34
YEARS’ (1988-2021)
Chapterwise Topicwise
NEET
Solved Papers
CHEMISTRY
34
YEARS’ (1988-2021)
Chapterwise Topicwise
NEET
Solved Papers
CHEMISTRY
Complete Collection of all Questions
asked in last 34 years’ in NEET & CBSE AIPMT
Arihant Prakashan (Series), Meerut
Arihant Prakashan
(Series), Meerut
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PREFACE
Whenever a student decides to prepare for any examination his/her
first and foremost curiosity is to know about the type of questions that
are expected in the exam. This becomes more important in the context
of competitive entrance examinations where there is neck-to-neck
competition.
We feel great pleasure in presenting before you this book containing
Error Free Chapterwise Topicwise Solutions of CBSE AIPMT/NEET
Chemistry Questions from the years 1988 to 2021.
It has been our efforts to provide correct solutions to the best of our
knowledge and opinion. Detailed explanatory discussions follow the
answers. Discussions are not just sketchy–rather, have been drafted in a
manner that the students will surely be able to answer some other
related questions too ! Going through this book, the students would be
able to have the complete idea of the questions being asked in the test.
We hope this chapterwise solved papers would be highly beneficial to
the students. We would be grateful if any discrepancies or mistakes in
the questions or answers are brought to our notice so that these could
be rectified in subsequent editions.
Publisher
CONTENTS
1. Some Basic Concepts in Chemistry
2. Atomic Structure
1-8
9-16
3. Chemical Bonding and Molecular Structure
17-32
4. Chemical Thermodynamics
33-43
5. States of Matter
44-49
6. Solid State
50-55
7. Solutions
56-64
8. Chemical Equilibrium
65-69
9. Ionic Equilibrium
70-79
10. Redox Reactions and Electrochemistry
80-89
11. Chemical Kinetics
90-99
12. Surface Chemistry
100-103
13. Classification of Elements and Periodicity in Properties
104-108
14. General Principles and Processes of Isolation of Metals
109-112
15. Hydrogen
113-114
16. s-Block Elements
115-118
17. p- Block Elements
119-134
18. d and f-Block Elements
135-144
19. Co-ordination Compounds
145-156
20. Purification and Characterisation of Organic Compounds
157-159
21. Some Basics Principles of Organic Chemistry
160-174
22. Hydrocarbons
175-184
23. Organic Compounds Containing Halogens
185-190
24. Organic Compounds Containing Oxygen
191-216
25. Organic Compounds Containing Nitrogen
217-226
26. Polymers
227-231
27. Biomolecules , Chemistry in Everyday Life and Environmental Chemistry
232-242
SYLLABUS
CLASS 11th
UNIT I Some Basic Concepts of Chemistry
General Introduction Important and scope of chemistry. Laws of chemical combination, Dalton's atomic theory
concept of elements, atoms and molecules. Atomic and molecular masses. Mole concept and molar mass,
percentage composition and empirical and molecular formula, chemical reactions, stoichiometry and
calculations based on stoichiometry.
UNIT II Structure of Atom
Atomic number, isotopes and isobars. Concept of shells and subshells, dual nature of matter and light, deBroglie's relationship, Heisenberg’s uncertainty principle, concept of orbital, quantum numbers, shapes of s,p
and d orbitals, rules for filling electrons in orbitals- Aufbau principle, Pauli exclusion principles and Hund's rule,
electronic configuration of atoms, stability of half-filled and completely filled orbitals.
UNIT III Classification of Elements and Periodicity in Properties
Modern periodic law and long form of periodic table, periodic trends in properties of elements- atomic radii,
ionic radii, ionisation enthalpy, electron gain enthalpy, electronegativity, valence.
UNIT IV Chemical Bonding and Molecular Structure
Valence electrons, ionic bond, covalent bond, bond parameters, Lewis structure, polar character of covalent
bond, valence bond theory, resonance, geometry of molecules, VSEPR theory, concept of hybridization
involving s, p and d orbitals and shapes of some simple molecules, molecular orbital theory of homonuclear
diatomic molecules (qualitative idea only). Hydrogen bond.
UNIT V States of Matter: Gases and Liquids
Three states of matter, intermolecular interactions, types of bonding, melting and boiling points, role of gas
laws of elucidating the concept of the molecule, Boyle's law, Charles’ law, Gay Lussac's law, Avogadro's law, ideal
behaviour of gases, empirical derivation of gas equation. Avogadro number, ideal gas equation. Kinetic energy
and molecular speeds (elementary idea), deviation from ideal behaviour, liquefaction of gases, critical
temperature. Liquid State- Vapour pressure, viscosity and surface tension (qualitative idea only, no
mathematical derivations).
UNIT VI Thermodynamics
First law of thermodynamics internal energy and enthalpy, heat capacity and specific heat, measurement of U
and H, Hess's law of constant heat summation, enthalpy of : bond dissociation, combustion, formation,
atomization, sublimation, phase transition, ionization, solution and dilution. Introduction of entropy as state
function, Second law of thermodynamics Gibbs’ energy change for spontaneous and non-spontaneous process,
criteria for equilibrium and spontaneity. Third law of thermodynamics Brief introduction.
UNIT VII Equilibrium
Equilibrium in physical and chemical processes, dynamic nature of equilibrium, law of chemical equilibrium,
equilibrium constant, factors affecting equilibrium-Le Chatelier's principle, ionic equilibrium- ionization of acids
and bases, strong and weak electrolytes, degree of ionization, ionization of polybasic acids, acid strength,
concept of pH, Hydrolysis of salts (elementary idea), buffer solutions, Henderson equation, solubility product,
common ion effect (with illustrative examples).
UNIT VIII Redox Reactions
Concept of oxidation and reduction, redox reactions oxidation number, balancing redox reactions in terms of
loss and gain of electron and change in oxidation numbers.
UNIT IX Hydrogen
Occurrence, isotopes, preparation, properties and uses of hydrogen, hydrides-ionic, covalent and interstitial,
physical and chemical properties of water, heavy water, hydrogen peroxide-preparation, reactions, uses and
structure.
UNIT X s-Block Elements (Alkali and Alkaline Earth Metals)
Group 1 and group 2 elements General introduction, electronic configuration, occurrence, anomalous
properties of the first element of each group, diagonal relationship, trends in the variation of properties (such
as ionization enthalpy, atomic and ionic radii), trends in chemical reactivity with oxygen, water, hydrogen and
halogens, uses. Preparation and Properties of Some important Compounds. Sodium carbonate, sodium
chloride, sodium hydroxide and sodium hydrogencarbonate, biological importance of sodium and potassium.
Industrial use of lime and limestone, biological importance of Mg and Ca.
UNIT XI Some p-Block Elements
General Introduction to p-Block Elements.
Group 13 elements General introduction, electronic configuration, occurrence, variation of properties,
oxidation states, trends in chemical reactivity, anomalous properties of first element of the group; Boron, some
important compounds borax, boric acids, boron hydrides. Aluminium, uses, reactions with acids and alkalies.
General 14 elements General introduction, electronic configuration, occurrence, variation of properties,
oxidation states, trends in chemical reactivity, anomalous behaviour of first element. Carbon, allotropic forms,
physical and chemical properties, uses of some important compounds, oxides.
Important compounds of silicon and a few uses, silicon tetrachloride, silicones, silicates and zeolites, their uses.
UNIT XII Organic Chemistry- Some Basic Principles and Techniques
General introduction, methods of purification qualitative and quantitative analysis, classification and IUPAC
nomenclature of organic compounds. Electronic displacements in a covalent bond: inductive effect,
electromeric effect, resonance and hyper conjugation. Homolytic and heterolytic fission of a covalent bond free
radials, carbocations, carbanions, electrophiles and nucleophiles, types of organic reactions.
UNIT XIII Hydrocarbons
Alkanes Nomenclature, isomerism, conformations (ethane only), physical properties, chemical reactions
including free radical mechanism of halogenation, combustion and pyrolysis. Alkenes Nomenclature, structure
of double bond (ethene), geometrical isomerism, physical properties, methods of preparation, chemical
reactions, addition of hydrogen, halogen, water, hydrogen halides (Markovnikov's addition and peroxide
effect), ozonolysis, oxidation, mechanism of electrophilic addition.
Alkynes Nomenclature, structure of triple bond (ethyne), physical properties, methods of preparation, chemical
reactions, acidic character of alkynes, addition reaction of- hydrogen, halogens, hydrogen halides and water.
Aromatic hydrocarbons Introduction, IUPAC nomenclature, Benzene, resonance, aromaticity, chemical
properties, mechanism of electrophilic substitution- Nitration sulphonation, halogenation, Friedel Craft's
alkylation and acylation, directive influence of functional group in mono-substituted benzene, carcinogenicity
and toxicity.
UNIT XIV Environmental Chemistry
Environmental pollution Air, water and soil pollution, chemical reactions in atmosphere, smogs, major
atmospheric pollutants, acid rain ozone and its reactions, effects of depletion of ozone layer, greenhouse effect
and global warming-pollution due to industrial wastes, green chemistry as an alternative tool for reducing
pollution, strategy for control of environmental pollution.
CLASS 12th
UNIT I Solid State
Classification of solids based on different binding forces, molecular, ionic covalent and metallic solids,
amorphous and crystalline solids (elementary idea), unit cell in two dimensional and three dimensional lattices,
calculation of density of unit cell, packing in solids, packing efficiency, voids, number of atoms per unit cell in a
cubic unit cell, point defects, electrical and magnetic properties, Band theory of metals, conductors,
semiconductors and insulators.
UNIT II Solutions
Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids,
solid solutions, colligative properties- relative lowering of vapour pressure, Raoult's law, elevation of boiling
point, depression of freezing point, osmotic pressure, determination of molecular masses using colligative
properties abnormal molecular mass. Van Hoff factor.
UNIT III Electrochemistry
Redox reactions, conductance in electrolytic solutions, specific and molar conductivity variation of conductivity
with concentration, kohlrausch's Law, electrolysis and Laws of electrolysis (elementary idea), dry cellelectrolytic cells and Galvanic cells; lead accumulator, EMF of a cell, standard electrode potential, Relation
between Gibbs’ energy change and EMF of a cell, fuel cells, corrosion.
UNIT IV Chemical Kinetics
Rate of a reaction (average and instantaneous), factors affecting rates of reaction, concentration, temperature,
catalyst, order and molecularity of a reaction, rate law and specific rate constant, integrated rate equations and
half-life (only for zero and first order reactions), concept of collision theory ( elementary idea, no mathematical
treatment). Activation energy, Arrhenious equation.
UNIT V Surface Chemistry
Adsorption physisorption and chemisorption, factors affecting adsorption of gases on solids, catalysis
homogeneous and heterogeneous, activity and selectivity, enzyme catalysis, colloidal state, distinction
between true solutions, colloids and suspensions, lyophillic, lyophobic multimolecular and macromolecular
colloids, properties of colloids, Tyndall effect, Brownian movement, electrophoresis, coagulation, emulsionstypes of emulsions.
UNIT VI General Principles and Processes of Isolation of Elements
Principles and methods of extraction concentration, oxidation, reduction electrolytic method and refining,
occurrence and principles of extraction of aluminium, copper, zinc and iron.
UNIT VII p- Block Elements
Group 15 elements General introduction, electronic configuration, occurrence, oxidation states, trends in
physical and chemical properties, preparation and properties of ammonia and nitric acid, oxides of nitrogen
(structure only), Phosphorous allotropic forms, compounds of phosphorous, preparation and properties of
phosphine, halides (PCI3, PCI5) and oxoacids (elementary idea only).
Group 16 elements General introduction, electronic configuration, oxidation states, occurrence, trends in
physical and chemical properties, dioxygen, preparation, properties and uses, classification of oxides, ozone.
Sulphur allotropic forms, compounds of sulphur, preparation, preparation, properties and uses of sulphur
dioxide, sulphuric acid, industrial process of manufacture, properties and uses, oxoacids of sulphur
(structures only).
Group 17 elements General introduction, electronic configuration, oxidation states, occurrence, trends in
physical and chemical properties, compounds of halogens, preparation, properties and uses of chlorine and
hydrochloric acid, interhalogen compounds oxoacids of halogens (structures only).
Group 18 elements General introduction, electronic configuration, occurrence, trends in physical and chemical
properties, uses.
UNIT VIII d-and f-Block Elements
General introduction, electronic configuration, characteristics of transition metals, general trends in properties
of the first row transition metals metallic character, ionization enthalpy, oxidation states, ionic
radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy formation. Preparation and
properties of K2Cr2O7 and KMnO4. Lanthanoids electronic configuration, oxidation states, chemical reactivity
and lanthanoid contraction and its consequences. Actinoids Electronic configuration, oxidation states and
comparison with lanthanoids.
UNIT XI Coordination Compounds
Coordination compounds Introduction, ligands, coordination number, colour, magnetic properties and
shapes, IUPAC nomenclature of mononuclear coordination compounds, isomerism (structural and stereo)
bonding, Werner's theory VBT,CFT, importance of coordination compounds (in qualitative analysis, biological
systems).
UNIT X Haloalkanes and Haloarenes
Haloalkanes Nomenclature, nature of C –X bond, physical and chemical properties, mechanism of substitution
reactions. Optical rotation. Haloarenes Nature of C-X bond, substitution reactions (directive influence of
halogen for monosubstituted compounds only). Uses and environment effects of – dichloromethane,
trichloromethane, tetrachloromethane, iodoform, freons, DDT.
UNIT XI Alcohols, Phenols and Ethers
Alcohols Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only),
identification of primary, secondary and tertiary alcohols, mechanism of dehydration, uses with special
reference to methanol and ethanol. Phenols, Nomenclature, methods of preparation, physical and chemical
properties, acidic nature of phenol, electrophillic substitution reactions, uses of phenols. Ethers, Nomenclature,
methods of preparation, physical and chemical properties uses.
UNIT XII Aldehydes, Ketones and Carboxylic Acids
Aldehydes and Ketones Nomenclature, nature of carbonyl group, methods of preparation, physical and
chemical properties, and mechanism of nucleophilic addition, reactivity of alpha hydrogen in aldehydes, uses.
Carboxylic Acids Nomenclature, acidic nature, methods of preparation, physical and chemical properties, uses.
UNIT XIII Organic Compounds Containing Nitrogen
Amines Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses,
identification of primary secondary and tertiary amines. Cyanides and Isocyanides will be mentioned at
relevant places. Diazonium salts Preparation, chemical reactions and importance in synthetic
organic chemistry.
UNIT XIV Biomolecules
Carbohydrates Classification (aldoses and ketoses), monosaccharide (glucose and fructose), D.L. configuration,
oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen): importance. Proteins
Elementary idea of – amino acids, peptide bond, polypeptides, proteins, primary structure, secondary structure,
tertiary structure and quaternary structure (qualitative idea only), denaturation of proteins, enzymes. Hormones
-Elementary idea (excluding structure). Vitamins Classification and function. Nucleic Acids DNA and RNA
UNIT XV Polymers
Classification Natural and synthetic, methods of polymerisation (addition and condensation),
copolymerization. Some important polymers natural and synthetic like polyesters, bakelite, rubber,
Biodegradable and non-biodegradable polymers.
UNIT XVI Chemistry in Everyday Life
Chemicals in medicines analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs,
antibiotics, antacids, antihistamines. Chemicals in food preservatives, artificial sweetening agents, elementary
idea of antioxidants. Cleansing agents soaps and detergents, cleansing action.
1
Some Basic
Concepts in Chemistry
TOPIC 1
Nature of Matter, Significant
Figures and Laws of Chemical
Combinations
01 The number of significant figures
for the three numbers 161 cm, 0.161
cm, 0.0161 cm are
[CBSE AIPMT 1998]
(a) 3,4 and 5 respectively
(b) 3,4 and 4 respectively
(c) 3,3 and 4 respectively
(d) 3,3 and 3 respectively
Ans. (d)
(i) All non-zero digits are significant.
(ii) Non-zero digits to the right of the
decimal point are significant.
(iii) Zeroes to the left of the first
non-zero digit in a number are not
significant.
So, the number of significant figures
for the numbers 161 cm, 0.161 cm and
0.0161 cm are same, i.e. 3.
02 0.24 g of a volatile gas, upon
vaporisation, gives 45 mL vapour at
NTP. What will be the vapour
density of the substance?
(Density of H2 = 0.089)
[CBSE AIPMT 1996]
(a) 95.93
(c) 95.39
Ans. (b)
(b) 59.93
(d) 5.993
Weight of gas = 0.24 g
Volume of gas (V ) = 45 mL = 0.045 L
Density of H2 (d ) = 0.089
Weight of 45 mL ofH2 = V × d
= 0.045 × 0.089
= 4.005 × 10 −3 g
Therefore, vapour density
Weight of certain volume of substance
=
Weight of same volume of hydrogen
0.24
= 59.93
=
.
4005
× 10 – 3
03 In the final answer of the
expression
(29.2 − 20.2) (1.79 × 10 5 )
1.37
the number of significant figures
is
[CBSE AIPMT 1994]
(a) 1
(c) 3
Ans. (c)
(b) 2
(d) 4
On calculation we find
(29.2 − 20.2) (1.79 × 10 5 )
= 1.17 × 10 6
1.37
As the least precise number contains
3 significant figures, therefore answer
should also contains 3 significant
figures.
04 The molecular weight of O 2 and
SO 2 are 32 and 64 respectively. At
15°C and 150 mmHg pressure, 1 L of
O 2 contains ‘N ’ molecules. The
number of molecules in 2L of SO 2
under the same conditions of
temperature and pressure will be
[CBSE AIPMT 1990]
(a) N/2
(c) 2 N
(b) N
(d) 4 N
Ans. (c)
According to Avogadro’s law “equal
volumes of all gases contain equal
number of molecules under similar
conditions of temperature and
pressure.” Thus, if 1 L of one gas
contains N molecules, 2 L of any gas
under similar conditions will contain
2 N molecules.
TOPIC 2
Atomic Mass, Molecular
Mass and Formulae of
Compounds
05 An organic compound contains
78% (by wt.) carbon and remaining
percentage of hydrogen. The right
option for the empirical formula of
this compound is [At. wt. of C is 12,
H is 1]
[NEET 2021]
(a) CH
(c) CH 3
(b) CH2
(d) CH 4
Ans. (c)
Element %
Simplest
Relative Simple
Atomic
whole
number ratio of
mass
number
of moles moles
ratio
C
78
12
78
6.5
= 6.5
=1
12
6.5
1
H
22
1
22
= 22
1
3
22
=
6.5
3.3
The empirical formula of the organic
compound is CH3.
2
NEET Chapterwise Topicwise Chemistry
(a) 104, 71 and 71
(c) 175, 104 and 71
Ans. (d)
In
(b) 71, 71 and 104
(d) 71, 104 and 71
175
71 Lu,
Mass number (A) = 175 = n + p
Atomic number (Z) = 71 = p = e −
∴ Number of protons = 71
Number of neutrons
= A − Z = 175 − 71 = 104
Number of electrons = 71
Thus, the empirical formula of the
compound is CH3O.
09 An element, X has the following
isotopic composition:
200
X : 90%, 199 X : 8.0%,
202
X :2.0%
The weighted average atomic
mass of the naturally occurring
element X is closest to
[CBSE AIPMT 2007]
(a) 201 u
(c) 199 u
Ans. (d)
07 Suppose the elements X and Y
combine to form two compounds
XY 2 and X 3Y 2 . When 0.1 mole of XY
2 weighs 10 g and 0.05 mole of
X 3Y 2 weighs 9 g, the atomic
weights of X and Y are
[NEET Phase II 2016]
Weight of 200 X = 0.90 × 200 = 180.00 u
Weight of 199 X = 0.08 × 199 = 15.92u
Weight of 202 X = 0.02 × 202 = 4.04u
Total weight = 199.96 ≈ 200 u
(a) CO2 , NO2
(b) NO2–, CO2
(c) CN–, CO
(d) SO2 , CO2
Ans. (c)
CN− and CO are isoelectronic because
they have equal number of electrons.
InCN− the number of electrons
= 6 + 7 + 1 = 14
In CO the number of electrons
Let atomic masses of X and Y be AX and
AY , respectively
10
For XY2 ,
nXY2 = 0.1 =
AX + 2AY
or
…(i)
AX + 2AY = 100
9
For X3Y2 , nX Y2 = 0.05 =
3
3AX + 2AY
A X = 40 g mol
−1
⇒ A Y = 30 g mol
…(ii)
−1
08 An organic compound contains
carbon, hydrogen and oxygen. Its
elemental analysis gave C, 38.71%
and H, 9.67%. The empirical
formula of the compound would be
[CBSE AIPMT 2002, 1999, 98]
(a) C2 H7 N2
(c) CH4 N
Ans. (c)
C
H
O
38.71
9.67
[100 – (38.71
+ 9.67)]
= 51.62
Simple
ratio
12 38.71 = 3.23 3.23 = 1
12
3.23
1
9.67
9.67
= 9.67
=3
1
3.23
16 51.62 = 3.23 3.23 = 1
16
3.23
(b) CH5 N
(d) C2 H7 N
Table for empirical formula
Element
Element
(a) CH3O (b) CH2O (c) CHO (d) CH4O
Ans. (a)
Molar
ratio
C
40
12
40
= 3.33
12
3.33
= 1
3.33
H
6.60
1
6.60
= 6.60
1
6.60
=2
3.33
O
53.3
4
16
53.34
= 3.33
16
3.33
=1
3.33
%
At.
wt.
Molar ratio
Simple
ratio
C
40.00
12
40
= 3.33
12
3.33
=1
3.33
H
13.33
1
13.33
= 13.33
1
13.33
=4
3.33
N
46.67
14
46.67
= 3.33
14
3.33
=1
3.33
Hence, empirical formula is CH4N.
At. wt. Molar ratio
Simple
ratio
Hence, empirical formula is
C : H : O = 1 : 2 : 1 = CH2O
13 Boron has two stable isotopes, 10 B
(19%) and 11 B (81%). Calculate
average atomic weight of boron in
the periodic table.
[CBSE AIPMT 1990]
(a) 10.8
(c) 11.2
Ans. (a)
(b) 10.2
(d) 10.0
Average of atomic weight
% of
11 An organic compound containing C,
H and N gave the following results
on analysis C = 40%,
H = 13.33%, N = 46.67%. Its
empirical formula would be
(b) CHO
(d) C2H2O
%
= 6 + 8 = 14
[CBSE AIPMT 2008]
%
At.
abundance wt.
[CBSE AIPMT 1994]
(a) CH2O
(c) CH4O2
Ans. (a)
10 Which of the following is
isoelectronic? [CBSE AIPMT 2002]
(a) 40, 30(b) 60, 40(c) 20, 30(d) 30, 20
Ans. (a)
or
3AX + 2AY = 180
On solving Eqs. (i) and (ii), we get,
(b) 202 u
(d) 200 u
12 An organic compound contains
C = 40%,O = 53.34% and
H =6.60%. The empirical formula
of the compound is
Element
06 The number of protons, neutrons
and electrons in 175
71 Lu,
respectively, are [NEET (Sep.) 2020]
=
10
B × atomic mass of 10 B + % of 11B
× atomic mass of 11B
% of 10 B + % of 11B
19 × 10 + 81 × 11
=
19 + 81
=
190 + 891
= 10.81
100
14 While extracting an element from
its ore, the ore is grind and leached
with dil. KCN solution to form the
soluble product potassium
argento- cyanide. The element is
[CBSE AIPMT 1989]
(a) lead
(b) chromium
(c) manganese
(d) silver
Ans. (d)
Silver metal is extracted from the
argentite ore Ag2S by cyanide process. In
this method, the concentrated ore is
treated with dilute solution of potassium
3
Some Basic Concepts in Chemistry
cyanide, then a soluble complex
potassium dicyanoargentate (I) is formed
which when reacted with zinc, silver is
extracted as a ppt.
Ag2S + 4KCN → 2K[Ag(CN) 2 ] + Na2S
2K[Ag(CN)2 ] + Zn → K2 [Zn(CN) 4 ]
+ 2Ag ↓
15 A metal oxide has the formula
Z 2O 3. It can be reduced by
hydrogen to give free metal and
water. 0.1596 g of the metal oxide
requires 6 mg of hydrogen for
complete reduction. The atomic
weight of the metal is
[CBSE AIPMT 1989]
(a) 27.9
(c) 79.8
Ans. (d)
(b) 159.6
(d) 55.8
Z2O3 + 3H2 → 2 Z + 3H2O
Q 6 × 10 −3 g H2 reduces = 0.1596 g of Z2O3
0.1596
g Z2O 3
∴ 1 g H2 reduces =
6 × 10 –3
= 26.6 g of Z2O 3
∴ Equivalent weight of Z2O 3 =26.6
Equivalent weight of Z + Equivalent
weight of O = 26.6
Equivalent weight of Z + 8 = 26.6
Equivalent weight of Z = (26.6 – 8) = 18.6
Valency of metal inZ2O3 =3
Equivalent weigh
Atomic weight
=
Valency
Atomic weight
= 18.6 × 3 = 55.8
TOPIC 3
Mole Concept and
Concentration Terms
16 One mole of carbon atom weighs 12
g, the number of atoms in it is
equal to, (Mass of carbon -12 is
1.9926 × 10 −23 g) [NEET (Oct.) 2020]
(a) 1.2 × 1023
(c) 12 × 1022
Ans. (d)
(b) 6.022 × 1022
(d) 6.022 × 1023
1 mole of carbon atoms weight 12 g, its
contains Avogadro number of carbon
atoms, i.e. 6022
.
× 1023 number of carbon
atoms.
17 Which one of the followings has
maximum number of atoms?
[NEET (Sep.) 2020]
(a) 1 g of Mg(s ) [Atomic mass of Mg = 24]
(b) 1 g of O2 (g) [Atomic mass of O = 16]
(c) 1 g of Li(s ) [Atomic mass of Li = 7]
(d) 1 g of Ag(s ) [Atomic mass of Ag = 108]
Ans. (c)
Number of atoms (n)
Mass in g (1 g) × Atomicity of
the molecule
=
× NA
Gram molar mass (M)
[QN A = Avogadro’s number]
Atomicity
⇒ n∝
M
1
(a) nMg =
24
2
1
(b) nO =
=
32 16
1
(c) nLi =
7
1
(d) nAg =
108
So, nLi > nO > nMg > nAg
18 In which case is the number of
molecules of water maximum?
[NEET 2018]
(a) 0.00224 L of water vapours at 1 atm
and 273 K
(b) 0.18 g of water
(c) 18 mL of water
(d) 10 −3 mol of water
Ans. (c)
Number of molecules = Mole ×
Avogadro’s number (N A )
The number of molecules of water in
each of the given options is calculated
as
(i) 18 mL of water
Number of moles (nH 2 O )
Mass of substance in g (wH 2 O )
=
Molar mass in g mol −1 (MH 2 O )
wH 2 O = 18g
[QDensity of water (dH 2 O ) = 1 g L−1]
18
nH 2 O = = 1
∴
18
Number of molecules of water
= 1 × NA
(ii) 0.18 g of water
wH O 0.18
= 0.01
nH 2 O = 2 =
18
MH 2 O
Number of molecules of water
= 0.01 × N A
(iii) 0.00224 L of water vapours at 1 atm
and 273 K. At STP [1 atm and 273 K],
Number of moles [with reference to
volume]
Volume of gas in litres
=
22.4
0.00224
=
= 0.0001
22.4
Number of molecules of water
= 0.0001 × NA
(iv) 10 −3 mol of water
Number of molecules of water
= 10 −3 × N A
∴ Among the given options, option (i)
contains the maximum number of water
molecules.
19 If Avogadro number N A , is changed
from 6.022 × 10 23 mol −1 to
6.022 × 10 20 mol −1 this would
change
[CBSE AIPMT 2015]
(a) the definition of mass in units of
grams
(b) the mass of one mole of carbon
(c) the ratio of chemical species to each
other in a balanced equation
(d) the ratio of elements to each other in
a compound
Ans. (b)
If Avogadro numberN A , is changed from
6.022 × 1023 mol − 1 to 6.022 × 1020 mol − 1,
this would change the mass of one mole
of carbon.
Q 1 mole of carbon has mass = 12 g
or 6.022 × 1023 atoms of carbon have
mass = 12 g
∴6.022 × 1020 atoms of carbon have
mass
12
=
× 6.022 × 1020 = 0.012 g
6.022 × 1023
20 How many grams of concentrated
nitric acid solution should be used
to prepare 250 mL of 2.0 M HNO 3 ?
The concentrated acid is 70%
[NEET 2013]
HNO 3 .
(a) 45.0 g conc. HNO 3
(b) 90.0 g conc. HNO 3
(c) 70.0 g conc. HNO 3
(d) 54.0 g conc. HNO 3
Ans. (a)
Given, molarity of solution = 2
Volume of solution = 250mL
4
NEET Chapterwise Topicwise Chemistry
=
250 1
= L
1000 4
Molar mass of
HNO3 = 1 + 14 + 3 × 16 = 63 g mol −1
Q Molarity
Weight of HNO3
=
Molecular mass of HNO3
× volume of solution (L)
∴Weight of
HNO3 = molarity × molecular mass
× volume (L)
1
= 2 × 63 × g
4
= 31.5 g
It is the weight of 100%HNO3
But the given acid is 70%HNO3
100
g
∴ Its weight = 31. 5 ×
70
= 45 g
21 6.02 × 10 20 molecules of urea are
present in 100 mL of its solution.
The concentration of solution is
[NEET 2013]
(a) 0.02 M
(c) 0.001 M
Ans. (b)
(b) 0.01 M
(d) 0.1 M
602
. × 1020
6023
.
× 10
23
= 0.999 × 10 −3
~− 1 × 10 −3 mol
Volume of the solution
100
L = 0.1 L
= 100 mL =
1000
Concentration of urea solution (in molL−1)
1 × 10 −3
=
mol L−1
0.1
= 1 × 10 −2 mol L−1
= 0.01 mol L−1
22 The number of atoms in 0.1 mole of
a triatomic gas is
(N A = 6.023 × 10 23 mol −1 )
[CBSE AIPMT 2010]
(a) 6.026 × 1022
(c) 3.600 × 1023
Ans. (b)
[CBSE AIPMT 2008]
(a) 9.0 × 10−23 cm3
(b) 6.023 × 10−23 cm3
(c) 3.0 × 10−23 cm3
(d) 5.5 × 10−23 cm3
Ans. (c)
1 mole = 6.023 × 1023 molecule
18 g = 6.02 × 1023 molecule
18 g = mass of 6.02 × 1023
water molecules
Mass of one water molecule
18
=
g
6.023 × 1023
Density = 1g cm–3
Mass of one water molecule
Volume =
Density
18
=
cm3
6.023 × 1023 × 1
≈ 3.0 × 10 −23 cm3
24 The maximum number of
molecules are present in
[CBSE AIPMT 2004]
Given, number of molecules of urea
= 602
. × 1020
602
. × 1020
∴ Number of moles =
NA
=
23 Volume occupied by one molecule
of water (density = 1 g cm −3 ) is
(b) 1.806 × 1023
(d) 1.800 × 1022
Number of atoms = number of moles
× NA × atomicity
= 0.1 × 6.023 × 1023 × 3
= 1.806 × 1023 atoms
(a) 15 L of H2 gas at STP
(b) 5 L of N2 gas at STP
(c) 0.5 g of H2 gas
(d) 10 g of O2 gas
Ans. (a)
In 15 L of H2 gas at STP,
the number of molecules
=
6.023 × 1023
× 15
22.4
= 4.033 × 1023
In 5 L of N2 gas at STP,
the number of molecules
=
6.023 × 1023 × 5
22.4
= 1.344 × 1023
In 0.5 g of H2 gas,
the number of molecules
=
6.023 × 1023 × 0.5
2
= 1.505 × 1023
In 10 g of O2 gas,
the number of molecules
=
6.023 × 1023 × 10
32
= 1.882 × 1023
Hence, maximum number of molecules
are present in 15 L ofH2 at STP.
25 Percentage of Se in peroxidase
anhydrase enzyme is 0.5% by weight
(at. weight = 78.4), then minimum
molecular weight of peroxidase
anhydrase enzyme is
[CBSE AIPMT 2001]
(a) 1.568 × 103
(c) 2.168 × 104
Ans. (d)
(b) 15.68
(d) 1.568 × 104
Suppose the molecular weight of
enzyme = x
0.5% by weight means in 100 g of
enzyme weight of Se = 0.5 g
∴ In x g of enzyme weight ofSe =
Hence,
784
. =
∴
0.5
×x
100
0.5 × x
100
x = 15680
= 1.568 × 10 4
26 The number of atoms in 4.25 g of
NH3 is approximately
[CBSE AIPMT 1999]
(a) 4 × 1023
(c) 1 × 1023
Ans. (d)
(b) 2 × 1023
(d) 6 × 1023
Weight of NH3 = 4.25 g
Number of moles of
Weight
NH3 =
Molecular weight
4.25
=
= 0.25 mol
17
Number of molecules in 0.25 mole ofNH3
= 0.25 × 6.023 × 1023
So, number of atoms
= 4 × 0.25 × 6.023 × 1023
= 60
. × 1023
27 Haemoglobin contains 0.33% of iron
by weight. The molecular weight of
haemoglobin is approximately
67200 g. The number of iron atoms
(at. weight of Fe is 56) present in one
molecule of haemoglobin are
[CBSE AIPMT 1998]
(a) 1
(b) 6
Ans. (c)
(c) 4
(d) 2
Q 0.33 % of iron by weight means 100 g
of haemoglobin has 0.33 g of iron
100 g of haemoglobin contains iron
= 0.33 g
∴ 67200 g of haemoglobin contains iron
0.33 × 67200
g
=
100
= 221.76 g of Fe
5
Some Basic Concepts in Chemistry
221.76
56
= 3.96 ≈ 4
Number of Fe-atoms =
28 The number of moles of oxygen in 1
L of air containing 21% oxygen by
volume, under standard conditions,
is
[CBSE AIPMT 1995]
(a) 0.0093 mole
(c) 0.186 mole
Ans. (a)
(b) 2.10 moles
(d) 0.21 mole
Volume of oxygen in 1 L of air
21
=
× 1000 = 210 mL
100
Q 22400 mL volume at STP is occupied
by oxygen = 1 mole
Therefore, number of moles occupied
by 210 mL
210
= 0.0093 mol
=
22400
29 The percentage weight of Zn in
white vitriol [ZnSO4 ⋅ 7H2O] is
approximately equal to
(at. mass of Zn = 65, S = 32,O = 16
and H= 1)
[CBSE AIPMT 1995]
(a) 33.65%
(c) 23.65%
Ans. (d)
(b) 32.56%
(d) 22.65%
Molecular weight of
ZnSO4 ⋅ 7H2O= 65 + 32+ (4 × 16) + 7 (18)
= 287
65
∴ Percentage weight ofZn =
× 100
287
= 22.65%
30 The total number of valence
electrons in 4.2 g of N −3 ion is (NA is
the Avogadro’s number)
[CBSE AIPMT 1994]
(a) 2.1 N A (b) 4.2 N A (c) 1.6 N A (d) 3.2 N A
Ans. (c)
Moles of N−3 ion =
4.2
= 0.1
42
Each nitrogen atom has 5 valence
electrons. Therefore, total number of
electrons inN−3 ion = 16
Total number of electrons in 0.1 mole or
4.2 g of N−3 ion = 0.1 × 16 × N A = 1.6 N A
31 The number of gram molecules of
oxygen in 6.02 × 10 24 CO molecules
is
[CBSE AIPMT 1990]
(a) 10 g molecules (b) 5 g molecules
(c) 1 g molecule
(d) 0.5 g molecule
Ans. (b)
Ans. (d)
6.023 × 1023 molecules of CO
= 1 mole of CO
6.02 × 1024 molecules of CO
= 10 moles of CO
= 10 g atoms of O = 5 g molecules of O2
At NTP 22400 cc ofN2O contains
= 6.02 × 1023 molecules
∴1 cc N2O will contain
6.02 × 1023
molecules
=
22400
In N2O molecule, number of atoms
=2+ 1=3
Thus, number of atoms
32 The number of oxygen atoms in
4.4 g of CO 2 is [CBSE AIPMT 1990]
(a) 1.2 × 10 23
(c) 6 × 10 23
Ans. (a)
(b) 6 × 10 22
(d) 12 × 10 23
1 mole of CO2 = 44 g of CO2
= 6.023 × 1023 molecules
∴ 4.4 g of CO2 = 0.1 mole of CO2
= 6.023 × 0.1 × 1023 molecules
= 6.023 × 1022 molecules
= 6.023 × 1022 molecules of O2
= 2 × 6.023 × 1022 atoms of O
≈ 1.2 × 1023 atoms of O
33 Ratio of C p and C V of a gas ‘ X ’ is
1:4. The number of atoms of the
gas ‘ X ’ present in 11.2 L of it at NTP
will be
[CBSE AIPMT 1989]
(a) 6.02 × 1023
(c) 3.01 × 1023
Ans. (a)
(b) 1.2 × 1023
(d) 2.01 × 1023
For the gas X ratio of C p /C V = 1 : 4
So, the gas X is diatomic.
At NTP, volume of 1 mole of a gas
= 22.4 L
1 mole of a gas = 6.023 × 1023 molecules
Thus, at NTP 22.4 L contains
= 6.023 × 1023 molecules
So, at NTP 11.2 L contains
=
6.023 × 1023 × 11.2
molecules
22.4
= 301
. × 1023 molecules
Hence, number of atoms of gas ‘X ’
(diatomic)
= 3.01 × 1023 × 2 atoms
= 6.02 × 1023 atoms
34 1 cc N 2O at NTP contains
[CBSE AIPMT 1988]
1.8
(a)
× 1022 atoms
224
6.02
(b)
× 1023 molecules
22400
1.32
(c)
× 1023 electrons
224
(d) All of the above
=
3 × 6.02 × 1023
atoms
22400
1.8 × 1022
atoms
224
In N2O molecule, number of electrons
=
= 7 + 7 + 8 = 22
Hence, number of electrons
=
6.02 × 1023
× 22 electrons
22400
=
1.32 × 1023
electrons
224
35 At STP, the density of CCl 4 vapour
in g/L will be nearest to
[CBSE AIPMT 1988]
(a) 6.87 (b) 3.42
Ans. (a)
(c) 10.26 (d) 4.57
1 mole CCl 4 vapours
= 12 + 4 × 35.5 = 154 g
At STP, volume of 1 mole of a gas
= 22.4 L
Thus,
154 g = 22.4 L
154
g L–1
∴ Density of CCl 4 vapours =
22.4
= 6.87 g L–1
TOPIC 4
Stoichiometric and
Volumetric Calculations
36 The number of moles of hydrogen
molecules required to produce 20
moles of ammonia through Haber’s
process is
[NEET (National) 2019]
(a) 20
(b) 30
Ans. (b)
(c) 40
(d) 10
According to Haber’s process,
N2 (g) + 3H2 (g)
2NH3 (g)
Now, according to above equations
2 moles of ammonia (NH3) require = 3
moles of H2
∴ 1 mole of NH3 require = 3 /2 moles of H2
-
6
NEET Chapterwise Topicwise Chemistry
or, 20 moles of NH3 require =
3
× 20
2
moles of H2 = 30 moles of H2 .
Note Involvement of any limiting
reagent is not mentioned in question.
37 20.0 g of a magnesium carbonate
sample decomposes on heating to
give carbon dioxide and 8.0 g
magnesium oxide. What will be the
percentage purity of magnesium
carbonate in the sample?
(Atomic weight of Mg = 24)
[CBSE AIPMT 2015]
(a) 75
(b) 96
Ans. (d)
(c) 60
(d) 84
Key Concept In the given problem we
have provided practical yield of MgO. For
calculation of percentage yield of MgO,
we need theoretical yield of MgO. For
this we shall use mole concept.
…(i)
MgCO3( s ) → MgO( s ) + CO2 ( g )
Weight in gram
Moles of MgCO3 =
Molecular weight
20
=
= 0.238 mol
84
From Eq. (i)
1 mole of MgCO3 gives = 1 mol MgO
∴0.238 moleMgCO3 will give
= 0.238 mol MgO
= 0.238 × 40 g
= 9.52 g MgO
Now, practical yield ofMgO = 8 g
8
∴ % purity =
× 100 = 84%
9.52
Alternate Method
MgCO3 → MgO + CO2
84 g
40 g
∴8 g MgO will be form from
∴
% purity =
84
g
5
84 100
×
= 84%
5
20
38 What is the mass of precipitate
formed when 50 mL of 16.9%
solution of AgNO 3 is mixed with 50
mL of 5.8% NaCl solution?
(Ag = 107.8,N = 14, O = 16,
Na = 23,Cl = 35.5)
[CBSE AIPMT 2015]
(a) 28 g
(c) 7 g
Ans. (c)
(b) 3.5 g
(d) 14 g
Plan For the calculation of mass of AgCl
precipitated, we find mass of AgNO3 and
NaCl in equal volume with the help of
mole concept.
16.9% solution of AgNO3 means 16.9 g
AgNO3 is present in 100 mL solution.
∴8.45 g AgNO3 will be present in 50 mL
solution.
Similarly,
5.8 g NaCl is present in 100 mL solution
∴2.9 g NaCl is present in 50 mL solution
AgNO3 + NaCl → AgCl + NaNO3
Initial mole
8.45
2.9
0
0
58.5
169.8
=0.049 = 0.049
After reaction
0
0
0.049 0.049
∴Mass of AgCl precipitated
= 0.049 × 143.5 = 7 g
39 When 22.4 L of H2 (g) is mixed with
11.2 L of Cl 2 (g), each at STP, the
moles of HCl(g) formed is equal to
[CBSE AIPMT 2014]
(a) 1 mole of HCl (g)
(b) 2 moles of HCl (g)
(c) 0.5 mole of HCl (g)
(d) 1.5 moles of HCl (g)
Ans. (a)
The given problem is related to the
concept of stoichiometry of chemical
equations. Thus, we have to convert the
given volumes into their moles and then,
identify the limiting reagent [possessing
minimum number of moles and gets
completely used up in the reaction]. The
limiting reagent gives the moles of
product formed in the reaction.
H2 ( g ) + Cl2 ( g ) → 2HCl ( g )
Initial vol. 22.4 L
11.2 L
2 mol
Q 22.4 L volume at STP is occupied by
Cl2 = 1 mole
∴ 11.2 L volume will be occupied by
1 × 11.2
mol = 0.5 mol
Cl2 =
22.4
22.4 L volume at STP is occupied byH2
= 1 mol
Thus, H2 (g) + Cl2 (g) → 2HCl (g)
1 mol 0.5 mol
Since, Cl2 possesses minimum number
of moles, thus it is the limiting reagent.
As per equation,
1 mole of Cl2 = 2 moles of HCl
∴ 0.5 mole of Cl2 = 2 × 0.5 mole of HCl
= 1.0 mole of HCl
Hence, 1.0 mole ofHCl (g) is produced by
0.5 mole of Cl2 [or 11.2 L].
40 1.0 g of magnesium is burnt with
0.56 g of oxygen in a closed vessel.
Which reactant is left in excess and
how much?
[CBSE AIPMT 2014]
(At. weight of Mg = 24, O = 16)
(a) Mg, 0.16 g
(b) O2 , 0.16 g
(c) Mg, 0.44 g
(d) O2 , 0.28 g
Ans. (a)
The balanced chemical equation is
1
Mg + O2 → MgO
2
24 g
16 g
40 g
From the above equation, it is clear that,
24 g of Mg reacts with 16 g ofO2 .
Thus, 1.0 g of Mg reacts with
16
g of O2 = 0.67 g of O2 .
24
But only 0.56 g ofO2 is available which is
less than 0.67 g. Thus,O2 is the limiting
reagent.
Further, 16 g ofO2 reacts with 24 g of Mg.
∴ 0.56 g of O2 will react with Mg
24
=
× 0 . 56
16
= 0 . 84 g
∴ Amount of Mg left unreacted
= (1.0 − 0.84) g Mg
= 0.16 g Mg
41 10 g of hydrogen and 64 g of
oxygen were filled in a steel vessel
and exploded. Amount of water
produced in this reaction will be
[CBSE AIPMT 2009]
(a) 2 moles
(c) 4 moles
Ans. (c)
H2 +
1
O2 → H2O
2
1
mol
2
10
64
mol
mol
2
32
1 mol
(b) 3 moles
(d) 1 mole
1 mol
?
5 mol 2mol
1
Q mole of O2 gives = 1 mole of H2O
2
∴ 2 moles of O2 will give = 1 × 2 × 2
= 4 moles of water
42 How many moles of lead (II)
chloride will be formed from a
reaction between 6.5 g of PbO and
3.2 g of HCl?
[CBSE AIPMT 2008]
(a) 0.044 (b) 0.333 (c) 0.011 (d) 0.029
7
Some Basic Concepts in Chemistry
Ans. (d)
PbO + 2HCl
→ PbCl2 + H2O
207 + 16 2 × 36.5
= 223
= 73
(1 mol) (2 mol)
(1 mol)
6.5
= 0.029
223
3.2
Mole of HCl =
= 0.087
36.5
Since, 1 mole of PbO reacts with 2 moles
of HCl, thus in this reaction PbO is the
limiting reagent.
Hence, 1 mole of PbO forms
= 1 mole of PbCl2
0.029 mole of PbO will form = 0.029 mole
of PbCl2
Mole of PbO =
43 What volume of oxygen gas (O 2 )
measured at 0°C and 1 atm, is
needed to burn completely 1L of
propane gas (C 3H8 ) measured
under the same conditions?
[CBSE AIPMT 2008]
(a) 7 L
(c) 5 L
(b) 6 L
(d) 10 L
Ans. (c)
C3H8 +
5O2
→ 3CO2 + 4H2O
22.4 L 5 × 22.4 L
For the combustion of 22.4 L propane,
oxygen required = 5 × 22.4 L
For the combustion of 1 L of propane
oxygen required
5 × 22.4
L = 5L
=
22.4
44 Number of moles of MnO–4 required
to oxidise one mole of ferrous
oxalate completely in acidic
medium will be [CBSE AIPMT 2008]
(a) 0.6 mole
(c) 7.5 moles
Ans. (b)
(b) 0.4 mole
(d) 0.2 mole
In acidic mediumMnO−4 oxidises ferrous
oxalate as follows:
2MnO–4 + 5C2O24– + 16H+ → 2Mn2 +
+ 10CO2 + 8H2O
Q 5 moles of oxalate ions are oxidised by
2 moles of MnO−4 .
∴ 1 mole of oxalate ion is oxidised by
2
= mole of MnO−4 = 0.4 mole of MnO–4
5
45 The number of moles of KMnO 4
that will be needed to react with
one mole of sulphite ion in acidic
solution is
[CBSE AIPMT 2007]
(a) 4/5
(c) 1
Ans. (b)
(b) 2/5
(d) 3/5
2KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4
+ 3H2O + 5[O]
[MnO–4 + 8H+ + 5e – → Mn2 + + 4H2O] × 2
[SO23– + H2O → SO24– + 2H+ + 2 e − ] × 5
2 MnO–4 + 6H+ + 5SO23– → 2Mn2 + + 5 SO24–
+ 3H2O
5 moles of sulphite ions react with
= 2 moles of MnO–4
So, 1 mole of sulphite ions react with
2
= moles of MnO−4 .
5
46 The number of moles of KMnO 4
reduced by one mole of KI in
alkaline medium is
[CBSE AIPMT 2005]
(a) one fifth
(c) one
Ans. (c)
(b) five
(d) two
In alkaline medium,KMnO4 is reduced to
K2MnO4
KI + H2O → KOH + HI
+7
+6
2KMnO4 + 2KOH → 2K2 MnO4
+ H2O + [O]
Hence, one mole ofKMnO4 is reduced by
one mole of KI.
47 The mass of carbon anode
consumed (giving only carbon
dioxide) in the production of 270 kg
of aluminium metal from bauxite by
the Hall process is
(at. mass of Al = 27)
[CBSE AIPMT 2005]
(a) 180 kg (b) 270 kg (c) 540 kg(d) 90 kg
Ans. (d)
In Hall and Heroult process,
2 Al2O3 + 4C → 4 Al + 2CO2 + 2CO
but for the removal of only CO2 , following
equation is possible.
2 Al2O3 + 3C
3 × 12
= 36
→ 4Al + 3CO2
4 × 27
= 108
Q For 108 g of Al, 36 g of C is required in
above reaction.
∴For270 × 10 3 g of Al required amount of C
36
=
× 270 × 10 3
108
= 90 × 10 3 g = 90 kg
48 In Haber process 30L of dihydrogen
and 30L of dinitrogen were taken
for reaction which yielded only 50%
of the expected product. What will
be the composition of gaseous
mixture under the aforesaid
condition in the end?
[CBSE AIPMT 2003]
(a) 20 L ammonia, 10 L nitrogen, 30 L
hydrogen
(b) 20 L ammonia, 25 L nitrogen, 15 L
hydrogen
(c) 20 L ammonia, 20 L nitrogen, 20 L
hydrogen
(d) 10 L ammonia, 25 L nitrogen,
15 L hydrogen
Ans. (d)
N2 + 3H2 → 2NH3
1V
3V
10 L 30 L
2V
20 L
As only 50% of the expected product is
formed, hence only 10 L ofNH3 is formed.
Thus, for the production of 10 L ofNH3, 5
L of N2 and 15 L of H2 are used and
composition of gaseous mixture under
the aforesaid condition in the end is
H2 = 30 − 15 = 15 L
N2 = 30 − 5 = 25 L
NH3 = 10 L
49 Which has maximum number of
molecules?
[CBSE AIPMT 2002]
(a) 7 g N2
(c) 16 g NO2
Ans. (b)
(b) 2 g H2
(d) 16 g O2
In 7 g nitrogen, number of molecules
7.0
mol
=
28
= 0.25 × NA molecules
where, N A = Avogadro numbe
= 6.023 × 1023
2.0
In 2 g of H2 =
mol
2
= 1 × N A molecules
8
NEET Chapterwise Topicwise Chemistry
In 16 g of NO2 =
16.0
mol
46
Ans. (c)
= 0.348 × NA molecules
In 16 g of
16
O2 = mol = 0.5 × NA molecules
32
Hence, maximum number of molecules
are present in 2 g ofH2 .
50 Assuming fully decomposed, the
volume of CO 2 released at STP on
heating 9.85 g of BaCO 3 (at. mass
of Ba = 137) will be
4NH3 (g) + 5O2 (g) → 4NO(g) +6H2O(l )
4 mol
5 mol
4 mol
6 mol
According to equation,
5 moles of O2 required = 4 moles of NH3
1 mole of O2 requires
4
= = 0.8 mole of NH3
5
5
While 1 mole ofNH3 requires =
4
= 1.25 moles of O2
As there is 1 mole ofNH3 and 1 mole ofO2 ,
so all the oxygen will be consumed.
[CBSE AIPMT 2000]
(a) 1.12 L
(c) 2.24 L
Ans. (a)
(b) 0.84 L
(d) 4.96 L
On decomposition,BaCO3 liberates CO2 as
BaCO3 → BaO +
197 g
CO2 ↑
22.4 L at STP
Q 197 g of BaCO3 gives
= 22.4 L of CO2 at STP
9.85
g
of
BaCO3 will give
∴
22.4 × 9.85
=
= 1.12 L
197
51 In the reaction,
4NH3 (g) + 5O 2 (g) →
4NO(g) + 6H2O(l)
When 1 mole of ammonia and 1
mole of O 2 are made to react to
completion, then
[CBSE AIPMT 1998]
(a) 1.0 mole of H2O is produced
(b) 1.0 mole of NO will be produced
(c) all the oxygen will be consumed
(d) all the ammonia will be consumed
52 Liquid benzene (C 6H6 ) burns in
oxygen according to the equation,
2C 6H6 (l) + 15O 2 (g) →
12CO 2 (g) + 6H2O(g)
How many litres of O 2 at STP are
needed to complete the
combustion of 39 g of liquid
benzene? (Mol. weight of O 2 = 32,
[CBSE AIPMT 1996]
C 6H6 = 78)
(a) 74 L (b) 11.2 L (c) 22.4 L (d) 84 L
Ans. (d)
2C6H6 + 15O2 (g) → 12CO2 (g)
2 × 78
= 156
15 × 32
= 330
+ 6H2O( g )
Q 156 g of benzene required oxygen
= 15 × 22.4 L
∴ 1 g of benzene required oxygen
15 × 22.4
=
L
156
∴ 39 g of benzene required oxygen
15 × 22.4 × 39
=
156
= 84.0 L
53 What is the weight of oxygen
required for the complete
combustion of 2.8 kg of ethylene?
[CBSE AIPMT 1989]
(a) 2.8 kg
(c) 9.6 kg
Ans. (c)
(b) 6.4 kg
(d) 96 kg
C2H4
12 × 2 + 4 × 1
=28 g
+ 3O2
→ 2CO2
16 × 6
= 96 g
+ 2H2O
Q For the combustion of 28 × 10 −3 kg of
ethylene oxygen required = 96 × 10 −3 kg
∴ For the combustion of 2.8 kg of
ethylene oxygen required
=
96 × 10 −3 × 2. 8
28 × 10 −3
= 9.6 kg
54 One litre hard water contains 12.00
mg Mg 2+ . Milliequivalents of
washing soda required to remove
its hardness is [CBSE AIPMT 1988]
(a) 1
(b) 12.16
(c) 1 × 10−3
(d) 12.16 × 10−3
Ans. (a)
Mg2 + +Na2 CO3 → MgCO3 + 2Na+
1 g -eq
1 g -eq
1 g-equivalent of Mg2 + = 12 g of Mg2 +
= 12000 mg of Mg2 +
Now, 12000 mg ofMg2 + ≡ 1000
milliequivalent ofNa2 CO3
12 mg of Mg2 + ≡ 1 milliequivalent of
Na2 CO3
2
Atomic Structure
TOPIC 1
Preliminary Models
01 Which of the following is never true
for cathode rays?
[CBSE AIPMT 1994]
(a) They possess kinetic energy
(b) They are electromagnetic waves
(c) They produce heat
(d) They produce mechanical pressure
Ans. (b)
Cathode rays are not electromagnetic
wave because they do not have electric
and magnetic components
perpendicular to each other.
TOPIC 2
Bohr’s Model and
Hydrogen Spectrum
02 The energies E 1 and E 2 of two
radiations are 25 eV and 50 eV
respectively. The relation between
their wavelengths, i.e. λ 1 and λ 2
will be
[CBSE AIPMT 2011]
(a) λ 1 = 2 λ 2
1
(c) λ 1 = λ 2
2
Ans. (a)
(b) λ 1 = 4 λ 2
(d) λ 1 = λ 2
E 1 = 25 eV, E2 = 50 eV
hc
hc
and E2 =
E1 =
λ1
λ2
E1 λ2
=
E2 λ 1
25 λ 2
or
=
50 λ 1
or
or
λ 1 = 2 λ2
03 The energy absorbed by each
molecule (A 2 ) of a substance is
4.4 × 10 −19 J and bond energy per
molecule is 4.0 × 10 −19 J. The
kinetic energy of the molecule per
atom will be
[CBSE AIPMT 2009]
(a) 2.0 × 10−20 J
(c) 2.0 × 10−19 J
Ans. (a)
(b) 2.2 × 10−19 J
(d) 4.0 × 10−20 J
Kinetic energy (KE) of molecule
= energy absorbed by molecule
− bond energy per molecule
= (44
. × 10 −19 ) − (40
. × 10 −19 ) J
= 0.4 × 10 −19 J
KE per atom
0.4 × 10 –19
J
=
2
= 2.0 × 10 −20 J
04 The energy of second Bohr orbit of
the hydrogen atom is
–328 kJ mol–1 , hence the energy of
fourth Bohr orbit would be
[CBSE AIPMT 2005]
(a) – 41 kJ mol–1
(b) –1312 kJ mol–1
(c) –164 kJ mol–1
(d) – 82 kJ mol–1
Ans. (d)
[CBSE AIPMT 2004]
(a) 1.54 × 1015 s–1
(c) 3.08 × 1015 s–1
(b) 1.03 × 1015 s–1
(d) 2.00 × 1015 s–1
Ans. (c)
Ionisation energy of H
= 2.18 × 10 − 18 J atom–1
∴ E 1 ( Energy of Ist orbit of H-atom)
= − 2.18 × 10 −18 J atom–1
∴
En =
− 2 .18 × 10 −18
J atom–1
n2
Z = 1 for H-atom
∆E = E 4 − E 1
− 2 .18 × 10 −18 − 2 .18 × 10 −18
=
−
42
12
1
1
= − 2 .18 × 10 −18 ×  2 − 2 
 4
1 
15
∆E = − 2 .18 × 10 −18 × −
16
∴
= + 2 .0437 × 10 −18 J atom–1
∆E
ν=
h
=
The energy of second Bohr orbit of
hydrogen atom (E2 ) is – 328 kJ mol –1
1312
E n = – 2 kJ mol –1
n
1312
E2 = − 2 kJ mol –1
∴
2
2 .0437 × 10 −18 J atom–1
6 .625 × 10 −34 J s
= 3 .084 × 10 15 s− 1 atom–1
06 In hydrogen atom, energy of first
excited state is –3.4 eV. Then, KE
of same orbit of hydrogen atom is
[CBSE AIPMT 2002]
If n = 4
∴
05 The frequency of radiation emitted
when the electron falls from n = 4
to n = 1 in a hydrogen atom will be
(Given ionisation energy of
H = 2.18 ×10–18 J atom–1 and
h = 6.625 × 10–34 Js)
1312
kJ mol –1
42
= − 82 kJ mol –1
E4 = –
(a) + 3.4 eV
(c) – 13.6 eV
Ans. (a)
(b) + 6.8 eV
(d) + 13.6 eV
10
NEET Chapterwise Topicwise Chemistry
Q Total energy ( E n ) =KE + PE
1
In first excited state = mv2 +
2
 Ze2 

–
 r 
1 Ze2 Ze2
–
r
2 r
Energy of first excited state is 3.4 eV
1 Ze2
– 3 .4 eV = –
2 r
1 Ze2
= + 3 .4 eV
KE =
∴
2 r
=+
07 Who modified Bohr’s theory by
introducing elliptical orbits for
electron path? [CBSE AIPMT 1999]
(a) Hund
(c) Rutherford
Ans. (d)
(b) Thomson
(d) Sommerfeld
08 Bohr radius for the hydrogen atom
(n = 1) is approximately 0.530Å. The
radius for the first excited state
[CBSE AIPMT 1998]
(n = 2) is (in Å)
(c) 4.77
(d) 2.12
r ∝ n /Z
where, n = number of orbit
Z = atomic number
Q
r1 ∝ n21
r2 ∝ n22 (Z = 1 for H-atom)
r1 n21
So,
=
r2 n22
2
2
0 . 530 1
= 2
r2
2
∴
r2 = 0 .530 × 4 = 2 .120 Å
09 The radius of hydrogen atom in the
ground state is 0.53 Å. The radius
of Li 2+ ion (at. no. = 3) in a similar
state is
[CBSE AIPMT 1995]
(a) 0.17 Å
(c) 0.265 Å
Ans. (a)
(a)
Z2 e2
r
(b) −
Ze2
Ze2
(c)
r
r
(d)
mv 2
r
Ans. (b)
Potential energy = work done
r
Ze2 dr
Ze2
=∫ −
=−
2
∞
r
r
11 If ionisation potential for hydrogen
atom is 13.6 eV, then ionisation
potential for He + will be
[CBSE AIPMT 1993]
Sommerfeld modified Bohr’s theory.
According to him electrons move in
elliptical orbits in addition to circular
orbits.
(a) 0.13 (b) 1.06
Ans. (d)
10 When an electron of charge e and
mass m moves with a velocity v
about the nuclear charge Ze in
circular orbit of radius r, the
potential energy of the electrons is
given by
[CBSE AIPMT 1994]
(b) 0.53 Å
(d) 1.06 Å
We know that rn (H-like)
r (H- atom) × n2
= n
Z
For ground state, n = 1
0.53 Å × (1)2
rn (Li2 + ) =
∴
3
(Li,Z = 3) = 0.17 Å
(a) 54.4 eV
(c) 13.6 eV
Ans. (a)
(b) 6.8 eV
(d) 24.5 eV
For hydrogen atom Z = 1
∴ Ionisation energy, EH =
2 π2 me 4
n2h2
For He+ ion, (He+ = 1s 1 )
so, (He+ = H) ionisation energy,
2 π2 me 4 Z2
EHe + =
n2h2
Eq (i)/Eq (ii), we get
EHe + = EH × Z2 = 136
. × 4 = 54.4 eV
…(i)
…(ii)
12 The energy of an electron in the nth
Bohr orbit of hydrogen atom is
(a) Energy of the electrons in the orbits
are quantised
(b) The electron in the orbit nearest the
nucleus has the lowest energy
(c) Electrons revolve in different orbits
around the nucleus
(d) The position and velocity of
electrons in the orbit cannot be
determined simultaneously
Ans. (d)
The main postulates of Bohr model of
atom are
(i) The electrons in an atom revolve
around the nucleus only in certain
selected circular paths, called orbits.
(ii) The energy is emitted or absorbed
only when the electrons jump from
one energy level to another.
(iii) Only those orbits are permitted in
which the angular momentum of the
electron is a whole number multiple
h
of
(where, h is Planck’s constant)
2π
that's why only certain fixed orbits
are allowed, i.e. the momentum of an
electron is quantised.
14 If r is the radius of the first orbit,
the radius of nth orbit of H-atom is
given by
[CBSE AIPMT 1988]
(a) rn2
r
(c)
n
Ans. (a)
rn =
13.6
(b) − 3 eV
n
13.6
(d) −
eV
n
Energy of an electron in an orbit,
1311.8 Z2
2 π2 me 4 Z2
=
−
kJ mol –1
En = −
n2h2
n2
21. 8 × 10 −12 Z2
erg atom–1
En = −
n2
21. 8 × 10 −19 Z2
J atom–1
En = −
n2
−13.6
13.6 Z2
En = −
eV atom−1 = 2 eV atom
n
n2
(Q Z = 1 atomic number
for hydrogen atom)
13 Which of the following statements
do not form a part of Bohr’s model
of hydrogen atom?
[CBSE AIPMT 1989]
(d) r2n2
Radius of an orbit,
[CBSE AIPMT 1992]
13.6
(a) − 4 eV
n
13.6
(c) − 2 eV
n
Ans. (c)
(b) rn
=
n2h2
4π2 me2 Z
0.529n2
Å
Z
For H-atom, Z = 1
If
r1 = r
(according to question r1 = r)
∴
rn =
r × n2
= rn2
1
15 The spectrum of helium is
expected to be similar to that of
[CBSE AIPMT 1988]
(a) H
(c) Li+
Ans. (c)
(b) Na
(d) He +
The spectrum of an atom depends on
the number of electrons present in it.
Here, helium has two electrons, so the
spectrum of Li+ (Z = 3) is similar to that of
helium because both He andLi+ have
two electrons.
11
Atomic Structure
TOPIC 3
18 Which one is the wrong statement?
[NEET 2017]
Wave Particle and
Quality of Matter
16 A particular station of All India
Radio, New Delhi, broadcasts on a
frequency of 1,368 kHz (kilohertz).
The wavelength of the
electromagnetic radiation emitted
by the transmitter is
[Speed of light, c = 3.0 × 10 3 ms −1 ]
[NEET 2021]
(a) 219.3 m
(c) 2192 m
Ans. (a)
(b) 219.2 m
(d) 21.92 cm
Ans. (d)
Frequency of electromagnetic radiation
ν = 1368 kHz
= 1368 × 10 3 s−1
Speed of light, c = 3 × 10 8 ms −1
Wavelength of electromagnetic
c
radiation, λ =
ν
3 × 10 8 ms−1
= 219.3 m
λ=
1368 × 10 3 s−1
17 In hydrogen atom, the de-Broglie
wavelength of an electron in the
second Bohr orbit is
[Given that, Bohr radius,
a 0 = 52.9 pm] [NEET (Odisha) 2019]
(a) 211.6 pm
(c) 52.9 π pm
Ans. (b)
According to Bohr,
nh
mvr =
2π
nh
= nλ
2πr =
mv
(a) de-Broglie’s wavelength is given by
h
λ=
, where
mv
m = mass of the particle,
v = group velocity of the particle
(b) The uncertainty principle is
∆E × ∆t ≥ h /4π
(c) Half-filled and fully filled orbitals
have greater stability due to greater
exchange energy, greater symmetry
and more balanced arrangement
(d) The energy of2s-orbital is less than
the energy of2p-orbital in case of
hydrogen like atoms
(b) 211.6 π pm
(d) 105.8 pm
h 

…(i) Qλ =

mv 
where, r = radius,
λ = wavelength
n = number of orbit
a n2
Also,
…(ii)
r= 0
Z
where,a 0 = Bohr radius = 52.9 pm
Z = atomic number
On substituting the value of ‘r ’ from Eq.
(ii) to Eq. (i), we get
2 πn2a 0
nλ =
Z
2 πna 0
λ=
Z
[Q n = 2, Z = 1]
λ = 2 π × 2 × 52.9
= 211.6π pm
(a) According to de-Broglie’s equation,
h
Wavelength (λ) =
mv
where,h = Planck’s constant.
Thus, statement (a) is correct.
(b) According to Heisenberg uncertainty
principle, the uncertainties of
position (∆x) and momentum
(p = m∆v) are related as
h
∆x . ∆p ≥
4π
h
or,
∆x . m∆v ≥
4π
h
∆x . m. ∆a . ∆t ≥
4π
 ∆v = ∆a, a = acceleration

 ∆t
h
[QF = m⋅ ∆a]
or,
∆x ⋅ F ⋅ ∆t ≥
4π
h
or,
∆E ⋅ ∆t ≥
4π
[Q∆E = F ⋅ ∆x, E = energy]
Thus, statement (b) is correct.
(c) The half and fully filled orbitals have
greater stability due to greater
exchange energy, greater symmetry
and more balanced arrangement.
Thus statement (c) is correct.
(d) For a single electronic species like H,
energy depends on value of n and
does not depend onl. Hence energy
of 2s-orbital. and 2p-orbital is equal in
case of hydrogen like species.
Therefore, statement (d) is incorrect.
19 How many electrons can fit in the
orbital for which n = 3 and l = 1?
[NEET (Phase II) 2016]
(a) 2
(c) 10
Ans. (a)
(b) 6
(d) 14
According to Hund’s rule of maximum
multiplicity, An orbital can accommodate
a maximum number of 2 electrons of
exactly opposite spin. Hence, option (a)
is correct.
Caution Remember, maximum number
of electrons in an orbital do not depend
upon the quantum numbers as given in
the question.
20 The number of d-electrons in
Fe 2+ (Z=26) is not equal to the
number of electrons in which one
of the following? [CBSE AIPMT 2015]
(a) s-electrons in Mg (Z = 12)
(b) p-electrons in CI (Z = 17)
(c) d-electrons in Fe (Z = 26)
(d) p-electrons in Ne (Z = 10)
Ans. (b)
Electronic configuration of Fe2 + is
[Ar]3d 6 4s 0 .
∴Number of electrons = 6
Mg – 1s 2 2s 2 2p6 3s 2 (6s electrons)
It matches with the6d electrons of Fe2 +
Cl – 1s 2 2s 2 2p6 3s 2 3p5 (11p electrons)
It does not match with the6d electrons
of Fe2 + .
Fe – [Ar]3d 6 4s 2 (6d electrons)
It matches with the6d electrons of Fe2 + .
Ne – 1s 2 2s 2 2p6 (6p electrons)
It matches with the6d electrons of Fe2 + .
Hence, Cl has 11p electrons which does
not matches in number with6d electrons
of Fe2 + .
21 The angular momentum of
electrons in d orbital is equal to
[CBSE AIPMT 2015]
(a) 6 h (b) 2 h
Ans. (a)
(c) 2 3 h (d) 0 h
Angular momentum of electron in
d-orbital is
h
;for d-orbital, l = 2
= l (l + 1)
2π
h 

