogen
(d) Just as hydrogen
Which of the following is heavy water
[AFMC 1997]
(b) Fission of U 235
(d) First (a) then (b)
[MP PMT/PET 1988; CPMT 1985, 82]
 1 H 3  2 He 4  0 n1  energy represents
37.
(b) Neutrons
(d) Protons
47.
hours. How much time 48 gm of same radioactive sodium will
36.
Liquid sodium finds use in nuclear reactors. Its function is
(a) To collect the reaction products
(b) To act as a heat exchanger or coolant
(c) To absorb the neutrons in order to control the chain reaction
(d) To act as a moderator which slows down the neutrons
Which is least effective for artificial transmutation
(a) Tarapur
(b) Narora
(c) Pokhran
(d) Pushkar
Energy required to separate neutron and proton from the nucleus is
called
[RPMT 1999]
(a) Bond energy
(b) Nuclear energy
(c) Chemical energy
(d) Radiation energy
55.
56.
(a) Proton
(b) Deuteron
(c) Helium nuclei
(d) Neutron
Which of the following cannot be accelerated
(a)  -particle
(b)  -particle
(c) Protons
(d) Neutrons
For the fission reaction
92 U
235
[KCET 2005]
 0 n1  56 Ba40  y E x  2 0 n1
The value of x and y will be
x  93 and y  34
(b)
x  92 and y  35
(c) x  89 and y  44
Heavy water is used as
(d)
x  94 and y  36
(a)
57.
Nuclear Chemistry
[Bihar MEE 1996; UPSEAT 1999, 2000, 02]
58.
(a) Control rods
(c) Fuel
(e) None of these
Unit for radioactive constant is
(a)
Time -1
(b) Moderator
(d) Coolant
[MP PET 1990]
[MP PET 1989; CPMT 1993]
1
(c) Mole  time
(d) Time  mole
Which of the following is used in dating archeological findings or In
a method of absolute dating of fossils a radioactive element is used.
It is
(a) Isotope H  1.000
1
60.
61.
(a)
92 U
(c)
6
C 12
(b)
6C
(d)
20
71.
14
Ca 40
72.
A radioactive isotope has a half-life of 20 days. If 100 gm of the
substance is taken, the weight of the isotope remaining after 40 days
is
[NCERT 1979]
(a) 25 gm
(b) 2.5 gm
(c) 60 gm
(d) 40 gm
In a fission reaction the nucleus of an element
73.
62.
63.
64.
65.
[CPMT 1972; BHU 1984; KCET 1999]
66.
(a) Seaberg
(b) Rutherford
(c) Einstein
(d) Irene Curie & Juliot
The half-life period of a radioactive element is 140 days. After 560
days, one gram of the element will reduce to
67.
1/2g
(b) 1 / 4 g
(c)
1/8 g
(d) 1 / 16 g
69.
14
(b)
6C
13
(c)
6C
12
(d)
6C
15
Artificial transmutation was discovered by
[Pb.CET 2003]
(a) Pauli
(b) Rutherford
(c) Soddy
(d) Curie
Which of the following is an example of nuclear fusion
1H
2
(b)
92 U
(c)
13
 1 H  2 He  energy
2
235
4
 o n1  56 Ba141  36 Kr 92  30 n1  energy
Al 27  1 H 1  12 Mg 24  2 He 4
74.
75.
76.
The radioactivity isotope 60
27 Co which is used in the treatment of
cancer can be made by (n, p) reaction. For this reaction the target
nucleus is
[CBSE PMT 2004]
(a)
60
28
Ni
(b)
60
27 Co
(c)
59
28
Ni
(d)
59
27 Co
Fusion bomb involves
[AFMC 2004]
(a) Combination of lighter nuclei into bigger nucleus
(b) Destruction of heavy nucleus into smaller nuclei
(c) Combustion of oxygen
(d) Explosion of TNT
The element used for dating the ancient remains is
[AFMC 2004]
(a) Ni
(c) C-12
77.
78.
(b) C-14
(d) Rd
If radium and chlorine combine to from radium chloride the
compound is
[Kerala PMT 2004]
(a) No longer radioactive
(b) Twice as radioactive as radium
(c) Half as radioactive as radium
(d) As radioactive as radium
(e) Thrice as radioactive as radium
Which of the following is an example of nuclear fission
A device used for the measurement of radioactivity is
[Pb. CET 2002]
[BHU 1979]
68.