= 2(2 + 1
Q h =


2π 
h= 6h
22 Calculate the energy in joule
corresponding to light of
wavelength 45 nm (Planck’s
constant, h = 6.63 × 10 −34 Js; speed
of light, c = 3 × 10 8 ms −1 ).
[CBSE AIPMT 2014]
(a) 6.67 × 1015
(c) 4 .42 × 10−15
(b) 6 . 67 × 1011
(d) 4 .42 × 10− 18
12
NEET Chapterwise Topicwise Chemistry
Ans. (d)
The wavelength of light is related to its
hc
energy by the equation, E = . (E = hv)
λ
Given, λ = 45 nm = 45 × 10 −9 m
[Q1 nm = 10 −9 m]
Hence, E =
6.63 × 10 −34 Js × 3 × 10 8 m s−1
45 × 10 −9 m
−18
= 4.42 × 10 J
Hence, the energy corresponds to light
of wavelength 45 nm is 4.42 × 10 −18 J.
23 The value of Planck’s constant is
6.63 × 10 −34 Js. The speed of light is
3 × 10 17 nm s −1 . Which value is
closest to the wavelength in
nanometer of a quantum of light
with frequency of 6 × 10 15 s −1 ?
[NEET 2013]
(a) 10
(c) 50
Ans. (c)
(b) 25
(d) 75
[CBSE AIPMT 2008]
1
h
(a)
2m π
1 h
(c)
m π
Ans. (a)
h
2π
h
(d)
π
(b)
Wavelength, λ = ?
c
c
We know that, ν = or λ =
λ
ν
3 × 10 17
=
6 × 10 15
= 0.5 × 102 nm = 50 nm
24 The measurement of the electron
position is associated with an
uncertainty in momentum, which is
equal to 1 × 10 −18 g cm s −1 . The
uncertainty in electron velocity is
(mass of an electron is 9 × 10 −28 g)
[CBSE AIPMT 2008]
(a) 1 × 109 cm s−1
(b) 1 × 106 cm s−1
(c) 1 × 105 cm s−1
(d) 1 × 1011 cm s−1
Ans. (a)
Given, ∆p = 1 × 10 −18 g cm s–1 (uncertainty
in momentum)
Mass = 9 × 10 −28 g
∆p = m∆v
−18
= 9 × 10 −28 × ∆v
(uncertainty in velocity)
∆v = 1 × 10 9 cm s−1
27 The value of Planck’s constant is
6.63 × 10 −34 Js. The velocity of light
is 3 .0 × 10 8 ms −1 . Which value is
closest to the wavelength in
nanometers of a quantum of light
with frequency of 8 × 10 15 s −1 ?
[CBSE AIPMT 2003]
(a) 4 × 101
(c) 2 × 10−25
According to Heisenberg’s uncertainty
principle
h
∆x ⋅ ∆p =
4π
Given, ∆x = ∆p (∆x = uncertainty in
position)
h
(∆p)2 =
(∆p = m × ∆v)
4π
h
m2 ∆v2 =
m = mass
4π
1
h
h
⇒ ∆v =
∆v2 = 2
2m π
m 4π
(∆v = uncertainty in velocity)
Given, Planck’s constant,
h = 6.63 × 10 −34 Js
Speed of light, c = 3 × 10 17 nm s−1
Frequency of quantam light
ν = 6 × 10 15 s−1
1 × 10
25 If uncertainty in position and
momentum are equal, then
uncertainty in velocity is
26 Given, the mass of electron is
9.11 × 10–31 kg, Planck’s constant is
6.626 × 10–34 Js, the uncertainty
involved in the measurement of
velocity within a distance of 0.1 Å is
[CBSE AIPMT 2006]
(a) 5.79 × 106 ms–1 (b) 5.79 × 107 ms–1
(c) 5.79 × 108 ms–1 (d) 5.79 × 105 ms–1
Ans. (a)
Frequency (ν) =
28 The energy of photon is given as :
∆e/atom= 3.03 × 10 −19 J atom–1 ,
then the wavelength (λ) of the
photon is
[CBSE AIPMT 2000]
(Given,h(Planck’s constant)
= 6.63 × 10 −34 Js, c (velocity of light)
= 3.00 × 10 8 ms −1 )
(a) 6.56 nm
(c) 656 nm
Ans. (c)
ms–1
= 5.785 × 10 ms
= 5.79 × 10 6 ms–1
–1
(b) 65.6 nm
(d) 0.656 nm
According to formula, E =
hc 
c
ν = 
λ 
λ
Energy E = hν
hc
λ
6.63 × 10 –34 × 3 .0 × 10 8
303
. × 10 – 19 =
Ans. (a)
6
c 3 × 10 8 m s–1
=
λ 8 × 1015 s–1
=0.375 × 10 –7 m
= 3.75 × 10 1 nm ≈ 4 × 10 1 nm
λ=
By Heisenberg’s uncertainty principle
h
h
or ∆x × ∆ (mvx ) ≥
∆x × ∆px ≥
4π
4π
h
∆x × ∆vx ≥
4πm
∆p = uncertainty in momentum
∆x = uncertainty in position
∆v = uncertainty in velocity
m= mass of particle
Given that,
∆x = 0.1 Å = 0.1 × 10 −10 m
m = 9.11 × 10 −31 kg
h = Planck’s constant = 6626
.
× 10 −34 Js
π = 3 .14
Thus,
6.626 × 10 −34
∆v × 0.1 × 10 −10 =
4 × 3.14 × 9.11 × 10 −31
6.626 × 10 −34
∆v =
4 × 3.14 × 9.11 × 10 −31 × 0.1 × 10 −10
(b) 3 × 107
(d) 5 × 10−18
3 .03 × 10 –19
= 6 .56 × 10 –7 m
= 6.56 ×10 –7 × 10 9 nm
= 6.56 × 102 nm
= 656 nm
29 The de-Broglie wavelength of a
particle with mass 1 g and velocity
100 m/s is
[CBSE AIPMT 1999]
(a) 6 . 63 × 10–33 m
(c) 6 . 63 × 10–35 m
(b) 6 . 63 × 10–34 m
(d) 6 . 65 × 10–36 m
Ans. (a)
p=
h
(de-Broglie equation)
λ
h
λ=
mv
h = 6625
.
× 10 −34
≈ 663
. × 10 −34 kg/s
. × 10 –34 kg m2 / s
663
λ=
10 –3 kg × 100 m/ s
= 663
. × 10 –33 m
(Qp = mv)
13
Atomic Structure
30 The uncertainty in momentum of
an electron is 1 × 10– 5 kg m/s. The
uncertainty in its position will be
(Given, h = 6.62 × 10–34 kg m 2 /s)
[CBSE AIPMT 1999]
(a) 1.05 × 10–28 m
(c) 5 . 27 × 10–30 m
Ans. (c)
(b) 1.05 × 10–26 m
(d) 5 . 25 × 10–28 m
According to Heisenberg’s uncertainty
principle
h
∆p × ∆ x ≥
4π
Uncertainty in momentum
∆p = 1 × 10 −5 kg m/s
6.62 × 10 –34
1 × 10 –5 × ∆ x =
22
4×
7
6.62 × 10 –34 × 7
∆x =
1 × 10 –5 × 4 × 22
(Given)
= 5 .265 × 10 –30 m
≈ 5 .27 × 10 –30 m
31 The position of both, an electron
and a helium atom is known within
1.0 mm. Further the momentum of
the electron is known within
5.0 × 10 −26 kg ms–1 . The minimum
uncertainty in the measurement of
the momentum of the helium atom
is
[CBSE AIPMT 1998]
(a) 50 kg ms–1
(b) 80 kg ms–1
(c) 80 × 10–26 kg ms–1
(d) 5.0 × 10–26 kg ms–1
Ans. (d)
By Heisenberg’s uncertainty principle
h
∆ x × ∆p ≥
4π
when the position of electron and helium
atom is same and momentum of
electron is known within a range,
therefore the momentum of helium
atom is also equal to the momentum of
electron, i.e.
5 × 10 −26 kg m s–1
32 The momentum of a particle having
a de-Broglie wavelength of 10 −17 m
is
[CBSE AIPMT 1996]
(Given, h = 6.625 × 10 −34 m)
(a) 3.3125 × 10−7 kg m s–1
(b) 26.5 × 10−7 kg m s–1
(c) 6.625 × 10–17 kg m s–1
(d) 13.25 × 10−17 kg m s−1
Ans. (c)
According to de-Broglie relation,
h
h
λ=
=
mv p
where,
λ = wavelength
h = Planck’s constant
p = momentum
Here,
h = 6.625 × 10 −34 J s
λ = 10 −17 m
h 6.625 × 10 –34
∴
p= =
λ
10 −17
= 6.625 × 10 −34 × 10 17
= 6.625 × 10 −17 kg m s–1
33 Uncertainty in position of an
electron
(mass of an
electron is = 9.1 × 10 −28 g) moving
with a velocity of 3 × 10 4 cm/ s
accurate upto 0.001% will be (use
h
in uncertainty expression where
4π
h = 6.626 × 10 −27 erg s)
[CBSE AIPMT 1995]
(a) 1.93 cm
(c) 5.76 cm
Ans. (a)
(b) 3.84 cm
(d) 7.68 cm
According to Heisenberg’s uncertainty
principle
h
∆ x × ∆v =
4π m
Here, ∆ x = uncertainty in position
∆v = uncertainty in velocity
h = Planck’s constant (6.626 × 10 −27 Js)
m = mass of electron (9.1 × 10 −28 kg)
Here, ∆v = 0.001% of 3 × 10 4
0.001
=
× 3 × 10 4 =0.3 cm / s
100
h
∆x =
∴
4πm ∆v
=
6.626 × 10 −27
4 × 3.14 × 9.1 × 10 −28 × 0.3
= 1.93 cm
34 In the photoelectron emission, the
energy of the emitted electron is
[CBSE AIPMT 1994]
(a)
(b)
(c)
(d)
greater than the incident photon
same as that of the incident photon
smaller than the incident photon
proportional to the intensity of
incident photon
Ans. (c)
In the photoelectric effect, the energy of
the emitted electron is smaller than that
of the incident photon because some
energy of photon is used to eject the
electron and remaining energy is used to
increase the kinetic energy of ejected
electron.
35 The electron was shown
experimentally to have wave
properties by
[CBSE AIPMT 1994]
(a) de-Broglie
(b) N Bohr
(c) Davisson and Germer
&&
(d) Schrodinger
Ans. (c)
The wave nature of an electron is proved
by Davisson and Germer experiment. In
this experiment the scattering pattern of
an electron is similar to that of X-rays.
TOPIC 4
Quantum Mechanical,
Model and Electronic
Configuration
36 The number of angular nodes and
radial nodes in 3s orbital are
[NEET (Oct.) 2020]
(a) 0 and 2, respectively
(b) 1 and 0, respectively
(c) 3 and 0, respectively
(d) 0 and 1, respectively
Ans. (a)
For 3 s-orbital, n = 3, l = 0
Number of radial nodes
= (n − l − 1) = 3 − 0 − 1 = 2
Number of angular nodes = l = 0.
Hence, option (a) is correct.
37 4d, 5p, 5f and 6p-orbitals are
arranged in the order of decreasing
energy. The correct option is
[NEET (National) 2019]
(a) 6 p > 5 f > 5 p > 4 d
(b) 5 p > 5 f > 4 d > 5 p
(c) 5 f > 6 p > 4 d > 5 p
(d) 5 f > 6 p > 5 p > 4 d
Ans. (d)
The order of energy of orbitals can be
calculated from (n + l ) rule. The lower the
value of (n + l ) for an orbital, lower is its
energy. If two orbitals have same (n + l )
value, the orbital with lower value of n
has the lower energy.
(i) 6p = 6 + 1 = 7
(ii) 5f = 5 + 3 = 8
(iii) 4d = 4 + 2 = 6
(iv) 5p = 5 + 1 = 6
∴ The order of decreasing energy will be
5f > 6p > 5p > 4d.
14
NEET Chapterwise Topicwise Chemistry
38 Orbital having 3 angular nodes and
3 total nodes is [NEET (Odisha) 2019]
(a) 5 p
(c) 4 f
Ans. (c)
(b) 3 d
(d) 6 d
[CBSE AIPMT 2015]
Angular node (l ) = 3
Total node = radial node + angular node
3 = n− 1
⇒
n=4
∴Orbital having 3 angular nodes and 3
total nodes is = nl = 4f [Ql = 3 for forbital]
39 Which one is a wrong statement?
[NEET 2018]
(a) The electronic configuration of
N-atom is
2p1x 2p1y 2p1z
2s2
(a) 3s 4s 3p 3d
(c) 3s 3p 3d 4s
Ans. (c)
(b) 4s 3s 3p 3d
(d) 3s 3p 4s 3d
According to Aufbau rule
3s < 3p < 3d < 4s
3 = (n − l − 1) + l
1s2
41 Which is the correct order of
increasing energy of the listed
orbitals in the atom of titanium?
(b) An orbital is designated by three
quantum numbers while an electron
in an atom is designated by four
quantum numbers
(c) Total orbital angular momentum of
electron in ‘s’ orbital is equal to zero
(d) The value of m for d 2 is zero
z
Ans. (a)
[NEET 2013]
(b) 6
(d) 2
The orbital of the electron having n = 3,
l = 1 and m = − 1 is 3pz (as nl m ) and an
orbital can have a maximum number of
two electrons with opposite spins.
∴ 3pz orbital contains only two electrons
or only 2 electrons are associated with
n = 3 , l = 1, m = − 1.
43 Maximum number of electrons in a
subshell with l = 3 and n = 4 is
[CBSE AIPMT 2012]
According to Hund’s rule “the pairing of
electrons in the orbitals of a particular
subshell does not takes place until all the
orbitals of a subshell are singly occupied.
Moreover, the singly orbitals must have
the electrons with parallel spin. i.e.
1s2
2s2
1s2
2s2
2p 1x 2p1y 2p1z
or
2p 1x 2p1y 2p1z
∴ Option (a) is the incorrect option.
40 Two electrons occupying the same
orbital are distinguished by
[NEET (Phase I) 2016]
(a) Magnetic quantum number
(b) Azimuthal quantum number
(c) Spin quantum number
(d) Principal quantum number
Ans. (c)
Two electrons occupying the same
orbital has equal spin but the directions
of their spin are opposite. Hence, spin
quantum number, s, (represented +1 /2
and − 1 /2) distinguishes them.
(a) 14
(b) 16
Ans. (a)
(c) 10
(d) 12
n represents the main energy level and
l represents the subshell.
If n = 4 and l = 3, the subshell is 4f.
In f-subshell, there are 7 orbitals and
each orbital can accommodate a
maximum number of two electrons, so
maximum number of electrons in
4f subshell = 7 × 2 = 14 .
44 The correct set of four quantum
numbers for the valence electron
of rubidium atom (at. no. = 37) is
[CBSE AIPMT 2012]
(a) 5, 1, 1, +
1
2
1
(c) 5, 0, 0, +
2
Ans. (c)
37 Rb = 36
[CBSE AIPMT 2011]
(a) ns → (n − 1) d → (n − 2) f → np
(b) ns → (n − 2) f → np → (n − 1) d
(c) ns → np → (n − 1) d → (n − 2) f
(d) ns → (n − 2) f → (n − 1) d → np
Ans. (d)
6s → 4f → 5d → 6p for n = 6
42 What is the maximum numbers of
electrons that can be associated
with the following set of quantum
numbers?
n = 3, l = 1 and m = − 1
(a) 10
(c) 4
Ans. (d)
45 If n = 6, the correct sequence for
filling of electrons will be
1
2
1
(d) 5, 1, 0, +
2
(b) 6, 0, 0, +
[Kr] 5 s 1
Its valence electron is 5 s 1.
n= 5
(For s-orbital)
l =0
m = 0 (As m = − l to + l)
1
s =+
2
46 Which of the following is not
permissible arrangement of
electrons in an atom?
[CBSE AIPMT 2009]
(a) n = 4,l = 0, m = 0, s = −1/2
(b) n = 5,l = 3, m = 0, s = + 1/2
(c) n = 3,l = 2, m = − 3, s = −1/2
(d) n = 3,l = 2, m = −2, s = −1/2
Ans. (c)
If n = 3,
l = 0 to (3 − 1) = 0, 1, 2
m = − l to + l = − 2, − 1, 0, + 1, + 2
1
s =±
2
Therefore, option (c) is not a permissible
set of quantum numbers.
47 Maximum number of electrons in a
subshell of an atom is determined
by the following [CBSE AIPMT 2009]
(a) 4 l + 2
(c) 4 l − 2
Ans. (a)
(b) 2 l + 1
(d) 2 n2
Total number of subshells = (2l + 1)
∴ Maximum number of electrons in the
subshell
= 2 (2l + 1) = 4l + 2
48 Consider the following sets of
quantum numbers.
n
l
m
s
(i)
3
0
0
+1 / 2
(ii)
2
2
1
+1 / 2
(iii)
4
3
–2
−1 / 2
(iv)
1
0
–1
−1 / 2
(v)
3
2
3
+1 / 2
Which of the following sets of
quantum number is not possible?
[CBSE AIPMT 2007]
(a) (ii), (iii) and (iv)
(c) (ii), (iv) and (v)
(b) (i), (ii), (iii) and (iv)
(d) (i) and (iii)
15
Atomic Structure
Ans. (d)
The value of l varies from 0 to (n − 1) and
the value of m varies from − l to + l
through zero.
1
The value of ‘ s ’ ± which signifies the
2
spin of electron. The correct sets of
quantum number are following
n l
m
s
1
(ii) 2 1 1
−
2
1
(iv) 1 0 0
−
2
1
(v) 3
2 2
+
2
49 The orientation of an atomic orbital
is governed by [CBSE AIPMT 2006]
(a) azimuthal quantum number
(b) spin quantum number
(c) magnetic quantum number
(d) principal quantum number
Ans. (c)
50 The following quantum numbers
are possible for how many orbital(s)
n = 3, l = 2 and m = + 2?
[CBSE AIPMT 2001]
(b) 2
(d) 4
+1
0
(a) [Xe] 4f 8 , 5 d 9 , 6 s 2
(b) [Xe] 4 f 7 , 5 d 1, 6 s 2
(c) [Xe] 4 f 6 , 5 d2 , 6 s 2
(d) [Xe] 4 f 3, 5 d 5 , 6 s 2
–1
–2
51 Which of the following
configuration is correct for iron?
[CBSE AIPMT 1999]
(a) 1s 2 , 2 s 2 2 p 6 , 3s 2 3p 6 3d 5
(b) 1s 2 , 2 s 2 2 p 6 , 3s 2 3p 6 , 4 s 2 , 3d 5
(c) 1s 2 , 2 s 2 2 p 6 , 3s 2 3p 6 , 4 s 2 , 3d 7
(d) 1s 2 , 2 s 2 2 p 6 , 3s 2 3p 6 3d 6 , 4 s 2
Ans. (d)
Firstly the electrons are filled in
increasing order of energy and then
rearrange the subshells in increasing
order as
2
2
6
2
6
6
2
26 Fe = 1s , 2s 2p , 3s 3p 3d , 4s
Ans. (b)
For n = 3,l = 2 the subshell is 3d (n + l = 5)
n = 4, l = 2 the subshell is 4d (n + l = 6)
n = 4, l = 1 the subshell is 4p (n + l = 5)
n = 5, l = 0, the subshell is 5s (n + l = 5)
According to (n + l ) rule greater the (n + l )
value, greater the energy that is 6.
Ans. (b)
Gd 64 = 1s 2 , 2s 2 2p6 , 3s 2 3p6 3d 10,
4s 2 4p6 4d 10 4f 7, 5s 2 5p6 5d 1, 6s 2
= [Xe] 4f 7 , 5d 1, 6s 2
53 The orbitals are called degenerate
when
[CBSE AIPMT 1996]
(a) they have the same wave functions
(b) they have the same wave functions
but different energies
(c) they have different wave functions
but same energy
(d) they have the same energy
The orbitals having the same energy
energy but different in orientation, are
called degenerate orbitals. e.g.3d-orbital,
l = 2, m = −2, –1, 0, +1, +2, i.e. there are five
different orientations represented byd xy ,
d yz ,d zx ,d 2 2 and d 2 .
x
−y
z
54 If an electron has spin quantum
1
number + and magnetic quantum
2
number –1, it cannot be present in
[CBSE AIPMT 1994]
n = 3,l = 2, m = + 2, s = ± 1 /2
These values of quantum numbers are
possible for only one of the five
3d-orbitals as +2 value of m is possible
only for one orbital.
m = +2
[CBSE AIPMT 1997]
Ans. (d)
The orientation of an atomic orbital is
governed by magnetic quantum number.
(a) 1
(c) 3
Ans. (a)
52 The electronic configuration of
gadolinium (at. no. = 64) is
(a) d-orbital
(c) p-orbital
Ans. (d)
(b) f-orbital
(d) s-orbital
Spatial orientation of the orbital with
respect to standard set of cordinate
axis. Magnetic quantum number –1 is
possible only when the azimuthal
quantum number have valuel = 1, which
is possible for p, d and f-subshells but
not for s-subshell because the value of l
for s-subshell is zero.
55 For which one of the following sets
of four quantum numbers, an
electron will have the highest
energy?
[CBSE AIPMT 1994]
n
l
(a) 3
2
(b) 4
2
(c) 4
1
(d) 5
0
m
s
1
1
2
1
–1
2
1
0 −
2
1
0 −
2
56 Electronic configuration of calcium
atom can be written as
[CBSE AIPMT 1992]
(a) [Ne] 4p2
(b) [Ar] 4s 2
(c) [Ne] 4s 2
(d) [Kr] 4p2
Ans. (b)
To write the electronic configuration of
an atom, it is better if we remember the
atomic number of noble gases and the
orbitals follow the noble gas. The atomic
number of Ca is 20 and its nearest noble
gas is argon (Ar = 18).
Hence, the electronic configuration of
Ca = [Ar] 4s 2 .
57 The electronic configuration of
Cu (at.no. = 29) is [CBSE AIPMT 1991]
(a) 1s 2 , 2 s 2 2 p 6 , 3 s 2 3p 6 , 4 s 2 , 3d 9
(b) 1s 2 , 2 s 2 2 p 6 , 3 s 2 3p 6 3d 10, 4 s 1
(c) 1s 2 , 2 s 2 2 p 6 , 3 s 2 3p 6 , 4 s 2 4 p 6 ,
5 s2 5 p 1
2
2
6
2
6
2
6
(d) 1s , 2 s 2 p , 3 s 3p , 4 s 4 p , 3d 3
Ans. (b)
The electronic configuration of Cu (29) is
an exceptional case due to exchange of
energy and symmetrical distribution of
electrons in orbital to acquire more
stability.
Cu(29)=
½¾ ½¾ ½¾ ½¾ ½¾ ½¾
1s
2s
2p
3s
½¾ ½¾ ½¾ ½¾ ½¾ ½¾ ½¾ ½¾ ½
3p
3d
4s
= 1s 2 , 2 s 2 2p6 , 3 s 2 3p6 3d 10 , 4s 1
58 The order of filling of electrons in
the orbitals of an atom will be
[CBSE AIPMT 1991]
(a) 3d, 4 s , 4 p, 4 d, 5 s
(b) 4 s , 3d, 4 p, 5 s , 4 d
(c) 5 s , 4 p, 3d, 4 d, 5 s
(d) 3d, 4 p, 4 s , 4 d, 5 s
Ans. (b)
The sequence of energy level can be
remembered by the systematic diagram
as shown below
16
NEET Chapterwise Topicwise Chemistry
(a) Hund’s rule
(b) Aufbau principle
(c) Uncertainty principle
(d) Pauli’s exclusion principle
Ans. (d)
1s
2s
2p
3s
3p
3d
4s
4p
4d
4f
5s
5p
5d
5f
6s
6p
6d
7s
7p
Hence, the correct order is
1s , 2 s , 2p, 3s , 3p, 4s , 3d, 4p, 5s , 4d, 5p, 4f,
5d, 6p, 7s……
59 For azimuthal quantum number
l = 3,the maximum number of
electrons will be [CBSE AIPMT 1991]
(a) 2
(b) 6
Ans. (d)
(c) 0
(d) 14
When azimuthal quantum number is 3
m = (2l + 1)
l =3
m = (2 × 3 + 1)
= 7 orbitals
then total values of m = (2 × 3 + 1) = 7
orbitals. We know that, one orbital
contains two electrons. Hence, total
number of electrons = 7 × 2 = 14.
Alternative
Total number of electrons = 4l + 2
= 4 × 3 + 2 = 12 + 2 = 14electrons
60 In a given atom no two electrons
can have the same values of all the
four quantum numbers. This is
called
[CBSE AIPMT 1991]
According, to Pauli’s exclusion principle
“no two electrons in an atom can have
the same values of all the four quantum
numbers.”
In 1s 2
1
for I electron n = 1, l = 0, m = 0, s = +
2
for II electron n = 1,
1
l = 0, m = 0, s = −
2
It means if the values of n, l, and m are
same, then the value of spin quantum
number must be different, i.e. +1/2 and
−1/2.
61 The total number of electrons that
can be accommodated in all the
orbitals having principal quantum
number 2 and azimuthal quantum
number 1 are
[CBSE AIPMT 1990]
(a) 2
(c) 6
Ans. (c)
(b) 4
(d) 8
When n = 2 and l = 1, then subshell is 2p.
The number of orbitals in p-subshell
= (2l + 1) = (2 × 1 + 1)
=3
Total (maximum) number of electrons
= 2 × number of orbitals
=2×3=6
(as each orbital contains 2 electrons)
62 The maximum number of electrons
in a subshell is given by the
expression
[CBSE AIPMT 1989]
(a) 4l − 2
(c) 2 l + 2
Ans. (b)
(b) 4l + 2
(d) 2 n2
The number of orbitals in a subshell
= (2l + 1)
where,l = azimuthal quantum number
Since, each orbital contains maximum
two electrons, the number of electrons
in any subshell
= 2 × number of orbitals
= 2 (2l + 1)
= 4l + 2
63 Number of unpaired electrons in
[CBSE AIPMT 1989]
N 2+ is/are
(a) 2
(c) 1
Ans. (c)
(b) 0
(d) 3
The electronic configuration of
7N
=
½¾ ½¾
½½ ½
½¾ ½¾
½
1s
2+
∴ 7N
=
2s
2p
Hence, the number of unpaired
electron in N2+ is 1.
64 The number of spherical nodes in
3p-orbital is/are [CBSE AIPMT 1988]
(a) one
(b) three
(c) two
(d) None of the above
Ans. (a)
The number of spherical nodes in any
orbital ( = n − l − 1)
For 3p-orbital, n = 3 and l = 1
∴ Number of spherical nodes = n − l − 1
=3− 1− 1
= 3 − 2 = 1 node
3
Chemical Bonding
and Molecular Structure
TOPIC 1
Type of Bonds, Bond
Parameter and Resonance
01 The correct sequence of bond
enthalpy of ‘CX’ bond is
[NEET 2021]
(a) CH3 F < CH3  Cl < CH3 Br< CH3 I
(b) CH3 F> CH3  Cl > CH3 Br > CH3 I
(c) CH3 F < CH3  Cl > CH3 Br >
CH3 I
(d) CH3  Cl > CH3 F > CH3 Br >
CH3 I
Ans. (b)
(b) CH 2O Hybridisation =
Shape = Trigonal planar.
H
02 Which of the following molecules is
non-polar in nature?
[NEET 2021]
(a) POCl3
(c) SbCl5
(b) CH2O
(d) NO2
Ans. (c)
(a) POCl 3 Hybridisation =
1
× 8 = 4 (sp3)
2
P
Cl
Cl
Cl
Shape = Tetrahedral
Dipole moment, µ ≠ 0
POCl3 is polar in nature.
(d) Ammonia, beryllium difluoride,
water, 1,4-dichlorobenzene
Ans. (c)
In option (c), for all molecules,
[i, vi, iii, viii]µ =0.
C
H
The structure of all compounds are as
follows :
Dipole moment,µ ≠ 0
CH2O is polar in nature.
1
× 10 = 5 (sp3d)
2
Shape = Trigonal bipyramidal
(c) SbCl 5 Hybridisation =
(i) Boron trifluoride (BF3)
F
⇒ F
; µ=0
B
Cl
F
Cl
Sb
Cl
On moving down the group from F to I,
the size of atom increases. Order of the
size of halogen atoms is I > Br > Cl > F.
So, the bond length of C—X bond also
increases from F to I and hence, the
bond enthalpy decreases from F to I.
Correct order of bond length of C—X
bond is
H3C  I > H3C Br > H3C  Cl > H3C F.
Correct order of bond enthalpy is
H3C  F > H3C  Cl > CH3 Br > H3C I.
1
(6) = 3 (sp2 )
2
Cl
Cl
Dipole moment, µ = 0
SbCl 5 is non-polar in nature.
(d) NO2 Hybridisation
1
1
= × (4 + 2) = × 6 = 3 (sp2 ).
2
2
(ii) Hydrogen fluoride (HF)
⇒ H
F; µ ≠ 0
(iii) Carbon dioxide (CO2)
⇒ O
O ; µ=0
C
(iv) 1,3-dichloro benzene
(m-C6H4Cl2) ⇒ Cl
; µ≠0
Cl
N
Shape = Trigonal planar
Dipole moment,µ ≠ 0
NO2 is polar in nature.
03 Which of the following set of
molecules will have zero dipole
moment ?
[NEET (Sept.) 2020]
(a) Boron trifluoride, hydrogen fluoride,
carbon dioxide, 1,3-dichlorobenzene
(b) Nitrogen trifluoride, beryllium
difluoride, water,
1,3-dichlorobenzene
(c) Boron trifluoride, beryllium
difluoride, carbon dioxide, 1,
4-dichlorobenzene
(v) Nitrogen trifluoride (NF3)
⇒
;µ ≠ 0
N
F
F
F
(vi) Beryllium difluoride (BeF2)
⇒
FBeF ;µ=0
(vii) Water (H2O)
⇒
O
H
; µ≠0
H
18
NEET Chapterwise Topicwise Chemistry
Key concept The species that have same
number of electrons have same bond
order.
(viii) 1,4-dichloro benzene
(p=C6H4Cl2) ⇒ Cl
Species
; µ=0
Cl
(ix) Ammonia (NH3)
CO
6 + 8 = 14
NO
7 + 8 = 15
O2
8 + 8 = 16
+
7 + 8 − 1 = 14
−
CN
6 + 7 + 1 = 14
O −2
8 + 8 + 1 = 17
NO
⇒
; µ≠0
N
H
H
H
[Q µ = Dipole moment]
04 Which of the following is the
correct order of dipole moment?
[NEET (Odisha) 2019]
(a) NH3 < BF3 < NF3 < H2O
(b) BF3 < NF3 < NH3 < H2O
(c) BF3 < NH3 < NF3 < H2O
(d) H2O < NF3 < NH3 < BF3
Ans. (b)
BF3 has zero dipole moment as it is
symmetrical in nature.H2O has maximum
dipole moment as it possess two lone
pair of electrons. BetweenNH3 and NF3,
NH3 has greater dipole moment though
in NH3 and NF3, both N possesses one
lone pair of electrons.
This is beacuse in case ofNF3, the net
N—H bond dipole is in the same direction
as the direction of dipole of lone pair.
But in case ofNF3, the direction of net
dipole moment of three —N—F bonds is
opposite to that of the dipole moment of
the lone pair. Thus, the correct of dipole
moment is
Number of electrons
Thus, both CN− and CO have equal
number of electrons. So, their bond
order will be same.
06 Predict the correct order among
the following. [NEET 2016, Phase I]
(a) lone pair-lone pair > bond pair-bond
pair > lone pair-bond pair
(b) bond pair-bond pair > lone pair-bond
pair > lone pair-lone pair
(c) lone pair-bond pair > bond pair-bond
pair > lone pair-lone pair
(d) lone pair-lone pair > lone pair-bond
pair > bond pair-bond pair
Ans. (d)
According to the postulate of VSEPR
theory, a lone pair occupies more space
than a bonding pair, since it lies closer to
the central atom. This means that the
repulsion between the different electron
pairs follow the order.
lp − lp > lp − bp > bp − bp
07 Which of the following molecules
has the maximum dipole moment?
[CBSE AIPMT 2014]
H
N
>
O
H
(a) CO2
(c) NH 3
Ans. (d)
>
H
H
H
(NH3)
(H2O)
F
N
F
> F
B
F
F
(NF3)
F
(BF3)
05 Which one of the following pairs of
species have the same bond order?
[NEET 2017]
(a) CO, NO
(b) O2 , NO +
(c) CN− , CO
(d) N2 , O2−
Ans. (c)
(b) CH 4
(d) NF 3
CO2 and CH4 have zero dipole moment as
these are symmetrical in nature.
BetweenNH3 and NF3,NF3 has greater
dipole moment though inNH3 and NF3
both, N possesses one lone pair of
electrons.
O
C
net = 0
O
H
µ1
µres
µ4
C
µ1
µH
H
H 3
∴ µ = µ + µ + µ = –µ
res
1
2
3
4
and µnet = µres + µ4
∴ µnet = µ1 + µ2 + µ3 + µ4
= – µ4 + µ4 = 0
net = 0
µ3
H
µ4
µ1
N
µ2H
H
Resultant of
3N  H
bond lie in the same
direction as µ4
Hence, µnet = µres + µ4
µ4
µ3
F
N
F
µ1
µ2 F
Resultant of
3N  F
bond lie opposite to µ4
This is because in case ofNH3, the net
N H bond dipole is in the same direction
as the direction of dipole of lone pair but
in case of NF3, the direction of net bond
dipole of three N F bonds is opposite
than that of the dipole of the then lone
pair.
08 Which one of the following
molecules contain no π-bond?
[NEET 2013]
(a) CO2
(c) SO2
Ans. (b)
(b) H2O
(d) NO2
All the molecules have O-atom with lone
pairs, but inH2O the H-atom has no
vacant orbital for π-bonding. That’s why
it does not have any π-bond.
In all other given molecules, the central
atom because of the presence of vacant
orbitals is capable to form π-bonds.
09 Which of the following is least likely
to behave as Lewis base?
[CBSE AIPMT 2011]
(a) NH3
(c) OH−
Ans. (b)
(b) BF3
(d) H2O
BF3 is an electron deficient species, thus
behaves like a Lewis acid.
N − Na
Q Bond order = b
2
10 The electronegativity difference
between N and F is greater than
that between N and H yet the
dipole moment of NH3 (1.5 D) is
larger than that of NF3 (0.2 D). This
is because
[CBSE AIPMT 2006]
(a) in NH3 as well as inNF3 , the atomic
dipole and bond dipole are in the
same direction
(b) in NH3 , the atomic dipole and bond
dipole are in the same direction
whereas inNF3 these are in opposite
directions
(c) in NH3 as well asNF3 , the atomic
dipole and bond dipole are in
opposite directions
19
Chemical Bonding and Molecular Structure
(d) in NH3 the atomic dipole and bond
dipole are in the opposite directions
whereas inNF3 these are in the same
directions
F
F
Ans. (b)
B
Si
In NH+
4 bond angle is maximum (nearer
109°) due to its tetrahedral geometry.
F
Ans. (b)
F
F
µ=0
N
N
H
F
F
F
µ = 0.24 D
F
H
H
µ = 1.47 D
F is more electronegative than N,
therefore direction of bond is from N to
F whereas N is more electronegative
than H, the direction of the bond is from
H to N. Thus whereas resultant moment
of N-H bonds adds up to the bond
moment of lone pair, that of 3N-F bonds
partly cancel the resultant moment of
lone pair. Hence, the net dipole moment
of NF3 is less than that ofNH3.
11 In which of the following molecules
are all the bonds not equal?
[CBSE AIPMT 2006]
(a) ClF3
(c) AlF3
Ans. (a)
(b) BF3
(d) NF3
In ClF3 all bonds are not equal due to its
trigonal-bipyramidal (sp3d hybridisation)
geometry
F
F
Cl
F
F
F
F
F
F
F
B
F
F
F
and
Al
F
NF3 shows pyramidal geometry due to
sp3 hybridisation.
µ=0
(Permanent dipole moment)
Q SF 4 have µ > 0
∴ It has permanent dipole moment.
13 In BrF3 molecule, the lone pairs
occupy equatorial positions to
minimise
[CBSE AIPMT 2004]
(a) lone pair-bond pair repulsion
(b) bond pair-bond pair repulsion
(c) lone pair-lone pair repulsion and lone
pair-bond pair repulsion
(d) lone pair-lone pair repulsion
Ans. (d)
In BrF3 molecule, Br is sp 3d hybridised,
but its geometry is T-shaped due to
distortion of geometry from trigonal
bipyramidal to T-shaped by the
involvement of lone pair-lone pair
repulsion.
F
Br—
F
Here, lp – lp repulsion = 0
lp – bprepulsion = 4
bp – bprepulsion = 2
14 In NO −3 ion number of bond pair and
lone pair of electrons on nitrogen
atom are
[CBSE AIPMT 2002]
(a) 2, 2 (b) 3, 1
Ans. (d)
(c) 1, 3
(d) 4, 0
O
F
F
3bp + 1lp
12 Which of the following would have
a permanent dipole moment?
[CBSE AIPMT 2005]
(b) SiF4
(d) XeF4
–
O–—N
O
O
N
O
O
Nitrogen has four bond pair and zero
lone pair of electrons, due to the
presence of one coordination bond.
15 In which of the following, bond
angle is maximum?
[CBSE AIPMT 2001]
(a) NH3
(b) NH+4
(c) PCl3
16 In PO 3−
4 ion, the formal charge on
each
oxygen atom and P—O
bond order respectively are
[CBSE AIPMT 1998]
(a) – 0.75, 0.6
(c) – 0.75, 1.25
Ans. (c)
(b) – 0.75, 1.0
(d) – 3, 1.25
P—O bond order
Total Number of bonds in all possible
direction between two atoms
=
Total number of resonating structures
2+ 1+ 1+ 1 5
= = 1.25
=
4
4
∴ Bond order = 1.25
Resonating structures are
–
O
O
|
|
|
–O—
–
P—O–
O== P—O
|
|
O–
O–
O–
|
O–— P==O
|
O–
O–
–O— |
P—O–
||
O
Total charge onPO3–
4 ion is –3
Total charge
=
Total entity of O- atom
So, the average formal charge on each
3
O-atom is = − = − 0.75
4
F
In NO3− ion
N
F
F
µ = 0.632 D
F
BF3 and AlF3 show trigonal symmetric
structure due to sp2 hybridisation.
F
Xe
S
Trigonal bipyramidal geometry
(a) BF3
(c) SF4
Ans. (c)
F
F
µ=0
(d) SCl2
17 Which one is not paramagnetic
among the following? [at. no. of
Be = 4,Ne = 10, As = 33, Cl = 17]
[CBSE AIPMT 1998]
(a) Cl−
(c) Ne2+
Ans. (a)
(b) Be
(d) As+
Paramagnetic character is based upon
presence of unpaired electron.
–
2
2
6
2 2
2
2
17 Cl = 1s , 2 s 2p , 3 s 3px 3py 3pz
In Cl − no unpaired electron, so it is in
nature diamagnetic.
2
1
1
4 Be = 1s , 2 s 2px
2+
2
2 2
= 1s , 2 s 2px 2py1 2pz1
10 Ne
= 1s 2 , 2 s 2 2p6 , 3 s 2 3p6 3d 10 ,
4s 2 4px1 4py1 4pz0
While all others have unpaired electron,
so they are paramagnetic in nature.
+
33 As
20
NEET Chapterwise Topicwise Chemistry
18 The molecule which does not
exhibit dipole moment is
[CBSE AIPMT 1997]
(a) NH3
(c) H2O
Ans. (d)
(b) CHCl3
(d) CCl4
CCl 4 does not show dipole moment
because it has tetrahedral symmetrical
structure.
19 For two ionic solids CaO and KI,
identify the wrong statement among
the following.
[CBSE AIPMT 1997]
(a) Lattice energy of CaO is much larger
than that of KI
(b) KI is soluble in benzene
(c) KI has lower melting point
(d) CaO has higher melting point
Ans. (d)
Ans. (b)
H-bond is weakest bond because its
bond dissociation energy is very low as
compared to other given bonds (10 kJ
mol −1).
NaF has high lattice energy because Na +
is smallest in size and lattice energy
increases as the size of cation
decreases. (In the given question anion
is common in all compound)
23 Which of the following pairs will
form the most stable ionic bond?
[CBSE AIPMT 1994]
(a) Na and Cl
(c) Li and F
Ans. (b)
(b) Mg and F
(d) Na and F
The ionic bond between Mg and F is most
stable because in these the electrostatic
force of attraction is maximum. As Mg
has high electropositive character and F
has high electronegative character
among all other options that are given in
question.
Ans. (b)
KI is ionic compound, so it is not soluble
in non-polar solvent (i.e. dipole moment
(µ) for benzene = 0).
20 Which one of the following has the
highest dipole moment?
[CBSE AIPMT 1997]
(a) AsH3
(c) PH3
Ans. (d)
(b) SbH3
(d) NH3
In the given molecules nitrogen has
greater electronegativity. So, it has
greater dipole moment and correct order
of dipole moment is
NH3 > PH3 > AsH3 > SbH3
21 The BCl 3 is a planar molecule
whereas NCl 3 is pyramidal because
[CBSE AIPMT 1995]
(a) B — Cl bond is more polar than N — Cl
bond
(b) N — Cl bond is more covalent than
B — Cl bond
(c) nitrogen atom is smaller than boron
atom
(d) BCl 3 has no lone pair butNCl 3 has a
lone pair of electrons
Ans. (d)
BCl 3 have sp2 hybridisation and no lone
pair of electron on central atom butNCl 3
have sp3 hybridisation and also contains
one lone pair of electron on nitrogen, so
BCl 3 is planar.
22 The weakest among the following
types of bond is [CBSE AIPMT 1994]
(a) ionic
(c) metallic
(b) covalent
(d) H-bond
24 Which of the following statements
is not correct?
[CBSE AIPMT 1993, 1990]
(a) Double bond is shorter than a single
bond
(b) Sigma bond is weaker than a π-bond
(c) Double bond is stronger than a single
bond
(d) Covalent bond is stronger than
hydrogen bond
Ans. (b)
Sigma bond is always stronger than
π-bond because the extent of
overlapping is maximum in sigma bond
formation.
25 Which one of the following is the
correct order of interactions?
[CBSE AIPMT 1993]
(a) Covalent < hydrogen bonding < van
der Waals’ < dipole-dipole
(b) van der Waals’ < hydrogen bonding
< dipole-dipole < covalent
(c) van der Waals’ < dipole-dipole <
hydrogen bonding < covalent
(d) Dipole-dipole < van der Waals’ <
hydrogen bonding < covalent
Ans. (b)
The van der Waals’ forces are weakest
forces and covalent bond is strongest,
so the order of interactions is van der
Waals’ < H-bonding < dipole-dipole <
covalent.
26 Among the following which
compound will show the highest
lattice energy? [CBSE AIPMT 1993]
(a) KF
(c) CsF
(b) NaF
(d) RbF
27 Strongest hydrogen bonding is
shown by
[CBSE AIPMT 1992]
(a) H2O
(c) HF
Ans. (c)
(b) NH3
(d) H2 S
HF have strongest hydrogen bond
because the electronegativity of F-atom
is high and produce strong electrostatic
force of attraction.
28 Which one of the following
formulae does not correctly
represent the bonding capacities
of the atoms involved?
[CBSE AIPMT 1991]
+
H





(a) H — P —H 





H
F
F
(b)
O
O
(c) O ← N
O —H
O
(d) H —C ==C
O —H
Ans. (d)
1
2
In H— C==C
O
carbon number 2
O —H
have five valency which is not possible,
so it does not correctly represent the
bonding capacities of C atom.
29 Among LiCl, BeCl 2 , BCl 3 and CCl 4 ,
the covalent bond character
follows the order
[CBSE AIPMT 1990]
(a) LiCl <BeCl2 > BCl3 > CCl4
(b) LiCl <BeCl2 <BCl3 > CCl4
(c) LiCl <BeCl2 <BCl3 <CCl4
(d) LiCl > BeCl2 > BCl3 > CCl4
Ans. (c)
The electronegativity increases from left
to right in any period, so the
electronegativity follows the order
Li < Be < B < C
21
Chemical Bonding and Molecular Structure
Cl
and hence, the covalent character of
chlorides of these elements increase
from Li to C because size of cation left to
right decreases and according to Fajans’
rule covalent character increases.
30 H2O has a net dipole moment while
BeF2 has zero dipole moment
because
[CBSE AIPMT 1989]
(a) H2O molecule is linear whileBeF2 is
bent
(b) BeF2 molecule is linear whileH2O is
bent
(c) fluorine has more electronegativity
than oxygen
(d) beryllium has more electronegativity
than oxygen
Ans. (c)
Cl
P
Cl
Cl
Hybridisation of P → sp3d
Structure of PCl 5 → Trigonal bipyramidal
(B) SF6
S (Ground state) = 3 s 2 3 p4
S* (excited state)
3p
3s
F
F F F
3d
F F
F
F
F
S
Ans. (b)
F
H2O have bent structure in which the two
O—H bonds are oriented at an angle of
104.5°, so water have a net dipole
moment whereasBeF2 have linear
geometry, so the dipole moment of one
bond is cancelled by another bond, so it
have zero dipole moment.
TOPIC 2
Hybridisation and
VSEPR Theory
Hybridisation of a central atom can be
calculate by using the formula :
1
Hybridisation = [number of valence
2
electrons + Number of side atoms –
Positive charge + Negative charge]
Electronic configuration of B
= 1s 2 , 2s 2 , 2p1
Number of valence electrons in B = 3
electrons in last shell, n = 2
Number of side atoms in
BF3 = 3F-atoms.
1
1
So, hybridisation = (3 + 3) = × 6 = 3.
2
2
Hybridisation of B in BF3 is sp2 .
Number of electrons around central
atom, B inBF3 is equal to the number of
electrons in three sigma bonds (B—F) i.e.
= 3 B—F bonds × 2 electrons in one
σ-bond.
= 6 electrons
Cl
F
F
Hybridisation of S→ sp3d2
Structure of SF6 → Octahedral
(C) BrF5
Br (Ground state) = 3 s 2 3 p5
Br* (excited state)
3p
3s
Lone
pair
33 Match the coordination number
and type of hybridisation with
distribution of hybrid orbitals in
space based on valence bond theory.
3d
[NEET (Oct.) 2020]
F F F
F F
Coordination
number and
type of
hybridisation
F
31 Match List-I with List-II.
List-I
F
F
Br
List-II
F
F
A.
PCl5
I.
Square
pyramidal
B.
SF6
II.
Trigonal planar
C.
BrF5
III.
Octahedral
Hybridisation of Br → sp d
Structure of BrF5 → Square pyramidal
(D) BF3
B (Ground state) = 2 s 2 2 p1
D.
BF3
IV.
Trigonal
bipyramidal
B* (excited state)
2p
2s
A B C
(a) IV III I
D
II
A B
(b) II III
C D
IV I
(c) III
II
(d) IV III
II
IV
Ans. (a)
(A) PCl 5
P (Ground state) = 3 s 2 3 p3
P* (excited state)
3p
3s
3d
Cl Cl Cl
Cl
4, sp 3
I.
trigonal
bipyramidal
B.
4, dsp2
II.
octahedral
C.
5, sp 3d
III.
tetrahedral
D.
6, d2 sp 3
IV.
square planar
I
F
Select the correct option.
B
F
F F
F
F
Hybridisation of B→ sp2
Structure of BF3 → Trigonal planar
∴ Correct match is
(A)-(IV); (B)-(III); (C)-(I); (D)-(II)
32 BF 3 is planar and electron deficient
compound. Hybridisation and
number of electrons around the
central atom, respectively are
[NEET 2021]
Cl
A.
3 2
Choose the correct answer from
the options given below [NEET 2021]
I
Distribution of
hybrid orbitals
in space
(a) sp 3 and 4
(c) sp2 and 6
(b) sp 3 and 6
(d) sp2 and 8
A B C
(a) II III IV
(c) IV I II
D
I
III
A B
(b) III IV
(d) III I
Ans. (b)
(A) CN==4, sp3
or tetrahedral ⇒ (III)
(B) CN==4, dsp2
or square planar ⇒ (IV)
C D
I II
IV II
22
NEET Chapterwise Topicwise Chemistry
Column I
(C) CN==5, sp3d
or trigonal bipyramidal ⇒ (I)
(D) CN ==6, d2sp3
or octahedral ⇒ (II)
Hence, option (b) is correct.
34 Identify the wrongly matched pair.
[NEET (Oct.) 2020]
Molecule Shape of geometry of
molecule
(a)
PCl5
Trigonal planar
(b)
SF6
Octahedral
(c)
BeCl2
Linear
(d)
NH3
Trigonal pyramidal
C.
XeOF4
(iii)
Distorted
octahedral
D.
XeO 3
(iv)
Square
pyramidal
Codes
A B C D
(a) (ii) (iii) (iv) (i)
(c) (iii) (iv) (i) (ii)
F
(A) XeF4
F
F
F
F
F
F
sp3d2 F
⇒ Geometry = shape = octahedral
sp
(C) BeCl2 ⇒ Cl Be Cl
⇒ Geometry = shape = linear
N
H
3
sp
H
⇒electrongeometry = tetrahedral
⇒ Shape = trigonal pyramidal
Hence, option (a) is wrongly matched
pair.
35 Match the xenon compounds in
Column I with its structure in
Column II and assign the correct
code :
[NEET (National) 2019]
Column I
Column II
A.
XeF4
(i)
Pyramidal
B.
XeF6
(ii)
Square planar
(Square pyramidal)
38 The hybridisations of atomic
orbitals of nitrogen in NO +2 , NO −3
and NH+4 respectively are
[NEET 2016, Phase II]
F
Xe
O
F
S
F
Xe
(C) XeOF4
(D) XeO3
F
(D) NH3 ⇒ H
(Distorted octahedral)
F
F
F
(B) SF6 ⇒
F
Xe
O
P
Cl
⇒ Geometry=shape=trigonal
bipyramidal
[Nottrigonalplanar]
(Square planar)
F
F
Cl
Cl
As the number of lone pair of electrons
on central element increases, repulsion
between those lone pair of electrons
increases and therefore, bond angle
decreases.
Molecules Bond angle
CH 4 (no lone pair of electrons) 109.5°
NH 3 (one lone pair of electrons) 107.5°
H2 O (two lone pair of electrons) 104.45°
F
Xe
F
sp3d
Ans. (a)
The given xenon compounds with its
structures are as follows:
Cl
(A) PCl5 ⇒ Cl
A B C D
(b) (iii) (iii) (i) (iv)
(d) (i) (ii) (iii) (iv)
Ans. (a)
(B) XeF6
Ans. (a)
(a) The H—O—H bond angle in H2 O is
larger than the H—C—H bond angle in
CH 4
(b) The H—O—H bond angle in H2 O is
smaller than the H—N—H bond angle
in NH 3
(c) The H—C—H bond angle in CH 4 is
larger than the H—N—H bond angle in
NH 3
(d) The H—C—H bond angle in CH 4 , the
H—N—H bond angle in NH 3 and the
H—O—H bond angle in H2 O are all
greater than 90°
Column II
(a) sp, sp 3 and sp2 (b) sp2 , sp 3 and sp
(c) sp, sp2 and sp 3 (d) sp2 , sp and sp 3
Ans. (c)
(Pyramidal)
Ion
O
Structure
O
NO 2+
O==N==O
Hence, the correct match is
(A) → (ii), (B) → (iii), (C)→ (iv), (D) → (i)
NO −3
O
NH 4+
37 Consider the molecules CH4 , NH3
and H2O. Which of the given
statements is false?
[NEET 2016, Phase I]
N→O
sp 3
H
+
N
H
H
H
Thus, option (c) is correct.
39 Which of the following species
contains equal number of σ and
[CBSE AIPMT 2015]
π-bonds?
Ans. (d)
According to the postulate of VSEPR
theory, a lone pair occupies more space
than a bonding pair, since it lies closer to
the central atom. This means that the
repulsion between the different electron
pairs follow the order.
lp − lp > lp − bp > bp − bp
sp
sp 2
O
36 Predict the correct order among
the following. [NEET 2016, Phase I]
(a) lone pair-lone pair > bond pair-bond
pair > lone pair-bond pair
(b) bond pair-bond pair > lone pair-bond
pair > lone pair-lone pair
(c) lone pair-bond pair > bond pair-bond
pair > lone pair-lone pair
(d) lone pair-lone pair > lone pair-bond
pair > bond pair-bond pair
Hybridisation
(a) HCO–3
(b) XeO4
(c) (CN)2
Ans. (b)
(d) CH2 (CN)2
σ and π bonds
Structure
(a)
O

C
O
(b)
O
O H
O

Xe

O O
σ bond-4
π bond-1
σ bond-4
π bond-4
23
Chemical Bonding and Molecular Structure
σ and π bonds
Structure
N ≡≡ C C ≡≡ N
(c)
σ bond-3
π bond-4
(d)
H

N ≡≡ C C C ≡≡ N

H
σ bond-6
π bond-4
40 Which of the following pairs of ions
are isoelectronic and isostructural?
[CBSE AIPMT 2015]
(a) CO23– , SO23–
(b) CIO–3, CO23–
–
(c) SO2–
3 , NO 3
Ans. (d)
ClO−3 , == SO23−
–
O
O
O
O
O
O–
C
O
N
–
O
–
O
O
[CBSE AIPMT 2014]
(b) NO −3
(d) CO2
(a) N3
(c) NO2−
Ans. (b)
Species with sp2 hybridisation are planar
triangular in shape. Among the given
speciesNO−3 is sp2 hybridised with no
lone pair of electrons on central atom, N.
Whereas,N3, NO2− and CO2 are
sp hybridised with a linear shape.
+
N
O
N
N
–
O
–
N–
O
O
C
O
42 Which of the following is a polar
molecule?
[NEET 2013]
(b) SF4
H
(2 bp + 2 lp)
[bp = bond pair and lp = lone pair]
F

(b) BF3 ⇒
B
F
F
(c) SiF4
(d) XeF4
–
(c) NH2 ⇒
–
N
P
Cl
Cl
Cl
(3 bp + 1 lp)
Thus, in PCl3 , the central P-atom is
surrounded by three bond pairs and one
lone pair.
44 Considering the state of
hybridisation of carbon atoms, find
out the molecule among the
following which is linear?
[CBSE AIPMT 2011]
(a) CH3 —C ≡≡ C—CH3
(b) CH2 == CH—CH2 — C ≡≡ CH
(c) CH3 — CH2 — CH2 — CH3
(d) CH3 — CH == CH— CH3
Ans. (a)
3
2
1
H3 C  C ≡≡ C  CH3 is linear because C2
and C3 are sp hybridised carbon atom.
45 Which of the two ions from the list
given below, have the geometry
that is explained by the same
hybridisation of orbitals,
NO–2 , NO–3 , NH–2 , NH+4 , SCN– ?
[CBSE AIPMT 2011]
(a) NH+4 and NO –3
(c) NO2– and NH2–
NO2− → sp2
NO−3 → sp2
NH2+ → sp3
NH−4 → sp3
SCN+ → sp
−
NO2 and NO−3 both have the same
hybridisation, i.e. sp2 .
46 In which of the following pairs of
molecules/ions, the central atoms
have sp 2 hybridisation?
[CBSE AIPMT 2010]
(a) NO2− and NH3
(b) BF3 and NO2−
(c) NH2− and H2O
(d) BF3 and NH2−
Ans. (b)
Key Idea For sp2 hybridisation, there
must be 3 σ-bonds or 2σ-bonds along
with a lone pair of electrons.
(i) NO2− ⇒2 σ + 1 lp = 3, i.e. sp2
hybridisation
(ii) NH3 ⇒3 σ + 1 lp = 4, i.e. sp3
hybridisation
(iv) NH2− ⇒2 σ + 2 lp = 4, i.e. sp3
hybridisation
(2bp + 2lp)
(d) PCl3 ⇒
Ans. (d)
iii) BF3 ⇒3 σ + 0 lp = 3, i.e. sp2
hybridisation
H
H
4
–
N
(d) PCl3
(3 bp + 0 lp)
41 Which one of the following species
has plane triangular shape?
(a) BF3
(a) H2O ⇒
O
Number of electrons
CO23− = 6 + 2 + 24 = 32
SO23− = 16 + 2 + 24 = 42
ClO−3 = 4 + 24 + 1 = 42
CO23− = 6 + 24 + 2 = 32
NO−3 = 7 + 2 + 24 = 33
Hence, ClO−3 and SO23− are isoelectronic
and are pyramidal in shape.
O
(c) NH2−
H
–
–
O
(a) H2O (b) BF3
Ans. (d)
S
O
O
43 Which of the following species
contains three bond pairs and one
lone pair around the central atom?
[NEET 2013]
(d) ClO –3, SO2–
3
Cl
Ans. (b)
Symmetrical molecules are generally
non-polar although they have polar
bonds. This is because bond dipole of
one bond is cancelled by that of the
other. BF3, SiF4 and XeF4 being
symmetrical as non-polar. SF4 is
unsymmetrical because of the
presence of a lone pair of electrons.
Due to which it is a polar molecule.
−
(b) SCN and NH2–
(d) NO2– and NO –3
(v) H2O ⇒2 σ + 2 lp = 4, i.e. sp3
hybridisation
Thus, among the given pairs, only BF3
and NO2− have sp2 hybridisation.
47 In which one of the following
species the central atom has the
type of hybridisation which is not
the same as that present in the
other three?
[CBSE AIPMT 2010]
(a) SF4
(c) SbCl2−
5
Ans. (c)
(b) I−3
(d) PCl5
Key Idea Molecules having same
hybridisation have same number of hybrid
orbitals,
1
H = [V + X − C + A]
2
where,
V = number of valence electrons of
central atom
X = number of monovalent atoms
C = charge on cation
A = charge on anion
SbCl25− = sp 3d2 , PCl 5 = sp 3d
SF 4 = sp 3d , I −3 = sp 3d
24
NEET Chapterwise Topicwise Chemistry
48 In which of the following
molecules/ions BF3 , NO −2 , NH−2 and
H2O, the central atom is
sp 2 hybridised? [CBSE AIPMT 2009]
(a) NO2− and NH2−
(c) NO2− and H2O
Ans. (d)
(b) NH2− and H2O
(d) BF3 and NO2−
BF3
F
σ
σ
F ⇒3 σ-bonds, i.e. sp2 hybridisation
B—
σ
F
Planar structure
NO2−
••
s
O==•• N •• ==O ⇒2σ-bonds +1 lone pair of
electrons, i.e. sp2 hybridisation
NH2–
••
H σ—N σ—H ⇒2σ-bonds +2 lone pairs, i.e.
•×
Ans. (d)
Ans. (a)
(a) SF4 = irregular tetrahedral (sp3d,one
lone pair)
XeF4 = square planar (sp3d2 , two lone
pairs)
(b) SO23− = pyramidal (sp3, one lone pair)
NO−3 = trigonal planar (sp2 )
(c) BF3 = trigonal planar (sp2 )
NF3 = pyramidal (sp3)
(d) BrO−3 = pyramidal (sp3, one lone pair)
XeO3 = pyramidal (sp3, one lone pair)
SCl 4 is not isostructural with SiCl 4
because it shows square planar
structure due to involvement of
repulsion between lone pair and bond
pair of electrons.
SO24− shows tetrahedral structure due to
sp3 hybridisation.
PO3−
4 shows tetrahedral structure due
to sp3 hybridisation.
NH+4 shows tetrahedral structure due to
sp3 hybridisation.
51 Which of the following species has
a linear shape? [CBSE AIPMT 2006]
(a) NO2–
(c) NO2+
Ans. (c)
(b) SO2
(d) O 3
54 Which of the following molecules
has trigonal planar geometry?
[CBSE AIPMT 2005]
(a) IF3
(c) NH3
Ans. (d)
+
NO2 has linear shape due to sp
+
hybridisation of N in NO2
(b) PCl3
(d) BF3
IF3 has bent-T geometry
+
F
O==N == O
sp3 hybridisation
H2O =σ
H
••
•O
•
σ
WhileSO2 , NO2– and O3 have angular
shape
S
H
⇒ 2σ-bonds +2 lone pairs, i.e. sp3
hybridisation, Thus, inBF3 and NO2− ,
central atom is sp2 hybridised, while
NH2 , NH 3 and H2 O are sp 3 hybridised.
O
O
V-shape
(a) 1 sigma and 2 pi-bonds
(b) 2 sigma and 2 pi-bonds
(c) 1 sigma and 1 pi-bonds
(d) 2 sigma and 1 pi-bonds
Ans. (d)
In case of single bond, there is only one
σ-bond, in case of double bond, there is
one σ and one π-bonds while in case of
triple bond, there is one σ and two
π-bonds. Thus, angular shape of ozone
(O3) contains 2σ and 1 π-bonds as shown
below
–
O
50 In which of the following pairs, the
two species are isostructural?
[CBSE AIPMT 2007]
(a) SF4 and XeF4
−
(b) SO2−
3 and NO 3
(c) BF3 and NF3
(d) BrO −3 and XeO 3
F
2lp + 3 bp = sp3d hybridisation
PCl3 has pyramidal geometry
O
V-shape
+
P
O–
O
Angular shape (due to sp 2 hybridisation
of central atom or ion)
Cl
1lp + 3 bp = sp3 hybridisation
NH3 has trigonal pyramidal geometry
52 Which of the following is not a
correct statement?
N
H
[CBSE AIPMT 2006]
(a) The electron deficient molecules can
act as Lewis acids
(b) The canonical structures have no
real existence
(c) Every AB 5 molecule does infact have
square pyramid structure
(d) Multiple bonds are always shorter
than corresponding single bond
Generally, AB 5 molecules have trigonal
bipyramidal structure due to sp3d
hybridisation but in some cases due to
presence of lone pair of electrons, its
geometry becomes distorted.
53 Which of the following is not
isostructural with SiCl 4 ?
[CBSE AIPMT 2006]
(a) SCl4
(c) PO 3–
4
(b) SO2–
4
(d) NH4+
Cl
Cl
V-shape
Ans. (c)
2 sigma and one π-bonds.
F
O
49 The angular shape of ozone
molecule (O 3 ) consists of
[CBSE AIPMT 2008]
N
I
H
H
1lp + 3bp = sp3 hybridisation
BF3 has trigonal planar geometry
F
F
B
(µ = 0)
F
3bp only = sp2 (hydridisation)
55 H2O is dipolar, whereas BeF2 is not.
It is because
[CBSE AIPMT 2004]
(a) the electronegativity of F is greater
than that of O
(b) H2O involves hydrogen bonding
whereas BeF2 is a discrete molecule
(c) H2O is linear andBeF2 is angular
(d) H2O is angular andBeF2 is linear
Ans. (d)
The structure of H2O is angular or
V-shape and has sp3-hybridisation and
25
Chemical Bonding and Molecular Structure
104.5° bond angle. Thus, its dipole
moment is positive or more than zero.
O
104.5°
14 Si =
1s 2 , 2 s 2 , 2p6 , 3s 2 3p2 (in ground
state)
2
2
6
1
3
14 Si = 1s , 2 s 2p , 3s 3p (in excited
state)
3s
H
H
3p
But in BeF2 , structure is linear due to
sp hybridisation (µ = 0)
59 Which of the following has pπ - dπ
bonding?
[CBSE AIPMT 2002]
(a) NO –3
(b) SO2–
3
3–
(d) CO2–
(c) BO 3
3
Ans. (b)
3
In SO2–
3 , S is sp hybridised, so
sp3 hybridisation
180°
σ
π
O σ S
σ
3
Be
F
F
Thus, due toµ > 0, H2O is dipolar and due
to µ = 0, BeF2 is non-polar.
56 In an octahedral structure, the pair
of d orbitals involved in
d 2 sp 3 -hybridisation is
Hence, four equivalent sp hybrid
orbitals are obtained and they are
overlapped by four p-orbitals of four
fluorine atoms on their axis. Thus, it
shows following structure :
(b) dxz , dx 2 − y 2
(c) dz 2 , dxz
(d) dxy , dyz
(a) 3
(c) 6
Ans. (a)
(b) 2
(d) 4
In octahedral structure M X6 , the six
hybrid orbitals (sp3d2 ) are directed
towards the cornes of a regular
octahedral with an angle of90°.
According to following structure of M X6 ,
the number of X — M — X bonds at 180°
must be three.
F
F
F
σ
or
σ
Si
σ
σ
F
F
1
= 1s 2 , 2 s 2 2p6 , 3 s 2 3px1 3py1 3pz1, 3d xy
144424443
sp 3d hybridisation
(in first excitation state)
F
X
F
F
58 Among the following, the pair in
which the two species are not
isostructural, is [CBSE AIPMT 2004]
(a) SiF4 and SF4
(c) BH−4 and NH+4
Ans. (a)
F
X
X
(b) IO –3 and XeO 3
(d) PF6− and SF6
SiF4 and SF4 are not isostructural
because SiF4 is tetrahedral due to sp3
hybridisation of Si.
3px1
sp 3hybridisation
In ‘S’ the three p-orbitals forms σ-bonds
with three oxygen atoms and
unhybridised d-orbital is involved in
π-bond formation.
O8 = 1s 2 , 2 s 2 2p2x 2py1 2pz1
In oxygen two unpaired p-orbitals are
present, one is involved in σ-bond
formation while other is used in π-bond
formation.
Thus inSO23– , p and d-orbitals are involved
for pπ- dπ bonding.
or
F
F
σ
σ
60 Which of the following two are
isostructural? [CBSE AIPMT 2001]
(a) XeF2 , and IF2–
2–
(c) CO2–
3 , and SO 3
Ans. (a)
(b) NH3, and BF3
(d) PCl5 , and ICl5
Compounds having same structure and
same hybridisation are known as
isostructural species. e.g.
XeF2 and IF2– are sp3d hybridised and
both have linear shape.
F — I– — F
F — Xe — F
61 Which one of the following is
planar?
[CBSE AIPMT 2000]
S
M
X
excited state)
2
F
WhileSF4 is not tetrahedral but it is
arranged in trigonal bipyramidal
geometry (has see saw shape) because
in itS is sp3d hybrid.
2
2
6
2 2
1
1
16 S = 1 s , 2 s 2p , 3s 3px 3py 3pz
(in ground state)
X
X
6
Unhybridised
Si
F
57 In a regular octahedral molecule,
MX 6 the number of X M  X
bonds at 180° is [CBSE AIPMT 2004]
2
1
3d xy
Ans. (a)
In the formation of d2 sp3 hybrid orbitals,
two (n − 1) d-orbitals of e g set, i.e.
(n– 1) dz 2 and (n– 1) d x 2 –y 2 orbitals, one n s
and three np (npx , npy and npz .) orbitals
combine together.
O–
= 1s , 2 s 2p , 3 s
3py1 3pz1
1
4
4
2
443
(Sulphur atom in
16 S
2
F
[CBSE AIPMT 2004]
(a) dx 2 − y 2 , dz 2
O–
F
σ
(a) XeF4
(b) XeO4
(c) XeO3F
(d) XeO3F2
Ans. (a)
Structure of XeF4 is as follows
S
σ
F
F
F
Hence, five sp3d hybrid orbitals are
obtained. One orbital is already paired
and rest four are overlapped with four
p-orbitals of four fluorine atoms on their
axis in trigonal bipyramidal form.
This structure is distorted from trigonal
bipyramidal to tetrahedral due to
involvement of repulsion between lone
pair and bond pair.
Xe
F
F
It involves sp3d2 hybridisation in
Xe-atom. The molecules has square
planar structure. Xe and four F-atoms
are coplanar. The lone pairs are present
on axial positions, minimise electron pair
repulsion.
26
NEET Chapterwise Topicwise Chemistry
62 The type of hybridisation of boron
in diborane is
[CBSE AIPMT 1999]
(a) sp hybridisation
(b) sp2 hybridisation
(c) sp 3 hybridisation
(d) sp 3d2 hybridisation
Ans. (c)
Each boron atom in diborane (B2H6 ) is
sp3 hybridised. In the structure of
diborane four H-atoms, two on the left
and two on the right, known as terminal
hydrogens, are in different environments
from the other two hydrogen atoms
which are known as bridging atoms. The
two boron atoms and the four terminal
hydrogen atoms lie in the same plane
while the two boron atoms and the two
bridging hydrogen atoms, one above and
the other below, lie in a plane
perpendicular to this plane.
H
H
H
B
B
H
H
H
Structure of diborane molecule
63 AsF5 molecule is trigonal
[CBSE AIPMT 1997]
(a) dx 2 − y 2 , dz 2 , s , px , py
(b) dxy , s , px , py , pz
66 When the hybridisation state of
carbon atom changes from sp 3 to
sp 2 and finally to sp, the angle
between the hybridised orbitals
[CBSE AIPMT 1993]
(a) decreases gradually
(b) decreases considerably
(c) is not affected
(d) increases progressively
Ans. (d)
In sp3 hybridisation bond angle is 109°28′.
In sp2 hybridisation bond angle is 120°.
In sp hybridisation bond angle is 180°.
[CBSE AIPMT 1992]
(a) SO2
(c) CO2−
3
Ans. (b)
64 Which of the following does not
have a tetrahedral structure?
[CBSE AIPMT 1994]
(d) H2O
BH3 have sp2 hybridisation, so it have
trigonal planar structure, not tetrahedral
structure.
65 Among the following orbital bonds,
the angle is minimum between
Ans. (b)
BeF2 — sp hybridisation
BF3 — sp2 hybridisation
C2H2 — sp hybridisation
NH3 — sp3 hybridisation
70 Which of the following molecule
does not have a linear arrangement
of atoms?
[CBSE AIPMT 1989]
(a) H2 S
(c) BeH2
Ans. (a)
(b) C2H2
(d) CO2
H2S have sp3 hybridisation while
remaining all have sp hybridisation, so
H2S have bent structure and other have
linear geometry.
71 Equilateral shape has
[CBSE AIPMT 1988]
(a) sp hybridisation
(b) sp2 hybridisation
(c) sp 3 hybridisation
(d) dsp2 hybridisation
Ans. (a)
Equilateral or triangular planar geometry
is formed by sp2 hybridisation.
+
and CO2 , CO2 have
sp hybridisation, thus have linear
geometry.
As = 1s 2 , 2 s 2 , 2p6 , 3s 13px1 3py1 3pz1 3d 1
14442444
3
sp3d hybridisation
Due to this hybridisation, geometry of
the AsF5 molecule is trigonal bipyramidal
and the hybrid orbitals used by As-atom
are s , px , py , pz and d xy .
s-orbital
+
p-orbital
p-orbital
68 An sp 3 hybrid orbital contains
120°
120°
[CBSE AIPMT 1991]
1
(a) s-character
4
1
(b) s-character
2
1
(c) s-character
3
2
(d) s-character
3
Ans. (a)
120°
sp2 hybrid
orbital
72 The angle between the overlapping
of one s-orbital and one p-orbital is
[CBSE AIPMT 1988]
In sp3 hybrid orbital one part, out of four
1
orbital is s-orbital, so it have 25% or
4
s-character.
69 In which one of the following
molecules, the central atom said to
adopt sp 2 hybridisation?
[CBSE AIPMT 1994]
(a) sp 3 bonds
(b) px and py -orbitals
(b) CO2
(d) SO2−
4
Out of SO2 , CO23− ,SO24−
(c) s , px , py , pz , dxy
(d) dx 2 − y 2 , s , px , py , pz
Ans. (c)
(c) NH+4
When px and py form bond, then the
bond angle is minimum and it is only 90°.
NOTE
Bond angle in sp3 bonds, H − O − H in
water and in sp bonds are 109°28, 180 °
and 180° respectively.
67 Which structure is linear?
bipyramidal. The hybrid orbitals
used by As-atoms for bonding are
(a) BH−4 (b) BH3
Ans. (b)
(c) H — O — H in water
(d) sp bonds
Ans. (b)
(a) 180°
(c) 109 °28′
Ans. (a)
(b) 120°
(d) 120° 60′
When s-orbital and p-orbital overlap
each other, then the bond angle formed
is 180° as given below
+
s-orbital
p-orbital
[CBSE AIPMT 1989]
(a) BeF2
(c) C2H2
(b) BF3
(d) NH3
180°
27
Chemical Bonding and Molecular Structure
TOPIC 3
Molecular Orbital Theory,
Hydrogen and Metallic
Bonding
73 Identify a molecule which does not
exist.
[NEET (Sep.) 2020]
(a) Li2
(b) C2
Ans. (d)
(c) O2
(d) He2
If bond order of a molecules becomes
zero, the molecule will not exist.
Bond order
Number of bonding electrons −
=
=
Number of antibonding electrons
2
N b −N a
2
Li2 (6e − ) ⇒σ21s σ *1s2
4−2
σ22 s ; BO =
=1
2
*2
2
σ2 s σ2 s π2 p x = π22 py ;
C2 (12e − ) ⇒σ21s σ *1s2
8−4
BO =
=2
2
O2 (16e − ) ⇒σ21s σ 1*s2 σ22 s σ2*2s σ22 p z π 22 p x
= π22 p y π2*1p x = π2*1p y ;
10 − 6
=2
BO =
2
2−2
He2 (4e − ) = σ21s σ *1s2 ; BO =
=0
2
So, He2 does not exist.
74 The potential energy (y) curve for
H2 formation as a function of
internuclear distance (x) of the
H-atoms is shown below.
[NEET (Oct.) 2020]
75 Which of the following diatomic
molecular species has only
π-bonds according to molecular
orbital theory? [NEET (National) 2019]
(a) N2
(c) Be2
Ans. (b)
(b) C2
(d) O2
The molecular orbital configuration of C2
is
C2 (Z = 12 ) = σ1s 2 , σ * 1s 2 , σ2s 2 , σ * 2s 2 ,
π2p2x = π2p2y .
Double bond in C2 consists of both
π-bonds because of the presence of last
(valence) four electrons in two
π-molecular orbitals.
The configuration of N2 , Be2 and O2 are as
follows:
N2 (Z = 14) − σ1s 2 , σ * 1s 2 , σ2s 2 , σ * 2s 2 ,
π2p2x = π2p2y , σ2p2z (1 σ and 2π-bonds)
Be2 (Z = 8) − σ1s 2 , σ * 1s 2 , σ2s 2 , σ * 2s 2 (σ
bonds only)
O2 (Z = 16) − σ1s 2 , σ * 1s 2 , σ2 s 2 , σ * 2 s 2 ,
σ * 2p2z , π2p2x = π2p2y , π * 2px1 = π * 2pz1 (σ, π
and 1π* bond)
Hence, option (b) is correct.
76 Which of the following is
paramagnetic? [NEET (Odisha) 2019]
(a) N2
(c) Li2
Ans. (d)
(b) H2
(d) O2
Key Idea If all the electron in a molecule
are paired, the molecule is diamagnetic
and if there are unpaired electrons in a
molecule, it is paramagnetic.
Molecular orbital configuration of given
molecules are as follows:
a
y
b
c
x
The bond energy of H2 is
(a) (b − a)
(b)
(b − a)
2
Ans. (a)
(c − a)
2
(d) (c − a)
(c)
[Repulsion between two H-atoms]
+ve
H H
(0.00) a
y
(PE)
From the information mentioned in the
above curve, bond energy (BE) of H2
molecule is (b − a) or | b − a |.
–ve
(–ve) b
C
BE
H
H H
Infinite separation of two
H-atom (No attraction)
(No repulsion)
H
Ovarlapping of two H-atom to
form a covalent bond (σ)
x (Internuclear distance)
(a) N2 (Z = 14) − σ 1s 2 , σ * 1s 2 , σ2s 2 ,
σ * 2s 2 , π2p2x = π2p2y , σ2p2z
It is a diamagnetic molecule due to
absence of unpaired electron.
(b) H2 (Z = 2) − 1σ2
It is a diamagnetic molecule.
(c) Li2 (Z = 6) − σ 1s 2 , σ * 1s 2 , σ2s 2
It is a diamagnetic molecule.
(d) O2 (Z = 16) − σ 1s 2 , σ * 1s 2 , σ2s 2 , σ * 2s 2 ,
σ2p2z , π2p2x = π2p2y , π * 2px1 = π * 2py1
It is a paramagnetic molecule due to
presence of unpaired electrons in
each orbital of degenerate levels.
Thus, option (d) is correct.
77 Consider the following species
CN + , CN − , NO and CN
Which one of these will have the
highest bond order? [NEET 2018]
(a) CN+
(c) NO
Ans. (b)
(b) CN−
(d) CN
The formula of bond order is given as
B.O.
No. of electrons  No. of electrons 
in bonding
 − in antibonding 
 

=
2
Energy level pattern for molecular
orbitals of different molecules depends
upon their central atom.
NO : Central atom is N
(Total number of electrons = 15)
∴
σ 1s 2 , σ * 1s 2 , σ2s 2 , σ * 2s 2 ,
( π2p2x ~− π2p2y ), σ2p2z ,( π2*px1 ≈ π2*py0 )
10 − 5
B.O. =
= 2.5
2
CN − : Central atom is C
[Total number of electrons= 14]
σ 1 s 2 , σ * 1 s 2 , σ2s 2 , σ * 2s 2 , ( π2px2
≈ π2py2 σ2p2z )
10 − 4
B.O. =
=3
2
CN : Central atom is C
[Total number of electrons = 13]
σ 1s 2 , σ * 1s 2 , σ2s 2 , σ * 2s 2 ,( π2p2x
≈ π2p2y ), σ2pz1
9−4
B.O =
= 2.5
2
CN + : Central atom is C
[Total number of electrons = 12]
σ 1s 2 , σ * 1s 2 , σ2s 2 , σ * 2s 2 , ( π2p2x ≈ π2p2y )
8−4
B.O =
=2
2
Therefore, option (b) is correct.
78 Which one of the following
compounds shows the presence of
intramolecular hydrogen bond?
[NEET 2016, Phase II]
(a) H2O2
(b) HCN
(c) Cellulose
(d) Concentrated acetic acid
Ans. (c)
Intermolecular hydrogen bonding is
present in concentrated acetic acid,H2O2
and HCN while cellulose has intramolecular hydrogen bonding as shown
below :
28
NEET Chapterwise Topicwise Chemistry
δ+
δ–
O
H
C
H3 C
C
δ–
δ+
O
Ans. (b)
O
H
Bond order of O2−
O2− = σ 1s 2 , σ* 1s 2 ,σ 2s 2 ,σ* 2s 2
CH3
O
σ 2p2z ( π2p2x = π2p2y ) ( π *2px1 = π *2py1 )
Concentrated Acetic Acid
H δ+
O
Bond order
number of electrons in BMO − number
O
δ–
H δ+
δ–
O
H δ+
=
O δ–
10 − 7 3
=
= = 1.5
2
2
H δ+
δ–
C
δ+
N
H
( π2p2x = π 2p2y ) ( π *2px1 = π *2py0 )
δ–
C
N
BO =
HCN
HO
5
( π2p2x = π 2p2y ) ( π* 2px1 = π* 2py1 )
O
O
2
3
HO
1
2
4
1
OH
δ–
6
HO
n
Cellulose
In above molecules, dotted lines
represent hydrogen bonding.
79 Which of the following pairs of ions
is isoelectronic and isostructural?
[NEET 2016, Phase II]
(b) CIO −3 , CO23−
(d) CIO −3 , SO23−
(a) CO23− , NO −3
(c) SO23− , NO −3
Ans. (a) and (d)
Species
ClO –3
Total
number of Structure
electrons
42
Shape
Pyramidal
Cl
–
SO 2–
3
Pyramidal
S
–
O
–
O
O
CO 2–
3
32
Trigonal
planar
O
C
–
32
= σ 1s 2 , σ* 1 s 2 , σ2 s 2 , σ* 2s 2 , σ2p2z , π2p2x
* 2p2 ≈ π
* 2py1
≈ π2p2y , π
x
Trigonal
planar
N
O
≈ π2p2y , σ2p2z
(All the electrons are paired so, it is
diamagnetic).
O2− (8 + 8 + 1 = 17)
O
O
(b) O2−
(d) NO +
Paramagnetic species contains unpaired
electrons in their molecular orbital
electronic configuration.
Molecular orbital configuration of the
given species is as
CO (6 + 8 = 14)
–
O
NO –3
(a) CO
(c) CN−
Ans. (b)
= σ 1s , σ* 1s 2 , σ2 s 2 , σ* 2 s 2 , π2p2x
O
42
81 Which of the following is
paramagnetic?
[NEET 2013]
2
O
O
10 − 6 4
= =2
2
2
So, the correct sequence is
O2− < O2 < O2+
BO =
O
5
10 − 5 5
= = 2. 5
2
2
O2 = σ 1s 2 σ* 1s 2 , σ 2s 2 , σ* 2s 2 σ 2p2z
OH
3
HO
O
4
O
δ+
δ+
δ–
6
2
O2+ = σ 1s 2 , σ* 1s 2 , σ 2s 2 , σ* 2s 2 σ 2p2z
H2O 2
H
of elections ABMO
–
O
Hence both options (a) and (d) are
correct.
(It contains one unpaired electron so, it
is paramagnetic.)
CN− (6 + 7 + 1 = 14) = same as CO
NO+ ( 7 + 8 − 1 = 14) = same as CO
Thus, among the given species onlyO2− is
paramagnetic.
Molecular orbital configuration of
O2+ ( 8 + 8 − 1 = 15)
= σ 1 s 2 , σ* 1s 2 , σ2 s 2 , σ* 2 s 2 , σ2p2z ,
π2p2x ≈ π2p2y , π* 2px1 ≈ π* 2py0
Bond order (BO) =
Nb − Na
2
(where,N b = number of electrons in
bonding molecular orbital,
N a = number of electrons in antibonding
molecular orbital)
10 − 5
∴ BO =
= 2.5
2
Similarly,
(b) O2− (8 + 8 + 1 = 17)
N − N a 10 − 7
so, BO = b
=
= 1.5
2
2
(c) O22 − (8 + 8 + 2 = 18)
N − N a 10 − 8
BO = b
=
=1
2
2
(d) O2 (8 + 8 = 16)
10 − 6
BO =
=2
2
Thus, O2− shows the bond order 1.5.
83 The pair of species with the same
bond order is
[NEET 2013]
(a) O22 − , B2
(c) NO, CO
Ans. (a)
(b) O2+ , NO +
(d) N2 , O2
According to molecular orbital theory,
O22 − (8 + 8 + 2 = 18)
= σ 1 s 2 , σ* 1 s 2 , σ2 s 2 , σ* 2 s 2 , σ2p2z , π2p2x
≈ π2p2y , π* 2p2x ≈ π* 2p2y
Bond order (BO) =
Nb − Na
2
=
10 − 8
=1
2
B2 (5 + 5 = 10) = σ 1s , σ 1s 2 , σ2 s 2 , σ* 2 s 2 ,
2 *
π2px1 ≈ π2py1
6−4
BO =
=1
2
Thus, O22 − and B2 have the same bond
order.
NOTE
BO of O2+ = 2.5 , NO + = 3 , NO = 2 .5 , CO = 3 ,
N2 = 3 and O2 = 2
84 Which of the following has the
minimum bond length?
[CBSE AIPMT 2011]
80 Which of the following options
represents the correct bond order?
[CBSE AIPMT 2015]
(a) O2– > O2 > O2+
(c) O2– > O2 < O2+
(b) O2–
(d) O2–
< O2 < O2+
< O2 > O2+
82 Bond order of 1.5 is shown by
[NEET 2013]
(a) O2+
(b) O2−
(c) O2−
2
(d) O2
Ans. (b)
(a) O2−
(c) O2
Ans. (d)
Bond order of O2+ =
(b) O2–
2
(d) O2+
10 − 5
= 2.5
2
29
Chemical Bonding and Molecular Structure
10 − 7
= 1.5
2
10 − 8
Bond order of O22 – =
=1
2
10 − 6
Bond order of O2 =
=2
2
Q Maximum bond order = minimum bond
length.
∴ Bond length is minimum forO2+
Bond order of O2– =
85 Which one of the following species
does not exist under normal
conditions?
[CBSE AIPMT 2010]
(a) Be +2
(c) B2
Ans. (b)
(b) Be 2
(d) Li 2
Hence, it is the intermolecular
H-bonding that must be overcome in
converting liquid CH3OH to gas.
87 According to molecular orbital
theory which of the following lists
rank the nitrogen species in terms
of increasing bond order?
[CBSE AIPMT 2009]
(a) N2− < N2 < N22 −
(c) N2 < N22 − < N2−
Ans. (b)
(b) N22 − < N2− < N2
(d) N2− < N22 − < N2
According to the molecular orbital
theory (MOT),
N2 ( 7 + 7 = 14) = σ 1s 2 , σ* 1s 2 , σ 2 s 2 ,
Key Idea Molecules with zero bond order,
do not exist.
According to molecular orbital theory,
(a) Be 2+ (4 + 4 − 1 = 7)
σ* 2 s 2 , π2p2x ≈ 2p2y , σ2p2z
Bond order =
10 − 4
=3
2
N2− ( 7 + 7 + 1 = 15)
= σ 1s 2 , σ* 1s 2 , σ 2 s 2 , σ* 2 s 2 ,
=
σ 1s 2 , σ* 1 s 2 , σ2 s 2 , σ* 2 s 1
Bond order (BO) =
σ2p2z ,
4− 3
= 0.5
2
10 − 5
= 2.5
2
N22 − ( 7 + 7 + 2 = 16)
= σ 1 s 2 , σ* 1s 2 , σ2 s 2 , σ* 2 s 2
4−4
BO =
=0
2
= σ 1s 2 , σ* 1s 2 , σ2 s 2 , σ* 2 s 2 ,
σ2p2z , π2p2x ≈ π2p2y , π* 2px1 ≈ π* 2py1
(c) B2 (5 + 5 = 10)
=
π2px1
≈
π2py1
6−4
Bond order (BO) =
=1
2
(d) Li2 (3 + 3 = 6) = σ 1s 2 , σ* 1s 2 , σ2s 2
4−2
BO =
=1
2
Thus, Be2 does not exist under normal
conditions.
86 What is the dominant
intermolecular force on bond that
must be overcome in converting
liquid CH3OH to a gas?
[CBSE AIPMT 2009]
(a) Hydrogen bonding
(b) Dipole-dipole interaction
(c) Covalent bonds
(d) London or dispersion force
Ans. (a)
In betweenCH3OH molecules
intermolecular H-bonding exist.
δ+
δ−
δ+
δ−
δ+
≈ 2p2y , π* 2px1
BO =
(b) Be 2 (4 + 4 = 8)
σ 1s 2 , σ* 1s 2 , σ2 s 2 , σ* 2 s 2 ,
π2p2x
δ−
- - - H — O- - - H — O- - - H — O- - 


CH3
CH3
CH3
Intermolecular H-bond
BO =
BO =
10 − 5
= 2.5
2
C22 − (6 + 6 + 2 = 14) = σ 1s 2 , σ* 1s 2 ,
σ2 s 2 , σ* 2 s 2 , π2p2y ≈ π2p2z , σ2p2x
10 − 4
BO =
=3
2
He2+ (2 + 2 − 1 = 3) = σ 1s 2 , σ* 1s 1
2− 1 1
BO =
= = 0.5
2
2
Hence, order of increasing bond order is
He2+ < O2– < NO < C22 –
89 The correct order of C—O bond
length among CO, CO 2−
3 , CO 2 is
[CBSE AIPMT 2007]
(a) CO2 <CO2–
3 <CO
(b) CO <CO2–
3 <CO2
(c) CO2–
<CO
3
2 <CO
(d) CO <CO2 <CO2–
3
Ans. (d)
A bond length is the average distance
between the centres of nuclei of two
bonded atoms. A multiple bond (double
or triple bonds) is always shorter than
the corresponding single bond.
The C-atom inCO23 – is sp2 hybridised as
shown:
O–
O
O=
=C
←→ O – —C
←→
10 − 6
=2
2
Hence, the increasing order of bond
order is,
N22 − < N2− < N2
88 Four diatomic species are listed
below in different sequences.
Which of these presents the
correct order of their increasing
bond order?
[CBSE AIPMT 2008]
+
(a) O2– <NO <C2–
2 <He2
2–
–
+
(b) NO <C 2 <O2 <He2
+
–
(c) C2–
2 <He2 <NO <O2
+
–
2–
(d) He2 <O2 <NO <C2
Ans. (d)
The molecular orbital configuration of
O2− (8 + 8 + 1 = 17) = σ 1s 2 , σ* 1s 2 ,
σ2 s 2 , σ* 2 s 2 ,
2
2
2 *
2
*
σ2px , π2py ≈ π2pz , π 2py ≈ π 2pz1
N − N a 10 − 7
Bond order (BO) = b
=
= 1.5
2
2
NO (7 + 8 = 15) = σ 1s 2 , σ* 1s 2 , σ2s 2 ,
σ* 2 s 2 , σ2p2x
2
2 *
π2py ≈ π2pz , π 2py1 ≈ π* 2pz0
O
–
O
–
O–
O– —C
O
The C-atom inCO2 is sp hybridised with
bond distance carbon-oxygen is 122 pm.
O== C==O ←→
+
–
O ≡≡ C —O ←→
–
+
O— C ≡≡ O
The C-atom in CO is sp hybridised with
C—O bond distance is 110 pm.
• ≡≡ + •
O •
•C
So, the correct order is
CO < CO2 < CO23–
90 The number of unpaired electrons
in a paramagnetic diatomic
molecule of an element with
atomic number 16 is
[CBSE AIPMT 2006]
(a) 2
(c) 4
Ans. (a)
(b) 3
(d) 1
Suppose the diatomic molecule is X.
Then, molecular orbital electronic
configuration of
30
NEET Chapterwise Topicwise Chemistry
16 X
= σ 1s 2 , σ* 1s 2 , σ2 s 2 , σ* 2 s 2 , σ2p2z , π2p2x
≈ π2p2y , π* 2px1 ≈ π* 2py1
Due to presence of two unpaired
electrons, it shows paramagnetic
character.
91 Main axis of a diatomic molecule is
z molecular orbital, p x and p y
overlaps to form which of
the following orbitals?
[CBSE AIPMT 2001]
(a) π-molecular orbital
(b) σ-molecular orbital
(c) δ-molecular orbital
(d) No bond will form
Ans. (a)
For π-overlap the lobes of the atomic
orbitals are perpendicular to the line
joining the nuclei.
+
+
+
–
–
π-overlapping
+
–
92 In X — H− − − Y , X and Y both are
electronegative elements, then
[CBSE AIPMT 2001]
(a) electron density on X will increase
and on H will decrease
(b) in both electron density will increase
(c) in both electron density will decrease
(d) on X electron density will decrease
and on H increase
94 Among the following group which
represents the collection of
isoelectronic species?
[CBSE AIPMT 2000]
(a) NO, CN–, N2 , O2–
–
(b) NO + , C2–
2 , O2 , CO
2–
(c) N2 , C2 , CO, NO
(d) CO, NO + , CN–, C2–
2
Ans. (d)
Species having equal number of
electrons are known as isoelectronic
species.
Number of electrons,
In CO = 6 + 8 = 14
In NO+ = 7 + 8 − 1 = 14
In CN– = 6 + 7 + 1 = 14
In C22 − =12 + 2 = 14
Hence, all have 14 electrons, so they are
isoelectronic species.
95 Which one of the following is not
paramagnetic? [CBSE AIPMT 2000]
(a) NO
(c) CO
Ans. (c)
Paramagnetic character is shown by
those atoms or molecules which have
unpaired electrons.
In the given compounds CO is not
paramagnetic since, it does not have
unpaired electrons. The configuration of
CO molecule is
*
*
CO(14)= σ 1s 2 , σ
1s 2 , σ2 s 2 , σ
2 s 2 , σ2p2x ,
π2p2y ≈ π2p2z
Ans. (a)
In X — H ----- Y, X and Y both are
electronegative elements, then electron
density on X will increase and on H will
decrease.
93 A compound contains atoms of
three elements 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 AIPMT 2000]
(a) A2 (BC 3)2
(c) A3 (B4C)2
Ans. (b)
(b) A3 (BC 4 )2
(d) ABC2
In A3 (BC 4 )2
3 × oxidation number of A
+ 2 [oxidation number ofB + 4 × oxidatio
oxidation number ofC ] = 0
3 × (+2) + 2 [5 + 4 × (–2)] = 0
6 + 2 [–3] = 0
(b) N2+
(d) O2−
96 The relationship between the
dissociation energy of N 2 and N 2+ is
[CBSE AIPMT 2000]
(a) dissociation energy ofN2+ >
dissociation energy ofN2
(b) dissociation energy ofN2 =
dissociation energy ofN2+
(c) dissociation energy of
N2 > dissociation energy ofN2+
(d) dissociation energy ofN2 can either
be lower or higher than the
dissociation energy ofN2+
So, bond order of
N − N a 10 − 4
N2 = b
=
=3
2
2
9−4
and bond order ofN2 + =
=2.5
2
As the bond order ofN2 is greater than
N2 + so, the dissociation energy ofN2 will
be greater thanN2 + .
97 Which one of the following
molecules will form a linear
polymeric structure due to
hydrogen bonding?
[CBSE AIPMT 2000]
(a) NH3
(c) HCl
Ans. (d)
(b) H2 O
(d) HF
HF molecules have linear polymeric
structure due to hydrogen bonding.
H — F --- H — F --- H — F --- H — F
↑
H- bond
98 The number of antibonding
electron pairs in O 2−
2 molecular ion
on the basis of molecular orbital
theory is (at. no. of O is 8)
[CBSE AIPMT 1998]
(a) 5
(b) 2
Ans. (c)
(c) 4
(d) 6
Total number of electrons in
O 22 – = 19 + 2 = 18
According to MOT, the configuration of
O 22 − is
σ 1s 2 , σ* 1s 2 , σ2 s 2 , σ* 2 s 2 , σ2p2x , π2p2y ≈
π 2p2z , π* 2p2y ≈ π* 2p2z
So, the number of antibonding electron
pairs = 4
99 The high density of water
compared to ice is due to
[CBSE AIPMT 1997]
(a)
(b)
(c)
(d)
hydrogen bonding interactions
dipole-dipole interactions
dipole-induced dipole interactions
induced dipole-induced dipole
interactions
Ans. (c)
Ans. (a)
The dissociation energy will be more
when the bond order will be greater and
bond order ∝ dissociation energy
Molecular orbital configuration of
Due to polar nature, water molecules
show intermolecular hydrogen bonding
as
- - - H —O- - - H —O - - - H —O



H
H
H
*
*
2 s 2 , π 2py 2
N2 (14) = σ 1s 2 , σ
1s 2 , σ 1 s 2 , σ
≈ π2pz2 , σ2px 2
Hydrogen bonding
31
Chemical Bonding and Molecular Structure
whereas the ice has open structure with
large number of vacant spaces. So,
density of ice is lower than water.
103 The ground state electronic
configuration of valence shell
electrons in nitrogen molecule (N 2 )
is written as
100 N 2 and O 2 are converted into
KK
, σ2 s 2 , σ * 2s 2 , σ2p x2 , π2p y2 ≈ π2p z2
−
−
monoanions N 2 and O 2
Bond order in nitrogen molecule is
respectively. Which of the following
[CBSE AIPMT 1995]
statements is wrong?
[CBSE AIPMT 1997]
(a) In N2 , the N—N bond weakens
(b) In O2− , O—O bond length increases
(c) In O2–, bond order decreases
(d) N2− , becomes diamagnetic
Ans. (d)
In N2− total electrons = 14 + 1 = 15
Electronic configuration of N2− is
σ 1s , σ 1s
2 *
2
, σ2 s , σ 2 s , σ2p2x ,
π2p2y ≈ π2p2z , π* 2py1
2 *
2
Due to presence of one unpaired
electron, it shows paramagnetic
character.
101 The ion that is isoelectronic with
CO is
[CBSE AIPMT 1997]
(b) N2+
(a) O2−
(c) O2+
(d) CN−
Ans. (d)
Isoelectronic species are having same
number of electrons.
Number of electrons in CO = 6 + 8 = 14
Number of electrons inO2− = 16 + 1 = 17
Number of electrons inN2+ = 14 − 1 = 13
Number of electrons inO2+ = 16 − 1 = 15
Number of electrons in
CN− = 6 + 7 + 1 = 14
Hence, CO isoelectronic withCN− ion.
102 The correct order of N — O bond
lengths in NO, NO −2 , NO −3 and N 2O 4
is
[CBSE AIPMT 1996]
(a) N2O 4 > NO2– > NO –3 > NO
(b) NO > NO –3 > N2O 4 > NO2–
(c) NO –3 > NO2– > N2O 4 > NO
(d) NO > N2O 4 > NO2– > NO –3
Ans. (c)
As the bond order increases, bond length
decreases and bond order is highest for
NO, i.e. 2.5 and least forNO−3 , i.e. 1.33. So,
the order of bond length is
NO–3 > NO2– > N2O4 > NO
1.33
1.5
1.5
2.5
(a) 0
(c) 0
Ans. (d)
(b) 1
(d) 3
Ans. (c)
The boiling point of p-nitrophenol is
higher than that of o-nitrophenol
because p-nitrophenol have
intermolecular hydrogen bonding
whereas o-nitrophenol have
intramolecular H-bonding as given below
O
The MO configuration of N2 is
KK , σ2 s 2 , * σ2 s 2 , σ2p2x , π2p2y ≈ π2p2z
Bond order of
1
N2 = [N b − N a ]
2
1
= [8 − 2]
2
6
= =3
2
104 The correct order of the O — O
bond length in O 2 , H2O 2 and O 3 is
[CBSE AIPMT 1995]
(a) O2 > O 3 > H2O2
(c) O2 > H2O2 > O 3
Ans. (d)
(d) p-nitrophenol has a higher molecular
weight than o-nitrophenol
(b) O 3 > H2O2 > O2
(d) H2O2 > O 3 > O2
The bond length of O—O inH2O2 is 147.5
pm, inO3 is 128 pm and inO2 it is 121 pm,
so the correct order isO2 < O3 < H2O2 .
H
O
N
O
o-nitrophenol
---HO—
O
—N
O---HO—
O
—N
p-nitrophenol
O
107 Linus Pauling received the Nobel
Prize for his work on
[CBSE AIPMT 1994]
(a) atomic structure
(b) photosynthesis
(c) chemical bonds
(d) thermodynamics
Ans. (c)
Linus Pauling contributed to chemical
bonding, so, he received the Nobel Prize
for his work in chemical bonding.
108 Mark the incorrect statement in the
105 Which of the following species is
following.
[CBSE AIPMT 1994]
paramagnetic? [CBSE AIPMT 1995]
(a) O2−
2
(c) CO
Ans. (b)
(b) NO
(d) CN−
The molecular orbital configuration of
NO is
KK (σ 2s ) 2 (σ* 2s ) 2 (σ 2px ) 2 ( π2py ) 2
( π2pz )2 ( π* 2py ) 1
So, NO is paramagnetic because it
contains one unpaired electron.
(a) The bond order in the speciesO2 ,O2+
and O2− decreases asO2+ > O2 > O2–
(b) The bond energy in a diatomic
molecule always increases when an
electron is lost
(c) Electrons in antibonding MO
contribute to repulsion between two
atoms
(d) With increase in bond order, bond
length decreases and bond strength
increases
Ans. (b)
106 The boiling point of p-nitrophenol is
higher than that of o-nitrophenol
because
[CBSE AIPMT 1994]
(a) NO2 group at p-position behave in a
different way from that at o-position
(b) intramolecular hydrogen bonding
exists in p-nitrophenol
(c) there is intermolecular hydrogen
bonding in p-nitrophenol
When a diatomic molecule lost electron,
then its bond order may increase or
decrease, so its bond energy may
decrease or increase.
109 The dielectric constant of H2O is
80. The electrostatic force of
attraction between Na + and Cl −
will be
[CBSE AIPMT 1994]
32
NEET Chapterwise Topicwise Chemistry
1
in water than in air
40
1
in water than in air
(b) reduced to
80
(c) will be increased to 80 in water than
in air
(d) will remain unchanged
(a) reduced to
Ans. (b)
Water is a polar solvent and have
dielectric constant 80. As NaCl is a polar
compound and like dissolves like so,
forces of attraction betweenNa+ and
1
in water.
Cl − ion will reduce to
80
110 Linear combination of two
hybridised orbitals belonging to the
two atoms, each having one
electron leads to a
[CBSE AIPMT 1990]
Since, F is most electronegative and
has smaller size, HF shows maximum
strength of hydrogen bond.
(a) sigma bond
(b) double bond
(c) coordinate bond
(d) pi-bond
Ans. (a)
When two hybridised orbitals of two atoms
undergoes linear combination, they form
sigma bond.
111 Which one shows maximum hydrogen
bonding?
[CBSE AIPMT 1990]
(a) H2O
(c) H2 S
Ans. (d)
(b) H2 Se
(d) HF
Hydrogen bonding ∝ electronegativity
1
∝
Size of atom to which
H is covalently bonded
112 Which of the following does not
apply to metallic bond?
[CBSE AIPMT 1989]
(a) Overlapping valence orbitals
(b) Mobile valence electrons
(c) Delocalised electrons
(d) Highly directed bonds
Ans. (d)
Metallic bond have force of attraction
on all sides between the mobile
electrons and the positive kernels.
Metals having free electrons as a
mobile electrons. So, the metallic
bond does not have directional
property.
4
Chemical
Thermodynamics
TOPIC 1
Generalised Terms and First
Law of Thermodynamics
01 Which one among the following is
the correct option for right
relationship between C p and C V for
one mole of ideal gas? [NEET 2021]
(a) C p + C V = R
(c) C p = RC V
Ans. (b)
(b) C p − C V = R
(d) C V = RC p
For an ideal gas,
C p − C V = nR
(where, n = number of moles of gas,
C p , C V = specific heat at constant
pressure and volume, R = universal gas
constant)
As n = 1, so C p − C V = R
02 The correct option for free
expansion of an ideal gas under the
adiabatic condition is
[NEET (Sept.) 2020]
(a) q = 0, ∆T < 0 and w > 0
(b) q < 0, ∆T < 0 and w = 0
(c) q > 0, ∆T < 0 and w > 0
(d) q = 0, ∆T = 0 and w = 0
Ans. (d)
For adiabatic process,q = 0
For adiabatic free expansion of an ideal
gas, no work will be done, because,
dV = 0.
∴
w = pdV = 0
For adiabatic free expansion of an ideal
gas, no change in internal energy.
So, dU = 0 and dT =0
[QdU = nC V dT ]
So, option (d) is correct.
03 Under isothermal condition, a gas
at 300 K expands from 0.1 L to 0.25
L against a constant external
pressure of 2 bar. The work done
by the gas is
(Given that 1 L bar = 100 J)
[NEET (National) 2019]
(a) 5 kJ (b) 25 J
Ans. (d)
(d) −30 J
(c) 30 J
Key Idea For an isothermal irreversible
expansion, Wirrev = − pext (V2 − V1)
Given,V1 = 0.1 L, V2 = 0.25 L, pext = 2 bar
We know that,
Wirrev = − pext (V2 − V1)
On substituting the given values in the
above equation, we get
Wirrev = − 2 bar (0.25 − 0.1) L
= − 2 × 0.15 L bar = − 0.3 L bar
[Q1 L bar = 100 J]
= − 0.3 × 100 J
= − 30 J
04 An ideal gas expands isothermally
from10 −3 m 3 to10 −2 m 3 at 300 K
against a constant pressure of 10 5
Nm −2 . The work done on the gas is
[NEET (Odisha) 2019]
(a) + 270 kJ
(c) + 900 kJ
Ans. (b)
(b) − 900 J
(d) − 900 kJ
For an isothermal irreversible expansion,
Work done (W) = − pext (V2 − V1)
where,V1 = initial volume
V2 = final volume
Given, pext = 10 5 Nm −2 ,
−3
−2
V1 = 10 m , V2 = 10 m
On substituting the given values in Eq, (i),
We get,
3
3
W = −10 5 Nm−2 (10 −2 m3 − 10 −3m3)
= −10 5 Nm−2 × 10 −3 (10 − 1) m3
= −900 Nm = −900 J
05 A gas is allowed to expand in a well
insulated container against a
constant external pressure of 2.5
atm from an initial volume of 2.50 L
to a final volume of 4.50 L. The
change in internal energy ∆U of the
gas in joules will be
[NEET 2017]
(a) 1136.25 J
(c) − 505 J
Ans. (c)
(b) − 500 J
(d) + 505 J
Key concept According to first law of
thermodynamics,
∆U = q + w
where, ∆U = internal energy
q = heat absorbed or evolved,w = work
done.
Also, work done against constant
external pressure (irreversible process).
w = − pext ∆V .
Work done in irreversible process,
w = − pext ∆V = − pext (V2 − V1)
= − 2.5 atm (4.5 L − 2.5 L)
= − 5 L atm = − 5 × 1013
. J
= − 505 J
Since, the system is well insulated,q = 0
∴
∆U = w = − 505 J
Hence, change in internal energy, ∆U of
the gas is − 505 J.
06 Which one of the following is
correct option for free expansion
of an ideal gas under adiabatic
condition?
[CBSE AIPMT 2011]
34
NEET Chapterwise Topicwise Chemistry
(a) q ≠ 0, ∆T = 0, W = 0
(b) q = 0, ∆T = 0, W = 0
(c) q = 0, ∆T < 0, W ≠ 0
(d) q = 0, ∆T ≠ 0, W = 0
Ans. (b)
In adiabatic process, heat exchange is
constant, so q = 0 and for free expanion,
W = 0,∴∆T = 0.
07 Which of the following are not
state functions?
Given that, c = 75 JK–1 mol –1
Ans. (b)
q = 1.0 kJ = 1000 J
Mass = 100 g water
Molar mass of water = 18 g
1000
75 =
5.55 × ∆T
For endothermic reactions standard
heat of reaction (∆H) is positive because
in these reactions total energy of
reactants is lower than that of products,
i.e. E R < E P
So,
∆H = E P − E R = + ve
∴
100


= 5 .55
 Number of moles =


18
1000
∆T =
= 2 .4 K
5.55 × 75
[CBSE AIPMT 2008]
I. q + W
III. W
II. q
IV. H − TS
(a) I and IV
(b) II, III and IV
(c) I, II and III
(d) II and III
Ans. (d)
[CBSE AIPMT 2002]
The thermodynamic parameters which
depend only upon the initial and final
states of system, are called state
functions, such as enthalpy (H = q + W),
Gibbs free energy (G = H − TS ), etc. While
those parameters which depend on the
path by which the process is performed
rather than on the initial and final states,
are called path functions, such as work
done, heat, etc.
08 The work done during the
expansion of a gas from a volume
of 4 dm 3 to 6dm 3 against a
constant external pressure of 3
atm, is
[CBSE AIPMT 2004]
(a) − 6 J
(c) + 304 J
Ans. (b)
(b) – 608 J
(d) – 304 J
(a) ∆ E = W ≠ 0, q = 0
(b) ∆ E = W = 0, q ≠ 0
(c) ∆ E = 0, W = q ≠ 0
(d) W = 0, ∆ E = q ≠ 0
Ans. (a)
09 The molar heat capacity C of water
at constant pressure is 75 JK −1 mol
−1
, when 1.0 kJ of heat is supplied
to 100 g of water which is free to
expand, the increase in
temperature of water is
[CBSE AIPMT 2003]
(b) 6.6 K
(d) 2.4 K
According to heat capacity rule,
q
q = mc∆T , c =
m(T2 − T1)
In closed insulated container a liquid is
stirred with a paddle to increase the
temperature, therefore it behaves as
adiabatic process, so for it q = 0.
Hence, from first law of thermodynamics
∆E =q + W
if, q = 0
∴ ∆ E = W but not equal to zero.
11 When 1 mole gas is heated at
constant volume, temperature is
raised from 298 to 308 K. Heat
supplied to the gas is 500 J. Then,
which statement is correct?
(a) q = W = 500 J, ∆ E = 0
(b) q = ∆ E = 500 J, W = 0
(c) q = − W = 500 J, ∆ E = 0
(d) ∆ E = 0, q = W = − 500 J
Ans. (b)
We know that, ∆ H = ∆ E + p∆V
When,
∆V = 0
∴
∆H = ∆ E
From first law of thermodynamics
∆E =q –W
In given problem ∆H = 500 J
–W = – p∆V , ∆V = 0
So,
∆E = q = 500 J
12 In an endothermic reaction, the
value of ∆H is [CBSE AIPMT 1999]
(a) zero
(c) negative
[CBSE AIPMT 1998]
(a) 163.7 cal
(c) 1381.1 cal
Ans. (b)
(b) zero
(d) 9 L atm
Isothermal process means temperature
remains constant. At constant
temperature, internal energy (∆E) also
remains constant. So, ∆E = 0
[CBSE AIPMT 2001]
Work done ( W) = − pext (V2 − V1)
= − 3 × (6 − 4) = − 6 L atm
= − 6 × 101.32 J (Q 1 L atm = 101.32 J)
= − 607.92 ≈ –608 J
(a) 4.8 K
(c) 1.2 K
Ans. (d)
10 In a closed insulated container a
liquid is stirred with a paddle to
increase the temperature, which of
the following is true?
13 One mole of an ideal gas at 300 K is
expanded isothermally from an
initial volume of 1 L to 10 L. The ∆E
for this process is
(R = 2cal mol–1K –1 )
(b) positive
(d) constant
14 During isothermal expansion of an
ideal gas, its [CBSE AIPMT 1991, 94]
(a) internal energy increases
(b) enthalpy decreases
(c) enthalpy remains unaffected
(d) enthalpy reduces to zero
Ans. (c)
We know that,
H=E+ W
Enthalpy = internal energy + pressure ×
volume
H = E + pV
∆H = ∆E + ∆ (pV )
∆H = ∆E + ∆ (ng RT ) (Q pV = nRT )
For isothermal expansion of ideal gas,
∆T = 0
∴
∆H = ∆E
TOPIC 2
Reactions Related to
Enthalpies and Hess’s Law
15 At standard conditions, if the
change in the enthalpy for the
following reaction is
−109 kJ mol −1 .
H2 (g) + Br 2 (g) → 2HBr(g)
Given that, bond energy of H2 and
Br 2 is 435 kJ mol −1 and 192 kJ mol
−1
respectively, what is the bond
energy (in kJ mol −1 ) of HBr?
[NEET (Oct.) 2020]
(a) 368
(b) 736
(c) 518
(d) 259
35
Chemical Thermodynamics
Ans. (c)
Ans. (a)
H2 (g) + Br2 (g) → 2HBr(g)
[H H]
[Br  Br]
[H  Br]
⇒
∆ r H = (ΣBE)Reactants − (ΣBE) Products
[QBE = bond energy]
⇒− 109 = [(BE)H 2 + (BE) Br2 ] − (BE) HBr × 2
= (435 + 192) − (BE)HBr × 2
⇒ (BE)HBr = 368 kJ mol − 1
16 The bond dissociation energies
of X 2 , Y 2 and XY are in the ratio of
1 : 0.5 : 1 . ∆H for the formation of
XY is −200 kJ mol −1 . The bond
dissociation energy of X 2 will be
[NEET 2018]
(a) 800 kJ mol−1
(c) 200 kJ mol−1
Ans. (a)
(b) 100 kJ mol−1
(d) 400 kJ mol−1
Key Concept Relation between heat of
reaction (∆ r H) and bond energies (BE) of
reactants and products is given by
∆ r H = Σ BEReactants − Σ BE Products
The reaction of formation for XY is
1
1
X2 (g) + Y2 (g) → XY (g);
2
2
∆H = − 200kJ mol −1
Given, the bond dissociation energies of
X2 , Y2 and XY are in the ratio 1 : 0.5 : 1. Let
the bond dissociation energies of X2 , Y2
and XY are a kJ mol −1, 0.5a kJ mol −1 and
a kJ mol −1, respectively.
∴ ∆ r H = ΣBEReactants − ∆ BEProducts
1
1

= × a + × 0.5a − [1 × a]
 2

2
a a
−200 = + − a
2 4
2a + a − 4a −a
−200 =
=
4
4
a = 800 kJ mol −1
∴ The bond dissociation energy of
X2 = a kJ mol −1 = 800 kJ mol −1
The given phase equilibria is
Liquid r Vapour
This equilibrium states that, when liquid
is heated, it converts into vapour but on
cooling, it further converts into liquid,
which is derived by Clausius Clapeyron
and the relationship is written as,
d ln p
∆H
= − v2
dT
RT
where, ∆Hv = Heat of vaporisation
18 The heat of combustion of carbon
to CO 2 is −393.5kJ / mol. The heat
released upon the formation of 35.2
g of CO 2 from carbon and oxygen
gas is
[CBSE AIPMT 2015]
(a) −315 kJ
(c) −630 kJ
Ans. (a)
(b) +315 kJ
(d) −315
. kJ
Given, C( s ) + O2 ( g ) → CO2 ( g ) ;
∆fH = − 393.5 kJ mol − 1
Q Heat released on formation of 44 g or 1
mole
CO2 = −395.5kJ mol
Q Heat released on formation of 35.2 g
of CO2
− 393.5 kJ mol − 1
=
× 35.2 g
44g
= − 315 kJ mol −1
19 A reaction having equal energies of
activation for forward and reverse
reactions has
[NEET 2013]
(a) ∆S = 0
(b) ∆G = 0
(c) ∆ H = 0
(d) ∆ H = ∆G = ∆S = 0
Ans. (c)
Energy profile diagram for a reaction is
as
17 Consider the following
liquid-vapour equilibrium
Energy
[NEET 2016, Phase I]
Liquid
Vapour
Which of the following relations is
correct?
!
dlnP − ∆Hv
=
dT
RT
dlnP − ∆Hv
(b)
=
dT 2
T2
dlnP − ∆Hv
(c)
=
dT
RT 2
dln G − ∆Hv
(d)
=
dT 2
RT 2
(a)
(Ea)f
(Ea)b
∆H
Progress of the reaction
From the figure, it is clear that
(E a ) b = (E a ) f + ∆H
[Here (E a ) b = activation energy of
backward reaction and (E a )f = activation
energy of forward reaction].
If
then
(E a ) b = (E a ) f
∆H = 0
20 Standard enthalpy of vaporisation
∆ vap H ° for water at 100°C is 40.66
kJ mol −1 . The internal energy of
vaporisation of water at 100°C
(in kJ mol −1 ) is (assume water
vapour to behave like an ideal gas).
[CBSE AIPMT 2012]
(a) + 37.56
(c) + 43.76
Ans. (a)
(b) − 43.76
(d) + 40.66
100 ° C
H2O(l ) →
H2O(g)
∆ vapH ° = ∆ vap E ° + ∆ng RT
∆ vapH ° = enthalpy of vaporisation
= 40.66 kJ mol − 1
For the above reaction,
∆ng = np − nr = 1 − 0 = 1
R = 8314
.
T = 100 ° C = 273 + 100 = 373 K
∴ 40.66 kJ mol −1 = ∆ vap E ° + 1 × 8.314
∆ vap
E°
× 10 −3 × 373
= 40.66 kJ mol − 3.1 kJ mol −1
= + 37.56 kJ mol −1
−1
21 Enthalpy change for the reaction,
4H(g) → 2H2 (g) is – 869.6 kJ
The dissociation energy of H—H
bond is
[CBSE AIPMT 2011]
(a) – 869.6 kJ
(c) + 217.4 kJ
Ans. (b)
(b) + 434.8 kJ
(d) – 434.8 kJ
4H(g) → 2 H2 (g) , ∆H = − 869.6 kJ
2H2 (g) → 4H(g) , ∆H = 869.6 kJ
H2 (g) → 2 H(g) ,
Dissociation energy of H—H bond
869.6
=
= 434.8 kJ
2
22 From the following bond energies
H—H bond energy : 431.37 kJ mol −1
C==C bond energy:606.10 kJ mol −1
C—C bond energy : 336.49 kJ mol −1
C—H bond energy : 410.50 kJ mol −1
Enthalpy for the reaction,
H
H
H H


 
C == C + H — H → H —C—C— H


 
H
H
H H
will be
[CBSE AIPMT 2009]
(a) 1523.6 kJ mol −1
(b) –243.6 kJ mol −1
(c) –120.0 kJ mol −1
(d) 553.0 kJ mol −1
36
NEET Chapterwise Topicwise Chemistry
Ans. (c)
For reaction,
H H
H H


 
C == C + H — H → H — C— C— H


 
H H
H H
∆Hreaction
= ∑ BE (reactant) − ∑ BE (product) ,
[BE = bond energy]
∆Hr = [4 × BE (C—H) + 1 × BE (C= =C) + 1
× BE (H—H) ] − [6 × BE (C —H) + 1 × BE (C  C) ]
= (4 × 410.50 + 1 × 606.10 + 1 × 431.37)
− [(6 × 410.50) + (1 × 336.49)] kJmol −1
= [1642 + 606.1 + 43137
. ]
− [2463 + 33649
. ] kJmol −1
= [267947
. ] − [279949
. ] kJmol −1
−1
= − 120.0 kJ mol
23 Bond dissociation enthalpy of H2 ,
Cl 2 and HCl are 434, 242 and 431 kJ
mol −1 respectively. Enthalpy of
formation of HCl is
[CBSE AIPMT 2008]
(a) 93 kJ mol −1
(c) –93 kJ mol −1
Ans. (c)
Given,
(b) –245 kJ mol −1
(d) 245 kJ mol −1
∆HH H = 434 kJ / mol
∆HCl  Cl = 242 kJ / mol
∆HH  Cl = 431 kJ / mol
1
1
H2 + Cl2 → HCl , ∆Hr = ?
2
2
1
1
∆Hr = × ∆HH H +
2
2
=
× ∆HCl  Cl − ∆HH  Cl
1
1
× 434 + × 242 − 431
2
2
= 217 + 121 − 431 = − 93 kJ / mol
24 Given that bond energies of H—H
and Cl—Cl are 430 kJ mol −1 and 240
kJ mol −1 respectively and ∆Hf for
HCl is –90 kJ mol −1 . Bond enthalpy
of HCl is
[CBSE AIPMT 2007]
(a) 290 kJ mol−1
(b) 380 kJ mol−1
(c) 425 kJ mol−1
(d) 245 kJ mol−1
Ans. (b)
∆Hreaction = ∆H—H + ∆HCl — Cl − 2∆HHCl
or
430 + 240 − (–90)
2
760
=
= 380 kJ mol −1
2
∆HH— Cl =
25 Consider the following reactions,
(i) H+ (aq) + OH− (aq) →H 2O(l),
− x 1kJmol −1
1
(ii) H 2 (g) + O 2 (g) → H 2O(l),
2
x 2 kJ mol–1
(iii) CO 2 (g) + H 2 (g) → CO(g) + H 2O(l)
− x 3 kJ mol −1
5
(iv) C 2H2 (g) + O 2 (g) → 2 CO 2 (g)
2
+ H 2O(l), + x 4 kJ mol −1
Enthalpy of formation of H2O(l) is
[CBSE AIPMT 2007]
(a) − x2 kJ mol−1
(c) − x4 kJ mol−1
Ans. (a)
(b) + x3 kJ mol−1
(d) + x1 kJ mol−1
Enthalpy of formation The amount of
heat evolved or absorbed during the
formation of 1 mole of a compound from
its constituent elements is known as
heat of formation. So, the correct
answer is
1
H2 ( g) + O2 ( g) → H2O(l ),
2
∆H = − x2 kJ mol −1
26 The enthalpy of combustion of H2 ,
cyclohexene (C 6H10 ) and
cyclohexane (C 6H12 ) are –241,
–3800 and –3920 kJ per mol
respectively. Heat of hydrogenation
of cyclohexene is
[CBSE AIPMT 2006]
(a) –121 kJ per mol (b) + 121 kJ per mol
(c) + 242 kJ per mol(d) –242 kJ per mol
Ans. (a)
Ans. (a)
As we know that,
∆ H = ∆ E + p∆ V
or
∆ H = ∆ E + ∆ ng RT
where, ∆ng → number of gaseous moles
of product – number of gaseous moles of
reactant
If ∆ng = 0 (for reactions in which the total
number of moles of gaseous products
are equal to total number of moles of
gaseous reactants), therefore ∆ H = ∆ E
So, for reaction (a) ∆ n = 2 − 2 = 0
Hence, for reaction (a), ∆ H = ∆E
28 The absolute enthalpy of
neutralisation of the reaction
[CBSE AIPMT 2005]
MgO(s) +2HCl(aq) → MgCl 2 (aq)
+ H2O( l ) will be
(a) less than – 57.33 kJ mol–1
(b) – 57.33 kJ mol–1
(c) greater than –57.33 kJ mol–1
(d) 57.33 kJ mol–1
Ans. (a)
Heat of neutralisation of strong acid and
strong base is − 57.33 kJ. MgO is weak
base while HCl is strong acid, so the heat
of neutralisation of MgO and HCl is lower
than –57.33 kJ because MgO requires
some heat for ionisation, therefore the
net released amount of heat is
decreased.
29 If the bond energies of
H — H, Br — Br and H — Br are 433,
192 and 364 kJ mol −1 respectively,
then ∆ H° for the reaction
H2 (g) + Br 2 (g) → 2HBr (g) is
[CBSE AIPMT 2004]
+ H2
Cyclohexene
Cyclohexane
∆H = [∆H of combustion of cyclohexane
– (∆H of combustion of cyclohexene
+ ∆H of combustion of H2 )]
= − [− 3920 − (−3800 − 241)] kJ
= − [−3920 + 4041] kJ
= − [121] kJ = − 121kJ
27 Assume each reaction is carried
out in an open container. For which
reaction will ∆ H = ∆ E?
[CBSE AIPMT 2006]
(a) H2 (g ) + Br2 (g ) → 2HBr (g )
(b) C ( s ) + 2 H2O ( g) → 2 H2 ( g) + CO2 ( g)
(c) PCl5 (g) → PCl3 (g) + Cl2 ( g)
(d) 2 CO ( g) + O2 ( g) → 2 CO2 ( g)
(a) − 261kJ
(c) +261kJ
Ans. (d)
(b) +103kJ
(d) −103kJ
For reaction,
H2 ( g) + Br2 ( g) → 2HBr( g) ∆H ° = ?
∆H° = − [(2 × bond energy of HBr) – (bond
energy of H2 + bond energy of Cl2 )]
∆H° = − [2 × (364) − (433 + 192)] kJ
= − [728 − (625)] kJ = − 103 kJ
30 For the reaction,
C 3H8 ( g) + 5O 2 ( g) → 3CO 2 ( g)
+ 4H2O(l)
at constant temperature, ∆H − ∆E
is
[CBSE AIPMT 2003]
(a) +3RT
(c) + RT
(b) −RT
(d) −3RT
37
Chemical Thermodynamics
Ans. (d)
Ans. (b)
For the reaction,
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O(l )
∆ng = number of gaseous moles of
products – number of gaseous moles of
reactants = 3 − 6 = − 3
∴
∆ H = ∆ E + ∆nRT
or
∆ H − ∆ E = ∆nRT
∴
∆ H − ∆ E = − 3 RT
1
CH4 ( g) + O2 ( g ) → CH3OH(l )
2
∴ ∆H = –[(∆Hc of CH3OH) – (∆Hc of CH4 )]
= –[(–y) – (–x)] = – [–y + x] = y – x
∴
x< y
31 For which one of the following
equations ∆H °r equal to ∆Hf° for
the product? [CBSE AIPMT 2003]
(a) Xe( g) + 2F2 ( g) → XeF4 ( g)
(b) 2CO (g) + O2 (g) → 2CO2 ( g)
(c) N2 ( g) + O 3 ( g) → N2O 3 ( g)
(d) CH4 ( g) + 2Cl2 (g) →
CH2Cl2 (l ) + 2HCl ( g)
Ans. (a)
When one mole of a substance is directly
formed from its constituent elements,
then the enthalpy change is called heat
of formation.
For the reaction,
Xe( g) +2F2 ( g) → XeF4 ( g)
1mol
∆H ° react = ∆Hf°
32 Heat of combustion ∆H° for C(s),
H2 (g) and CH4 (g) are –94, –68 and
–213 kcal/mol. Then, ∆H° for
C(s) +2H2 (g) → CH4 (g) is
[CBSE AIPMT 2002]
(a) – 17 kcal/mol
(c) – 170 kcal/mol
Ans. (a)
(b) – 111 kcal/mol
(d) – 85 kcal/mol
For reaction,
C(s ) + 2H2 ( g) → CH4 ( g), ∆H ° = ?
...(i)
C + O2 → CO2 , ∆H = −94kcal
2H2 + O2 → 2H2O,
∆H = − 68 × 2kcal ...(ii)
CH4 + 2O2 → CO2 + 2H2O,
∆H = −213 kcal ...(iii)
On adding Eqs. (i) and (ii) and then
subtracting Eq. (iii)
= (−94) + (−2 × 68) − (−213)
= −230 + 213 = −17 k cal / mol
34 Change in enthalpy for reaction,
2H2O 2 (l) → 2H2O(l) + O 2 (g)
if heat of formation ofH2O 2 (l) and
H2O(l) are –188 and –286 kJ/mol
respectively is [CBSE AIPMT 2001]
(a) – 196 kJ/mol
(c) + 948 kJ/mol
Ans. (a)
(b) + 196 kJ/mol
(d) – 948 kJ/mol
2H2O2 (l ) → 2H2O(l ) + O2 ( g) ∆H = ?
∆H = [(2 × ∆Hf of H2O(l )) + (∆ Hf of O2 (g)]
–(2 × ∆Hf of H2O2 (l ))]
= [(2 × –286) + (0) –(2 × – 188)]
= [–572 + 376] = –196 kJ / mol
35 If ∆E is the heat of reaction for
C 2H5OH(l) + 3O2 (g) → 2CO2 (g)
+ 3H2O(l) at constant volume, the
∆H (heat of reaction at constant
pressure), then the correct
relation is
[CBSE AIPMT 2000]
(a) ∆H = ∆ E + RT
(b) ∆H = ∆ E − RT
(c) ∆H = ∆ E − 2 RT
(d) ∆H = ∆ E + 2 RT
Ans. (b)
We know that, ∆H = ∆E + ∆ng RT
where, ∆ng = total number of moles of
gaseous product – total number of moles
of gaseous reactant
= 2 – 3 = –1
So,
∆ H = ∆ E – RT
36 From the given reactions,
3
S(s) + O 2 (g) → SO 3 (g) + 2x kcal
2
1
SO 2 (g) + O 2 (g) → SO 3 (g) + y kcal,t
2
he heat of formation ofSO 2 is
[CBSE AIPMT 1999]
33 Enthalpy of the reaction,
1
CH4 + O 2 → CH3OH, is
2
negative. If enthalpy of combustion
of CH4 and CH3 OH are x and y
respectively, then which relation is
correct?
[CBSE AIPMT 2001]
(a) x > y (b) x < y (c) x = y (d) x ≥ y
(a) (x + y)
(c) (2x + y)
Ans. (d)
(b) (x − y)
(d) (2x − y)
3
S(s ) + O2 ( g) → SO3 ( g) +2x kcal …(i)
2
By inverting second equation we get,
1
SO3 (g) → SO2 ( g) + O2 ( g) – y kcal …(ii)
2
On addition Eqs. (i) and (ii)
S(s ) + O2 ( g) → SO2 ( g) + (2 x − y) kcal
Hence, heat of formation ofSO2 is (2x − y)
kcal.
37 Given that,
C (s) +O 2 (g) → CO 2 (g) ,
∆H ° = − x kJ
2CO(g) +O 2 (g) → 2CO 2 (g) ,
∆H ° = − y kJ
The enthalpy of formation of
carbon monoxide will be
[CBSE AIPMT 1997]
(a) y − 2 x (b) 2x − y (c)
y − 2x
2x − y
(d)
2
2
Ans. (c)
C+ O2 → CO2 , ∆H ° = − x kJ ...(i)
On reversing given second equation we
get,
2CO2 → 2CO + O2 ,
∆H ° = + y kJ
CO2 → CO+ 1 / 2O2 ,
∆H ° = + y /2kJ ...(ii)
From Eqs. (i) and (ii) (by addition)
1
C+ O2 → CO,
2
y
y − 2x
kJ
∆H ° = − x =
2
2
or
38 If enthalpies of formation of
C 2H4 (g), CO 2 (g) and H2O(l) at 25°C
and 1 atm pressure are 52, –394
and –286 kJ/mol, the enthalpy of
combustion of ethene is equal to
[CBSE AIPMT 1995]
(a) –141.2 kJ/mol
(c) +14.2 kJ/mol
Ans. (b)
(b) –1412 kJ/mol
(d) +1412 kJ/mol
Combustion of hydrocarbon,
C2H4 (g) + 3O2 (g) → 2CO2 (g) + 2H2O(l )
∆ r H = Σ∆fHp − Σ∆fHr
= {2 × ∆fH (CO2 ) + 2 × ∆fH (H2O)}
− { ∆fH (C2H4 ) + 3∆fH(O2 )}
= (2 × − 394 + 2 × − 286) − (52 + 3 × 0)
= − 788 − 572 − 52 = − 1412kJ mol −1
39 For the reaction, [CBSE AIPMT 1991]
N 2 + 3H2
2NH3 , ∆H = ?
(a) ∆E + 2 RT
(c) ∆H = RT
Ans. (b)
3
(b) ∆E − 2 RT
(d) ∆E − RT
According to enthalpy equation
∆H = ∆E + ∆ng RT
Enthalpy
change
Internal
energy
38
NEET Chapterwise Topicwise Chemistry
∆n = 2 − (1 + 3)
[∆n = nP
∴
= 2 − 4 = −2
∆H = ∆E + (− 2) RT
∆H = ∆E − 2 RT
Product
mole
− nR ]
Reactant
mole
40 Equal volumes of molar
hydrochloric acid and sulphuric
acid are neutralised by dilute NaOH
solution and x kcal and y kcal of
heat are liberated respectively.
Which of the following is true?
(a) x = y
[CBSE AIPMT 1991]
1
(b) x = y
2
(c) x = 2 y
(d) None of the above
Ans. (b)
HCl + NaOH → NaCl + H2O + x kcal
H2SO4 + 2NaOH → Na2SO4 + 2H2O
+ y kcal
1 molar HCl = 1 g - equivalent of HCl
1 molar H2SO4 = 2 g - equivalent of H2SO4
1
so,
y =2x ⇒ x = y
2
41 If ∆H is the change in enthalpy and
∆E, the change in internal energy
accompaning a gaseous reaction,
then
[CBSE AIPMT 1990]
(a) ∆ H is always greater than ∆ E
(b) ∆ H < ∆ E only if the number of moles
of products is greater than the
number of moles of the reactants
(c) ∆H is always less than ∆ E
(d) ∆ H < ∆ E only if the number of moles
of products is less than the number
of moles of the reactants
Ans. (d)
Reactions in which there is a decrease in
the number of moles of the gaseous
components,
i.e. ∆ng is negative, the enthalpy change
(∆H) is lesser than the internal energy
change (∆E).
Reaction in which there is a increase in
the number of moles of gaseous
components i.e. ∆ng is positive, the
enthalpy change is greater than the
internal energy change.
∆H = ∆E + ∆ng RT
TOPIC 3
Entropy, Free Energy
Change and Spontaneity
42 For irreversible expansion of an
ideal gas under isothermal
condition the correct option is
[NEET 2021]
(a) ∆U = 0, ∆S total = 0
(b) ∆U ≠ 0, ∆S total ≠ 0
(c) ∆U = 0, ∆S total ≠ 0
(d) ∆U ≠ 0, ∆S total = 0
Ans. (c)
The change in internal energy depends
on the temperature. For isothermal
process, ∆T = 0.
So,
∆U = 0.
With an expansion of an ideal gas, more
space is available for the gaseous
particles.
∴Entropy of gas increases so, entropy of
system is not zero.
i.e.
∆S ≠ 0
43 If for a certain reaction ∆ r H is 30
kJ mol −1 at 450 K, the value of ∆ r S
(in JK −1 mol −1 ) for which the same
reaction will be spontaneous at the
same temperature is
[NEET (Oct.) 2020]
(b) −33
(d) −70
(a) 70
(c) 33
Ans. (a)
∆G = ∆H − T∆S (Gibbs equation)
For a spontaneous reaction,
∆G < 0, i.e.T∆S > ∆H
∆H 30 × 1000 J mol − 1
=
⇒ T>
∆S
∆S
[Given,T = 450 K]
30 × 1000 J mol − 1
⇒∆S >
450 K
−1
−1
= 6667
. J mol K
When ∆S > 6667
. J mol − 1 K− 1, the
reaction will be spontaneous.
So, from the options,
∆ r S = 70 J mol − 1K− 1 indicates
spontaneity of the reaction.
44 Hydrolysis of sucrose is given by
the following reaction.
Sucrose + H2O
Glucose +
Fructose
If the equilibrium constant (K C ) is
2 × 10 13 at 300 K, the value of ∆ r G È
at the same temperature will be
c
[NEET (Sept.) 2020]
(a) 8.314 J mol −1 K −1 × 300 K ×ln(2 × 10 13)
(b) 8.314 J mol −1 K −1 × 300 K ×ln(3 × 10 13)
(c) − 8.314 J mol −1 K −1 × 300 K ×ln(4 × 10 13)
(d) − 8.314 J mol −1 K −1 × 300 K ×ln(2 × 10 13)
Ans. (d)
Sucrose + H2O r Glucose + Fructose
For this equilibrium,
∆ r G s = − RT ln K C
Given,K C = 2 × 10 13 andT = 300 K
⇒∆ r G s = − (8.314 J mol − 1K− 1) × (300 K) ×
ln (2 × 10 13)
45 For the reaction, 2Cl(g) → Cl 2 (g),
the correct option is
[NEET (Sep.) 2020]
(a) ∆ r H > 0 and ∆ r S < 0
(b) ∆ r H < 0 and ∆ r S > 0
(c) ∆ r H < 0 and ∆ r S < 0
(d) ∆ r H > 0 and ∆ r S > 0
Ans. (c)
In this reaction, combination of two Cl
atoms takes place to give more stable
Cl2 molecule. So, the reaction is
exothermic, i.e, ∆ r H < 0.
Here, ∆ng = 1 − 2 = − 1
So, entropy change of the reaction will
be, ∆ r S < 0.
46 Reversible expansion of an ideal
gas under isothermal and adiabatic
conditions are as shown in the
figure.
[NEET (Odisha) 2019]
A ( pA, VA, TA)
B ( pB, VB, TB)
p
C ( pC, VC, TC)
V
AB → Isothermal expansion
AC → Adiabatic expansion
39
Chemical Thermodynamics
Which of the following option is
not correct?
(a) ∆S isothermal > ∆S adiabatic
(b)TA = TB
(c) Wisothermal > Wadiabatic
(d) Tc > TA
Ans. (d)
From first law of thermodynamics,
∆U = q + W
In adiabatic expansion,q = 0
∴
∆U = W
During expansion of a gasw is negative
i.e w < 0 or ∆U < 0.
We know that, ∆U = nC V ∆T
∴ nC V ∆T < 0
or
∆T < 0
∴
TC −TA < 0
or
TC < TA
Thus, option (d) is incorrect while the
remaining options are correct.
47 For an ideal solution, the correct
option is
[NEET (National) 2019]
(a) ∆ mix V ≠ 0 at constantT and p
(b) ∆ mix H = 0 at constantT and p
(c) ∆ mix G = 0 at constantT and p
(d) ∆ mix S = 0 at constantT and p
Ans. (b)
Ideal solutions are those which obey
Raoult’s law over all concentration
ranges at a given temperature, e.g.
benzene-toluene, n-hexane-n-heptane,
etc.
For an ideal solution,
∆Vmix = 0, ∆Hmix = 0,
∆G mix < 0, ∆S mix > 0.
Hence, option (b) is correct.
(c) Entropy is negative in case of
2H(g) → H2 (g) as the number of
moles of gaseous reactants are more
than that of gaseous products.
(d) Entropy is positive in case of
evaporation of water as gas is much
disordered than a liquid.
Hence, option (c) is correct.
49 For a given reaction, ∆H = 35.5 kJ
mol −1 and ∆S = 83.6 JK −1 mol −1 . The
reaction is spontaneous at :
(Assume that ∆H and ∆S do not
vary with temperature) [NEET 2017]
(a)T < 425 K
(b)T > 425 K
(c) all temperatures
(d)T > 298 K
Ans. (b)
According to Gibbs-Helmholtz equation,
Gibbs energy (∆G) = ∆H − T∆S
Where,
∆H = Enthalpy change
∆S = Entropy change
T = Temperature
For a reaction to be spontaneous
∆G < 0.
∴ Gibbs -Helmholtz equation becomes,
∆G = ∆H − T∆S < 0
or,
∆H < T∆S
∆H 35.5 kJ mol −1
or,
=
T>
∆S 83.6 JK−1mol −1
35.5 × 1000
= 425 K
=
836
.
T > 425 K
50 The correct thermodynamic
conditions for the spontaneous
reaction at all temperatures is
[NEET 2016, Phase I]
48 In which case change in entropy is
negative?
[NEET (National) 2019]
(a) Expansion of a gas at constant
temperature
(b) Sublimation of solid to gas
(c) 2H(g )→ H2 (g )
(d) Evaporation of water
Ans. (c)
The explanation of given statements are :
(a) Entropy is positive in case of
expansion of a gas at constant
temperature. It is because during
expansion of gas volume increases
and hence randomness increases.
(b) Entropy is positive in case of
sublimation of solid to gas as gas is
much disordered than a solid.
(a) ∆H > 0 and ∆S < 0
(b) ∆H < 0 and ∆S > 0
(c) ∆H < 0 and ∆S < 0
(d) ∆H < 0 and ∆S = 0
Ans. (b,d)
We have the Gibbs Helmholtz reaction
for spontaneity as
∆G = ∆H − T∆S
For reaction to be spontaneous, ∆G must
be negative.
For this, ∆H should be negative and ∆S
should be positive.
∴
∆H < 0
and
∆S > 0.
and also ∆S = 0 shows ∆G a negative
quantity.
51 For a sample of perfect gas when
its pressure is changed
isothermally from pi to pf , the
entropy change is given by
[NEET 2016, Phase II]
p 
(a) ∆S = nR ln  f 
 pi 
p 
(b) ∆S = nR ln  i 
 pf 
p 
(c) ∆S = nRT ln  f 
 pi 
 pi 
(d) ∆S = RT ln  
 pf 
Ans. (b)
Entropy change is given as,
T
p
∆S = nC p ln f + nR ln i
pf
Ti
K(i)
For isothermal process,Ti = Tf
T
T
nC p ln f = nC p ln i = 0 [ln 1 = 0]
∴
Ti
Ti
From Eq. (i)
∆S = nR ln
pi
pf
52 Which of the following statements
is correct for a reversible process
in a state of equilibrium?
[CBSE AIPMT 2015]
(a) ∆G = – 2.30 RT log K
(b) ∆G = 2.30 RT log K
(c) ∆G ° = – 2.30 RT log K
(d) ∆G ° = 2.30 RT log K
Ans. (a)
Mathematical expression of the
thermodynamic equilibrium is
∆G = ∆G ° + 2 .303RT log Q
At equilibrium when ∆G = 0 and Q = K
then ∆G = ∆G ° + 2.303 RT log K = 0
∆G ° = − 2.303 RT log K
53 Which of the following statements
is correct for the spontaneous
absorption of a gas?
[CBSE AIPMT 2014]
(a) ∆S is negative and therefore, ∆H
should be highly positive
(b) ∆S is negative and therefore, ∆H
should be highly negative
(c) ∆S is positive and therefore, ∆H
should be negative
(d) ∆S is positive and therefore, ∆H
should also be highly positive
40
NEET Chapterwise Topicwise Chemistry
Ans. (b)
∆S [change in entropy] and ∆H [change
in enthalpy] are related by the equation
∆G = ∆H − T ∆S
[Here, ∆G = change in Gibbs free energy]
For adsorption of a gas, ∆S is negative
because randomness decreases. Thus,
in order to make ∆G negative [for
spontaneous reaction], ∆H must be
highly negative. Hence for the
adsorption of a gas, if ∆S is negative,
therefore, ∆H should be highly negative.
54 For the reaction,
X 2 O 4 (l) → 2XO 2 (g),
∆U = 2.1kcal,
∆S = 20cal K −1 at 300 K.
Hence, ∆G is
[CBSE AIPMT 2014]
(a) 2.7 kcal
(c) 9.3 kcal
Ans. (b)
(b) − 2.7 kcal
(d) − 9.3 kcal
The change in Gibbs free energy is given
by
∆G = ∆H − T∆S
where, ∆H = change enthalpy of the
reaction
∆S = change entropy of the reaction
Thus, in order to determine ∆G, the values
of ∆H must be known. The value of ∆H can
be calculated by using equation
∆H = ∆U + ∆ng RT
K(i)
where, ∆U = change in internal
energy using
∆ng = number of moles of gaseous
products − number of moles of gaseous
reactants
= 2−0 = 2
R = gas constant = 2 cal
given, ∆U = 2.1 kcal
= 2.1 × 10 3 cal
[Q1 kcal = 10 3 cal]
By putting the values in eq. (i) we get,
∴
∆H = (2.1 × 10 3) + (2 × 2 × 300)
= 3300 cal
Hence,
∆G = ∆H − T∆S
⇒
∆G = (3300) − (300 × 20)
∴
∆G = −2700 cal
∆G = − 2.7 kcal
55 In which of the following reactions,
standard reaction entropy changes
(∆S° ) is positive and standard Gibbs
energy change (∆G° ) decreases
sharply with increasing
temperature? [CBSE AIPMT 2012]
1
(a) C (graphite) + O2 ( g) → CO ( g)
2
1
(b) CO ( g) + O2 ( g) → CO2 ( g)
2
1
(c) Mg (s ) + O2 ( g) → MgO (s )
2
1
1
(d) C (graphite) + O2 ( g) →
2
2
1
CO2 ( g)
2
Ans. (a)
Among the given reactions only in the
case of
1
C (graphite) + O2 ( g) → CO ( g)
2
entropy increases because randomness
(disorder) increases. Thus, standard
entropy change (∆S ° ) is positive.
Moreover, it is a combustion reaction
and all the combustion reactions are
generally exothermic, i.e. ∆H° = −ve
We know that,
∆G ° = ∆H ° − T∆S °
∆G ° = − ve − T (+ ve)
Thus, as the temperature increases, the
value of ∆G ° decreases.
56 The enthalpy of fusion of water is
1.435 kcal/mol. The molar entropy
change for the melting of ice at 0°C
is
[CBSE AIPMT 2012]
(a) 10.52 cal/mol K (b) 21.04 cal/mol K
(c) 5.260 cal/mol K (d) 0.526 cal/mol K
Ans. (c)
Molar entropy change for the melting of
ice,
∆Hfusion 1435
.
kcal / mol
∆S melt =
=
T
(0 + 273) K
= 526
. × 10 −3 kcal/mol K
= 526
. cal/mol K
57 If the enthalpy change for the
transition of liquid water to steam
is 30 kJ mol −1 at 27°C, the entropy
change for the process would be
[CBSE AIPMT 2011]
(a) 1.0 J mol −1K −1 (b) 0.1 J mol −1K −1
(c) 100 J mol −1K −1 (d) 10 J mol −1K −1
Ans. (c)
∆G ° = ∆H ° − T∆S °
Given, ∆Hvap. = 30kJ mol −1
T = 27 + 273 = 300k
∆G ° = 0 at equilibrium,
∆Hvap 30 × 10 3 J mol –1
∆S vap =
=
T
300 K
= 100 J mol –1 K–1
58 The values of ∆ H and ∆S for the
reaction,
C (graphite) + CO 2 (g) → 2CO(g) are
170 kJ and 170 JK −1 , respectively.
This reaction will be spontaneous
at
[CBSE AIPMT 2009]
(a) 710 K
(c) 1110 K
Ans. (c)
(b) 910 K
(d) 510 K
Given, ∆H = 170 kJ = 170 × 10 3 J
∆S = 170 JK−1 ; T = ?
∆G = ∆H –T∆S
For spontaneous reaction,
∆G < 0
⇒
0 < 170 × 10 3 − T × 170 ; T > 1000
∴ T = 1110 K
59 For the gas phase reaction,
PCl 5 (g)
PCl 3 (g) + Cl 2 (g)
which of the following conditions
are correct? [CBSE AIPMT 2008]
º
(a) ∆H = 0 and ∆S < 0
(b) ∆H > 0 and ∆S > 0
(c) ∆H < 0 and ∆S < 0
(d) ∆H > 0 and ∆S < 0
Ans. (b)
From enthalpy equation,
∆H = ∆E + ng RT
For the reaction,
PCl 5 ( g)
PCl 3 ( g) + Cl2 ( g)
∆ng = product mole − reactant mole
∆n = 2 − 1 = 1
Thus, the value of ∆H is positive or > 0.
∆G = ∆H − T∆S
For a spontaneous reaction, ∆G must be
negative. Since in this reaction ∆H is
positive, so for the negative value of ∆G,
∆S must be positive or > 0.
Hence,
∆H > 0, ∆S > 0
3
60 The enthalpy and entropy change
for the reaction,
Br 2 (l) + Cl 2 (g) → 2BrCl (g)
are 30 k J mol–1 and 105 JK –1 mol–1
respectively. The temperature at
which the reaction will be in
equilibrium is [CBSE AIPMT 2006]
(a) 285.7 K
(c) 450 K
Ans. (a)
(b) 273 K
(d) 300 K
At equilibrium Gibbs free energy change (
∆G) is equal to zero.
∆G = ∆H − T∆S
41
Chemical Thermodynamics
0 = 30 × 10 3 (J mol –1)
− T × 105 (J K–1 mol –1)
3
30 × 10
K = 285.71 K
∴ T=
105
61 Identify the correct statement for
change of Gibbs free energy for a
system (∆G system ) at constant
temperature and pressure.
[CBSE AIPMT 2006]
(a) If ∆G system > 0, the process is
spontaneous
(b) If ∆G system = 0, the system has
attained equilibrium
(c) If ∆G system = 0, the system is still
moving in a particular direction
(d) If ∆G system < 0, the process is not
spontaneous
Ans. (b)
If the Gibbs free energy for a system (
∆G system ) is equal to zero, then system is
present in equilibrium at a constant
temperature and pressure.
∆G = ∆ r G o + RT ln K ; ∆G = 0
∆ r G o = − RT ln K
K = equilibrium constant
If ∆G system < 0, then the process is
spontaneous
62 Which of the following pairs of a
chemical reaction is certain to
result in a spontaneous reaction?
[CBSE AIPMT 2005]
(a) Exothermic and decreasing disorder
(b) Endothermic and increasing disorder
(c) Exothermic and increasing disorder
(d) Endothermic and decreasing disorder
Ans. (c)
If reaction is exothermic, therefore ∆H is
negative and on increasing disorder, ∆S
is positive thus, at these condition, ∆G is
negative according to following
equation.
∆G = ∆H − T∆S
∆G = negative, and for spontaneous
reaction ∆G must be negative.
following equation (Gibbs-Helmholtz
equation)
∆G = ∆ H − T∆ S
If the magnitude of ∆ H − T∆ S is negative,
then the reaction is spontaneous.
whenT∆ S > ∆ H or we can say that ∆H
and ∆ S are positive, then ∆G is
negative.
64 Considering entropy (S) as a
thermodynamic parameter, the
criterion for the spontaneity of any
process is
[CBSE AIPMT 2004]
(a) ∆S system + ∆S surrounding >0
(b) ∆S system – ∆S surrounding >0
(c) ∆S system >0
(d) ∆S surrounding >0
Ans. (a)
For spontaneous process, ∆S must be
positive. In reversible process
∆ S system + ∆ S surrounding = 0
Hence, system is present in equilibrium.
(i.e. it is not spontaneous process)
While in irreversible process
∆S system + ∆S surrounding > 0
Hence, in the process ∆ S is positive.
65 Standard enthalpy and standard
entropy changes for the oxidation
of ammonia at 298 K are
−382.64 kJ mol −1 and
−145.6 JK –1mol–1 , respectively.
Standard Gibbs energy change for
the same reaction at 298 K is
[CBSE AIPMT 2004]
(a) − 221.1 kJmol–1 (b) − 339.3 kJ mol–1
(c) − 439.3 kJ mol–1 (d) − 523.2 kJ mol–1
Ans. (b)
∆G ° = ∆H ° − T∆S °
K(i)
Given that, ∆H ° = − 382 .64 kJ mol –1
∆S ° = − 1456
. J K–1 mol –1
= − 145.6 × 10 −3 kJ K–1
T = 298 K
On putting the given values in eq. (i)
we get,
or
63 A reaction occurs spontaneously if
[CBSE AIPMT 2005]
(a)T∆S < ∆H and both ∆H and ∆S are +ve
(b)T∆S > ∆H and both ∆H and ∆S are +ve
(c)T∆S = ∆H and both ∆H and ∆S are +ve
(d)T∆S > ∆H and ∆H is +ve and ∆S is –ve
Ans. (b)
The spontaneity of a reaction is based
upon the negative value of ∆G and ∆G is
based upon T, ∆S and ∆H according to
∆G ° = − 382 .64
− [298 × (− 145.6 × 10 −3)]
= − 339.3 kJ mol –1
66 What is the entropy change (in
J K −1 mol −1 ) when one mole of ice
is converted into water at 0°C?
(The enthalpy change for the
conversion of ice to liquid water is
6.0 kJ mol −1 at 0°C)
[CBSE AIPMT 2003]
(a) 2.198 JK−1mol −1 (b) 21.98 JK−1mol −1
(c) 20.13 JK−1mol −1 (d) 2.013 JK−1mol −1
Ans. (b)
Given, ∆Hf = 60
. kJ mol −1
T = 0 + 273 = 273 K
∆Hf 60
.
∆S =
=
= 0.02198 kJ K–1 mol –1
T
273
(T = 0 ° C + 273 = 273 K)
= 0.02198 × 1000 JK–1 mol –1
= 21.98 JK–1 mol –1
67 The densities of graphite and
diamond at 298 K are 2.25 and 3.31
g cm −3 , respectively. If the
standard free energy difference
(∆G° ) is equal to 1895 J mol −1 , the
pressure at which graphite will be
transformed into diamond at 298 K
is
[CBSE AIPMT 2003]
(a) 9.92 × 106 pa
(b) 9.92 × 105 pa
(c) 9.92 × 108 pa
(d) 9.92 × 107 Pa
Ans. (c)
Mass
12
=
Density 2.25
12
Volume of diamond =
3.31
Change in volume,
12 
 12
−3
∆V = 
−
 × 10 L
 331
.
. 
225
Volume of graphite =
= − 1.91 × 10 −3 L
∆G ° = work done= − p∆V
∆G °
p= −
∆V
1895 J mol –1
=
1.91 × 10 −3 × 101.3
= 9794 atm
[Q 1 atm = 10 5 × 1.013 Pa]
= 992
. × 10 8 Pa
68 2 moles of an ideal gas at 27°C
temperature is expanded reversibly
from 2 L to 20 L. Find entropy
change (R = 2 cal/mol K).
[CBSE AIPMT 2002]
(a) 92.1 (b) 0
Ans. (d)
(c) 4
(d) 9.2
∆S (entropy change) = 2 .303 nR log 10
V2
V1
20
= 2 .303 × 2 × 2 × log 10
2
= 2 .303 × 2 × 2 × 1 = 9.212 cal
42
NEET Chapterwise Topicwise Chemistry
Ans. (b)
69 Unit of entropy is
[CBSE AIPMT 2002]
(a) JK −1 mol −1
(b) J mol −1
(c) J −1 K −1 mol −1
(d) JK mol −1
Ans. (a)
2 SO2 = –544 kJ
2 ZnS = –293 kJ
2ZnO = –480 kJ
For the reaction,
Entropy change equal to change in heat
q
per degree.
∆S =
T
q = required heat per mol
T = constant absolute temperature
Thus, unit of entropy is JK–1mol –1
70 PbO 2 → PbO, ∆G 298 < 0
SnO 2 → SnO, ∆G 298 > 0
Most probable oxidation state of
Pb and Sn will be
(a) Pb 4+ , Sn4+
(b) Pb 4+ , Sn2+
(c) Pb2+ , Sn2+
(d) Pb2+ , Sn4+
Ans. (d)
+4
[CBSE AIPMT 2001]
+2
Pb O2 → Pb O, ∆G298 < 0
For this reaction ∆G is negative, hence
Pb2 + is more stable thanPb4+ .
+4
+2
Sn O2 → Sn O,
∆G298 > 0
For this reaction ∆G is positive, hence
Sn4+ is more stable thanSn2 + because
for spontaneous change ∆G must be
negative.
71 The factor of ∆G values is
important in metallurgy. The ∆G
values for the following reactions
at 800°C are given as
S 2 (s) +2O 2 (g) → 2SO 2 (g),
∆G = − 544 kJ
2Zn(s) +S 2 (s) → 2ZnS(s),
∆G = − 293 kJ
2Zn(s) + O 2 (g) → 2ZnO(s),
∆G = − 480 kJ
The ∆G for the reaction,
2ZnS(s) + 3O 2 ( g) → 2ZnO(s)
+ 2SO 2 (g)
will be
[CBSE AIPMT 2000]
(a) –357 kJ
(c) –773 kJ
∆G of formation of different substances
are as
(b) –731 kJ
(d) –229 kJ
[CBSE AIPMT 1996]
2ZnS + 3 O2 ( g) → 2ZnO(s ) + 2 SO2 ( g)
∆G = [(∆G (products) —∆G (reactants) ]
= [(− 480) + (− 544) − ( − 293)]
= − 1024 + 293
= − 731kJ
72 The entropy change in the fusion of
one mole of a solid melting at
27°C (latent heat of fusion is
2930 J mol −1 ) is
[CBSE AIPMT 2000]
(a) 9.77J K –1 mol–1
(b) 10.73 J K –1 mol–1
(c) 2930 J K –1 mol–1
(d) 108.5 J K –1 mol–1
Ans. (a)
Entropy, ∆Sf =
∆ Hf
Tf
Fusion enthalpy
=
Temperature
∆Sf =
74 Given the following entropy values
(in JK −1 mol −1 ) at 298 K and 1 atm :
H2 (g) : 130.6, Cl 2 (g ) : 223.0, HCl(g) :
186.7. The entropy change (in
JK −1 mol −1 ) for the reaction
H2 (g) + Cl 2 (g) → 2HCl(g), is
2930 J mol –1
300 K
= 9.77 J K–1 mol –1
73 Identify the correct statement
regarding entropy.
[CBSE AIPMT 1998]
(a) At absolute zero temperature, entropy
of a perfectly crystalline substance is
taken to be zero
(b) At absolute zero temperature, the
entropy of a perfectly crystalline
substance is positive
(c) At absolute zero temperature, the
entropy of all crystalline substances
is to be zero
(d) At 0°C, the entropy of a perfectly
crystalline substance is taken to be
zero
Ans. (a)
“At absolute zero temperature, entropy
of a perfectly crystalline substance is
taken to be zero.” It is called third law of
thermodynamics.
(a) +540.3
(c) –166.9
Ans. (d)
(b) +727.0
(d) +19.8
H2 ( g) + Cl2 ( g) → 2HCl( g)
∆ r S = Σ S m ° (P) − Σ S m ° (R)
∆ r S ° = 2 × S m ° (HCl) − [S m ° (Cl2 )
+ S m ° (H2 )]
= (2 × 186.7) − (223 + 130.6)
= 373.4 − 353.6
= + 19.8 JK−1 mol −1
75 According to the third law of
thermodynamics which one of the
following quantities for a perfectly
crystalline solid is zero at absolute
zero?
[CBSE AIPMT 1996]
(a) Free energy
(b) Entropy
(c) Enthalpy
(d) Internal energy
Ans. (b)
Entropy is the degree of randomness or
disorder of the system. When the
temperature of the system is zero kelvin,
then all the motion of molecules ceases.
According to third law of
thermodynamics “At absolute zero the
entropy of a perfectly crystalline
substance is taken as zero.”
76 The correct relationship between
free energy and equilibrium
constant K of a reaction is
[CBSE AIPMT 1996]
(a) ∆G ° = − RT ln K
(b) ∆G = RT ln K
(c) ∆G ° = RT ln K
(d) ∆G = − RT ln K
Ans. (a)
The Gibbs free energy of a reaction, ∆ r G
is related to the composition of the
reaction mixture and the standard
reaction Gibbs free energy ∆ r G ° as
∆ r G = ∆ r G ° + RT ln Q
where, Q = reaction quotient
43
Chemical Thermodynamics
At equilibrium Q = K and ∆ r G = 0.
Therefore, the above reaction becomes
0 = ∆ r G ° + RT ln K
∆ r G ° = − RT ln K
or ∆ r G ° = − 2.303 RT log K
K = equilibrium constant
77 Standard Gibb's free energy change
for the isomerisation reaction
cis-2-pentene
trans-2-pent
ene is –3.67 kJ/mol at 400 K. If
more trans-2-pentene is added to
the reaction vessel, then
3
[CBSE AIPMT 1995]
(a)
(b)
(c)
(d)
more cis-2-pentene is formed
equilibrium remains unaffected
additional trans-2-pentene is formed
equilibrium is shifted in forward
direction
Ans. (a)
According to Le-Chatelier’s principle,
when we increase the concentration of
trans-2-pentene, then the reaction
shifts in backward direction and hence,
the concentration of cis-2-pentene
increase to maintain the equilibrium
constant K constant at given
temperature.
78 Consider the following reaction
occurring in an automobile
2C 8H18 (g) + 25O 2 (g) →
16CO 2 (g) + 18H2O(g)
The sign of ∆H, ∆S and ∆G would
be
[CBSE AIPMT 1994]
(a) +, –, + (b) –, +, – (c) –, +, + (d) +, +, –
Ans. (b)
The given reaction is combustion
reaction, so it takes place by evolution of
heat and hence, the sign of
∆H = negative and there is a increase in
the number of moles of gaseous
products, so entropy also increases and
hence, ∆S = positive.
Thus,
∆G = ∆H − T∆S = − ve − T (+ ve)
= − ve at any temperature
79 A chemical reaction will be
spontaneous if it is accompanied
by a decrease in [CBSE AIPMT 1994]
(a) entropy of the system
(b) enthalpy of the system
(c) internal energy of the system
(d) free energy of the system
Ans. (d)
Gibbs free energy of a system will decide
the spontaneity of a process.
If ∆G is negative, then the process is
spontaneous.
5
States of Matter
TOPIC 1
Gaseous State
Pressure (p)
(bar)
01 Choose the correct option for
graphical representation of Boyle’s
law, which shows a graph of
pressure vs volume of a gas at
different temperatures [NEET 2021]
(a)
200
K
400
K
600 K
Volume (V)
(dm3)
Pressure (p)
(bar)
1
V
λ
p = , (λ is a constant)
V
pV = λ
p∝
(200 K, 400 K, 600 K)
p
V
As temperature increases, gas expands
i.e. volume of gas increases.
So, the product pV increases and the p
vs V curve of Boyle’s law shifts upwards.
Mass of H2
Molar mass of H2
2
= 1 mol
2
Total number of moles of mixture,
nT = nO 2 + nH2
9
1
nT = + 1 = mol
8
8
Ideal gas equation;
pV = nRT
 Given, R = 0.082 L atm mol –1 K–1 





T = 0 ° C = 273 K, V = 1 L
=
p× 1=
⇒
9
× 0.082 × 273
8
p = 25.18 atm
03 The minimum pressure required to
compress 600 dm 3 of a gas at 1 bar
to 150 dm 3 at 40°C is
p
(bar)
600 K
400 K
200 K
(b)
Volume (V)
(dm3)
nH 2 =
V
(dm3)
[NEET (Oct.) 2020]
(a) 4.0 bar
(c) 1.0 bar
Ans. (a)
(b) 0.2 bar
(d) 2.5 bar
20
0
40 K
0
K
Pressure (p)
(bar)
p1 = 1 bar, p2 = ?
0
60
K
(c)
Pressure (p)
(bar)
Volume (V)
(dm3)
(d)
600
K
400
200 K
K
Volume (V)
(dm3)
Ans. (d)
Boyle’s law states, that at constant
temperature, pressure of a gas of fixed
amount varies inversely with volume.
02 Choose the correct option for the
total pressure (in atm) in a mixture
of 4g O 2 and 2 g H2 confined in a
total volume of one litre at 0°C is
[Given, R = 0.082 L atm mol − 1 K − 1 ,
T = 273K]
[NEET 2021]
(a) 2.518
(c) 25.18
Ans. (c)
(b) 2.602
(d) 26.02
Given, mass of O2 = 4 g
Molar mass of O2 = 32 g mol −1
Number of moles of O2 ,
Mass of O2
4 1
=
= mol
nO 2 =
Molar mass of O2 32 8
Given, mass of H2 = 2 g
Molar mass of H2 = 2 g mol −1
Number of moles of H2 ,
V1 = 600 dm3
V2 = 150 dm3
⇒From Boyle’s law (Temperature, 40ºC
is constant)
p1V1 = p2V2
p V 1 × 600
= 4 bar
⇒ p2 (min) = 1 1 =
150
V2
04 A gas at 350 K and 15 bar has molar
volume 20 percent smaller than
that for an ideal gas under the
same conditions. The correct
option about the gas and its
compressibility factor (Z) is
[NEET (National) 2019]
(a) Z > 1 and repulsive forces are
dominant
(b) Z < 1 and attractive forces are
dominant
45
States of Matter
(c) Z < 1 and repulsive forces are
dominant
(d) Z > 1 and attractive forces are
dominant
Ans. (b)
Compressibility factor (Z) is the factor
which decides the extent of deviation of
real gases from ideal gases.
V
Z = real
Videal
Real gases deviate from ideal gas
behaviour because for ideal gas it is
considered that there is no force of
attraction between gas molecules.
Also, for ideal gas, volume of gas
molecules is negligible as compared to
volume of gas container.
Given,
Vreal < Videal
∴
Z< 1
If Z < 1, the gas is more compressible
than expected from ideal behaviour. As a
result, attractive forces are present
between molecules and are dominant.
05 The volume occupied by 1.8 g of
water vapour at 374°C and 1 bar
pressure will be
[Use R = 0.083 bar L K −1 mol −1 ]
[NEET (Odisha) 2019]
(a) 96.66 L
(c) 3.10 L
Ans. (d)
(b) 55.87 L
(d) 5.37 L
According to ideal gas equation,
pV = nRT
nRT
w RT
w 
…(i) Qn =
or V =
=

p
M.wt p
M.wt 

Given, w = 1.8 g,T = 374º C
= (374 + 273) K = 647 K
p = 1 bar, R = 0.083 bar LK −1 mol −1
On substituting the given values in Eq. (i),
we get
1.8 g
V=
18 g mol -1
0.083 bar LK–1 mol –1 × 647 K
×
1 bar
= 537
. L
06 Given van der Waals’ constant of
NH3 , H2 , O 2 and CO 2 are
respectively 4.17, 0.244, 1.36 and
3.59, which one of the following
gases is most easily liquefied?
[NEET 2018]
(a) O2
(c) NH3
Ans. (c)
(b) H2
(d) CO2
In the van der Waals’ equation,

an2 
 p + 2  (V − nb) = nRT
V 

‘a’ and ‘b’ are known as van der Waals’
constant.
‘a’ is the measure of force of attraction
between gas molecules. Greater the
value of a, easier the liquefaction of the
gas.
Thus, amongNH3 (4.17),H2 (0.244),O2 (1.36)
and CO2 (3.59), the value ofa is greatest
in NH3, hence it is most easily liquefied.
07 The correction factor ‘a’ to the ideal
gas equation corresponds to
[NEET 2018]
(a) electric field present between the
gas molecules
(b) volume of the gas molecules
(c) density of the gas molecules
(d) forces of attraction between the gas
molecules
Ans. (d)
According to van der Waals’ equation,

an2 
 P + 2  (V − nb) = nRT
V 

where,a and b are called van der Waals’
constant.
an2
is called internal pressure of the gas
V2
where, ‘a’ is a measure of force of
attraction between gas molecules.
‘b’ is also called co-volume or excluded
volume.
The constants ‘a’ and ‘b’ are expressed in
atm L2 mol −2 and L mol −1, respectively.
08 Equal moles of hydrogen and
oxygen gases are placed in
container with a pin-hole through
which both can escape. What
fraction of the oxygen escapes in
the time required for one-half of
the hydrogen to escape?
[NEET 2016, Phase I]
(a) 1/4
(c) 1/2
Ans. (d)
(b) 3/8
(d) 1/8
Given, number of moles of hydrogen
(nH 2 ) and that of oxygen (nO 2 ) are equal.
∴ We have, the relation between ratio of
number of moles escaped and ratio of
molecular mass.
nO 2
MH 2
=
nH 2
MO 2
where, M = Molecular mass of the
molecule.
nO 2
⇒
nH 2
nO 2
⇒
nH 2
nO 2
⇒
0.5
⇒
nO 2
=
2
32
=
1
16
1
4
0.5 1
=
=
4
8
=
09 A mixture of gases contains H2 and
O 2 gases in the ratio of 1 : 4 (w/w).
What is the molar ratio of the two
gases in the mixture?
[CBSE AIPMT 2015]
(a) 1 : 4 (b) 4 : 1
Ans. (b)
(c) 16 : 1 (d) 2 : 1
Let the mass of H2 gas be xg and mass of
O2 gas 4xg
Molar
mass
i.e.
H2 : O2
2 : 32
1 : 16
nH
∴ Molar ratio = 2
nO 2
x /2
4x /32
x × 32 4
= = 4: 1
=
2 × 4x 1
=
10 Equal masses of H2 , O 2 and
methane have been taken in a
container of volume V at
temperature 27°C in identical
conditions. The ratio of the
volumes of gases H2 : O 2 : CH4
would be
[CBSE AIPMT 2014]
(a) 8 : 16 : 1
(c) 16 : 1 : 2
Ans. (c)
(b) 16 : 8 : 1
(d) 8 : 1 : 2
According to Avogadro’s hypothesis,
Volume of a gas (V ) ∝number of moles (n)
Therefore, the ratio of the volumes of
gases can be determined in terms of
their moles.
∴ The ratio of volumes of H2 : O2 :
methane (CH4 ) is given by
VH 2 :VO 2 :VCH 4 = nH 2 : nO 2 : nCH 4
mH 2 mO 2 mCH
4
⇒ VH 2 :VO 2 :VCH 4 : =
:
:
MH 2 MO 2 MCH 4
Given, mH 2 = mO 2 = mCH 4 = m
mass 

Qn =

molar mass
46
NEET Chapterwise Topicwise Chemistry
m m m
: :
2 32 16
= 16 : 1 :2
Thus, VH 2 :VO 2 :VCH 4 =
11 Maximum deviation from ideal gas
is expected from
[NEET 2013]
(a) H2 (g) (b) N2 (g) (c) CH4 (g) (d) NH3 (g)
Ans. (d)
The extent to which a real gas deviates
from ideal behaviour can be understood
by a quantity ‘Z’ called the
compressibility factor. Easily liquifiable
gases likeNH3, SO2 etc. exhibit maximum
deviation from ideal gas as for them
Z < < < 1.
CH4 also exhibits deviation but it is less
as compared to NH3.
12 50 mL of each gas A and of gas B
takes 150 and 200 s respectively
for effusing through a pin hole
under the similar conditions. If
molecular mass of gas B is 36, the
molecular mass of gas A will be
[CBSE AIPMT 2012]
(a) 96
(b) 128
Ans. (*)
(c) 32
(d) 64
VA = VB = 50 mL
TA = 150 s
TB = 200 s
MB = 36
MA = ?
From Graham’s law of effusion
Given,
rB
rA
=
MB
=
VBTA
TBVA
⇒
MA
36
=
VA × 150
200 × VA
M
15 3
9
= ⇒ A=
36 20 4
36 16
9 × 36 9 × 9 81
MA =
=
= = 20.2
16
4
4
* According to question, no option is
correct in this condition, answer any
option.
NOTE
IfTA = 200 s andTB = 150 s then MA = 64
or
MA
MA
=
13 A gaseous mixture was prepared by
taking equal moles of CO and N 2 . If
the total pressure of the mixture
was found 1 atom, the partial
pressure of the nitrogen (N 2 ) in the
mixture is
[CBSE AIPMT 2011]
(a) 0.8 atm
(c) 1 atm
Ans. (d)
(b) 0.9 atm
(d) 0.5 atm
Equal moles of CO andN2
nCO = nN 2
then, according to ideal gas equation,
pressure of both gases CO andN2
becomes equal
∴
pCO = pN 2
Given, pCO + pN 2 = Total pressure of
mixture.
or 2pN 2 = 1 atm or pN 2 = 0.5 atm
14 Two gases A and B having the
same volume diffuse through a
porous partition in 20 and 10s
respectively. The molecular mass
of A is 49 u. Molecular mass of B
will be
[CBSE AIPMT 2011]
(a) 12.25 u
(c) 25.00 u
Ans. (a)
(b) 6.50 u
(d) 50.00 u
According to Graham’s law of diffusion,
rate of diffusion
1
V
,r ∝
r∝
t
M
where, V is the volume of the gas diffused
in time t.
MB
t
MB
rA
V
or A × B =
=
rB
MA
t A VB
MA
Given, VA = VB
MB
10
1 M
=
⇒ = B
∴
4 49
20
49
49
MB =
= 12.25 u
4
15 If a gas expands at constant
temperature, it indicates that
[CBSE AIPMT 2008]
(a) kinetic energy of molecules
decreases
(b) pressure of the gas increases
(c) kinetic energy of molecules remains
the same
(d) number of the molecules of gas
increases
Ans. (c)
KE =
3
RT (for one mole of a gas)
2
As, the kinetic energy of a gaseous
molecule depends only on temperature,
thus at constant temperature, the
kinetic energy of the molecules remains
the same.
16 van der Waals’ real gas, act as an
ideal gas, at which condition?
[CBSE AIPMT 2002]
(a) High temperature, low pressure
(b) Low temperature, high pressure
(c) High temperature, high pressure
(d) Low temperature, low pressure
Ans. (a)
At higher temperature and low pressure
real gas acts as an ideal gas and obey pV
= nRT relation.
17 The beans are cooked earlier in
pressure cooker, because
[CBSE AIPMT 2001]
(a) boiling point increases with
increasing pressure
(b) boiling point decreases with
increasing pressure
(c) extra pressure of pressure cooker,
softens the beans
(d) internal energy is not lost while
cooking in pressure cooker
Ans. (b)
The beans are cooked earlier in pressure
cooker because boiling point decreases
with increasing pressure.
18 Which of the following expressions
correctly represents the
relationship between the average
molar kinetic energy, KE of CO and
N 2 molecules at the same
temperature?
(a) KECO < KE N 2
[CBSE AIPMT 2000]
(b) KECO > KE N 2
(c) KECO = KE N 2
(d) Cannot be predicted unless volumes
of the gases are given
Ans. (c)
3
KE= RT (for one mole of a gas)
2
The temperature is constant and kinetic
energy is independent on molecular
weights. So,
KECO =KEN 2
19 Which one of the following
statements is wrong for gases?
[CBSE AIPMT 1999]
(a) Gases do not have a definite shape
and volume
(b) Volume of the gas is equal to volume
of container confining the gas
(c) Confined gas exerts uniform
pressure on the walls of its container
in all directions
(d) Mass of gas cannot be determined by
weighing a container in which it is
enclosed
Ans. (d)
Mass of gas can be determined by
weighing a container in which it is
enclosed as follows:
47
States of Matter
Mass of the gas = mass of the cylinder
including gas – mass of empty cylinder
So, it is a wrong statement.
20 At 25°C and 730 mm pressure, 380
mL of dry oxygen was collected. If
the temperature is constant, what
volume will the oxygen occupy at
760 mm pressure?
[CBSE AIPMT 1999]
(a) 365 mL
(c) 10 mL
Ans. (a)
(b) 2 mL
(d) 20 mL
As the temperature is constant, Boyle’s
law is applicable.
p 1V1 = p2V2
V1 = 380 mL, p1 = 730 mm ,V2 = ?,
p2 = 760 mm
730 × 380 = 760 × V2
730 × 380
= 365 mL
V2 =
760
21 At which one of the following
temperature pressure conditions,
the deviation of a gas from ideal
behaviour is expected to be
minimum?
[CBSE AIPMT 1996]
(a) 350 K and 3 atm
(b) 550 K and 1 atm
(c) 250 K and 4 atm
(d) 450 K and 2 atm
Ans. (b)
A real gas behave as an ideal gas at low
pressure and high temperature. Among
the given option 550K temperature is the
highest and 1atm pressure is the lowest
pressure.
22 Cyclopropane and oxygen at partial
pressures 170 torr and 570 torr are
mixed in a gas cylinder. What is the
ratio of the number of moles of
cyclopropane to the number of
moles of oxygen?
p2V = n2 RT
From Eqs. (i) and (ii)
p1V n1RT
p
n
=
⇒ 1 = 1
p2V n2 RT
p2 n2
n1 p1 170
=
=
= 0.30
n2 p2 570
23 600 cc of a gas at a pressure of 750
mm is compressed to 500 cc.
Taking the temperature to remain
constant,the increase in pressure
is
[CBSE AIPMT 1995]
(a) 150 mm
(c) 350 mm
Ans. (a)
24 500 mL of nitrogen at 27°C is
cooled to –5°C at the same
pressure. The new volume
becomes
[CBSE AIPMT 1995]
(a) 326.32 mL
(b) 446.66 mL
(c) 546.32 mL
(d) 771.56 mL
Ans. (b)
Initial volume,V1 = 500 mL
Initial temperature,
T1 = 27 ° C = 27 + 273 = 300 K
Final temperature,
T2 = − 5 + 273 = 268 K
V2 = ?
V1 V2
=
T1 T2
V2 =
T1
=
500 × 268
300
25 The temperature of the gas is
raised from 27°C to 927°C the root
mean square velocity is
[CBSE AIPMT 1994]
927
times of the earlier value
27
(b) same as before
(c) halved
(d) doubled
(a)
Ans. (d)
According to ideal gas equation,
p1V = n1RT
VT
1 2
= 446.66 mL
(a)
…(i)
Ans. (d)
Root mean square velocity atT1
temperature,
3RT1
3R (27 + 273)
=
U1 =
M
M
Root mean square velocity atT2
temperature,
3RT2
3R (927 + 273)
=
U2 =
M
M
Eq. (i) divided by Eq. (ii)
U1
27 + 273
=
927 + 273
U2
(b) 250 mm
(d) 450 mm
According to Boyle’s law,
p 1V1 = p2V2
750 × 600 = p2 × 500
750 × 600
p2 =
500
= 900 mm
So, the increase in pressure
= 900 − 750 = 150 mm
[CBSE AIPMT 1996]
170 × 42
= 0.39
570 × 32
170  170 570
(b)
+
 ≈0.19

42  42 32 
170
(c)
= 0.23
740
170
(d)
= 0.30
570
…(ii)
=
…(i)
…(ii)
300
1
=
1200 2
U2 = 2 U 1
26 When is the deviation more in the
behaviour of a gas from the ideal gas
equation pV = nRT ?
[CBSE AIPMT 1993]
(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 pressure
Ans. (b)
Gases show deviation from ideal gas
behaviour when the temperature is low
and pressure is high. At low
temperature, the volume of one
molecule is not negligible in comparison
to total volume and intermolecular force
of attraction is maximum at low
temperature and high pressure.
27 The ratio among most probable
velocity, mean velocity and root
mean square velocity is given by
[CBSE AIPMT 1993]
(a) 1 :2 : 3
(b) 1 : 2 : 3
8
(d) 2 :
: 3
π
8
(c) 2 : 3 :
π
Ans. (d)
2 RT
M
8RT
Mean velocity, U av =
πM
Most probable velocity, U mp =
Root mean square velocity, U rms =
3RT
M
so,
2RT
8RT
3RT
:
:
πM
M
M
8
= 2:
: 3
π
U mp : U av : U rms =
48
NEET Chapterwise Topicwise Chemistry
28 Internal energy and pressure of a
gas per unit volume are related as
[CBSE AIPMT 1993]
2
(a) p = E
3
1
(c) p = E
2
Ans. (a)
(b) p =
3
E
2
(d) p = 2 E
The average translational kinetic energy
1
of a gas molecule is mu− 2 at a
2
temperature, T. The total energy of the
whole of the gas containingN molecules
is
1
…(i)
E k = mN u− 2
2
The kinetic gas equation is
1
…(ii)
pV = mN u− 2
3
2 1
pV = × mN u− 2
3 2
2
…(iii)
pV = E k
3
2
so, p = E k per unit volume.
3
29 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 mol L−1 ?
(R=0.082 L atm mol–1 deg–1 )
(a) At STP
[CBSE AIPMT 1993]
(b) When V = 22.4 L
(c) WhenT = 12 K
(d) Impossible under any conditions
Ans. (c)
According to ideal gas equation,
pV = nRT
n
p = RT
V
1atm = 1mol L–1 × 0.082 × T
1
T=
=12 K
0.082
30 Select the correct statement. In
the gas equation pV = nRT
[CBSE AIPMT 1992]
(a) n is the number of molecules of a gas
(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 the gas is present
Ans. (c)
The ideal gas equation is pV = nRT
where, V is the volume of n moles of a gas.
31 An ideal gas cannot be liquefied
because
[CBSE AIPMT 1992]
(a) its critical temperature is always
above 0°C
(b) its molecules are relatively smaller in
size
(c) it solidifies before becoming a liquid
(d) forces operating between its
molecules are negligible
Ans. (d)
Gases can be liquefied by lowering the
temperature and increasing the
pressure. An ideal gas have no
intermolecular force of attraction, so it
cannot be liquefied by applying high
pressure and decreasing temperature.
32 The correct value of the gas
constant ‘ R ’ is close to
[CBSE AIPMT 1992]
(a) 0.082 L atm K
(b) 0.082 L atm K −1 mol −1
(c) 0.082 L atm −1 K mol −1
(d) 0.082 L −1 atm −1 K mol
Ans. (b)
The numerical value of R depends upon
the units in which pressure and volume
are expressed. When pressure is
expressed in atmosphere and volume in
litres, then the value of R at STP for
1 mole of gas is
1 atm × 22.414 L
R=
1 mole × 273.15 K
= 0.082 L atm K–1 mol –1
34 The root mean square speeds at
STP for the gases H2 , N 2 ,O 2 and
HBr are in the order
[CBSE AIPMT 1991]
(a) H2 < N2 < O2 < HBr
(b) HBr < O2 <N2 <H2
(c) H2 <N2 == O2 <HBr
(d) HBr <O2 <H2 <N2
Ans. (b)
The root mean square velocity of gas
molecule at STP is given by
3RT
U rms =
M
∴ As the molar mass of gas increases,
then U rms will decrease, so the order of
U rms of these gases is
HBr < O2 < N2 < H2
35 At constant temperature, in a given
mass of an ideal gas
[CBSE AIPMT 1991]
(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
Ans. (d)
According to ideal gas equation,
pV = nRT
If T is constant and mass is constant, so
number of moles (n) also constant
mass 

,
n=

molar mass
therefore
33 Which is not true in case of an ideal
gas?
[CBSE AIPMT 1992]
(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
Ans. (c)
A gas is a collection of tiny particles
separated from one another by large
empty space and moving rapidly at
random in all the directions. In the
course of their motion, they collide with
one another and also with the walls of
the container. Due to frequent collisions,
speeds and direction of motion of
molecules keeps on changing. Thus, all
the molecules in a sample of a gas do not
have same speeds.
pV = constant
36 In a closed flask of 5 L, 1.0 g of H2 is
heated from 300 to 600 K. Which
statement is not correct?
[CBSE AIPMT 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
Ans. (c)
Here, volume is constant and mass of
hydrogen gas is also fixed, therefore the
number of moles remains same. Now as
the temperature increases, then
pressure also increases. This will lead to
more collisions among the gaseous
molecules and hence, the energy of
molecules increases.
49
States of Matter
37 A gas is said to behave like an ideal
gas when the relation
pV
= constant. When do you
T
expect a real gas to behave like an
ideal gas?
[CBSE AIPMT 1991]
(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
Ans. (d)
As we know that the van der Waals’
equation is
 p + a  (V − b) = RT



V2 
The real gases show ideal behaviour
when pressure approaches zero and
temperature is high. At this condition
there is no force of attraction and
repulsion between the molecules of gas.
a
Thus, the effect of 2 and b is negligible,
V
i.e.
pV = RT
pV
=1
RT
pV
We also know
= Z (for ideal gas Z = 1)
RT
(Z is compressibility factor)
Therefore an real gas behaves like ideal
gas when the temperature is high and
pressure is low.
38 In van der Waals’ equation of state
for a non-ideal gas, the term that
accounts for inter molecular forces
is
[CBSE AIPMT 1990]
(a) (V − b )
a

(c)  p + 2 

V 
(b) (RT ) −1
(d) RT
Ans. (c)
The van der Waals’ equation is
 p + a  (V − b) = RT



V2 
a
The term  p + 2  is used for pressure

V 
correction, it measures the
intermolecular forces between the
molecules of gas.
39 Absolute zero is defined as the
temperature
[CBSE AIPMT 1990]
(a) at which all molecular motion ceases
(b) at which liquid helium boils
pM =
Ans. (a)
pM = dRT
pM
d=
RT
At absolute zero temperature, the
entropy of substance becomes zero, it
means that all the molecular motions are
stopped or ceased.
40 Root mean square velocity of a gas
molecule is proportional to
[CBSE AIPMT 1990]
(a) m 1/2
(c) m −1/2
Ans. (c)
(b) m 0
(d) m
Root mean square speed is given by the
expression
3RT
U rms =
[Q M = mN ]
M
1
∴
U rms ∝
m
⇒
U rms ∝ (m) −1/ 2
41 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
[CBSE AIPMT 1989]
(a) 0°C
(b) absolute zero
(c) its critical temperature
(d) its Boyle’s temperature
Ans. (a)
According to Charle’s law “the volume of
a fixed mass of a gas increases or
1
of its volume at 0°C
decreases by
273.15
for each degree rise or fall of
temperature, if pressure is kept
constant”.
42 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 AIPMT 1989]
RT
pM
M
(c)
V
Ans. (d)
(a)
p
RT
pM
(d)
RT
(b)
We know that ideal gas equation is
pV = nRT
w
pV = RT
M
⇒
Q w = d 
 V

w
RT
V
(c) at which ether boils
(d) All of the above
TOPIC 2
Liquid State
43 The surface tension of which of the
following liquids is maximum?
[CBSE AIPMT 2005]
(a) H2O
(c) CH3OH
Ans. (a)
(b) C 6H6
(d) C2H5OH
Surface tension of H2O is maximum due
to maximum hydrogen bonding in
comparison to C6H6 , CH3OH, C2H5OH . The
order of H-bonding is
H2O > CH3OH > C2H5OH
(Benzene does not form H-bond).
44 A liquid can exist only
[CBSE AIPMT 1994]
(a) between triple point and critical
temperature
(b) at any temperature above the
melting point
(c) between melting point and critical
temperature
(d) between boiling and melting
temperature
Ans. (d)
A liquid below its melting point, is
present in solid state and above its
melting point, is present in vapour
(gaseous) state, so a liquid can exist
between melting point and boiling point.
45 A closed flask contains water in all
its three states solid, liquid and
vapour at 0°C. In this situation, the
average kinetic energy of water
molecules will be
[CBSE AIPMT 1992]
(a) the greatest in all the three states
(b) the greatest in vapour state
(c) the greatest in the liquid state
(d) the greatest in the solid state
Ans. (b)
In the three states of matter, the
maximum kinetic energy is possessed by
the gaseous molecules, so water vapour
state has maximum kinetic energy in this
situation.
6
Solid State
TOPIC 1
Classification of Solid
01 A pure crystalline substance on
being heated gradually first forms a
turbid liquid at constant
temperature and still at higher
temperature turbidity completely
disappears. The behaviour is a
characteristic of substance forming
[CBSE AIPMT 1993]
(a) allotropic crystals
(b) liquid crystals
(c) isomeric crystals
(d) isomorphous crystals
Ans. (b)
Such type of phenomenon is only
exhibited by liquid crystals.
02 Most crystals show good cleavage
because their atoms, ions or
molecules are [CBSE AIPMT 1991]
(a) weakly bonded together
(b) strongly bonded together
(c) spherically symmetrical
(d) arranged in planes
Ans. (d)
In crystals the constituents (atoms, ions
or molecules) are arranged in definite
orderly arrangement. When these
crystals are cleaved they cut into regular
patterns.
03 For orthorhombic system axial
ratios are a ≠ b ≠ c and the axial
angles are
[CBSE AIPMT 1991]
(a) α = β = γ ≠ 90°
(b) α = β = γ = 90°
(c) α = γ = 90°,β ≠ 90°
(d) α ≠ β ≠ γ ≠ 90°
Ans. (b)
Orthorhombic crystal has three unequal
axis which are at right angle to each
other a ≠ b ≠ c, all angles = 90 °
06 The correct option for the number
of body centred unit cells in all 14
types of Bravais lattice unit cell is
[NEET 2021]
(a) 7
(b) 5
Ans. (d)
So, axial distancesa ≠ b ≠ c and axial
angles α = β = γ = 90 °.
04 The ability of a substance to
assume in two or more crystalline
structure is called
[CBSE AIPMT 1990]
(a) isomerism
(c) isomorphism
Ans. (b)
(b) polymorphism
(d) amorphism
Some substances adopt different
structural arrangement under different
conditions. Such arrangements are
called polymorphs and this phenomenon
is called polymorphism.
TOPIC 2
Unit Cell and
Packaging and Solid
05 Right option for the number of
tetrahedral and octahedral voids in
hexagonal primitive unit cell are
[NEET 2021]
(a) 8, 4
(c) 2, 1
Ans. (d)
(b) 6, 12
(d) 12, 6
Number of octahedral and tetrahedral
voids are equal to N and 2N respectively,
where N is the number of atoms in unit
cell.
Number of atoms per unit cell in
hexagonal primitive unit cell = 6.
Number of tetrahedral voids
= 2N = 2 × 6 = 12.
Number of octahedral voids = N = 6.
(c) 2
(d) 3
Crystal system Bravais lattice
Cubic
Primitive, body centred,
face centred, end
centred
Orthorhombic Primitive, body centred,
face centred, end
centred
Tetragonal
Primitive, body centred
Monoclinic
Primitive, end centred
Triclinic
Primitive
Rhombohedral Primitive
Hexagonal
Primitive
Body centred unit cell exists in three
systems, i.e. cubic, orthorhombic and
tetragonal.
07 An element has a body centered
cubic (bcc) structure with a cell
edge of 288 pm. The atomic radius
is
[NEET (Sep.) 2020]
2
× 288 pm
4
4
(c)
× 288 pm
2
Ans. (d)
4
× 288 pm
3
3
(d)
× 288 pm
4
(a)
(b)
In bcc crystal,
Q 4r = 3a; r =
3 × 288
3a
pm
=
4
4
r
Corner
2r
a
r
Body centre
51
Solid State
where, r = radius of atoms,
a = edge length of the unit cell.
08 A compound is formed by cation C
and anion A. The anions form
hexagonal close packed (hcp)
lattice and the cations occupy 75%
of octahedral voids.The formula of
the compound is
[NEET (National) 2019]
(a) C 3A2 (b) C 3A4
Ans. (b)
(c) C 4 A3
(d) C2 A3
Anions (A) form hexagonal close packed
(hcp) lattice, so
Number of anions (A) = 6
Number of octahedral voids = Number of
atoms in the close packed structure = 6.
Cations (C ) occupy 75% of octahedral
75
voids, so number of cations (C ) = 6 ×
100
= 6 × 3 /4 = 9 /2
∴The formula of compound = C 9 / 2 A6
= C 9 A12 = C 3A4
Thus, option (b) is correct.
09 Iron exhibits bcc structure at room
temperature. Above 900°C , it
transforms to fcc structure. The
ratio of density of iron at room
temperature to that at 900°C
(assuming molar mass and atomic
radii of iron remains constant with
temperature) is
[NEET 2018]
3 3
4 3
(b)
(c)
4 2
3 2
Ans. (a)
(a)
Density of unit celld =
3
2
(d)
1
2
Z×M
NA × a
3
where, Z = Number of atoms per unit cell
M = Molar mass
a 3 = Volume of unit cell [a = edge length]
NA = Avogadro’s number = 6.022 × 1023
3a
For bcc, Z = 2, radius (r) =
4
4r
a=
3
a
For fcc, Z = 4, r =
⇒ a =2 2r
2 2
According to question
 ZM 


d room temp.  N Aa 3  bcc
=
 ZM 
d 900 ° C


N a3 
 A  fcc
On substituting the given values, we get
d room temp.
2× M
4× M
=
3
d 900 ° C
×
(2 2r) 3
N
 4r 
A
NA ×  
 3
[QGiven, M and r of iron remains
constant with temperature]
2 × 3 3 16 2 r 3
=
×
4
64 r 3
d bcc 3 3
=
d fcc 4 2
10 The ionic radii of A + and B − ions
are 0.98 × 10 −10 m and 1.81 × 10 −10 m.
The coordination number of each
ion in AB is
[NEET 2016, Phase I]
(a) 4
(c) 2
Ans. (d)
(b) 8
(d) 6
Given, ionic radius of cation (A + ) = 0.98
× 10 −10 m
Ionic radius of anion (B − ) = 1.81 × 10 −10 m
∴ Coordination number of each ion in AB
=?
Now, we have
Radius of cation
Radius ratio =
Radius of anion
0.98 × 10 −10 m
=
1.81 × 10 −10 m
= 0.541
If radius ratio range is in between 0.441 –
0.732, ion would have octahedral
structure with coordination number ‘six’.
11 In calcium fluoride, having the
fluorite structure, the coordination
numbers for calcium ion (Ca 2+ ) and
fluoride ion (F − ) are
[NEET 2016, Phase II]
(a) 4 and 2
(c) 8 and 4
Ans. (c)
(b) 6 and 6
(d) 4 and 8
In CaF2 (Fluorite structure), Ca2 + ions
are arranged in ccp arrangement (Ca2 +
ions are present at all corners and at the
centre of each face of the cube) whileF −
ions occupy all the tetrahedral sites.
Ca2+
F–
From the above figure, you can clearly
see that coordination number ofF − is 4
while that of Ca2 + is 8.
12 The vacant space in bcc lattice cell
is
[CBSE AIPMT 2015]
(a) 26%
(c) 23%
Ans. (d)
(b) 48%
(d) 32%
Q Packing efficiency in bcc lattice = 68%.
∴ Vacant space in bcc lattice
= 100 − 68 = 32%
13 A given metal crystallises out with
a cubic structure having edge
length of 361 pm. If there are four
metal atoms in one unit cell, what
is the radius of one atom?
[CBSE AIPMT 2015]
(a) 40 pm
(c) 80 pm
Ans. (b)
(b) 127 pm
(d) 108 pm
Given, edge length = 361 pm
Four metal atoms in one unit cell
i.e. effective number in unit cell (z) = 4
(given)
∴ It is a FCC structure
∴ Face diagonal = 4r
2 a = 4r
2 × 361
r=
4
= 127 pm
14 Lithium metal crystallises in a body
centred cubic (bcc) crystal. If the
length of the side of the unit cell of
lithium is 351 pm, the atomic radius
of the lithium will be
[CBSE AIPMT 2009]
(a) 240.8 pm
(c) 75.5 pm
Ans. (b)
(b) 151.8 pm
(d) 300.5 pm
In case of body centred cubic (bcc)
crystal,
a 3 = 4r
Given, edge length,a = 351 pm
Hence, atomic radius of lithium,
a 3 351 × 1.732
r=
=
4
4
= 151.98 pm
15 Copper crystallises in a face
centred cubic (fcc) lattice with a
unit cell length of 361 pm. What is
the radius of copper atom in pm?
[CBSE AIPMT 2009]
(a) 128 pm
(c) 181 pm
(b) 157 pm
(d) 108 pm
52
NEET Chapterwise Topicwise Chemistry
Ans. (a)
=
In case of face centred cubic (fcc)
lattice,
2a
radius =
4
∴ Radius of copper atom (fcc lattice)
2 × 361
= 128 pm
=
4
16 Percentage of free space in body
centred cubic (bcc) unit cell is
[CBSE AIPMT 2008]
(a) 30% (b) 32%
Ans. (b)
(c) 34% (d) 28%
In bcc unit cell, the number of atoms = 2
Thus, volume of atoms in unit cell
4
(v) = 2 × πr 3
3
3
For bcc structure (r) =
a
4
a
Ans. (d)
2 2
In a face centred cubic (fcc) lattice, a
unit cell is shared equally by six unit
cells.
Hence, the ratio of radii
1
3
1
a:
a
= a:
2
4
2 2
18 Which one of the following
statements is an incorrect?
[CBSE AIPMT 2008]
(a) The fraction of the total volume
occupied by the atoms in a primitive
cell is 0.48
(b) Molecular solids are generally volatile
(c) The number of carbon atoms in an
unit cell of diamond is 4
(d) The number of Bravais lattices in
which a crystal can be categorised is
14
Ans. (a)
Volume of atoms in a unit cell (V ) =
3
(V ) = 2 ×
3
4  3 
π
π a3
a =
8
3  4 
For primitive cell, r =
Volume of unit cell (V ) = a 3
Percentage of volume occupied by unit
cell
Volume of the atoms in unit cell
=
Volume of unit cell
3
π a3
3
8
π × 100 = 68%
=
× 100 =
3
8
a
Hence, the free space in bcc unit cell
= 100 − 68 = 32%
17 If ‘ a ’ stands for the edge length of
the cubic systems : simple cubic,
body centred cubic and face
centred cubic, then the ratio of
radii of the spheres in these
systems will be respectively,
[CBSE AIPMT 2008]
1
3
1
(a) a :
a:
a
2
4
2 2
1
1
(b) a : 3 a :
a
2
2
1
3
2
(c) a :
a:
a
2
2
2
(d) 1a : 3 a : 2 a
Ans. (a)
a
2
3
V=
4 a 
πa 3
π  =
3  2
6
Volume of the unit cell (V ) = a 3
Thus, total volume occupied by the
atoms
Volume of the atoms in unit cell
Volume of unit cell
1 π
πa 3
=
× 3 = = 0.52 = 100 -0.52 = 0.48
6
6
a
=
19 The fraction of total volume
occupied by the atoms present in a
simple cube is [CBSE AIPMT 2007]
π
(a)
6
π
(c)
4 2
Ans. (a)
π
(b)
3 2
π
(d)
4
For simple cube,
a
Radius (r) =
2
[a = edge length]
Volume of the atom =
4 a 
π 
3  2
3
3
If a = edge length of cubic systems
For simple cubic structure, radius =
4 3
πr
3
a
2
For body centred cubic structure, radius
3
a
=
4
For face centred cubic structure, radius
4 a 
π 
 