6C
(d) None of these
[CPMT 1989; IIT 1986; EAMCET 1992;
MP PET 1997; UPSEAT 1999]
(a)
(a)
(a)
[CPMT 1982; Kurukshetra CEE 1998]
(a) Mass defect
(b) Energy of protons
(c) Energy of neutrons
(d) Total energy of nucleons
The first controlled artificial disintegration of an atomic nucleus was
achieved by
[BHU 1987]
(a) Geiger
(b) Wilson
(c) Cockcroft
(d) Rutherford
Artificial radioactivity was first discovered by
(c) Isotope O  16.000
(d) Isotope C 12  12.000
Which radioactive carbon has been helpful in understanding the
mechanism of photosynthesis in plants
[MP PMT 1989; DCE 2004]
[NCERT 1977]
(a) Loses only some elementary nuclear particles from another
nucleus
(b) Captures some elementary nuclear particles from another
nucleus
(c) Breaks up into several smaller nuclei
(d) Breaks up into two smaller nuclei with the loss of same
elementary nuclear particles
The huge amount of energy which is released during atomic fission
is due to
[CPMT 1990]
(a) Loss of mass
(b) Loss of electrons
(c) Loss of protons
(d) Loss of  -particles
The measure of binding energy of a nucleus is the
(b) Oxygen = 16.000
16
[CPMT 1983, 85; NCERT 1978; BHU 1981;
MP PMT 1993; AFMC 1997]
235
(a) Lubricant
(b) Moderator to slow down neutrons
(c) Fuel
(d) Liner of the reactor
The modern basis of atomic weight is
(b) Time
1
59.
70.
281
(a) Mass spectrometer
(b) Cyclotron
(c) Nuclear reactor
(d) G.M. counter
In a nuclear reactor, chain reaction is controlled by introducing
(a) Iron rod
(b) Cadmium rod
(c) Graphite rod
(d) Platinum rod
In atomic reactors, graphite is used as a
[NCERT 1980; MP PET 1989]
2
 1 H  2 He  
2
(a)
1H
(b)
A  B  C  energy
(c)
92 U
(d)
13
4
[EAMCET 1984]
235
 0 n1  56 Ba141  36 Kr 92  3 0 n1  energy
Al 27  2 He 4  15 P 30  0 n1
282 Nuclear Chemistry
79.
80.
81.
The C 14 to C 12 ratio in a wooden article is 13% that of the fresh
wood. Calculate the age of the wooden article. Given that the halflife of C 14 is 5770 years
[Pb.CET 2004]
(a) 16989 years
(b) 16858 years
(c) 15675 years
(d) 17700 years
Hydrogen bomb is based on the principle of
[AIEEE 2005]
(a) Nuclear fission
(b) Natural radioactivity
(c) Nuclear fusion
(d) Artificial radioactivity
Match List –I and List-II and choose right one by using code given
in list
[Kerala CET 2005]
List – I
List –II
Nuclear reactor
Used substance
Component
1. Moderator
(A) Uranium
2. Control rods
(B) Graphite
3. Fuel rods
(C) Boron
4. Coolent
(D) Lead
(E) Sodium
8.
(a)
(b)
(c)
(d)
(e)
2
A
C
B
D
C
3
C
A
A
A
B
4
E
E
E
B
A
3.
4.
9.
10.
11.
12.
5.
(a) Isothermals
(b) Isotopes
(c) Isentropus
(d) Elementary particles
Tritium is an isotope of
[DPMT 1985]
(a) Hydrogen
(b) Titanium
(c) Tantalum
(d) Tellurium
O – 18 isotope of oxygen will have
[CPMT 1972, 79]
(a) 18 protons
(b) 9 protons and 9 neutrons
(c) 8 neutrons and 10 protons
(d) 10 neutrons and 8 protons
Which of the following is an isobaric pair
[CPMT 1987, 93]
13.
14.
15.
7
16.