π
∴ Packing fraction = 3 32 =
6
a
20 In a face centred cubic (fcc) lattice,
a unit cell is shared equally by how
many unit cells? [CBSE AIPMT 2005]
(a) 8
(c) 2
(b) 4
(d) 6
21 A compound formed by elements X
and Y crystallises in a cubic
structure in which the X-atoms are
at the corners of a cube and the
Y-atoms are at the face centres.
The formula of the compound is
[CBSE AIPMT 2004]
(a) X Y3
(c) X Y
Ans. (a)
(b) X 3Y
(d) X Y2
In unit cell, X-atoms at the corners
1
= ×8=1
8
1
Y-atoms at the face centres = × 6 = 3
2
Ratio of X and Y = 1 :3.
Hence, formula is XY3.
22 Zn converts from its melted state
to its solid state, it has hcp
structure, then find out the number
of nearest atoms.
[CBSE AIPMT 2001]
(a) 6
(c) 12
Ans. (c)
(b) 8
(d) 4
HCP is a closed packed arrangement, in
which the unit cell is hexagonal and
coordination number is 12.
23 A compound formed by elements A
and B crystallises in the cubic
structure, where A atoms are
present at the corners of a cube
and B atoms are present at the
face centres. The formula of the
compound is [CBSE AIPMT 2000]
(a) A2B2
(c) AB
Ans. (b)
(b) AB3
(d) A3B
A-atoms are present at the corners of a
cube. So, the number of A-atoms per
1
unit cell = 8 × = 1
8
Similarly, B-atoms are present at face
centres of a cube.
So, the number of B-atoms per unit cell
1
=6× =3
2
Hence, the formula of compound is A B 3.
53
Solid State
24 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
[CBSE AIPMT 1998]
(a) 288 pm
(c) 144 pm
Ans. (c)
(b) 398 pm
(d) 618 pm
Edge length (a) = 2 r + + 2 r −
a = 2(r + + r − )
a = 508 pm
r + = 110 pm
508
= r+ + r−
2
254 = 110 + r −
r − = 254 − 110 = 144 pm
25 The intermetallic compound LiAg
crystallises in cubic lattice in which
both lithium and silver have
coordination number of eight. The
crystal class is [CBSE AIPMT 1997]
(a) simple cube
(b) body centred cube
(c) face centred cube
(d) None of the above
Ans. (b)
In body centered cubic, each atom/ion
has a coordination number of 8.
26 The edge length of a centred unit
cubic cell is 508 pm. If the radius of
the cation is 100 pm, the radius of
the anion is
[CBSE AIPMT 1996]
(a) 288 pm
(b) 398 pm
(c) 154 pm
(d) 618 pm
Ans. (c)
For centred unit cell,
2(r + + r − ) = a
r = radii of cation
r − = radii of anion
2 (100 + r − ) = 508
508
100 + r − =
2
508
−
r =
− 100
2
+
= 254 − 100
= 154 pm
27 In the fluorite structure, the
coordination number of Ca 2+ ion is
[CBSE AIPMT 1993]
(a) 4
(c) 8
(b) 6
(d) 3
Ans. (c)
Ans. (a)
In fluorite structure each Ca2 + ion is
surrounded by eightF − ions. Thus, the
coordination number of Ca2 + is eight.
Let, in the given crystalNi0.98O
Ni2 + = x and Ni3+ = 0.98 − x
Total charge on M2 + and M3+
= (+2 ) x + (+3) (0.98 − x)
= 2x + 2.94 − 3x = 2.94 − x
As metal oxide is neutral. Therefore,
total charge on cations = total charge on
anions.
28 The number of atoms contained in
a fcc unit cell of a monoatomic
substance is
[CBSE AIPMT 1993]
(a) 1
(c) 4
Ans. (c)
(b) 2
(d) 6
Face centred cubic is also called cubic
close packed arrangement. It has
points at all the corners as well as at the
centre of each of the six faces.
The number of atoms present at corners
1
per unit cell = 8 × = 1.
8
The number of atoms present at faces
1
per unit cell = 6 × = 3
2
∴ Total number of atoms in ccp or
fcc arrangement = 1 + 3 = 4
TOPIC 3
Density and Imperfection
in Solid
29 Which one of the following
compounds show both, Frenkel as
well as Schottky defects?
[NEET (Oct.) 2020]
(a) AgBr
(c) NaCl
Ans. (a)
(b) Agl
(d) ZnS
Frenkel defect is shown by ionic
substances in which there is a large
difference in size of ions.
e.g.,
AgBr, AgI, ZnS
Schottky defect is shown by ionic
substances in which the cation and
anion are of almost similar sizes.
e.g. : AgBr, NaCl, KCl, CsCl
So, AgBr shows both, Frenkel as well as
Schottky defects.
30 Formula of nickel oxide with metal
deficiency defect in its crystal is
Ni 0.98O. The crystal contains Ni 2+
and Ni 3+ ions. The fraction of nickel
existing as Ni 2+ ions in the crystal
is
[NEET (Odisha) 2019]
(a) 0.96
(c) 0.50
(b) 0.04
(d) 0.31
2.94 − x = 2
[Q Charge of oxygen atom = −2]
x = 2.94 − 2 = 0.94
So, the fraction of Ni2 + ions in the crystal
0.94
=
= 0 .96
0.98
31 Which is the incorrect statement?
[NEET 2017]
(a) FeO0.98 has non-stoichiometric metal
deficiency defect
(b) Density decreases in case of crystals
with Schottky’s defect
(c) NaCl(s) is insulator, silicon is
semiconductor, silver is conductor,
quartz is piezoelectric crystal
(d) Frenkel defect is favoured in those
ionic compounds in which sizes of
cation and anions are almost equal
Ans. (a,d)
(a) FeO0.98 has non-stoichiometric metal
excess defect. It occurs due to
missing of a negative ion from its
lattice site, thus leaving a hole which
is occupied by an electron.
Non-stoichiometric ferrous oxide is
FeO0.93 − 0.96 and it is due to metal
deficiency defect. Thus, statement
(a) is incorrect.
(b) In an ionic crystal of A + B − type, if
equal number of cations and anions
are missing from their lattice sites,
the defect is called Schottky defect.
Due to such defect, density of solid
decreases. Thus, statement (b) is
correct.
(c) NaCl-insulator; Silicon (Si) semiconductor, Silver (Ag) conductor; Quartz - piezoelectric
crystal.
Thus, statement (c) is correct.
(d) In an ionic crystal when an ion is
missing from its lattice site and
occupies interstitial site, the defect
is called Frenkel’ defect. This type of
defect is seen in those crystals
where the difference in the size of
cations and anions is very large and
their coordination number is low.
Thus, statement (d) is incorrect.
54
NEET Chapterwise Topicwise Chemistry
32 Lithium has a bcc structure. Its
density is 530 kg m − 3 and its
atomic mass is 6.94 g
mol − 1 . Calculate the edge length of
a unit cell of lithium metal.
(N A = 6.02 × 10 23 mol − 1 )
[NEET 2016, Phase I]
(a) 352 pm
(c) 264 pm
Ans. (a)
(b) 527 pm
(d) 154 pm
Given, Li has a bcc structure.
Density (ρ) = 530 kg-m −3
Atomic mass (M) = 6.94 g mol −1
Avogadro’s number (N A )
= 6.02 × 1023 mol −1
We know that, number of atoms per unit
cell in bcc (Z) = 2.
∴ We have the formula for density,
ZM
ρ=
N Aa 3
where a = edge-length of a unit cell.
ZM
or a = 3
ρN A
=3
. g mol −1
2 × 694
. × 1023 mol −1
0.53 g cm−3 × 602
= 3 435
. × 10 −23 cm−3
= 3.52 × 10 −8 cm
a = 352 pm
33 The correct statement regarding
defects in the crystalline solid is
[CBSE AIPMT 2015]
(a) Schottky defects have no effect on
the density of crystalline solids
(b) Frenkel defects decreases the
density of
crystalline solids
(c) Frenkel defect is a dislocation defect
(d) Frenkel defect is found in halides of
alkaline metals
Ans. (c)
In Frenkel defect, ions in solids dislocate
from their positions. Hence, Frenkel
defect is a dislocation defect.
34 If NaCl is doped with 10 −4 mol % of
SrCl 2 , the concentration of cation
vacancies will be
(NA = 6.023 × 10 23 mol −1 )
[CBSE AIPMT 2007]
(a) 6.023 × 1015 mol−1
(b) 6.023 × 1016 mol−1
(c) 6.023 × 1017 mol−1
(d) 6.023 × 1014 mol−1
Ans. (c)
Doping of NaCl with 10 −4 mol% of SrCl2
means, 100 moles of NaCl are doped with
10 −4 mol of SrCl2 .
∴ 1 mol of NaCl is doped with
10 −4
= 10 −6 mole
100
As each Sr2 + ion introduces one cation
vacancy.
∴ Concentration of cation vacancies
SrCl2 =
= 10 −6 mol/mol of NaCl
= 10 −6 × 6023
.
× 1023 mol −1
= 6023
.
× 10 17 mol −1
35 The appearance of colour in solid
alkali metal halides is generally due
to
[CBSE AIPMT 2006]
(a) F-centres
(b) Schottky defect
(c) Frenkel defect
(d) Interstitial positions
Ans. (a)
F-centres are the sites where anions are
missing and instead electrons are
present and the appearance of colour in
solid alkali metal halides is generally due
to F-centres.
36 CsBr crystallises in a body centred
cubic lattice. The unit cell length is
436.6 pm. Given that the atomic
mass of Cs = 133 u and that of Br =
80 u and Avogadro number being
6.023 × 10 23 mol–1 , the density of
CsBr is
[CBSE AIPMT 2006]
(a) 42.5 g /cm 3
(b) 0.425 g /cm 3
(c) 8.25 g /cm 3
(d) 4.25 g /cm 3
Ans. (d)
Density of CsBr =
Z×M
a3 × N0
Z → number of atoms in the bcc unit cell
=2
M → molar mass of CsBr = 133 + 80 = 213
a → edge length of unit cell = 436.6 pm
= 436.6 × 10 −10 cm
2 × 213
∴ Density =
(436.6 × 10 −10 ) 3 × 6.023 × 1023
= 8.49 × 10 −7 × 10 7 g / cm3
= 8.50 g / cm 3
8 .50
For a unit cell =
2
= 4 .25 g /cm3
37 The pyknometric density of sodium
chloride crystal is 2.165
× 10 3 kg m–3 while its X-ray density
is 2.178 × 10 3 kg m–3 . The fraction
of unoccupied sites in sodium
chloride crystal is
[CBSE AIPMT 2003]
(a) 5.96 × 10−1
(b) 5.96 × 10−3
(c) 5.96
(d) 5.96 × 10−2
Ans. (b)
The fraction of unoccupied site in
sodium chloride crystal
X - ray density – pyknometric density
=
X - ray density
=
=
2.178 × 10 3 − 2.165 × 10 3
2.178 × 10 3
0.013 × 10 3
2.178 × 10 3
13
=
2178
= 5 .96 × 10 −3.
38 The second order Bragg diffraction
of X-rays with λ = 1.0 Å from a set
of parallel planes in a metal occurs
at an angle 60°. The distance
between the scattering planes in
the crystals is [CBSE AIPMT 1998]
(a) 0.575 Å
(c) 2.00 Å
Ans. (d)
(b) 1.00 Å
(d) 1.17 Å
According to Bragg’s equation,
nλ = 2 d sinθ
n=2
λ=1
deflected angleθ = 60 °
d=?
Distance between two plane of crystal.
2 × 1 = 2 × d × sin 60 °
3
2× 1=2×d ×
2
2
d=
3
2
=
= 1.17 Å
1.7
39 Schottky defect in a crystal is
observed when [CBSE AIPMT 1998]
(a) an ion leaves its normal site and
occupies an interstitial site
(b) unequal number of cations and
anions are missing from the lattice
55
Solid State
(c) density of the crystal is increased
(d) equal number of cations and anions
are missing from the lattice
Ans. (d)
Schottky defect in crystals is observed
when equal number of cations and
anions are missing from the lattice. So,
the crystal remains neutral, e.g. NaCl.
40 When electrons are trapped into
the crystalline anion vacancy, the
defect is known as
[CBSE AIPMT 1994]
(a) Schottky defect
(b) Stoichiometric defect
(c) Frenkel defect
(d) F-centres
Ans. (d)
When the electrons are trapped in anion
vacancy, then the defect is called
F-centres. This defect is observed in
LiCl, NaCl, KCl, etc.
Ans. (b)
TOPIC 4
Electrical and Magnetic
Properties of Solids
41 With which one of the following
elements silicon should be doped so
as to give p-type of semiconductor?
[CBSE AIPMT 2008]
(a) Germanium
(c) Selenium
Ans. (d)
(b) Arsenic
(d) Boron
p-type of semiconductor are obtained by
doping silicon or germanium with elements
of group 13 like B, Al, Ga or In so silicon is
doped with boron.
42 A solid with high electrical and
thermal conductivity from the
following is
[CBSE AIPMT 1994]
(a) Si
(b) Li
(c) NaCl (d) ice
In lithium crystal free electrons are
present so, Li have high thermal and
electrical conductivity.
43 On doping Ge metal with a little
of In or Ga, one gets
[CBSE AIPMT 1993]
(a) p-type semiconductor
(b) insulator
(c) n-type semiconductor
(d) rectifier
Ans. (a)
When we add some of In or Ga (group
13 element) into Ge (group number 14
element), then electron vacancy is
created which is called holes. In this
type of semiconductor the charge
carrier are holes (positive), so it is
called p-type semiconductor.
7
Solutions
TOPIC 1
Alternate Method N =
Expression of
Concentration
01 Which of the following is
dependent on temperature?
[NEET 2017]
(a) Molality
(b) Molarity
(c) Mole fraction
(d) Weight percentage
Ans. (b)
Molarity and normality are temperature
dependent because they involve volume
of solutions. Volume is dependent on
temperature.
Molarity (M) =
Moles of solute
Volume of solution (in L)
Molality, mole fraction and weight
percentage does not depend on
temperature because they involve
masses of solute and solvent.
02 What is the mole fraction of the
solute in a 1.00 m aqueous
solution?
[CBSE AIPMT 2015]
(a) 0.177
(c) 0.0354
Ans. (d)
(b) 1.770
(d) 0.0177
1000 × n
Molality (m) =
N×M
where, n = number of moles of solute
N = number of moles of solvent
M = molar mass of solvent
Given, m = 1
1000 × n
1=
∴
N × 18
⇒
or
18
n
=
N 1000
18
n
=
= 0.0177
n + N 1018
1000
= 55.5 mol
18
n= 1
[Q 1 m solution implies that 1 mole of
solute is present in 1 kg or 1000 g water]
∴Mole fraction of solute
n
1
=
=
= 0.0177
n + N 1 + 55.5
03 25.3 g of sodium carbonate,
Na 2CO 3 is dissolved in enough
water to make 250 mL of solution.
If sodium carbonate dissociates
completely, molar concentration of
sodium ion, Na + and carbonate ion,
CO 2−
3 are respectively (Molar mass
of Na 2CO 3 = 106 g mol −1 )
[CBSE AIPMT 2010]
(a) 0.955 M and 1.910 M
(b) 1.910 M and 0.955 M
(c) 1.90 M and 1.910 M
(d) 0.477 M and 0.477 M
Ans. (b)
Molarity
Number of moles of solute
=
× 1000
Volume of solution (in mL)
25. 3 × 1000
=
= 0.9547 ≈ 0.955 M
106 × 250
Na2 CO3 in aqueous solution remains
dissociated as
Na2 CO3
2 Na+ + CO23−
x
a
2x
x
Since, the molarity ofNa2 CO3 is 0.955 M,
the molarity of CO23− is also 0.955 M and
that of Na+ is 2 × 0.955 = 1.910 M
04 Concentrated aqueous sulphuric
acid is 98% H2SO 4 by mass and has
a density of 1.80 g mL−1 . Volume of
acid required to make one litre of
0.1 M H2 SO4 solution is
[CBSE AIPMT 2007]
(a) 11.10 mL
(c) 22.20 mL
(b) 16.65 mL
(d) 5.55 mL
Ans. (d)
Normality
Weight percentage × density × 10
=
Equivalent weight
98 × 1.8 × 10
=
= 36 N
49
N2 = 2 × 0.1 N = 0.2 N
N 1V1 = N2V2
36 × V = 0.2 × 1000
0.2 × 1000
36
= 5.55 mL (dissociation of acid)
V=
05 The mole fraction of the solute in
one molal aqueous solution is
[CBSE AIPMT 2005]
(a) 0.027
(c) 0.018
Ans. (c)
(b) 0.036
(d) 0.009
Molality of solution = mole of solute per
kg of solvent
So, 1 m = 1 mole of solute per 1000 g of
solvent
Hence,
moles of solute in 1 m aqueous solution = 1
moles of solvent in 1 m aqueous solution
1000
= 55 .55
=
18
Mole fraction of solute in 1 m solution
1
1
=
=
1 + 55.55 56.55
= 0.0176 ≈ 0.018
06 1 M and 2.5 L NaOH solution is
mixed with another 0.5 M and 3 L
NaOH solution. Then, find out the
molarity of resultant solution.
[CBSE AIPMT 2002]
(a) 0.80 M
(c) 0.73 M
Ans. (c)
(b) 1.0 M
(d) 0.50 M
Moles of 2.5 L of 1M NaOH = 2 .5 × 1 = 2 .5
57
Solutions
Moles of 3.0 L of 0.5 M NaOH
= 3.0 × 0.5 = 1.5
Total moles of NaOH in solution
= 2 .5 + 1.5 = 40
.
(Total volume of solution
= 2 .5 + 3.0 = 5 .5 L)
Thus,
M1 × V1 = M2 × V2
4 = M2 × 5 .5
∴Molarity of resultant solution,
4
M
M2 =
5.5
≈0.73 M
07 Molarity of liquid HCl, if density of
solution is 1.17 g/cc is
[CBSE AIPMT 2001]
(a) 36.5
(c) 32.05
Ans. (c)
(b) 18.25
(d) 42.10
Density = 1.17 g /cc = 1170 g/L
Strength in g / L
Molarity of solution =
Molecular weight
1170
M
=
36.5
= 32.05 M
08 How many gram of a dibasic acid
(mol. wt. 200) should be present in
100 mL of the aqueous solution to
give 0.1 N?
[CBSE AIPMT 1999]
(a) 1 g
(c) 10 g
Ans. (a)
(b) 2 g
(d) 20 g
Equivalent weight of dibasic acid
Molecular weight
=
2
200
= 100
E=
2
Strength = 0.1 N, m = ?V = 100 mL
Mass 1000
Normality (N ) =
×
E
V (L)
E NV 100 × 100 × 0.1
w=
=
= 1g
1000
1000
09 The volume strength of 1.5 N H2O 2
solution is
(a) 4.8
(c) 8.4
Ans. (c)
[CBSE AIPMT 1997]
(b) 5.2
(d) 8.8
Normality = 1.5 N
Equivalent weight ofH2O2 = 17
So,, strength of the solutions, S = E × N
=17 × 1.5 = 25.5
2H2O2 → 2H2O + O2
= 2 × 34 = 68 g
Q 68 g of H2O2 produce O2 at NTP = 22 .4 L
∴ 25.5 g of H2O2 will produce
22.4
=
× 25.5
68
= 8.4 L of O2
10 Which one of the following modes
of expressing concentration is
independent of temperature?
[CBSE AIPMT 1995, 92]
(a) Molarity
(c) Formality
Ans. (b)
(b) Molality
(d) Normality
Molality is the best method of expressing
concentration of solution because in
molality we take mass of solvent which is
independent of temperature, so molality
of solution is independent of
temperature.
TOPIC 2
Henry’s Law and
Raoult’s Law
11 The correct option for the value of
vapour pressure of a solution at
45°C with benzene to octane in
molar ratio 3 : 2 is
[At 45°C vapour pressure of
benzene is 280 mm Hg and that of
octane is 420 mm Hg. Assume
ideal gas].
[NEET 2021]
(a) 160 mm of Hg
(c) 336 mm of Hg
Ans. (c)
(b) 168 mm of Hg
(d) 350 mm of Hg
n
3
Molar ratio of benzene to octane, B =
nO 2
Let nB = 3x mol, nO = 2x mol
Total number of moles
= nB + nO = 3x + 2x = 5x mol
Mole fraction of benzene,
nB
3x 3
χB =
=
= .
nB + nO 5x 5
Mole fraction of octane,
nO
2x 2
χO =
=
=
nB + nO 5x 5
Vapour pressure of benzene,
pB° = 280 mm Hg
Vapour pressure of octane,
pO° = 420 mm Hg
Total vapour pressure of solution,
pS = χ BpB° + χ O pO°
3
2
× 280 + × 420
5
5
= 3 × 56 + 2 × 84
= 168 + 168
= 336 mm of Hg
=
12 A mixture of N 2 and Ar gases in a
cylinder contains 7 g of N 2 and 8 g
of Ar. If the total pressure of the
mixture of the gases in the cylinder
is 27 bar, the partial pressure of N 2
is
[Use atomic masses (in g mol −1 ) :
N = 14, Ar =40] [NEET (Sep.) 2020]
(a) 12 bar
(c) 18 bar
Ans. (b)
(b) 15 bar
(d) 9 bar
From Dalton’s law of partial pressure of
gases. We know,
pi = χ i × p
where,
pi = partial pressure ofith component
χ i = mole-fraction of ith component
p = total pressure = 27 bar
(partial pressure) N 2
= (mole-fraction) N 2 × p
nN 2
=
×p
nN 2 + nAr
=
7
28
7
8
+
28 40
× 27 bar = 15 bar
∴ Partial pressure of N2 in the mixture is
15 bar.
13 The mixture which shows positive
deviation from Raoult’s law is
[NEET (Sep.) 2020]
(a) Benzene + toluene
(b) Acetone + chloroform
(c) Chlorethane + bromoethane
(d) Ethanol + acetone
Ans. (d)
Let us study the nature of binary
solutions made of two volatile liquids:
(a) (Benzene + Toluene) Ideal solution,
obeys Raoult’s law.
(b) (Acetone + Chloroform) Non-ideal
solution, shows negative deviation
from Raoult’s law.
(c) (Chloroethane + Bromoethane) Ideal
solution, obeys Raoult’s law.
(d) (Ethanol + Acetone) Non-ideal
solution, shows positive deviation
from Raoult’s law.
58
NEET Chapterwise Topicwise Chemistry
14 The mixture that forms maximum
boiling azeotrope is
[NEET (National) 2019]
(a) ethanol + water
(b) acetone + carbon disulphide
(c) heptane + octane
(d) water + nitric acid
Ans. (d)
Key Idea The binary liquid mixtures
having the same composition in liquid and
vapour phase and boil at constant
temperature are called azeotropic
mixtures or azeotropes.
The solutions that show large negative
deviation from Raoult’s law forms
maximum boiling azeotrope. For, e.g.
nitric acid and water.
The remaining option containing
different mixtures forms minimum
boiling azeotrope.
15 Which of the following statements
is correct regarding a solution of
two compounds A and B exhibiting
positive deviation from ideal
behaviour?
[NEET (Odisha) 2019]
(a) Intermolecular attractive forces
between A—A and B—B are stronger
than those between A—B.
(b) ∆mix H = 0 at constant T and p
(c) ∆mix V = 0 at constant T and p
(d) Intermolecular attractive forces
between A—A and B—B are equal to
those between A—B.
Ans. (a)
Solution exhibits positive deviation from
ideal behaviour if the intermolecular
interactions, i.e., A  B interactions are
more stronger than in pure components,
i.e. between A  B or B  B. Thus.
option (a) is correct. The remaining
options are valid only for ideal solutions.
16 In water saturated air the mole
fraction of water vapour is 0.02. If
the total pressure of the saturated
air is 1.2 atm, the partial pressure
of dry air is
[NEET (Odisha) 2019]
(a) 1.18 atm
(c) 1.176 atm
Ans. (c)
(b) 1.76 atm
(d) 0.98 atm
Partial pressure of dry air = total
pressure × mole fraction of dry air
⇒ pdry air = ptotal × λ dry air
Given, λ saturated air = 0.02
λ dry air = 1 − 0.02 = 0.98
ptotal = 12
. atm
∴ pdry air = 12
. atm × 0.98 = 1. 176 atm
17 Which of the following statements
about the composition of the
vapour over an ideal 1 : 1 molar
mixture of benzene and toluene is
correct? Assume that the
temperature is constant at 25°C.
[NEET 2016, Phase I]
(Given, vapour pressure data at
25°C, benzene = 12.8 kPa, toluene
= 3.85 kPa)
(a) The vapour will contain a higher
percentage of toluene
(b) The vapour will contain equal
amounts of benzene and toluene
(c) Not enough information is given to
make a prediction
(d) The vapour will contain a higher
percentage of benzene
Ans. (d)
Since, component having higher vapour
pressure will have higher percentage in
vapour phase. Benzene has vapour
pressure 12.8 kPa which is greater than
toluene 3.85 kPa.
Therefore, the vapour will contain a
higher percentage of benzene.
18 Which one of the following is
incorrect for ideal solution?
[NEET 2016, Phase II]
(a) ∆H mix = 0
(b) ∆Umix = 0
(c) ∆P = Pobs. − Pcalculated by Raoult’ s law = 0
(d) ∆G mix = 0
Ans. (d)
Key Idea For this problem, the following
expression can be used.
∆G mix = ∆Hmix −T∆S mix
For an ideal gas
∆Hmix = 0; ∆U mix = 0; ∆S mix ≠ 0
Putting all these values in the
expression,
∆G mix = ∆Hmix −T∆S mix
⇒ ∆G mix = 0 −T∆S mix
∴
∆G mix ≠ 0
Thus, option (d) is incorrect.
19 A gas such as carbon monoxide
would be most likely to obey the
ideal gas law at [CBSE AIPMT 2015]
(a) high temperatures and low pressures
(b) low temperatures and high pressures
(c) high temperatures and high pressures
(d) low temperatures and low pressures
Ans. (a)
Real gases show ideal gas behaviour at
high temperatures and low pressures.
20 Which one is not equal to zero for
an ideal solution?
[CBSE AIPMT 2015]
(a) ∆H mix
(b) ∆S mix
(c) ∆Vmix
(d) ∆P = Pobserved − PRaoult
Ans. (b)
For an ideal solution
(i) There will be no change in volume on
mixing the two components i.e.
∆Vmixing = 0
(ii) There will be no change in enthalpy
so ∆Hmixing = 0
So, ∆S mix ≠ 0 for an ideal solution.
21 p A and p B are the vapour pressure
of pure liquid components A and B,
respectively of an ideal binary
solution. If χ A represents the mole
fraction of component A, the total
pressure of the solution will be
[CBSE AIPMT 2012]
(a) pA + χ A (pB − pA) (b) pA + χ A (pA − pB)
(c) pB + χ A (pB − pA) (d) pB + χ A (pA − pB)
Ans. (d)
According to Raoult’s law, if volatile
liquid added in pure solvent, then total
pressure equal to sum of the partial
pressure of volatile liquid and solvent.
Total pressure, pT = pA′ + pB′
…(i)
We know that, pA′ = pA χ A
pB′ = pBχ B
Substituting the values of pA′ and pB′ in
Eq. (i)
pT = pA χ A + pBχ B
Mole fraction, χ A + χ B = 1 ⇒ χ B = 1 − χ A
On substituing in above equation,
pT = pA χ A + pB (1 − χ A )
= pA χ A + pB − pBχ A
∴
pT = pB + χ A (pA − pB)
22 A solution of acetone in ethanol
[CBSE AIPMT 2006]
(a) shows a negative deviation from
Raoult’s law
(b) shows a positive deviation from
Raoult’s law
(c) behaves like a near ideal solution
(d) obeys Raoult’s law
Ans. (b)
A solution of acetone in ethanol shows a
positive deviation from Raoult’s law due
to miscibility of these two liquids with
difference of polarity and length of
hydrocarbon chain.
59
Solutions
23 A solution has 1 : 4 mole ratio of
pentane to hexane. The vapour
pressure of the pure hydrocarbons
at 20°C are 440 mm of Hg for
pentane and 120 mm of Hg for
hexane. The mole fraction of
pentane in the vapour phase would
be
[CBSE AIPMT 2005]
(a) 0.549
(c) 0.786
Ans. (d)
(b) 0.200
(d) 0.478
Total vapour pressure of mixture
= Vapour pressure of
pentane in mixture
+ vapour pressure of hexane
in mixture
Since, the ratio of pentane to hexane
= 1 :4
1
∴ Mole fraction of pentane =
5
4
Mole fraction of hexane =
5
= (mole fraction of pentane × vapour
pressure of pentane) + (mole fraction of
hexane × vapour pressure of hexane)
4
1
=  × 440 + × 120  = 184 mm
5

5
Q Vapour pressure of pentane in mixture
= Vapour pressure of mixture × mole
fraction of pentane in vapour phase
88 = 184 × mole fraction of pentane in
vapour phase
∴Mole fraction of pentane in vapour phase
88
=
= 0.478
184
24 The vapour pressure of two liquids
P and Q are 80 and 60 torr,
respectively. The total vapour
pressure of solution obtained by
mixing 3 moles of P and 2 moles of
Q would be
[CBSE AIPMT 2005]
(a) 140 torr
(c) 68 torr
Ans. (d)
(b) 20 torr
(d) 72 torr
3
3
=
3+2 5
2
2
Mole fraction of Q =
=
3+2 5
Mole fraction of P =
Hence, total vapour pressure
= Mole fraction of
P × vapour pressure of P + mole fraction
of Q × vapour pressure of Q
2
3

=  × 80 + × 60  = 48 + 24 = 72 torr
5

5
25 Formation of a solution from two
components can be considered as
[CBSE AIPMT 2003]
I. pure solvent → separated
solvent molecules, ∆H 1
II. pure solute → separated solute
molecules, ∆H 2
III. separated solvent and solute
molecules → solution, ∆H 3
Solution so formed will be ideal, if
(a) ∆H sol. = ∆H 1 − ∆H2 − ∆H 3
(b) ∆H sol. = ∆H 3 − ∆H 1 − ∆H2
(c) ∆H sol. = ∆H 1 + ∆H2 + ∆H 3
(d) ∆H sol. = ∆H 1 + ∆H2 − ∆H 3
Ans. (c)
For an ideal solution,
∆Hsol = ∆H1 + ∆H2 + ∆H3
26 A solution containing components
A and B follows Raoult’s law, when
[CBSE AIPMT 2002]
(a) A - B attraction force is greater than
A - A and B - B
(b) A - B attraction force is less than A - A
and B - B
(c) A - B attraction force remains same
as A - A and B - B
(d) volume of solution is different from
sum of volumes of solute and
solvent
Ans. (c)
Raoult’s law is valid for ideal solution
only. The two components A and B
follows the condition of Raoult’s law only
when the force of attraction between A
and B is equal to the force of attraction
between A – A and B – B.
While non-ideal solutions exhibit either
positive or negative deviations from
Raoult’s law. When A – B attraction force
is greater than A – A and B – B, then
solution shows negative deviation and
when A–B attraction force is less than
A–A and B–B, the solution shows positive
deviation.
27 According to Raoult’s law, relative
lowering of vapour pressure of a
solution is equal to [CBSE AIPMT 1995]
(a) moles of solute
(b) moles of solvent
(c) mole fraction of solute
(d) mole fraction of solvent
Ans. (c)
According to Raoult’s law, the relative
lowering of vapour pressure is equal to
the mole fraction of solute, i.e.
p° − p
= χB
p°
χ B = mole fraction of solute
28 The relative lowering of vapour
pressure is equal to the ratio
between the number of
[CBSE AIPMT 1991]
(a) solute molecules to the solvent
molecules
(b) solute molecules to the total
molecules in solution
(c) solvent molecules to the total
molecules in the solution
(d) solvent molecules to the total
number of ions of the solute
Ans. (b)
According to Raoult’s law the relative
lowering of vapour pressure is equal to
mole fraction of solute, i.e. the ratio of
number of moles of solute to total
number of moles of all component in
solution.
29 All form ideal solution except
[CBSE AIPMT 1988]
(a) C 6H6 and C 6H5CH3
(b) C2H5Cl and C2H5I
(c) C 6H5Cl and C 6H5Br
(d) C2H5I and C2H5OH
Ans. (d)
When we mix C2H5I and C2H5OH, there is
change in enthalpy and volume, so it is
an example of non-ideal solution.
30 An ideal solution is formed when its
components
[CBSE AIPMT 1988]
(a) have no volume change on mixing
(b) have no enthalpy change on mixing
(c) have both the above characteristics
(d) have high solubility
Ans. (c)
In ideal solution there is no change in
enthalpy and volume after mixing, i.e.
∆Vmixing = 0,
∆Hmixing = 0
and
Tmix = 0
60
NEET Chapterwise Topicwise Chemistry
TOPIC 3
Colligative Properties
31 The following solutions were
prepared by dissolving 10 g of
glucose (C 6H12O 6 ) in
250 mL of water (p 1 ), 10 g of urea
(CH4N 2O) in 250 mL of water (p 2 )
and 10 g of sucrose
(C 12H22O 11 ) in 250 mL of water (p 3 ).
The right option for the decreasing
order of osmotic pressure of these
solutions is
[NEET 2021]
(a) p2 > p1> p3
(c) p2 > p3 > p1
(b) p1 > p2 > p3
(d) p3 > p1 > p2
Ans. (a)
van’t Hoff factor for glucose, urea and
sucrose is one as all are non-ionic.
Mass of glucose = 10 g
Molar mass of glucose = 180 g mol −1
Number of moles of glucose
10
1
mol = 0.056 mol
=
=
180 18
Mass of urea = 10 g
Molar mass of urea = 60 g mol −1
10 1
Number of moles of urea =
= mol
60 6
= 0.167 mol
Mass of sucrose = 10 g
Molar mass of sucrose = 342 g mol −1
Number of moles of sucrose
10
=
= 0.029 mol
342
Osmotic pressure is a colligative
property which depends on the amount
of solute present in the solution. As the
amount of urea is more than that of
glucose and sucrose.
Amount of solute : Urea > Glucose >
Sucrose
∴Osmotic pressure, p2 > p1 > p3
32 Isotonic solutions have same
[NEET (Oct.) 2020]
(a) vapour pressure
(b) freezing temperature
(c) osmotic pressure
(d) boiling temperature
Ans. (c)
Isotonic solutions have same osmotic
pressure ( π) at a given temperature.
π = CRT
When, two solution have same molar
concentration (C) their osmotic pressure
will be equal.
33 If 8 g of a non-electrolyte solute is
dissolved in 114 g of n-octane to
reduce its vapour pressure to 80%,
the molar mass (in g mol −1 ) of the
solute is [Given that, molar mass of
n-octane is 114 g mol −1 ]
[NEET (Oct.) 2020]
(a) 40
(b) 60
Ans. (*)
(c) 80
(d) 20
Does not match with the options.
Here, relative lowering of given vapour
pressure,
∆p p° − p° × 0.8
=
= 0.2
p°
p°
wB
Now,
nB
MB
∆p
= xB =
=
p°
nB + nA w B + w A
MB MA
p° = vapour pressure of pure solvent
(octane)
Mass of solute (w B) = 8 g
Mass of solvent (w A ) = 114g
Molar mass of solute (MB) = ?
Molar mass of solvent (octane) (MA )
= 114g mol − 1.
8 / MB
⇒ MB = 32g mol − 1
⇒ 0.2 =
8
114
+
MB 114
34 If molality of the dilute solution is
doubled, the value of molal
depression constant (K f ) will be
[NEET 2017]
(a) doubled
(c) tripled
Ans. (d)
(b) halved
(d) unchanged
For a dilute solution, the depression in
freezing point (∆Tf ) is directly
proportional to molality (m) of the
solution.
∆Tf ∝ m or ∆Tf = K f m
Where,K f is called molal depression
constant or freezing point depression
constant or cryoscopic constant. The
value of K f depends only on nature of the
solvent and independent of composition
of solute particles, i.e. does not depend
on the concentration of solution.
35 At 100°C the vapour pressure of a
solution of 6.5 g of a solute in 100 g
water is 732 mm. If K b = 0.52, the
boiling point of this solution will be
[NEET 2016, Phase I]
(a) 100°C
(c) 103°C
(b) 102°C
(d) 101°C
Ans. (d)
From Raoult’s law of partial pressure,
pA° − pS nB
=
pS
nA
⇒
760 − 732 WB × MA
=
732
MB × WA
⇒
6.5 × 18
28
=
732 MB × 100
⇒
MB = 30.6
∴
∆Tb = 0.52 ×
6.5 × 1000
= 1.10
30.6 × 100
∴ Boiling point = 100 + 1.10 = 101.1°C
≈ 101° C
36 An aqueous solution is 1.00 molal in
KI. Which change will cause the
vapour pressure of the solution to
increase?
[CBSE AIPMT 2010]
(a) Addition of NaCl
(b) Addition of Na2 SO 4
(c) Addition of 1.00 molal KI
(d) Addition of water
Ans. (d)
Key Idea Vapour pressure depends upon
the surface area of the solution. Larger
the surface area, higher is the vapour
pressure.
Addition of solute decreases the vapour
pressure as some sites of the surface
are occupied by solute particles,
resulting in decreased surface area.
However, addition of solvent, i.e.
dilution, increases the surface area of
the liquid surface, thus results in
increased vapour pressure.
Hence, addition of water to the aqueous
solution of (1 molal) KI, results in
increased vapour pressure.
37 A solution of sucrose (molar mass
= 342 g mol −1 ) has been prepared
by dissolving 68.5 g of sucrose in
1000 g of water. The freezing point
of the solution obtained will be
(k f for water = 1.86 K kg mol −1 )
[CBSE AIPMT 2010]
(a) − 0.372 °C
(c) + 0.372 °C
Ans. (a)
(b) − 0.520°C
(d) − 0.570°C
Depression in freezing point,
∆Tf = kf × m
W × 1000
where, m = molality = B
MB ⋅ WA
=
68.5 × 1000 68. 5
=
342 × 1000 342
61
Solutions
68.5
= 0.372 °C
342
0
Tf = TF − ∆TF
= 0 − 0.372° C
= − 0.372° C
∆Tf = 1.86 ×
38 A solution containing 10 g per dm 3
of urea (molecular mass = 60 g
mol–1 ) is isotonic with a 5% solution
of a non-volatile solute. The
molecular mass of this non-volatile
solute is
[CBSE AIPMT 2006]
(a) 250 g mol–1
(c) 350 g mol–1
Ans. (b)
(b) 300 g mol–1
(d) 200 g mol–1
10 g per dm 3 of urea is isotonic with 5%
solution of a non-volatile solute. Hence,
between these solutions osmosis is not
possible, so their molar concentrations
are equal to each other.
Thus, molar concentration of urea
solution
10 g / dm3
=
Molecular weight of urea
10
1
M= M
=
60
6
Molar concentration of 5% non-volatile
solute
50 g / dm3
=
Molecular weight of non −
volatile solute
50
=
M
m
Both solutions are isotonic to each
other, therefore
1 50
=
6 m
or m = 50 × 6 = 300 g mol −1
39 During osmosis, flow of water
through a semipermeable
membrane is [CBSE AIPMT 2006]
(a) from solution having higher
concentration only
(b) from both sides of semipermeable
membrane with equal flow rates
(c) from both sides of semipermeable
membrane with unequal flow rates
(d) from solution having lower
concentration only
Ans. (d)
During osmosis, flow of water through a
semipermeable membrane is from
solution having lower concentration
only.
40 1.00 g of a non-electrolyte solute
(molar mass 250 g mol–1 ) was
dissolved in 51.2 g of benzene. If
the freezing point depression
constant, k f of benzene is 5.12 K kg
mol–1 , the freezing point of
benzene will be lowered by
[CBSE AIPMT 2006]
(a) 0.4 K
(c) 0.5 K
Ans. (a)
(b) 0.3 K
(d) 0.2 K
Molality of non-electrolyte solute
Weight of solute in (g)
molecular
weight of solute
=
Weight of solvent in (kg)
1
250
=
0.0512
1
=
= 0.0781 m
250 × 0.0512
∆Tf = kf × molality of solution
= 5.12 × 0.0781 ≈ 0.4 K
41 A solution of urea (mol. mass 56 g
mol–1 ) boils at 100.18°C at the
atmospheric pressure. If k f and k b
for water are 1.86 and 0.512 K kg
mol–1 respectively, the above
solution will freeze at
[CBSE AIPMT 2005]
(a) –6.54°C
(c) 0.654°C
Ans. (d)
(b) 6.54°C
(d) –0.654°C
According to depression of freezing
point,
∆Tf = kf × molality of solution
According to elevation of boiling point,
∆Tb = kb × molality of solution
∆Tf
k
or
= f
∆Tb kb
Given that
∆Tb = T2 − T1 = 100.18 − 100 = 0.18
kb for water = 1.86 K kg mol –1
kb for water = 0.512 K kg mol –1
∆Tf
1.86
=
∴
0.18 0.512
1.86 × 0.18
∆Tf =
= 0.6539 ≈ 0.654
0.512
∆Tf = T1 − T2
0.654 = 0 ° C − T2
∴
T2 = − 0.654° C
(T2 → freezing point of aqueous urea
solution)
42 A solution contains non-volatile
solute of molecular mass, M 2 .
Which of the following can be used
to calculate the molecular mass of
solute in terms of osmotic
pressure?
[CBSE AIPMT 2002]
m 
 m  RT
(a) M2 =  2  VRT (b) M2 =  2 
π 
V  π
 m2 
m  π
(c) M2 =   πRT (d) M2 =  2 
V
 
 V  RT
Ans. (b)
For dilute solution,
pV = nRT (p = π, osmotic pressure)
n
or πV = nRT or π = RT
V
m
m RT
⇒ πV = 2 RT ⇒ M2 = 2
M2
πV
where, π =osmotic pressure
V = volume of solution
n = number of moles of solute
m2 =mass of solute
M2 = molecular mass of solute
43 Pure water can be obtained from
sea water by
[CBSE AIPMT 2001]
(a) centrifugation (b) plasmolysis
(c) reverse osmosis (d) sedimentation
Ans. (c)
Reverse osmosis The minimum
external pressure applied to a solution
separated from a solvent by
semipermeable membrane to prevent
osmosis, is called osmotic pressure.
When the pressure applied to solution is
more than osmotic pressure, solute will
pass from the solution into solvent
through the semipermeable membrane.
This phenomenon is known as reverse
osmosis.
The osmotic pressure of sea water is 25
atm at 15°C. When pressure greater than
26 atm is applied on sea water separated
by a rigid semipermeable membrane,
pure water is obtained. This is also called
desalination of sea water.
44 Which of the following colligative
property can provide molar mass of
proteins (or polymers or colloids)
with greatest precision?
[CBSE AIPMT 2000]
(a)
(b)
(c)
(d)
Osmotic pressure
Elevation in boiling point
Depression in freezing point
Relative lowering of vapour pressure
62
NEET Chapterwise Topicwise Chemistry
Ans. (a)
Ans. (c)
Ans. (c)
Osmotic pressure is a colligative
property which is used to find the
molecular weight of polymer because
other colligative properties give so low
value of molecular weight that it cannot
be measured accurately.
Isotonic solutions are the solutions
having same osmotic pressure.
Osmotic pressure of 5% cane sugar
solution
50 g / L
( π 1) = C × R × T =
× 0.0821 × T
342
Osmotic pressure of 1% solution of
substance
10 g / L
× 0.0821 × T
X ( π2 ) =
M
Both are isotonic
So,
π 1 = π2
10
50
or
× 0.0821 × T =
× 0.0821 × T
342
M
342
= 68.4
∴ M (molecular weight of X) =
5
We know that, according to Raoult’s law
p° − p
= χB
p°
nB
p° − 60
0.2
=
=
p°
nA + nB 0.2 + 0.8
45 The vapour pressure of benzene at
a certain temperature is 640
mmHg. A non-volatile and
non-electrolyte solid, weighing
2.175 g is added to 39.08 g of
benzene. If the vapour pressure of
the solution is 600 mm Hg, what is
the molecular weight of solid
substance?
[CBSE AIPMT 1999]
(a) 49.50 (b) 59.60 (c) 69.40 (d) 79.82
Ans. (c)
According to Raoult’s law,
p° − p w × M
=
p°
m× W
640 − 600 2 .175 × 78
=
640
m × 3908
.
(M for C6H6 = 78)
2 .175 × 78 × 640
m=
40 × 3908
.
= 6945
. ≈ 694
.
46 If 0.15 g of solute, dissolved in 15 g
of solvent, is boiled at a
temperature higher by 0.216°C,
than that of the pure solvent, the
molecular weight of the substance
is (molal elevation constant for the
solvent is 2.16°C)
[CBSE AIPMT 1999]
(a) 1.01
(c) 10.1
Ans. (d)
(b) 10
(d) 100
48 The vapour pressure of a solvent
decreased by 10 mm in two
columns 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 the decrease in the
vapour pressure is to be 20 mm of
mercury
[CBSE AIPMT 1998]
(a) 0.8
(b) 0.6
Ans. (b)
∆Tb = 0.216° C kb = 2 .16, m = ?
1000 × kb w
m=
×
W
∆Tb
=
1000 × 2 .16 × 0.15
= 100
0.216 × 15
47 A 5% solution of cane sugar (mol.
wt. = 342) is isotonic with 1%
solution of a substance X. The
molecular weight of X is
For other solution of same solvent
20
n
20
n
or
=
=
p° n + N
50 n + N
n
(mole fraction of solute)
n+ N
Q Mole fraction of solvent + mole fraction
of solute = 1
So, mole fraction of solvent = 1 − 0.4 = 0.6
49 The vapour pressure, at a given
temperature, of an ideal solution
containing 0.2 mole of a
non-volatile solute and 0.8 mole of
solvent is 60 mm of Hg. The vapour
pressure of the pure solvent at the
same temperature is
[CBSE AIPMT 1998]
(a) 34.2
(c) 68.4
(b) 171.2
(d) 136.8
(d) 0.2
According to Raoult’s law, the relative
lowering of vapour pressure is equal to
the mole fraction of solute, i.e.
p° − p
n
∆p
n
or
=
=
p°
n+ N
p° n + N
10
= 0.2 ⇒ p° = 50 mm
p°
0.4 =
w = 0.15 g, W = 15 g,
(c) 0.4
[CBSE AIPMT 1996]
(a) 150 mm of Hg
(c) 75 mm of Hg
(b) 60 mm of Hg
(d) 120 mm of Hg
0.2 2 1
= =
1.0 10 5
p°
p° − 60 =
5
p°
⇒
p° −
= 60
5
5p° − p°
= 60
5
4p° = 60 × 5
60 × 5 300
p° =
=
= 75 mm of Hg
4
4
=
50 Vapour pressure of benzene at
30°C is 121.8 mm. When 15 g of a
non-volatile solute is dissolved in
250 g of benzene its vapour
pressure decreased to 120.2 mm.
The molecular weight of the solute
is (mol. weight of solvent = 78)
[CBSE AIPMT 1995]
(a) 356.2 (b) 456.8 (c) 530.1 (d) 656.7
Ans. (a)
Given, vapour pressure of pure benzene
( p° ) = 121.8 mm
Vapour pressure of solution
(p) = 120.2 mm
Mass of solute (w B) = 15 g
Mass of benzene (w A ) = 250g
Molar mass of solvent (mA ) = 78
Molecular weight of solute (mB) = ?
Hence, according to Raoult’s law,
p° − p w B × mA
=
p°
mB × w A
p° = vapour pressure of solvent
p = vapour pressure of the solution
15 × 78
121.8 − 120.2
=
121.8
mB × 250
mB =
121.8 × 15 × 78
250 × (121.8 − 120.2)
=
142506 142506
=
= 356.26 g mol −1
250 × 1.6
400
51 Which one is a colligative property?
[CBSE AIPMT 1992]
(a) Boiling point
(b) Vapour pressure
(c) Osmotic pressure
(d) Freezing point
63
Solutions
Ans. (c)
Osmotic pressure is an example of
colligative property because its value
depends only on the number of moles of
solute, not on their chemical nature.
52 Blood cells retain their normal
shape in solutions which are
[CBSE AIPMT 1991]
(a) hypotonic to blood
(b) isotonic to blood
(c) hypertonic to blood
(d) equinormal to blood
Ans. (b)
When blood cells are placed in a solution
of similar concentration as that of blood,
then they neither swell nor shrink it
means the concentration of solution is
same as that of inside the blood cells,
i.e. they are isotonic to each other.
∴ By elevation in boiling point relation
∆Tb = iK b m or ∆Tb ∝ i
where,i is van’t Hoff factor
Since, ∆Tb of solution X is greater than
∆Tb of solution Y.
(Observed colligative property is greater
than normal colligative property).
∴ i of solution X > i of solution Y
∴ Solution X undergoing dissociation
55 Which one of the following
electrolytes has the same value of
van't Hoff's factor (i) as that of
Al 2 (SO 4 ) 3 (if all are 100% ionised)?
[CBSE AIPMT 2015]
TOPIC 4
Abnormal Molecular Masses
and Distribution Law
53 The van’t Hoff factor (i) for a dilute
aqueous solution of the strong
electrolyte barium hydroxide is
[NEET 2016, Phase II]
(a) 0
(b) 1
Ans. (d)
to the molal concentration of the
solution i.e.
∆Tb ∝ m or ∆Tb = K b m
where, m is the molality of the solution
and K b is molal boiling point constant or
ebullioscopic constant.
(c) 2
(d) 3
Key Idea Strong electrolytes dissociate
completely in their solutions.
van’t Hoff factor = total number of ions
after dissociation
So for Ba(OH)2 q Ba2 + (aq ) + 2OH− (aq)
Number of ions
1
2
van’t Hoff factor, i = 1 + 2 = 3
−1
54 The boiling point of 0.2 mol kg
solution of X in water is greater
than equimolal solution of Y in
water. Which one of the following
statements is true in this case?
[CBSE AIPMT 2015]
(a) X is undergoing dissociation in water.
(b) Molecular mass of X is greater than
the molecular mass ofY.
(c) Molecular mass of X is less than the
molecular mass of Y.
(d) Y is undergoing dissociation in water
while X undergoes no change.
Ans. (a)
Molality of solution X = molality of
solution Y = 0 .2 mol/kg
We know that, elevation in the boiling
point (∆Tb ) of a solution is proportional
(a) K2 SO 4
(c) AI (NO 3) 3
Ans. (d)
(b) K 3 [Fe(CN) 6]
(d) K 4 [Fe(CN) 6]
Al2 (SO4 ) 3
2 Al 3+ + 3 SO24−
Value of van’t Hoff’s factor (i) = 5
(a) K2SO4
2K+ + SO24− (i = 3)
(b) K3 [Fe(CN) 6 ]
3 K+ + [Fe(CN) 6 ] 3−
(i = 4)
(c) Al(NO3) 3
Al 3+ + 3 NO−3
(i = 4)
(d) K4 [Fe(CN) 6 ]
4 K+ + [Fe(CN) 6 ] 3−
(i = 5)
Therefore, K4 [Fe(CN) 6 ] has same value
of i that ofAl2 (SO4 ) 3 i.e. i = 5.
a
a
a
a
a
56 Of the following 0.10 m aqueous
solutions, which one will exhibit the
largest freezing point depression?
[CBSE AIPMT 2014]
(a) KCl
(c) Al2 (SO 4 ) 3
Ans. (c)
(b) C 6 H 12O 6
(d) K2 SO 4
∆Tf (freezing point depression) is a
colligative property and depends upon
the van’t Hoff factor (i), i.e. number of
ions given by the electrolyte in aqueous
solution.
∆Tf = i × kf × m
where, kf = molal freezing point
depression constant
m = molality of the solution
∴ kf and m are constant,
∴
∆Tf ∝ i
van’t Hoff factor for ionic solution.
(a) KCl (aq)
K+ (aq) + Cl − (aq),
(Total ions =2 thus, i = 2)
s
(b) C6H12O6
no ions [i = 0]
because glucose does not gives ions.
(c) Al2 (SO4 ) 3 (aq)
2Al 3+ + 3SO24−
[Total ions = 5, thus,i = 5 ]
(d) K2SO4 (aq)
2K+ + SO24−
[Total ions = 3, thus, i = 3 ]
Hence, Al2 (SO4 ) 3 will exhibit largest
freezing point depression due to the
highest value ofi.
s
s
s
57 The van’t Hoff factor, i for a
compound which undergoes
dissociation in one solvent and
association in other solvent is
respectively.
[CBSE AIPMT 2011]
(a) less than one and less than one
(b) greater than one and less than one
(c) greater than one and greater than one
(d) less than one and greater than one
Ans. (b)
Abnormality present due to dissociation
and association of the solution.
For dissociation, van’t Hoff factor is
greater than one and for association,
van’t Hoff factor is less than one.
For dissociation, i > 1
For association, i < 1
58 The freezing point depression
constant for water is – 1.86°C m −1 .
If 5.00 g Na 2 SO4 is dissolved in
45.0 g H2O, the freezing point
is changed by – 3.82°C. Calculate
the van’t Hoff factor for Na 2 SO4 .
[CBSE AIPMT 2011]
(a) 2.63 (b) 3.11
Ans. (a)
(c) 0.381 (d) 2.05
According to depression in freezing point,
∆Tf = i × kf ⋅ m
where, kf = cryoscopic constant
∆Tf × Wsolvent
or i =
kf × nsolute × 1000
=
3.82 × 45
= 2.63
 5 
1.86 × 
×
1000

 142 
59 A 0.0020 m aqueous solution of an
ionic compound Co(NH3 ) 5 (NO 2 )Cl
freezes at – 0.00732°C. Number of
moles of ions which 1 mole of ionic
compound produces on being
dissolved in water will be
(k f = − 1.86° C / m)
[CBSE AIPMT 2009]
(a) 2
(b) 3
(c) 4
(d) 1
64
NEET Chapterwise Topicwise Chemistry
Ans. (a)
Given, molality, m = 0.0020 m
∆Tf = 0 ° C − 0.00732° C
= − 0.00732° C
kf = − 1.86° C / m
∆Tf = i ⋅ kf × m
∆Tf
i=
kf × m
0.00732
=
1.86 × 0.0020
= 1.96 ≈ 2
Since, the compound is ionic, so number
of moles produced is equal to van’t Hoff
factor, i.
Hence, 2 moles of ions are produced.
[Co(NH3) 5 NO2 ]Cl →
1 mol
[Co(NH3) 5 NO2 ] + + Cl –
1444
424444
3
2 ions
60 0.5 molal aqueous solution of a
weak acid (HX) is 20% ionised. If k f
for water is 1.86K kg mol −1 , the
lowering in freezing point of the
solution is
[CBSE AIPMT 2007]
(a) –1.12 K
(c) 1.12 K
Ans. (c)
HX
Weak acid
1
1−α
(b) 0.56 K
(d) – 0.56 K
ºH
0
α
+
+ X−
0 (initially)
α (at equilibrium)
Total = 1 − α + α + α = 1 + α
1+ α
i=
1
α = 20%dissociation
i.e. α = 0.2
i = 1 + α = 1 + 0.2 = 1.2
∆Tf = i × K f × m
= 1.2 × 1.86 K kg mol –1 × 0.5
= 1.12 K
61 Which of the following 0.10 M
aqueous solution will have the
lowest freezing point?
[CBSE AIPMT 1997]
(a) Al2 (SO 4 ) 3
(c) KI
Ans. (a)
(b) C 5H10O 5
(d) C 12H22O 11
Depression in freezing point ∝ number of
particles.
In colligative properties ions behave like
particles.
Al2 (SO4 ) 3 provides five ions on ionisation
as
Al2 (SO4 ) 3 → 2 Al 3+ + 3SO24–
KI
K + + I–
while KI provides two ions and C5 H10O5
and C12H22O11 are not ionised, so they
have single particle
Hence, lowest freezing point is possible
for Al2 (SO4 ) 3.
º
62 At 25°C, the highest osmotic
pressure is exhibited by 0.1 M
solution of
[CBSE AIPMT 1994]
(a) CaCl2
(c) glucose
Ans. (a)
(b) KCl
(d) urea
CaCl2 is an electrolyte and dissociates to
give three ions as
CaCl2 → Ca2 + + 2Cl −
While KCl gives two ions and glucose and
urea are non-electrolytes, so remains
undissociated.
As osmotic pressure is a colligative
property (i.e. depend only on number of
particles), so highest for CaCl2 .
63 Which one of the following salts
will have the same value of van’t
Hoff factor (i) as that of
K 4 [Fe(CN) 6] ? [CBSE AIPMT 1994]
(a) Al2 (SO 4 ) 3
(b) NaCl
(c) Al(NO 3) 3
(d) Na2 SO 4
Ans. (a)
Both K4 [Fe(CN) 6 ] and Al2 (SO4 ) 3 gives 5
ions after dissociation, so they have
same value of van’t Hoff factor (i).
K4 [Fe(CN) 6 ] → 4K+ + [Fe(CN) 6 ] 4–
Al2 (SO4 ) 3 → 2Al 3+ + 3SO24–
64 In a pair of immiscible liquids, a
common solute dissolves in both
and the equilibrium is reached.
Then, the concentration of the
solute in upper layer is
[CBSE AIPMT 1994]
(a) in fixed ratio with that in the lower
layer
(b) same as the lower layer
(c) lower than the lower layer
(c) higher than the lower layer
Ans. (a)
According to Nernst distribution law
when we mixed a common solute in a
pair of immiscible liquids, then the ratio
of amount of solute in both liquids is
fixed at a fixed temperature.
65 Which of the following aqueous
solutions has minimum freezing
point?
[CBSE AIPMT 1991]
(a) 0.01 m NaCl
(b) 0.005 m MgI2
(c) 0.005 m C2H5OH
(d) 0.005 m MgSO 4
Ans. (a)
0.01 molal NaCl solution have minimum
freezing point because its molality is
more and it gives two ions after
dissociation of one molecule.
8
Chemical Equilibrium
TOPIC 1
The Chemical Equilibrium,
Basic Law of Mass Action
and Equilibrium Constant
01 A 20 litre container at 400 K
contains CO 2 (g) at pressure 0.4 atm
and an excess of SrO (neglect the
volume of solid SrO). The volume of
the container is now decreased by
moving the movable piston fitted in
the container. The maximum
volume of the container, when
pressure of CO 2 attains its
maximum value, will be [NEET 2017]
(Given that :
SrCO 3 (s) q SrO(s) + CO 2 (g),
K p = 1.6 atm)
(a) 5 L
(b) 10 L
Ans. (a)
(c) 4 L
(d) 2 L
For the reaction,
SrCO3 (s ) q S rO(s ) + CO2 (q),
K p = 16
. atm = pCO 2 = maximum
pressure of CO2
Given, p1 = 0.4 atm, V1 = 20 L,T1 = 400 K
p2 = 16
. atm, V2 = ?,T2 = 400 K
At constant temperature, p1V1 = p2V2
0.4 × 20 = 16
. × V2
0.4 × 20
= 5L
V2 =
.
16
02 For a given exothermic reaction,
K p and Kp′ are the equilibrium
constants at temperaturesT 1 and
T 2 , respectively. Assuming that
heat of reaction is constant in
temperature range betweenT 1 and
T 2 , it is readily observed that
[CBSE AIPMT 2014]
(a) K p > K ′p
(c) K p = K ′p
(b) K p < K ′p
1
(d) K p =
K′p
Ans. (a)
Ans. (a)
The equilibrium constant at two
different temperatures for a
thermodynamic process is given by
K
∆H °  1
1
log 2 =
 − 
K 1 2.303R T1 T2 
Here,K 1 and K 2 are replaced byK p andK ′p .
K′
∆H °  1
1
Therefore, log p =
 − 
K p 2.303R T1 T2 
For exothermic reaction,
T2 > T1 and H = −ve
⇒ K p > K ′p
Plan As we can see the reaction for
which we have to find out equilibrium
constant is different only in
stoichiometric coefficient as compared
to the given reaction. Hence, we can
find equilibrium constant for the
required reaction with the help of
mentioned equilibrium constant in the
problem.
Given, equilibrium constant for the
reaction,
N22 ( g ) + O2 (g )
2NO(g ) is K
[NO]2
i.e.
…(i)
K=
[N2 ][O2 ]
03 If the value of an equilibrium
constant for a particular reaction is
1.6 × 10 12 , then at equilibrium the
system will contain
(a)
(b)
(c)
(d)
all reactants [CBSE AIPMT 2015]
mostly reactants
mostly products
similar amounts of reactants and
products
Ans. (c)
For a reaction,
A
Product
K=
[B] eq
1.6 × 10 12 =
[B] eq
[A] eq
[A] eq
∴
[B] eq >> [A] eq
So, mostly the product will be present in
the equilibrium mixture.
04 If the equilibrium constant for
N 2 (g ) + O 2 (g )
2NO(g ) is K, the
equilibrium constant for
1
1
NO(g ) will be,
N 2 (g )+ O 2 (g )
2
2
a
a
[CBSE AIPMT 2015]
(a) K 1/2
1
(b) K
2
(c) K
Let equilibrium constant for the
reaction,
1
1
N2 (g ) + O2 (g )
NO(g ) is K ′
2
2
[NO]
i.e. K ′ =
[N2 ] 1/2 [O2 ] 1/2
-
On squaring both sides
K ′2 =
[NO]2
[N2 ][O2 ]
…(ii)
On comparing Eqs. (i) and (ii), we get
K = K ′2
or
K′= K
eB
Reactant
-
(d) K 2
05 Using the Gibbs energy change,
∆G° = + 63.3 kJ for the following
reaction,
Ag 2 CO 3 (s)
2Ag + (aq )
+ CO 23− (aq)
the K sp of Ag 2 CO 3 (s) in water at
25°C is
(R = 8.314 JK −1mol −1 )
r
[CBSE AIPMT 2014]
(a) 3.2 × 10−26
(c) 2.9 × 10−3
Ans. (b)
(b) 8.0 × 10−12
(d) 7.9 × 10−2
∆G ° is related toK sp by the equation,
∆G ° = −2.303RT log K sp
66
NEET Chapterwise Topicwise Chemistry
Given, ∆G ° = + 63.3 kJ
= 63.3 × 10 3 J
Thus, substitute ∆G ° = 63.3 × 10 3 J,
R = 8.314 JK−1 mol −1 andT = 298K [25 +
273 K] from the above equation we get,
63.3 × 10 3 = − 2.303 × 8.314 × 298 log K sp
∴ log K sp = − 11.09
⇒
K sp = antilog (− 11.09)
K sp = 8.0 × 10 −12
06 For the reaction,
N 2 (g) + O 2 (g)
2NO(g), the
equilibrium constant is K 1 . The
equilibrium constant is
K 2 for the reaction,
2NO(g) + O 2 (g)
2 NO 2 (g). What is
K for the reaction,
1
NO 2 (g)
N 2 (g) + O 2 (g)?
2
r
r
r
where, [∆ng gaseous = np − nR ]
(b) nP = nR = 2, thus, K p = K C
(c) nP = nR = 2, thus, K p = K C
(d) nP = 2, nR = 1, thus, K p ≠ K C
s
s
s
s
∴
1/ 2
07 In which of the following
equilibrium K c and K p are not
equal?
[CBSE AIPMT 2010]
r N ( g) + O ( g)
rSO ( g) + NO( g)
(c) H ( g) + I ( g) r2HI ( g)
(d) 2C(s ) + O ( g) r2CO ( g)
(a) 2NO( g)
2
2
(b) SO2 ( g) + NO2 ( g)
2
3
2
2
2
Ans. (d)
Key Idea The reaction for which the
number of moles of gaseous products (np )
is not equal to the number of moles of
gaseous reactants (nR ), has different
value of K C and K p .
From the equation,K p = K C × (RT )
H2 ( g ) + I 2 (g )
[CBSE AIPMT 2009]
(a) 3.0 × 105
(b) 3.0 × 10−5
(c) 3.0 × 10−4
Ans. (d)
(d) 3.0 × 104
–
+
3
−5
a
∆n g
For
CN− + CH3COOH
1
8
2
[I2 ] 1/2 [H2 ] 1/2
K=
–
09 If the concentration of OH− ions in
the reaction,
Fe(OH) 3 (s)
Fe 3+ (aq)
+ 3OH− (aq)
is decreased by 1/4 times, then
equilibrium concentration of Fe 3+
will increase by [CBSE AIPMT 2008]
r
(b) 16 times
(d) 4 times
(aq) + 3OH− (aq)
[Fe ][OH– ] 3
…(i)
K=
[Fe(OH) 3]
3+
sFe
3+
To maintain equilibrium constant, let the
concentration of Fe3+ is increased x
times, on decreasing the concentration
1
of OH− by times
4
1
[x Fe3+ ][ × OH− ] 3
4
…(ii)
K=
[Fe(OH) 3]
…(i)
[HI]
H2 ( g) + I2 ( g)
s2HI( g)
3
10 5
= 3.33 × 10 4 ≈ 3 × 10 4
3
Fe(OH) 3 (s )
2
–
sHCN + CH COO
s
(d)
s 12 H (g) + 21 I (g)
K′=
CH3COOH + CN
HCN + CH3COO
Ka
1.5 × 10 −5
K=
=
K a 1 4.5 × 10 −10
–
(c) 16
–
K =?
On subtracting Eq. (ii) from Eq. (i), we get
(a) 8 times
(c) 64 times
Ans. (c)
1
(b)
64
Ans. (b)
s CH COO + H ,
…(i)
K = 1.5 × 10
HCN sH + CN ,
=
1
(a)
16
HI( g)
Given, CH3COOH
r2HI(g) will be
[CBSE AIPMT 2008]
K a 1 = 4.5 × 10 −10 …(ii)
…(i)
N2 ( g) + O2 ( g)
2NO( g);K 1
…(ii)
2NO( g) + O2 ( g)
2NO2 ( g);K 2
On adding Eqs. (i) and (ii)
N2 (g) + 2O2 ( g)
2NO2 (g); K = K 1 × K 2
…(iii)
1
On dividing (iii) by and on reversing we
2
get,
1
NO2 ( g)
N2 ( g) + O2 ( g);
2
(N ) 1/2 (O2 )
So,
K= 2
(NO2 )
 1 
K =

 K 1K 2 
r
r
+
(b) [1/K 1K 2] 1/2
(d) 1/(2 K 1K 2 )
10 The value of equilibrium constant
of the reaction,
1
1
HI(g)
H2 (g) + I 2 (g) is 8.0.
2
2
The equilibrium constant of the
reaction,
08 The dissociation constants for
acetic acid and HCN at 25°C are
1. 5 × 10 −5 and 4.5 × 10 −10 ,
respectively. The equilibrium
constant for the equilibrium,
CN– + CH3COOH
HCN
+ CH3COO– would be
[CBSE AIPMT 2011]
(a) 1/(4 K 1K 2 )
(c) 1/(K 1K 2 )
Ans. (b)
By dividing eq. (ii) by (i) we get
1
×x=1
64
⇒
x = 64 times
(a) nP = nR = 2, thus, K p = K C
[HI]2
[H2 ][I2 ]
…(ii)
From Eqs. (i) and (ii)
K × K′ =1
K′=
1
K2
=
1
(8)2
=
1
64
11 The equilibrium constants of the
following are
[NEET 2017, CBSE AIPMT 2007]
N 2 + 3H2 q 2NH3 ;
N 2 + O 2 q 2NO;
1
H2 + O 2 → H2O;
2
K1
K2
K3
The equilibrium constant (K) of the
reaction
K
5
2NH3 + O 2 q 2NO + 3H2O, will
2
be
(a) K 1K 33 / K 2
(c) K 2K 3 / K 1
Ans. (b)
(b) K 2K 33 / K 1
(d) K 23K 3 / K 1
Given,N2 + 3H2 q
2NH3, K 1
…(i)
N2 + O2 q 2NO, K 2
1
H2 + O2 → H2O, K 3
2
…(ii)
…(iii)
To calculate,
K
5
O2 q 2NO + 3H2O,
2NH3 +
2
K =?
…(iv)
67
Chemical Equilibrium
On reversing the equation (i) and
multiplying the equation (iii) by 3, we get
1
…(v)
2NH2 q N2 + 3H2 ,
K1
3
…(vi)
3H2 + O2 → 3H2O, K 33
2
Now, add equation. (ii), (v) and (vi), we get
the resultant equation. (iv).
K
5
2NH3 + O2 q 2NO + 3H2O
2
K K3
K= 2 3
∴
K1
rCO (g)
2
+ 2H2O(l),
∆ r H = –170.8kJ mol–1
Which of the following statement is
not true ?
[CBSE AIPMT 2006]
(a) At equilibrium, the concentrations of
CO2 ( g) and H2O(l ) are not equal
(b) The equilibrium constant for the
[CO2 ]
reaction is given byK p =
[CH4 ] [O2 ]
(c) Addition of CH4 ( g) or O2 ( g) at
equilibrium will cause a shift to the
right
(d) The reaction is exothermic
Ans. (b)
s CO (g) +2H O(l)
2
2
∆Hr = − 170.8 k J mol −1
This equilibrium is an example of
heterogeneous chemical equilibrium.
Hence, for it
[CO2 ]
...(i)
Kc =
[CH4 ][O2 ]2
and
13 In the two gaseous reactions (i) and
(ii) at 250°C
1
(i) NO(g) + O 2 (g)
NO 2 (g),K 1
2
(ii) 2NO 2 (g)
2NO(g) + O 2 (g),K 2
the equilibrium constants K 1 and
K 2 are related as
r
r
[CBSE AIPMT 2005, 1994]
12 For the reaction,
CH4 (g) + 2 O 2 (g)
For the reaction,
CH4 ( g) + 2O2 ( g)
the concentration of CO2 will increased
in same order. Hence, on addition of CH4
or O2 equilibrium will cause to the right.
Combustion reaction is an example of
exothermic reaction.
(equilibrium constant on the
basis of concentration)
pCO 2
...(ii)
Kp =
pCH 4 × p 2
O
2
1
(a) K 2 =
K1
1
(c) K 2 = 2
K1
(b) K 2 = K 11/2
(d) K 2 = K 21
Ans. (c)
For equation (i),
1
NO ( g) + O2 ( g)
NO2 ( g)
2
[NO2 ]
K1 =
[NO][O2 ] 1/2
s
For equation (ii),
2NO2 (g)
2NO(g) + O2 (g)
[NO]2 [O2 ]
K2 =
[NO2 ]2
…(i)
…(ii)
Now, on reversing equation (i), we get,
1
1
=
[NO2 ]
K1
[NO][O2 ] 1/2
[NO][O2 ] 1/2
[NO2 ]
2
r2
According to law of mass action,
the rate of forward reaction=r1
r1 ∝ [BaO2 ]
or r1 = k1 [BaO2 ]
BaO2 is solid substance in pure state
concentration = 1m
then,
r1 = k1
Similarly the rate of backward reaction
=r2
r2 ∝ [BaO] [O2 ]
or r2 = k2 [BaO][O2 ]
Q Concentration of solid [BaO] = 1 [O2 (g)]
∴
r2 = k2 [O2 ]
At equilibrium,
r1 = r2
K 1 =K 2 [O2 ]
or K 1 =K 2 ⋅ pO 2
where,
pO 2 = partial pressure ofO2
K1
(equilibrium
or
= pO 2
K2
constant)
K1
=K
K2
or
K = pO 2
So, from the above it is clear that
pressure of O2 does not depend upon the
concentration of reactants. The given
equation is an endothermic reaction. If
the temperature of such reaction is
increased, then dissociation ofBaO2
would increase and moreO2 is produced.
15 For the equilibrium,
 [NO][O2 ] 1/2 
 1 
  =

K 1 
 [NO2 ] 
[NO]2 [O2 ]
=
= K2
[NO2 ]2
K 21
r1
sBaO(s) +O (g), ∆H = + ve
2
2
1
BaO2 (s )
Q
s
=
Ans. (c)
= K2
(equilibrium constant according
to partial pressure)
Thus, in this concentration of CO2 ( g) and
H2O (l ) are not equal at equilibrium.
The equilibrium constant
[CO2 ]
is not correct
(K p ) =
[CH4 ] [O2 ]
14 Reaction,
BaO 2 (s)
BaO(s) +O 2 (g),
∆H = +ve
In equilibrium condition, pressure
of O 2 depends on
expression.
On adding CH4 ( g) or O2 ( g) at equilibrium,
K C will be decreased according to
expression (i) but K C remains constant at
constant temperature for a reaction, so
for maintaining the constant value ofK C ,
(a) increased mass ofBaO2
(b) increased mass of BaO
(c) increased temperature of
equilibrium
(d) increased mass ofBaO2 and BaO
both
r
[CBSE AIPMT 2002]
MgCO 3 (s)
∆
rMgO(s) + CO (g)
2
which of the following expressions
is correct?
[CBSE AIPMT 2000]
(a) K p = pCO 2
[MgO][CO2]
(b) K p =
[MgCO 3]
p MgO ⋅ p CO 2
(c) K p =
p MgCO 3
(d) K p =
p MgO + pCO 2
p MgCO 3
Ans. (a)
In heterogeneous system,K C and K p are
not depend upon the concentration or
pressure of solid substance. Hence, at
equilibrium their concentration or
pressure are assumed as one.
MgCO3 ( s )
MgO (s ) + CO2 ( g)
∴
K p = p CO 2
s
68
NEET Chapterwise Topicwise Chemistry
16 If K 1 and K 2 are the respective
equilibrium constants for the two
reactions,
XeF6 (g ) + H2O(g)
XeOF4 (g )
+2HF (g)
XeO 4 (g) + XeF6 (g)
XeOF4 (g)
+XeO 3 F2 (g)
The equilibrium constant of the
reaction,
XeO 4 (g ) + 2HF (g)
XeO 3F2 (g )
+ H2O(g)
will be
[CBSE AIPMT 1998]
r
r
r
(b) K 1 ⋅ K 2
(d) K 2 /K 1
(a) K 1 / (K 2 )2
(c) K 1 /K 2
Ans. (d)
XeF6 ( g) + H2O( g)
XeOF4 ( g) + 2HF ( g)
[XeOF4 ][HF]2
...(i)
K1 =
[XeF6 ][H2O]
s
XeO4 ( g) + XeF6 ( g)
sXeOF ( g)
4
+ XeO3F2 ( g)
[XeOF4 ] [XeO3F2 ]
...(ii)
K2 =
[XeO4 ] [XeF6 ]
For the reaction,
XeO4 ( g) +2HF ( g)
s XeO F ( g)
3 2
[XeO3F2 ][H2O]
[XeO4 ][HF]2
…(iii)
By dividing eq. (ii) by (i) we get,
K
K= 2
K1
17 The equilibrium constants for the
reaction, A 2
2 A at 500 K and
700 K are 1 × 10 −10 and 1 × 10 −5 . The
given reaction is [CBSE AIPMT 1996]
r
(a) exothermic
(c) endothermic
Ans. (b)
For the reaction,
A2
(b) slow
(d) fast
s2 A
K=
[A]2
[A2 ]
The value of equilibrium constant is very
less and hence, the product
concentration is also very less. So, the
reaction is slow.
18 If α is the fraction of HI dissociated
at equilibrium in the reaction,
2HI(g)
H2 (g) + I 2 (g) starting
with the 2 moles of HI, then the
r
[NO]
1
1
=
=
K 2 [N2 ] 1/2 [O2 ] 1/2 [N2 ] 1/2 [O2 ] 1/2
[NO]
[CBSE AIPMT 1996]
(a) 2 + 2 α
(c) 1 + α
Ans. (b)
(b) 2
(d) 2 − α
2HI(g)
In initial
2 mol
At equilibrium (2 − 2α)
mol
H2 (g)
0 mol
α mol
s
+ I2 (g)
0 mol
α mol
r
(b) 5.33
(d) 7.33
kf
Forward rate constant
Backward rate constant
1.1 × 10 −2 1.1 × 10 11
= 7.33
K=
=
=
1.5
1.5
1.5 × 10 −3
20 K 1 and K 2 are equilibrium constant
for reactions (i) and (ii)
…(i)
N 2 (g) + O 2 (g)
2NO(g)
1
1
…(ii)
NO(g)
N 2 (g) + O 2 (g)
2
2
Then,
[CBSE AIPMT 1989]
r
r
2
(b) K 1 = K 22
(d) K 1 = (K 2 ) 0
Consider reaction (i),
K1 =
2
s2NO( g)
s
K2 =
K(i)
[NO]
[N2 ] [O2 ]
Now, consider reaction (ii),
1
1
NO(g)
N2 (g) + O2 (g)
2
2
[N2 ] 1/2 [O2 ] 1/2
[NO]
21 Which one of the following conditions
will favour maximum formation of the
product in the reaction, [NEET 2018]
A 2 (g) + B2 (g) r X 2 (g);
∆ r H = − X kJ?
(a)
(b)
(c)
(d)
High temperature and high pressure
Low temperature and low pressure
Low temperature and high pressure
High temperature and low pressure
Key Concept The given question is based
upon Le-Chatelier’s principle. According
to this principle, if a stress is applied to a
reaction mixture at equilibrium, reaction
proceeds in such a direction that relieves
the stress.
The given reaction is
A2 (g) + B2 (g) r X2 (g); ∆ r H = − X kJ
According to Le-Chatelier’s principle,
with increase in temperature the
equilibrium shifts in the direction of
endothermic reaction (i.e., heat is
absorbed).
Alternatively, the decrease in
temperature shifts the equilibrium
towards the direction of exothermic
reaction (i.e. heat is produced).
∴
Ans. (a)
N2 ( g) + O2 ( g)
Factors Affecting
Equilibrium and
Le-Chatelier’s Principles
Ans. (c)
kb
=
 1 
(a) K 1 =  
K2 
1
(c) K 1 =
K2
 1 


[NO]
  = 
1/2
1/2 
K 
 [N2 ] [O2 ] 
 2
[NO]2
=
=K1
[N2 ] [O2 ]
TOPIC 2
19 The rate constants for forward and
backward reaction of hydrolysis of
ester are 1.1 × 10 −2 and 1.5 × 10 −3 per
minute. Equilibrium constant for the
reaction,
[CBSE AIPMT 1995]
CH3COOC 2H5 + H+
CH3COOH + C 2H5OH is
(a) 4.33
(c) 6.33
Ans. (d)
2
2
So, at equilibrium total moles
= 2 − 2α + α + α
= 2 − 2α + 2α = 2
Equilibrium constant,K =
+ H2O ( g)
K=
total number of moles of reactants
and products at equilibrium are
K(ii)
A2 (g) + B2 (g)
Endothermic
r
Exothermic
X2 (g)
Similarly, an increase in pressure will
shifts the equilibrium to that direction
which leads to decrease in total number
of gaseous moles. Whereas, a decrease
in the pressure will shift the equilibrium
to that direction which leads to an
increase in total number of gaseous
moles.
For, A2 (g) + B2 (g) r X2 (g)
∆ng = 1 − 2 = − 1
Thus, low temperature and high
pressure will favour maximum formation
of the product in the given reaction.
69
Chemical Equilibrium
22 For the reversible reaction,
N 2 (g) + 3H2 (g)
2NH 3 (g) + heat
the equilibrium shifts in forward
direction
[CBSE AIPMT 2014]
r
(a) by increasing the concentration of
NH3 ( g)
(b) by decreasing the pressure
(c) by decreasing the concentrations of
N2 ( g) and H2 ( g)
(d) by increasing pressure and
decreasing temperature
Ans. (d)
Any change in the concentration,
pressure and temperature of the
reaction results in change in the
direction of equilibrium. This change in
the direction of equilibrium is governed
by Le-Chatelier’s principle. According to
this equilibrium shifts in the opposite
direction to undo the change.
N2 ( g) + 3H2 ( g)
2NH3 ( g) + Heat
s
(a) Increasing the concentration of
NH 3 (g) On increasing the
concentration of NH3 ( g), the
equilibrium shifts in the backward
direction where concentration of
NH3 ( g) decreases.
(b) Decreasing the pressure Since, p ∝ n
(number of moles), therefore,
equilibrium shifts in the backward
direction where number of moles are
increasing.
(c) Decreasing the concentration of
N 2 (g) and H2 (g) Equilibrium shifts in
the backward direction when
concentration of N2 ( g) and H2 ( g)
decreases.
(d) Increasing pressure and decreasing
temperature On increasing pressure,
equilibrium shifts in the forward
direction where number of moles
decreases. It is an example of
exothermic reaction therefore
decreasing temperature favours the
forward direction.
23 KMnO 4 can be prepared from
K 2MnO 4 as per reaction,
3MnO 24− + 2H2O
2MnO −4 + MnO 2 + 4OH−
The reaction can go to completion
by removing OH− ions by adding
r
[NEET 2013]
(a) HCl
(c) CO2
(b) KOH
(d) SO2
Ans. (c)
Since,OH− are generated from weak acid
(H2O), and a weak acid (like CO2 ) should be
used to remove it. Because if we add
strong acid like (HCl) it reverse the
reaction. KOH increases the
concentration of OH− , thus again shifts
the reaction in backward side.
CO2 combines withOH− to give carbonate
which is easily removed.
SO2 reacts with water to give strong acid,
so it cannot be used.
24 The value of ∆H for the reaction,
X 2 (g) + 4Y 2 (g)
2 XY 4 (g)
is less than zero. Formation of
XY 4 (g) will be favoured at
r
[CBSE AIPMT 2011]
(a)
(b)
(c)
(d)
low pressure and low temperature
high temperature and low pressure
high pressure and low temperature
high temperature and high pressure
Ans. (c)
X2 (g) + 4Y2
∆H < 0
26 For a reversible reaction, if the
concentrations of the reactants are
doubled, the equilibrium constant
will be
[CBSE AIPMT 2000]
(a) one-fourth
(c) doubled
Ans. (d)
(b) halved
(d) the same
Consider a hypothetical change,
A+B
C +D
[C] [D]
For this reaction,K eq =
[A] [B]
s
For the above reaction if concentration
of reactants are doubled then the rate of
forward reaction increases for a short
time but after sometime equilibrium will
established. So, concentration has no
effect on equilibrium constant. It
remains unchanged after increasing the
concentration of reactants.
27 According to Le-Chatelier’s
principle, adding heat to a solid
liquid equilibrium will cause
the
[CBSE AIPMT 1993]
r
s2 XY (g); where
4
and ∆n< 0 [∆n = nP − nR ]
∴ The forward reaction is favoured at
high pressure and low temperature.
(According to Le-Chatelier’s principle)
25 The reaction quotient (Q) for the
reaction,
N 2 (g) +3H2 (g)
2NH3 (g) is
given by
[NH3] 2
.
Q=
[N 2] [H2] 3
r
The reaction will proceed towards
right side, if
[CBSE AIPMT 2003]
(a) Q > K c
(c) Q = K c
(b) Q = 0
(d) Q < K c
where, K c is the equilibrium
constant.
(a) temperature to increase
(b) temperature to decrease
(c) amount of liquid to decrease
(d) amount of solid to decrease
Ans. (d)
When we add heat to the equilibrium
between solid and liquid, then the
equilibrium shifts towards liquid and
hence, the amount of solid decrease and
amount of liquid increase.
28 Which one of the following
information can be obtained on the
basis of Le-Chatelier’s principle?
[CBSE AIPMT 1992]
(a) Dissociation constant of a weak acid
(b) Entropy change in a reaction
(c) Equilibrium constant of a chemical
reaction
(d) Shift in equilibrium position on
changing value of a constant
Ans. (a)
Ans. (d)
For the reaction,
Le-Chatelier’s and Braun French
chemists made certain generalisations
to explain the effect of changes in
concentrations, temperature or
pressure on the state of system in
equilibrium. When a system is subjected
to a change in one of these factors, the
equilibrium gets disturbed and the
system re-adjusts itself until it return to
equilibrium.
N2 ( g) +3H2 ( g)
Q (Quotient) =
s2NH (g)
3
[NH3]2
[N2 ][H2 ] 3
,
∆ ng = 2 − 4 = − 2
At equilibrium Q is equal toK c but for the
progress of reaction towards right side,
Q >Kc
9
Ionic Equilibrium
TOPIC 1
Ans. (c)
Ostwald’s Dilution Law
that forms pyridinium ion
(C 5H5N + H) in a 0.10M aqueous
pyridine solution (K b for
C 5H5N = 1.7 × 10 −9 ) is
[NEET 2016, Phase II]
(b) 0.013%
(d) 1.6%
The percentage of pyridine can be equal
to the percentage of dissociation of
pyridinium ion and pyridine solution as
shown below:
+ H2O
r
N
H
N
+OH
In equation,
X
Y + Z
Initial moles
1
0
0
At equil.
( 1 − α)
α
α
where,α = degree of dissociation
Total number of moles
= 1 − α + α + α = (1 + α)
 1−α
pX = 
 p1
 1+ α
3
01 The percentage of pyridine (C 5H5N)
(a) 0.0060%
(c) 0.77%
Ans. (b)
K p1 =
[pY ][pZ ]
[pX ]
–
= 1. 7 × 10
= 1.3 × 10
−4
or, percentage of dissociation
= (α × 100)%
= (1.3 × 10 −4 ) × 100 = 0 .013%
02 The values of Kp and K p for the
1
2
reactions
…(i)
X
Y +Z
and
…(ii)
A
2B
are in ratio of 9 : 1. If degree of
dissociation of X and A be equal,
then total pressure at equilibrium
(i) and (ii) are in the ratio
3
3
[CBSE AIPMT 2008]
(a) 3 :1
(c) 36 :1
(b) 1 :9
(d) 1 :1
p1 36
=
= 36 : 1
1
p2
 α 
 α 

 p1 × 
 p1
1
+
α


 1 + α
=
 1−α

 p1
 1 + α
 α 

 p1
 1 + α
=
 1−α


 1 + α
…(i)
3
 1−α
pA = 
 p2
 1 + α
3
2 AB (g) + B2 (g)
1
0
0
2 ( 1 − x)
2x
x
xp
(2 + x)
2
 2x p    x  

 
p
(pAB) (pB )  2 + x    2 + x  
2


Kp =
=
2
(pAB )2




2 (1 − x)
2
 p

  (2 + x)  
2
2
K p2
(b) (2 K p / p) 1/ 3
(d) (K p / p)
where, x = degree of dissociation
Total moles at equilibrium
= 2 − 2x + 2 x + x = (2 + x)
2 (1 − x) p
2x p
So, pAB =
, pAB =
2
(2 + x)
(2 + x)
2
  2α  
 p2 