17
[NCERT 1983]
16
(b)
7
(d)
6C
N
14
14
e  emission
(b)
2 He
4
emission
(c) e  emission
(d) K -electron capture
Atomic weights of carbon, nitrogen and oxygen are 12, 14 and 16
respectively. An atom of atomic weight 14 and nuclear charge + 6 is
an isotope of
(a) Oxygen
(b) Carbon
(c) Nitrogen
(d) None of these
Isotopes of an element have
[MNR 1985]
(a) Similar chemical properties but different physical properties
(b) Similar chemical and physical properties
(c) Similar physical properties but different chemical properties
(d) Different chemical and physical properties
Whose number is common in isotopes
[AIIMS 1988]
(a) Proton
(b) Neutron
(c) Proton and neutron
(d) Nucleon
In
the
following
radioactive
transformation



R 
X 
Y 
Z ; the nuclei R and Z are
(a) Isotopes
(b) Isobars
(c) Isomers
(d) Isotones
Which one of the following pairs represents isobars
[CPMT 1988]
(c)
(d)
39
19
Isotopes are atoms having the same
17.
4
2
He and
Mg and
He
25
12
Mg
K and
40
20
Ca
K and
40
19
K
Nuclei of isotopes differ in
[CPMT 1986, 90; MP PMT 1987]
(a) The number of protons
(b) The number of neutrons
(c) The number of protons and neutrons both
(d) None of these
An isotope of ‘parent’ is produced, when its nucleus loses
(a) Isobar
(b) Isomorph
18.
(c) Isotope
(d) Isomer
In treatment of cancer, which of the following isotope is used[DPMT 1985; BHU 1995;
AMU 1999;
 -particle
(a)KCET
One1999;
Pb.CET 2001; MP PET 2002; Kurukshetra CET 2002]
8O
Emission of  -particle by an atom of an element results in the
formation of its
[BHU 1979; DPMT 1985; KCET 1999]
(a) Isotope
(b) Isomer
(c) Isomorph
(d) Isobar
Radioactive isotopes that have an excessive neutron/proton ratio
generally exhibit
40
19
(a) Atomic mass
(b) Mass number
(c) Atomic number
(d) Number of neutrons
Successive emission of an  -particle and two  -particles by an
atom of an element results in the formation of its
(b) Isobars
(d) Isotones
N 15
24
12
7
[MP PMT/PET 1988; BHU 1979]
7.
(c)
(b)
(d)
13
2
[BHU 1987]
[EAMCET 1978, 79; MP PMT 1980; CPMT 1973;
BHU 2001; AFMC 2003]
6.
8O
N 13 , 8 O15
N 14 , 8 O15
7
(a)
(a)
6C
1H
Which isotope on bombardment with  -particles will give
, 7 N 14
(b)
(c)
13
(d)
Elements having different nuclear charge but the same mass number
are called
[NCERT 1974; MP PMT 1991;
3
2
, 7 N 13
6C
60
15
and 1 H 1
Substances which have identical chemical properties but differ in
atomic weights are called
(a)
27 Co
P 32
(b)
CBSE PMT 1991; CPMT 1989; EAMCET 1992]
[EAMCET 1980, 83; DPMT 1985; MNR 1982]
2.
(c)
(a)
Isotopes-Isotones and Nuclear isomers
1.
53
(a) Isotopes
(c) Isomers
Code :
1
B
B
C
C
D
I131
(a)
(b) One  -particle
[CPMT 1987; MP PET 1991]
Nuclear Chemistry
(c) One  and two  -particles
19.
(d) One  and two  - particles
Which of the following isotopes is likely to be most stable
[EAMCET 1982]
(a)
(c)
20.
30
30
Zn
71
Zn
64
(b)
30
Zn
66
31.
(d) None of these
Which of the following statement is false
[Manipal MEE 1995]
35
(a)
(b)
(c)
(d)
37
21.
In chlorine gas, the ratio of Cl and Cl is 1 : 3
The hydrogen bomb is based on the principle of nuclear fusion
The atom bomb is based on the principle of nuclear fission
The penetrating power of a proton is less than that of an
electron
Isotones are elements having
22.
(a) Same mass number but different neutrons
(b) Same atomic number but different neutrons
(c) Same atomic number, mass number and neutrons
(d) Different atomic and mass number but same neutrons
Isobaric atoms may contain
32.
33.