2
[p ]
 1 + α 
= B =
 1−α
[pA ]

 p2
 1 + α
 2α 

 p2
 1 + α
=
 1−α


 1 + α
Initial
moles
At equil.
pB =
2
2
[CBSE AIPMT 2008]
(a) (2K p / p)
(c) (2 K p / p) 1/2
Ans. (b)
2 AB2 (g)
For equation, A
2B
Initial moles 1
0
At equil.
(1 − α)
2α
Total number of moles at equilibrium
= (1 + α)
 2α 
pB = 
 p2
 1 + α
Kp
03 The dissociation equilibrium of a
gas AB2 can be represented as
2 AB2 (g)
2 AB(g) + B2 (g)
The degree of dissociation is x and
is small compared to 1. The
expression relating the degree of
dissociation (x) with equilibrium
constantK p and total pressure p is
3
2
Kb
1. 7 × 10 −9
=
C
0 . 10
−8
2
9 p1
=
1 4p2
 α 
pY = 
 p1
 1+ α
α
pZ =
p1
(1 − α)
As pyridinium is a weak base, so degree
of dissociation is given as
α=
Eq. (i) divide by Eq. (ii)
Kp
α2 × p
1
= 2 1
Kp
4α × p2
…(ii)
=
4x 3p3
(2 + x)
3
×
(2 + x)2
p 4(1 − x)
2
2
=
x 3p
(2 + x)(1 − x)2
71
Ionic Equilibrium
x 3p
2
1/ 3
 2K p 
x=

 p 
=
and 2
[Q x <<< 1
so, (1 − x) ≈ 1
(2 + x) ≈ 2
04 A weak acid, HA, has a K a of
1.00 × 10 −5 . If 0.100 mole of this acid
is dissolved in one litre of water,
the percentage of acid dissociated
at equilibrium is closest to
[CBSE AIPMT 2007]
(a) 99.0%
(c) 99.9%
Ans. (b)
(b) 1.00%
(d) 0.100%
HA
H+ + A −
At equilibrium [H+ = A − ]
[H+][ A – ] [H+]2
Ka =
=
[HA]
[HA]
3
[H+ ] = K a [HA] = 1 × 10 −5 × 0.1
= 1 × 10 −6 = 1 × 10 −3
Actual ionisation
α=
Molar concentration
10 −3
=
= 10 −2
0.1
% of acid dissociated = 10 −2 × 1.00
= 1% = 100%
05 At 25°C, the dissociation constant
of a base, BOH is 1.0 × 10 −12 . The
concentration of hydroxyl ions in
0.01 M aqueous solution of the base
would be
[CBSE AIPMT 2005]
(a) 2.0 × 10–6 mol L–1
(b) 1.0 × 10–5 mol L–1
(c) 1.0 × 10–6 mol L–1
(d) 1.0 × 10–7 mol L–1
Ans. (d)
Base, BOH is dissociated as follows
BOH
B + + OH–
So, the dissociation constant of BOH
base
[B + ] [OH– ]
…(i)
Kb =
[OH]
3
At equilibrium [B + ] = [OH– ]
[OH– ] 2
Kb =
∴
[BOH]
Given that K b = 1.0 × 10 −12
and [BOH] = 0.01 M
[OH– ] 2
Thus, 1.0 × 10 –12 =
0.01
[OH– ]2 = 1 × 10 –14
[OH– ] = 1.0 × 10 –7 mol L–1
06 Ionisation constant of CH3COOH is
1. 7 × 10 −5 and concentration of H+
ions is 3.4 × 10 −4 . Then, find out
initial concentration of CH3COOH
molecules.
[CBSE AIPMT 2001]
(a) 3 .4 × 10−4
(c) 6 .8 × 10−4
Ans. (a)
H2O
Acid
HF
3
Given that, [CH3COO– ] = [H+ ] = 3.4 × 10 –4 M
K a for CH3COOH = 1.7 × 10 –5
CH3COOH is weak acid, so in it
[CH3COOH] is equal to initial
concentration. Hence,
(3.4 × 10 −4 ) (3.4 × 10 −4 )
1.7 × 10 −5 =
[CH3COOH]
3.4 × 10 × 3.4 × 10
–4
[CH3COOH] =
An acid on losing a proton produces a
species which has the tendency to
accept H+ .
It is called conjugate base of that acid.
(b) 3 .4 × 10−3
(d) 6 .8 × 10−3
CH3COO– + H+
[CH3COO– ] [H+ ]
Ka =
[CH3COOH]
CH3COOH
Ans. (b)
–4
1.7 × 10 –5
= 6.8 × 10 −3M
Acid
−
OH + H+ ,
-
Conjugate
base
F − + H+
Conjugate
base
Water (H2O) is amphoteric in nature and
thus act both as an acid and base. e.g.
HF + H2O
Acid
Base
-
F−
Conjugate
base
+ H3O+
Conjugate
acid
09 Which of the following
fluoro-compounds is most likely to
behave as a Lewis base?
[NEET 2016, Phase II]
(a) BF3
(c) CF4
Ans. (b)
(b) PF3
(d) SiF4
Key Idea The molecule with lone pair at
centre atom, will behave as Lewis base.
In the given molecules, onlyPF3 has lone
pair at P as shown below:
TOPIC 2
Acid Base Concepts
07 Which of the following cannot act
both as Bronsted acid and as
Bronsted base?
[NEET (Odisha) 2019]
(a) HCO −3
(b) NH3
(c) HCl
(d) HSO −4
Ans. (c)
Key Idea Bronsted acid is a substance
which has a tendency to donate proton.
Bronsted base is a substance which has
a tendency to accept proton.
HCl can act as Bronsted acid becuase it
can only donate proton.
HCl + H2O
Acid
Base
H3O+ + Cl −
-
The remaining options contains
substances which act both as Bronsted
acid and Bronsted base.
HCO–3 + HCO–3
H2 CO3 + CO23–
NH3 + NH3
NH+4 + NH2−
–
–
HSO4 + HSO4
H2SO4 + SO24–
Thus, option (c) is correct.
-
08 Conjugate base for Bronsted acids
H2O and HF are
[NEET (National) 2019]
(a) H3O + and F − , respectively
(b) OH− and F − , respectively
(c) H3O + and H2F + , respectively
(d) OH− and H2F + , respectively
Thus, PF3 acts as a Lewis base
(electron-pair donor) due to presence of
lone pair on P-atom.
10 Which is the strongest acid in the
following?
[NEET 2013]
(a) H2 SO 4
(c) HClO 4
Ans. (c)
(b) HClO 3
(d) H2 SO 3
The strength of oxyacids can also be
decided with the help of the oxidation
number of central atom. Higher the
oxidation number of central atom, more
acidic is the oxyacid.
+6
+5
+7
+4
H2SO4 , HClO3, HClO4 , H2SO3
Order of acidic nature
HClO4 > H2SO4 > HClO3 > H2SO3
Since, inHClO4 , oxidation number of Cl is
highest, so,HClO4 is the strongest acid
among the given acids.
11 Which of these is least likely to act
as a Lewis base?
[NEET 2013]
(a) CO
(c) BF3
(b) F −
(d) PF3
72
NEET Chapterwise Topiewise Chemistry
Ans. (c)
Electron rich species are called Lewis
base. Among the given,BF3 is an
electron deficient species, so have a
capacity of electron accepting instead of
donating. That’s why it is least likely to
act as a Lewis base. It is a Lewis acid.
12 Which of the following is electron
deficient?
[NEET 2013]
(a) (CH3)2
(c) (BH3)2
Ans. (c)
CH3COOH is weak acid while NaOH is
strong base, so one equivalent of NaOH
cannot be neutralised with one
equivalent of CH3COOH. Hence, one
equivalent of each does not have pH
value 7. As the NaOH is a strong base,
the solution will be basic having a pH
more than 7.
15 The conjugate acid of NH2− is
[CBSE AIPMT 2000]
(b) (SiH3)2
(d) PH3
Boron is an element of 13 group and
contains three electrons in its valence
shell. When its compoundBH3 dimerises,
each boron atom carry only 6 electrons
that is their octet is incomplete. Hence,
(BH3)2 is an electron deficient
compound.
In all other given molecules octet of
central atom is complete.
(a) N2H4
(c) NH2OH
Ans. (d)
(b) NH4+
(d) NH3
The species formed after adding a
proton to the base is known as
conjugate acid of the base and the
species formed after losing a proton is
known as conjugate base of acid. So,
NH2– + H+ →
NH3
Base
Conjugate acid
16 The strongest conjugate base is
[CBSE AIPMT 1999]
13 Which of the following molecules
acts as a Lewis acid?
[CBSE AIPMT 2009]
(a) (CH3) 3B
(c) (CH3) 3P
Ans. (a)
CH3
ו
CH3 ו B
ו
CH3
Incomplete
octet
(Lewis acid)
CH3
•×
CH3 ו P ••
ו
CH3
complete octet
and presence
of lp of e –
(Lewis base)
(b) (CH3)2 O
(d) (CH3) 3N
••×
•
••
CH3 ו O
(b) Cl–
(d) CH3COO –
Weak acid forms strong conjugate base.
In HNO3 , HCl, H2SO4 and CH3COOH ,
CH3COOH is weakest acid, so its
conjugate base is strongest.
CH3COOH
CH3COO– + H+
CH3
Complete octet and
presence of lp of e –
(Lewis base)
CH3
ו
CH3 •× N ••
•×
CH3
complete octet
and presence
of lp of e –
(Lewis base)
14 Which of the following statements
about pH and H + ion concentration
is incorrect?
[CBSE AIPMT 2000]
(a) Addition of one drop of
concentrated HCl in NH4 OH solution
decreases pH of the solution
(b) A solution of the mixture of one
equivalent of each of CH3COOH and
NaOH has a pH of 7
(c) pH of pure neutral water is not zero
(d) A cold and concentratedH2SO4 has
lowerH+ ion concentration than a
dilute solution ofH2SO4
Ans. (b)
(a) NO –3
(c) SO2–
4
Ans. (d)
3
17 The hydride ion H− is stronger base
than its hydroxide ion OH− . Which
of the following reactions will occur
if sodium hydride (NaH) is dissolved
in water?
[CBSE AIPMT 1997]
(a) 2H– (aq) + H2O(l ) → H2O + H2 + 2 e –
(b) H– (aq) + H2O (l ) → OH– + H2
(c) H– + H2O (l ) → No reaction
(d) None of the above
Ans. (b)
Sodium hydride dissolved in water as
NaH +H2O → NaOH + H2
or H− (aq) +H2O(l ) → OH– + H2 ↑
In the above reaction hydride ion take
proton from water molecule and
hydrogen gas is evolved.
18 0.1M solution of which one of these
substances will be basic?
[CBSE AIPMT 1992]
(a) Sodium borate
(b) Calcium nitrate
(c) NH4Cl
(d) Sodium sulphate
Ans. (a)
On hydrolysis sodium borate form
sodium hydroxide and boric acid, so the
solution will show basic character
because sodium hydroxide is strong
base and boric acid is weak acid. While
solution of sodium sulphate is neutral
and that of NH4 Cl and calcium nitrate is
acidic.
19 Aqueous solution of acetic acid
contains
[CBSE AIPMT 1991]
(a) CH3COO – and H+
(b) CH3COO –,H3O + and CH3COOH
(c) CH3COO –, H3O + and H+
(d) CH3COOH,CH3COO – and H+
Ans. (b)
The aqueous solution of acetic acid
ionise as follows:
H2O+ CH3COOH
CH3COO– +H3O+
Base
Acid
So, the aqueous solution of acetic acid
contains CH3COO– , H3O+ and CH3COOH.
!
TOPIC 3
Solubility Product and
Common Ion Effect
20 Find out the solubility of Ni(OH) 2 in
0.1 M NaOH. Given, that the ionic
product of Ni(OH) 2 is 2 × 10 −15 .
[NEET (Sep.) 2020]
(a) 2 × 10−8 M
(c) 1 × 108 M
Ans. (d)
(b) 1 × 10−13 M
(d) 2 × 10−13 M
NaOH(aq) → Na+ (aq) + OH− (aq)
(0. 1 M )
Ni(OH)2 (s ) w
(0. 1 M)
2+
Ni (aq) + 2OH− (aq)
S′
0.1 + 2S ′
Ionic product = (S ′) (0.1 + 2S ′)2 (Q2S ′ is
very small)
2 × 10 −15 = S ′ (0.1)2
S ′ = 2 × 10 −13 M
21 The molar solubility of CaF2 (
K sp = 5.3 × 10 −11 ) in 0.1 M solution of
NaF will be
[NEET (Odisha) 2019]
(a) 5.3 × 1011 mol L −1
(b) 5.3 × 10−8 mol L −1
(c) 5.3 × 10−9 mol L −1
(d) 53
. × 10 −10 mol L −1
Ans. (c)
Let the solubility of CaF2 in 0.1 M NaF is
‘S’ mol L −1
73
Ionic Equilibrium
CaF2 (s )
Ca2 + (aq) + 2F − (aq)
S
2S
-
NaF (aq)
Na+ + F − (aq)
−
0.
[F ] = 2 S + 0.1 1 M 0. 1 M
K sp of CaF2 = [Ca2 + ] [F − ]2
= [S ] [2S + 0.1]2
= 53
. × 10 −11 = [S ] [2 S + 0. 1]2
−11
⇒
53
. × 10 = [S ] [0. 1]2 [Q2S << 0.1 ]
53
. × 10 −11
[S ] =
= 5.3 × 10 −9 mol L−1
(0. 1)2
22 pH of a saturated solution of
Ca(OH) 2 is 9. The solubility product
(K sp ) of Ca(OH) 2 is
[NEET (National) 2019]
(a) 025
. × 10−10
(b) 0125
.
× 10−15
(c) 05
. × 10−10
(d) 05
. × 10−15
−
Ca2 + (aq) + 2OH(aq)
-
2S
S
[where, S = solubility]
… (i)
K sp = [Ca2 + ] [OH− ]2 = S (2 S )2
Given, pH = 9
We know that, pH + pOH = 14
∴ pOH = 14 − 9 = 5
−
pOH = − log[OH]
−
5 = − log [OH]
[OH− ] = 10 − 5
From above equation,
−
[OH] = 2 S = 10 − 5
∴
S=
10 −5
2
= 1.04 × 10 −5 mol L−1
On substituting the value of S in Eq. (i),
we get
K sp = (1.04 × 10 −5 mol L−1)2
= 1.08 × 10 −10 mol2 L−2
24 Concentration of the Ag ions in a
saturated solution of Ag 2C 2O 4 is
2.2 × 10 −4 mol −1 solubility product
of Ag 2C 2O 4 is
[NEET 2017]
(a) 2.42 × 10−8
(c) 4.5 × 10−11
Ans. (d)
(b) 2.66 × 10−12
(d) 5.3 × 10−12
Key concept For a sparingly soluble salt,
if S is the molar solubility,
Ax B y (s ) + H2O q xA y + + yB x −
At saturation,
K [Ax B y ] = [A y + ] x × [B x − ] y = [xS ] x [yS ] y
or K sp = x y . y y S x + y
Where, the constantK sp is called
solubility product.
Ag2 C2O4 ( s ) q 2Ag + + C2O24−
2S
On substituting the value of ‘S’ in eqn. (i),
we get
3
K sp
S
+
For the reaction,
or,
S
K sp = [Ba2 + ] [SO24− ] = (S ) (S ) = S 2 …(i)
[where, S =
Solubility]
Given, S = 2.42 × 10 −3 gL−1
Molar mass of BaSO4 = 233 g mol −1
∴Solubility of BaSO4
2.42 × 10 −3
mol L−1
(S ) =
233
∴
K sp = 1.6 × 10 −10 = [Ag + ][Cl − ]
Given,
or 1.6 × 10 −10 = s (0 . 1 + s ) = 0.1 s + s 2
Q K sp is small, so s is very less in
comparison with 0.1. Hence, s 2 can be
neglected.
Thus, 1.6 × 10 −10 = 0 . 1 s
or
s = 1.6 × 10 −9 M
26 MY and NY 3 , two nearly insoluble
salts, have the same K sp values of
6.2 × 10 −13 at room temperature.
Which statement would be true in
regard to MY and NY 3 ?
[NEET 2016, Phase I]
Ans. (d)
Ca(OH)2 (s )
NaCl(aq ) → Na+ (aq ) + Cl – (aq )
0.1 M
0
0
0
0.1M
0.1+s
AgCl(s ) q Ag + (aq ) + Cl – (aq )
s
s+0.1
Solubility product (K sp ) = [A y + ] x [B x − ] y
For BaSO4 (binary solute giving two ions)
BaSO4 (s ) r Ba2 + (aq) + SO24− (aq)
 10 −5 
 = 0.5 × 10 − 15
= 4S = 4 
 2 


3
23 The solubility of BaSO 4 in water is
2.42 × 10 −3 g L−1 at 298 K. The
value of its solubility product (K sp )
will be
(Given molar mass of BaSO 4 = 233
g mol −1 )
[NEET 2018]
(a) 1.08 × 10−14 mol2L−2
(b) 1.08 × 10−12 mol2L−2
(c) 1.08 × 10−10 mol2L−2
(d) 1.08 × 10−8 mol2L−2
Ans. (c)
For a general reaction,
Ax B y r xA
y+
+ yB
x−
S
K sp = [Ag + ]2 [C2O24− ] = [2S ]2 [S ]
Given, 2S = 22
. × 10 −4 or S = 1.1 × 10 −4 M
∴
K sp = [22
. × 10 −4 ]2 [1.1 × 10 −4 ]
= 53
. × 10 −12
25 The solubility of AgCl(s) with
solubility product 1.6 × 10 −10 in 0.1 M
NaCl solution would be
[NEET 2016, Phase II]
(a) 1.26 × 10−5 M
(b) 1. 6 × 10−9 M
(c) 1. 6 × 10−11 M
Ans. (b)
(d) zero
Key Idea As solubility of AgCl(s) is asked
in 0.1 M NaCl solution, so in the
calculation, solubility of Cl – (from NaCl)
must be added to the solubility of
Cl – (from AgCl).
Let s be the solubility of Ag + and Cl – in
AgCl before the addition of NaCl.
(a) The molar solubility of MY in water is
less than that of NY 3.
(b) The salts MY and NY 3 are more
soluble in 0.5M KY than in pure water
(c) The addition of the salt of KY to
solution of MY and NY 3 will have no
effect on their solubilities
(d) The molar solubilities of MY and NY 3
in water are identical.
Ans. (a)
MY r M+ + Y −
For MY,
0
S
S
where, s = solubility andK sp = solubility
product.
∴ K sp = [M+ ] [Y − ] = S 2
S = K sp = 6.2 × 10 −13 = 7.874 × 10 −7
Similarly, for NY3,
NY3 r N + + 3Y −
∴ s =4
S
0
3S
K sp = [N + ] [Y − ] 3 = s × (3s ) 3
K sp = 27 S 4
∴
K sp
27
=4
6.2 × 10 −13
= 3.89 × 10 −4
27
Therefore, molar solubility of MY in water
is less than that ofNY3.
27 The K sp of Ag 2CrO 4, AgCl, AgBr
and AgI are respectively,
1.1 × 10–12 , 1.8 × 10 −10 , 5.0 × 10 −13 ,
8.3 × 10 −17 . Which one of the
following salts will precipitate last
if AgNO 3 solution is added to the
solution containing equal moles of
NaCl,NaBr,NaI and Na 2CrO 4 ?
[CBSE AIPMT 2015]
(a) AgI
(c) AgBr
(b) AgCI
(d) Ag2CrO 4
74
NEET Chapterwise Topiewise Chemistry
Ans. (d)
Ans. (c)
Ag2 CrO4
2Ag + + CrO24−
Solubility product
K sp = (2s )2 × S = 4s 3
K sp = (1.1 × 10 −12 ) (given)
K
S = 3 sp = 0 .65 × 10 − 4
4
s
−
AgCl
Ag + Cl
K sp = S × S
(K sp = 1. 8 × 10 −10 )
S = K sp = 1.34 × 10 −5
s
+
AgBr
Ag + + Br −
K sp = S × S
(K sp = 5 × 10 −13)
−6
S = K sp = 0.71 × 10
s
AgI
Ag + + I−
K sp = S × S (K sp = 83
. × 10 −17 )
−8
S = K sp = 0.9 × 10
Q Solubility of Ag2 CrO4 is highest.
So, it will precipitate last.
s
28 H2S gas when passed through a
solution of cations containing HCl
precipitates the cations of second
group in qualitative analysis but not
those belonging to the fourth
group. It is because
[CBSE AIPMT 2005]
(a) presence of HCl decreases the
sulphide ion concentration
(b) presence of HCl increases the
sulphide ion concentration
(c) solubility product of group II
sulphides is more than that of group
IV sulphides
(d) sulphides of group IV cations are
unstable in HCl
Ans. (a)
In qualitative analysis of cations of
second group H2S gas is passed in
presence of HCl, therefore due to
common ion effect, lower concentration
of sulphide ions is obtained which is
sufficient for the precipitation of second
group cations in the form of their
sulphides due to lower value of their
solubility product (K sp ). Here, fourth
group cations are not precipitated
because it require more sulphide ions for
exceeding their ionic product to their
solubility products which is not obtained
here due to common ion effect.
29 The solubility product of a sparingly
soluble salt AX 2 is 3.2 × 10 − 11 . Its
solubility (in mol/L) is
[CBSE AIPMT 2004]
(a) 5.6 × 10−6
(c) 2 × 10− 4
(b) 3.1 × 10− 4
(d) 4 × 10− 4
32 Solubility of a M 2 S type salt is
3. 5 × 10 −6 , then find out its
solubility product.
AX2 is ionised as follows
A X2
A2 + + 2 X –
3
S mol L–1
2S
S
[CBSE AIPMT 2001]
(a) 1. 7 × 10−6
(c) 1. 7 × 10−18
Ans. (b)
Solubility product of A X2 ,
2+
− 2
K sp = [A ] [X ]
= S × (2 S )2 = 4S 3
Q
∴
K sp of A X2 = 3.2 × 10 − 11
Solubility of M2 S salt is 3.5 × 10 – 6 M
M2 S
2 M+ + S 2–
3 .2 × 10 − 11 = 4S 3
S 3 = 0.8 × 10 −11
= 8 × 10
3.5 × 10 –6 M
−12
Solubility = 2 × 10 − 4 mol/L
30 The solubility product of AgI at
25°C is 1.0 × 10 −16 mol 2 L−2 . The
solubility of AgI in 10 −4 N solution
of KI at 25°C is approximately
[CBSE AIPMT 2003]
(in mol L−1 )
−10
(a) 1 .0 × 10
(b) 1 .0 × 10−8
(c) 1 .0 × 10−16
(d) 1 .0 × 10−12
Ans. (d)
K sp = S
where, S = solubility in mol/L
=S
2
or
S =1 × 10 −8 mol/L
Normality of KI solutiuon = 10 −4 N
Here change is one
[n = 1]
∴
M = 10 −4 M
or S for KI solution = 10 −4 M
Solubility of AgI in K I solution
= 1 × 10 –8 × 10 – 4
= 1 × 10 –12 mol /L
31 Solubility of MX 2 type electrolytes
is 0.5 × 10 −4 mol/L, then find out
K sp of electrolytes.
[CBSE AIPMT 2002]
−12
(b) 25 × 10−10
(d) 5 × 10−13
(a) 5 × 10
(c) 1 × 10−13
Ans. (d)
MX2
Solubility 0.5 × 10 – 4M
→
2 × 3. 5 × 10– 6 M 3.5 × 10 –6 M
(on 100% ionisation)
∴ K sp (solubility product of M2 S )
= [M+] 2 [S 2– ]
=(7.0 × 10 –6 )2 (3.5 × 10 –6 )
= 171. 5 × 10 –18
= 1. 71 × 10 –16 [M] 3
33 The solubility of a saturated
solution of calcium fluoride is
2 × 10–4 mol/L. Its solubility
product is
[CBSE AIPMT 1999]
CaF2
2 × 10 –4 M
2
1.0 × 10
3
(a) 12 × 10–2
(c) 22 × 10–11
Ans. (d)
AgI → Ag + + I–
For binary electrolyte
–16
(b) 1. 7 × 10−16
(d) 1. 7 × 10−12
(b) 14 × 10–4
(d) 32 × 10–12
2+
3 Ca
2 × 10 –4 M
+
2F –
2 × 2 × 10 –4 M
K sp of CaF2 = [Ca2 + ] [F – ]2
= [2 × 10 –4 ] [4 × 10 –4 ]2
= 32 × 10 –12 (mol/L)2
34 Which of the following is most
soluble?
[CBSE AIPMT 1994]
(a) Bi2 S3 (K sp = 1 × 10−70)
(b) MnS (K sp = 7 × 10−16 )
(c) CuS (K sp = 8 × 10−37 )
(d) Ag2 S (K sp = 6 × 10−51)
Ans. (b)
Higher the value of solubility product,
higher is its solubility. In all these
compounds the MnS is most soluble
because its solubility product is
maximum.
35 In which of the following the
solubility of AgCl will be minimum?
M2 +
[CBSE AIPMT 1993]
0.5 × 10 – 4M
+
2X –
2 × 0.5 × 10 –4M
(on 100% ionisation)
∴ K sp of MX2 = [M2 + ] [X – ]2
= (0.5 × 10 – 4 ) (1.0 × 10 – 4 )2
=0.5 × 10 –12
= 5 × 10 –13 [M] 3
(a) 0.1 M NaNO 3
(b) Water
(c) 0.1 M NaCl
(d) 0.1 M NaBr
Ans. (c)
In 0.1 M NaCl, the solubility of AgCl is
minimum due to the phenomenon of
common ion effect.
75
Ionic Equilibrium
TOPIC 4
Let us consider all the options,
pH, Buffer and Indicator
36 The pK b of dimethyl amine and pK a
of acetic acid are 3.27 and 4.77
respectively atT(K). The correct
option for the pH of dimethyl
ammonium acetate solution is
[NEET 2021]
(a) 8.50 (b) 5.50
Ans. (c)
(c) 7.75
(d) 6.25
Dimethyl ammonium acetate
[CH3COONH2 (CH3)2 ] is a salt of weak acid
(CH3COOH) and weak base [(CH3)2 NH].
pH of dimethyl ammonium acetate salt
solution can be calculated using
formula :
1
pH = 7 + (pK a − pK b )
2
pK a of acetic acid = 4.77
pK b of dimethyl amine = 3.27
1
pH = 7 + (4.77 − 3.27)
2
1
pH = 7 + × 1.50
2
⇒
pH = 7 + 0.75
pH = 7.75
37 The pH of 0.01 M NaOH (aq) solution
will be
[NEET (Odisha) 2019]
(a) 7.01
(c) 12
Ans. (c)
(b) 2
(d) 9
NaOH is a strong base, thus
[OH− ] = 0.01M = 10 −2 M
pOH = − log [OH− ]
= − log(10 −2 ) = 2
We know that, pH + pOH = 14
∴ pH = 14 − 2 = 12
Thus, option (c) is correct.
38 Which will make basic buffer?
[NEET (National) 2019]
(a) 100 mL of 0.1 M CH3COOH + 100 mL
of 0.1 M NaOH
(b) 100 mL of 0.1 M HCl + 200 mL of 0.1 M
NH4OH
(c) 100 mL of 0.1 M HCl + 100 mL of 0.1 M
NaOH
(d) 50 mL of 0.1 M NaOH + 25 mL of 0.1 M
CH3COOH
Ans.
Key idea A buffer solution having pH
more than 7 is known as basic buffer. It is
obtained by mixing weak base and its salt
with strong acid in a fixed proportion.
(a) 100 mL of 0.1 M CH3COOH + 100 mL
of 0.1M NaOH
CH3COOH + NaOH → CH3COONa + H2O
Initial
conc.
100 mL ×
0.1 M
100 mL ×
0 mmol
=10mmol =10mmol
0
10 mmol
It is not basic buffer because hydrolysis
of salt takes place and final solution
contains salt of weak acid with strong
base only.
Hence, option (a) is incorrect.
(b) 100 mL of 0.1 M HCl + 200 mL of 0.1 M
NH4OH
HCl + NH4OH → NH4 Cl + H2O
Initial conc. 100 mL× 200 mL ×
0.1 M HCl
0 mmol
0.1 M
=10 mmol =20mmol
Final conc.
0
10 mmol
10 mmol
It is basic buffer because final solution
contains weak base and its salt with
strong acid. Hence, option (b) is correct.
(c) 100 mL of 0.1 M HCl + 100 mL of 0.1 M
NaOH
HCl + NaOH → NaCl + H2O
Initial conc. 100 mL 100 mL × 0 mmol
×01
.M
0
0
10 mmol
It is a neutral solution. Hence, option (c)
is incorrect.
(d) 50 mL of 0.1 M NaOH + 25mL of 0.1 M
CH3COOH
CH3COOH + NaOH → CH3COONa + H2O
Initial
conc.
25mL
× 0.1 M
=2 .5 mmol
Final conc.
0
50mL
× 0.1 M
0 mmol
= 5mmol
2.5 mol
(b) I
(c) II
(d) III
M
M
HCl + 25 mL NaOH
5
5
Milliequivalent of HCl
M
1
= 75 mL of HCl = × 75 = 15
5
5
Milliequivalent of NaOH
M
= 25 mL of NaOH
5
1
= × 25 = 5
5
∴ Milliequivalent of HCl left unused
= 15 − 5 = 10
Volume of solution = 100 mL
∴ Molarity of [H+ ] in the resulting mixture
10
1
=
=
100 10
1
∴
pH = log + = log(10) = 1
[H ]
40 What is the pH of the resulting
solution when equal volumes of 0.1
M NaOH and 0.01 M HCl are mixed?
[CBSE AIPMT 2015]
(a) 12.65 (b) 2.0
Ans. (a)
0.1 M
=10 mmol =10 mmol
Final conc.
(a) IV
Ans.
75 mL
0.1 M
Final conc. 0
pH of which one of them will be
equal to 1?
[NEET 2018]
2.5 mmol
It is basic solution. Hence, option (d) is
incorrect.
39 Following solutions were prepared
by mixing different volumes of
NaOH and HCl of different
concentrations :
M
M
I. 60 mL HCl + 40 mL NaOH
10
10
M
M
II. 55 mL HCl + 45 mL NaOH
10
10
M
M
III. 75 mL HCl + 25mL NaOH
5
5
M
M
IV. 100 mL HCl + 100mL NaOH
10
10
(c) 7.0
(d) 1.04
Key Concept When equal volumes of
acid and base are mixed, then resulting
solution become alkaline if concentration
of base is taken high.
Let normality of the solution after mixing
0.1 M NaOH and 0.01 M HCl is N.
∴
N 1V1 − N2V2 = NV
or
0.1 × 1 − 0.01 × 1 = N × 2
Since, normality of NaOH is more than
that of HCl.
Hence, the resulting solution is alkaline.
0.09
or [OH] = N =
= 0.045 N
2
or
∴
pOH = − log (0.045) = 1.35
pH = 14 − pOH = 14 − 1.35 = 12.65
41 Which one of the following pairs of
solution is not an acidic buffer?
[CBSE AIPMT 2015]
(a) HClO 4 and NaClO 4
(b) CH3COOH and CH3COONa
(c) H2CO 3 and Na2CO 3
(d) H3 PO 4 and Na3 PO 4
Ans. (a)
Strong acid with its salt cannot form
buffer solution. Hence,HClO4 and NaClO4
is not an acidic buffer.
76
NEET Chapterwise Topiewise Chemistry
42 pH of a saturated solution of
Ba(OH) 2 is 12. The value of
solubility product K sp of Ba(OH) 2 is
[CBSE AIPMT 2012]
(a) 3.3 × 10−7
(c) 4.0 × 10−6
Ans. (b)
(b) 5.0 × 10−7
(d) 5.0 × 10−6
Given, pH ofBa(OH)2 = 12
∴
pOH = 14 − pH
= 14 − 12 = 2
We know that,
pOH = −log[OH− ]
2 = − log[OH− ]
−
[OH ] = antilog (−2)
[OH− ] = 1 × 10 −2
Ba(OH)2 dissolves in water as
Ba(OH)2 (s )
Ba2 + + 2OH−
S mol L−1
∴
a
−
S
2S
−2
[OH ] = 2 S = 1 × 10
[OH− ]
S=
2
[OH− ] 1 × 10 −2
=
[Ba2 + ] =
2
2
[Ba2 + = S ]
K sp = [Ba2 + ] [OH− ]2
 1 × 10 −2 
 (1 × 10 −2 )2
= 
2 

= 0.5 × 10 −6
= 5 × 10 −7
43 Buffer solutions have constant
acidity and alkalinity because
[CBSE AIPMT 2012]
(a) these give unionised acid or base on
reaction with added acid or alkali
(b) acids and alkalies in these solutions
are shielded from attack by other
ions
(c) they have large excess ofH+ or OH−
ions
(d) they have fixed value of pH
Ans. (a)
If small amount of an acid or alkali is
added to a buffer solution, it converts
them into unionised acid or base. Thus,
its pH remains unaffected or in other
words its acidity/alkalinity remains
constant. e.g.
H3O+ + A −
H2O + HA
−
OH + H A → H2O + A −
If acid is added, it reacts with A − to form
undissociated H A. Similarly, if
base/alkali is added,OH− combines with
H A to give H2O and A − and thus,
maintains the acidity/ alkalinity of buffer
solution.
a
44 A buffer solution is prepared in
which the concentration of NH3 is
0.30 M and the concentration of
NH+4 is 0.20 M. If the equilibrium
constant, K b for NH3 equals
1.8 × 10 −5 , what is the pH of this
solution? (log 2.7 = 0.43)
[CBSE AIPMT 2011]
(a) 9.43 (b) 11.72 (c) 8.73
Ans. (a)
(d) 9.08
[salt]
[base]
[salt]
=− log K b + log
[base]
0.20
= − log 1.8 × 10 −5 + log
0.30
= 5 − 0.25 + (− 0.176)
= 4. 75 − 0.176 = 4. 57
∴ pH = 14 − 4.57 = 9.43
pOH = pK b + log
45 What is [H+ ] in mol/L of a solution
that is 0.20 M in CH3COONa and
0.10 M in CH3COOH ?
(K a for CH3COOH = 1.8 × 10 −5 )
[CBSE AIPMT 2010]
(a) 3.5 × 10−4
(b) 1.1 × 10−5
(c) 1.8 × 10−5
(d) 9.0 × 10−6
Ans. (d)
Key Idea CH3COOH (weak acid) and
CH3COONa (conjugated salt) form acidic
buffer and for acidic buffer,
[salt]
pH = pK a + log
[acid]
and [H+ ] = − antilog pH
pH = − log Ka + log
[salt]
[acid]
[QpK a = − log K a ]
(0.20)
= − log (1.8 × 10 −5 ) + log
(0.10)
= 4.74 + log2
= 4.74 + 0.3010 = 5.041
Now, [H+ ] = antilog (− 5.045)
= 9.0 × 10 −6 mol/L
46 If pH of a saturated solution of
Ba (OH) 2 is 12, the value of its K sp is
[CBSE AIPMT 2010]
(a) 4.00 × 10−6 M3
(c) 5.00 × 10−7 M3
Ans. (d)
(b) 4.00 × 10−7 M3
(d) 5.00 × 10−6 M3
Given, pH ofBa(OH)2 = 12
So, pOH = 2
∴ [H+ ] = [1 × 10 −12 ]
K w = (H+ ) (OH− )
K w = 1 × 10 −14
K
OH− = +ω
H
1 × 10 −14
−
and [OH ] =
1 × 10 −12
[Q[H+ ] [OH− ] = 1 × 10 −14 ]
= 1 × 10 −2 mol/L
Ba(OH)2 → Ba2 + + 2 OH−
S
2S
2+
− 2
K sp = [Ba ] [OH ] = [S ] [2 S ]2
 1 × 10 −2 
−2 2
=
 (1 × 10 )
 2 
= 0.5 × 10 −6 = 5.0 × 10 −6 M3
47 In a buffer solution containing
equal concentration of B − and H B,
the K b for B − is 10 −10 . The pH of
buffer solution is
[CBSE AIPMT 2010]
(a) 10
(c) 6
Ans. (d)
(b) 7
(d) 4
Key Idea (i) For basic buffer,
[salt]
pOH = pKb + log
[base]
(ii) pH + pOH = 14
Given, K b = 1 × 10 −10 , [salt] = [base]
[salt]
pOH = − log K b + log
[base]
∴ pOH = − log (1 × 10 −10 ) + log 1 = 10
pH + pOH = 14
[Qconcentration of [B − ] = [HB]
pH = 14 − 10 = 4
48 What is the [OH− ] in the final
solution prepared by mixing 20.0
mL of 0.050 M HCl with 30.0 mL of
0.10 M Ba(OH) 2 ?
[CBSE AIPMT 2009]
(a) 0.10 M
(c) 0.0050 M
Ans. (a)
(b) 0.40 M
(d) 0.12 M
Number of milliequivalents of HCl
= 20 × 0.050 × 1 = 1
Number of milliequivalents of Ba(OH)2
= 2 × 30 × 0.10 = 6
[OH− ] of final solution
Milliequivalents of Ba(OH)2
=
– milliequivalents of HCl
Total volume
= 0.1 M
=
6−1
50
77
Ionic Equilibrium
49 Equal volumes of three acid
solutions of pH 3, 4 and 5 are mixed
in a vessel. What will be the H+ ion
concentration in the mixture?
[CBSE AIPMT 2008]
(a) 1.11 × 10−4 M
(c) 3.7 × 10−3 M
Ans. (b)
(b) 3.7 × 10−4 M
(d) 1.11 × 10−3 M
Total [H+ ] concentrated of mixture
M V + M2V2 + M3V3
(M) = 1 1
V1 + V2 + V3
=
1 × 10 −3 × V + 1 × 10 −4 × V + 1 × 10 −5 × V
V +V +V
=
1 × 10 −3 × V (1 + 0.1 + 0.01)
3V
1.11 × 10−3
= 3.7 × 10−4 M
3
50 Calculate the pOH of a solution at
25°C that contains 1 × 10 −10 M of
hydronium ion. [CBSE AIPMT 2007]
(a) 7.00
(c) 9.00
Ans. (b)
(b) 4.00
(d) 1.00
[H3O+ ] = [H+ ] =10 −10
pH + pOH = 14
and pH = − log [H+ ]
pH = − log [10 −10 ]
pH = 10
from eq. (i) and (ii), we get
pOH + 10 = 14
pOH = 14 − 10 = 4
K(i)
K(ii)
53 What is the correct relationship
between the pH of isomolar
solutions of sodium oxide (pH1 ),
sodium sulphide (pH2 ) , sodium
selenide (pH3 ) and sodium telluride
[CBSE AIPMT 2005]
(pH4 ) ?
(a) pH1 > pH2 ≈ pH3 > pH4
(b) pH1 < pH2 < pH3 < pH4
(c) pH1 < pH2 < pH3 ≈ pH4
(d) pH1 > pH2 > pH3 > pH4
54 The rapid change of pH near the
stoichiometric point of an acid
base titration is the basis of
indicator detection. pH of the
solution is related to ratio of the
concentrations of the conjugate
acid (HIn) and base (In − ) forms of
the indicator given by the
expression
[CBSE AIPMT 2004]
[In–]
[HIn]
[HIn]
(b) log
[In–]
[HIn]
(c) log
[In–]
[In–]
(d) log
[HIn]
(a) log
The correct order of pH of isomolar
solution of sodium oxide (pH1), sodium
sulphide (pH2 ),sodium selenide (pH3) and
sodium telluride (pH4 ) is
pH1 > pH2 > pH3 > pH4 because in
aqueous solution, they are hydrolysed as
follows.
Na2O + 2H2O → 2NaOH + H2O
2NaOH
(b) 1.0525 × 10 M
(d) 1.0 × 10–8 M
+
In aqueous solution of 10 M HCl, [H ]
ion concentration is based upon the
concentration of H+ ion of 10 −8 M HCl
and concentration ofH+ ion of water.
K w of H2O = 10 –14 = [H+ ][OH− ]
or
[H+ ] = 10 –7 M
(due to its neutral behaviour)
= pK In – pH
= pK In – pH
= pH –pK In
= pH – pK In
Ans. (d)
Acid indicators are generally weak acid.
The dissociation of indicator HIn takes
place as follows
HIn
H+ + In–
+
[H ] [In– ]
∴
K In =
[HIn]
[HIn]
+
or [H ] =K In ⋅
K(i)
[In– ]
3
Ans. (d)
Q
pH = – log [H+ ]
From eq. (i) and (ii) we get,

[HIn] 
pH = − log  KIn ⋅
∴


[In− ] 
+
= − log KIn + log
H2 S
Strong base Weak acid
Na2Se + 2H2O →
–7
−8
A pair constituent withHNO2 and NaNO2
because HNO2 is weak acid andNaNO2 is
a salt of weak acid (HNO2 ) with strong
base (NaOH). Hence, it is an example of
acidic buffer solution.
neutralisation of NaOH with H2O,H2S,
H2Se and H2Te.
Na2O > Na2S > Na2Se > Na2Te
(Q pH of basic solution is higher than
acidic or least basic solution)
Base
[CBSE AIPMT 2006]
(a) 1.0 × 10 M
(c) 9 .525 × 10–8 M
Ans. (b)
(a) HNO2 and NaNO2
(b) NaOH and NaCl
(c) HNO 3 and NH4NO 3
(d) HCl and KCl
Ans. (a)
Na2S + 2H2O →
51 The hydrogen ion concentration of
a 10–8 M HCl aqueous solution at
298 K (K w = 10–14 ) is
–6
52 Which of the following pairs
constitutes a buffer?
[CBSE AIPMT 2006]
Let the volume of each acid = V
pH of first, second and third acids = 3, 4
and 5 respectively
[H+ ] of first acid (M1) = 1 × 10 −3
[Q H+ = 1 × 10 − pH ]
+
[H ] of second acid (M2 ) = 1 × 10 −4
[H+ ] of third acid(M3) = 1 × 10 −5
=
So, in aqueous solution of 10 –8 M HCl,
[H+ ] = [H+ ] of HCl + [H+ ] of water
= 10 –8 + 10 –7
= 11 × 10 –8 M ≈ 1.10 × 10 –7 M
= pK In + log
2NaOH
Strong base
+
H2 Se
Weak acid
Na2Te + 2H2O →
2NaOH
Strong base
+
H2Te
Weak acid
On moving down the group acidic
character of oxides increases.
Order of acidic strength
H2Te > H2Se > H2S > H2O
Order of neutralisation of NaOH
H2Te > H2Se > H2S > H2O
Hence, their aqueous solutions have the
following order of basic character due to
or log
K(ii)
[In– ]
[HIn]
[In– ]
[HIn]
[In– ]
= pH − pK ln
[HIn]
55 Solution of 0.1 N NH4 OH and 0.1 N
NH4 Cl has pH 9.25, then find out
pK b of NH4 OH. [CBSE AIPMT 2002]
(a) 9.25
(c) 3.75
Ans. (b)
(b) 4.75
(d) 8.25
Solution of NH4OH and NH4 Cl acts as a
basic buffer solution. For basic buffer
solution
[salt]
pOH = pK b + log
[base]
78
NEET Chapterwise Topiewise Chemistry
pOH = 14 – pH
= 14 – 925
. = 4.75
0.1
4.75 = pK b + log
0.1
pK b = 4.75
56 The concentration of [H+ ] and
concentration of [OH– ] of a 0.1 M
aqueous solution of 2% ionised
weak monobasic acid is
[CBSE AIPMT 1999]
[ionic product of water = 1 × 10–14 ]
(a) 002
. × 10–3 Mand 5 × 10–11 M
(b) 1 × 10–3 M and 3 × 10–11 M
(c) 2 × 10–3 M and 5 × 10–12 M
(d) 3 × 10–2 M and 4 × 10–13 M
Ans. (c)
[H + ] in monobasic acid
= molarity × degree of ionisation
2
= 0.1 ×
100
= 2 × 10 –3 M
ionisation constant of water
K w = (H+ ) (OH− )
1 × 10 –14
K
[OH– ] = w+ =
[H ] 2 × 10 –3
= 5 × 10 –12 M
57 A physician wishes to prepare a
buffer solution at pH= 3.58 that
efficiently resist changes in pH yet
contains only small concentration
of the buffering agents. Which one
of the following weak acid together
with its sodium salt would be best
to use?
[CBSE AIPMT 1997]
(a) m-chlorobenzoic acid
(pK a = 3 .98)
(b) p-chlorocinnamic acid
(pK a = 441
. )
(c) 2, 5-dihydroxy benzoic acid
(pK a = 2 .97)
(d) Acetoacetic acid (pK a = 3 .58)
Ans. (d)
By the use of Henderson’s equation
[salt]
pH = pKa + log 10
[acid]
When, [salt] = [acid]
∴ pH = pK a
Q pK a = 3.58, thus at this state
pH = 3.58
So, acetoacetic acid (pK a = 3.58) is best
to use.
58 The pH value of blood does not
change appreciably by a small
addition of an acid or base,
because the blood
[CBSE AIPMT 1995]
(a) is a body fluid
(b) can be easily coagulated
(c) contains iron as a part of the
molecule
(d) contains serum protein that acts as
buffer
Ans. (d)
Blood is an example of buffer solution,
which contains serum protein, so its pH
does not change appreciably by adding
small amount of an acid or a base to it.
59 The pH value of a 10 M solution of
HCl is
[CBSE AIPMT 1995]
(a) less than 0
(c) equal to 1
Ans. (a)
(b) equal to 0
(d) equal to 2
+
[Where ⇒WB= Weak base,SB =
Strong base WA = Weak acid,SA =
Strong acid.]
Hence, option (d) is the correct.
61 Which of the following salts will
give highest pH in water?
[CBSE AIPMT 2014]
(a) KCl
(c) Na2CO 3
Ans. (c)
(b) NaCl
(d) CuSO4
The highest pH refers to the basic
solution containingOH− ions. Therefore,
the basic salt releasing moreOH− ions on
hydrolysis will give highest pH in water.
Only the salt of strong base and weak
acid would release moreOH− ion on
hydrolysis. Among the given salts,
Na2 CO3 corresponds to the basic salt as
it is formed by the neutralisation of
NaOH [strong base] and H2 CO3 [weak
acid].
CO23− + H2O
HCO3− + OH−
s
−
HCl(aq) → H (aq) + Cl (aq)
[S = K sp ]
[HCl] = 10 M
⇒ [H+ ] = 10 mol / L
pH = − log [H+ ] = − log 10
= − 1, so the pH is less than zero.
TOPIC 5
Hydrolysis of Salts
60 Which among the following salt
solutions is basic in nature?
[NEET (Oct.) 2020]
(a) Ammonium chloride
(b) Ammonium sulphate
(c) Ammonium nitrate
(d) Sodium acetate
Ans. (d)
Nature of a salt solution depends on the
nature of constituent acid and base
whether they are strong or weak.
(a) NH4 Cl is made of
[NH4OH(WB) + HCl(SA)] → Acidic
solution.
(b) (NH4 )2 SO4 is made of
[NH4OH(WB) + H2SO4 (SA)] → Acidic
solution.
(c) NH4NO3 is made of
[NH4OH(WB) + CH3COOH(WA)] →
Basic solution.
(d) CH3COONa is made of
[NaOH(SB) + CH3COOH(WA)] → Basic
solution
62 Equimolar solutions of the
following substances were
prepared separately. Which one of
these will record the highest pH
value?
[CBSE AIPMT 2012]
(a) BaCl2 (b) AlCl3 (c) LiCl
Ans. (a)
(d) BeCl2
BaCl2 is a salt of strong acid HCl and
strong base Ba(OH)2 . So, its aqueous
solution is neutral with pH 7. All other
salts give acidic solution due to cationic
hydrolysis, so their pH is less than 7.
Thus, pH value is highest for the solution
of BaCl2
63 The ionisation constant of
ammonium hydroxide is 1.77 × 10 −5
at 298 K. Hydrolysis constant of
ammonium chloride is
[CBSE AIPMT 2009]
(a) 5.65 × 10−10
(c) 5.65 × 10−13
Ans. (a)
(b) 6.50 × 10−12
(d) 5.65 × 10−12
Given,K a (NH4OH) = 1.77 × 10 −5
NH4OH
NH+4 + OH–
+
[NH4 ][OH– ]
Ka =
= 1.77 × 10 −5
[NH4OH]
3
Hydrolysis of NH4 Cl takes place as,
NH4 Cl + H2O → NH4OH + HCl
or
NH+4 + H2O → NH4OH + H+
Hydrolysis constant,
…(i)
79
Ionic Equilibrium
Kh =
or K h =
[NH4OH][H+ ]
[NH+4 ]
…(ii)
[NH4OH][H+ ][OH– ]
[NH+4 ][OH– ]
…(iii)
From Eqs. (i), (ii) and (iii)
K
[Q [H+ ][OH− ] = K w ]
Kh = w
Ka
=
10 −14
1.77 × 10 −5
NH4 Cl + H2O →
NH4OH
64 Which has highest pH?
[CBSE AIPMT 2002]
(b) Na2CO3
(d) NaNO3
pH=log
1
[H+ ]
pH is inversely proportional to hydrogen
ion concentration. As concentration of
H+ decreases pH increases and
vice-versa.
+
HCl
Strong acid
Weak base
So, pH of NH4 Cl is less than 7.
Sodium nitrate (NaNO3 ) is the salt of
strong acid and strong base. So, its
aqueous solution is neutral as
NaNO3 + H2O →
= 5.65 × 10 −10
(a) CH3CO −OK +
(c) NH4 Cl
Ans. (b)
Ammonium chloride (NH4 Cl) is a salt of
weak base and strong acid. So, its
aqueous solution will be acidic as
NaOH
Strong base
+
HNO3
Strong acid
So, pH of NaNO3 is 7.
Potassium acetate (CH3COOK) is a salt of
strong base and weak acid. Its aqueous
solution will be basic and pH value will be
greater than 7 ≈ 8.8
CH3COO−K+ + H2O → CH3COOH
Weak acid
+
KOH
Strong base
Sodium carbonate (Na2 CO3) is a salt of
strong base and weak acid. Its aqueous
solution is also basic and its pH value will
be more than 10,
i.e. highest among them.
Na2 CO3 + H2O →
2NaOH
Strong base
+ H2 CO3
Weak acid
65 The compound whose aqueous
solution has the highest pH is
[CBSE AIPMT 1988]
(a) NaCl
(c) Na2CO3
Ans. (c)
(b) NaHCO3
(d) NH4Cl
The hydrolysis of NaCl gives neutral
solution because it is salt of strong acid
and strong base and hence, its pH is 7.
NH4 Cl is salt of weak base and strong
acid, so its pH is less than 7.NaHCO3 is
also acidic whereasNa2 CO3 is salt of
strong base and weak acid, so its pH is
more than 7.
10
Redox Reactions and
Electrochemistry
TOPIC 1
TOPIC 2
O
Redox Reactions
O
Cr
01 The oxidation state of Cr in CrO 5 is
[NEET (Odisha) 2019]
(a) − 6
(c) + 6
Ans. (c)
115°
O
(b) + 12
(d) + 4
O
O
Oxidation state of Cr in CrO5 is +6 due to
the presence of two peroxide linkages
which can be calculated as
x + (−1) × 4 + 1 × (−2)
[For Cr]
[For 0 − 0]
[For 0]
x − 6 = 0; x = +6
The structure of CrO5 is
O
O
03 Oxidation state of Fe in Fe 3 O 4 is
O
Cr
O
[CBSE AIPMT 1999]
O
Oxidation state of Cr is +6 due to the
presence of two peroxide linkages,
which can be calculated as.
In CrO(O2 )2 , let the oxidation state of Cr
be x
x + (−1) 4 + (−2) = 0
x − 6 = 0 ⇒ x = +6
02 In acidic medium, H2O 2 changes
Cr 2O 2−
7 to CrO 5 which has two
(—O—O—) bonds. Oxidation state of
Cr in CrO 5 is
[CBSE AIPMT 2014]
(a) +5
(b) +3
Ans. (c)
(c) +6
(d) −10
When H2O2 is added to an acidified
solution of a dichromate Cr2O27− , a deep
blue coloured complex, chromic
peroxide CrO5 [or CrO(O2 )2 ] is formed.
Cr2O27− + 2 H+ + 4H2O → 2 CrO(O2 )2
14243
Chromic peroxide
[blue coloured complex] + 5H2O
This deep blue coloured complex has the
following structure
3
(a)
2
4
(b)
5
(c)
5
4
(d)
8
3
Ans. (d)
Oxidation state of Fe inFe3O4 is
calculated as
x
–2
Electrolytic Conductance,
Electrolysis and Its
Quantitative Aspect
05 The molar conductance of NaCl,
HCl and CH3COONa at infinite
dilution are 126.45, 426.16 and 91.0
S cm 2 mol − 1 respectively. The
molar conductance of CH3COOH at
infinite dilution is
Choose the right option for your
answer.
[NEET 2021]
(a) 201.28 S cm2 mol − 1
(b) 390.71 S cm2 mol − 1
(c) 698.28 S cm2 mol − 1
(d) 540.48 S cm2 mol − 1
Ans. (b)
Fe3 O4
Molar conductance of NaCl,
λ°NaCl = 126.45 S cm2 mol −1
Molar conductance of HCl,
3x + (–2 × 4) = 0
3x = + 8
8
x=+
3
λ°HCl = 426.16 S cm2 mol −1
− +
04 The oxidation state of Cr in
[CBSE AIPMT 1988]
K 2Cr 2O 7 is
(a) +5
(b) +3
Ans. (c)
(c) +6
(d) +7
Let the oxidation state of Cr is x
K2 Cr2O7
∴
2 (+1) + 2x + 7 (−2) = 0
2 + 2x − 14 = 0
2x − 12 = 0
2x = 12
12
x = = +6
2
Molar conductance of CH3COONa,
λ°CH COONa = 91.0 S cm2 mol −1
3
According to Kohlrausch’s law;
…(i)
λ°HCl = λ°H + + λ°Cl−
λ°CH
λ°NaCl = λ°Na + + λ°Cl−
3 COONa
λ°CH
+
3 COOH
λ°CH
3 COOH
= λ°Na + + λ°CH
= λ°CH
3 COO
= (λ°CH
3 COO
−
…(iii)
+ λ°H +
−
3 COO
…(ii)
−
+ λ°Na + )
+ (λ°H + + λ°Cl− ) − (λ°Na + + λ°Cl− )
λ°CH
3 COOH
= λ°
−
CH 3 CO O Na +
+ λ°HCl − λ°NaCl
81
Redox Reactions and Electrochemistry
λ°CH
λ°CH
3 COOH
3 COOH
= 91.0 + 426.16 − 126.45
= 390.71 S cm2 mol −1
06 The molar conductivity of 0.007 M
acetic acid is 20 S cm 2 mol − 1 .
What is the dissociation constant
of acetic acid ? Choose the correct
option.
[λ °H = 350S cm 2 mol − 1 ,
λ °CH COO− = 50 S cm 2 mol − 1 ]
3
[NEET 2021]
−4
−1
−1
(a) 1.75 ×10 mol L
(b) 2.50 ×10− 4 mol − 1 L − 1
(c) 1.75 ×10− 5 mol − 1 L − 1
(d) 2.50 ×10− 5 mol − 1 L − 1
Ans. (c)
−
= 50 S cm2 mol −1
3
λ°CH
3
3 COOH
= 350 + 50
= 400 S cm2 mol −1
Degree of dissociation,
λ CH COOH
3
α=
λ°CH COOH
3
α=
20
1
=
= 0.05 = 5 × 10 −2
400 20
CH3COOH q
t=0
t = teq
CH3COO– + H+
–
Cα
C
C − Cα
–
Cα
Equilibrium constant,
C2α2
Cα2
=
C (1 − α) 1 − α
As α < < 1;1 − α ≈ 1
k=
k = Cα2 = 0.007 × (0.05)2
= 0.007 × 25 × 10 −4
= 1.75 × 10 −5 mol L−1
07 On electrolysis of dilute sulphuric
acid using platinum (Pt) electrode,
the product obtained at anode will
be
[NEET (Sep.) 2020]
(a) oxygen gas
(c) SO2 gas
Ans. (a)
−
4OH − 4e − → O2 ↑ + 2H2O
(b) H2 S gas
(d) hydrogen gas
Dilute H2SO4 contains H2SO4 and H2O and
they show electrolytic dissociation as,
+7
+4
+7
08 The number of Faradays (F )
required to produce 20 g of
calcium from molten CaCl 2 (Atomic
mass of Ca = 40 g mol −1 ) is
(c) 4
(d) 1
CaCl2 r Ca2 + + 2Cl −
Molten
2+
At cathode Ca
(Reduction)
+4
(ii) 3MnO24 − + 4H+ → 2MnO4− + MnO2
+ 2H2O
The above reaction is a
disproportionation reaction as Mn
(+ 6) is oxidised to MnO−4 (Mn+ 7 ) and
reduced toMnO2 (Mn).
We get,O2 gas at anode.
+ 2e − → Ca
According to Faraday’s first law,
Charge passed in Faraday = Gram
equivalent of product
Given mass
=
× no. of e − released
Atomic mass
20
=
× 2= 1 F
40
So, one Faraday (F) is required for the
process.
According to Kohlrausch’s law,
λ°CH COOH = λ°H + + λ°CH COO −
The above reaction is a
disproportionation reaction as Cu(+1)
is oxidised to Cu(+2 ) and reduced to
Cu(0).
+6
[NEET (Sep.) 2020]
λ°H + = 350 S cm2 mol −1
3 COO
H2O r H+ + OH−
Reaction at cathode (Pt)
2H+ + 2e − → H2 ↑
Reaction at anode (Pt)
(a) 2
(b) 3
Ans. (d)
λ CH COOH = 20 S cm2 mol −1
3
Concentration of CH3COOH , C = 0.007 M
λ°CH
→ 2H+ + SO24 −
H2SO4
Strong electrolyte
as it is dilute
09 Which of the following reactions
are disproportionation reaction?
(i) 2Cu + → Cu 2+ + Cu 0
(ii) 3MnO 24− + 4H+ → 2MnO 4−
+ MnO 2 + 2H2O
∆
(iii) 2KMnO 4 → K 2MnO 4
+ MnO 2 + O 2
(iv) 2MnO −4 + 3Mn 2+ + 2H2O →
5MnO 2 + 4H⊕
Select the correct option from the
following.
[NEET (National) 2019]
(a) (i), (ii) and (iii)
(b) (i), (iii) and (iv)
(c) (i) and (iv) only
(d) (i) and (ii) only
Ans. (d)
The reaction in which the same species
is simultaneously oxidised and reduced
are called disporportionation reactions.
Let us, consider the given reaction one
by one:
(i) 2Cu+ → Cu2 + + Cu0
+6
+4
(iii) 2KMnO4 →
∆ K2 MnO4 + MnO2 + O2
The above reaction is not a
disproportionation reaction as Mn
(+7) is only reduced toK2MnO4 (Mn+ 6 )
and MnO2 (Mn+ 4 ).
+7
(iv) 2MnO−4 + 3Mn2 + + 2H2O + 4
→ 5MnO2 + 4H+
The above reaction is not a
disproportionation reaction as Mn(+7 ) is
only reduced toMnO2 (+ 4).
Hence, option (d) is correct.
10 Following limiting molar
conductivities are given as
λ °m (H SO ) = x S cm 2mol −1
λ °m (K
2
4
2 SO4 )
= y S cm 2mol −1
λ °m (CH COOK) = z S cm 2mol −1
2
λ °m (in S cm 2mol −1 ) for CH3COOH
will be
[NEET (Odisha) 2019]
(a) x − y + 2 z
(b) x + y − z
(x − y)
(d)
+z
2
(c) x − y + z
Ans. (d)
Key Idea According to Kohlrausch’s law,
λºm for Ax B y = xλº y + + yλº(B) x −
(A )
where, λºm = limiting molar conductivity
of electrolyte and λº y + and λº(B) x − are
(A )
the limiting molar conductivities of
y+
cation (A ) and anion (B x − ),
respectively.
Given
λºm
= 2λ°H + + λº 2 − = x S c2mol −1
SO
(H 2 SO 4 )
4
…(i)
λºm
(K 2 SO 4 )
= 2λºK + + λº
SO 2 −
4
= y S cm2mol −1 …(ii)
λºm (CH3COOK) = λºCH
3 COO
−
+ λºK +
= z S cm2mol −1 …(iii)
To find :
= λºCH COO − + λºH +
3
The above equation can be obtained by
Eq. (i) Eq. (ii)
Eq. (iii) +
−
2
2
λºm(CH COOH)
3
82
NEET Chapterwise Topicwise Chemistry
∴ λºm(CH
3 COOH)
= λºm (CH
3 COOK)
+
λºm(H 2 SO 4 )
−
2
λºm(K 2 SO 4 )
2
x y

2
=  z + −  S cm mol −1

2 2
 x−y

=
+ z  S cm2 mol −1
 2

11 The molar conductivity of a 0.5
mol/dm 3 solution of AgNO 3 with
electrolytic conductivity of
5.76 × 10 −3 S cm −1 at 298 K is
[CBSE AIPMT 2016, Phase II]
(a) 2.88 S cm2 /mol
(b) 11.52 S cm2 /mol
(c) 0.086 S cm2 /mol
(d) 28.8 S cm2 /mol
Ans. (b)
Key Idea The relation between molar
conductivity (λm) and electrolytic
conductivity (κ) is given as
κ × 1000
λm =
M
where, M is molarity of solution.
Given, concentration of solution,
M = 0 . 5 mol / dm3
Electrolytic conductivity,
κ = 5. 76 × 10 −3 S cm–1
Temperature,T = 298 K
∴ Molar conductivity,
κ × 1000
λm =
M
5. 76 × 10 −3 × 1000
=
0. 5
= 11. 52 S cm2 /mol
12 During the electrolysis of molten
sodium chloride, the time required
to produce 0.10 mol of chlorine gas
using a current of 3 amperes is
[CBSE AIPMT 2016, Phase II]
(a) 55 minutes
(c) 220 minutes
Ans. (b)
(b) 110 minutes
(d) 330 minutes
Key Idea This problem is based on
Faraday’s first law of electrolysis which
states that when an electric current is
passed through an electrolytic solution,
the amount of substance (w), deposited at
the electrode, is proportional to the
electric charge (q) passed through the
electrolytic solution.
The formula used in the problem is
Eit
…(i)
w=
96500
where, E = gram-equivalent mass of Cl −
i = current,t = time, it = q
Given,w = 0.10 mol = (0 . 10 × 71) g,
i = 3 A, E = 35.5
The following reactions occured,
At cathode :
+2e–
2H2O → H2 + 2OH−
At anode :
15 When 0.1 mole of MnO 2−
4 is
oxidised, the quantity of electricity
required to completely oxidise
−
MnO 2−
4 to MnO 4 is
[CBSE AIPMT 2014]
−2e–
2Cl − → Cl2
35.5g
MnO24−
71 g
13 The number of electrons delivered
at the cathode during electrolysis
by a current of 1 ampere in 60
seconds is (charge on electron =
1.60 × 10 −19 C)
[CBSE AIPMT 2016, Phase II]
(a) 6 × 10
(c) 3.75 × 1020
Ans. (c)
[1 mole e − = 1F]
Ç
(+6)
Putting all values in expression (i) we get
35.5
(0.10 × 71) =
×3 ×t
96500
or t = 6433 s
or t = 107 .22 min
 1 s = 1 min
~− 110 min
60