First isotope
Second isotope
(a) 34
36
(b) 44
46
(c) 45
47
(d) 79
81
Isotopes are those which contain
[RPMT 1997]
(a) Same number of neutrons
(b) Same physical properties
(c) Same chemical properties
(d) Different atomic mass
An element ' A' emits an  -particle and forms ' B'.' A' and ' B'
are
[DPMT 1990]
(a) Isotopes
(b) Isobars
(c) Isotones
(d) Isodiasphere
Which of the following properties are different for neutral atoms of
isotopes of the same element
[EAMCET 1987; NCERT 1971; CPMT 1976; MP PET 1994]
[Bihar MEE 1996; Bihar CEE 1995]
(a) Same number of p

and different number of n
0
(b) Same number of n and different number of p
(a)
(b)
(c)
(d)
34.
0
24.
25.
X 40 and
21
X 40 are
[CPMT 1996]
(a) Isobars
(b) Isotopes
(c) Isotones
(d) Isostereomers
Which property is different for neutral atoms of the two isotopes of
the same element
[JIPMER 2001]
(a) Number of protons
(b) Atomic number
(c) Number of neutrons
(d) None of these
Which of the following species is isotonic with
37 Rb
(a)
26.
36 Kr
(b)
37 Rb
85
(c) 38 Sr 87
(d) 39 Y 89
The maximum sum of the number of neutrons and protons in an
isotope of hydrogen is
[Pb. PMT 2001]
(a) 4
(b) 5
(c) 6
(d) 3
35
and
37
27.
Difference in
28.
(a) Atomic number
(b) Number of protons
(c) Number of neutrons
(d) Number of electrons
Which of the following is an isotonic pair
17 Cl
17 Cl
is of
35.
36.
86
[BHU 2001]
84
37.
30.
40
19
K, 40
20 Ca
(b)
39
19
K, 40
20 Ca
(c)
33
18
Ar, 40
18 Ar
(d)
40
18
Ar, 4020 Ca
6
C 11 and
5
B 11 are referred as
U 235 decays in a number of steps to an isotope of
[NCERT 1978]
(d) 10 , 8 
7 , 4 
Addition of two neutrons in an atom A would
(a) Change the chemical nature of A
[AMU 1984]
(b) Produce an isobar of A
(c) Produce an isotope of A
(d) Produce another element
Atomic weight of the isotope of hydrogen which contains 2 neutrons
is the nucleus would be
[CPMT 1980]
(a) 2
(b) 3
(c) 1
(d) 4
If a radioactive isotope with atomic number A and mass number M
emits an  -particle, the atomic number and mass number of that
new isotope will become
[NCERT 1980]
38.
39.
The symbol of an isotope is
[AFMC 2000]
(a)
92
(a) A – 2, M – 4
(b) A – 2, M
(c) A, M – 2
(d) A – 4, M – 2
Which character is different of the two isotopes of an element[NCERT 1971; EAM
(a) Atomic mass
(b) Atomic number
(c) Number of electrons
(d) Number of protons
32
X 65 , this reveals that
[MP PET 1991]
[AMU (Engg.) 2000]
29.
The isotope
(c)
(d) Different numbers of both p  and n 0
20
Mass
Atomic number
General chemical reactions
Number of electrons
lead 82 Pb 207 . The groups of particles emitted in this process will
be
[MP PMT 1987]
(a) 4 , 7 
(b) 6 , 4 

(c) Same number of both p  and n 0
23.
283
40.
(a) Its atomic number is 32 and atomic weight is 65
(b) Its atomic number is 65
(c) It has 65 electrons
(d) It has 32 neutrons
Two atoms have the same atomic mass but different atomic
numbers. Such atoms are called as
(a) Nuclear isomers
(b) Isobars
(c) Isotopes
(d) Fission products
(a) Isotopes
The atomic number of bromine is 35 and its atomic weight is 79.
(c) Isomer
Two isotopes of bromine are present in equal amounts. Which of
the following statements represents the correct number of neutrons[NCERT41.1983] 18 Ar 40 , 20 Ca 40 and
[NCERT 1971, 76; IIT 1983]
(b) Isobars
(d) Isoelectronic
19
K 40 are
284 Nuclear Chemistry
[MNR 1983; DPMT 1991; EAMCET 1992;
RPMT 1997; Pb.CET 2000]
42.
(a) Isomers
(b) Isotopes
(c) Isobars
(d) Isotones
Atoms in hydrogen gas have preponderance of
(c)
55.
43.
44.