23
(b) 2 × 96500 C
(d) 96.50 C
(a) 96500 C
(c) 9650 C
Ans. (c)
(b) 6 × 10
(d) 7.48 × 1023
20
From Faraday’s first law of electrolysis,
w
it
…(i)
=
E 96500
Given,i = 1 A;t = 60 s
Putting these values in Eq. (i), we get
w 1 × 60
=
E 96500
6
w
or
=
E 9650
= Number of mole of electrons
∴ Number of electrons
6
=
× 6.022 × 1023
9650
16 The weight of silver (at. wt. = 108)
displaced by a quantity of
electricity which displaces
5600 mL of O 2 at STP will be
[CBSE AIPMT 2014]
(a) 5.4 g
(c) 54.0 g
Ans. (d)
Since, HCl is strong acid and dissociates
completely. Hence, it conducts
electricity best in its aqueous solution.
(b) 10.8 g
(d) 108.0 g
Since, 22400 mL volume is occupied by 1
mole of O2 at STP.
Thus, 5600 mL O2 means
5600
1
mol O2 = mol O2
=
4
22400
1
∴ Weight of O2 = × 32 = 8 g
4
According to problem,
Equivalents of Ag = Equivalents of O2
Weight of Ag
=
Equivalent weight of Ag
WO2
=
Equivalent weight of O2
WO 2
WAg
=
MAg
MO 2
1
∴
[CBSE AIPMT 2015]
(a) Acetic acid, C2 H4O2
(b) Hydrochloric acid, HCl
(c) Ammonia, NH3
(d) Fructose, C 6 H12O 6
Ans. (b)
(+7)
As per the equation, for 1 mole ofMnO24− , 1
F of electricity is required. Thus, for 0.1
mole of MnO24− , 0.1 F of electricity is
required.
Since, 1 F = 96500 C
0.1 F = 0.1 × 96500 C = 9650 C
∴
Hence, 9650 C of electricity is required
to completely oxidise MnO24− to MnO−4 .
= 3.75 × 1020
14 Aqueous solution of which of the
following compounds is the best
conductor of electric current?
MnO−4
⇒
4
8
×1=
×4
108
32
WAg
[Q2H2O → O1 + 4H+ + 4e − ]
WAg = 108 g
17 At 25° C molar conductance of
0.1 molar aqueous solution of
ammonium hydroxide is
9.54 Ω −1 cm 2 mol −1 and at infinite
dilution its molar conductance is
83
Redox Reactions and Electrochemistry
238 Ω −1 cm 2 mol −1 . The degree of
ionisation of ammonium hydroxide
at the same concentration and
temperature is
[NEET 2013]
(a) 2.080 %
(c) 4.008 %
Ans. (c)
(b) 20.800 %
(d) 40.800 %
[CBSE AIPMT 2009]
Given, molar conductance at 0.1 M
concentration,
λ c = 9.54 Ω −1 cm2 mol −1
Molar conductance at infinite dilution,
λ∞c = 238 Ω −1 cm2 mol −1
We know that, degree of ionisation,
λ
α = ∞c × 100
λc
=
M
32
solution of a weak monobasic acid
is 8.0 mho cm 2 and at infinite
dilution is 400 mho cm 2 . The
dissociation constant of this acid is
20 The equivalent conductance of
9.54
× 100 = 4.008%
238
18 Limiting molar conductivity of
NH4OH
° m (NH OH) ) is equal to
(i.e. Λ
4
(a) 125
. × 10−5
(b) 125
. × 10−6
(c) 6.25 × 10−4
Ans. (a)
(d) 125
. × 10−4
Λc
Λ∞
c
∞
where, Λ and Λ are equivalent
conductances at a given concentration
and at infinite dilution respectively.
8.0
⇒
α=
= 2 × 10 −2
400
From Ostwald’s dilution law (for weak
monobasic acid),
Degree of dissociation,α =
Kc =
Cα2
(1 − α)
(a) Λ°m(NH 4Cl) + Λ°m(NaCl) − Λ°m(NaOH)
(b) Λ °m (NaOH) + Λ°m(NaCl) − Λ °m (NH 4Cl)
= 1.25 × 10 −5
(c) Λ °m(NH 4OH) + Λ °m(NH 4Cl) − Λ °m(HCl)
(d) Λ °m(NH 4Cl) + Λ °m(NaOH) − Λ°m(NaCl)
Ans. (d)
According to Kohlrausch’s law,
limiting molar conductivity ofNH4OH
Λ°m (NH OH) = Λ°m (NH Cl) + Λ°m (NaOH)
4
4
− Λ°m (NaCl)
19 An increase in equivalent
conductance of a strong
electrolyte with dilution is mainly
due to
[CBSE AIPMT 2010]
(a) increase in ionic mobility of ions
(b) 100% ionisation of electrolyte at
normal dilution
(c) increase in both, i.e. number of ions
and ionic mobility of ions
(d) increase in number of ions
Ans. (a)
Key Idea λ eq = κ × V =
k × 1000
.
Normality
On dilution, the number of current
carrying particles per cm3 decreases but
the volume of solution increases.
Consequently, the ionic mobility
increases, which in turn increases the
equivalent conductance of strong
electrolyte.
or
21 Al 2O 3 is reduced by electrolysis at
low potentials and high currents. If
4.0 × 10 4 A of current is passed
through molten Al 2O 3 for 6 h, what
mass of aluminium is produced?
(Assume 100% current efficiency,
atomic mass of Al = 27 g mol −1 )
[CBSE AIPMT 2009]
(a) 9.0 × 103 g
(c) 2.4 × 105 g
(b) 8.1 × 104 g
(d) 1.3 × 104 g
Ans. (b)
Al2O3 ionises as,
Al2O3
º
Al 3+ + AlO3–
3
Cathode
Anode
At cathode
Al 3+ +3 e − → Al
3F
27g
Q Mass of aluminium deposited by 3 F of
electricity = 27 g
∴ Mass of aluminium deposited by
4.0 × 10 4 × 6 × 3600 C of electricity
=
27 × 4.0 × 10 4 × 6 × 3600
g
3F
=
27 × 4.0 × 10 4 × 6 × 3600
g
3 × 96500
= 8.1 × 10 4 g
[CBSE AIPMT 2008]
(a) infinite dilution, each ion makes definite
contribution to equivalent conductance
of an electrolyte, whatever be the
nature of the other ion of the
electrolyte
(b) infinite dilution, each ion makes
definite contribution to equivalent
conductance of an electrolyte
depending on the nature of the other
ion of the electrolyte
(c) infinite dilution, each ion makes
definite contribution to conductance
of an electrolyte, whatever be the
nature of the other ion of the
electrolyte
(d) infinite dilution, each ion makes
definite contribution to equivalent
conductance of an electrolyte,
whatever be the nature of the other
ion of the electrolyte
Ans. (d)
= Cα2
(Q 1>>> α)
1
−2 2
=
(2 × 10 )
32
[CBSE AIPMT 2012]
22 Kohlrausch’s law states that at
Kohlrausch’s law states that “the
equivalent conductance of an electrolyte
at infinite dilution is equal to the sum of
the equivalent conductances of the
component ions.”
λ∞ = λa + λc
where, λ a = equivalent conductance of
the anion
λ c = equivalent conductance of the
cation
Each ion has the same constant ionic
conductance at a fixed temperature, no
matter of which electrolyte it forms a
part.
23 4.5 g of aluminium (atomic mass
27u) is deposited at cathode from
Al 3+ solution by a certain quantity
of electric charge. The volume of
hydrogen produced at STP from H+
ions in solution by the same
quantity of electric charge will be
[CBSE AIPMT 2005]
(a) 44.8 L (b) 22.4 L (c) 11.2 L (d) 5.6 L
Ans. (d)
From second law of Faraday
mAl E Al
=
mH EH
4.5 27 /3
=
1
mH
or
mH = 0.5 g
Q Volume of 2 gH2 at STP = 22.4 L
∴ Volume of 0.5 gH2 at STP
22.4 × 0.5
=
L = 5.6 L
2
84
NEET Chapterwise Topicwise Chemistry
24 In electrolysis of NaCl when Pt
electrode is taken then H2 is
liberated at cathode while with Hg
cathode it forms sodium amalgam
because
[CBSE AIPMT 2002]
(a) Hg is more inert than Pt
(b) more voltage is required to reduce
H+ at Hg than at Pt
(c) Na is dissolved in Hg while it does not
dissolved in Pt
(d) concentration of H+ ions is larger
when Pt electrode is taken
Ans. (b)
Sodium chloride in water dissociates as
NaCl
Na+ + Cl –
H2O
º
ºH
+
25 Cell reaction is spontaneous when
[CBSE AIPMT 2000]
° is negative (b) Ered
° is positive
(a) Ered
(c) ∆G° is negative (d) ∆G° is positive
Ans. (c)
When the value of ∆G ° is negative, the
cell reaction is spontaneous.
∆G ° = − nFE °
where, n = number of electrons take part
F = Faraday constant
E ° = EMF of the cell
Thus, for a spontaneous reaction, the
EMF of the cell must be positive.
26 The equivalent conductances of
Ba 2+ and Cl − are 127 and
76 Ω −1 cm −1 eq −1 respectively at
infinite dilution. The equivalent
conductance of BaCl 2 at infinite
dilution will be [CBSE AIPMT 2000]
(c) 279
27 The specific conductance of a 0.1 N
KCl solution at 23°C is 0.012
Ω −1 cm −1 . The resistance of cell
containing the solution at the same
temperature was found to be 55 Ω.
The cell constant will be
[CBSE AIPMT 1999]
(a) 0.142 cm−1
+ OH–
When electric current is passed through
this solution using platinum electrodes,
Na+ and H+ move towards cathode
whereas Cl – and OH– ions move towards
anode.
At cathode
H+ + e – → H
H + H → H2
At anode
Cl – → Cl + e −
Cl + Cl → Cl2
If mercury is used as cathode,H+ ions
are not discharged at mercury cathode
because mercury has a high hydrogen
over voltage. Na+ ions are discharged at
cathode in preference ofH+ ions yielding
sodium, which dissolves in mercury to
form sodium amalgam.
(a) 139.52 (b) 203
Ans. (a)
The equivalent conductance ofBaCl2 at
infinite dilution,
1
λ ∞ of BaCl2 = λ ∞ of Ba2 + + λ ∞ of Cl –
2
127
=
+ 76 = 139.5
2
(d) 101.5
(b) 0.66 cm−1
(c) 0.918 cm−1
(d) 1.12 cm−1
Ans. (b)
Specific conductivity,
k = 0.012 Ω −1 cm−1 ;
Resistance = 55 Ω
1
l 
1
× G=
k=
resistance a 
R 
l
= cell constant
a
l
= 55 × 0.012 = 0.66 cm−1
a
29 A 5A current is passed through a
solution of zinc sulphate for 40
min. The amount of zinc deposited
at the cathode is
[CBSE AIPMT 1996]
(a) 40.65 g
(c) 4.065 g
Ans. (c)
(b) 0.4065 g
(d) 65.04 g
Current, I = 5 A
time,t = 40 min = 40 × 60 = 2400 s
Amount of electricity passed
Q = It
Q = 5 × 2400
Q = 12000 C.
Zn2 + + 2e – → Zn
n = 2e −
From Faraday first law
W = ZI t
Z = equivalent mass
[65.39 = mass of zinc]
Mass
65.39
g of zinc
= −
=
e × F 2 × 96500
therefore, 12000 C charge will deposite
65.39 × 12000
=
2 × 96500
= 4.065 g of zinc
TOPIC 3
Electronium Series,
28 Equivalent conductance of NaCl,
Electrode Potential and Emf
HCl and C 2H5COONa at infinete
dilution are 126.45, 426.16 and
30 Identify the reaction from following
91 Ω −1 cm 2 , respectively. The
having top position in EMF series
equivalent conductance of
(Standard reduction potential)
[CBSE AIPMT 1997]
C 2H5COOH is
according to their electrode
(a) 201.28 Ω −1 cm2
potential at 298 K.
(b) 390.71 Ω −1 cm2
−1
[NEET (Oct.) 2020]
(a) Mg2+ + 2 e − → Mg(s )
(b) Fe2+ + 2 e − → Fe(s )
(c) Au 3+ + 2 e − → Au(s )
(d) K + +1 e − → K(s )
Ans. (c)
(c) 698.28 Ω cm
(d) 540.48 Ω −1 cm2
Ans. (b)
2
By Kohlrausch’s law
λ ∞ for NaCl =λ Na + + λ Cl−
…(i)
λ ∞ for HCl =λH + + λ Cl−
…(ii)
λ ∞ for C2H5 COONa =λ Na + + λ C
2 H 5 COO
–
…(iii)
So, λ ∞ for C2H5 COOH can be obtained by
adding Eqs. (ii) and (iii) and then
subtracting Eq. (i)
= λ ∞ of C2H5 COONa + λ ∞ of HCl
− λ ∞ for NaCl
= (91 + 426.16 − 126.45) Ω −1 cm2
= 390.71 Ω –1 cm2
Let us consider the segment of EMF
series (standard reduction potential) of
the given reduction half-call reactions:
Reduction half reaction
E/V at 298K
(a) Mg 2 + + 2e → Mg
2+
– 2.36
+ 2e → Fe
– 0.44
(c) Au3 + + 3 e → Au
+ 140
.
(d) K + + e → K
– 2.93
(b) Fe
Hence, option (c) is correct.
85
Redox Reactions and Electrochemistry
31 The standard electrode potential
(E − ) values of
(Al 3+ / Al, Ag + / Ag, K + /K and
Cr 3+ /Cr are − 1.66 V, 0.80V, 2.93 V
and − 0.74 V, respectively. The
correct decreasing order of
reducing power of the metal is
[NEET (Odisha) 2019]
(a) Ag > Cr > Al > K
(b) K > Al > Cr > Ag
(c) K > Al > Ag > Cr
(d) Al > K > Ag > Cr
Ans. (b)
More negative the value of standard
reduction potential, higher is the
reduction power.
i.e. Reducing power
1
∝
standard reduction potential
Thus, the correct decreasing order of
reducing power of the metal is
K
>
Al
>
(EKº + /K = −2.93V)
(E º 3 + = −166
. V)
Al / Al
Cr
>Ag
º
(E º 3 +
= −0.74 V) (E Ag
+ / Ag = 0.80 V)
Cr
/ Cr
32 For the cell reaction,
2Fe 3+ (aq) + 2I − (aq) → 2Fe 2+ (aq)
+I 2 (aq)
È
E cell = 0. 24 V at 298 K. The
standard Gibbs energy (∆ r G È ) of
the cell reaction is
[Given that Faraday constant
F = 96500 C mol −1 ]
[NEET (National) 2019]
(a) −23.16 kJ mol −1 (b) 46.32 kJ mol −1
(c) 23.16 kJ mol −1 (d) −46.32 kJ mol −1
Ans. (d)
º
The relation between E cell
and ∆ r G º is as
follows:
º
∆ r G º = − nFE cell
For the cell reaction,
2Fe3 + (aq) + 2 I− (aq) → 2Fe2 + (aq) + I2 (aq)
n=2
−1
Given, F = 96500 C mol
= 0.24 V
º
Now, we know that ∆ r G º = − nFE cell
On substituting the given values in above
equation we get
∆ r G º = − 2 mol × 96500 C mol − 1
× 0.24 J mol − 1
−1
= − 46320 J mol
= − 4632
. KJ mol − 1
Hence, option (d) is correct.
º
, E cell
33 In the electrochemical cell
Zn||ZnSO 4 (0.01 M)|| CuSO 4 (1.0M)
Cu, the emf of this Daniel cell is
E 1 . When the concentration
ZnSO 4 is changed to 1.0 M and
that of CuSO 4 changed to 0.01 M,
the emf changes to E 2 . From the
followings, which one is the
relationship between E 1 and E 2 ?
RT
( Given,
= 0.059)
F
[NEET 2017, 2003]
(a) E1 = E2
(c) E1 > E2
Ans. (c)
(b) E1 < E2
(d) E2 = 0 ≠ E1
Thinking process Calculate the value of
E cell i.e. E 1 and E2 by substituting the
respective given values in the Nernst
equation,
0.059
[Zn2 + ]
log
E cell = E ° −
n
[Cu2 + ]
Compare the calculated values of E 1 and
E2 and find the correct relation.
For the electrochemical cells,
Zn|ZnSO4 (0.01M)| | CuSO4 (1M)| Cu
Cell reaction :
Zn + Cu2 + → Zn2 + + Cu; n = 2
Zn2 +
0.059
log 2 +
E1 = E ° −
2
Cu
0.059
0.01
= E° −
log
2
1
0.059
1
E1 = E ° −
log
= (E ° + 0.059)
2
100
For cell, Zn|ZnSO4 (1M)| | CuSO4 (0.01M)| Cu
0.059
1
E2 = E ° −
log
2
0.01
0.059
E2 = E ° −
log 100
2
∴
= (E °− 0.059) ⇒ E 1 > E2
34 The pressure of H2 required to
make the potential of H2 -electrode
zero in pure water at 298 K is
[CBSE AIPMT 2016, Phase I]
(a) 10−12 atm
(c) 10−4 atm
Ans. (d)
(b) 10−10 atm
(d) 10−14 atm
From the question, we have an equation
2H+ + 2e − → H2 (g)
According to Nernst equation,
pH
0.0591
E = E° −
log + 2 2
2
[H ]
pH 2
0.0591
=0 −
log
2
(10 −7 )2
[Q [H+ ] = 10 −7 ]
∴ For potential of H2 electrode to be
zero, pH 2 should be equal to [H+ ]2 , i.e.
10 −14 atm.
10 −14
=0
log
∴
(10 −7 )2
°
35 If the E cell
for a given reaction has
a negative value, which of the
following gives correct
relationships for the values of ∆G°
and K eq ?
[CBSE AIPMT 2016, Phase II, 2011]
(a) ∆G ° > 0; K eq < 1 (b) ∆G ° > 0; K eq > 1
(c) ∆G ° < 0; K eq > 1 (d) ∆G ° < 0; K eq < 1
Ans. (a)
Given,
° = −ve
E cell
° is
The relation between ∆G ° and E cell
given as
... (i)
∆G ° = − nF E °
cell
° is negative , so ∆G ° comes out to
If E cell
be positive. Again, relation between ∆G °
and K eq is given as
... (ii)
∆G ° = −2 . 303 nRT log K eq
From Eq. (i) we get that ∆G ° is positive.
Now, if ∆G ° is positive thenK eq comes
out to be negative from eq (ii).
i.e. ∆G ° >1 and K eq < 1
Short trick As E ° is negative so
cell
reaction is non-spontaneous or you can
say reaction is moving in backward
direction. For non-spontaneous
reaction, ∆G ° is always positive andK eq
is always less than 1.
36 A hydrogen gas electrode is made
by dipping platinum wire in a
solution of HCl of pH=10 and by
passing hydrogen gas around the
platinum wire at 1 atm pressure.
The oxidation potential of
electrode would be
[NEET 2013]
(a) 0.059 V
(c) 0.118 V
Ans. (b)
(b) 0.59 V
(d) 1.18 V
For hydrogen electrode, oxidation half
reaction is
H2 → 2H+ + 2e−
(1 atm)
(At pH 10)
pH = 10
H+ = 1 × 10 − pH = 1 × 10 −10
From Nernst equation,
0.0591
[H+ ]2
E cell = E ° cell −
log
2
pH 2
If
For hydrogen electrode, E ° cell = 0
86
NEET Chapterwise Topicwise Chemistry
0.0591
( 10 −10 )2
log
2
1
0.0591 × 2
1
=+
log −10
2
10
E cell = −
= 0 .0591 log 10 10
= 0 .0591 × 10
= 0.591 V
37 Standard electrode potential of
three metals X, Y and Z are –1.2V, +
0.5V and –3.0V respectively. The
reducing power of these metals will
be
[CBSE AIPMT 2011]
(a) Y > X > Z
(c) X > Y > Z
Ans. (b)
(b) Z > X > Y
(d) Y > Z > X
. V,
E X° = − 12
E Y° = + 0.5 V,
E Z° = − 3.0 V,
∴
Z> X>Y
[Q higher the reduction potential,
lesser the reducing power.]
38 The electrode potentials for
Cu 2+ (aq ) + e – → Cu + (aq )
and Cu + (aq ) + e – → Cu( s )
are +0.15 V and +0.50 V
respectively. The value of
will be
E ° 2+
Cu
/Cu
[CBSE AIPMT 2011]
(a) 0.325 V
(c) 0.150 V
Ans. (a)
2+
Cu
(b) 0.650 V
(d) 0.500 V
+ e → Cu ,
–
+
E 1° = 0.15 V, ∆G 1° = − n1E 1°F
Cu+ + e – → Cu, E2° = 0.50 V,
∆G2° = − n2 E2°F
Cu2 + + 2e – → Cu, E ° = ?,
∆G ° = − nE °F
∆G ° = ∆G 1° + ∆G2°
− nE ° F = − n1E 1°F − n2 E2°F
or − 2 E ° F = − 1 F × 0.15 + (− 1 F × 0.50)
or − 2 E ° F = − 0.15 F − 0.50 F
or − 2 FE ° = − F(0.15 + 0.50)
0.65
E° =
= 0.325 V
∴
2
39 Standard electrode potential for
Sn 4+ / Sn 2+ couple is +0.15 V and
that for the Cr 3+ /Cr couple is –0.74.
These two couples in their
standard state are connected to
make a cell. The cell potential will
be
[CBSE AIPMT 2011]
(a) + 0.89 V
(c) + 1.83 V
Ans. (a)
E°
Sn 4+ / Sn 2 +
E ° 3+
Cr
/ Cr
(b) + 0.18 V
(d) + 1.199 V
= + 0.15 V
= −0.74 V
° = E cathode
°
°
E cell
(RP) − E anode(RP)
= 0.15 − ( − 0.74)
= + 0.89 V
40 For the reduction of silver ions with
copper metal, the standard cell
potential was found to be + 0.46 V
at 25°C. The value of standard
Gibbs energy, ∆G° will be
(F = 96500C mol −1 )
[CBSE AIPMT 2010]
(a) − 89.0 kJ
(c) − 44.5 kJ
Ans. (a)
(b) − 89.0 J
(d) − 98.0 kJ
We know that, standard Gibbs energy,
°
∆G ° = − nFE cell
For the cell reaction,
2 Ag + + Cu → Cu2 + + 2 Ag
° = + 0.46 V
∆E cell
°
∆G ° = − nF E cell
n=2
∆G ° = − 2 × 96500 × 0.46
= − 88780 J
= − 88.7 kJ ≈ − 89.0 kJ
41 Given,
(i) Cu 2+ + 2 e − → Cu,
E° = 0.337 V
(ii) Cu 2+ + e − → Cu + ,
E° = 0.153 V
Electrode potential, E° for the
reaction,
Cu + + e − → Cu, will be
[CBSE AIPMT 2009]
(a) 0.52 V
(c) 0.30 V
(b) 0.90 V
(d) 0.38 V
Ans. (a)
∆G ° = − nFE °
For reaction,
Cu2 + + 2e − → Cu,
K(i)
∆G ° = − 2 × F × 0.337
For reaction, Cu+ → Cu2 + + e − , K(ii)
∆G ° = − 1 × F × (−0.153)
= + 0.153 F
Adding Eqs. (i) and (ii), we get
Cu+ + e − → Cu, ∆G ° = −0.521 F
∆G ° = − nFE °
∴
− 0.521 F = − nFE °
∴
E ° = 0.52 V
42 Standard free energies of
formation (in kJ/mol) at 298 K are
–237.2, –394.4 and –8.2 for H2O(l),
CO 2 (g) and pentane (g),
° for
respectively. The value of E cell
the pentane-oxygen fuel cell is
[CBSE AIPMT 2008]
(a) 1.968 V
(c) 1.0968 V
Ans. (c)
(b) 2.0968 V
(d) 0.0968 V
∆G of H2O(l ) = − 237.2kJ / mol
∆G of CO2 ( g) = − 394.4kJ / mol
∆G of pentane ( g) = − 8.2kJ / mol
In pentane-oxygen fuel cell following
reaction takes place
C5H12 + 10H2O(l ) → 5CO2 + 32 H+ + 32 e −
8O2 + 32H+ + 32 e − → 16 H2O(l )
C5H12 + 8O2 → 5CO2 + 6H2O(l ), E ° = ?
∆G reaction = Σ∆G product − Σ ∆G reactant
= 5 × ∆G (CO 2 ) + 6∆G (H 2 O) − [∆G (C H 12 )
5
+8 × ∆GO 2 ]
= 5 × (−394.4) + 6 × (−237.2) − (−8.2 + 0)
= − 1972 − 1423.2 + 8.2
= − 3387 kJ / mol
= −3387 × 10 3 J / mol
°
∆G = − nFE cell
°
−3387 × 10 3 = − 32 × 96500 × E cell
° =
E cell
−3387 × 10 3
= 1.0968 V
–32 × 96500
43 On the basis of the following E°
values, the strongest oxidising
agent is
[CBSE AIPMT 2008]
[Fe(CN) 6] 4– → [Fe(CN) 6] 3– + e − ;
E° = − 0.35 V
87
Redox Reactions and Electrochemistry
Fe 2+ → Fe 3+ + e − ;
E° = − 0.77 V
(a) [Fe(CN) 6] 4–
(c) Fe 3+
Ans. (c)
(b) Fe2+
(d) [Fe(CN) 6] 3−
Substance which have higher reduction
potential are stronger oxidising agent.
[Fe(CN) 6 ] 4– → [Fe(CN) 6 ] 3– + e − ,
E ° = − 0.35 V
Fe2 + → Fe3+ + e − , E ° = − 0.77 V
Q
° = − E red
°
E oxi
∴ [Fe(CN) 6 ] 3− + e − → [Fe(CN) 6 ] 4 − ,
E ° = 0.35 V
Fe3+ + e − → Fe2 + , E ° = 0.77 V
Hence,Fe3+ has maximum tendency to
reduced, so it is the strongest oxidising
agent.
44 The equilibrium constant of the
reaction,
Cu(s) + 2 Ag + (aq) → Cu 2+ (aq)
+ 2Ag( s ), E° = 0.46 V at 298 K
is
[CBSE AIPMT 2007]
(a) 2.0 × 10
(c) 4.0 × 1015
(b) 4.0 × 10
(d) 2.4 × 1010
10
10
(aq) → Cu (aq)
+ 2 Ag(s )
E ° = 0.46 V at 298 K
RT ln K = nFE °
nFE °
lnK =
RT
2 × 0.46
lnK =
0.0591
K = 3.68 × 10 15 ~− 4 × 10 15
E ° 3+
Fe
= − 0.441 V and
/Fe2+
= 0.771V
the standard emf of the reaction
Fe + 2Fe 3 + → 3Fe 2 + will be
[CBSE AIPMT 2006]
(a) 0.111 V
(c) 1.653 V
Ans. (d)
(b) 0.330 V
(d) 1.212 V
Given that E °
Fe 2 + /Fe
So,
and
So,
° = E cathode
°
°
So, E cell
− E anode
= (+ 0.771) − (− 0.441) = + 1.212 V.
46 A hypothetical electrochemical cell
is shown below
A | A + (xM ) || B + (yM) | B
The EMF measured is + 0.20V.
The cell reaction is
[CBSE AIPMT 2006]
(a) A + B + → A+ + B
(b) A+ + B → A + B +
(d) the cell reaction cannot be predicted
2+
Cu(s ) + 2 Ag
/Fe
Alternative On the basis of cell reaction
following half-cell reactions are written
At anode
(oxidation)
Fe → Fe2 + + 2e −
At cathode
2Fe3+ + 2e − → 2Fe2 + (reduction)
Ans. (a)
2+
Fe
° = 1.212 V
Fe + 2Fe3+ → 3Fe2 + , E cell
(c) A+ + e – → A, B + + e – → B
Ans. (c)
45 E ° 2+
Cell reaction
(i)
Fe → Fe2 + + 2e − ,
E ° = 0.441 V
(ii) 2Fe3+ + 2 e − → 2Fe2 + , E ° = + 0.771 V
= − 0.441 V
E°
Fe 3+ /Fe 2 +
…(i)
= 0.771 V
2Fe3+ + 2 e − → 2 Fe2 + ,
E ° = 0.771 V
A | A + (xM) | | B + ( yM) | B
The EMF of cell is + 0.20 V. So, cell
reaction is possible. The half-cell
reactions are given as follows
(i) At negative pole
(oxidation)
A → A + + e –
(ii) At positive pole
B + + e – → B
(reduction)
Hence, cell reaction is
° = + 0.20 V
A + B + → A + + B,
E cell
47 The standard EMF of a galvanic cell
involving cell reaction with n = 2 is
found to be 0.295 V at 25° C. The
equilibrium constant of the
reaction would be
(Given F = 96500 C mol–1 ,
R = 8.314 JK –1 mol–1 )
[CBSE AIPMT 2004]
−
Fe → Fe + 2e ,
E ° = + 0.441 V
2+
Electrochemical cell,
…(ii)
(a) 2.0 × 1011
(c) 10
. × 102
Ans. (d)
(b) 4.0 × 1012
(d) 10
. × 1010
By Nernst equation,
2 .303 RT
E cell = E ° cell −
log 10 K
nF
At equilibrium, E cell = 0
Given that,
R = 8.314 JK– 1 mol – 1
T = 25° C + 273 = 298 K
F = 96500 C and n = 2
. × 298
2 .303 × 8314
log 10 K
∴ E ° cell =
2 × 96500
0.0591
=
log 10 K
2
Given that E ° cell = 0.295 V
0.0591
∴
0.295 =
log 10 K
2
0.295 × 2
= 10
log 10 K =
0.0591
antilog log 10 K = antilog 10
K = 1 × 10 10
48 On the basis of the information
available from the reaction.
4
2
Al + O 2 → Al 2O 3 ,
3
3
∆G = −827kJ mol–1 of O 2 , the
minimum EMF required to carry
out the electrolysis of Al 2O 3 is
(F = 96500 C mol −1 ) [CBSE AIPMT
2003]
(a) 2.14 V
(c) 6.42 V
Ans. (a)
(b) 4.28 V
(d) 8.56 V
2
4
Al + O2 → Al2O3 ,
3
3
∆G =–827 kJ mol –1
−
2−
12e + 6O → 3 O2
4 Al 3+ → Al + 12e −
4 3+
4
or
Al → Al + 4e −
3
3
4e − + 2O−2 → O2
(n = 4)
∆G = − nEF °
−827 × 10 3 J = − 4 × E ° × 96500
827 × 10 3
E=
4 × 96500
E° = 2.14 V
49 The most convenient method to
protect the bottom of the ship
made of iron is [CBSE AIPMT 2001]
(a) coating it with red lead oxide
(b) white tin plating
(c) connecting it with Mg block
(d) connecting it with Pb block
Ans. (b)
The most convenient method to protect
the bottom of ship made of iron is white
tin plating which prevents the build up of
barnacles.
88
NEET Chapterwise Topicwise Chemistry
50 Standard electrode potentials are
Fe 2+ / Fe, E° = − 0.44 V
Fe 3+ / Fe 2+ , E° = 0.77 V
Fe 2+ , Fe 3+ and Fe block are kept
together, then [CBSE AIPMT 2001]
(a) Fe 3+ increases
(b) Fe2+ decreases
Fe2 +
(c) 3+ remains unchanged
Fe
(d) Fe2+ decreases
Ans. (b)
The metals have higher negative values
of their electrode potential can displace
metals having lower values from their
salt solution.
So, Fe3+ decreases.
51 Cu + (aq ) is unstable in solution and
undergoes simultaneous oxidation
and reduction according to the
reaction
[CBSE AIPMT 2000]
2Cu + (aq )
Cu 2+ (aq ) + Cu( s )
Choose the correct E° for above
reaction if
E ° 2+
= 0.34 V
º
Cu
and
E°
/Cu
Cu2+ /Cu+
(a) – 0.38 V
0.38 V
Ans. (c)
= 0.15 V
(b) + 0.49 V (c) +
(d) – 0.19 V
∆G ° = − nFE °
From given data,
(i) Cu(s ) → Cu2 + (aq) + 2e − ,
∆G 1° = − 2 (− 0.34) × F
(ii) Cu2 + (aq) + e − → Cu + (aq),
∆G2° = − 1 (0.15) × F
On addition,
Cu(s ) → Cu+ (aq) + e − ,
∆G 3° = − 1 × E ° × F
and ∆G 3° = ∆G 1° + ∆G2°
− n3FE ° = − n 1FE 1° − n2FE2
− E ° = − 2(−0.34) − 1(0.15)
= (−2 × − 0.34) + ( − 1 × 0.15)
− E ° = + 0.68 − 0.15 = 0.53
or
E ° = − 0.53 V
Cu+ (aq)
Cu2 + (aq) + e − ;
∆G 1 = − 1 × (−0.15) × F
Cu+ (aq) + e −
Cu(s ) ;
∆G2 = − 1 × (−0.53) × F
º
º
°
E cell
= 0.00 V
− RT C2
ln
E=
nF C 1
RT C2
ln
E=
nF C 1
On adding above equation we get,
2Cu+
Cu2+ + Cu ; ∆G
∆G = ∆G 1 + ∆G2
− nFE ° = 0.15F + (−0.53F )
−FE ° = = 0.38F
E ° = 0.38V
Thus, for the result reaction E ° value is
0.38 V.
º
52 Which one of the following pairs of
substances on reaction will not
evolve H2 gas? [CBSE AIPMT 1998]
C2 > C 1
RT C2
ln
E=
nF C 1
(R, T, n and F are constant)
C
therefore, E ° is based upon ln 2
C1
∆G = − nFE °
(a) Iron and H2 SO 4 (aq)
(b) Iron and steam
(c) Copper and HCl (aq)
(d) Sodium and ethyl alcohol
Ans. (c)
= − nF ×
= − RT ln
Since copper is placed below hydrogen
in the electrochemical series, thus
copper does not give hydrogen with
dilute acids. While all other pairs give
hydrogen on reaction.
Fe + dil. H2SO4 → FeSO4 + H2 ↑
3Fe + 4H2O → Fe3O4 + 4H2 ↑
Steam
2Na + C2H5OH → 2C2H5ONa + H2 ↑
Cu + dil. HCl → No reaction
53 Without losing its concentration
ZnCl 2 solution cannot be kept in
contact with
[CBSE AIPMT 1998]
(a) Au
(b) Al
Ans. (b)
(c) Pb
(d) Ag
In electrochemical series, Al is placed
above Zn and all other are present below
Zn. So, aluminium displaces zinc from
ZnCl2 solution. Hence, it cannot keep in
contact with Al.
54 For the cell reaction,
Cu 2+ (C 1 , aq ) + Zn( s )
Zn 2+ (C 2 , aq ) + Cu(s)
of an electrochemical cell, the
change in free energy (∆G) at a
given temperature is a function of
º
[CBSE AIPMT 1998]
(a) ln (C 1)
(c) ln (C2 )
Ans. (b)
(b) ln (C2 /C 1)
(d) ln (C 1 + C2 )
∆G = − nFE °
For concentration cell, from Nernst
equation,
RT C 1
°
E = E cell
−
ln
nF C2
RT C2
ln
nF C 1
C2
C1
Hence, at constant temperature Gibbs
C
free energy ∆G depends uponln 2 .
C1
55 E° for the cell,
Zn|Zn 2+ (aq )||Cu 2+ (aq )|Cu is 1.10 V
at 25°C. The equilibrium constant
for the reaction,
Zn( s ) + Cu 2+ (aq )
Cu( s )
+ Zn 2+ (aq ) is of the order
º
[CBSE AIPMT 1997]
(a) 10−37 (b) 10−28 (c) 1018
Ans. (a)
Zn(s ) + Cu2 + (aq)
(d) 1017
2+
º Cu(s) + Zn
(aq),
E ° = + 1.10 V
0.0591
log 10 K eq
n
because at equilibrium, E cell = 0
(n = number of electrons exchanged = 2)
0.0591
1.10 =
log 10 K eq
2
2.20
= log 10 K eq
0.0591
∴
E° =
K eq = antilog 37.225 = 1.66 × 10 −37
56 Reduction potential for the
following half-cell reactions are
Zn → Zn 2+ + 2 e − ,
= − 0.76 V)
(E ° 2+
(Zn
Fe → Fe
/Zn)
+ 2 e− ,
(E ° 2+ = − 0.44 V)
2+
Fe
/Fe
The EMF for the cell reaction,
Fe 2+ + Zn → Zn 2+ + Fe will be
[CBSE AIPMT 1996]
89
Redox Reactions and Electrochemistry
TOPIC 4
(a) + 0.32 V
(b) – 0.32 V
(c) +1.20 V
(d) – 1.20 V
Ans. (a)
E°
= − 0.76 V
E°
= − 0.44 V
Zn 2 + / Zn
Fe 2 + /Fe
Arrhenius Theory, Activation
Energy and Collision Theory
of Biomolecular Gaseous
Reaction
59 In a typical fuel cell, the reactants
(R) and product (P) are
Cell reaction,
[NEET (Oct.) 2020]
+ Zn → Zn
2+
2+
Fe
+ Fe
° = E °(cathode) − E °(anode)
E cell
= − 0.44 – (–0.76)
= + 0.32 V
(a) R = H2 (g), O2 (g); P = H2O2 (l)
(b) R = H2 (g), O2 (g); P = H2O(l)
(c) R = H2 (g), O2 (g), Cl2 (g) ; P = HClO 4 (aq)
(d) R = H2 (g), N2 (g);P = NH3 (aq)
Ans. (b)
The reaction takes place in a fuel cell is
57 An electrochemical cell is shown
below Pt, H2 (1atm) | HCl (0.1 M) |
CH3COOH (0.1 M) | H2 (1 atm), Pt
The EMF of the cell will not be
zero, because [CBSE AIPMT 1995]
(a) EMF depends on molarities of acids
used
(b) pH of 0.1 M HCl and 0.1 M CH3COOH is
not same
(c) the temperature is constant
(d) acids used in two compartments are
different
Ans. (b)
The EMF of the cell will not be zero
because concentration ofH+ ions in two
electrolytic solutions is different. Mean
HCl is strong acid where, acetic acid is
weak acid and gives different pH.
58 Standard reduction potentials at
25°C of Li + / Li, Ba 2+ / Ba, Na + / Na
and Mg 2+ / Mg are –3.05, –2.90,
–2.71 and –2.37 V respectively.
Which one of the following is the
strongest oxidising agent?
[CBSE AIPMT 1994]
(a) Mg2+
(b) Ba2+
(c) Na+
(d) Li+
Ans. (a)
A cation having maximum (positive) value
of standard reduction potential is the
strongest oxidising agent. Hence,Mg2 +
is the strongest oxidising agent.
0
+1 –2
0
2H2 (g) + O2 (g)
2H2O(l) +
[at anode] [at cathode]
Reactants R
Product (P)
Heat energy [gets converted
into electrical energy]
Hence, option (b) is the correct.
60 Zinc can be coated on iron to
produce galvanised iron but the
reverse is not possible. It is because
[CBSE AIPMT 2016, Phase II]
(a) zinc is lighter than iron
(b) zinc has lower melting point than iron
(c) zinc has lower negative electrode
potential than iron
(d) zinc has higher negative electrode
potential than iron
Ans. (d)
The metal with higher negative standard
reduction potential, have higher
tendency to get reduced.
Zn2 + + 2e – → Zn; E ° = −0 . 76 V
Fe2 + + 2e – → Fe; E ° = −0 .44 V
Here, in galvanised iron, Zn has higher
negative reduction potential means Zn
takes electrons given by iron and itself
gets reduced.
Thus, Zn works as anode and protects
iron from rusting by making iron as
cathode.
61 A device that converts energy of
combustion of fuels like hydrogen
and methane, directly into
electrical energy is known as
[CBSE AIPMT 2015]
(a) fuel cell
(b) electrolytic cell
(c) dynamo
(d) Ni-Cd cell
Ans. (a)
Fuel cell is a device that converts energy
of combustion of fuels like hydrogen and
methane, directly into electrical energy.
Electrolytic cell converts electrical
energy into chemical energy.
Dynamo is an electrical generator that
produces direct current with the use of a
commutator.
Ni-Cd cell is a type of rechargeable
battery which consists of a cadmium
anode and a metal grid containingNiO2
acting as a cathode.
62 A button cell used in watches
functions as following [NEET 2013]
Zn(s) + Ag 2O (s) + H2O (l)
2Ag (s) + Zn 2+ (aq) + 2OH− (aq)
If half-cell potentials are
º
Zn 2+ (aq) + 2e − → Zn (s),
E° = − 0.76 V
Ag 2O (s) + H2O (l) + 2e −
→ 2Ag (s) + 2OH− (aq),E° = 0.34 V
The cell potential will be
(a) 1.10 V
(c) 0.84 V
Ans. (a)
(b) 0.42 V
(d) 1.34 V
Anode is always the site of oxidation
thus, anode half-cell is
Zn2 + (aq) + 2e − → Zn(s ), E ° = − 0.76 V
Cathode half-cell is
Ag2O( s ) + H2O(l ) + 2e − →
2Ag( s ) + 2OH− (aq ) ,
E ° = 0.34 V
E ° cell = E ° cathode − E ° anode
= 0.34 − (− 0.76) = + 1.10 V
63 The efficiency of a fuel cell is given
by
[CBSE AIPMT 2007]
(a) ∆G / ∆ S
(c) ∆ S / ∆G
Ans. (b)
(b) ∆G / ∆ H
(d) ∆ H / ∆G
∆G
× 100
∆H
Fuel cells are expected to have an
efficiency of 100%.
Efficiency of a fuel cell (φ) =
11
Chemical Kinetics
TOPIC 1
Rate of Chemical Reaction,
Rate Expression
01 Mechanism of a hypothetical
reaction X 2 + Y 2 → 2XY is given
below
[NEET 2017]
(i) X 2 q X + X(fast)
(ii) X + Y 2 → XY + Y (slow)
(iii) X + Y → XY (fast)
The overall order of the reaction
will be
(a) 1
(c) 0
Ans. (d)
(b) 2
(d) 1.5
We know that, slowest step is the rate
determining step.
Rater (r) = K 1 [X] [Y2 ]
… (i)
∴
Now, from equation. (i), i.e.
X2 → 2X [fast]
[X]2
K eq =
[X2 ]
… (ii)
[X] = {K eq [X2 ]} 1/ 2
Now, substitute the value of [X] from
equation. (ii) in equation. (i), we get
Rate (r) = K 1 (K eq ) 1/ 2 [X2 ] 1/2 [Y2 ]
= K [X2 ] 1/ 2 [Y2 ]
3
1
∴ Order of reaction = + 1 = = 1.5
2
2
02 In a reaction, A + B → Product, rate
is doubled when the concentration
of B is doubled and rate increases
by a factor of 8 when the
concentrations of both the
reactants (A and B) are doubled.
Rate law for the reaction can be
written as
[CBSE AIPMT 2012]
(a) rate = k [A][B] 2
(c) rate = k [A][B]
(c) 6.25 × 10 −3molL−1s−1
and 3.125 × 10 −3mol L−1s−1
Ans. (d)
(b) rate = k [A] 2 [B] 2
(d) rate = k [A] 2 [B]
Let the order of reaction with respect to
A and B is x and y respectively. So, the
rate law can be given as
…(i)
R = k [A] x [B] y
When the concentration of only B is
doubled,
the rate is doubled, so
…(ii)
R 1 = k [A] x [2B] y = 2R
If concentrations of both the reactants A
and B are doubled, the rate increases by
a factor of 8, so
…(iii)
R ′ ′ = k [2A] x [2B] y = 8R
…(iv)
⇒
k2x 2y [A] x [B] y = 8R
From Eqs. (i) and (ii), we get
2R [A] x [2B] y
=
⇒
R
[A] x [B] y
(d) 1.25 × 10 −2 mol L−1s−1
and 6.25 × 10 −3mol L−1s−1
Ans. (b)
Key Idea Rate of disappearance of
reactant = rate of appearance of product
or
d [reactant]
1
−
Stoichiometric coefficient
dt
of reactant
1
=+
Stoichiometric coefficient
of product
For the reaction,
N2O5 (g) → 2NO2 (g) +
2 = 2y ⇒ y = 1
rom Eqs. (i) and (iv), we get
8R 2x 2y [A] x [B] y
or 8 = 2x 2y
=
⇒
R
[A] x [B] y
− d [N2O5 ]
∴
Substitution of the value of y gives,
8 = 2x 21
4 = 2x
(2)2 = (2) x
∴
x =2
Substitution of the value of x and y in Eq.
(i) gives,
R = k [A]2 [B]
dt
d [NO2 ]
dt
1 d [NO2 ]
=+
2 dt
d [N2O5 ]
= −2
dt
=+
1
O2 (g)
2
2 d [O2 ]
dt
= 2 × 6.25 × 10 −3 mol L−1 s−1
= 12.5 × 10 −3 mol L−1 s−1
= 1.25 × 10 −2 mol L−1 s−1
d [N2O5 ] 1
d [O2 ]
=−
×
dt
dt
2
6.25 × 10 −3 mol L−1s−1
=
2
= 3.125 × 10 −3 mol L−1s−1
03 For the reaction,
1
N 2O 5 (g) → 2NO 2 (g) + O 2 (g)
2
The value of rate of disappearance
of N 2O 5 is given as 6.25 × 10 −3 mol
L−1 s −1 . The rate of formation of NO 2
and O 2 is given respectively as
04 During the kinetic study of the
reaction, 2A + B → C + D,
following results were obtained
Run
[CBSE AIPMT 2010]
(a) 6.25 × 10 −3mol L−1s−1
and 6.25 × 10 −3mol L−1s−1
−2
d [product]
dt
−1 −1
(b) 1.25 × 10 mol L s
and 3.125 ×10 −3mol L−1s−1
Initial rate of
A/mol L−1 B/mol L−1 formation of
D/mol L−1 min−1
I
0.1
0.1
6.0 × 10 −3
II
0.3
0.2
7.2 × 10 −2
III
IV
0.3
0.4
0.4
0.1
2.88 × 10 −1
2.40 × 10 −2
91
Chemical Kinetics
Based on the above data which one
of the following is correct?
[CBSE AIPMT 2010]
(a) Rate = k [A] 2 [B]
(b) Rate = k [A] [B]
(c) Rate = k [A] 2 [B] 2
(d) Rate = k [A] [B] 2
Ans. (d)
Let the order of reaction with respect to
A is x and with respect to B is y.Thus,
rate = k [A] x [B] y
(x and y are stoichiometric coefficient )
For the given cases,
I. rate = k (0.1) x (0.1) y = 6.0 × 10 −3
II. rate = k (0.3) x (0.2) y = 7.2 × 10 −2
III. rate = k (0.3) x (0.40) y = 2.88 × 10 −1
IV. rate = k (0.4) x (0.1) y = 2.40 × 10 −2
Dividing Eq. (I) by Eq. (IV), we get
−3
 0.1   0.1  = 6.0 × 10

  
−
 0.4   0.1 
2.4 × 10 2
x
x
or
y
 1 =  1
 
 
 4
 4
1
∴
x=1
On dividing Eq. (II) by Eq. (III), we get
7.2 × 10 −2
 0.3   0.2 

 
 =
 0.3   0.4 
2.88 × 10 −1
x
y
y
 1
 1
  = 
 2
 2
2
∴
y =2
Thus, rate law is,
rate = k [A] 1 [B]2
= k [A] [B]2
05 In the reaction,
BrO −3 (aq) + 5Br − (aq) + 6H+ →
3Br 2 (l) + 3H2O(l)
the rate of appearance of bromine
(Br 2 ) is related to rate of
disappearance of bromide ions as
following.
[CBSE AIPMT 2009, 2000]
d [Br2]
(a)
dt
d [Br2]
(b)
dt
d [Br2]
(c)
dt
d [Br2]
(d)
dt
3 d [Br −]
5 dt
5 d [Br −]
=−
3 dt
5 d [Br −]
=
3 dt
3 d [Br −]
=
5 dt
=−
06 For the reaction,
N 2 + 3H2 → 2NH3 , if
d [NH3]
= 2 × 10 −4 mol L−1 s −1 , the
dt
−d [H2]
would be
value of
dt [CBSE AIPMT 2009]
The rate of this reaction is, given
by
(a) rate =k [A] 2 [B]
(b) rate =k [A][B] 2
(c) rate = k [A] 2 [B] 2
(d) rate = k [A][B]
Ans. (b)
For the reaction,
A + B → Products
On doubling the initial concentration of A
only, the rate of the reaction is also
doubled, therefore
…(i)
Rate ∝ [A] 1
Let initial rate law is
…(ii)
Rate = k [A][B] y
If concentration of A and B both are
doubled, the rate gets changed by a
factor of 8.
…(iii)
8 × rate = k [2 A] [2 B] y
[Q Rate ∝ [A] 1]
Dividing Eq. (iii) by Eq. (ii), we get
8 = 2 × 2y
4 = 2y
(2)2 = (2) y
∴
y =2
Hence, rate law is,rate = k [A][B]2
y
1
 1
or   =
 2
4
or
Ans. (a)
Rate of appearance/disappearance
1
=±
×
stoichiometric coefficient
[reactant or product]
time taken
For the reaction,
BrO−3 (aq) + 5Br − (aq) + 6H+ → 3Br2 (l )
+ 3H2O(l )
Rate of appearance of bromine (Br2 )
1 d [Br2 ]
=+
3 dt
Rate of disappearance of bromide ion
(Br − )
1 d [Br − ]
=−
5 dt
d [Br2 ]
3 d [Br − ]
or
=−
5 dt
dt
(a) 3 × 10−4 mol L−1 s−1
(b) 4 × 10−4 mol L−1 s−1
(c) 6 × 10−4 mol L−1 s−1
(d) 1 × 10−4 mol L−1 s−1
Ans. (a)
For the reaction,
N2 + 3H2 → 2NH3
1 d [H2 ]
1 d [NH3]
=−
=+
dt
3 dt
2 dt
1 d [H2 ]
1 d [NH3]
or −
=+
3 dt
2 dt
d [H2 ] 3
−
= × 2 × 10 −4 mol L−1s−1
dt
2
Rate = −
d [N2 ]
= 3 × 10
−4
−1 −1
mol L s
07 For the reaction,
A + B → products, it is observed
that
I. On doubling the initial
concentration of A only, the
rate of reaction is also doubled
and
II. On doubling the initial
concentrations of both A and
B, there is a change by a factor
of 8 in the rate of the reaction.
[CBSE AIPMT 2009]
08 The bromination of acetone that
occurs in acid solution is
represented by this equation
CH3COCH3 (aq) + Br 2 (aq) →
CH3COCH2Br(aq) + H+ (aq) + Br − (aq)
These kinetic data were obtained
for given reaction concentrations.
Initial concentrations, M
[H + ]
[CH3COCH3]
[Br2 ]
0.30
0.05
0.05
0.30
0.10
0.05
0.30
0.10
0.10
0.40
0.05
0.20
Initial rate, disappearance of Br 2 ,
Ms–1
5.7 × 10−5
5.7 × 10−5
1.2 × 10−4
3.1 × 10−4
Based on these data, the rate
equation is
[CBSE AIPMT 2008]
(a) rate = k [CH3COCH3][H+]
(b) rate = k [CH 2 == COCH3][Br2]
(c) rate = k [CH3COCH3][Br2][H+] 2
(d) rate = k [CH3COCH3][Br2][H+]
92
NEET Chapterwise Topicwise Chemistry
Ans. (a)
Let the order of reaction wrt CH3COCH3,
Br2 and H+ are x, y and z respectively.
Thus,
Rate (r) = [CH3COCH3] x [Br2 ] y [H+ ] z
…(i)
5.7 × 10 −5 = (0.30) x (0.05) y (0.05) z
5.7 × 10 −5 = (0.30) x (0.10) y (0.05) z
 1
1=  
 2
y
1
or 1° =  
 2
…(iv)
y
y =0
From Eqs. (ii) and (iii)
z=1
From Eqs. (i) and (iv)
x=1
Thus, rate law ∝ [CH3COCH3] 1 [Br2 ] 0 [H+ ] 1
= k [CH3COCH3][H+ ]
09 Consider the reaction,
N 2 (g) + 3H2 (g) → 2NH3 (g)
The equality relationship between
d [NH3]
d [H2]
and –
is
dt
dt
[CBSE AIPMT 2006]
d [NH3]
1 d [H2 ]
(a)
=–
dt
3 dt
d [NH3]
2 d [H2 ]
(b) +
=–
dt
3 dt
d [NH3]
3 d [H2 ]
(c) +
=–
dt
2 dt
d [H2 ]
d [NH3]
(d)
=–
dt
dt
Ans. (b)
For the reaction,
N2 ( g) + 3H2 ( g) → 2NH3 ( g)
The rate of reaction wrt
d [N2 ]
N2 = −
dt
[Rate of disappearance]
The rate of reaction with respect to
1 d [H2 ]
H2 = −
3 dt
[Rate of disappearance]
The rate of reaction with respect to
1 d [NH3]
NH3 = +
2 dt
[Rate of appearance]
Hence, at a fixed time
or
or
d [N2 ]
1 d [NH3]
1 d [H2 ]
=−
=+
dt
3 dt
2 dt
d [NH3]
2 d [H2 ]
+
=−
dt
3 dt
2d [N2 ]
d [NH3]
+
=−
dt
dt
…(ii)
1.2 × 10 −4 = (0.30) x (0.10) y (0.10) z …(iii)
3.1 × 10 −4 = (0.40 )x (0.05)y (0.20 )z
From Eqs. (i) and (ii)
−
10 For the reaction,
2A + B → 3 C + D
which of the following does not
express the reaction rate?
[CBSE AIPMT 2006]
d [C]
3 dt
d [D]
(c)
dt
(a) –
d [B]
dt
d [A]
(d) –
2 dt
(b) –
Ans. (a)
For the reaction,
2A + B → 3 C + D
The reaction rate is written as follows:
The reaction rate with respect to
1 d [A]
A=−
2 dt
The reaction rate with respect to
d [B]
B=−
dt
The reaction rate with respect to
1 d [C]
C=+
3 dt
The reaction rate with respect to
d [D]
D=
dt
Hence, the answer (a) is not correct
expression to represent the rate of the
reaction.
11 The rate of reaction between two
reactants A and B decreases by a
factor of 4, if the concentration of
reactant B is doubled. The order of
this reaction with respect to
reactant B is [CBSE AIPMT 2005]
(a) −1
(c) 1
Ans. (b)
(b) −2
(d) 2
A + B → Product
Rate, r ∝ [A] x [B] y
…(i)
The rate decreases by a factor 4 if the
concentration of reactant B is doubled
r
…(ii)
∝ [A] x [2 B] y
4
From Eqs. (i) and (ii)
 1
4=  
 2
y
y = −2
Hence, order of reaction with respect to
B is –2.
12 3A → 2B, rate of reaction +
is equal to
3 d [A]
(a) −
2 dt
1 d [A]
(c) −
3 dt
d [B]
dt
[CBSE AIPMT 2002]
2 d [A]
3 dt
d [A]
(d) + 2
dt
(b) −
Ans. (b)
For reaction,
3A → 2 B
1 d [A]
Rate = −
3 dt
[Rate of disappearance]
1 d [B]
=+
2 dt
[Rate of appearance]
d [B]
2 d [A]
=−
∴ +
dt
3 dt
13 3A → B + C It would be a zero
order reaction, when
[CBSE AIPMT 2002]
(a) the rate of reaction is proportional to
square of concentration of A
(b) the rate of reaction remains same at
any concentration of A
(c) the rate remains unchanged at any
concentration of B and C
(d) the rate of reaction doubles if
concentration of B is increased to
double
Ans. (b)
For reaction,
3A → B + C
If it is zero order reaction, then the rate
remains same at any concentration of A
dx
or
= k [A 0 ]
[A 0 = 1].
dt
It means that for zero order reaction,
rate is independent of concentration of
reactants.
14 For the reaction,
2N 2O 5 → 4NO 2 + O 2 , rate and
rate constant are 1.02 × 10 −4 and
3 .4 × 10 −5 s −1 respectively, then
concentration of N 2O 5 at that time
will be
[CBSE AIPMT 2001]
(a) 1.732
(c) 102
. × 10−4
Ans. (b)
(b) 3
(d) 3 .4 × 105
93
Chemical Kinetics
2N2O5 → 4NO2 + O2
–d [N2O5 ]
= k ⋅ [N2O5 ]
dt
102
. × 10 –4 = 3 .4 × 10 –5 s–1 × [N2O5 ]
∴
[N2O5 ] =
1.02 × 10 –4
3.4 × 10 –5
=3
15 The experimental data for the
reaction 2 A+B2 → 2 AB is
Exp.
[A]
[B2]
Rate (M s −1)
1.
0.50
0.50
. × 10 −4
16
2.
0.50
1.00
3. 2 × 10 −4
3.
1.00
1.00
3. 2 × 10 −4
The rate equation for the above
data is
[CBSE AIPMT 1997]
Rate constant (k) = 4.606 × 10 − 3 s− 1
Initial amount (a 0 ) = 2g
Final amount (a) = 0.2 g
So, time required,
a
2.303
t=
log 0
k
a
2
2.303
=
× log
= 500 s
0.2
4.606 × 10 − 3
(a) 1.0 × 10−4
(c) 2.0 × 10−3
Ans. (a)
(b) 2.0 × 10−4
(d) 1.0 × 10−2
For a zero order reaction,t =
Consider the following rate law equation,
dx
= k [A] m [B2 ] n
dt
1 .6 × 10 −4 = k [0.50] m [0.50] n ...(i)
...(ii)
3 .2 × 10 −4 = k [0.50] m [10
. ]n
−4
m
...(iii)
3 . 2 × 10 = k [100
. ] [10
. ]n
By dividing Eq. (iii) by (ii) we get,
3 .2 × 10 −4 k [1.00] m [1.0] n
=
3 .2 × 10 −4 k [0.50] m [1.0] n
⇒
∴ m=0
By dividing Eq. (ii) by (i)
[0.50] m [1.0] n
3 .2 × 10 −4
=
1.6 × 10 −4 [0.50] m [0.50] n
2 = 2n or 21 = 2n
n= 1
∴
Hence rate,
 dx 
0
1
  = k [A] [B2 ] = k [B2 ]
 dt 
TOPIC 2
⇒
[NEET (Sep.) 2020]
(a) 200 s
(c) 1000 s
Ans. (b)
(b) 500 s
(d) 100 s
For a first order reaction,
1
(a 0 − a)
k
1 a0
[Qat t 50 , a = a 0 /2]
×
k 2
a0
0.02 M
k=
=
2 × t 50 2 × 100 s
t 50 =
= 1 × 10 − 4 Ms− 1 = 1 × 10 − 4 mol L− 1 s− 1
18 A first order reaction has a rate
constant of2.303 × 10 −3 s −1 . The
time required for 40 g of this
reactant to reduce to 10 g will be
[Given that log 10 2 = 0.3010]
[NEET (Odisha) 2019]
(a) 230.3 s
(c) 2000 s
Ans. (d)
(b) 301 s
(d) 602 s
For first order reaction,
2.303
a
t=
log
k
a−x
…(i)
Given : k = 2.303 × 10 −3 s−1,
a = 40 g, a − x = 10 g
On substituting the given values in Eq. (i),
we get
2.303
40
log
t=
10
2.303 × 10 −3
= 10 3 log 22 = 2 × 10 3 × log 2
= 2 × 10 3 × 0.3010 = 602 s
Order and Molecularity
16 The rate constant for a first order
reaction is 4.606 × 10 −3 s −1 . The
time required to reduce 2.0 g of
the reactant to 0.2 g is
[NEET (National) 2019]
(a) t = 6.909/k
(c) t = 2.303/k
Ans. (b)
17 The half-life for a zero order
reaction having 0.02 M initial
concentration of reactant is 100 s.
The rate constant (in mol L −1 s −1 )
for the reaction is [NEET (Oct.) 2020]
(a) rate = k [B2]
(b) rate = k [B2] 2
2
2
(c) rate = k [A] [B] (d) rate = k [A] 2 [B]
Ans. (a)
1 = 2m or 20 = 2m
19 If the rate constant for a first order
reaction is k, the time (t) required
for the completion of 99% of the
reaction is given by
Alternative method
For first order reaction,
0.693
t 1/ 2 =
k
0.693
t 1/ 2 (t 50%) =
= 301s
2.303 × 10 −3
Also,
t 75% = 2t 50%
∴
t 75% = 2 × 301 = 602 s
(b) t = 4.606/k
(d) t = 0.693/k
Key Idea For first order reaction,
2 .303
a
t=
log
K
a−x
where,a = initial concentration,a − x =
final concentration.
Let the initial concentration (a) = 100
After timet, final concentration
(a − x) = 100 − 99 = 1
2 .303
a
We know that, t =
log
K
a−x
On substituting the given values in above
eqn. we get
2.303
100 2 .303
t=
log
=
log 102
K
1
K
2.303
4606
.
=
× 2log 10 =
K
K
Thus, option (b) is correct.
20 The correct difference between
first-and second- order reactions is
that
[NEET 2018]
(a) a first-order reaction can be
catalysed; a second-order reaction
cannot be catalysed
(b) the half-life of a first-order reaction
does not depend on [A] 0 ; the
half-life of a second-order reaction
does depend on [A] 0
(c) the rate of a first-order reaction does
not depend on reactant
concentrations; the rate of a
second-order reaction does depend
on reactant concentrations
(d) the rate of a first-order reaction does
depend on reactant concentrations;
the rate of a second-order reaction
does not depend on reactant
concentrations
Ans. (b)
For first order reactions, the rate of
reaction is proportional to the first
power of the concentration of the
reactant .
For, A → B
d [A]
Rate = −
= k [A]
dt
[where, k = constant]
0.693
Half-life (t 1/ 2 ) =
k
94
NEET Chapterwise Topicwise Chemistry
∴ Rate of first order reaction depends
upon reactant concentrations and half
life does not depend upon initial
concentration of reactant, [A] 0 .
For second order reactions, the rate of
reaction is proportional to the second
power of the concentration of the
reactant.
For, 2A → B
Rate = k [A]2
1
Half-life (t 1/ 2 ) =
k [A] 0
∴ Rate of second order reaction
depends upon reactant concentration
and half life also does depend on [A] 0 .
21 When initial concentration of the
reactant is doubled, the half-life
period of a zero order reaction
[NEET 2018]
Two half-lives are required for the
reduction of 20 g of reactant into 5 g.
t1/2
t1/2
20 g → 10 g → 5 g.
∴ Total time = 2t 1/ 2
= 2 × 693
. = 1386
. s
23 The rate constant of the reaction
A → B is 0.6 × 10 −3 mole per second.
If the concentration of A is 5 M then
concentration of B after 20 min is
[CBSE AIPMT 2015]
(a) 1.08 M
(c) 0.36 M
Ans. (d)
(b) 3.60 M
(d) 0.72 M
Key Concept For a zero order reaction
unit of rate constant is mole per second.
Hence, we can easily calculate
concentration of B after 20 min by the
following formula,
x = Kt
x = Kt = 0.6 × 10 −3 × 20 × 60 = 0.72 M
(a) is tripled
(b) is doubled
(c) is halved
(d) remains unchanged
Ans. (b)
For zero order reaction,
[R] 0
t 1/ 2 =
2k
where, [R] 0 = Initial concentration of the
reactant.
k = Rate constant.
Thus, t 1/ 2 for zero order reaction is
directly proportional to the initial
concentration of the reactant.
t 1/ 2 ∝ [R] 0
∴ For zero order reaction, when the
concentration of reactant is doubled, the
half-life (t 1/ 2 ) will also get doubled.
22 A first order reaction has a specific
reaction rate of 10 −2 s −1 . How much
time will it take for 20 g of the
reactant to reduce to 5 g?
[NEET 2017]
(a) 238.6 s
(c) 346.5 s
Ans. (b)
Alternatively,
Half-life for the first order reaction,
t 1/ 2 0.693 0.693
= −2 = 69.3 s
=
2
k
10
(b) 138.6 s
(d) 693.0 s
For a first order reaction,
.
2303
a
Rate constant (k) =
log
t
a−x
where,a = initial concentration
a − x = concentration after time ‘t’
t = time in sec.
Given, a = 20 g, a − x = 5 g, k = 10 −2
2303
.
20
= 1386
. s
∴ t = −2 log
5
10
24 The rate of a first-order reaction is
0.04 mol L−1 s −1 at 10 sec and 0.03
mol L−1 s −1 at 20 sec after initiation
of the reaction. The half-life period
of the reaction is
[NEET 2016, Phase I]
(a) 34.1 s (b) 44.1 s (c) 54.1 s (d) 24.1 s
Ans. (d)
Given, order of reaction = 1
Rate of reaction at 10 s = 0.04 mol L −1 s −1
Rate of reaction at 20 s = 0.03 mol L −1 s −1
∴ Half-life period (t 1/ 2 ) = ?
We have the equation for rate-constant ‘
k’ in first order reaction.
A 2.303
0.04
2303
.
k=
log
log t =
t
A0
105
0.03
2.303
=
× 0.124
105
k = 0.028 s−1
We know that,
0.693
0.693
t 1/ 2 =
=
k
0.028773391 s−1
= 24.14 s ≈ 24.1 s
25 The decomposition of phosphine
(PH 3 ) on tungsten at low pressure is
a first-order reaction. It is because
the
[NEET 2016, Phase II]
(a) rate is proportional to the surface
coverage
(b) rate is inversely proportional to the
surface coverage
(c) rate is independent of the surface
coverage
(d) rate of decomposition is very slow
Ans. (a)
W
3
PH3 → P + H2
2
This is an example of surface catalysed
unimolecular decomposition.
For the above reaction, rate is given as
kαp
Rate =
1 + αp
where, p = partial pressure of absorbing
substrate.
At low pressure, αp<< 1 or Rate = kαp
So, (αp + 1) can be neglected.
hus, the decomposition is predicted to
be first order.
26 When initial concentration of a
reactant is doubled in a reaction,
its half-life period is not affected.
The order of the reaction is
[CBSE AIPMT 2015]
(a) zero
(b) first
(c) second
(d) more than zero but less than first
Ans. (b)
For a zero order reactiont 1/ 2 is directly
proportional to the initial concentration
of the reactant [R]0
t 1/ 2 ∝ [R]0
For a first order reaction
2303
.
[R]
k=
log 0
[R]
t
[R] 0
at t 1/ 2 , [R] =
2
So, the above equation becomes
K=
[R] 0
2303
.
log
[R]0 /2
t 1/ 2
2303
.
2303
.
= log 2=
× .3010
K
K
.693
t 1/ 2 =
K
i.e., half life period is independent of
initial concentration of a reactant.
t 1/ 2 =
27 Which one of the following
statements for the order of a
reaction is incorrect?
[CBSE AIPMT 2011]
95
Chemical Kinetics
(a) Order is not influenced by
stoichiometric coefficient of the
reactants
(b) Order of reaction is sum of power to
the concentration terms of
reactants to express the rate of
reaction
(c) Order of reaction is always whole
number
(d) Order can be determined only
experimentally
Ans. (c)
Order of reaction may be zero, whole
number or fraction number.
28 Half-life period of a first order
reaction is 1386s. The specific rate
constant of the reaction is
[CBSE AIPMT 2009]
−3 −1
−2
(a) 5.0 × 10 s
(c) 0.5 × 10−3 s−1
Ans. (c)
−1
(b) 0.5 × 10 s
(d) 5.0 × 10−3 s−1
Specific rate constant,
0.693 0.693
k=
=
1386
t 1/ 2
= 0.5 × 10 −3 s−1
29 The reaction of hydrogen and
iodine monochloride is given as
H2 (g) + 2ICl(g) → 2HCl(g) + I 2 (g)
This reaction is of first order with
respect to H2 (g) and ICl(g),
following mechanisms were
proposed
Mechanism A
H2 (g) + 2ICl(g) → 2HCl(g) + I 2 (g)
Mechanism B
H2 (g) + ICl(g) → HCl(g) + HI(g),
slow
HI(g) + ICl(g) → HCl(g) + I 2 (g), fast
Which of the above mechanism(s)
can be consistent with the given
information about the reaction?
[CBSE AIPMT 2007]
(a) Only B
(b) Both A and B
(c) Neither A nor B
(d) Only A
Ans. (a)
In the reactions which take place in a
number of steps, the slowest step is
known as the rate determining step.
Hence, rate of reaction always depends
on slow step.
H2 (g) + ICl(g) → HCl(g) + HI (g)
is first order reaction with respect toH2
and ICI. Thus, the mechanism B will be
more consistent with the given
information.
30 If 60% of a first order reaction was
completed in 60 min, 50% of the
same reaction would be completed
in approximately (log 4 = 0.60,
[CBSE AIPMT 2007]
log 5 = 0.69)
(a) 50 min
(c) 60 min
Ans. (b)
(b) 45 min
(d) 40 min
From first order reaction,
Rate constant,
2.303
a
k=
log 10
t
(a − x)
a1
2.303
k1 =
log
t1
a1 − x1
a2
2.303
k2 =
log
t2
a2 − x2
60
x1 =
a 1,t 1 = 60
100
50
x2 =
a2 ,t2 = ?
100
From Eqs. (i) and (ii)
a2
a1
2.303
2.303
log
=
log
t1
a1 − x1
t2
a2 − x2
…(i)
…(ii)
[CBSE AIPMT 2007]
log 2
k
ln2
(d)
k
a

x = 

2
[CBSE AIPMT 2005]
31 In a first order reaction, A → B,
if k is rate constant and initial
concentration of the reactant A is
0.5 M, then the half-life is
(b)
2.303
a
log
a
k
a−
2
ln2
2.303
=
log 2 or =
k
k
t 1/ 2 =
32 For a first order reaction, A → B,
the reaction rate at reactant
concentration of 0.01 M is found to
be 2.0 × 10–5 mol L–1 s–1 . The half-life
period of the reaction is
a1
2.303
2.303
=
log
log
 a − 60 a 
60
t2
 1
1
a2

100 
50 

a2 
 a2 −

100 
100a2
100a 1 2.303
2.303
=
log
log
t2
60
40a 1
50a2
1
100 1
100
log
= log
60
40 t2
50
100
60 log
50
t2 =
100
log
40
60 (log 10 − log 5)
=
(log 10 − log 4)
60(1 − 0.69) 60 × 0.31
=
=
(1 − 0.60)
0.40
= 1.5 × 31 = 46.5 ≈ 45 min
0.693
(a)
0.5k
log 2
(c)
k 0.5
Ans. (d)
For first order reaction,
2.303
a
Rate constant, k =
log
t
(a − x)
(a) 220 s
(c) 300 s
Ans. (d)
(b) 30 s
(d) 347 s
For first order reaction,
A → B
rate = k × [A]
Rate = 2.0 × 10 –5 mol L–1s–1
[A] = 0.01 M
So, 2 .0 × 10 −5 = k × 0.01
2.0 × 10 –5 –1
s
0.01
−3 −1
= 2 .0 × 10 s
For first order reaction,
0.693
t 1/ 2 =
k
0.693
=
2 .0 × 10 −3
k=
=346.5 ≈ 347 s
33 The rate of first order reaction is
1.5 × 10 −2 mol L–1 min–1 at 0.5 M
concentration of the reactant. The
half-life of the reaction is
[CBSE AIPMT 2004]
(a) 0.383 min
(c) 8.73 min
Ans. (b)
(b) 23.1 min
(d) 7.53 min
For the first order reaction,
 dx 
Rate   = k [A]
 dt 
[A] = concentration of reactant
k = rate constant
Given that,
dx
= 1.5 × 10 − 2 mol L–1 min–1
dt
k=?
and
[A] = 0.5 M
∴ 1.5 × 10 − 2 = k × 0.5
96
NEET Chapterwise Topicwise Chemistry
∴
1.5 × 10 − 2
0.5
= 3 × 10 − 2 min−1
(c) first power of final concentration
(d) square root of final concentration
Ans. (a)
k=
For first order reaction,
0.693
0.693
half-life period,t 1/ 2 =
=
k
3 × 10 −2
t 1/ 2 of nth order reaction ∝
= 23.1 min
34 If the rate of a reaction is equal to
the rate constant, the order of the
reaction is
[CBSE AIPMT 2003]
(a) 2
(b) 3
Ans. (c)
(c) 0
(d) 1
Rate of reaction is equal to the rate
constant for zero order reaction.
Let us consider the following
hypothetical change.
A → B + C
Suppose this reaction is zero order, then
rate ∝ [A] 0
rate = k [A] 0 = k
∴
35 The reaction, A→ B follows first
order kinetics. The time taken for
0.8 mole of A to produce 0.6 mole
of B is 1h. What is the time taken
for the conversion of 0.9 mole of A
to 0.675 mole of B ?
[CBSE AIPMT 2003]
(a) 0.25 h (b) 2 h
Ans. (c)
(c) 1 h
(d) 0.5 h
Rate constant of first order reaction
(A)
2 .303
k=
log 10 0
(A) t
t
or
k=
2 .303
0.8
× log 10
1
0.2
...(i)
(because 0.6 mole of B is formed)
Suppose t 1 hour are required for
changing the concentration of A from
0.9 mole to 0.675 mole of B.
Remaining mole of A = 0.9 − 0.675 = 0.225
2 .303
0.9
...(ii)
log 10
k=
∴
0.225
t1
From Eqs. (i) and (ii)
2 .303
0.9
0.8 2 .303
=
log 10
log 10
t1
0.225
1
0.2
2 .303 log 10 4 =
2 .303
log 10 4 ⇒ t 1 = 1 h
t1
36 For a first-order reaction, the
half-life period is independent of
[CBSE AIPMT 1999]
(a) initial concentration
(b) cube root of initial concentration
1
a n –1
where,a = initial concentration of
reactant
n= order of reaction
∴t 1/ 2 for first order reaction (n = 1)
1
t 1/ 2 ∝ 1– 1
a
1
or
t 1/ 2 ∝ 0
(a 0 = 1)
a
So, for a first order reaction half-life is
independent on initial concentration of
reactants.
For first order reaction
A0 = Initial concentration
A = Final concentration
[A] 0
2.303
Rate constant, k =
log
t
[A]
2.303
2.0
=
log
200
0.15
=
2.303
(log 200 − log 15)
200
=
2.303
× (2.3010 − 1.1761)
200
=
2.303 × 1.1249
200
= 0.01295 min−1
0.6932
k
0.6932
=
0.01295
Now, half life,t 1/ 2 =
37 The plot of concentration of the
reactant versus time for a reaction
is a straight line with a negative
slope. This reaction follows
[CBSE AIPMT 1996]
(a) zero order rate equation
(b) first order rate equation
(c) second order rate equation
(d) third order rate equation
Ans. (b)
For first order reaction, we know that
kt
log [A] = −
+ log [A] 0
2.303
On comparing it with the equation of
straight line,
i.e.
y = mx + c
Plot of log [A] versus time
→ straight line
−k
(negative)
slope =
2.303
= 53.50 min
TOPIC 3
Activation Energy, Arrhenius
and Collision Theory, Factors
Affecting Rate of Reaction
39 For a reaction, A → B, enthalpy of
reaction is − 4.2 kJ mol −1 and
enthalpy of activation is 9.6 kJ mol
−1
. The correct potential energy
profile for the reaction is shown in
option.
[NEET 2021]
(a)
PE
B
A
log [A]
Reaction progress
log [A]0
slope (m) = –k
2.303
t
(b)
PE
B
38 A substance A decomposes by a
first order reaction starting initially
with [A] = 2.00 m and after 200 min,
[A] becomes 0.15 m. For this
reaction t 1/ 2 is [CBSE AIPMT 1995]
(a) 53.49 min
(c) 48.45 min
Ans. (a)
(b) 50.49 min
(d) 46.45 min
[A] 0 = 2.0 m, [A] = 0.15 m, t = 200 min
A
Reaction progress
(c)
PE
A
B
Reaction progress
97
Chemical Kinetics
Ans. (b)
PE A
Reaction progress
Ans. (b)
The enthalpy of reaction is negative,
– 4.2 kJ mol −1. The reaction is an
exothermic reaction i.e. the energy of
product B, is less than the energy of
reactant A. So, the potential energy
profile for the reaction is
9.6 kJ mol –1
PE
A
–4.2 kJ mol –1
B
Reaction process
40 The slope of Arrhenius plot
1

ln k vs  of first order reaction is

T
− 5 × 10 3 K. The value of E a of the
reaction is
[Given, R = 8.314 JK − 1 mol − 1 ]
[NEET 2021]
−1
(a) 41.5 kJ mol
(c) 166 kJ mol − 1
Ans. (a)
−1
(b) 83.0 kJ mol
(d) − 83 kJ mol − 1
Arrhenius equation,
− E / RT
k = Ae a
E
1
ln k = ln A − a  
R T 
−E a
is the slope of Arrhenius plot
R
 ln k vs 1 



T
ln A is the intercept
Slope = − 5 × 10 3 K
R = 8.314 J K–1 mol –1
Ea
So, −
= − 5 × 10 3 K
R
⇒ E a = 5 × 10 3 × R = 5 × 10 3 × 8.314
= 41.57 × 10 3 J mol −1 = 41.5 kJ mol −1
41 In collision theory of chemical
reaction, Z AB represents
[NEET (Oct.) 2020]
(a) the fraction of molecules with
energies greater than E a
(b) the collision frequency of reactants,
A and B
(c) steric factor
(d) the fraction of molecules with
energies equal to E a
where, k = rate constant,
− E / RT
e a
= fraction of molecules with
energy greater than E a atT K. (option-(a))
A = Arrhenius factor or frequency factor
= P × Z AB
when, P = steric factor (option-(c))
Z AB = collision frequency of reactants, A
and B
= number of effective binary collisions
between A and B in one second in unit
volume (option-(b)).
Hence, option (b) is the correct.
Ans. (b)
For a reversible reaction, it accelerates
the speed of forward as well as
backward reaction to the same extent.
Hence, it does not disturb the
equilibrium, i.e. does not change the
equilibrium constant of the reaction but
helps to attain the equilibrium faster.
44 The addition of a catalyst during a
chemical reaction alters which of
the following quantities?
[NEET 2016, Phase I]
(a) Internal energy
(b) Enthalpy
(c) Activation energy
(d) Entropy
Ans. (c)
42 For a reaction, activation energy
E a = 0 and the rate constant at
200K is 1.6 × 10 6 s −1 . The rate
constant at 400 K will be
[Given that gas constant
R = 8.314 JK −1 mol −1 ]
A catalyst is a substance which alters
the reaction rate but itself remains
unchanged in amount and chemical
composition at the end of the reaction. It
provides a new reaction path with a
lower energy barrier (lowering activation
energy).
[NEET (Odisha) 2019]
(a) 3.2 × 104 s−1
(c) 1.6 × 103 s−1
Ans. (b)
(b) 1.6 × 106 s−1
(d) 3.2 × 106 s−1
Key Idea Rate constants at two
different temperatures is given by
Arrhenius equation as follows :
k
Ea  1 1 
log 2 =
 − ,T2 > T1
k1 2303
. R T1 T2 
Given, E a = 0, T1 = 200 K, k1 = 16
. × 10 6 s−1
T2 = 400 K,
R = 8314
. JK −1 mol −1
According to Arrhenius equation,
k
Ea  1
1
log 2 =
 − 
k1 2.303R T1 T2 
On substituting the given values in above
equation, we get
k2
0
1 
 1
log
=
−
 200 400 
.
. × 10 6 2.303 × 8314
16
 k2 
log 
 =0
 16
. × 10 6 
k2
= 10 0 = 1
16
. × 10 6
k2 = 16
. × 10 6 s−1 at 400K
Thus, option (b) is correct.
43 Which one of the following
statements is not correct?
[NEET 2017]
(a) Catalyst does not initiate any
reaction
(b) The value of equilibrium constant is
changed in the presence of a
catalyst in the reaction equilibrium
Activated complex
B without catalyst
Energy
B
(d)
(c) Enzymes catalyse mainly
biochemical reaction
(d) Coenzymes increase the catalytic
activity of enzyme
From collision theory of chemical
reaction, Arrhenius equation.
− E / RT
We get, k = A × e a
Activated complex with catalyst
Lowest energy
of activation
provided by the
Ea
catalyst through new
reaction pathway
ER
Reactants
EP
Products
Progress of reaction
45 The activation energy of a reaction
can be determined from the slope
of which of the following graphs?
[CBSE AIPMT 2015]
(a) In K vs T
(b)
In K
vs T
T
I
T
(d)
T
I
vs
In K T
(c) In K vs
Ans. (c)
By Arrhenius equation
K = Ae − Ea / RT
where, E a = energy of activation
Applyinglog on both the side,
E
ln k = ln A − a
RT
Ea
or
log k = −
+ log A
2303
. RT
…(i)
…(ii)
98
NEET Chapterwise Topicwise Chemistry
lnk
Compare the above equation w.r.t.
straight line equation of y = mx + c.
1
Thus, if a plote oflnk vs is a straight
T
line, the validity of the equation is
confirmed.
E
Slope of the line = − a
R
Thus, measuring the slope of the line,
the value of
E a can be calculated.
slope = –
Ea
R
47 In a zero order reaction for every
10°C rise of temperature, the rate is
doubled. If the temperature is
increased from 10°C to 100°C, the
rate of the reaction will become
[CBSE AIPMT 2012]
(a) 256 times
(c) 64 times
Ans. (b)
For 10° rise in temperature, n = 1
n=9
rate = 29 = 512 times
So,
46 What is the activation energy for a
reaction if its rate doubles when
the temperature is raised from
20°C to 35°C?
(R = 8.314J mol −1 K −1 ) [ NEET 2013]
(a) 342 kJ mol−1
(b) 269 kJ mol−1
(c) 34.7 kJ mol−1
(d) 15.1 kJ mol−1
Ans. (c)
Given, initial temperature,
T1 = 20 + 273 = 293 K
Final temperature,
T2 = 35 + 273 = 308 K
R = 8314
. J mol −1 K−1
Since, rate becomes double on raising
temperature,
r
∴
r2 = 2r1 or 2 = 2
r1
As rate constant, k ∝ r
k2
=2
∴
k1
From Arrhenius equation, we know that
k
E a T1 − T2 
log 2 = −


k1
2.303 R  TT
1 2 
 293 − 308 
Ea
log 2 = −
.  293 × 308 
2.303 × 8314
 −15 
Ea
0.3010 = −
2.303 × 8314
.  293 × 308 
∴
0.3010 × 2.303 × 8.314 × 293 × 308
15
= 34673.48 J mol −1
= 34.7 kJ mol −1
Ea =
Alternate method with every 10° rise in
temperature, rate becomes double,
so
 100 − 10 

10 

r′
=2
r
= 29 = 512 times.
(b)
Ans. (d)
Given,
(a) less than ∆H
(b) equal to ∆H
(c) more than ∆H
(d) equal to zero
Ans. (c)
k1 = 10 16 ⋅ e −2000 / T
k2 = 10 15 ⋅ e −1000 / T
On taking log of both the equations we
get
2000
log k1 = 16 −
2.303T
log k2 = 15 −
1000
2.303T
At k1 = k2
16 −
48 For an endothermic reaction,
energy of activation is E a and
enthalpy of reaction is ∆H (both of
these in kJ/mol). Minimum value of
[CBSE AIPMT 2010]
E a will be
2000
1000
= 15 −
2.303T
2.303T
1000
K
T=
2.303
50 The activation energy for a simple
chemical reaction, A → B is E a in
forward direction. The activation
energy for reverse reaction
[CBSE AIPMT 2003]
Key Idea In endothermic reactions,
energy of reactants is less than that of
the products.
Potential energy diagram for endothermic
reactions is
Ea'
Ea
Potential
Energy
P
∆H
R
Progress of the reaction
where,
E a = activation energy of forward
reaction
activation energy of backward reaction
∆H = enthalpy of the reaction.
From the above diagram,
E a = E a′ + ∆H
Thus,
2000
K
2.303
1000
(d)
K
2.303
(a) 1000 K
(c) 2000 K
so
rate = 2n = 21 = 2
When temperature is increased from
10°C to 100°C, change in temperature
= 100 − 10 = 90 ° C
i.e.
1/T
(b) 512 times
(d) 128 times
49 The rate constants k 1 and k 2 for
two different reactions are
10 16 ⋅ e −2000 /T and 10 15 ⋅ e −1000 /T ,
respectively. The temperature at
which k 1 = k 2 is [CBSE AIPMT 2008]
E a > ∆H
(a) can be less than or more than Ea
(b) is always double of Ea
(c) is negative of Ea
(d) is always less than Ea
Ans. (a)
The energy of activation of reverse
reaction is less than or more than energy
of activation (E a ) of forward reaction.
∴
∆H = (E a ) F − (E a ) R
Because it depends upon the nature of
reaction.
If (E a ) F > (E a ) R , reaction is endothermic
or (E a ) F < (E a ) R , reaction is exothermic
51 The temperature dependence of rate
constant (k ) of a chemical reaction is
written in terms of Arrhenius equation,
k = Ae −E */ RT . Activation energy (E*) of
the reaction can be calculated by
plotting
[CBSE AIPMT 2003]
(a) log k vs
(c) k vs T
1
T
(b) log k vs
(d) k vs
1
log T
1
log T
99
Chemical Kinetics
Arrhenius equation k = Ae − E * / RT
E*
ln k = ln A −
RT
(E * = energy of activation)
E*
or log k = log A −
2 .303 RT
Compare this equation with the straight
line equation,
i.e.
y = mx + c
where ‘m’ is slope and ‘c’ is intercept
Hence, E * can be calculated with the
help of following slope
log A
Slope () = m
log k
E*
2.303 R
1
T
52 When a biochemical reaction is
carried out in laboratory from
outside of human body in the
absence of enzyme, the rate of
reaction obtained is 10 −6 times,
then activation energy of the
reaction in the presence of enzyme
is
[CBSE AIPMT 2001]
6
RT
(b) P is required
(c) different from E a obtained in
laboratory
(d) cannot say any things
(a)
Ans. (c)
When a biochemical reaction is carried
out in laboratory from outside of human
body in the absence of enzyme, then rate
of reaction obtained is 10 –6 times than
activation energy of reaction in the
presence of enzyme. It is different from
E a obtained in laboratory because for a
given chemical reaction.
− E / RT
(Arrhenius equation)
k = Ae a
Also activation energy have different
value in absence or presence of
enzyme.
53 Activation energy of a chemical
reaction can be determined by
55 A chemical reaction is catalysed by
a catalyst X. Hence, X
[CBSE AIPMT 1998]
[CBSE AIPMT 1995]
(a) evaluating rate constant at standard
temperature
(b) evaluating velocities of reaction at
two different temperatures
(c) evaluating rate constants at two
different temperatures
(d) changing concentration of reactants
(a) reduces enthalpy of the reaction
(b) decreases rate constant of the
reaction
(c) increases activation energy of the
reaction
(d) does not affect equilibrium constant
of the reaction
Ans. (c)
Ans. (d)
Activation energy can be calculated by
using Arrhenius equation. The Arrhenius
equation is
k
E a T2 − T1 
log 2 =


k1 2 .303 R  TT
1 2 
where, k1 and k2 = rate constants at two
different temperatures, i.e.T1 andT2
respectively.
E a = Activation energy
R = Gas constant
So, activation energy of a chemical
reaction can be determined by
evaluating rate constants at two
different temperatures.
Although a catalyst speeds up the
reaction but it does not shift the position
of equilibrium. This is due to the fact
that the presence of catalyst reduces
the height of barrier by providing an
alternative path for the reaction and
lowers the activation energy. However,
the lowering in activation energy is to
the same extent for the forward as well
as the backward reaction.
54 In a reversible reaction, the energy
of activation of the forward
reaction is 50 kcal. The energy of
activation for the reverse reaction
will be
[CBSE AIPMT 1996]
(a) < 50 kcal
(b) 50 kcal
(c) either greater than or less than 50 kcal
(d) > 50 kcal
56 For exothermic reaction, the
energy of activation of the
reactants is
[CBSE AIPMT 1994]
(a) equal to the energy of activation of
products
(b) less than the energy of activation of
products
(c) greater than the energy of activation
of products
(d) sometimes greater and sometimes
less than that of the products
Ans. (b)
The plot of activation energy versus
reaction coordinates is given below for
exothermic reaction.
Ans. (c)
The activation energy of a reverse
reaction decide whether the given
reaction is exothermic or endothermic,
so, the energy of activation of reverse
reaction is either greater or less than
50 kcal. In case of exothermic reaction,
the activation energy for reverse
reaction is more than activation energy
of the forward reaction and in case of
endothermic reaction, the activation
energy for reverse reaction is less than
activation energy of the forward
reaction.
Activation
energy
(Ea )R
Ans. (a)
Reactant
(Ea )P
∆ rH
Product
Reaction coordinates
It is clear from the above plot that the
activation energy of reactant is less than
the activation energy of products.
12
Surface Chemistry
TOPIC 1
Adsorption and
Various Isotherms
01 The correct option representing a
Freundlich adsorption isotherm is
[NEET (Odisha) 2019]
x
(a)
= kp 0. 3
m
x
(c) = kp −0. 5
m
Ans. (a)
x
= kp2. 5
m
x
(d) = kp −1
m
(b)
According to Freundlich adsorption
isotherm,
x
= kp1/ n
m
x
where, = amount of the gas adsorbed
m
per unit mass of adsorbent
p = pressure k and n = constants.
The value of n lies in between 0 to 1.
x
Thus, = kp0. 3 and option (a) is correct.
m
02 Which one of the following
characteristics is associated with
adsorption?
[NEET 2016, Phase I]
(a) ∆G, ∆H and ∆S all are negative
(b) ∆G and ∆H are negative but ∆S is
positive
(c) ∆G and ∆S are negative but ∆H is
positive
(d) ∆G is negative but ∆H and ∆S are
positive
Ans. (a)
Adsorption is a spontaneous process
that occurs with release in energy and
decrease in the randomness (i.e.
entropy) of the adsorbed substance.
For a spontaneous process, ∆G must be
negative.
∆G = ∆H −T∆S
As the process is exothermic and
randomness of molecule (entropy)
decreases hence, both ∆H and ∆S will be
negative as well.
03 Which of the following statements
is correct for the spontaneous
adsorption of a gas?
[CBSE AIPMT 2014]
(a) ∆S is negative and therefore, ∆H
should be highly positive
(b) ∆S is negative and therefore, ∆H
should be highly negative
(c) ∆S is positive and therefore, ∆H
should be negative
(d) ∆S is positive and therefore, ∆H
should also be highly positive
Ans. (b)
∆S [change in entropy] and ∆H [change
in enthalpy] are related by the equation
∆G = ∆ H − T ∆S
[Here, ∆G = change in Gibbs free energy]
For adsorption of a gas, ∆S is negative
because randomness decreases. Thus,
in order to make ∆G negative [for
spontaneous reaction], ∆H must be
highly negative because reaction is
exothermic. Hence, for the adsorption of
a gas, if ∆S is negative, therefore, ∆H
should be highly negative.
04 In Freundlich adsorption isotherm,
the value of 1/n is
[CBSE AIPMT 2012]
(a) between 0 and 1 in all cases
(b) between 2 and 4 in all cases
(c) 1 in case of physical adsorption
(d) 1 in case of chemisorption
Ans. (a)
In Freundlich adsorption isotherm,
x
= kp1/ n
m
Where, x = amount of adsorbent
m = amount of adsorbate
The value of n is always greater than 1.
So, the value of 1/n lies between 0 and 1
in all cases.
05 If x is amount of adsorbate and m is
amount of adsorbent, which of the
following relations is not related to
adsorption process?
[CBSE AIPMT 2011]
x
(a) = f (T ) at constant p
m
 x
(b) p = f (T ) at constant  
 m
x
(c) = p × T
m
x
(d) = f (p) at constant T
m
Ans. (c)
x
= p × T is the incorrect relation. The
m
correct relation is amount of adsorption
x p
∝
m T
06 The Langmuir adsorption isotherm
is deduced by using the
assumption that
[CBSE AIPMT 2007]
(a) the adsorption takes place in
multilayers
(b) the adsorption sites are equivalent in
their ability to adsorb the particles
(c) the heat of adsorption varies with
coverage
(d) the adsorbed molecules interact
with each other
Ans. (b)
The main points of Langmuir’s theory of
adsorption are as
(i) Adsorption takes place on the
surface of the solid only till the whole
of the surface is completely covered
with a unimolecular layer of the
adsorbed gas.
101
Surface Chemistry
(ii) Adsorption consist of two opposing
processes (a) condensation and (b)
evaporation.
(iii) The rate of condensation depend
upon the uncovered surface of the
adsorbent available for
condensation.
07 For adsorption of a gas on a solid,
x
the plot of log vs log p is linear
m
with slope equal to (n being a whole
number) [CBSE AIPMT 2006, 1994]
(a) k
(b) log k
1
(d)
n
(c) n
x
If we plot a graph betweenlog   and
 m
log p, a straight line will be obtained. The
1
slope of the line is equal to and the
n
intercept is equal tolog k.
Slope =
1
n
Intercept log k
log p
x
where, = amount of adsorption
m
According to Freundlich adsorption
isotherm
x
= kp1/ n
m
Taking log of both sides,
1
x
log = log k + log p
m
n
from
TOPIC 3
Adsorption is the ability of a substance
to concentrate or hold gases, liquids
upon its surface.
Solids adsorb greater amounts of
substances at lower temperature. In
general adsorption decreases with
increasing temperature.
Colloids, Micelles and
Emulsions
y = zx + c
1
z = (slope)
n
08 Which is not correct regarding the
adsorption of a gas on surface of
solid?
[CBSE AIPMT 2001]
(a) On increasing temperature
adsorption increases continuously
(b) Enthalpy and entropy change is
negative
(c) Adsorption is more for some specific
substance
(d) Reversible
11 The right option for the statement
“Tyndall effect is exhibited by”, is
[NEET 2021]
(a) NaCl solution (b) glucose solution
(c) starch solution (d) urea solution
Ans. (c)
TOPIC 2
Catalyst
09 Which one of the following
statements is incorrect about
enzyme catalysis?
Ans. (d)
x
log
m
Ans. (a)
[CBSE AIPMT 2012]
(a) Enzymes are mostly proteinous in
nature
(b) Enzyme action is specific
(c) Enzymes are denaturated by UV-rays
and at high temperature
(d) Enzymes are least reactive at
optimum temperature
Ans. (d)
Most of the enzymes have proteinous
nature. They are highly specific and get
denaturated by high temperature or
UV-rays. At optimum temperature,
which is generally in between 25°-35°C,
enzyme activity is maximum.
10 According to the adsorption theory
of catalysis, the speed of the
reaction increases because
[CBSE AIPMT 2003]
(a) adsorption produces heat which
increases the speed of the reaction
(b) adsorption lowers the activation
energy of the reaction
(c) the concentration of reactant
molecules at the active centres of
the catalyst becomes high due to
adsorption
(d) in the process of adsorption, the
activation energy of the molecules
becomes large
Ans. (b)
According to adsorption theory of
catalysis, the speed (rate) of the reaction
increases because adsorption lowers
the activation energy of the reaction.


1
∴ Rate of reaction ∝


activation energy 
Tyndall effect is shown by colloidal
solution in which particles having larger
size scatters the light.
NaCl, glucose and urea do not form a
colloidal solution. Only starch solution is
colloidal solution. So, Tyndall effect is
exhibited by starch solution.
12 In which of the sols, the colloidal
particles are with negative charge?
[NEET (Oct.) 2020]
(a) TiO2
(c) Starch
Ans. (c)
(b) Haemoglobin
(d) Hydrated Al2O 3
(a) In TiO2 sol, TiO2 particles are
positively charged.
(b) In blood, haemoglobin is positively
charged.
(c) In starch sol, starch is negatively
charged.
(d) In hydrated Al2O3, Al2O3 ⋅ xH2O is
positively charged.
Hence, option (c) is the correct.
13 Measuring zeta potential is useful
in determining which property of
colloidal solution? [NEET (Sep.) 2020]
(a) Solubility
(b) Stability of the colloidal particles
(c) Size of the colloidal particles
(d) Viscosity
Ans. (b)
Stability of a colloidal solution is
explained by zeta potential.
The particles of colloid carry an electric
charge, i.e. acquire positive or negative
charge by preferential adsorption of
positive or negative ions from dispersion
medium. The combination of two layer of
positive and negative charges around
the colloid is called Helmholtz electrical
double layer and the potential difference
of this double layer is called zeta
potential.
102
NEET Chapterwise Topicwise Chemistry
14 Which mixture of the solutions will
lead to the formation of negatively
charged colloidal [AgI]I − sol ?
[NEET (National) 2019]
(a)
(b)
(c)
(d)
50 mL of 1 M AgNO3 + 50 mL of 2 M KI
50 mL of 2 M AgNO3 + 50 mL of 1.5 M KI
50 mL of 1 M AgNO3 + 50 mL of 0.1 M KI
50 mL of 1 M AgNO3 + 50 mL of 1.5 M KI
Ans. (d)
Key Idea The colloidal particles acquire
positive or negative charge by
preferential adsorption of positive ions
on negative ions.
When silver nitrate (AgNO3) solution is
added to potassium iodide (KI) solution,
the precipitated silver iodide (AgI)
adsorbs iodide ions (I− ) from the
dispersion medium and negatively
charged colloidal solution results.
Among the given options, in option (a)
and option (b) millimole of KI is higher
than AgNO3. Hence, both the mixture will
lead to the formation of negatively
charged colloidal [AgI]I− sol.
Note The most appropriate option is (a)
because millimole of KI
(50 mL × 2 M = 100 mmol)is maximum in
this option.
15 On which of the following
properties does the coagulating
power of an ion depend?
[NEET 2018]
(a) Both magnitude and sign of the
charge on the ion
(b) Size of the ion alone
(c) The magnitude of the charge on the
ion alone
(d) The sign of charge on the ion alone
Ans. (a)
The process of settling of colloidal
particles due to the neutralisation of
their charge by any means is called
coagulation.
Coagulation power of an ion depends
both on magnitude and sign of the
charge (positive or negative) on the ion.
This fact can be explained by Hardy
Schulze rule.
According to this rule “greater the
valency of the coagulating
ion/flocculating ion (oppositely charged
ion) added, the greater is its power to
cause coagulation.
To coagulate a positively charged sol,
the order of coagulating power of
negative ion is
I− < SO24− < PO34− < [Fe(CN) 6 ] 4 − .
Similarly, to coagulate a negatively
charged sol, the order of coagulating
power of positive ions is
Ag + < Pb2 + < Fe3+ < Si4+
16 Fog is a colloidal solution of
[NEET 2016, Phase I]
(a) Gas in liquid
(c) Gas in gas
Ans. (d)
(b) Solid in gas
(d) Liquid in gas
Fog is a colloidal solution of liquid in a
gas, in which liquid is the dispersed
phase whereas gas is the dispersion
medium. Examples of other options are
as follows:
Gas in liquid : Shaving cream, soda
water, froth
Solid in gas : Dust in air
Gas in gas : Atmospheric air.
17 The coagulation values in
millimoles per litre of the
electrolytes used for the
coagulation of As 2 S 3 are given
below
I. (NaCl) = 52,
II. (BaCl 2 ) = 0.69
III. (MgSO 4 ) = 0.22
The correct order of their
coagulating power is
[NEET 2016, Phase II]
(a) I > II > III
(c) III > II > I
Ans. (c)
(b) II > I > III
(d) III > I > II
Lower the coagulating power, higher is
the coagulation value in millimoles per
litre, i.e. coagulating power is inversely
proportional to coagulation values. Thus,
correct order of coagulating power is
MgSO4 > BaCl2 > NaCl or III > II > I
18 The suspension of slaked lime in
water is known as
[NEET 2016, Phase II]
(a) limewater
(b) quicklime
(c) milk of lime
(d) aqueous solution of slaked lime
Ans. (c)
Aqueous solution of slaked lime is called
lime water whereas suspension solution
of slaked lime is called milk of lime.
19 Which property of colloidal solution
is independent of charge on the
colloidal particles?
[CBSE AIPMT 2015, 2014]
(a) Coagulation
(b) Electrophoresis
(c) Electroosmosis
(d) Tyndall effect
Ans. (d)
Coagulation is generally brought about
by the addition of electrolytes. When an
electrolyte is added to a colloidal
solution, the particles of the sol take up
the ions which are oppositely charged.
As a result their charge gets neutralised.
Electrophoresis The movement of
colloidal particles under an applied
electric potential is called
electrophoresis.
Electroosmosis may be defined as a
phenomenon in which the molecules of
the dispersion medium are allowed to
more under the influence of an electric
field whereas colloidal particles are not
allowed to more.
Tyndall effect is the scattering of light
by sol particles, which cannot be
affected by charge on them.
20 The protecting power of lyophilic
colloidal sol is expressed in terms
of
[CBSE AIPMT 2012]
(a) coagulation value
(b) gold number
(c) critical micelle concentration
(d) oxidation number
Ans. (b)
Lyophobic sols are unstable, so they are
stabilised by adding some lyophilic
colloids which protect them from
precipitation. Thus, lyophilic colloids are
called protecting colloids. Their
protecting power is expressed in terms
of gold number. In other words, gold
number can be defined as the minimum
amount of lyophilic colloid in milligrams,
which prevent the flocculation of 10 mL
gold sol by the addition of 1 mL of 10%
NaCl solution.
NOTE Lesser the gold number, higher is
the protecting power.
21 Which one of the following forms
micelles in aqueous solution above
certain concentration?
[CBSE AIPMT 2005]
103
Surface Chemistry
(a) Urea
(b) Dodecyl trimethyl ammonium
chloride
(c) Pyridinium chloride
(d) Glucose
Ans. (b)
Surfactants like detergents, form
micelles in aqueous solution above to
their critical micelle concentration
(CMC). Dodecyl trimethyl ammonium
chloride is an example of surfactant
(cationic surfactant),
+
CH3
CH3  (CH2 ) 11  N  CH3
14
4244
3 –
Cl
CH3
Non- polar part
14
4244
3
Polar part
22 Which of the following forms
cationic micelles above certain
concentration? [CBSE AIPMT 2004]
(a) Sodium ethyl sulphate
(b) Sodium acetate
(c) Urea
(d) Cetyl trimethyl ammonium bromide
Ans. (d)
Cetyltrimethylammonium bromide forms
cationic micelles above a certain
concentration. In the molecule of
detergents and soap, the negative ions
aggregate to form a micelle of colloidal
size. In polar medium (like water), the
negative ion has a long hydrocarbon
chain and a polar group (—COO− ) at one
end and on other end, it hasN+ ion, thus
cationic micelle is formed.
23 Position of non-polar and polar
parts in micelle is
[CBSE AIPMT 2002]
(a) polar at outer surface but non-polar
at inner surface
(b) polar at inner surface but non-polar
at outer surface
(c) distributed all over the surface
(d) present in the surface only
Ans. (a)
Ans. (d)
Micelles are the clusters formed by the
association of colloids. They are formed
by lyophilic and lyophobic groups. As the
concentration increases, the lyophobic
parts receding away from the solvent
approach each (non-polar part) other and
form a cluster, the lyophobic ends are in
the interior, lyophilic groups (polar part)
projecting outward in contact with the
solvent.
According to Hardy-Schulze law,
coagulation ∝ charge of ion
So, coagulation is affected by both
magnitude of charge and nature of
charge. For coagulation of a positive sol,
negative ions are required and for
coagulation of negative sol, positive ions
are required.
Greater the magnitude of charge,
quicker will be the coagulation.
The ions having opposite charge to sol
particle causes coagulation.
Coagulating power of an electrolyte is
directly proportional to the fourth power
of the valence of the ions causing the
coagulation.
24 The method usually employed for
the precipitation of a colloidal
solution is
(a) dialysis
[CBSE AIPMT 2000]
(b) addition of electrolytes
(c) diffusion through animal
membrane
(d) condensation
Ans. (b)
On addition of electrolyte charge of
colloidal particles will neutralise and
hence, coagulation or precipitation of
colloidal solution will occur.
25 At the critical micelle
concentration (CMC) the surfactant
molecules
[CBSE AIPMT 1998]
(a) decompose
(b) dissociate
(c) associate
(d) become completely soluble
Ans. (c)
The soap concentration at which
micelles first appear is called critical
micelle concentration (CMC). At this
condition the surfactant molecules
associate with each other.
26 The ability of anion to bring about
coagulation of a given colloid
depends upon
(a) its charge
[CBSE AIPMT 1997]
(b) the sign of the charge alone
(c) the magnitude of its charge
(d) both magnitude and sign of its charge
27 During dialysis
[CBSE AIPMT 1996]
(a) only solvent molecules can diffuse
(b) solvent molecules, ions and colloidal
particles can diffuse
(c) all kinds of particles can diffuse
through the semipermeable
membrane
(d) solvent molecules and ions can
diffuse
Ans. (d)
The principle of dialysis is based upon
the fact that colloidal particles cannot
pass through a parchment or cellophane
membrane while the ions of the
electrolyte can pass through it.
28 If a beam of light is passed through
true solution, then it is
(a) visible
[CBSE AIPMT 1995]
(b) scatter
(c) not visible
(d) None of the above
Ans. (c)
When a beam of light is passed through
true solution, then the path of light is
invisible due to fact that the size of
particles in true solution is below 1 nm,
so they cannot scatter the light that
means cannot show Tyndall effect.
13
Classification of Elements
and Periodicity in Properties
TOPIC 1
Ans. (c)
Periodic Table and
Classification of Elements
01 From the following pairs of ion
which one is not an iso-electronic
pair?
[NEET 2021]
−
2−
(a) O , F
(c) Mn2+ , Fe 3+
Ans. (d)
Ion
O
2−
F−
+
Na
Mg 2 +
Mn
2+
+
2+
(b) Na , Mg
(d) Fe2 + , Mn2+
Number of electrons
Name
IUPAC official
name
101
(i) Mendelevium
(Md)
(B) Unniltrium
103
(ii) Lawrencium
(Lr)
(C) Unnilhexium
106
(iii) Seaborgium
(Sg)
(D) Unununnium
111
(iv) Roentgenium
(Rg) [but (iv)
Darmstadtium
(Ds), Given] [Z
= 110]
10
10
23
Fe 2 +
24
Mn2 +
23
O2 − and F − are iso-electronic pair.
Na+ and Mg2 + are iso-electronic pair.
Mn2 + and Fe3+ are iso-electronic pair.
Fe2 + and Mn2 + are not iso-electronic pair.
03 The element Z = 114 has been
discovered recently. It will belong
to which of the following
family/group and electronic
configuration?
[NEET 2017]
(a) Halogen family, [Rn] 5f 14 6d 10 7 s 2 7 p5
(b) Carbon family, [Rn] 5f 14 6d 10 7 s 2 7 p2
Name
IUPAC official
name
(A) Unnilunium
(i) Mendelevium
(B) Unniltrium
(ii) Lawrencium
(C) Unnihexium
(iii) Seaborgium
(D) Unununnium (iv) Darmstadtium
[NEET (Sep.) 2020]
(a) (B), (ii)
(c) (D), (iv)
(b) (C), (iii)
(d) (A), (i)
14
(d) Nitrogen family, [Rn] 5f 6d
10
(b) 3d 5 , 4 s 1
(c) 3d 5 , 4 s 2
(d) 3d2 , 4 s 2
The sum of number of electrons
(unpaired) in d-orbitals and number of
electrons in s-orbital gives the number
of oxidation states (os) exhibited by a
d-block element. Therefore,
(a) 3d 3, 4s 2 ⇒OS = 3 + 2 = 5
(b) 3d 5 , 4 s 1 ⇒OS = 5 + 1 = 6
(c) 3d 5 , 4 s 2 ⇒OS = 5 + 2 = 7
So, D-(iv) is the incorrect match.
(c) Oxygen family, [Rn] 5f 14 6d 10 7 s 2 7 p4
02 Identify the incorrect match.
(a) 3 d 3, 4 s 2
Ans. (c)
23
Fe 3 +
04 Which one of the elements with the
following outer orbital
configurations may exhibit the
largest number of oxidation states?
[CBSE AIPMT 2009]
(A) Unnilunium
10
10
Atomic
number
(Z)
2
7s 7p
6
Ans. (b)
The element with atomic number, Z = 114
is flerovium (F1). It is a super heavy
artificial chemical element. In the
periodic table of the elements, it is a
transactinide element in the p-block. It is
a member of the 7th period and is the
heaviest known member of the carbon
family.
Electronic configuration for Z = 114 is
[Rn] 86 5f 14 , 6d 10 , 7 s 2 , 7 p2
(d) 3d2 , 4 s 2 ⇒OS = 2 + 2 = 4
Hence, element with3d 5 ,4s 2
configuration exhibits largest number of
oxidation states.
05 An atom has electronic
configuration
1s 2 , 2s 2 2p 6 , 3s 2 3p 6 3d 3 , 4s 2 , you
will place it in [CBSE AIPMT 2002]
(a) fifth group
(c) second group
Ans. (a)
(b) fifteenth group
(d) third group
The amount of energy required to
remove an electron from the outermost
orbit of a gaseous atom is known as
ionisation potential. Elements having
half-filled or completely filled orbitals
are more stable than partially filled
orbitals.
In a period from left to right ionisation
potential decreases as the atomic
number increases. The given elements
(Be, B, C, N, O) are present in II period as
105
Classification of Elements and Periodicity in Properties
Be B C N O
Ionisation potential increases
→
But in case of Be and B, Be has higher
ionisation potential due to stable
configuration.
2
2
4 Be = 1s , 2s
2 s2
Stable configuration
B = 1s 2, 2s 2 2p1
5
Unstable
configuration
2p1
In the same way in case of N and O, N has
higher ionisation potential than O due to
stable configuration
2
2
3
7 N = 1s , 2s 2p
Stable configuration
(half-filled)
2p3
8 O = 1s
2p 4
2
, 2s 2 2p4
Unstable
configuration
So, the correct order of increasing
ionisation potential will be
B < Be < C < O < N
06 The element with the atomic
number 118, will be
(a) alkali
[CBSE AIPMT 1996]
(b) noble gas
(c) lanthanide
(d) transition element
Ans. (b)
The outermost electronic configuration
of element with atomic number 118 is
7 s 2 7 p6 , so it will be a noble gas.
07 The electronic configuration of an
element is 1s 2 , 2 s 2 2p 6 , 3s 2 3p 3 .
What is the atomic number of the
element, which is present just
below the above element in the
periodic table? [CBSE AIPMT 1995]
(a) 33
(b) 34
Ans. (a)
(c) 36
(d) 49
The element which present just below
the given element will have outermost
electronic configuration as4 s 2 4p3, so
its full electronic configuration is
1s 2 , 2 s 2 2p6 , 3s 2 3p6 , 4 s 2 , 3d 10 , 4p3 and
hence, its atomic number is 33.
08 If the atomic number of an element
is 33, it will be placed in the
periodic table in the
[CBSE AIPMT 1993]
(a) first group
(c) fifth group
(b) third group
(d) seventh group
Ans. (c)
The electronic configuration of element
with atomic number 33 is
1s 2 , 2 s 2 2p6 , 3 s 2 3p6 ,4s 2 , 3d 10 , 4p3. As, its
last shell have five electrons and hence,
its group is 10 + 5 = 15th or V A.
09 The electronic configuration of
four elements are given below.
Which element does not belong to
the same family as others?
[CBSE AIPMT 1989]
(a) [Xe] 4 f 14 , 5 d 10, 6 s 2
(b) [Kr] 4 d 10, 5 s 2
(c) [Ne] 3 s 2 , 3p 5
(d) [Ar] 3 d 10, 4 s 2
Ans. (c)
In a family, all elements have same
outermost electronic configuration.
Since [Ne] 3s 2 3p5 , chlorine belongs to
halogen family while the remaining three
are in same group i.e. group 12.
80 Hg
= [Xe]4f 14 5d 10 6s 2
48 Cd
= [Kr]4d 10 5s 2
30 Zn =
[Ar] 3d 10 4s 2
TOPIC 2
Periodic Properties
10 The correct order of atomic radii in
group 13 elements is [NEET 2018]
(a) B < Ga < Al < Tl < In
(b) B < Al < Ga < In < Tl
(c) B < Al < In < Ga < Tl
(d) B < Ga < Al < In < Tl
Ans. (d)
The atomic radii as well as ionic radii
increases on moving down the group 13
elements because of the successive
addition of one extra shell of electrons.
However, there is an anomaly at gallium
in case of atomic radii. Atomic radii of Ga
is lesser as compared to Al.
Gallium (Ga) with electronic
configuration, [Ar] 18 3d 10 4s 2 4p1 has an
extra d-electrons which do not screen
the nucleus effectively. Consequently,
electrons of Ga are more attracted by
nucleus.
Thus, the increasing order of atomic
radii of the group 13 elements is B (85
pm) < Ga (135 pm) < Al (143 pm) < In (167
pm) < Tl (170 pm).
11 In which of the following options
the order of arrangement does not
agree with the variation of property
indicated against it?
[NEET 2016, Phase I]
(a) B < C < N < O (increasing first
ionisation enthalpy)
(b) I < Br < Cl < F (increasing electron
gain enthalpy)
(c) Li < Na < K < Rb (increasing metallic
radius)
(d) Al 3+ < Mg2 + < Na + <F − (increasing
ionic size)
Ans. (a,b)
For option (a)
First ionisation energy is the energy
required to remove an electron from
outermost shell.
Hence, correct order is B < C < O < N.
For option (b)
Electron gain enthalpy is the energy
required to gain an electron in the
outermost shell.
Hence, the correct order is I < Br < F < Cl.
For option (c)
As we move down the group in alkali
metal, metallic radius increases Li < Na <
K < Rb.
For option (d)
In case of isoelectronic species, as
positive charge decreases or negative
charge increases the ionic size of the
species increases and vice-versa
Al 3+ < Mg2 + < Na+ < F − .
12 The species Ar, K + and Ca 2+
contain the same number of
electrons. In which order do their
radii increase? [CBSE AIPMT 2015]
(a) Ar < K+ < Ca2 +
(b) Ca2 + < Ar < K+
+
(d) K + < Ar < Ca2+
(c) Ca < K < Ar
Ans. (c)
2+
Ca2 + < K+ < Ar
Ar, K and Ca are isoelectronic i.e. with
same number of electrons, 18. For
isoelectronic species ionic radii decreases
with increase in effective (relative) positive
charge. Also Ar, K and Ca belong to the
same period (3rd period).
+
2+
13 Which of the following orders of
ionic radii is correctly represented?
[CBSE AIPMT 2014]
(a) H− > H+ > H
(b) Na+ > F − > O2 −
(c) F − > O2 − > Na+
(d) Al3+ > Mg2 + > N3−
106
NEET Chapterwise Topicwise Chemistry
Ans. (*)
(No option is correct.)
(a) H− > H+ > H
It is known that radius of a cation is
always smaller than that of a neutral
atom due to decrease in the number of
orbits. Whereas, the radius of anion is
always greater than a cation due to
decrease in effective nuclear charge.
Hence, the correct order is
H− > H > H+
(b) Na+ > F − > O2 −
The given species are isoelectronic as
they contain same number of electrons.
For isoelectronic species,
1
Ionic radii ∝
atomic number
Ion:
Na+ F − O2 −
Atomic number : 11 9 8
Hence, the correct order of ionic radii is
O2 − > F − > Na+
(c) Similarly, the correct option is
O2 − > F − > Na+
(d) Ions :
Al 3+ Mg2 + N3−
Atomic number : 13 12 7
Hence, the correct order is,
N3− > Mg2 + > Al 3+
14 Identify the wrong statement in the
following.
[CBSE AIPMT 2012]
(a) Amongst isoelectronic species,
smaller the positive charge on the
cation, smaller is the ionic radius
(b) Amongst isoelectronic species,
greater the negative charge on the
anion, larger is the ionic radius
(c) Atomic radius of the elements
increases as one moves down the
first group of the periodic table
(d) Atomic radius of the elements
decreases as one moves across
from left to right in the 2nd period of
the periodic table
Ans. (a)
Atomic radius of the elements
decreases across a period from left to
right due to increase in effective nuclear
charge. On moving down a group, since,
number of shells increases, so atomic
radius increases.
Amongst isoelectronic species, ionic
radius increases with increase in
negative charge or decrease in positive
charge.
15 The correct order of the
decreasing ionic radii among the
following isoelectronic species is
[CBSE AIPMT 2010]
(a) Ca2 + > K + > S2 − > Cl−
(b) Cl− > S2 − > Ca2 + > K +
(c) S2 − > Cl− > K + > Ca2 +
(d) K + > Ca2 + > Cl− > S2 −
Ans. (c)
Key Idea Ionic radii ∝ charge on anion
1
∝
charge on cation
During the formation of a cation, the
electrons are lost from the outer shell
and the remaining electrons experience
a great force of attraction by the
nucleus, i.e. attracted more towards the
nucleus. In other words, nucleus hold the
remaining electrons more tightly and
this results in decreased radii.
However, in case of anion formation, the
addition of electron(s) takes place in the
same outer shell, thus the hold of
nucleus on the electrons of outer shell
decreases and this results in increased
ionic radii.
Thus, the correct order of ionic radii is
S2 − > Cl − > K+ > Ca2 +
16 Which of the following represents
the correct order of increasing
electron gain enthalpy with
negative sign for the elements
O, S, F and Cl? [CBSE AIPMT 2010]
(a) Cl < F < O < S
(b) O < S < F < Cl
(c) F < S < O < Cl
(d) S < O < Cl < F
Ans. (b)
Key Idea Electron gain enthalpy,
generally, increases in a period from left
to right and decreases in a group on
moving downwards. However, members
of III period have somewhat higher
electron gain enthalpy as compared to
the corresponding members of second
period, because of their small size.
O and S belong to VI A (16) group and Cl
and F belong to VII A (17) group. Thus, the
electron gain enthalpy of Cl and F is
higher as compared to O and S.
Cl and F > O and S
Between Cl and F, Cl has higher electron
gain enthalpy then the F, since the
incoming electron experiences a greater
force of repulsion because of small size
of F-atom. Similar is true in case of
O and S, i.e. the electron gain enthalpy
of S is higher as compared to O due to
its small size. Thus, the correct order
of electron gain enthalpy of given
elements is
O < S < F < Cl
17 Amongst the elements with
following electronic configurations,
which one may have the highest
ionisation energy?
[CBSE AIPMT 2009]
(a) [Ne] 3 s 2 3p 3
(b) [Ne] 3 s 2 3p2
(c) [Ar] 3d 10, 4 s 2 4 p 3
(d) [Ne] 3 s 2 3p 1
Ans. (a)
Key Idea Across a period, increasing
nuclear charge outweighs the
shielding, hence the outermost
electrons are held more and more
tightly and ionisation energy. increases
across a period while as we move down
a group increase in shielding outweighs
the increasing nuclear charge and the
removal of the outermost electron
required less energy down a group.
Electronic configuration Group
V
[Ne] 3 s 2 3p3
IV
[Ne] 3 s 2 3p2
V
[Ar] 3d 10 , 4 s 2 4p3
III
[Ne] 3 s 2 3p1
Since, ionisation energy increases in a
period and decreases in a group,
[Ne] 3 s 2 3p3 configuration has the
highest ionisation energy among the
given elements.
18 Which of the following oxides is not
expected to react with sodium
hydroxide?
[CBSE AIPMT 2009]
(a) B2O 3 (b) CaO
Ans. (b)
(c) SiO2
(d) BaO
Sodium hydroxide, NaOH, being a strong
alkali never react with a basic oxide
(compound). Among the given options,
B2O3 and BeO are amphoteric oxides,SiO2
is an acidic oxide and CaO is a basic
oxide. Therefore, NaOH does not react
with CaO.
19 The correct order of decreasing
second ionisation enthalpy of
Ti(22), V(23), Cr(24) and Mn(25) is
[CBSE AIPMT 2008]
(a) Cr > Mn > V > Ti
(b) V > Mn > Cr > Ti
(c) Mn > Cr > Ti > V
(d) Ti > V > Cr > Mn
Ans. (a)
The amount of energy required to
remove an electron from unipositive ion
is referred as second ionisation
potential.
107
Classification of Elements and Periodicity in Properties
In Ti, V, Cr and Mn, generally second
ionisation energy increases with
increase in atomic number but second
ionisation potential of Cr is greater than
that of Mn due to the presence of exactly
half-filled d-subshell in Cr.
Thus, the order of second ionisation
enthalpy is
Cr > Mn > V > Ti
20 Which of the following electronic
configuration of an atom has the
lowest ionisation enthalpy?
[CBSE AIPMT 2007]
(b) 1s 2 , 2 s 2 2 p 3
(a) 1s 2 , 2 s 2 2 p 5
2
2
5
1
(c) 1s , 2 s 2 p , 3 s (d) 1s 2 , 2 s 2 2 p 6
Ans. (c)
The electronic configuration
1s 2 , 2 s 2 2p5 , 3 s 1 shows lowest ionisation
energy because this configuration is
unstable due to the presence of one
electron in s-orbital. Hence, less energy
is required to remove the electron.
21 Identify the correct order of the
size of the following.
[CBSE AIPMT 2007]
(a) Ca2+ < K + < Ar < S2– < Cl–
(b) Ca2+ < K + < Ar < Cl– < S2–
(c) Ar < Ca2+ < K + < Cl– < S2–
(d) Ca2+ < Ar < K + < Cl– < S2–
Ans. (b)
A cation has always the lesser ionic size
than a metal atom due to loss of
electrons and an anion has always the
greater size than metal atom due to gain
of electrons. The given species are
isoelectronic species as they contain
same number of electrons. For
isoelectronic species ionic radii
1
.
∝
atomic number
Ion
: Ca2 + K+ Ar S2 Cl
Atomic number : 20 19 18 16 17
So, the correct order of size is as
Ca2 + < K+ < Ar < Cl – < S2 –
22 Ionic radii are
[CBSE AIPMT 2004]
(a) inversely proportional to effective
nuclear charge
(b) inversely proportional to square of
effective nuclear charge
(c) directly proportional to effective
nuclear charge
(d) directly proportional to square of
effective nuclear charge
Unstable
configuration
Ans. (a)
Ionic radii ∝
1
2 p1
Z eff
Z eff → effective nuclear charge
This Z eff is calculated as follows :
Z eff = Z − screening constant (σ)
This value of screening constant is
based upon the number of electrons in
valence shell as well as in penultimate
shells.
23 The ions
O 2– , F – , Na + , Mg 2+ and Al 3+ are
isoelectronic. Their ionic radii show
[CBSE AIPMT 2003]
(a) an increase fromO2 − to F − and then
decrease fromNa+ to Al 3+
(b) a decrease fromO2 – to F − and then
increase fromNa+ to Al 3+
(c) a significant increase fromO2 – to Al 3+
(d) a significant decrease fromO2– to
Al 3+
Ans. (d)
On increasing atomic number of
isoelectronic species ionic radii
decreases due to increasing effective
nuclear charge (Z eff).
1
1
Radius ∝
∝
Atomic number Z eff
So, as the negative charge increases
ionic radii increases while on increasing
positive charge ionic radii decreases.
Anions having higher ionic radii than the
cation.
Hence, order of radii
O2 – > F – > Na+ > Mg2 + > Al 3+
24 Which of the following order is
wrong?
[CBSE AIPMT 2002]
(a) NH3 < PH3 < AsH3 – Acidic
(b) Li < Be < B < C–1st Ionisation potential
(c) Al2O3 < MgO < Na2O < K2O – Basic
(d) Li+ < Na+ < K+ < Cs+ – Ionic radius
Ans. (b)
Li, Be, B and C are present in IInd period.
In a period from left to right ionisation
potential increases.
Ionisation potention increases
Li Be B C
 →
* But in case of Be and B, Be has higher
ionisation potential than B due to stable
configuration of Be.
2
2
4 Be = 1s , 2s
Stable configuration
(due to fully-filled orbital)
So, the correct order of ionisation
potential of given elements is
Li < B < Be < C
25 Correct order of Ist ionisation
potential (IP) among following
elements Be, B, C, N, O is
[CBSE AIPMT 2001]
(a) B < Be < C < O < N
(b) B < Be < C < N < O
(c) Be < B < C < N < O
(d) Be < B < C < O < N
Ans. (a)
An atom has electronic configuration
1s 2 , 2s 2 , 2p6 , 3s 2 3p6 3d 3, 4s 2
It is a member ofd- block element
because the last electron is filled in
d- subshell as3d 3 and the following
electronic configuration is possible for
d- subshell as (n − 1)d (1 to 10 )
Group number IIIB IVB VB VIB VIIB
3 4
5
6
7
ns 2 (n − 1) s 2 p6 d 1 d2 d 3 d 4 d 5
VIII VIII VIII IB IIB
8
9 10 11 12
d 6 d 7 d 8 d 9 d 10
Hence, it is member of third group.
26 The first ionisation potential (in eV)
of Be and B, respectively are
[CBSE AIPMT 1998]
(a) 8.29, 9.32
(c) 8.29, 8.29
Ans. (d)
(b) 9.32, 9.32
(d) 9.32, 8.29
First ionisation potential of beryllium
(Be) is greater than boron (B) due to
stable configuration
2
2
4 Be = 1s , 2 s
2
2
B = 1s , 2 s 2p1
5
Order of attraction of electrons towards
nucleus is2 s > 2 p, so more amount of
energy is required to remove the
electron from 2 s orbital in comparison
to 2p orbital. So, ionisation potential of
Be is 9.32 eV and B is 8.29 eV.
27 In crystals of which of the following
ionic compounds would you expect
maximum distance between
centres of cations and anions?
[CBSE AIPMT 1998]
2 s2
5B
2
2
1
= 1s , 2 s 2p
(a) LiF
(c) CsI
(b) CsF
(d) LiI
108
NEET Chapterwise Topicwise Chemistry
Ans. (c)
On moving from top to bottom in a group
of periodic table distance between ions
in ionic compounds increases. Hence, it
is maximum in CsI.
28 Which one of the following ions will
be smallest in size?
[CBSE AIPMT 1996]
(a) Na+ (b) Mg2+ (c) F −
Ans. (b)
(d) O2−
Na+ ,Mg2 + , O2 − and F − all are isoelectronic
but Mg2 + have 12 protons in his nucleus,
so the attraction force on last shell is
maximum and hence, it have smallest
size.
29 Among the following, the one
which is most basic is
[CBSE AIPMT 1994]
(a) ZnO
(c) Al2O 3
Ans. (b)
(b) MgO
(d) N2O 5
ZnO and Al2O3 are amphoteric oxide,N2O5
is oxide of non-metal, so it is acidic and
hence, MgO is most basic among
Al2O3, ZnO, N2O5 and MgO.
30 Which electronic configuration of
an element has abnormally high
difference between second and
third ionisation energy?
[CBSE AIPMT 1993]
(a) 1s 2 , 2 s 2 2 p 6 , 3 s 1
(b) 1s 2 , 2 s 2 2 p 6 , 3 s 2 3p 1
(c) 1s 2 , 2 s 2 2 p 6 , 3 s 2 3p2
(d) 1s 2 , 2 s 2 2 p 6 , 3 s 2
Ans. (d)
When the element having
1s 2 , 2 s 2 2p6 , 3 s 2 configuration, loss two
electrons, then it acquire the electronic
configuration of noble gas (Ne), so to
remove third electron a large amount of
energy is required and hence, its second
and third ionisation energy have large
difference.
31 In the periodic table from left to
right in a period, the atomic volume
[CBSE AIPMT 1993]
(a) decreases
(b) increases
(c) remains same
(d) first decrease then increases
Ans. (d)
In the periodic table the atomic size first
decreases from left to right in period, so
the atomic volume first decreases and
then increases because atomic size in
last of any period increases.
32 One of the characteristic
properties of non-metals is that
they
[CBSE AIPMT 1993]
(a) are reducing agents
(b) form basic oxides
(c) form cations by electron gain
(d) are electronegative
Ans. (d)
Non-metals easily gain electrons and
hence, they form negative ions, so they
are electronegative in nature.
33 Na + , Mg 2+ , Al 3+ and Si 4+ are
isoelectronic. The order of their
ionic size is
[CBSE AIPMT 1993]
(a) Na+ > Mg2+ < Al3+ < Si4+
(b) Na+ < Mg2+ > Al3+ > Si4+
(c) Na+ > Mg2+ > Al3+ > Si4+
(d) Na+ < Mg2+ > Al3+ < Si4+
Ans. (c)
In isoelectronic species the number of
electrons are same but nuclear charge
is different. As the nuclear charge
increase, the attraction force on last
electron increases, so the size
decreases or in other words
1
and
Ionic size ∝
Charge on cation
hence, order is
Na+ > Mg2 + > Al 3+ > Si4+
→
Nuclear charge increase
→
Size decrease
34 One would expect proton to have
very large
[CBSE AIPMT 1993]
Ans. (c)
N, O, F are more electronegative
element, so they accept electrons more
easily and form negative ions (anions).
36 The ionisation of hydrogen atom
would give rise to
[CBSE AIPMT 1990]
(a) hydride ion
(c) proton
Ans. (c)
(b) hydronium ion
(d) hydroxyl ion
Hydrogen have one proton and one
electron, when it ionise, i.e. it lose one
electron, then only proton is left in the
nucleus, soH+ ion is formed during
ionisation which is also called proton.
H
→ H+ + e −
 e − = 1


 p= 1 
Proton
37 In the periodic table, with the
increase in atomic number, the
metallic character of an element
[CBSE AIPMT 1989]
(a) decrease in a period and increases in
a group
(b) ncreases in a period and decreases
in a group
(c) increases in a period as well as in the
group
(d) decreases in a period and also in the
group
Ans. (a)
In periodic table, the metallic character
increases down the group because the
ionisation enthalpy decreases down the
group and metallic character decreases
from left to right because the ionisation
enthalpy increases from left to right.
38 Pauling’s electronegativity values
for elements are useful in
predicting
[CBSE AIPMT 1989]
(a) charge
(b) ionisation potential
(c) hydration energy
(d) radius
Ans. (c)
(a) polarity of the molecules
(b) position in the emf series
(c) coordination numbers
(d) dipole moments
Ans. (a)
Proton have very small size, so have
large hydration energy. The degree of
hydration depends upon the size of the
cation. Smaller the size of a cation
greater the hydration energy.
Pauling’s electronegativity values are
useful in determination of polarity of the
bond in molecules. If electronegativity
difference is zero, then the molecule is
non-polar otherwise it is polar.
x A − x B = 0.028 ∆ E
35 Which of the following sets has
strongest tendency to form
anions?
[CBSE AIPMT 1993]
(a) Ga, In, TI
(c) N, O, F
(b) Na, Mg, Al
(d) V, Cr, Mn
x A and x B are electronegativities of the
atoms A and B respectively. While,
∆ E = actual bond energy
− EA − A × EB − B
14
General Principles and
Processes of Isolation
of Metals
TOPIC 1
Occurrence, Thermodynamic
and Electrochemical
Principles of Metallurgy
03 Considering Ellingham diagram,
which of the following metals can
be used to reduce alumina?
[NEET 2018]
(a) Mg
(c) Fe
Ans. (a)
01 Identify the incorrect statement.
[NEET (Odisha) 2019]
Key concept Ellingham diagrams help us
in predicting the feasibility of thermal
reduction of an ore. The criterion of
feasibility is that at a given temperature,
Gibbs energy of the reaction must be
negative.
Gibbs energy ∆G °vsT plots (schematic)
for formation of some oxides (Ellingham
diagram).
According to Ellingham diagram, the
temperature at which two lines intersect
shows that the metal will reduce the
oxide of other metals which lie above it
in Ellingham diagram.
(a) The scientific and technological
process used for isolation of the
metal from its ore is known as
metallurgy
(b) Minerals are naturally occurring
chemical substances in the earth’s
crust
(c) Ores are minerals that may contain a
metal
(d) Gangue is an ore contaminated with
undesired materials
Ans. (d)
02 Which one is malachite from the
following?
[NEET (National) 2019]
(a) Cu(OH)2
(b) Fe 3O 4
(c) CuCO 3 ⋅Cu(OH)2
(d) CuFeS2
Ans. (c)
Malachite is an ore of copper and its
composition is CuCO3 ⋅ Cu(OH)2 .
Azurite (Cu(OH)2 ) and copper pyrites
(CuFeS2 ) are also the ores of copper (Cu).
Mgnetite (Fe3O4 ) is an ore of iron (Fe).
Hence, option (c) is correct.
0
–100
4Cu+O 2
–200
DG°/kJ mol–1of O2
The earthly impurities like sand, clay,
mica, etc., associated with ores are
called gangue or matrix. In other words,
contaminated undesired materials
present in an ore is called gangue. Thus,
statement (d) is incorrect while other
options contain correct statements.
(b) Zn
(d) Cu
2Fe+O 2
–300
–400
C+O2
2CO 2
–500
+O 2
2CO
O2
–600
2Zn+
2Cu 2O
2FeO
2
2ZnO
I O3
2/3A 2
4/3
AI+O 2
2CO
A
2MgO
O2
–1000
2Mg+
–1100
–1200
0°C
273K
400°C
673K
800°C
1073K
1200°C 1600°C
1473K 1873K
(a) H2 S (b) SO2
Ans. (b)
(c) CO2
(d) SO 3
SO2 gas is obtained when any sulphide
ore is roasted.
2 M2S + 3 O2 →
∆ 2 M2O + 2 SO2
This gas exhibits all the characteristics
that are given in the question.
05 Which one of the following is a
mineral of iron? [CBSE AIPMT 2012]
(a) Malachite
(c) Pyrolusite
Ans. (d)
(b) Cassiterite
(d) Magnetite
Mineral
Chemical composition
Malachite
CuCO3.Cu(OH)2
Cassiterite
SnO2
Pyrolusite
MnO2
Magnetite
FeO4
Thus, magnetite is a mineral of iron.
2C+O
–700
–800
–900
CO2
04 Roasting of sulphides gives the gas
X as a by-product. This is a
colourless gas with choking smell
of burnt sulphur and causes great
damage to respiratory organs as a
result of acid rain. Its aqueous
solution is acidic acts as a reducing
agent and its acid has never been
insolated. The gas X is [NEET 2013]
2000°C
2273K
Temperature
In other words, the metal oxide having
more negative value of ∆Gf° can reduce
the oxide having less negative ∆Gf° . As,
Mg has more − ∆G ° value than alumina,
so it will be in lower part of Ellingham
diagram. Hence, Mg will be used to
reduce alumina.
06 Sulphide ores of metals are usually
concentrated by froth floatation
process. Which one of the
following sulphide ores offers an
exception and is concentrated by
chemical leaching?
[CBSE AIPMT 2007]
(a) Argentite
(c) Copper pyrite
(b) Galena
(d) Sphalerite
110
NEET Chapterwise Topicwise Chemistry
Ans. (d)
Galena (PbS), copper pyrites (CuFeS2 )
and argentite (Ag2S) are concentrated by
froth floatation process but sphalerite
(ZnS) is concentrated by chemical
leaching.
TOPIC 2
Extraction and
Isolation of Metals
11 The maximum temperature that
can be achieved in blast furnace is
07 Cassiterite is an ore of
[CBSE AIPMT 1999]
(a) Mn
(c) Sb
Ans. (d)
(b) Ni
(d) Sn
Cassiterite is an ore of Sn, its chemical
composition is SnO2 . It is also known as
tin stone.
08 Calcium is obtained by the
[CBSE AIPMT 1997]
(a) roasting of limestone
(b) electrolysis of solution of calcium
chloride inH2O
(c) electrolysis of molten anhydrous
calcium chloride
(d) reduction of calcium chloride with
carbon
Ans. (c)
Calcium is obtained by electrolysis of
molten anhydrous calcium chloride.
09 Cinnabar is an ore of
[CBSE AIPMT 1991]
(a) Hg
(c) Pb
Ans. (a)
(b) Cu
(d) Zn
Cinnabar is an ore of mercury which have
formula HgS.
10 Calgon used as a water softner, is
[CBSE AIPMT1989]
(a) Na2 [Na4 (PO 3) 6]
(b) Na4 [Na2 (PO 3) 6]
(c) Na4 [Na4 (PO 4 ) 5]
(d) Na4 [Na2 (PO 4 ) 6]
Ans. (a)
Sodium polymetaphosphate is used to
remove the permanent hardness of
water. The commercial name of sodium
polymetaphosphate is calgon meaning
calcium gone.
The molecular formula of calgon is
Na2 [Na4 (PO3) 6 ].
[NEET 2021]
(a) upto 1200 K
(b) upto 2200 K
(c) upto 1900 K
(d) upto 5000 K
Ans. (b)
(a) iron (II) sulphide
(b) carbon monoxide
(c) copper (I) sulphide
(d) sulphur dioxide
Ans. (c)
Cu2S + 2Cu2O → 6Cu + SO2 ↑
14 Aluminium is extracted from
alumina (Al 2O 3 ) by electrolysis of a
molten mixture of
[CBSE AIPMT 2012]
A blast furnace is generally used for
reduction of iron oxides but it can be
used for extraction of other metals like
Pb from PbO, etc.
In a blast furnace, hot air is blown from
the bottom of furnace. This bottom
surface has the maximum temperature
of upto 2200 K.
12 Extraction of gold and silver
involves leaching with CN − ion.
Silver is later recovered by
(a) liquation
[NEET 2017]
(b) distillation
(c) zone refining
(d) displacement with Zn
Ans. (d)
Extraction of gold and silver involves
leaching with CN− ion. Silver is later
recovered by distillation of Zn.
In the metallurgy of silver or gold, the
respective metal is leached with a dilute
solution of NaCN or KCN in the presence
of air to obtain the metal in solution as
complex. From the complex, metal is
obtained later by replacement.
In general,
4M( s ) + 8CN − (aq ) + 2H2O(aq )
+ O2 (g) → 4[ M( CN)2 ] − (aq ) + 4OH− (aq )
2[ M( CN)2 ] − (aq ) + Zn( s ) →
[Zn(CN) 4 ]2 − (aq ) + 2M(s)
M = Ag or Au
This method is known as Mac-Arthur
Forest cyanide process.
13 In the extraction of copper from its
sulphide ore, the metal finally
obtained by the reduction of
cuprous oxide with
[CBSE AIPMT 2015]
(a) Al2O 3 + HF + NaAlF4
(b) Al2O 3 + CaF2 + NaAlF4
(c) Al2O 3 + Na3AlF6 + CaF2
(d) Al2O 3 + KF + Na3AlF6
Ans. (c)
Alumina, Al2O3 is a bad conductor of
electricity and has very high melting
point, so before subjecting to
electrolysis, it is mixed with fluorspar
(CaF2 ) and cryolite (Na3AlF6 ), which lower
its melting point and make it more
conducting. Mainly CaF2 and Na3AlF6 are
mixed with Al2O3 for converting Al2O3 in
molten state.
15 In the extraction of copper from its
sulphide ore, the metal is finally
obtained by the reduction of
cuprous oxide with
[CBSE AIPMT 2012]
(a) copper (I) sulphide (Cu2 S)
(b) sulphur dioxide (SO2 )
(c) iron sulphide (FeS)
(d) carbon monoxide (CO)
Ans. (a)
In the extraction of copper from its
sulphide ore, when ore is subjected to
roasting, some of it is oxidised to Cu2O
which reacts with the remaining Cu2S
(sulphide ore) to give copper metal.
2Cu2S + 3O2 → 2Cu2O + 2SO2 ↑
2Cu2O + Cu2S → 6Cu + SO2 ↑
In this process Cu2S behaves as reducing
agent.
16 Which of the following elements is
present as the impurity to the
maximum extent in the pig iron?
[CBSE AIPMT 2011]
(a) Carbon
(c) Phosphorus
(b) Silicon
(d) Manganese
111
General Principles and Processes of Isolation of Metals
Ans. (a)
Ans. (c)
Pig iron contains about 4% carbon
(major impurity) and other impurities (S,
P, Si, Mn) in trace amounts.
Mercury is the only metal which is liquid
at room temperature. Impure mercury
metal is evaporated to obtain highly pure
mercury metal as distillate. This method
is known as distillation.
17 Which of the following statements,
about the advantage of roasting of
sulphide ore before reduction is
not true?
[CBSE AIPMT 2007]
(a) Carbon and hydrogen are suitable
reducing agents for metal
sulphides
(b) The ∆f G ° of the sulphide is greater
than those for CS2 and H2S
(c) The ∆f G ° is negative for roasting of
sulphide ore to oxide
(d) Roasting of the sulphide to the oxide
is thermodynamically feasible
Ans. (a)
Carbon and hydrogen are not suitable
reducing agents for metal sulphides.
18 Which one of the following
elements constitutes a major
impurity in pig iron?
[CBSE AIPMT 1998]
(a) Silicon
(c) Sulphur
Ans. (d)
(b) Oxygen
(d) Graphite
Graphite produces impurity in pig iron.
Pig iron contains 2.5 to 5.0% of carbon.
19 The reaction of H2O 2 with hydrogen
sulphide is an example of …………
reaction.
[CBSE AIPMT 1988]
(a) addition
(c) reduction
Ans. (b)
(b) oxidation
(d) acidic
When H2O2 is reacted with hydrogen
sulphide (H2S), it form S and water. In this
reaction H2S is oxidised to sulphur. H2O2
act as oxidising agent.
H2O2 + H2S → S + 2 H2O
TOPIC 3
Refining of Metals
20 Which one of the following
methods can be used to obtain
highly pure metal which is liquid at
room temperature?
[NEET 2021]
(a) Electrolysis
(b) Chromatography
(c) Distillation
(d) Zone refining
21 Match the elements in Column I
with methods of purification in
Column II.
[NEET (Oct.) 2020]
Column I
Column II
A. Boron
I. van-Arkel method
B. Tin
II. Mond’s process
C. Zirconium
III. Liquation
D. Nickel
IV. Zone refining
A B C
(a) IV III I
(c) II I IV
D
II
III
A B
(b) IV III
(d) III IV
C D
II I
I II
22 Identify the correct statement from
the following:
[NEET (Sep.) 2020]
(a) Blister copper has blistered
appearance due to evolution of CO2 .
(b) Vapour phase refining is carried out
for nickel by van Arkel method.
(c) Pig iron can be moulded into a variety
of shapes.
(d) Wrought iron is impure iron with 4%
carbon.
Ans. (c)
(a) Matte on auto-reduction produces
blister copper, which has blistered
appearance due to evolution ofSO2
(not CO2 ).
Cu2S + 2Cu2O →
6 Cu
+ SO2 ↑
Blister copper
(b) Vapour phase refining is carried out
for Ni by Mond’s process (not by van
Arkel method).
°C
Ni 4CO
→ Ni(CO) 4 200

→ Ni
− 4CO
Column II
A. Cyanide process
1.
Ultrapure Ge
B. Froth floatation
process
2. Dressing of ZnS
C. Electrolytic
reduction
3. Extraction of Al
D. Zone refining
4. Extraction of Au
5. Purification of Ni
C
1
5
D
5
1
A
(b) 1
(d) 4
B
2
2
C D
3 4
3 1
Ans. (d)
(A) Boron (and Ge, Si, Ga, In) get purified
by zone refining (IV).
(B) Tin (and Bi, Pb) get purified by
Liquation (III).
(C) Zirconium (and Ti, Hf) get purified by
van Arkel method (I).
(D) Nickel is purified by Mond’s process
(II).
Hence, option (a) is correctly match.
Impure 80 °C
Column I
Codes
A B
(a) 2 3
(c) 3 4
Ans. (a)
Matte
23 Match items of Column I with the
items of Column II and assign the
correct code. [NEET 2016, Phase I]
Pure
(c) Pig iron can be moulded into variety
of shapes.
(d) Wrought iron the purest form of iron
which contains 0.2-0.5% carbon (not
4%).
So, option (c) is correct.
A – 4, B – 2, C – 3, D – 1
• Cyanide process It is a metallurgical
technique for extracting Au (gold)
from low grade ore by converting the
Au to a water-soluble coordination
complex.
• Froth floatation process This
process is used for dressing of
sulphide ore, i.e. ZnS.
• Electrolytic reduction This process
is used for extraction of Al which is
carried out in a steel tank lined inside
with graphite. Here, graphite serves
as cathode. The electrolyte consists
of alumina dissolved in fused cryolite
(Na3 AlF6 ) and fluorspar (CaF2 ).
• Zone refining This process is used
for ultra pure Ge element. An ingot of
Ge is first purified by zone refining.
Then a small amount of antimony is
placed in the molten zone which is
passed through the pure Ge with the
proper choice of rate of heating and
other variables.
24 Which of the following pairs of
metals is purified by van-Arkel
method?
[CBSE AIPMT 2011]
(a) Zr and Ti
(b) Ag and Au
(c) Ni and Fe
(d) Ga and In
Ans. (a)
Zr and Ti are purified by van-Arkel
method.
600 °C
1800 °C
Zr + 2 I2 → Zr I4 → Zr + 2 I2
Impure
Pure
112
NEET Chapterwise Topicwise Chemistry
This method is useful for removing all
the oxygen and nitrogen present in the
form of impurity in certain metals like Zr
and Ti.
25 The method of zone refining of
metals is based on the principle of
[CBSE AIPMT 2003]
(a) greater noble character of the solid
metal than that of the impurity
(b) greater solubility of the impurity in
the molten state than in the solid
(c) greater mobility of the pure metal
than that of impurity
(d) higher melting point of the impurity
than that of the pure metal
of the impurity in the molten state than in
the solid.
Elements which are used as
semiconductors like Si, Ge, Ga etc, are
refined by this method. Gallium arsenide
and indium antimonide (also used as
semiconductor) are also refined by this
method.
26 Purification of aluminium by
electrolytic refining is known as
[CBSE AIPMT 1999]
Ans. (b)
(a) Hall’s process
(b) Baeyer's process
(c) Hoope's process
(d) Serpeck's process
Ans. (c)
The method of zone refining of metals is
based on the principle of greater solubility
Purification of aluminium by electrolytic
refining is known as Hoope’s process. By
this process 99.9% pure aluminium
metal is obtained.
The cell used for this process consist of
three layers. In this cell pure Al acts as
cathode and anode is made up of impure
Al.
27 Elemental silicon to be used as a
semiconductor is purified by
[CBSE AIPMT 1996]
(a) heating under vacuum
(b) floatation
(c) zone refining
(d) electrolysis
Ans. (c)
Zone refining is used for metals which
are required in very high purity.
Semiconductor grade silicon is purified
by this method (Si, Ge).
15
Hydrogen
TOPIC 1
Preparation and
Properties of Hydrogen
01 Tritium, a radioactive isotope of
hydrogen, emits which of the
following particles?
[NEET 2021]
(a) Beta (β − )
(c) Gamma (γ)
Ans. (a)
Protium is the most common isotopes of
hydrogen with an abundance of 99.98%.
03 ‘‘Metals are usually not found as
nitrates in their ores’’.
[CBSE AIPMT 2015]
Out of the following two (I and II)
reasons which is/are true for the
above observation?
I. Metal nitrates are highly
unstable.
II. Metal nitrates are highly soluble
in water.
(b) Alpha (α)
(d) Neutron (n)
Tritium ( 31 H) is an isotope of hydrogen. It
is a radioactive isotope and decays by
emitting beta-particle ( β − ) to form 23He+ .
3
→ 23He+ +
e−
1H
Tritium
(a) I and II are true
(b) I and II are false
(c) I is false but II is true
(d) I is true but II is false
Ans. (c)
β-particle (β – )
+ 18.6 keV
02 Which of the following statements
about hydrogen is incorrect?
KNO3
Ans. (c,d)
• For ionic salts, hydrogen never
behaves as cation, but behaves as
anion (H− ).
• H3 O + exists freely in solution.
• Dihydrogen acts as a reducing agent.
• Hydrogen has three isotopes.
• Protium (11H)
• Deuterium (21 H)
• Tritium (31 H)
+
−
3
eK (aq) + NO (aq)
The nitrate anion has three equivalent
oxygen surrounding a central nitrogen
atom.
This tends to spread the single negative
charge and make it easier for water
(using hydrogen bonds) to separate the
ions in solution.
N
–
O
–
–
O
–
O
–
O
O
O
N
N
O
04 Match the following and identify the
correct option. [NEET (Sep.) 2020]
A. CO(g ) + H2 (g )
(i) Mg(HCO 3) 2 +
B. Temporary
hardness of
water
(ii) An electron
deficient
hydride
C. B2H6
(iii) Synthesis gas
D. H2O 2
(iv) Non-planar
structure
Ca(HCO 3) 2
A B C D
(a) (iii) (ii) (i) (iv)
(c) (i) (ii) (ii) (iv)
Metals are usually not found as nitrates
in their ores, because metal nitrates are
highly soluble in water.
For example,KNO3 (salt peter) would be
classified as completely soluble.
Thus, KNO3 could be expected to
dissociate completely in aqueous
solution to give K+ and NO−3 ions.
[NEET 2016, Phase I]
(a) Hydrogen never acts as cation in
ionic salts
(b) Hydronium ion, H 3O + exists freely in
solution
(c) Dihydrogen does not act as a
reducing agent
(d) Hydrogen has three isotopes of
which tritium is the most common
TOPIC 2
Hydrides and Water
O
–
O
A B C D
(b) (iii) (iv) (ii) (i)
(d) (iii) (i) (ii) (iv)
Ans. (d)
So, correct combinations are:
(a)—(iii), (b)—(i), (c)—(ii), (d)—(iv)
(a) Water gas (CO + H2 ) is also known as
synthesis gas (iii).
(b) Temporary hardness of water is due
to the presence ofHCO−3 radicals of
Mg2 + and/or Ca2 + ion(s) (i).
(c) Boron atoms of B2H6 are
sp3-hybridised but due to the
presence of twoB H B bonds
(3C2e − ), B2H6 becomes as an
electron deficient hydride (ii).
(d) H2O2 has a book-shaped non-planar
structure (iv).
H
O
O
H
Non-planar structure
114
NEET Chapterwise Topicwise Chemistry
TOPIC 3
Heavy Water and H2O2
05 The method used to remove
temporary hardness of water is
[NEET (National) 2019]
(a) Clark’s method
(b) ion-exchange method
(c) synthetic resins method
(d) Calgon’s method.
Ans. (a)
Temporary hardness in water is due to
presence of magnesium and calcium
hydrogen carbonates.
Temporary hardness in water can be
removed by Clark’s method. In this
method calculated amount of lime is
added to hard water. It precipitates out
calcium carbonate and magnesium
hydroxide which can be filtered off.
Ca(HCO3)2 + Ca(OH)2 →
2CaCO3 ↓ + 2H2O
Mg(HCO3)2 + 2Ca(OH)2 →
2CaCO3 ↓ + Mg(OH)2 ↓ + 2H2O
Besides this, temporary hardness can
also be removed by boiling. All the other
given methods are used to remove
permanent hardness of water.
06 The structure of H2O 2 is
(c) spherical
(d) linear
Ans. (b)
H2O2 shows non-planar structure. It has a
half opened book like structure in which
the two O—H groups lie on the two pages
of the book.
0.97Å
O
1.48 Å
101.5°
H
H
97°
O
94°
[CBSE AIPMT 1999]
(a) planar
(b) non-planar
O—O single bond distance is 1.48 Å.
16
s-Block Elements
TOPIC 1
Group 1 Elements
(Alkali Metals)
01 The following metal ion activates
many enzymes, participates in the
oxidation of glucose to produce
ATP and with Na, is responsible for
the transmission of nerve signals
[NEET (Sep.) 2020]
(a) copper
(c) potassium
Ans. (c)
(b) calcium
(d) iron
The ionic-gradients between two sides
of a cell membrane, is operated by
Na K pump and consumes one-third of
the ATP used by a resting animal.
Potassium ion (K+ ) present in synapse
helps transmission of nerve signals from
one neuron to another.
02 Ionic mobility of which of the
following alkali metal ions is lowest
when aqueous solution of their
salts are put under an electric
field?
[NEET 2017]
(a) Na
(c) Rb
Ans. (d)
(b) K
(d) Li
Key concept More the extent of
hydration, lesser is the ionic mobility .
In all the alkali metals,Li+ ion is smallest.
Thus, extent of hydration is maximum in
Li+ ion.
i.e. the dissolution ofLi+ in water occurs
and get hydrated. Smaller the size of a
cation, greater is the extent of hydration
and lesser is the ionic mobility.
03 The alkali metals form salt like
hydrides by the direct synthesis at
elevated temperature. The thermal
stability of these hydrides
decreases in which of the following
orders?
[CBSE AIPMT 2008]
(a) CsH > RbH > KH > NaH > LiH
(b) KH > NaH > LiH > CsH > RbH
(c) NaH > LiH > KH > RbH > CsH
(d) LiH > NaH > KH > RbH > CsH
Ans. (d)
As the size of the alkali metal cation
increases, thermal stability of their
hydrides decreases.
Hence, the correct order of thermal
stability of alkali metal hydrides is
LiH > NaH > KH > RbH > CsH
04 The sequence of ionic mobility in
aqueous solution is
[CBSE AIPMT 2008]
(a) K + > Na+ > Rb + > Cs+
(b) Cs+ > Rb + > K + > Na+
(c) Rb + > K + > Cs+ > Na+
(d) Na+ > K + > Rb + > Cs+
Ans. (b)
The smaller the size of the ion, the
greater is the degree of hydration, thus
degree of hydration is highest forLi+ and
lowest for Cs+ . Thus, Li+ holds more
water molecules in its hydration sphere
and becomes largest in size among alkali
metals and Cs+ . ion hold least number of
water molecules.
Hence, ionic mobility is highest for Cs+ .
(due to its smallest size in aqueous
solution) and lowest for Li+ Here the
lowest is for Na+ . Thus, the order of ionic
mobility in aqueous solution is
Cs+ > Rb+ > K+ > Na+
05 The correct order of the mobility of
the alkali metal ions in aqueous
solution is
[CBSE AIPMT 2006]
(a) Li+ > Na+ > K + > Rb +
(b) Na+ > K + > Rb + > Li+
(c) K + > Rb + > Na+ > Li+
(d) Rb + > K + > Na+ > Li+
Ans. (d)
The correct order of the mobility of the
alkali metal ions in aqueous solution is
Rb+ > K+ > Na+ > Li+ due to following
order of hydration energy of these ions
Li+ > Na+ > K+ > Rb+ and as the hydration
of ion increases, mobility decreases.
Hydration enthalpy
1
∝
Size of cation
06 Sodium is made by the electrolysis
of a molten mixture of about 40%
NaCl and 60% CaCl 2 because
[CBSE AIPMT 1995]
(a) Ca2 + can reduce NaCl to Na
(b) Ca2 + can displace Na from NaCl
(c) CaCl2 helps in conduction of
electricity
(d) this mixture has a lower melting
point than NaCl
Ans. (d)
The melting point of sodium chloride is
high, so to reduce the melting point of
NaCl some CaCl2 is added to the
electrolytic mixture.
07 A certain compound X when
treated with copper sulphate
solution yields a brown precipitate.
On adding hypo solution, the
precipitate turns white. The
compound X is [CBSE AIPMT 1994]
(a) K2CO 3
(c) KBr
Ans. (b)
(b) KI
(d) K 3PO 4
When potassium iodide is reacted with
CuSO4 , it gives iodine gas which is brown
colour. This iodine reacted with sodium
thiosulphate and form white precipitate
of sodium tetrathionate.
2CuSO4 + 4 KI → K2SO4 + I2 ↑ + 2CuI
2Na2S2O3 + I2 → Na2S 4O6 ↓ + 2NaI
116
NEET Chapterwise Topicwise Chemistry
08 Which of the following is known as
fusion mixture? [CBSE AIPMT 1994]
(a) Mixture of Na2CO 3 + NaHCO 3
(b) Na2CO 3 ⋅10H2O
(c) Mixture of K2CO 3 + Na2CO 3
(d) NaHCO 3
Ans. (c)
A mixture of Na2 CO3 and K2 CO3 is used as
a fusion mixture.
09 Which of the following elements is
extracted commercially by the
electrolysis of an aqueous solution
of its compound?
[CBSE AIPMT 1993]
(a) Cl
(b) Br
Ans. (d)
(c) Al
(d) Na
Sodium is prepared by electrolysis of
molten NaCl as
Electrolysis
2NaCl → 2Na + Cl2
At cathode : 2Cl – + 2 e − → Cl2
At anode : 2Na → 2Na+ + 2e −
10 Which of the following has largest
size?
[CBSE AIPMT1993]
(a) Na
(b) Na+
(c) Na–
(d) Can’t be predicted
Ans. (c)
Na− has largest size because anion is
always larger than neutral atom and
cation is smaller than neutral atom. So
the order is given as
Na− > Na > Na+
Anion > Parental atom > Cation
11 Washing soda has formula
[CBSE AIPMT 1990]
(a) Na2CO 3 ⋅ 7H2O
(c) Na2CO 3 ⋅ 3H2O
Ans. (b)
(b) Na2CO 3 ⋅10H2O
(d) Na2CO 3
Washing soda is chemicaly named as
sodium carbonate decahydrate, so its
formula is Na2 CO3 ⋅ 10H2O.
12 Which one of the following
properties of alkali metals
increases in magnitude as the
atomic number rises?
[CBSE AIPMT 1989]
(a) Ionic radius
(b) Melting point
(c) Electronegativity
(d) First ionisation energy
Ans. (a)
The ionic radii of alkali metal increases
as the atomic number increases when
we move from top to bottom because on
moving down the group, there is a
increase in the number of shells and
therefore, ionic radii increases.
TOPIC 2
Group 2 Elements
(Alkaline Earth Metals)
13 Among the following alkaline earth
metal halides, one which is
covalent and soluble in organic
solvents is
[NEET 2021]
(a) calcium chloride
(b) strontium chloride
(c) magnesium chloride
(d) beryllium chloride
Ans. (d)
On moving down the group, the ionic size
of alkaline earth metals increases. So,
due to small size ofBe2 + ion, Be has
highest polarising power [ability to attract
the electron cloud of anion (Cl − )].
∴ BeCl2 is more covalent than other
alkaline earth metal halides.
Organic molecules are covalent in
nature. Rule for solubility is “Like
dissolves like”. So,BeCl2 is soluble in
organic solvents as both are covalent in
nature.
14 The structures of beryllium
chloride in solid state and vapour
phase, are
[NEET 2021]
(a) chain and dimer, respectively
(b) linear in both
(c) dimer and linear, respectively
(d) chain in both
Ans. (a)
Beryllium chloride (BeCl2 ) is an electron
deficient compound. So, it does not exist
in its monomer form.
In solid state and vapour phase,BeCl2
exists in chain and dimer forms
respectively.
Cl
Cl
Cl
Be
Be
Be
Cl
Cl
Cl
Chain form
Cl
Cl
Be
Be
Cl
Dimer form
Cl
15 HCl was passed through a solution
of CaCl 2 , MgCl 2 and NaCl. Which of
the following compound(s)
crystallise(s)?
[NEET (Sep.) 2020]
(a) Only NaCl
(b) Only MgCl2
(c) NaCl, MgCl2 and CaCl2
(d) Both MgCl2 and CaCl2
Ans. (a)
Let us explain the crystallisation process
by a flow-sheet diagram.
Aqueous solution : Filtration
CaCl2+MgCl2+NaCl
Insoluble
Filtrate
impurities
get removed
Crystals of
pure NaCl(s)
Excess of
HCl(g) is
passed
Aqueous solution
of CaCl2 and MgCl2
as these are more
soluble than NaCl
16 What is the role of gypsum,
CaSO 4 ⋅ 2H2O is setting of cement?
Identify the correct option from the
following.
[NEET (Oct.) 2020]
(a) to fasten the setting process
(b) to provide water molecules for
hydration process
(c) to help to remove water molecules
(d) to slow down the setting process
Ans. (d)
Gypsum (CaSO4 ⋅2H2O) is present in
cement (Portland cement) by a mass of
2-3%.
Gypsum slow down the process of
setting of cement so that it gets
sufficiently hardened.
17 Which of the following is an
amphoteric hydroxide?
[NEET (National) 2019]
(a) Ca(OH)2
(c) Be(OH)2
Ans. (c)
(b) Mg(OH)2
(d) Sr(OH)2
Be(OH)2 is amphoteric in nature as it
reacts with acid and alkali both as :
Be(OH)2 + 2HCl → BeCl2 + 2H2O
Be(OH)2 + 2NaOH → Na2 [Be(OH) 4 ]
This amphoteric nature of Be is due to
small size of Be. The other hydroxides of
alkaline earth metals are basic in nature.
18 The product obtained as a result of
a reaction of nitrogen with CaC 2 is
[NEET 2016, Phase I]
(a) CaCN
(c) Ca2 CN
(b) CaCN 3
(d) Ca(CN)2
117
s-Block Elements
Ans. (*)
When calcium carbide (CaC2 ) reacts with
nitrogen (N2 ) under high temperature, it
forms calcium cyanamide which is also
called nitrolim.
High
CaC2 + N2 →
temperature
CaCN2
Calcium cyanamide
+ C
Hence, option (d) should be CaCN2
instead of Ca(CN)2 . Thus no option is
correct.
19 Which of the following statements
is false?
[NEET 2016, Phase I]
2+
(a) Ca ions are important in blood
clotting
(b) Ca2 + ions are not important in
maintaining the regular beating of
the heart
(c) Mg2 + ions are important in the green
parts of plants
(d) Mg2 + ions form a complex with ATP
Ans. (b)
• Ca2 + ions are very important factor in
blood clotting.
• Ca2 + ions are very important for
maintaining the regular heart beating.
• Mg2 + ions is present in the green
parts of plants i.e., chlorophyll.
• Mg2 + can form a complex with ATP.
20 In context with beryllium, which
one of the following statements is
incorrect?
[NEET 2016, Phase II]
(a)
(b)
(c)
(d)
It is rendered passive by nitric acid
It forms Be2 C
Its salts rarely hydrolyse
Its hydride is electron-deficient and
polymeric
Ans. (c)
Beryllium salts are covalent in nature
because of very small size ofBe2 + ion
and its high polarising power, so it is
easily hydrolysed.
e.g. BeCl2 + 2H2O → Be(OH) 2 + 2HCl
21 Solubility of the alkaline earth's
metal sulphates in water decreases
in the sequence [CBSE AIPMT 2015]
(a) Mg > Ca > Sr > Ba
(b) Ca > Sr > Ba > Mg
(c) Sr > Ca > Mg > Ba
(d) Ba > Mg > Sr > Ca
Ans. (a)
Solubility of the sulphates. The
sulphates becomes less soluble as you
go down the group i.e.
Mg > Ca > Sr > Ba
The magnitude of the lattice energy
remains almost constant as the size of
the sulphate ion is so big that small
increase in the size of the cation from Be
to Ba does not make any difference.
However, the hydration energy
decreases fromBe2 + to Ba2 + appreciably
as the size of the cation increases down
the group. The significantly high
solubility of MgSO4 is due to high
enthalpy of solvation of the smallerMg2 +
ions.
22 On heating which of the following
releases CO 2 most easily?
[CBSE AIPMT 2015]
(a) K2CO 3
(c) MgCO 3
Ans. (c)
(b) Na2CO 3
(d) CaCO 3
Order of thermal stability is
K2 CO3 > Na2 CO3 > CaCO3 > MgCO3
Hence,MgCO3 releases CO2 most easily
MgCO3 →
∆ MgO + CO2
23 Which one of the following is
present as an active ingredient in
bleaching powder for bleaching
action?
[CBSE AIPMT 2011]
(a) Ca(OCl)2
(c) CaCl2
Ans. (a)
(b) CaO2Cl2
(d) CaOCl2
Ca(OCl)2 , calcium hypochlorite is the
active ingredient in bleaching powder
which releases chlorine.
24 Equimolar solutions of the
following were prepared in water
separately. Which one of the
solutions will record the highest
pH?
[CBSE AIPMT 2008]
(a) SrCl2
(c) MgCl2
Ans. (b)
(b) BaCl2
(d) CaCl2
All salts are soluble in water and give
strong acid and weak base
SrCl2 + 2H2O → Sr(OH) 2 + 2HCl
BaCl2 + 2H2O → Ba(OH) 2 + 2HCl
MgCl2 + 2H2O → Mg(OH) 2 + 2HCl
CaCl2 + 2H2O → Ca(OH) 2 + 2HCl
The basic nature of alkaline earth metals
generally increases from Be to Ra. Thus,
the order of basic nature of these
hydroxides is
Mg(OH)2 < Ca(OH)2 < Sr(OH) 2 < Ba(OH) 2
Hence, pH is highest forBaCl2 . (As pH
increases with basic nature)
25 In which of the following the
hydration energy is higher than the
lattice energy? [CBSE AIPMT 2007]
(a) BaSO 4
(c) RaSO 4
Ans. (b)
(b) MgSO 4
(d) SrSO 4
Hydration energy of sulphate decreases
from top to bottom in II group.Mg2 + is
smaller than other given ions of II group,
so Mg2 + is readily hydrated.MgSO4 has
higher hydration energy than lattice
energy.
26 The correct order of increasing
thermal stability of K 2CO 3 , MgCO 3 ,
CaCO 3 and BeCO 3 is
[CBSE AIPMT 2007]
(a) BeCO 3 < MgCO 3 < K2CO 3 < CaCO 3
(b) BeCO 3 < MgCO 3 < CaCO 3 < K2CO 3
(c) MgCO 3 < BeCO 3 < CaCO 3 < K2CO 3
(d) K2CO 3 < MgCO 3 < CaCO 3 < BeCO 3
Ans. (b)
Thermal stability of carbonates
increases in a group as we move from
top to bottom and decreases in a period
as we move from left to right, so the
correct order of thermal stability of
given carbonates is
BeCO3 < MgCO3 < CaCO3 < K2 CO3
Be, Mg and Ca are present in second
group and K is present in first group.
27 Which one is the correct statement
with reference to the solubility of
MgSO 4 in water? [CBSE AIPMT 1996]
(a) SO24− ions mainly contributes towards
hydration energy
(b) Sizes of Mg2 + and SO24− are similar
(c) Hydration energy of MgSO4 is higher
in comparison to its lattice energy
(d) Ionic potential (charge/radius ratio)
of Mg2 + is very low
Ans. (c)
MgSO4 is soluble in water because it
have hydration energy more than lattice
energy.
Mg2 + ions mainly contributes towards
hydration energy
size of SO24− ion is greation thanMg2 +
ions.
28 Identify the correct statement.
[CBSE AIPMT 1995]
(a) Gypsum is obtained by heating
plaster of Paris
(b) Plaster of Paris can be obtained by
hydration of gypsum
118
NEET Chapterwise Topicwise Chemistry
(c) Plaster of Paris is obtained by partial
oxidation of gypsum
(d) Gypsum contains a lower percentage
of calcium than plaster of Paris
Ans. (d)
The formula of gypsum is CaSO4 ⋅2H2O
and that of plaster of Paris is
(CaSO4 )2 ⋅2H2O, so the percentage of Ca
in plaster of Paris is more than gypsum.
29 Which of the following statement is
false?
[CBSE AIPMT 1994]
(a) Strontium decomposes water readily
than beryllium
(b) BaCO3 melts at a higher temperature
than CaCO3
(c) Barium hydroxide is more soluble in
water than Mg(OH)2
(d) Beryllium hydroxide is more basic
than barium hydroxide
Ans. (d)
The size of beryllium is very small and
the hydroxide formed by it, are
amphoteric in nature, so its hydroxide is
less basic than barium hydroxide.
(c) Ca and CaH2
(d) Ba and BaO2
Ans. (c)
36 Which of the following metal
evolves hydrogen on reacting with
cold dilute HNO 3 ?
Ca and CaH2 gives H2 and calcium
hydroxide when reacted with water.
Ca + H2O → Ca(OH)2 + H2 ↑
CaH2 + 2H2O → Ca(OH) 2 + 2H2 ↑
33 Which one of the following has
minimum value of size of
cation/anion ratio?
[CBSE AIPMT 1993]
(a) NaCl
(c) MgCl2
Ans. (c)
(b) KCl
(d) CaF2
The size of Mg2 + is minimum amongNa+ ,
Ca2 + , K+ and Mg2 + and size of Cl − is more
than F − , so the ratio of size of cation and
anion are minimum inMgCl2 .
34 When chlorine is passed over dry
slaked lime at room temperature,
the main reaction product is
[CBSE AIPMT 1992]
30 Which of the following metal ions
play an important role in muscle
contraction?
[CBSE AIPMT 1994]
(b) Na+
(a) K +
Ans. (d)
(c) Mg2+ (d) Ca2+
Calcium ion (Ca2 + ) play an important role
in muscle contraction.
31 The formula for calcium chlorite is
[CBSE AIPMT 1994]
(a) Ca(ClO 4 )2
(c) CaClO2
Ans. (d)
(b) Ca(ClO 3)2
(d) Ca(ClO2 )2
The formula of calcium chlorite is
Ca(ClO2 )2 .
32 All the following substances react
with water. The pair that gives the
same gaseous product is
[CBSE AIPMT 1994]
(a) K and CO2
(b) Na and Na2O2
(a) Ca(ClO2 )2
(c) CaOCl2
Ans. (c)
(b) CaCl2
(d) Ca(OCl)2
When chlorine gas is passed over dry
slaked lime, it form bleaching powder i.e.
CaOCl2 .
Ca(OH)2 + Cl2 → CaOCl2 + H2O
35 Compared with the alkaline earth
metals, the alkali metals exhibit
[CBSE AIPMT 1990]
(a) smaller ionic radii
(b) higher boiling points
(c) greater hardness
(d) lower ionisation energies
Ans. (d)
Alkali metals have the lowest ionisation
enthalpy in each period because alkali
metals are largest in their respective
periods and therefore, the valency
electrons are loosely held by the
nucleus.
[CBSE AIPMT 1989]
(a) Mg
(c) Fe
Ans. (a)
(b) Al
(d) Cu
Magnesium react with cold and dilute
nitric acid to form hydrogen.
Mg + 2HNO3 → Mg(NO3)2 + H2
37 Which one of the following atoms
will have the smallest size?
[CBSE AIPMT 1989]
(a) Mg
(c) Be
Ans. (c)
(b) Na
(d) Li
Atomic size increases down the group
and decreases in period from left to
right. So, Be is smallest in size in these
elements.
Alkali Alkaline earth
metal
metal
Group 1 Group 2
Li
Be
Na
Mg
Size increases
Size decreases
38 Bleaching powder is obtained by
the action of chlorine gas and
[CBSE AIPMT 1988]
(a) dilute solution of Ca(OH)2
(b) concentrated solution of Ca(OH)2
(c) dry CaO
(d) dry slaked lime
Ans. (d)
Bleaching powder is prepared by passing
chlorine gas over dry slaked lime Ca(OH)2
by Hesenclaver method or by Bachmann
method.
Ca(OH)2 + Cl2 → CaOCl2 + H2O
17
p-Block Elements
TOPIC 1
Group 13 Elements
01 Which one of the following
elements is unable to form MF63−
ion?
[NEET 2018]
(a) B
(c) Ga
Ans. (a)
(b) Al
(d) In
Boron belongs to 2nd period of the
periodic table with electronic
configuration 1s 2 , 2s 2 2p1. It does not
have vacant d-orbitals, thus cannot
increase its covalency above four.
Therefore, boron (B) cannot form MF63−
ion. In contrast, aluminium (Al), gallium
(Ga), indium (In) have the vacant
3d-orbitals, thus can increase their
covalence above four and form MF63− ion.
02 Boric acid is an acid because its
molecule
[NEET 2016, Phase II]
+
(a) contains replaceableH ion
(b) gives up a proton
(c) accepts OH − from water releasing
proton
(d) combines with proton from water
molecule
Ans. (c)
Boric acid can be considered as an acid
because its molecule acceptsOH− from
water, releasing proton.
H3BO3 + H2O q
Acid
Base
B(OH) –4 +
Conjugate
base
H+
Conjugate
acid
03 AlF 3 is soluble in HF only in
presence of KF. It is due to the
formation of [NEET 2016, Phase II]
(a) K 3 [AIF3H3]
(c) AIH 3
Ans. (b)
Key Idea Al 3+ shows maximum
coordination number 6, thus it will form
AlF63− .
AlF3 forms K3 [AlF6 ] when dissolved in HF
in the presence of KF as shown below:
HF
AlF3 + 3KF → K3 [AlF6 ]
04 The stability of +1 oxidation state
among Al, Ga, In and Tl increases
in the sequence [CBSE AIPMT 2015]
(a) Ga < ln < Al < Tl (b) Al < Ga < ln < Tl
(c) Tl< ln < Ga < Al (d) ln < Tl < Ga < Al
Ans. (b)
(c) B 4C
(d) B2H6
Boron nitride (BN)x resembles with
graphite in structure as shown below
N
B
N
B
B
N
C
N
C
B
C
N
B
Boron nitride
C
C
C
N
C
C
C
B
N
C
C
C
C
Graphite
(b) H2O
(d) CH4
Key Idea Electron deficient molecules
behave as Lewis acid.
Among the given molecules, only
diborane is electron deficient, i.e. does
not have complete octet. Thus, it acts as
a Lewis acid.
NH3 and H2O being electron rich
molecules behave as Lewis base.
07 The tendency of BF3 , BCl 3 and
BBr 3 behave as Lewis acid
decreases in the sequence
[CBSE AIPMT 2009]
05 Which of the following structure is
similar to graphite?
[NEET 2013]
(a) BN
(b) B
Ans. (a)
(a) NH3
(c) B2H6
Ans. (c)
(a) BCl3 > BF3 > BBr3
Al < Ga < In < Tl
This is due to inert pair effect or
tendency of ns 2 electrons do not
participate in bond formation. This
tendency decreases on moving down the
group.
B
Remember In the given options to the
question, (a), (b) and (c) are correct as all
of these sentences have more or less
similar meaning but here (c) option is the
most appropriate one as it gives
complete explanation of the fact that
how boric acid can be combined with an
acid.
(b) K 3 [AIF6]
(d) K [AIF3H]
06 Which one of the following
molecular hydrides acts as a Lewis
acid?
[CBSE AIPMT 2010]
(b) BBr3 > BCl3 > BF3
(c) BBr3 > BF3 > BCl3
(d) BF3 > BCl 3 > BBr3
Ans. (b)
As the size of halogen atom increases,
the acidic strength of boron halides
increases. Thus,BF3 is the weakest
Lewis acid. This is because of the
pπ − pπ back bonding between the fully
filled unutilised2p-orbitals of F and
vacant 2p-orbitals of boron which makes
BF3 less electron deficient. Such back
donation is not possible in case ofBCl 3 or
BBr3 due to larger energy difference
between their orbitals. Thus, these are
more electron deficient. Since on
moving down the group the energy
difference increases, the Lewis acid
character also increases. Thus, the
tendency to behave as Lewis acid
follows the order
BBr3 > BCl 3 > BF3
120
NEET Chapterwise Topicwise Chemistry
08 The stability of +1 oxidation state
increases in the sequence
[CBSE AIPMT 2009]
(a) Al < Ga < In < Tl (b) Tl < In < Ga < Al
(c) In < Tl < Ga < Al (d) Ga < In < Al < Tl
Ans. (a)
The given elements belong to third
group. These elements mainly exhibit +3
and +1 oxidation states. As we know, the
stability of lower oxidation state,
increases on moving down a group due
to inert pair effect. Thus, the sequence
of stability of +1 state is
Al < Ga < In < Tl
09 Al 2O3 can be converted into
anhydrous AlCl 3 by heating
[CBSE AIPMT 2006]
(a) Al2O3 with HCl gas
(b) Al2O3 with NaCl in solid state
(c) a mixture of Al2O3 and carbon in dry
Cl2 gas
(d) Al2O3 with Cl2 gas
3Cl2 → Al2 Cl 6
Hot and dry
AnhyAlCl3
(b) C2H6
(d) SiH4
97°
H
Å
B
97°
H
122°
3Å
1.3
1.77Å
In it two electrons of aB H bond are
involved in formation of three centre
bond, these bonds are represented as
dotted lines.
H
Ans. (a)
In diborane 3 centred 2 electron bond is
present.
H
H
H
B
B
H
H
H
Diborane
(3c -2 e bond present)
15 Which of the following statements
about H3BO 3 is not correct?
[CBSE AIPMT 1994]
(a) It is a strong tribasic acid
(b) It is prepared by acidifying an
aqueous solution of borax
(c) It has a layer structure in which
planar BO3 units are joined by
hydrogen bonds
(d) It does not act as proton donor but
acts as a Lewis acid by accepting
hydroxyl ion
Ans. (a)
Boric acid (H3BO3) is a weak monobasic
acid withK a = 1.0 × 10 −9 .It may be noted
that boric acid does not act as a protonic
acid (i.e. proton donor) but behaves as a
Lewis acid by accepting a pair of
electrons from OH− ion.
B(OH) 3 + 2H — O — H →
[B(OH) 4 ] – + H3O+
–10H2O
→
∆ 2NaBO2 + B2O3
Cu(BO2 ) 2
13 Among the following the electron
deficient compound is
[CBSE AIPMT 2000]
H
1.19
In borax bead test the coloured meta
borates are formed by transition metal
salts.
Cupric metaborate
(Blue bead)
B2H6 is electron deficient molecule
because boron atom has three half-filled
orbitals in excited state. The structure of
B2H6 is represented as follows:
H
(b) Meta borate
(d) Tetra borate
B2O3 + CuO →
[CBSE AIPMT 2005]
B
(a) Ortho borate
(c) Double oxide
Ans. (b)
∆
10 Which of the following is the
electron deficient molecule?
122°
Zeolites are aluminosilicates having
three dimensional open structure in
which four or six membered rings
predominates. Thus, due to open chain
structure, they have cavities and can
take up water and other small molecules.
Na2B4O7 ⋅ 10H2O → Na2B4O7
NOTE Anhydrous AlCl 3 exists in the form
of dimer as Al2 Cl 6
H
Ans. (d)
[CBSE AIPMT 2002]
Al2O3 may be converted into anhyd. AlCl 3
by heating a mixture of Al2O3 and carbon
in dry chlorine.
(a) B2H6
(c) PH3
Ans. (a)
(a) They are used as cation exchangers
(b) They have open structure which
enables them to take up small
molecules
(c) Zeolites are aluminosilicates having
three dimensional network
(d) Some of the SiO4–
4 units are replaced
by AlO54 – and AlO96 – ions in zeolites
12 In borax bead test which
compound is formed?
Ans. (c)
Al2O3 + 3C +
11 Which one of the following
statements about the zeolites is
false?
[CBSE AIPMT 2004]
(a) BCl3 (b) CCl4
Ans. (a)
(c) PCl5
(d) BeCl2

Cl
 boron contains six
In BCl 3  Cl —B



Cl
electrons in its valence shell so, it is
capable to accommodate one pair of
electrons. Hence, it acts as Lewis acid or
electron deficient (incomplete octate)
compound. As we know that Lewis acids
are the substances having a tendency to
accept a pair of electron.
14 Which of the following compound
has a 3-centre bond?
[CBSE AIPMT 1996]
(a) Diborane
(b) CO2
(c) Boron trifluoride
(d) Ammonia
TOPIC 2
Group 14 Elements
16 Which of the following oxide is
amphoteric in nature?
[NEET (Oct.) 2020]
(a) SnO2
(c) GeO2
Ans. (a)
(b) SiO2
(d) CO2
Nature of dioxides (MO2 ) of group 14
elements can be represented as.
CO2
SiO2
GeO2
Acidic oxides
SnO2
PbO2
Amphoteric oxides
Hence, option (a) is correct.
17 Identify the correct statements
from the following:
1. CO 2 (g) is used as refrigerant
for ice-cream and frozen
food.
2. The structure of C 60 contains
twelve six carbon rings and
twenty five carbon rings.
121
p-Block Elements
3. ZSM-5, a type of zeolite, is
used to convert alcohols into
gasoline.
4. CO is colourless and odourless
gas.
[NEET (Sep.) 2020]
(a) (1) and (3) only
(b) (2) and (3) only
(c) (3) and (4) only
(d) (1), (2) and (3) only
Ans. (c)
3 and 4 are correct statements, whereas
1 and 2 are incorrect.
1. Dry ice or CO2 (s ) [not CO2 (g)] is used
as refrigerant for ice-cream and
frozen food.
2. C60 contains 20 hexagones (not 12)
and 12 pentagones (not 20).
3. ZSM-5, a type of zeolite, is used to
convert alcohols directly into
gasoline. It is true.
4. CO is a colourless and odourless gas.
It is true.
18 Which of the following is incorrect
statement? [NEET (National) 2019]
(a) SiCl 4 is easily hydrolysed
(b) GeX4 (X =, F, Cl, Br, I) is more stable
than GeX2
(c) SnF4 is ionic in nature
(d) PbF4 is covalent in nature
(a) MeSiCl3
(c) Me 3SiCl
Ans. (c)
(b) Me2 SiCl2
(d) PhSiCl3
Straight chain polymer
(silicon)
Me3SiCl is not a monomer for a high
molecular mass silicon polymer because
it generatesMe3SiOH when subjected to
hydrolysis which contains only one
reacting site. Hence, the polymerisation
reaction stops just after first step.
21 Name the type of the structure of
silicate in which one oxygen atom
of [SiO 4] 4– is shared?
[CBSE AIPMT 2011]
(a) Sheet silicate
(b) Pyrosilicate
(c) Three dimensional silicate
(d) Linear chain silicate
Ans. (b)
In pyrosilicate, only one oxygen atom is
shared.
–
–
–
–
–
Pyrosilicate
19 The basic structural unit of
[NEET 2013]
silicates is
(a) SiO −
(b) SiO 4−
4
2−
(d) SiO2−
(c) SiO 3
4
Ans. (b)
The basic building unit of all silicates
is the tetrahedral SiO 4−
4 . It is
represented as
–
O
Si
–
–
O
–
O
O
Structure of SiO4–
4 unit
=O
= Si
Straight chain silanes are silicon oils.
These are more stable at high
temperature than mineral oils and have
less tendency to thicken at low
temperature.
23 Which one of the following anions
is present in the chain structure
silicates?
[CBSE AIPMT 2007]
(a) Si2O 76–
(c) (SiO2–
3 )n
Ans. (c)
(b) (Si2O2–
5 )n
(d) SiO 4–
4
[SiO23– ] n and [Si4O11] 6 − have chain
structure of silicates.
O=O
= Si
–
Ans. (d)
All the tetrahalides of group 14 elements
are covalent in nature and sp3-hybridised
with tetrahedral geometry. Exceptions
are SnF4 and PbF4 which are ionic in
nature. Thus, statement (d) is incorrect
while the remaining statements are
correct.

CH3  CH3
 


 Si  O
O  Si  O—
—
 



CH3  CH3
 n −1

20 Which of these is not a monomer
for a high molecular mass silicon
polymer?
[NEET 2013]
22 The straight chain polymer is
formed by
[CBSE AIPMT 2009]
(a) hydrolysis of (CH3) 3SiCl followed by
condensation polymerisation
(b) hydrolysis of CH3SiCl 3 followed by
condensation polymerisation
(c) hydrolysis of (CH3) 4 Si by addition
polymerisation
(d) hydrolysis of (CH3)2 SiCl2 followed by
condensation polymerisation
Ans. (d)
CH3
CH3


Hydrolysis
HO  Si  OH
Cl  Si  Cl →
–HCl


CH3
CH3
Dimethyl
dichlorosilane
CH3

nHO  Si  OH →
–H2O

(Condensation
CH3
polymerisation)
24 Which of the following oxidation
states are the most characteristics
for lead and tin respectively?
[CBSE AIPMT 2007]
(a) +4, +2
(c) +4, +4
Ans. (b)
(b) +2, +4
(d) +2, +2
The tendency to form +2 ionic state
increase on moving down the group due
to inert pair effect.
Most characteristic oxidation state for
lead and tin are +2, +4 respectively.
25 Percentage of lead in lead pencil is
[CBSE AIPMT 1999]
(a) zero (b) 20
Ans. (a)
(c) 80
(d) 70
In lead pencil graphite and clay is
present, so the percentage of lead is
zero.
26 Which of the following does not
show electrical conduction?
[CBSE AIPMT 1999]
(a) Potassium
(c) Diamond
Ans. (c)
(b) Graphite
(d) Sodium
Diamond does not show electrical
conductivity due to the absence of free
electrons. Sodium and potassium are
metallic conductors while graphite is a
non-metallic conductor.
122
NEET Chapterwise Topicwise Chemistry
27 A one litre flask is full of brown
bromine vapours. The intensity of
brown colour of vapours will not
decrease appreciably on adding to
the flask some [CBSE AIPMT 1998]
(a) pieces of marble
(b) animal charcoal powder
(c) carbon tetrachloride
(d) carbon disulphide
Ans. (a)
Bromine is soluble in CCl 4 and CS2 .
Animal charcoal also adsorbs on bromine
water. But marble has no action withBr2 .
So, after adding marble piece to the
flask, there will be no change in the
intensity of brown colour.
28 The structure and hybridisation of
[CBSE AIPMT 1996]
Si(CH3 ) 4 is
(a) octahedral, sp 3d
(b) tetrahedral, sp 3
(c) bent, sp
(d) trigonal, sp2
Ans. (b)
In tetramethyl silicane, i.e.Si(CH3) 4 , Si is
sp3 hybridised. Hence it has tetrahedral
structure.
29 In graphite, electrons are
[CBSE AIPMT 1993, 97]
(a) localised on each C-atom
(b) localised on every third C-atom
(c) spread out between the structure
(d) Both (b) and (c)
Ans. (a)
Ans. (a)
Water gas is produced when steam is
passed over red hot coke beds.
The given road map problem is
C(s ) + H2O( g ) → CO( g ) + H2 ( g )
14
4244
3
Structure of graphite consist of a two
dimensional sheet like network joined
together in hexagonal rings. These
layers are held together by weak van der
Waals’ forces. In graphite each carbon
atom is bonded to three others, forming
sp2 hybrid bonds. The fourth electron
forms a π-bond.
Graphite is a conductor of electricity
which is due to the fact that all the
carbon bonds being not satisfied. Thus,
some of the electrons are free to move
through the crystal.
30 Water gas is produced by
[CBSE AIPMT 1992]
(a) passing steam through a red hot
coke bed
(b) saturating hydrogen with moisture
(c) mixing oxygen and hydrogen in the
ratio of 1 :2
(d) heating a mixture of CO2 and CH4 in
petroleum refineries
‘Y’ (colourless)
Rotten fish smell
Water gas
31 Which of the following types of
forces bind together the carbon
atoms in diamond?
[CBSE AIPMT 1992]
(a) Ionic
(b) Covalent
(c) Dipolar
(d) van der Waals’
Ans. (b)
[CBSE AIPMT 1991]
Glass is an example of amorphous solid.
It is also known as supercooled liquid.
Glass have short range order of
constituents.
33 The substance used as a smoke
screen in warfare is
35 Which of the following oxoacids of
phosphorus has strongest reducing
property?
[NEET (National) 2019]
(a) H4P2O 7
(b) H3PO 3
(c) H3PO2
(d) H3PO 4
Ans. (c)
P H bonds found in phosphorus
acids have reducing properties. Thus,
reducing property is directly
proportional to number of P H
bonds. The structures of given
oxoacids of phosphorus are as
follows :
O
HO
HO
TOPIC 3
Group 15 Elements
34 A compound ‘X’ upon reaction with
H2O produced a colourless gas ‘Y’
with rotten fish smell. Gas ‘Y ’ is
absorbed in a solution of CuSO 4 to
give Cu 3P2 as one of the products.
Predict the compound ‘X’
[NEET (Odisha) 2019]
(b) NH4Cl
(d) Ca3 (PO 4 )2
O
O
P
P
O
P
H
(H3PO2)
OH
OH
(H3PO3)
O
O
H
P
H
OH
OH
(H4P2O7)
(b) PH3
(d) acetylene
Silicon chloride is easily hydrolysed to
give white fumes, so it is used as a
smoke screen in warfare.
SiCl 4 + 4H2O → Si(OH) 4 + 4HCl
(a) Ca3P2
(c) As2O 3
As compound ‘X’ produces a colorless
gas ‘Y ’ with rotten fish smell, it can be
Ca3P2 (calcium phosphide). The reaction
takes palce as follows :
'Y'
(a) liquid
(b) solid
(c) supercooled liquid
(d) transparent-organic polymer
Ans. (c)
(a) SiCl4
(c) PCl5
Ans. (a)
Cu3P2 + Product
CuSO4 +PH3 → Cu3P2 + H2SO4
In diamond, each carbon atom
undergoes sp3 hybridisation and is
covalently bonded to three other carbon
atoms by single bonds.
32 Glass is a
CuSO4
Ca3P2 + H2O → Ca(OH) 2 + PH3
Colorless gas with rotten fish smell. ‘Y ’
[CBSE AIPMT 1989]
Ans. (d)
‘X’ + H2O
OH
HO
P
OH
HO
(H3PO4)
H3PO2 contains 2P H bonds which is
maximum among given options. Thus,
H3PO2 has strongest reducing
property.
36 Identify the incorrect statement
related to PCl 5 from the following:
[NEET (National) 2019]
(a) Two axial P Cl bonds make an
angle of 180° with each other
(b) Axial P Cl bonds are longer than
equatorial
P Cl bonds
(c) PCl 5 molecule is non-reactive
(d) Three equatorial PCl bonds make
an angle of 120° with each other
123
p-Block Elements
Ans. (c)
In gaseous and liquid phases,PCl 5 has a
trigonal bipyramidal structure with
sp3d-hybridisation.
Cl
Equatorial
bond
240
pm
Axial
bond
202
pm
120°
90°
Cl—— P
Cl
Cl
Cl
Due to presence of longer and weaker
axial bonds PCl 5 is a reactive molecule.
Hence, statement in option (c) is
incorrect while the remaining options
contain correct statements.
37 The correct order of N-compounds
in its decreasing order of oxidation
states is
[NEET 2018]
(a) HNO 3, NH4Cl, NO, N2
(b) HNO 3, NO, NH4Cl, N2
(c) HNO 3, NO, N2 , NH4Cl
(d) NH4Cl, N2 , NO, HNO 3
Ans. (c)
+2
0
(a) PH3 (b) ClF3
Ans. (d)
(c) NCl3
(d) BCl3
The species having bond angles of 120°
is BCl 3. It is sp2 -hybridised and central
atom does not have any lone pair of
electrons.
Chemical
formula
Species
Bond angle
93.5°
PH 3
P
H H
H
F
ClF3
90°
Cl
−3
HN O3 > NO >N2 > NH4 Cl
38 Which oxide of nitrogen is not a
common pollutant introduced into
the atmosphere both due to natural
and human activity?
[NEET 2018]
(a) N2O
(c) N2O 5
Ans. (c)
39 The species, having bond angles of
120° is
[NEET 2017]
(Pyramidal)
Let the oxidation state of nitrogen in
each of the given N-compounds be x.
(i) HNO3 : + 1 + x + 3 (−2) = 0
x=+5
∴Oxidation state of N inHNO3 is +5.
(ii) NO : x + 1(−2) = 0
x = +2
∴Oxidation state of N in NO is +2.
(iii) NH4 Cl : x + 4(+1) + 1(−1) = 0
x = −3
∴ Oxidation state of N inNH4 Cl is −3.
(iv) N2 : x = 0 [QN2 is present in elemental
state]
∴Oxidation state of N inN2 is 0.
Thus, the correct decreasing order of
oxidation states of given Ncompounds will be
+5
exhaust fumes and when fossil fuels are
burnt as well as produced during
thunderstorms. In each case NO is
formed first and thenNO2 .
(b) NO2
(d) NO
Nitrous oxide (N2O) , nitrogen dioxide
(NO2 ) and nitric oxide (NO) are the
common pollutant introduced into the
atmosphere.
N2O occurs naturally in environment. NO
and NO2 causes considerable amount of
air pollution. They are given off in car
F
F
(T-shaped)
NCl 3
Due to the presence of one replaceable
proton in phosphinic acid, it is
monoprotic acid. And due to presence of
two replaceable proton in phosphonic
acid, it is diprotic acid.
41 Among the following, which one is
a wrong statement?
[NEET 2016, Phase II]
(a) PH5 and BiCl5 do not exist
(b) pπ-dπ bonds are present in SO2
(c) SeF4 and CH4 have same shape
(d) I+3 has bent geometry
Ans. (c)
PH5 does not exist due to very less
electronegativity difference between P
and H. Hydrogen is slightly more
electronegative than phosphorus, thus
could not hold significantly the sharing
electrons.
On the other hand,BiCl 5 does not exist
due to inert pair effect.
On moving down the group, +5 oxidation
state becomes less stable while +3
oxidation state becomes more stable.
In SO2 , pπ-dπ and pπ-pπ both types of
bonds are present
107.8°
N
Cl Cl
Cl
(Pyramidal)
Cl
BCl 3
120°
B
Cl
Cl
Thus, SeF4 and CH4 do not have same
shape.
(Trigonal planar)
r
40 Which is the correct statement for
the given acids?
I
[NEET 2016, Phase I]
(a) Phosphinic acid is a monoprotic acid
while phosphonic acid is a diprotic
acid
(b) Phosphinic acid is a diprotic acid
while phosphonic acid is a
monoprotic acid
(c) Both are triprotic acids
(d) Both are diprotic acids
Ans. (a)
Phosphinic acid
O
H
OH
H
Thus, option (c) is incorrect statement.
42 Maximum bond angle at nitrogen is
present in which of the following?
[CBSE AIPMT 2015]
(b) NO2–
(d) NO –3
(a) NO2
(c) NO2+
Ans. (c)
NO 2
Hybridisation
Bond angle
less than 120°
sp
NO −2
sp
O
NO 2+
sp ( linear)
180°
P
NO −3
sp 2
120°
Phosphonic acid
HO
HO
Geometry-Bent
Species
P
I
I
H
2
115.4°
So, NO2+ has maximum bond angle.
124
NEET Chapterwise Topicwise Chemistry
43 Strong reducing behaviour of
H3PO 2 is due to [CBSE AIPMT 2015]
(a) presence of one  OHgroup and two
P  H bonds
(b) high electron gain enthalpy of
phosphorus
(c) high oxidation state of phosphorus
(d) presence of two —OH groups and one
P  H bond
Ans. (a)
The oxy acid of phosphorus which
contain P—H bond act as a reducing
agent or reductant.
O
P
H H OH
In H3PO2 one —OH group and two P—H
bonds are present.
44 Which of the following statements
is not valid for oxoacids of
phosphorus?
[CBSE AIPMT 2012]
(a) Orthophosphoric acid is used in the
manufacture of triple
superphosphate
(b) Hypophosphorous acid is a diprotic
acid
(c) All oxoacids contain tetrahedral four
coordinated phosphorus
(d) All oxoacids contain at least one
P ==O unit and oneP OHgroup
Ans. (b)
∴
O
H P OH
H
As it contains only one replaceable
H-atom (that is attached with O, not with
P directly) so it is a monoprotic acid.
All other given statements are true.
45 Oxidation states of P in H4P2O 5 ,
H4P2O 6 and H4 P2O 7 , respectively
are
[CBSE AIPMT 2010]
(b) +5, + 3 and + 4
(d) +3, + 4 and + 5
Oxidation state of H is +1 and that of O is
–2.
Let the oxidation state of P in the given
compounds is x.
In H4P2O5
(+1) × 4 + 2 × x + (− 2) × 5 = 0
48 Nitrogen forms N 2 , but phosphorus
when form P2 readily converted
into P4 , reason is
[CBSE AIPMT 2001]
In H4P2O6
(+ 1) × 4 + 2 × x + (− 2) × 6 = 0
4 + 2x − 12 = 0
2x = 8
∴
x=+4
In H4P2O7
(+ 1) × 4 + 2 × x + (− 2) × 7 = 0
4 + 2x − 14 = 0
2x = 10
∴
x=+ 5
Thus, the oxidation states of P inH4P2O5 ,
H4P2O6 and H4P2O7 are +3, + 4 and + 5
respectively.
46 Which of the following is the most
basic oxide?
[CBSE AIPMT 2006]
(a) Al2O 3
(c) Bi2O3
Ans. (c)
(b) Sb2O3
(d) SeO2
In Al2O3, Sb2O3, Bi2O3 and SeO2 ,Bi2O3 is
most basic oxide. As across the period
basic nature of oxide decreases and on
moving down the group it increases.
Bi2O3 + 6HCl → 2BiCl3 + 3H2O
(a) triple bond present between
phosphorus atom
(b) pπ – pπ bonding is weak
(c) pπ – pπ bonding is strong
(d) multiple bond form easily
Ans. (b)
Nitrogen form N2 (i.e. N ≡≡ N) but
phosphorus form P2 , it is at a time
convert inP4 , because inP2 , pπ - pπ
bonding is present which is a weaker
bonding due to larger size.
49 Among the following ions the pπdπ overlap could be present in
[CBSE AIPMT 2000]
(a) NO2− (b) NO −3
Ans. (c)
(c) PO 3−
4
47 Zn gives H2 gas with H2 SO 4 and
HCl but not with HNO 3 because
(a) Zn act as oxidising agent when react
with HNO3
(b) HNO3 is weaker acid thanH2SO4 and
HCl
(c) In electrochemical series Zn is
placed above hydrogen
(d) NO−3 is reduced in preference to
hydronium ion
Ans. (d)
°
Zn have lower value of E cell
and easily
gives oxidation. Zn is present aboveH2 in
electrochemical series. So, it liberates
hydrogen gas from dilu. HCl/H2SO4 . But
HNO3 is an oxidising agent. The hydrogen
obtained in this reaction is converted
into H2O. In HNO3,NO–3 ion is reduced and
give NH4NO3,N2O, NO andNO2 (based
upon the concentration ofHNO3)
[Zn + 2HNO3 → Zn (NO3)2 + 2H] × 4
(Nearly 6%)
HNO3 + 8H → NH3 + 3H2O
NH3 + HNO3 → NH4NO3
4Zn + 10HNO3 → 4 Zn(NO3)2 + NH4NO3
+ 3H2O
(d) CO2−
3
In P—O bond, π-bond is formed by the
sidewise overlapping of d-orbital of P
and p-orbital of oxygen. Hence, it is
formed by pπ and dπ-overlapping.
O
σ π
O
–
P
O
[CBSE AIPMT 2002]
Hypophosphorous acid, H3PO2 , has the
following structure.
(a) +3, + 5 and + 4
(c) +5, + 4 and + 3
Ans. (d)
4 + 2x − 10 = 0
2x = 6
x=+3
O
–
–
In nitrogen and carbon, no vacant
d-orbital is present. So, they do not form
pπ-dπ bond.
50 Which of the following phosphorus
is the most reactive?
[CBSE AIPMT 1999]
(a) Red phosphorus
(b) White phosphorus
(c) Scarlet phosphorus
(d) Violet phosphorus
Ans. (b)
White phosphorus has low ignition
temperature. So, it is most reactive.
51 Which of the following is most
acidic?
[CBSE AIPMT 1999]
(a) N2O5 (b) P2O5
Ans. (a)
(c) As2O5 (d) Sb2O5
Acidic nature of oxides decreases down
a group. So, N2O5 is most acidic.
Another reason of acidic strength of
N2O5 is that the electronegativity of N is
maximum in the given Vth group
elements. As we know that on increasing
the electronegative character, acidic
nature increases.
125
p-Block Elements
52 Repeated use of which one of the
following fertilizers would increase
the acidity of the soil?
(a) Urea
[CBSE AIPMT 1998]
(b) Superphosphate of lime
(c) Ammonium sulphate
(d) Potassium nitrate
Ans. (c)
The fertilizer ammonium sulphate is a
salt of weak base and strong acid, thus
its aqueous solution is acidic, so it
produces acidity.
53 The structural formula of
hypophosphorous acid is
[CBSE AIPMT 1997]
H
O
||
P
|
H
H
O
||
P
|
OH
HO
O
||
P
|
H
HO
O
||
P
|
OH
(a)
(b)
(c)
(d)
OH
(a) NF5
(c) AsF5
Ans. (a)
(b) PF5
(d) SbF5
(a) CO2 and SO2
(c) (COOH)2
Ans. (c)
(b) 2HCOOH
(d) no reaction
When nitric acid reacts with cane sugar,
it forms oxalic acid.
COOH
+ 5H2O
C12H22O11 + 18[O] → 6 
COOH
From
Cane sugar
Oxalic acid
60 Number of electrons shared in the
formation of nitrogen molecule is
[CBSE AIPMT 1992]
(a) 6
(c) 2
Ans. (a)
(b) 10
(d) 8
Nitrogen molecule is formed by sharing
of three electrons by each nitrogen, so
total number of electrons shared are six.
N
N
⇒
N
N ⇒N
N
[CBSE AIPMT 1992]
(a) phosphorous acid and 2
(b) hypophosphorous acid and 2
(c) hypophosphorous acid and one
(d) hypophosphoric acid and two
Ans. (c)
OH
OH
[CBSE AIPMT 1996]
(b) As4O 10
(d) P4O 6
As the oxidation state of central atom
increases the acidic character
increases, so the correct order of acidic
character is
+5
59 Cane sugar on reaction with nitric
acid gives
[CBSE AIPMT 1992]
nitric acid
56 Which of the following fluorides
does not exist? [CBSE AIPMT 1993]
OH
54 Which of the following oxides will
be the least acidic?
+3
2Na2 CO3 + NO + 3NO2 → 4NaNO2 + CO
57 H3PO 2 is the molecular formula of
an acid of phosphorous. Its name
and basicity respectively are
Hypophosphorous acid (H3PO2 ) is
monobasic acid, so its structure is
O