1H
1
(d)
NO 2 and CO 2
[Bihar CEE 1982]
[CPMT 1972]
(a)
CN and CO
Which of the following are pairs of isotopes
56.
atoms
(b) Deuteron atoms
(c) Tritium atoms
(d) All the three (a), (b) and (c) are in equal proportion
Positron emission results from the transformation of one nuclear
proton into neutron. The isotope thus produced possesses
(a) Same mass number
(b) Higher nuclear charge
(c) Intense radioactivity
(d) No radioactivity
An isotope of oxygen has mass number 18. Other isotopes of oxygen
will have the same
57.
(a)
2

1H
and
3
1H
(b)
3
1H
and
(c)
3
2 He
and
4
2 He
(d)
12
6 C
and
4

2H
14
7
N
Which among the following isotope is not found in natural uranium
(a)
92 U
234
(b)
92 U
235
(c)
92 U
238
(d)
92 U
239
An isotone of
76
32 Ge
is (one or more are correct)
[MP PMT 1990]
[IIT 1984; MADT Bihar 1995; MP PMT 1995]
(a)
77
32 Ge
(b)
77
33
As
(c)
77
34
(d)
78
34
Se
Se
[MP PMT 1985; MADT Bihar 1981]
45.
46.
(a) Mass number
(b) Atomic weight
(c) Number of neutrons
(d) Number of protons
Two nuclei which are not identical but have the same number of
nucleons represent
(a) Isotopes
(b) Isobars
(c) Isotones
(d) None of the three
The  -decay of
11 Na
24
1.
produces an isotope of
23
11 Na
which
is the more stable isotope of Na. Find out the process by
24
11
Na can undergo radioactive decay
[NCERT 1978]
48.
(a) Mg
(b) Na
(c) Al
(d) Ne
Isotopes differ in
[NCERT 1973]
(a) Number of protons
(b) Valency
(c) Chemical reactivity
(d) Number of neutrons
The isobars are atoms with the same number of
49.
(a) Protons
(b) Neutrons
(c) Protons and neutrons
(d) Nucleons
Radioactive isotope of hydrogen is
50.
(a) Tritium
(b) Deuterium
(c) Para hydrogen
(d) Ortho hydrogen
Isotopes of same elements have the same number of
47.
[DPMT 1982; CPMT 1994]
[IIT Screening 2003]
(a)
(c)
2.
3.
[MP PMT 2001; MPPET 2003]
4.
[BHU 1984; DPMT 1983; CPMT 1972, 78; AFMC 2000, 01]
(a) Protons
(c) Deutrons
51.
5.
[BHU 1984; CPMT 1977, 80]
52.
(b) 3 : 1
(d) 1 : 4
[NCERT 1977]
(a) Only O  16 isotopes
(b) Only O  17 isotopes
(c) A mixture of O  16 and O  18 isotopes
53.
54.
 emission
(b)  emission

 emission
(d) K electron capture
16
Oxygen contains 90% of O and 10% of O 18 . Its atomic mass is
(a) 17.4
(b) 16.2
(c) 16.5
(d) 17
The
missing
particle
in
the
reaction,
235
1
146
1
is
[DPMT
2001]
U

n

Ba

...

3
n
92
0
56
0
(a)
87
32 Ge
(b)
89
35
Br
(c)
87
36
(d)
86
35
Br
Kr
Sulphur-35 (34.96903 amu) emits a   particle but no   rays,
the product is chlorine-35 (34.96885 amu). The maximum energy
emitted by the   particle is
[DPMT 2004]
(b) Neutrons
(d) None
In chlorine gas, ratio of Cl 35 and Cl 37 is
(a) 1 : 3
(c) 1 : 1
An ordinary oxygen contains

(d) A mixture of O – 16, O  17 and O  18 isotopes
Isotopes were discovered by
[AMU 1983; AFMC 1995]
(a) Aston
(b) Soddy
(c) Thomson
(d) Millikan
Which of the following are iso-electronic
[CBSE 2002]
(a) CO 2 and NO
(b) SO 2 and CO 2
6.
(a) 0.016767 MeV
(b) 1.6758 MeV
(c) 0.16758 MeV
(d) 16.758 MeV
A radioactive substance has a constant activity of 2000
disintegration/minute. The material is separated into two fractions,
one of which has an initial activity of 1000 disintegrations per
second while the other fraction decays with t1 / 2 = 24 hours. The
total activity in both samples after 48 hours of separation is
(a) 2000
(b) 1250
(c) 1000
(d) 1500
How many alpha particles are emitted per second by 1 microgram of
radium
(a)
7.
3.62  10 4 / sec
(b) 0.362  10 4 / sec
(c) 362  10 4 / sec
(d) 36.2  10 4 / sec
If 1 microgram of radium has disintegrated for 500 years, how many
alpha particles will be emitted per second
(a)
2.92  10 4 / sec
(b)
292  10 4 / sec
Nuclear Chemistry
(c)
8.
0.292  10 4 / sec
(d)
disintegration s g
1
. The activity of X
in millicuries g
1
(m ci g ) is
If
54
92 U
235
Xe 139 ,
1
19.
11.
nucleus absorbs a neutron and disintegrates in
38 Sr
94
20.
and X, then what will be the product X
17.281  10
2
years
(b) 172.81  10
2
years
(c) 1.7281  10 years
(d) 1728.1  10 years
The radium and uranium atoms in a sample of uranium mineral are
in the ratio of 1 : 2.8  10 6 . If half-life period of radium is 1620
years, the half-life period of uranium will be
2
13.
45.3  10 9 years
(b)
45.3  10 10 years
16.
17.
18.
2000 gm
The half-life of 6 C 14 , if its decay constant is 6.31  10 4 is
(a)
2.5  10 3 yrs
(b) 1.832  10 3 yrs
(c)
2.275  10 3 yrs
(d)
21.
22.
23.
The half-life of a radioactive isotope is three hours. If the initial mass
of the isotope were 256 g, the mass of it remaining undecayed after
18 hours would be
(b) 8.0 g
(d) 16.0 g
15
th of a radioactive sample decays in 40 days half-life of the
16
sample is
[DCE 2001]
(a) 100 days
(b) 10 days
(d) log e 2 days
A radioactive element with half-life 6.5 hrs has 48  1019 atoms.
Number of atoms left after 26 hrs
[BHU 2003]
(a)
24  10
19
(b) 12  10
19
(c) 3  10
(d) 6  1019
The half-life of 1 gm of radioactive sample is 9 hours. The
radioactive decay obeys first order kinetics. The time required for
the original sample to reduce to 0.2 gm is
19
[AMU (Engg.) 2002]
[DCE 2000]
A radioactive isotope has a t1 / 2 of 10 days. If today 125 gm of it is
left, what was its weight 40 days earlier
0.25 gm of the sample
(c) 1 day
8 gms of a radioactive substance is reduced to 0 .5 g after 1 hour.
What is the half-life of a radioactive substance if 87.5% of any given
amount of the substance disintegrates in 40 minutes [Kerala CET 1996]
(a) 160 min
(b) 10 min
(c) 20 min
(d) 13 min 20 sec
1 gm of the sample
[AIEEE 2003]
24.
Number of --particles emitted per second by a radioactive element
falls to 1/32 of its original value in 50 days. The half-life-period of
this elements is
[AMU 2001]
(a) 5 days
(b) 15 days
(c) 10 days
(d) 20 days
A radioactive sample has a half-life of 1500 years.A sealed tube
containing 1 gm of the sample will contain after 3000 years[MNR 1994; UPSEAT
(a) 4.0 g
(c) 12.0 g
8.825  10 2 yrs
(a) 15 min
(b) 30 min
(c) 45 min
(d) 10 min
A first order nuclear reaction is half completed in 45 minutes. How
long does it need 99.9% of the reaction to be completed [KCET 2001]
(a) 5 hours
(b) 7.5 hours
(c) 10 hours
(d) 20 hours
(b) 109.8 yrs
(d) 1.098 yrs
(d) 0.00 gm of the sample
(c) 4.53  10 9 years
(d) 4.53  10 10 years
Half-life of radium is 1580 yrs. Its average life will be
The t1 / 2 of the radioactive substance is
15.
(d)
(c)
[AIIMS 1999; AFMC 1999; CPMT 1999]
14.
1250 gm
(b) 0.5 gm of the sample
2
[MP PMT 1999]
(a)
(c)
(a)
[MP PMT 1999]
12.
(b) 1000 gm
(a) 1098 yrs
(c) 10.98 yrs
(b) 0.270  10 5
(d) 0.000270
(a) -particle
(b) -particle
(c) 2-neutrons
(d) 3-neutrons
The half-life of a radioactive isotope is 3 hours. Value of its
disintegration constant is
[BHU 2002]
(a) 0.231 per hr
(b) 2.31 per hr
(c) 0.2079 per hr
(d) 2.079 per hr
The activity of carbon-14 in a piece of an ancient wood is only 12.5%.
If the half-life period of carbon-14 is 5760 years, the age of the piece
of wood will be (log 2  0.3010)
(a)
600 gm
[CBSE PMT 2001]
[CBSE 2002]
10.
(a)
[MP PET 2001]
(a) 0.027
(c) 0.00270
9.
[EAMCET 1991]
29.2  10 4 / sec
A radioactive nucleide X decays at the rate of 1.00  10 5
disintegration s 1 g 1 . Radium decays at the rate of 3.70  10 10
1
285
(a) 15.6 hours
(c) 20.9 hours
(b) 156 hours
(d) 2.09 hours
25.
The half-life period of a radioactive substance is 140 days. After how
much time 15 g will decay from 16 g sample of it
26.
(a) 140 days
(b) 560 days
(c) 280 days
(d) 420 days
Percentage of a radioactive element decayed after 20 sec when halflife is 4 sec
[BHU 2003]
(a) 92.25
(b) 96.87
(c) 50
(d) 75
[AFMC 2002]
27.
Consider an -particle just in contact with a 92 U 238 nucleus.
Calculate the coulombic repulsion energy (i.e. the height of the
coulombic barrier between U 238 and alpha particle) assuming that
the distance between them is equal to the sum of their radii[UPSEAT 2001]
(a) 23.8517× 10 4 eV
(b)
26.147738  10 4 eV
286 Nuclear Chemistry
28.
(c)
25.3522  10 4 eV
(d)
20.2254  10 4 eV
3.
The half-life period of Pb 210 is 22 years. If 2 gm of Pb 210 is
taken, then after 11 years how much of Pb 210 will be left
[KCET 2001]
(a) 1.414 gm
(c) 3.442 gm
29.
(b) 2.428 gm
(d) 4.456 gm
4.
A wood specimen from an archeological centre shows a 14
6 C
activity of 5.0 counts/min/gm of carbon. What is the age of the
specimen (t1 / 2 for 14
6 C is 5000 years) and a freshly cut wood gives
15 counts/min/gm of carbon
30.
5.78  10 4 years
(b)
9.85  10 4 years
(c)
7.85  10 3 years
(d)
0.85  10 4 years
92 U
235
product+neutron  3.20  10
 n  fission
energy released when 1 g of
92
U
235
11
5.
undergoes fission is
31.
(b) 18.60  10 9 kJ
12.75  10 8 kJ
(c) 8.21  10 7 kJ
(d)
The triad of nuclei that is isotonic is
Assertion
:
The activity of 1 g pure uranium-235 will be
greater than the same amount present as U3 O8 .
Reason
:
In the combined state, the activity of the
radioactive element decreases.
Assertion
:
Nuclear forces are called short range forces.
Reason
:
Nuclear forces operate over very small distance
Assertion
:
7.
6.55  10 6 kJ
32.
15
(a)
6C
(b)
6C
12
(c)
6C
14
, 7N
(d)
6C
14
, 7 N 14 , 9 F 19
, 7N
, 7 N 14 , 9 F 19
14
, 9F
17
, 9F
8.
The relative abundance of two isotopes of atomic weight 85 and 87
is 75% and 25% respectively. The average atomic weight of element
is [DCE 2003]
(a) 75.5
(b) 85.5
(c) 40.0
(d) 86.0
10.
Ba  e

133
55
of
K-capture
is
Cs  X  ray.
The atomic number decreases by one unit as
result of K-capture.
Assertion
:
Radioactive heavy nuclei decay by a series of
  and / or   emission, to form a stable
isotope of lead.
Reason
:
Radioactivity is a physical phenomenon.
Assertion
:
Actinium series is so called because it starts with
an isotope of actinium.
Reason
:
Actinium is formed in the nature as such and is
not formed from the disintegration of any other
radioisotope.
Assertion
:
For maximum stability N/P ratio must be equal
to 1.
Reason
:
Loss of   and   particles has no role in
N/P ratio.
Assertion
:
Reason
:
The neutrons are better initiators of nuclear
reactions, than the protons, deutrons or particles of the same energy.
Neutrons are uncharged particles and hence, they
are not repelled by positively charged nucleus.
Assertion
:
17
9.
example
:
[IIT 1988; DCE 2000;MP PMT 2004]
14
An
Reason
The
[CBSE PMT 1997]
(a)
Nuclides of the same element of different mass
numbers are called isotopes of that element.
133
56
6.
J.
:
i.e., 10 15 m or 1 fermi.
[AMU (Engg.) 2002]
(a)
Reason
Breeder reactor produces fissile
94
Pu239 from
non-fissile uranium.
11.
Read the assertion and reason carefully to mark the correct option out of
the options given below :
(a)
(c)
(d)
(e)
If both assertion and reason are true and the reason is the correct
explanation of the assertion.
If both assertion and reason are true but reason is not the correct
explanation of the assertion.
If assertion is true but reason is false.
If the assertion and reason both are false.
If assertion is false but reason is true.
1.
Assertion
:
Mass number of an atom is equal to total
number of nucleons present in the nucleus.
Reason
:
Mass number defines the identity of an atom.
Assertion
:
1H
(b)
2.
1
, 1 H 2 and 1 H 3 are isotopes of hydrogen.
12.
13.
14.
Reason
:
A breeder reactor is one that produces more
fissionable nuclei that it consumes.
Assertion
:
The activation energies for fusion reactions are
very low.
Reason
:
They require very low temperature to overcome
electrostatic repulsion between the nuclei.
Assertion
:
The archeological studies are based on the
radioactive decay of carbon-14 isotope.
Reason
:
The ration of C-14 to C-12 in the animals and
plants is same as that in the atmosphere.
Assertion
:
Photochemical smog is produce by nitrogen
oxides.
Reason
:
Vehicular pollution is a major sources of nitrogen
oxides.
Assertion
:
A nuclear binding energy per nucleon is in the
order
9
4 Be
73 Li 42 He .
Nuclear Chemistry
Reason
15.
:
Assertion
:
Binding energy per nuclear increases linearly
with difference in number of neutrons and
protons.
21
a
22
c
23
d
24
b
25
c
26
d
27
d
28
c
29
c
30
a
Nuclear fission is always accompanied by release
of energy.
31
d
32
c
33
a
34
b
35
a
36
c
37
a
38
acd
39
a
Reason
:
Nuclear fission is a chain process.
Assertion
:
Protones are more effective than neutrons of
equal energy in causing artificial disintegration of
atoms.
[AIIMS 1994]
16.
Reason
17.
:
Assertion
Reason
A beam of electrons deflects more than a beam
of  -particles in an electric field.
Electrons possess negative charge while  particles possess positive charge.
:
[AIIMS 2002]
18.
Assertion
Reason
Causes of radioactivity and Group displacement law
1
b
2
d
3
d
4
d
5
c
6
c
7
c
8
a
9
b
10
a
11
a
12
c
13
d
14
b
15
a
16
a
17
a
18
a,b,c
19
c
20
c
21
b
22
d
23
d
24
a
25
b
26
d
27
b
28
b
29
b
30
a
31
a
32
a
33
c
34
b
35
a
36
b
37
c
38
b
39
c
40
d
41
a
42
b
43
c
44
c
45
d
46
b
47
b
48
d
49
a
50
d
51
a
52
a
53
d
54
d
55
b
56
a
57
d
58
c
59
c
60
a
61
a
62
c
63
d
64
d
65
a
66
d
67
b
68
c
69
b
70
a
71
c
72
c
73
d
74
a
75
b
Neutrons are neutral they penetrate the nucleus.[AIIMS 1998]
:
22
11
:
Na emits a position giving
22
12
Mg .

In  emission neutron is transformed into
proton.
[AIIMS 1994]
:
287
Nucleus (Stability and Reaction)
Rate of decay and Half-life
1
b
2
b
3
a
4
a
5
b
6
d
7
b
8
c
9
c
10
d
1
c
2
a
3
b
4
a
5
d
11
b
12
c
13
c
14
c
15
c
6
a
7
d
8
d
9
d
10
c
a
12
d
13
d
14
a
15
c
16
c
17
d
18
c
19
a
20
b
11
21
b
22
a
23
b
24
d
25
c
16
d
17
b
18
c
19
b
20
a
26
c
2