P

H
OH
H
+5
(a) CO2 + NaNO 3
(b) CO2 + NaNO2
(c) NaNO2 + CO
(d) NaNO 3 + CO
Ans. (c)
Nitrogen does not form pentahalide
because it does not have vacant
d-orbital.
Ans. (a)
(a) As4O 6
(c) P4O 10
Ans. (a)
55 An aqueous solution of sodium
carbonate absorbs NO and NO 2 to
give
[CBSE AIPMT 1996]
+3
P4 O10 >P4 O6 > A s4 O10 > A s4 O6
and hence, As4O6 is least acidic.
The name ofH3PO2 is hypophosphorous
acid when dissolve in water, it gives only
one H+ , so its basicity is one.
O
O


P
→ H+ +
P

H
OH
H  O–
H
H
58 Nitrogen is relatively inactive
element because
[CBSE AIPMT 1992]
(a) its atom has a stable electronic
configuration
(b) it has low atomic radius
(c) its electronegativity is fairly high
(d) dissociation energy of its molecule is
fairly high
Ans. (d)
Dinitrogen (N2 ) is chemically unreactive
at ordinary temperature. The N—N bond
in nitrogen molecule is a triple bond
(N ≡≡ N) with a bond distance of 109.8 pm
and bond dissociation energy of 946 kJ
mol −1. The low reactivity of nitrogen is
due to fairly high bond dissociation
energy of the molecule.
61 Which is used in the laboratory for
fast drying of neutral gases?
[CBSE AIPMT 1992]
(a) P2O 5
(b) Anhyd. CaCl2
(c) Activated charcoal
(d) Na3PO 4
Ans. (a)
P2O5 absorb moisture, so it is used as a
drying agent for neutral gases.
62 Pure nitrogen is prepared in the
laboratory by heating a mixture of
[CBSE AIPMT 1991]
(a) NH4OH + NaCl
(c) NH4Cl + NaOH
Ans. (d)
(b) NH4NO 3 + NaCl
(d) NH4Cl + NaNO2
In the laboratory dinitrogen is prepared
by heating an aqueous solution
containing an equivalent amount of
ammonium chloride and sodium nitrite.
Heat
NH4 Cl(aq) + NaNO2 (aq) →
N2 (g)
+ 2 H2O(l ) + NaCl
63 PH4I + NaOH forms
[CBSE AIPMT 1991]
(a) PH3
(c) P4O 6
(b) NH3
(d) P4O 10
126
NEET Chapterwise Topicwise Chemistry
Ans. (a)
When PH4I and NaOH react, phosphine
gas is obtained.
PH4I + NaOH → PH3 + NaI + H2O
64 PCl 3 reacts with water to form
[CBSE AIPMT 1991]
(a) PH3
(c) POCl3
Ans. (b)
(b) H3PO 3, HCl
(d) H3PO 4
PCl 3 is easily hydrolysed by water to give
POCl 3 and finally it givesH3PO3 and HCl.
PCl 3 + H2O → POCl 3 + 2HCl
POCl 3 + 3H2O → H3PO3 + 3HCl
65 Basicity of orthophosphoric acid is
[CBSE AIPMT 1991]
(a) 2
(c) 4
Ans. (b)
(b) 3
(d) 5
Orthophosphoric acid (H3PO4 ) have the
following structure
O

P
 OH
HO
OH
It is clear from the structure that it
contains three replaceable hydrogen
atoms, so it gives threeH+ ions on
dissolution in water. So, the basicity of
H3PO4 is three.
H3PO4 → 3H+ + PO34−
66 P2O 5 is heated with water to give
[CBSE AIPMT 1991]
(a) hypophosphorous acid
(b) phosphorous acid
(c) hypophosphoric acid
(d) orthophosphoric acid
Ans. (d)
When P2O5 (or P4O10 ) is heated with water,
it form orthophosphoric acid (H3PO4 ).
P4O10 + 6H2O → 4H3PO4
67 Aqueous solution of ammonia
consists of
[CBSE AIPMT 1991]
68 Which of the following statements
is not correct for nitrogen?
[CBSE AIPMT 1990]
(a)
(b)
(c)
(d)
Its electronegativity is very high
d-orbitals are available for bonding
It is a typical non-metal
Its molecular size is small
Ans. (b)
Nitrogen does not have vacant d-orbital
in its outermost shell.
N(7) = 1s 2 , 2 s 2 2p3
=
1s
2s
2p
69 Which one has the lowest boiling
point?
[CBSE AIPMT 1989]
(a) NH3
(c) AsCl3
Ans. (b)
(b) PH3
(d) SbH3
In the hydrides of group 15, the boiling
points changes as
NH3 > PH3 < AsH3 < SbH3 < BiH3
238.5 K 185.5 K 210.6 K 254.6 K 290 K
∴Generally mass increases then boiling
point increases.
The higher boiling point ofNH3 is due to
excessive hydrogen bonding, soPH3
have lowest boiling point among
hydrides of group number 15, i.e.
nitrogen family.
(a) NCl5
(c) SbCl5
Ans. (a)
(b) AsF5
(d) PF5
Nitrogen does not formNCl 5 (nitrogen
pentachloride) because nitrogen does
not have vacant d-orbital, so it can form
only NCl 3.
73 Which of the following is a nitric
acid anhydride? [CBSE AIPMT 1988]
(a) NO
(c) N2O 5
Ans. (c)
(b) NO2
(d) N2O 3
Dinitrogen pentaoxide (N2O5 ) is prepared
by dehydrating the concentrated nitric
acid with phosphorus pentoxide.
4HNO 3 + P4O 10 → 2N2O 5 + HPO 3
So, N2O5 is regarded as anhydride of
HNO3.
TOPIC 4
Group 16 Elements
74 In which one of the following
arrangements the given sequence
is not strictly according to the
properties indicated against it?
[NEET 2021]
70 When orthophosphoric acid is
heated to 600°C, the product
formed is
[CBSE AIPMT 1989]
(a) PH3 (b) P2O 5
Ans. (d)
(c) H3PO 3 (d) HPO 3
When orthophosphoric acid (H3PO4 ) is
heated and dehydration takes place. It
form metaphosphoric acid (HPO3).
∆
H3PO4 → HPO3 + H2O
600 ° C
71 Each of the following is true for
white and red phosphorus except
that they
[CBSE AIPMT 1989]
(a) H+
(b) OH−
(c) NH+4
(d) NH+4 and OH−
Ans. (d)
(a) both are soluble in CS2
(b) can be oxidised by heating in air
(c) consist of the same kind of atoms
(d) can be converted into one another
Ans. (a)
When ammonia dissolve in water, it form
ammonium hydroxide which is ionise as
given below
NH3 + H2O → NH4OH
NH+4 + OH–
White and red phosphorus are the main
allotropes of phosphorus. White
phosphorus is soluble in carbon
disulphide whereas red phosphorus is
insoluble in carbon disulphide.
º
72 Which of the following compound
does not exist? [CBSE AIPMT 1989]
(a) HF< HCI < HBr < HI
: Increasing
acidic strength
(b) H2O < H2S < H2Se < H2Te : Increasing
pK a values
(c) NH3 < PH3 < AsH3 < SbH3 : Increasing
acidic character
(d) CO2 < SiO2 < SnO2 < PbP2 : Increasing
oxidising power
Ans. (b)
HF < HCl < HBr < HI; Down the group,
size of atom increases, bond length
decreases and bond enthalpy decreases.
So, acidic strength increases.
∴ The given sequence is correct
H2O > H2S > H2Se > H2Te, pK a is inversely
proportional to acidic strength. So, pK a
decreases.
∴ The given sequence is incorrect.
NH3 < PH3 < AsH3 < SbH3; Acidic
character increases down the group,
∴ The given sequence is correct.
CO2 < SiO2 < SnO2 < PbP2 ; on moving
down the group oxidising power
increases.
∴ The given sequence is correct.
127
p-Block Elements
75 Which of the following oxoacid of
sulphur has O O  linkage?
[NEET (Sep.) 2020]
(a) H2 SO 4 , sulphuric acid
(b) H2 S2O 8 , peroxodisulphuric acid
(c) H2 S2O 7 , pyrosulphuric acid
(d) H2SO3, sulphurous acid
Ans. (b)
The structure of given sulphur oxyacids
are :
l
H2SO4 (Sulphuric acid) ⇒
O
HOSOH
l
O
H2S2O8 (Peroxodisulphuric acid) ⇒
Al2O3 + 6HCl → 2AlCl 3 + 3H2O
Al2O3 + 2NaOH → 2NaAlO2 + H2O
(D) Cl2O7 It is an acidic oxide and
produces a very strong acid,
perchloric acid inH2O.
Cl2O7 + H2O → 2HClO4
77 Identify the correct formula of
oleum from the following
[NEET (Odisha) 2019]
(a) H2 S2O 7
(c) H2 SO 4
Ans. (a)
(b) H2 SO 3
(d) H2 S2O 8
The correct formula of oleum isH2 S2O7
which is also known as pyrosulphuric
acid.
O
Peroxy linkage
O
O
HO—S—O—S—OH
HOSO OSOH
l
l
O
O
H2S2O7 (Pyrosulphuric acid) ⇒
H2S2O7 (Pyrosulphuric acid) ⇒
O
O
HOSO SOH
l
O
O
78 Match the oxide given in column A
with its property given in column B.
Which of the following options has
all correct pairs?
[NEET (Odisha) 2019]
O
O
H2SO3 (Sulphurous acid) ⇒
Column-A
Column-B
O
1.
Na 2O
i. Neutral
HOSOH
2.
Al 2O 3
ii. Basic
3.
N 2O
iii. Acidic
4.
Cl 2O 7
iv. Amphoteric
76 Match the following.
Oxide
Nature
A.
CO
(i) Basic
B.
BaO
(ii) Neutral
C.
Al 2O 3
(iii) Acidic
D.
Cl 2O 7
(iv) Amphoteric
Which of the following is correct
option?
[NEET (Sep.) 2020]
A
(a) (ii)
(c) (iv)
Ans.
O
B C D
(i) (iv) (iii)
(iii) (ii) (i)
A B C D
(b) (iii) (iv) (i) (ii)
(d) (i) (ii) (iii) (iv)
(a)
The correct matching is :
(A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)
(A) CO It is a neutral oxide and almost
insoluble in water.
(B) BaO It is basic oxide.
BaO + H2O → Ba(OH)2
(C) Al2O 3 It is an amphoteric oxide,
because it reacts with both acids and
alkalis.
1
(a) (ii),
(c) (i),
Ans.
2 3 4
(i), (iv), (iii)
(ii), (ii), (iii)
1 2 3 4
(b)(iii), (ii), (i), (iv)
(d)(ii), (iv), (i), (iii)
(d)
Key Idea Metal oxides are basic,
non-metal oxides are acidic while
semi-metal oxides are amphoteric in
nature. Thus, the basic character of
oxides decreases across the period and
increases down the group.
The correct match of oxide with its
property are as follows:
Na2O- Basic
Al2O3- Amphoteric
N2O- Neutral
Cl2O7 - Acidic
Thus, option (d) is correct.
79 Which is the correct thermal
stability order for H2E (E = O, S, Se,
Te and Po)? [NEET (National) 2019]
(a) H2O < H2 S < H2 Se < H2Te < H2Po
(b) H2Po < H2Te < H2 Se < H2 S < H2O
(c) H2 Se < H2Te < H2Po < H2O < H2 S
(d) H2 S < H2O < H2 Se < H2Te < H2Po
Ans. (b)
For group 16 elements, the hydrides with
high molar mass (e.g.H2 Po) are less
thermally stable than hydride with lower
molar mass (e.g.H2O). This is due to the
increase in size of central atom which
results to the weakening of MHbond
due to increased bond length.
Thus, the correct order of thermal
stability for
H2 E(E = O, S, Se, Te and Po) is as follows:
H2Po < H2Te < H2Se < H2S < H2O
80 In which pair of ions both the
species contain S S bond?
[NEET 2017]
(a) S2O27− , S2O23−
(b) S4O26− , S2O23−
(c) S2O27− , S2O28−
(d) S4O26− , S2O27−
Ans. (b)
S 4O26− and S2O23− have S—S bond
2–
O
S–
O–
S
S O
S
O S
S
O
O
O–
O
O
1
Thiosulphate ion
(S4O62–)
(S2O32–)
(tetrathionate)
81 Hot concentrated sulphuric acid is
a moderately strong oxidising
agent. Which of the following
reaction does not show oxidising
behaviour?
[NEET 2016, Phase II]
(a) Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O
(b) 3S + 2H2SO4 → 3SO2 + 2H2O
(c) C + 2H2SO4 → CO2 + 2SO2 + 2H2O
(d) CaF2 + H2SO4 → CaSO4 + 2HF
Ans. (d)
Key Idea An oxidising agent is a species,
which oxidises the other species and
itself gets reduced.
+2
0
(i) Cu+ 2H2SO4 → CuSO4 + SO2 + 2H2O
+4
0
(ii) 3S+ 2H2SO4 → 3 S O2 + 2H2O
+4
0
(iii) C+ H2SO4 → CO2 + 2SO2 + 2H2O
+2
–1
+2
–1
(iv) C aF2 + H2SO4 → C aSO4 + 2HF
In reaction (iv), oxidation number of
elements remains unchanged. Thus, in
this reaction,H2SO4 does not act as an
oxidising agent.
128
NEET Chapterwise Topicwise Chemistry
82 Nitrogen dioxide and sulphur
dioxide have some properties in
common. Which property is shown
by one of these compounds, but
not by the other? [CBSE AIPMT 2015]
(a) Forms’ acid-rain
(b) Is a reducing agent
(c) Is soluble in water
(d) Is used as a food-preservative
Ans. (a)
Nitrogen dioxide and sulphur dioxide
forms acid rain. ‘Acid rain’ is the rain
water containing sulphuric acid and
nitric acid.
HNO2 + 2H2O + O2 → 4HNO3
2SO2 + 2H2O + O2 → 2H2SO4
(b) NO2 and SO2 act as a good reducing
agent
e.g. SO2 reduces halogens to halogen
acid
Cl2 + 2H2O + SO2 → 2HCl + H2SO4
(c) NO2 and SO2 both are soluble in water.
(d) SO2 is used in the manufacture of
sodium bisulphite (NaHSO3) which is
used as a preservative for jams,
jellies and squashes.
83 Decreasing order of stability of
O 2 ,O −2 ,O +2 and O 2−
2 is
[CBSE AIPMT 2015]
(a) O2+ > O2 > O2− > O22 −
(b) O22 − > O2− > O2 > O2+
(c) O2 > O2+ > O22 − > O2−
(d) O2− > O22 − > O2+ > O2
Ans. (a)
Order of stability ∝ bond order
∴Order of the stability of given species,
O2+ > O2 > O2− > O22 −
Bond order 2.5 2 1.5 1
84 Which of the statements given
below is incorrect?
[CBSE AIPMT 2015]
(a)
(b)
(c)
(d)
Cl2O7 is an anhydride of perchloric acid
O3 molecule is bent
ONF is isoelectronic withNO2
OF2 is an oxide of fluorine
Ans. (d)
(a) Cl2O7 is an anhydride of perchloric acid
2HClO4 →
∆ Cl2O7
−H2 O
(b) Shape of O3 molecule is bent.
O
O
O
(c) Number of electrons in ONF = 24
Number of electrons inNO2 = 24
∴ ONF and NO2 both are
isoelectronic.
(d) OF2 is a fluoride of oxygen because
electronegativity of fluorine is more
than that of oxygen.
OF2 = Oxygen difluoride
85 The formation of the oxide ion
O 2− (g), from oxygen atom requires
first an exothermic and then an
endothermic step as shown below,
[CBSE AIPMT 2015]
O(g )+ e − → O−(g );
∆f H° = −141 kJmol−1
O−(g ) + e − → O2−(g );
∆f H° = +780 kJ mol−1
Thus, process of formation ofO 2−
in gas phase is unfavourable even
though O 2− is isoelectronic with
neon. It is due to the fact that
(a) electron repulsion outweighs the
stability gained by achieving noble
gas configuration
(b) O− ion has comparatively smaller
size than
oxygen atom
(c) Oxygen is more electronegative
(d) addition of electron in oxygen result
in large size of the ion
Ans. (a)
Since, electron repulsion predominate
over the stability gained by achieving
noble gas configuration. Hence,
formation of O2 − in gas phase is
unfavourable.
S to Te size increases, bond dissociation
enthalpy decreases and acidic nature
increases.
87 Sulphur trioxide can be obtained by
which of the following reaction?
[CBSE AIPMT 2012]
∆
(a) CaSO 4 + C →
∆
(b) Fe2 (SO 4 ) 3 →
∆
(c) S + H2 SO 4 →
∆
(d) H2 SO 4 + PCl5 →
Ans. (b)
∆
(a) CaSO4 + C → CaO + SO2 + CO
∆
(b) Fe2 (SO4 ) 3 → Fe2O3 + 3 SO3
∆
(c) S + 2H2SO4 → 3SO2 + 2H2O
∆
(d) H2SO4 + PCl 5 →
+ POCl 3 + HCl
Thus, SO3 is obtained by heating
Fe2 (SO4 ) 3.
88 Which one of the following oxides
is expected to exhibit
paramagnetic behaviour?
[CBSE AIPMT 2005]
(a) CO2
(c) ClO2
Ans. (c)
(b) SO2
(d) SiO2
ClO2 shows paramagnetic character due
to presence of unpaired electron in its
structure.
+
Cl
O
O
86 Acidity of diprotic acids in aqueous
solutions increases in the order
[CBSE AIPMT 2014]
(a) H2 S < H2 Se < H2Te
(b) H2 Se < H2 S < H2Te
(c) H2Te < H2 S < H2 Se
(d) H2 Se < H2Te < H2 S
Ans. (a)
Acidic strength of hydrides increases as
the size of central atom increases which
weakens the MH bond. Since, the size
increases from S to Te thus acidic
strength follows the order.
H2S < H2Se < H2Te
Acidic nature
1
∝
Bond dissociation enthalpy
SO3HCl
Chloro sulphonic acid
89 The oxidation states of sulphur in
2−
2–
the anions SO 2–
3 , S 2O 4 and S 2O 6
follows the order
[CBSE AIPMT 2003]
2–
2–
(a) S2O2–
4 < S2O 6 < SO 3
2–
2–
(b) S2O 6 < S2O 4 < SO2–
3
2–
2–
(c) S2O2–
4 < SO 3 < S2O 6
2–
2–
(d) SO2–
3 < S2O 4 < S2O 6
Ans. (c)
Oxidation state of S inSO23−
x + (− 2 × 3) = − 2.
x = + 6−2= + 4
Oxidation state of S inS2O24−
2 x + (–2 × 4) = –2
2x = + 8 − 2 = + 6
+6
x=
=+3
2
129
p-Block Elements
Oxidation state of S inS2O26−
2 x + (–2 × 6) = – 2
2x = +12 − 2 = 10
10
x=
=+5
2
Hence, increasing order of oxidation
states of S is
S2O24− < SO23− < S2O26−
+ I2 →
S—O—O—H
Pyrogallol absorb the oxygen gas and oil
of cinnamon absorb the ozone (O3).
[CBSE AIPMT 1994]
(a) nitrogen
(b) oxygen
(c) sulphur
(d) boron
Ans. (c)
Sulphur has more tendency to form
2−
2−
polyanion such as S2–
3 , S4 and S5
because sulphur has more tendency
for catenation among the given
elements.
94 Oleum is
S 4O26–
+ 2I
OH
93 Polyanion formation is maximum in
(a) tetrathionate ion
(b) sulphide ion
(c) sulphate ion
(d) sulphite ion
Ans. (a)
Thiosulphate
ion
Ans. (b)
O
90 Oxidation of thiosulphate by iodine
gives
[CBSE AIPMT 1996]
2S2O23–
O
–
Tetrathionate
ion
91 About 20 km above the earth, there
is an ozone layer. Which one of the
following statements about ozone
and ozone layer is true?
[CBSE AIPMT 1995]
(a) Ozone is a triatomic linear molecule
(b) It is harmful as it stops useful
radiation
(c) It is beneficial to us as it stops
UV-radiation
(d) Conversion of O3 to O2 is an
endothermic reaction
Ans. (c)
Ozone is an allotropic form of oxygen. It
is present in the upper atmosphere
(about 20 km above the surface of the
earth). It is believed to be formed in the
upper atmosphere by the action of
UV-rays on oxygen as
3O2 + UV - rays → 2O3,
∆ H = 142.7 kJ mol −1
Therefore, UV-rays, which are harmful to
human beings are absorbed by oxygen to
form ozone. The layer of ozone, thus
formed also prevents the remaining
UV-rays to reach the earth’s surface.
(a) castor oil
(c) fuming H2 SO 4
Ans. (c)
[CBSE AIPMT 1991]
(b) oil of vitriol
(d) None of these
(a) sulphurous acid
(b) pyrosulphuric acid
(c) dithionic acid
(d) Caro’s acid
Ans. (d)
Peroxomonosulphuric acid,H2SO5 is also
known as Caro’s acid. It has peroxy
linkage —O—O— which is confirmed by
X-ray studies.
Disulphuric acid (H2S2O7 ) is called oleum
or fuming sulphuric acid. It is strongest
oxidising agent and more powerful
dehydrating agent.
95 Which would quickly absorb
oxygen?
[CBSE AIPMT 1991]
(a) Alkaline solution of pyrogallol
(b) Conc ⋅H2 SO 4
(c) Lime water
(d) Alkaline solution of CuSO 4
Ans. (a)
Alkaline solution of pyrogallol is used to
absorb dioxygen gas (O2 ).
96 It is possible to obtain oxygen from
air by fractional distillation because
[CBSE AIPMT 1989]
(a) oxygen is in a different group of the
periodic table from nitrogen
(b) oxygen is more reactive than
nitrogen
(c) oxygen has higher boiling point than
nitrogen
(d) oxygen has a lower density than
nitrogen
Oxygen gas is prepared by fractional
distillation of air. During this process,
dinitrogen with less boiling point (78 K)
distills as vapours while dioxygen with
higher boiling point (90 K) remains in the
liquid state and can be separated.
97 The gases respectively absorbed
by alkaline pyrogallol and oil of
cinnamon are [CBSE AIPMT 1989]
(a) O 3,CH4
(c) SO2 ,CH4
[CBSE AIPMT 1989]
(a) P
(c) Na
Ans. (b)
(b) Cl
(d) S
Out of P, Na, S and Cl, chlorine does not
react directly but Na, P and S react with
oxygen directly.
P4 + 5O2 → P4O10
S + O2 → SO2
4Na + O2 → 2Na2O
99 Hypo is used in photography to
[CBSE AIPMT 1988]
Ans. (c)
92 The acid which has a peroxy
linkage is
[CBSE AIPMT 1994]
98 Oxygen will directly react with each
of the following elements except?
(b) O2 ,O 3
(d) N2O,O 3
(a) reduce AgBr grains to metallic silver
(b) convert metallic silver to silver salt
(c) remove undecomposed silver
bromide as a soluble complex
(d) remove reduced silver
Ans. (c)
After the developing, the sensitive
emulsion of silver bromide is still present
on the plate in the parts unaffected by
light. Therefore, it is necessary to
remove it in order to get the permanent
image. This process is called fixing of
image. The negative plate after washing
is dipped in a fixing solution of sodium
thiosulphate (hypo). It dissolves
unaffected silver bromide but leaves
metallic silver unaffected.
2Na2S2O3 + AgBr → Na3 [Ag(S2O3) 2 ]
+ NaBr
TOPIC 5
Group 17 Elements
100 Statement I Acid strength
increases in the order given as
HF << HCl << HBr << HI.
Statement II As the size of the
elements F, Cl, Br and HI
increases down the group. the
bond strength of HF, HCl, HBr and
HI decreases and so the acid
strength increases.
In the light of the above
statements, choose the correct
answer from the options given
below.
[NEET 2021]
130
NEET Chapterwise Topicwise Chemistry
(a) Both Statement I and Statement II
are true.
(b) Both Statement I and Statement II
are false.
(c) Statement I is true but Statement II
is false.
(d) Statement I is false but Statement II
is true.
Time Saver If someone know correct
match of B, i.e. (iii) then according to the
options (c) is the only correct answer as
this match is given in this option only.
102 The correct structure of
tribromooctaoxide is
[NEET (National) 2019]
Ans. (a)
O
The order of acidic strength is
HF << HCl << HBr << HI
As we move down the group, the size of
atom increases. Thus, the bond length
also increases and bond enthalpy
decreases. So, it becomes easier to
break H—X bond on moving down the
group. This results the increase in the
acidic strength.
So, both statements I and II are true.
O
O
(b)
A. Pure nitrogen
(i)
Chlorine
C. Contact
process
(iii) Ammonia
D. Deacon’s
process
(iv) Sodium azide
or barium
azide
A
(a) (ii)
(b) (iii
(c) (iv)
(d) (i)
Ans.
B C
(iv) (i)
(iv) (ii)
(iii) (ii)
(ii) (iii)
D
(iii)
(i)
(i)
(iv)
O–
O–
104 In the structure of ClF3 , the number
of lone pairs of electrons on central
atom ‘Cl’ is
[NEET 2018]
O Br  Br  Br == O–
–
O–
O
O–
O–
O
O
(a) four
(c) one
Ans. (b)
(c) O == BrBrBrO –
O–
O
O–
O
O
O
F
Ans. (d)
The correct structure of
tribromooctaoxide is
O
O
O
[NEET 2018]
-
Hence, the correct match is
(A) → (iv), (B) → (iii), (C)→ (ii), (D) → (i)
Thus, there are three bond pairs and two
lone pairs of electrons.
O
In this compound, Br exhibits variable
oxidation state. The oxidation state of Br
from left to right are + 6, + 4 and + 6,
respectively.
(c)
F
O
O=
=Br—Br—Br=
=O
O
Cl
F
103 Which of the following statements
is not true for halogens?
The correct match is as follows :
(A) Pure nitrogen Pure nitrogen can be
obtained by thermal decomposition
of sodium or barium azide.
Ba(N3)2 →
∆ Ba + 3N2
2NaN3 →
∆ 2Na + 3N2
(B) Haber’s process On large scale,
ammonia is manufactured by Haber’s
process.
N2 (g) + 3H2 (g)
2NH3 (g)
(C) Contact process Sulphuric acid is
manufactured by the contact
process.
(D) Deacon’s process Chlorine is
prepared by Deacon’s process.
CuCl
4HCl + O2  2→ 2Cl2 + 2H2O.
(b) two
(d) three
The central atom Cl has seven electrons
in the valence shell. Three of these will
form electron pair bonds with three
fluorine atoms leaving behind four
electrons.
O
(d) O == BrBrBr==O
Which of the following is the
correct option?
[NEET (National) 2019]
O–
O
–
O
B. Haber process (ii) Sulphuric acid
O
–
(a) O == BrBrBrO
O
101 Match the following :
O
to accept an electron. Thus, option
(b) is correct.
(c) All halogens form monobasic
oxyacids. Thus, option (c) is also
correct.
(d) Electron gain enthalpy of halogens
become less negative down the
group. However, the negative
electron gain enthalpy of fluorine is
less than chlorine due to small size of
fluorine atom.
Thus, option (d) is also correct.
(a) All but fluorine show positive
oxidation states
(b) All are oxidising agents
(c) All form monobasic oxyacids
(d) Chlorine has the highest
electron-gain enthalpy
Ans. (a)
Fluorine is the most electronegative
element and cannot exhibit any positive
oxidation state. Other halogens have
d-orbitals and therefore, can expand
their octets and show +1, +3, +5 and +7
oxidation states. Thus, option (a) is
incorrect.
Note Fluorine can form an oxoacid, HOF
in which oxidation state ofF is +1. But
HOF is highly unstable compound.
(b) All halogens are strong oxidising
agents as they have strong tendency
105 Match the interhalogen compounds
of Column I with the geometry in
Column II and assign the correct
code.
[NEET 2017]
Column I
A.
Column II
(i)
T- shape
B. XX′ 3
(ii)
Pentagonal
bipyramidal
C.
XX′ 5
(iii)
Linear
D.
XX′ 7
(iv)
Square-pyramidal
(v)
Tetrahedral
XX′
Code
A B
(a) (iii) (iv)
(b) (iii) (i)
(c) (v) (iv)
(d) (iv) (iii)
Ans. (b)
C
(i)
(iv)
(iii)
(ii)
D
(ii)
(ii)
(ii)
(i)
Two different halogens may react to
form interhalogen compounds as
XX′ (ClF, BrF, BrCl, IF, ICl) Linear
XX′ 3 (ClF3, BrF3, IF3, ICl 3)
Bent T-shaped
131
p-Block Elements
XX′ 5 (ClF5 , BrCl 5 , IF5 )
Squarepyramidal
Pentagonal
bipyramidal
XX′ 7 (IF7 )
106 Among the following, the correct
order of acidity is
[NEET 2016, Phase I, CBSE AIPMT 2005]
(a) HClO < HClO2 < HClO3 < HClO4
(b) HClO2 < HClO < HClO3 < HClO4
(c) HClO4 < HClO2 < HClO < HClO3
(d) HClO3 < HClO4 < HClO2 < HClO
Ans. (a)
As the oxidation state of halogen i.e. —Cl
in this case increases, acidity of oxyacid
increases.
HClO : Oxidation state of Cl = +1
HClO2 : Oxidation state of Cl = + 3
HClO 3 : Oxidation state of Cl = + 5
HClO 4 : Oxidation state of Cl = + 7
Therefore, the correct order of acidity
would be
HClO < HClO2 < HClO 3 < HClO
107 Which one of the following orders
is correct for the bond dissociation
enthalpy of halogen molecules?
[NEET 2016, Phase I]
(a) Cl2 > Br2 > F2 > I2
(b) Br2 > I2 > F2 > Cl2
(c) F2 > Cl2 > Br2 > I2
(d) l2 > Br2 > Cl2 > F2
Ans. (a)
As the size increases, bond dissociation
enthalpy becomes lower. Also, as the
size of atoms get smaller, ion pairs on
the two atoms get close enough
together to experience repulsion. In
case of F2 , this repulsion is bigger and
bond becomes weaker.
Hence, the correct order is
Cl2 > Br2 > F2 > I2
Ans. (b)
Ans. (c)
Since, there is a strong hydrogen
bonding between HF molecules. Hence,
boiling point is highest for HF.
HF > HI > HBr > HI
Generally as the size of the atom
increases, bond dissociation energy
decreases, so in halogensI2 have lowest
bond dissociation energy, but the bond
dissociation energy of chlorine is higher
than that of fluorine because in fluorine
there is a greater repulsion between
non-bonding electrons (2p). Hence, the
order of bond dissociation energy is
Cl2 > F2 > Br2 > I2
Bond dissociation 243 159 193 151
energy (kJ/mol)
109 When Cl 2 gas reacts with hot and
concentrated sodium hydroxide
solution, the oxidation number of
chlorine changes from
[CBSE AIPMT 2012]
(a) zero to +1 and zero to −5
(b) zero to −1 and zero to +5
(c) zero to −1 and zero to +3
(d) zero to +1 and zero to −3
Ans. (b)
112 Which one of the following ionic
species has the greatest proton
affinity to form stable compound?
When chlorine gas reacts with hot and
concentrated NaOH solution, it
disproportionates into chloride (Cl − ) and
chlorate (ClO−3 ) ions.
Oxidation
−1
0
+5
3 Cl2 + 6NaOH → 5Na Cl+ Na ClO3 + 3H2O
Hot and
concentrated
reduction
110 Among the following which is the
strongest oxidising agent?
[CBSE AIPMT 2009]
(a) F2
(c) I2
Ans. (a)
(b) Br2
(d) Cl2
Fluorine is the most electronegative
element because electronegativity
decreases on moving down the group.
Hence, it gets reduced readily intoF − ion
and is the strongest oxidising agent.
NOTE The electron gain enthalpy of
fluorine is less negative than that of
chlorine inspite of that flourine is the
strongest oxidising agent. This is due to
its low bond dissociation energy and high
heat of hydration as compared to those
of chlorine.
108 The variation of the boiling point of
the hydrogen halides is in the order
HF > HI > HBr > HCl. What explains
the higher boiling point of hydrogen 111 Which one of the following
fluoride?
[CBSE AIPMT 2015]
arrangements does not give the
(a) The electronegativity of fluorine is
correct picture of the trends
much higher than for other elements
indicated against it?
in the group
(b) There is strong hydrogen bonding
between HF molecules
(c) The bond energy of HF molecules is
greater than in other hydrogen
halides
(d) The effect of nuclear shielding is
much reduced in fluorine which
polarises the HF molecule
[CBSE AIPMT 2008]
(a) F2 > Cl2 > Br2
(b) F2 > Cl2 > Br2
enthalpy
(c) F2 > Cl2 > Br2
energy
(d) F2 > Cl2 > Br2
> I2 Oxidising power
> I2 Electron gain
> I2 Bond dissociation
> I2 Electronegativity
[CBSE AIPMT 2007]
(a) HS−
(c) F −
Ans. (c)
(b) NH2−
(d) I−
Fluorine is the most electronegative
element in the periodic table. So, it has
the greatest proton affinity to form
stable compounds.
113 Which one of the following orders
is not in accordance with the
property stated against it?
[CBSE AIPMT 2006]
(a) F2 > Cl2 > Br2 > I2 Oxidising power
(b) Hl > HBr > HCl > HF acidic property in
water
(c) F2 > Cl2 > Br2 > I2 Electronegativity
(d) F2 > Cl2 > Br2 > I2 Bond dissociation
energy
Ans. (d)
Incorrect order of bond dissociation
energy F2 > Cl2 > Br2 > I2 due to following
order of size I > Br > Cl > F.
114 Which one of the following
arrangements represents the
correct order of electron gain
enthalpy (with negative sign) of the
given atomic species?
[CBSE AIPMT 2005]
(a) Cl < F < S < O
(b) O < S < F < Cl
(c) S < O < Cl < F
(d) F < Cl < O < S
Ans. (b)
The correct order of electron gain
enthalpy or (electron affinity) is
O < S < F < Cl
Values of electron affinity are (in eV) 1.48,
2.07, 3.45 and 3.61.
132
NEET Chapterwise Topicwise Chemistry
115 Among K, Ca, Fe and Zn, the
element which can form more than
one binary compound with chlorine
is
[CBSE AIPMT 2004]
(a) Fe
(b) Zn
Ans. (a)
(c) K
(d) Ca
A binary compound is that compound
which is formed by two different
elements. Metals or elements which
shows variable oxidation states can form
more than one binary compound. In the
given elements only Fe shows +2 and +3
oxidation states. So, it can form two
binary compounds with chlorine asFeCl2
and FeCl 3.
116 Which is the best description of
behaviour of bromine in the
reaction given below?
H2O + Br 2 → HBr + HOBr
[CBSE AIPMT 2004]
(a) Only oxidised
(b) Only reduced
(c) Both oxidised and reduced
(d) Only proton accepted
Ans. (c)
In the reaction,
0
+1
–1
H2O + Br2 → HOBr + HBr
The oxidation number of bromine
increases from 0 to + 1 and decreases
from 0 to − 1, due to this reason bromine
is both oxidised as well as reduced in the
above reaction.
117 Which of the following statement is
not true?
[CBSE AIPMT 2003]
(a) HOCl is a stronger acid than HOBr
(b) HF is a stronger acid than HCl
(c) Among halide ions, iodide is the most
powerful reducing agent
(d) Fluorine is the only halogen that does
not show a variable oxidation state
Ans. (b)
Due to strong H—F bond,H+ ions are not
easily removed due to higher
electronegativity (EN) of F. Hence more
bond dissociation energy required.
1
Acidic nature ∝
Bond dissociation energy
So, HF is not a stronger acid than HCl.
(c) Mn(III) oxidation state is more stable
than Mn(II) in aqueous state
(d) Elements of 15th group shows only
+3 and +5 oxidation states
Ans. (b)
Bond energy ofF2 is less than Cl2
because inF2 molecule electron-electron
repulsion of 2p-orbital of two fluorine
atom is maximum in comparison to the
repulsion of 3p-orbitals of two chlorine
atom. So, less amount of energy is
required to break the bond ofF2 in
comparison to Cl2 .
119 Which reaction is not feasible?
[CBSE AIPMT 2002]
(a) 2KI + Br2 → 3KBr + I2
(b) 2KBr + I2 → 2KI + Br2
(c) 2KBr + Cl2 → 2KCl + Br2
(d) 2H2O + 2F2 → 4HF + O2
Ans. (b)
2KBr + I2 → 2KI + Br2
Reaction is not possible becauseBr − ion
is not oxidised in Br2 with I2 due to higher
electrode potential ofI2 than bromine.
In halogens, fluorine can displace chlorine
bromine and iodine, chlorine can displace
bromine and iodine and bromide can
displace iodine from their salts.
120 Which one of the following
arrangements does not truly
represent the property indicated
against it?
[CBSE AIPMT 2000]
(a) Br2 < Cl2 < F2 Oxidising power
(b) Br2 < Cl2 < F2 Electronegativity
(c) Br2 < F2 < Cl2 Electron affinity
(d) Br2 < Cl2 < F2 Bond energy
Ans. (d)
sbond dissociation enthalpy of halogens
follows the sequency as :
Cl2 > Br > F2 > I2
Enthalpy of dissociation decreass as the
bond distance increases fromF2 to I2 due
to a corresponding increase in size of
the atom as one move down the group
from F to I. However, the F-F bond
dissociation enthalpy is smaller than that
of Cl-Cl (even than that of Br-Br) because
F-atom is very small and hence
electron-electron repulsion between the
lone pairs of electrons are very large.
118 Which of the following statements
121 Which of the following is used in
is true?
[CBSE AIPMT 2002]
the preparation of chlorine?
(a) Silicon exhibits 4 coordination
number in its compounds
(b) Bond energy ofF2 is less than Cl2
[CBSE AIPMT 1999]
(a) Only MnO2
(b) Only KMnO 4
(c) Both MnO2 and KMnO 4
(d) Either MnO2 and KMnO 4
Ans. (c)
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
2KMnO4 + 16HCl → 2KCl + 2MnCl2
+ 8H2O + 5Cl2
122 Which of the following elements
has maximum electron affinity?
[CBSE AIPMT 1999]
(a) Cl
(c) I
Ans. (a)
(b) Br
(d) F
The electron affinity decreases from
Cl → Br → I, i.e. on moving down the
group. However, electron affinity of
fluorine is unexpected low. It cannot be
explained by any simple mechanism. It is
probably due to small size of the atom.
The addition of an extra electron
produces high electron charge density in
a relatively compact 2p subshell
resulting in strong electron-electron
repulsion. The repulsive forces between
electrons imply low electron affinity. So,
the correct order of electron affinity for
halogens is
I < Br < F < Cl
123 Regarding F − and Cl − which of the
following statement (s) is/are
correct?
I. Cl − can give up an electron
more easily than F − .
II. Cl − is a better reducing agent
than F − .
III. Cl − is smaller in size than F − .
IV. F − can be oxidised more
readily than Cl − .
[CBSE AIPMT 1996]
(a) I and II
(c) III and IV
Ans. (d)
(b) I, II and IV
(d) Only I
The electronegativity of F − ion is high, so
it accept an electron and Cl − ion can give
an electron more easily. Also the size of
F − is smaller than chloride ion (Cl − ).
124 HI can be prepared by all the
following methods except
[CBSE AIPMT 1994]
(a) PI3 + H2O
(b) KI + H2 SO 4
Pt
(c) H2 + I2 →
(d) I2 + H2 S
Ans. (b)
8KI+ 5H2SO4 → 4K2SO4
+4I2 + H2S + 4H2O
133
p-Block Elements
Bromide ion when treated with Cl2 gas
125 Which one of the following oxides
then form chloride ion and Br2 gas
of chlorine is obtained by passing
liberated.
dry chlorine over silver chlorate at
90°C?
[CBSE AIPMT 1994] 130 Strongest hydrogen bonding is
(a) Cl2O
(b) ClO 3
shown by
[CBSE AIPMT 1992]
(c) ClO2
Ans. (c)
(d) ClO 4
Heat
2 AgClO3 + Cl2 (dry) →
2AgCl
+ 2ClO2 + O2
126 Which among the following is
paramagnetic? [CBSE AIPMT 1994]
(a) Cl2O (b) ClO2
Ans. (b)
(c) Cl2O 7 (d) Cl2O 6
ClO2 contains total valence electrons 19,
7 valence electrons of Cl
6 valence electrons of one oxygen atom
So there must be unpaired electron, thus
it is paramagnetic in nature.
127 A solution of KBr is treated with
each of the following. Which one
would liberate bromine?
[CBSE AIPMT 1993]
(a) Hydrogen iodide
(b) Sulphur dioxide
(c) Chlorine
(d) Iodine
Ans. (c)
HF has strongest hydrogen bonding due
to the fact that it have small size and
high electronegativity.
131 Elements of which of the following
groups will form anions most
readily?
[CBSE AIPMT 1992]
(a) Oxygen family
(c) Halogens
Ans. (c)
(b) Nitrogen family
(d) Alkali metals
Halogens have maximum electron gain
enthalpy due to the fact that the atoms
of these elements have only one
electron less than the stable noble gas
(ns 2 np6 ) configuration. Therefore, they
have maximum tendency to accept the
electron to form negative ion (anion).
132 Which of the following bonds will
be most polar? [CBSE AIPMT 1992)
Chlorine is good oxidising agent than
bromine, so, Cl2 oxidise Br − to bromine.
2KBr + Cl2 → 2KCl + Br2
128 Which of the following species has
four lone pairs of electrons?
[CBSE AIPMT 1993]
(a) I
(c) Cl−
Ans. (c)
(a) water
(b) ammonia
(c) HF
(d) hydrogen sulphide
Ans. (c)
(b) O −
(d) He
Cl − has eight electrons in it valence shell,
•• –
so its Lewis dot structure is •• Cl •• thus, it
••
has four lone pairs of electrons.
129 In the manufacture of bromine
from sea water the mother liquor
containing bromide is treated with
[CBSE AIPMT 1992]
(a) carbon dioxide
(b) chlorine
(c) iodine
(d) sulphur dioxide
Ans. (b)
Cl2 + 2Br – → Br2 + 2Cl –
(a) N—Cl (b) O—F
Ans. (c)
(c) N—F (d) N—N
The polar character arises due to the
difference in electronegativity. The
electronegativity difference of N—F
bond is maximum, so it is more polar
bond.
133 The bleaching action of chlorine is
due to
[CBSE AIPMT 1992]
(a) reduction
(c) chloronation
Ans. (d)
(b) hydrogenation
(d) oxidation
When chlorine react with water it gives
nascent oxygen which bleaches the
coloured substances.
H2O + Cl2 → 2HCl + [O]
Ans. (a)
When bleaching powder reacts with HCl,
it form chlorine gas.
CaOCl2 + 2HCl → CaCl2 + H2O + Cl2
TOPIC 6
Group 18 Elements
135 Noble gases are named because of
their inertness towards reactivity.
Identify an incorrect statement
about them.
[NEET 2021]
(a) Noble gases are sparingly soluble in
water.
(b) Noble gases have very high melting
and boiling points.
(c) Noble gases have weak dispersion
forces.
(d) Noble gases have large positive
values of electron gain enthalpy.
Ans. (b)
(a) Noble gases or inert gases are
sparingly soluble in water as they are
non-polar in nature. For solubility the
thumb rule is “Like dissolves like”.
∴ Statement (a) is correct.
(b) Nobles gases have weak interatomic
forces (van der Waals’ forces). So,
they have low melting and boiling
points.
∴ Statement (b) is incorrect.
(c) Noble gases have weak London
dispersion forces which is weaker
than all other intermolecular forces.
∴ Statement (c) is correct.
(d) The last shell of noble gas have 8
electrons. They have stable
electronic configuration. So, addition
of an electron is difficult. Hence, they
have large positive values of electron
gain enthalpy.
∴ Statement (d) is correct.
136 Match the compounds of Xe in
column I with the molecular
structure in column II.
[NEET (Oct.) 2020]
Coloured substance +
[O] → Colourless substance
134 Bleaching powder reacts with a
few drops of conc. HCl to give
[CBSE AIPMT 1989]
(a) chlorine
(b) hypochlorous acid
(c) calcium oxide
(d) oxygen
Column I
Column II
A.
XeF2
I. Square planar
B.
XeF4
II. Linear
C.
XeO 3
III. Square pyramidal
D.
XeOF4
IV. Pyramidal
A B C
(a) II I III
(c) II III I
D
IV
IV
A B C D
(b) II IV III I
(d) II I IV III
134
NEET Chapterwise Topicwise Chemistry
Ans. (d)
(B)
(A) XeF2
F
sp3d
A B C
XeF4
F
Xe
F
sp3d2
Xe
F
F
F
Linear (II)
Square planar (I)
(C) XeO3
Xe
(D) XeOF4
O
sp3
O
O
O
Pyramidal (IV)
F
F
sp3d2
Xe
F
F
137 Which of the following pairs of
compounds is isoelectronic and
isostructural ?
[NEET 2017]
(a) BeCl2 , XeF2
(b) TeI2 , XeF2
(c) IBr2− ,XeF2
(d) IF3, XeF2
Ans. (c)
BeCl 2
2 + 14 = 16
Linear
2.
XeF2
8 + 14 = 22
Linear
6 + 14 = 20
Bent or
V-shape
4.
IBr2−
7 + 14 + 1 = 22 Linear
5.
IF3
7 + 21 = 28
(b) 4
3
1
2
(c) 4
1
2
3
(d) 1 3 4
Ans. (d)
2
[NEET 2016, Phase I]
Column II
1. Distorted
octahedral
B.
XeO 3
2. Square planar
C.
XeOF 4
3. Pyramidal
D.
XeF 4
4. Square pyramidal
lp + bp
4lp + 2bp
XeF 2
Structure
Linear
F
TeF 2
A – 1, B – 3, C – 4, D – 2
The structure of the xenon compounds
are represented below:
2lp + 2bp
Xe
F
Angular or
V-shape
Te
F
ICl –2
4lp + 2bp
F
Linear
F
F
Cl
F
I–
Cl
Xe
F
Xe
F
F
Distorted
octahedral XeF6
O
O
O
Pyramidal XeO3
O
F
F
F
F
Xe
F
F
F
Square planar
XeF4
139 The correct geometry and
hybridisation for XeF4 are
[NEET 2016, Phase II]
(a) octahedral, sp 3d2
(b) trigonal bipyramidal, sp 3d
(c) planar triangle, sp 3d 3
(d) square planar, sp 3d2
Ans. (a)
F
F
F
F
Xe
F
F
F
F
140 XeF2 is isostructural with
[NEET 2013]
(a) TeF2 (b) ICl2−
Ans. (b)
(c) SbCl3 (d) BaCl2
Species having the same number of
bond pairs and lone pairs are
isostructural (have same structure).
Cl—Ba—Cl (linear)
Thus, XeF2 is isostructural with ICl2− .
141 Which of the following statements
is false?
[CBSE AIPMT 1994]
(a) Radon is obtained from the decay of
radium
(b) Helium is inert gas
(c) Xenon is the most reactive among
the rare gases
(d) The most abundant rare gas found in
the atmosphere is helium
Ans. (d)
The amount of noble gases present in
atmosphere (in percent by) is given
below:
Element
Abundance (Volume%)
He
5.24 × 10 −4
Ne
1.82 × 10 −3
Ar
0.934
Kr
1.14 × 10 −3
Xe
8.7 × 10 −6
So, argon is most abundant, not helium.
142 Noble gases do not react with
other elements because
[CBSE AIPMT 1994]
(a)
(b)
(c)
(d)
Geometry – octahedral,
Hybridisation – sp3d2
Thus, option (a) is correct.
0lp + 2bp
BaCl 2
Key Idea Geometry is determined by
electron pair arrangement whereas
shape is determined by arrangement of
atoms around the centre atom.
T-shape
138 Match the compounds given in
column I with the hybridisation and
shape given in column II and mark
the correct option.
XeF 6
3
Square pyramidal
XeOF4
1.
A.
4
Xe
Number of
S.No. Compounds valence
Geometry
electrons
Column I
2
F
Key concept Isoelectronic species have
equal number of valence electrons .
Both IBr2− and XeF2 are linear and
number of valence electrons present in
both the species is same, i.e. they are
also isoelectronic.
TeI2
D
(a) 1
Square pyramidal (III)
Hence, option (d) is correct match.
3.
Species
Codes
they are monoatomic
they are found in abundance
the size of their atoms is very small
they are completely paired up and
have stable electron shells
Ans. (d)
In general, noble gases are not very
reactive. Their chemical inertness is due
to the fact that they have completely
filled ns 2 np6 electronic configuration of
their valence shells. The other reasons
are very high ionisation enthalpy and
almost zero electron affinity.
18
d- and f-Block Elements
TOPIC 1
Characteristics of
d-Block Elements
01 The calculated spin only magnetic
moment of Cr 2+ ion is
[NEET (Sep.) 2020]
(a) 4.90 BM
(c) 2.84 BM
Ans. (a)
(b) 5.92 BM
(d) 3.87 BM
Electronic configuration of Cr2 + ion is
2+
Cr
(22e–)
4
: [Ar]3d or [Ar]
3d 4
It has four (n = 4) unpaired electrons.
∴Spin only magnetic moment
(µ) = n(n + 2 ) BM
= 4 (4 + 2 ) BM = 4.90 BM
02 Identify the incorrect statement.
[NEET (Oct.) 2020]
(a) The transition metals and their
compounds are known for their
catalytic activity due to their ability
to adopt multiple oxidation states
and to form complexes.
(b) Interstitial compounds are those
that are formed when small atoms
like H, C or N are trapped inside the
crystal lattices of metals.
(c) The oxidation states of chromium in
CrO24− and Cr2O27− are not the same.
(d) Cr2 + (d 4 ) is a stronger reducing agent
than Fe2 + (d 6 ) in water.
Ans. (c)
Statements in options (a), (b) and (d) are
correct.
Statement in option (c) is incorrect,
because oxidation states of Cr in CrO24 −
(chromate) and Cr2O27 − (dichromate) are
same and it is +6.
x
i.e., CrO24 − ⇒ x + 4 (− 2) = 2 ⇒x = + 6
y
Cr2O27 −
⇒ 2y + 7 (− 2 ) = − 2 ⇒y = + 6
03 Match the followings aspects with
the respective metal.
Aspects
I.
Scandiu
m
B. The metal although
placed in 3d block is
considered not as a
transition element.
II.
Copper
C. The metal which does not
exhibit variable oxidation
states
III.
D. The metal which in +1
oxidation state in aqueous
solution undergoes
disproportionation
IV.
Mangan
ese
[NEET (Oct.) 2020]
A B C
(b) III IV
(d) II IV
D
I II
I III
(b)
(A) Mn shown six oxidation states in its
compounds, viz, + 2, + 3, + 4, + 5, + 6 and
+ 7 ⇒(III)
(B) Zn is a pseudo transition element like
Cd and Hg of group12 (d 10 -configuration)
⇒(IV)
(C) Sc does not exhibit variable oxidation
states. Only + 3 state is shown by to Sc in
its compounds ⇒(I)
(D) Cu+ disproportionaton in aqueous
solution as,
2Cu+ → Cu0 + Cu2 + ⇒(II)
04 Match the catalyst with the process
Catalyst
3. PdCl 2
iii. Oxidation of SO 2 in
the manufacture of
H2SO 4
4. Nickel complexes
iv. Polymerisation of
ethylene
Which of the following is the
correct option? [NEET (Odisha) 2019]
1 2 3 4
(a) (iii),(iv), (i), (ii)
(b) (i), (ii), (iii), (iv)
(c) (i), (iii), (ii), (iv)
Zinc
Select the correct option.
B C D
IV II III
I IV II
Process
Metal
A. The metal which reveals a
maximum number of
oxidation states
A
(a) I
(c) III
Ans.
Catalyst
Process
1. V2O 5
i. The oxidation of
ethyne to ethanal
2. TiCl 4 + Al(CH3) 3
ii. Polymerisation of
alkynes
(d) (iii), (i), (iv), (ii)
Ans. (a)
(1) V2 O5 –(iii) Oxidation ofSO2 in the
manufacture of H2SO4 by contact
process.
(2) TiCl 4 + Al(CH3) 3– (Ziegler-Natta
catalyst)– (iv) Polymerisation of
ethylene.
(3) PdCl2 –(i) Oxidation of ethyne to
ethanal.
(4) Nickel complexes–(ii) Polymerisation of
alkynes.
Thus, the correct match is
(1) → (iii), (2)→ (iv), (3) → (i), (4) → (ii)
05 When neutral or faintly alkaline
KMnO 4 is treated with potassium
iodide, iodide ion is converted into ‘
[NEET (Odisha) 2019]
X’. ‘X’ is
(b) IO −4
(d) IO −
(a) I2
(c) IO −3
Ans. (c)
When neutral or faintly alkalineKMnO4 is
treated with potassium iodide,KMnO4 is
converted intoMnO2 while iodide ( I− ) ion
is converted inotIO−3 (iodate). The
reaction takes place as follows:
–
KMnO4 + I– +OH → MnO2 +IO–3 + H2O
'X'
136
NEET Chapterwise Topicwise Chemistry
06 Match the metal ions given in
Column I with the spin magnetic
moments of the ions given in
Column II and assign the correct
code :
[NEET 2018]
Column I
Column II
1.
Co
3+
i.
8 BM
2.
Cr 3+
ii.
35 BM
3.
Fe 3+
iii.
3 BM
4.
Ni 2 +
iv.
24 BM
v.
15 BM
1
(a) iv
(c) iv
Ans.
2
i
v
3
ii
ii
4
iii
i
1 2
(b) i ii
(d) iii v
3 4
iii iv
i ii
The complex which contains unpaired
electrons exhibitd-d transition and
paramagnetism.
(i) In MnO−4 ,
The electronic configuration ofMn7 + is
[Ar]3d 0 .
Number of unpaired electrons = 0
Therefore, it will be diamagnetic and
will not showd-d transition.
(ii) In Cr2O27− ,
The electronic configuration of Cr 6 + is
[Ar]3d 0 .
Number of unpaired electrons = 0
So, it will be diamagnetic and will not
show d-d transition.
(iii)
Key Concept Spin magnetic moment can
be calculated as
µ = n(n + 2) BM
where, µ = magnetic moment
BM = Bohr Magneton (unit of µ)
n = number of unpaired electrons
in d-orbital.
The electronic configuration of Co3+ is
[Ar] 3d 6 .
Here, n = 4
µ = 4(4 + 2) = 24 BM
The electronic configuration of Cr 3+ is
[Ar]3d 3.
Here, n = 3
µ = 3 (3 + 2) = 15BM
(iv)
In MnO24− , The electronic
6+
configuration of Mn is [Ar] 3d 1.
Number of unpaired electrons = 1
Since, it contains one unpaired
electron so it will exhibit bothd-d
transition and paramagnetism.
08 Magnetic moment 2.84 BM is given
by (At. no.
Ni =28, Ti=22, Cr =24, Co = 27)
[CBSE AIPMT 2015, 2014]
(a) Ni2 +
(b) Ti3+
3+
(d) Co2+
The electronic configuration ofFe3+ is
[Ar]3d 5 .
(c) Cr
Ans. (a)
Here, n = 5
µ = 5 (5 + 2) = 35 BM
Magnetic moment,µ = n(n + 2) BM
where,
n = number of unpaired electrons
The electronic configuration ofNi2 + is
[Ar] 3d 8 .
Here, n = 2
µ = 2 (2 + 2) = 8 BM
So, the correct option is (c).
µ = 2. 84 (given)
∴
2.84 = n(n + 2) B.M
(2. 84)2 = n(n + 2)
8 = n2 + 2n
n + 2n − 8 = 0
2
07 Which one of the following ions
exhibits d-d transition and
paramagnetism as well? [NEET 2018]
2−
2−
(a) MnO −4 (b) Cr2O2−
7 (c) CrO 4 (d) MnO 4
Ans. (d)
Key Concept In d-d transition, an electron
in a d-orbital of the metal is excited by a
photon to another d-orbital of higher
energy.
Paramagnetism The complex compound
which contains unpaired electrons shows
paramagnetism while which contains
paired electrons shows diamagnetism.
n2 + 4n − 2n − 8 = 0
2+
Ni
(a) KClO 3
(c) K2Cr2O 7
Ans. (d)
n (n + 4) − 2(n + 4) = 0
n=2
= [Ar]3d 8 4s 0 (two unpaired
electrons)
∆
2KClO3 →
2KCl + 3 O2 ↑
∆
2Zn(ClO3)2 →
ZnCl2 + 3 O2
∆
4K2 Cr2O7 →
4K2 CrO4
Potassium
dichromate
Potassium
chromate
+ 2Cr2O3 + 3 O2 ↑
∆
Chromic
oxide
(NH4 )2 Cr2O7 → N2 ↑ + Cr2O3 + 4H2O
10 Which of the following statements
about the interstitial compounds is
incorrect?
[NEET 2013]
(a) They retain metallic conductivity
(b) They are chemically reactive
(c) They are much harder than the pure
metal
(d) They have higher melting points than
the pure metals
Ans. (b)
Interstitial compounds are obtained
when small atoms like H, B, C, N, etc., fit
into the interstitial space of lattice
metals. These retain metallic
conductivity. These resemble the parent
metal in chemical properties (reactivity)
but differ in physical properties like
hardness, melting point, etc.
11 The d-electron configurations of
Cr 2+ , Mn2+ Fe 2+ and Co 2+ are
d 4 , d 5 , d 6 and d 7 respectively.
Which one of the following will
exhibit minimum paramagnetic
behaviour?
[CBSE AIPMT 2011]
(At. no.
Cr = 24, Mn = 25, Fe = 26, Co = 27)
(a) [Fe(H2O) 6] 2+
(c) [Cr(H2O) 6] 2+
Ans. (b)
Ti3+ = [Ar] 3d14s 0 (one unpaired
electrons)
Cr2+ : d4
Cr 3+ = [Ar] 3d 3, (three unpaired
electrons)
Mn2+ : d5
Co2 + = [Ar], 3d 7 , 4s 0 (three unpaired
electrons)
So, only Ni2 + has 2 unpaired electrons.
(b) Zn(ClO 3)2
(d) (NH4 )2 Cr2O 7
Oxygen rich compounds like chlorate,
perchlorate, K2 Cr2O7 , etc. when heated
gives oxygen but ammonium dichromate
gives nitrogen gas when heated.
In CrO24− ,
The electronic configuration of Cr 6 + is
[Ar]3d 0 .
Number of unpaired electrons = 0
Therefore, it is also diamagnetic and
will not showd-d transition.
(c)
09 Which of the following does not
give oxygen on heating? [NEET 2013]
(b) [Co(H2O) 6] 2+
(d) [Mn(H2O) 6] 2+
Fe2+ : d6
(4 unpaired
electrons)
(5 unpaired
electrons)
(4 unpaired
electrons)
Co2+ : d7
(3 unpaired
electrons)
137
d- and f-Block Elements
∴ [Co(H2O) 6 ]2 + has minimum number of
unpaired electrons and thus, shows
minimum paramagnetic behaviour.
Higher the unpaired e – .
Higher the magnetic moment
µ = n(n + 2)
n = Number of unpaired e –
12 For the four successive transition
elements (Cr, Mn, Fe and Co), the
stability of +2 oxidation state will
be there in which of the following
order ?
[CBSE AIPMT 2011]
(At. no. Cr = 24, Mn = 25,
Fe = 26, Co = 27)
(a) Fe > Mn > Co > Cr
(b) Co > Mn > Fe > Cr
(c) Cr > Mn > Co > Fe
(d) Mn > Fe > Cr > Co
Ans. (d)
This can be understood on the basis of
E ° values for M2 +/M.
Cr
Mn
Fe
Co
E °/V
M2 +/M – 0.90 –1.18 –0.44 –0.28
E ° value for Mn is more negative than
expected from general trend due to
extra stability of half-filled electronic
configuration of Mn2 + ion.
Thus, the correct order should be
Mn > Cr > Fe > Co
An examination of E ° values for redox
couple M3 / M2 + shows that Cr2 + is strong
reducing agent (E ° 3 + 2 + = 0.41 V) and
M
/M
liberatesH2 from dilute acids.
2Cr2 + (aq) + 2H + (aq) → 2Cr 3+ (aq) + H2 ↑ (g)
∴ The correct order is Mn > Fe > Cr > Co.
13 Which one of the following ions has
electronic configuration [Ar]3 d 6 ?
(At. no. Mn = 25, Fe = 26,
Co = 27, Ni = 28)
[CBSE AIPMT 2010]
(a) Ni3+ (b) Mn3+
Ans. (d)
(d) Co 3+
(c) Fe 3+
Ni3+ (28) = [Ar] 3d 7
Mn3+ (25) = [Ar] 3d 4
Fe3+ (26) = [Ar] 3d 5
3+
Co (27) = [Ar] 3d
Hence, Co3+ has 3d 6 electronic
configuration correct answer is (d).
6
14 Which of the following pairs has
the same size? [CBSE AIPMT 2010]
(a) Fe2 + , Ni2 +
(c) Zr 4 + , Hf 4 +
(b) Zr 4 + , Ti4 +
(d) Zn2 + , Hf 4 +
Ans. (c)
In general, the atomic and ionic radii
increases on moving down a group but
the elements of second transition series
(Zr, Nb, Mo) have almost same radii as
the elements of third transition series
(Hf, Ta, W, etc). This is because of
lanthanoid contraction, i.e. imperfect
shielding of one4f e − by another.
15 Which one of the following ions is
the most stable in aqueous
solution?
[CBSE AIPMT 2007]
(At. no. Ti = 22, V = 23,Cr = 24,
Mn = 25)
(a) Cr 3+ (b) V 3+
Ans. (d)
(c) Ti3+
(d) Mn3+
Stability of transition metal ions is
directly proportional to the unpaired
electrons. The exactly half-filled and
completely filled d-orbitals are extra
stable.
Cr 3+ (21) = 3d 3, 4s 0 (3 unpaired electrons)
V 3+ (20) = 3d2 , 4s 0 (2 unpaired electrons)
Ti3+ (19) = 3d 1, 4s 0 (1 unpaired electron)
Mn3+ (22) = 3d 4 , 4s 0 (4 unpaired electrons)
So, Mn3+ ion is most stable in aqueous
solution.
16 The d electron configurations of
Cr 2+ , Mn 2+ , Fe 2+ and Ni 2+ are 3d 4 ,
3d 5 ,3d 6 and 3d 8 respectively.
Which one of the following aqua
complexes will exhibit the
minimum paramagnetic behaviour?
(At. no. of Cr = 24,Mn = 25, Fe = 26,
[CBSE AIPMT 2007]
Ni = 28)
(a) [Fe(H2O) 6] 2+
(c) [Cr(H2O) 6] 2+
Ans. (b)
(b) [Ni(H2O) 6] 2+
(d) [Mn(H2O) 6] 2+
As the number of unpaired electron
increases, the magnetic moment
increases and hence, the paramagnetic
behaviour increases.
So, Cr2 + (22) = 3d 4 (4 unpaired electrons)
Mn2 + (23) = 3d 5 (5 unpaired electrons)
Fe2 + (24) = 3d 6 (4 unpaired electrons)
Ni2 + (26) = 3d 8 (2 unpaired electrons)
So, [Ni(H2O) 6 ]2 + exhibit minimum
paramagnetic behaviour.
17 In which of the following pairs are
both the ions coloured in aqueous
solution?
(At. no. Sc = 21, Ti = 22, Ni = 28, Cu
= 29, Co = 27) [CBSE AIPMT 2006]
(a) Ni2+ , Ti3+
(c) Sc 3+ , Co2+
Ans. (a)
(b) Sc 3+ , Ti3+
(d) Ni2+ , Cu +
28 Ni =
2+
Ni
1s 2 , 2 s 2 2p6 ,3 s 2 3 p6 3d 8, 4s 2
= 1s 2 , 2 s 2 2p6 , 3 s 2 3p6 3d 8
8
3d
(2 unpaired electrons)
22 Ti =
3+
2
2
1s , 2 s 2p6 , 3 s 2 3p6 3d2 ,4s 2
Ti = 1s 2 , 2 s 2 2p6 , 3 s 2 3p6 3d 1
(1 unpaired electron)
2
2
6
2
Sc
=
1
s
,
2
s
2
p
,
3
s
3p6 3d 1,4s 2
21
3+
2
2
6
2
Sc = 1s , 2 s 2p , 3 s 3p6
(no unpaired electron)
2
2
6
Cu
=
1
s
,
2
s
2
p
,
3s 2 3p6 3d 10 ,4s 1
29
+
2
2
6
Cu = 1s , 2 s 2p , 3 s 2 3p6 3d 10
(no unpaired electron)
Hence, in the above ions,Ni2 + and Ti3+
are coloured in aqueous solution due to
the presence of unpaired electrons in d
subshell.
18 The aqueous solution containing
which one of the following ions will
be colourless ?
(At. no. Sc = 21, Fe = 26, Ti = 22,Mn
[CBSE AIPMT 2005]
= 25)
(a) Sc 3+
(c) Ti3+
Ans. (a)
(b) Fe2+
(d) Mn2+
21Sc =
3+
1s 2 , 2 s 2 2p6 , 3s 2 3p6 3d 1, 4s 2
So, Sc = 1s 2 , 2 s 2 2p6 , 3s 2 3p6
(It is colourless due to the absence of
unpaired electrons in d sub-shell)
2
2
6
2
6
6
2
26 Fe = 1s , 2 s 2p , 3s 3p 3d , 4s
2+
2
2
6
2
6
6
Fe = 1s , 2 s 2p , 3s 3p 3d
d
(It is coloured due to the presence of
four unpaired electrons in d sub-shell)
2
2
6
2
6 2
2
22 Ti = 1s , 2 s 2p , 3s 3p 3d , 4s
3+
2
2
6
2
6
1
Ti = 1s , 2 s 2p , 3s 3p 3d
d
(It is coloured due to the presence of an
unpaired electron ind sub-shell)
Mn = 1s 2 , 2 s 2 2 p6 , 3s 2 3p6 3d 5 , 4s 2
25
Mn2 + = 1s 2 , 2 s 2 2 p6 , 3s 2 3p6 3d 5
d
(It is coloured due to the presence of five
unpaired electrons in d sub-shell).
138
NEET Chapterwise Topicwise Chemistry
19 Four successive members of the
first row transition elements are
listed below with their atomic
numbers. Which one of them is
expected to have the highest third
ionisation enthalpy?
[CBSE AIPMT 2005]
(a) Vanadium (Z = 23)
(b) Chromium (Z = 24)
(c) Iron (Z = 26)
(d) Manganese (Z = 25)
Ans. (d)
In 23V = 1s 2 , 2s 2 2p6 , 3s 2 3p6 3d 3, 4s 2
Third electron which is removed to give
third ionisation potential, belongs to3d 3
subshell.
24 Cr
= 1s 2 , 2s 2 2p6 , 3s 2 3p6 3d 5 , 4s 1
Third electron which is removed to give
third ionisation potential, belongs to3d 5
subshell.
26 Fe = 1s , 2s 2p , 3s 3p 3d , 4s
Third electron which is removed to give
third ionisation potential, belongs to3d 6
subshell.
2
25
2
6
2
6
6
2
Mn = 1s 2 , 2s 2 2p6 , 3s 2 3p6 3d 5 , 4s 2
Third electron which is removed to give
third ionisation potential, belongs to3d 5
subshell.
In all elements shell and subshells are
same. Required amount of energy
(enthalpy) is based upon the stability of
d- subshell.
The 3d 5 -subshell has highest stability in
all because it is half-filled subshell. So,
Mn shows highest third ionisation
potential.
20 Among the following series of
transition metal ions, the one in
which all metal ions have 3 d 2
electronic configuration is
(At. no. Ti =22,V =23,Cr =24,
[CBSE AIPMT 2004]
Mn =25)
(a) Ti3+ , V2+ , Cr 3+ , Mn4+
(b) Ti+ , V 4+ , Cr 6+ , Mn7 +
(c) Ti4+, V 3+ , Cr2 + , Mn3+
(d) Ti2+ , V 3+ , Cr 4+ , Mn5+
Ans. (d)
Ti2 + = 1s 2 , 2s 2 2p6 , 3 s 2 3p6 3d2 , 4s 0
V 3 + = 1s 2 , 2s 2 2p6 , 3 s 2 3p6 , 3d2 , 4s 0
Cr 4 + = 1s 2 , 2s 2 2p6 , 3 s 2 3p6 3d2 , 4s 0
Mn5 + = 1s 2 , 2s 2 2p6 , 3 s 2 3p6 3d2 , 4s 0
21 Which one of the following
characteristics of the transition
metals is associated with their
catalytic activity?
[CBSE AIPMT 2003]
(a) Colour of hydrated ions
(b) Variable oxidation states
(c) High enthalpy of atomisation
(d) Paramagnetic behaviour
Ans. (b)
The catalytic activity of transitional
metals is due to their variable oxidation
states.
22 The basic character of the
transition metal monoxides follows
the order
[CBSE AIPMT 2003]
(At. no. Ti = 22, V = 23, Cr = 24,
Fe = 26)
(a) TiO > FeO > VO > CrO
(b) TiO > VO > CrO > FeO
(c) VO > CrO > TiO > FeO
(d) CrO > VO > FeO > TiO
Ans. (b)
[CBSE AIPMT 1999]
(a) Cu + Pb
(c) Cu + Zn
Ans. (b)
(b) Cu + Sn
(d) Cu + Ni
Bell-metal is an alloy of copper and tin.
Cu = 80% and Sn = 20%.
It is used for making bells, utensils, etc.
26 Which of the following has more
unpaired d-electrons?
(b) Fe2+
(d) Cu +
(a) Zn
(c) N3+
Ans. (b)
[CBSE AIPMT 2002, 2000]
(a) chromium (Z = 24)
(b) manganese (Z = 25)
(c) iron (Z = 26)
(d) titanium (Z = 22)
Ans. (b)
Manganese shows maximum number of
oxidation states because it has 5
unpaired electrons in 3d and also
contains 2 electrons in 4s sub-shell.
+ 2 +3 +4+ 5+6+ 7
Other given metals show the following
oxidation states
Cr = + 2 + 3 + 4 + 5 + 6
Fe = + 2 + 3 ; Ti = + 2 + 3 + 4
24 Which one of the following forms a
colourless solution in aqueous
medium?
(At. no. Sc = 21, Ti = 22, V = 23,
Cr = 24)
[CBSE AIPMT 2000]
(c) Ti3+
25 Bell-metal is an alloy of
[CBSE AIPMT 1999]
23 In the following transition metals,
the maximum number of oxidation
states are exhibited by
(b) Cr 3+
The electronic configuration of Sc is
1s 2 , 2 s 2 2 p6 , 3 s 2 3p6 3d 1, 4s 2
and configuration of Sc 3+ is
1 s 2 , 2 s 2 2 p6 , 3 s 2 3 p 6
3+
So, Sc is colourless due to absence of
electrons in d-orbital.
+
The order of basic character of the
transition metal monoxide is TiO > VO >
CrO> FeO because basic character of
oxides decreases with increase in
atomic number. Hence, oxides of
transitional metals in low oxidation
state, i.e. +2 and +3 are generally basic
except Cr2O3 .
(a) V2+
Ans. (d)
(d) Sc 3+
Zn+ (at. no.= 30) = 1s 2 ,2 s 2 2p6 ,
3 s 2 3p6 3d 10 ,4s 1
(no unpaired d-electrons)
Fe2 + (at .no.= 26) = 1s 2 , 2 s 2 2p6 ,
3 s 2 3 p6 3 d 6
3d 6
4-unpaired d-electrons
N3+ (at. no.= 7) = 1s 2 , 2 s 2 2p0
(no unpaired d-electrons)
Cu+ (at. no.= 29) = 1s 2 , 2 s 2 2p6 ,
3 s 2 3p6 3d 10
(no unpaired d-electrons)
So, maximum number of unpaired
electrons are present in Fe2 + .
27 Which one of the following ionic
species will impart colour to an
aqueous solution?
[CBSE AIPMT 1998]
(a) Ti4+ (b) Cu +
Ans. (d)
(c) Zn2 +
(d) Cr 3 +
Cr 3+ = 1s 2 , 2 s 2 2 p6 , 3s 2 3p6 3d 3 (coloured)
(Cr 3 + contains 3 unpaired e − , so it gives
colour)
Zn2 + = 1s 2 , 2 s 2 2p6 , 3s 2 3p6 3d10
(colourless)
Cu+ = 1s 2 , 2 s 2 2p6 , 3s 2 3p6 3d 10
(colourless)
Ti4+ = 1s 2 , 2 s 2 2p6 , 3s 2 3p6 (colourless)
(Colour is produced due to presence of
unpaired electrons.)
139
d- and f-Block Elements
28 Which of the following elements is
responsible for oxidation of water
to O 2 in biological processes?
[CBSE AIPMT 1997]
(a) Fe
(b) Cu
Ans. (c)
(c) Mn
(d) Mo
Manganese is responsible for oxidation
of water to O2 in biological processes.
29 Which of the following does not
represent the correct order of the
properties indicated?
[CBSE AIPMT 1997]
(a) Ni2 + > Cr2 + > Fe2 + > Mn2 + (size)
(b) Sc > Ti > Cr > Mn (size)
(c) Mn2 + > Ni2 + > Co2 + > Fe2 +
(unpaired electron)
(d) Fe2 + > Co2 + > Ni2 + > Cu2 +
(unpaired electron)
Ans. (a)
In a period on moving from left to right
ionic radii decreases. So, the order of
cationic radii is
Cr2 + > Mn2 + > Fe2 + > Ni2 +
In Sc > Ti > Cr > Mn (correct order of
atomic radii)
In
Mn2 + >
Ni2 + <
Co2 + < Fe2 +
↓
↓
Five
↓
↓
Two
Three
unpaired
unpaired
electrons
electrons
electrons electrons
↓
>
4
Co2 +
↓
3
>
Ni2 +
↓
unpaired
>
Cu2 +
2
↓
1
Number of unpaired electrons
30 The common oxidation states of Ti
are
[CBSE AIPMT 1994]
(a) +2, +3
(c) 3, −4
Ans. (b)
During electrolysis, copper is deposited
at the cathode while oxygen is liberated
at anode. The following reactions occur
at the electrodes
At anode
2H2O → 4H+ + O2 + 4e −
(Oxidation)
At cathode
Cu2 + + 2e − → Cu(s ) (Reduction)
32 The transition elements have a
general electronic configuration.
[CBSE AIPMT 1991]
(a) ns 2np 6nd 1− 10
(b) (n − 1) d 1− 10, ns 0−2 , np 0− 6
(c) (n − 1) d 1− 10, ns 1−2
(d) nd 1− 10ns 2
Ans. (c)
In the transition elements, the d-orbitals
are successively filled. The general
electronic configuration of d-block is
(n − 1)d 1− 10 ns 1−2
Where (n − 1) stands for inner shell and
d-orbitals may have one to ten electrons
and the s-orbital of the outermost shell
(n) may have one or two electrons.
33 Which of the following metals
corrodes readily in moist air?
Four
unpaired
In
Fe2 +
Ans. (b)
(b) +3, +4
(d) +2, +3, +4
The most common oxidation state of
titanium are +3 and +4.
31 When CuSO 4 is electrolysed using
platinum electrodes?
[CBSE AIPMT 1993]
(a) Copper is liberated at cathode,
sulphur at anode
(b) Copper is liberated at cathode,
oxygen at anode
(c) Sulphur is liberated at cathode,
oxygen at anode
(d) Oxygen is liberated at cathode,
copper at anode
[CBSE AIPMT 1988]
(a) Gold
(c) Nickel
Ans. (d)
∆
(b) Cr2O3 + 2Al →
Al2O3 + 2Cr;
(Metal displacement reaction)
∆
(c) Fe + 2HCl →
FeCl2 + H2 ;
(Displacement reaction)
(d) 2Pb(NO3)2 → 2PbO + 4NO2 + O2 ;
(Decomposition reaction)
Metal displacement reaction is a
displacement reaction in which a more
reactive metal displaces/replaces the
less reactive metal. In option (c),
hydrogen is a non-metal. So, it is
displacement reaction but not a metal
displacement reaction.
35 The manganate and permanganate
ions are tetrahedral, due to
[NEET (National) 2019]
(a) there is no π-bonding
(b) the π-bonding involves overlap of
p -orbitals of oxygen with p -orbitals
of manganese
(c) the π-bonding involves overlap of
d-orbitals of oxygen with d-orbitals
of manganese
(d) the π-bonding involves overlap of
p-orbitals of oxygen with d-orbitals
of manganese
Ans. (d)
The structures of manganate and
permanganate ions are as follows :
O–
(b) Silver
(d) Iron
Mn
O
Iron corrodes readily in moist air
because iron is more reactive than Ni, Au
and Ag. In other words the reduction
potential of iron is very less, so its
oxidation takes place readily.
TOPIC 2
Important Compounds
of d-Block Elements
34 Which of the following reactions is
the metal displacement reaction ?
Choose the right option. [NEET 2021]
∆
(a) 2KClO 3 →
2KCl + 3 O2
∆
(b) Cr2O 3 + 2Al →
Al2O 3 + 2Cr
(c) Fe + 2HCl→ FeCl2 + H2 ↑
(d) 2Pb(NO 3)2 → 2PbO + 4 NO2 + O2 ↑
Ans. (b)
∆
(a) 2KClO3 →
2KCl + 3O2 ;
(Decomposition reaction)
O
O–
Mn
O
O–
O
O
Manganate
2–
(MnO4 )
Permanganate
–
(MnO4 )
The manganate and permanganate
ions are tetrahedral and contain
π-bonds of dπ - pπ -type, oxygen does
not have any d-orbital.
Hence, the π-bonding takes place by
overlap of p-orbital of oxygen and
d-orbitals of manganese.
36 The number of hydrogen bonded
water molecule(s) associated with
CuSO 4 ⋅ 5H2O is [NEET (Odisha) 2019]
(a) 3
(c) 2
Ans. (b)
(b) 1
(d) 5
Cu2 + ion has 17 electrons in its
outermost orbital. 4H2O molecules
donate 4 pairs of electrons for
coordinate covalent sharing. The oneH2O
molecule is attached to the sulphate ion
by hydrogen bonding, the oxygen being
oriented to the hydrated cupirc ion.
140
NEET Chapterwise Topicwise Chemistry
H2O
OH2
H----O
Cu
H2O
O
OH2
O
S
H----O
O
Thus, the number of hydrogen bonded
water molecule associated with CuSO4 .
5H2O is 1.
37 Name the gas that can readily
decolourise acidified KMnO 4
solution.
[NEET 2017]
(a) CO2 (b) SO2
Ans. (b)
(c) NO2
(d) P2O 5
SO2 gas can readily oxidise acidified
KMnO4 solution becauseKMnO4 is an
oxidising agent andSO2 act as reducing
agent.
2MnO−4 + 5SO2 + 2H2O → 2Mn2 +
+ 5SO24− + 4H+
While other options such asNO2 (strong
oxidising agent), CO2 (neither oxidising
agent nor reducing agent) cannot
decolourise acidifiedKMnO4 Solution.
38 HgCl 2 and I 2 both when dissolved in
water containing I − ions the pair of
species formed is
[NEET 2017]
(b) HgI2 , I−
(d) Hg2I2 , I−
(a) HgI2 , I3−
(c) HgI24− , I−3
Ans. (c)
HgCl2 and I2 both when dissolved in water
containing I− ions, the pair of species
formed is HgI24− and I−3 .
In aqueous solution, I2 reacts with I − and
maintains the following equilibrium.
I2 + I− q
I−3
2+
Hg gives ppt. of HgI2 on reaction with
I− .
But HgI2 is soluble in excess of I −
Hg2 + + 2I− → HgI2 ↓ + 2Cl −
HgI2 + 2I− q
Red ppt.
[HgI4 ]2 −
39 When copper is heated with conc.
HNO 3 it produces
[NEET 2016, Phase I]
(a) Cu(NO 3)2 and NO
(b) Cu(NO 3)2 , NO and NO2
(c) Cu(NO 3)2 and N2 O
(d) Cu(NO 3)2 and NO2
Ans. (d)
Nitric acid acts as an oxidising agent
while reacting with copper. When dil.
HNO3 reacts, reaction proceeds as:
3Cu + 8HNO3 (dil.) → 3Cu(NO3) 2
+ 2NO + 4H2O
and when conc.HNO3 is used, reaction
proceeds as
Cu + 4HNO3 (conc.) → Cu(NO3) 2
+ 2NO2 + 2H2O
40 Which one of the following
statements is correct when SO 2 is
passed through acidified K 2 Cr 2 O 7
solution?
[NEET 2016, Phase I]
(a) The solution is decolourised
(b) SO2 is reduced
(c) Green Cr2 (SO 4 ) 3 is formed
(d) The solution turns blue
Ans. (c)
When SO2 is passed through acidified
K2 Cr2O7 solution, green chromium
sulphate is formed. In this reaction,
oxidation state of Cr changes from +6 to
+3.
K2 Cr2O7 + H2SO4 + 3SO2 → K2SO4
OS of Cr = +6
+ Cr2 (SO4 ) 3 + H2O
OS of Cr = + 3
Green
The appearance of green colour is due to
the reduction of chromium metal.
41 Assuming complete ionisation,
same moles of which of the
following compounds will require
the least amount of acidified
KMnO 4 for complete oxidation?
[CBSE AIPMT 2015]
(a) FeSO 4
(c) FeC2O 4
Ans. (a)
(b) FeSO 3
(d) Fe(NO2 )2
FeSO4 will require the least amount of
acidifiedKMnO4 for complete oxidation.
42 Which of the following processes
does not involve oxidation of iron?
[CBSE AIPMT 2015]
(a) Rusting of iron sheets
(b) Decolourisation of blue CuSO4
solution by iron
(c) Formation of Fe(CO) 5 from Fe
(d) Liberation of H2 from steam by iron a
high temperature
Ans. (c)
0
III
(a) Fe + H2O + O2 → Fe2 O3 ⋅x H2O]
1424
3
∴ Formation of Fe(CO) 5 from Fe does not
involve oxidation of iron because there is
no change in oxidation state.
43 The pair of compounds that can
exist together is [CBSE AIPMT 2014]
(a) FeCl3, SnCl2
(c) FeCl2 , SnCl2
Ans. (c)
The compounds with lower oxidation
number and which cannot reduced by
one another can exist together. Thus,
FeCl2 and SnCl2 can exist together as
Fe2 + cannot be reduced bySn2 + .
44 The reaction of aqueous KMnO4
with H2O 2 in acidic conditions gives
[CBSE AIPMT 2014]
(a) Mn4 + and O2
(c) Mn2 + and O 3
Ans. (b)
II
(b) Fe + CuSO4 → FeSO4 + Cu
0
0
(c) Fe + 5CO → Fe (CO) 5
0
III
(d) Fe + H2O → Fe2 O3 + H2
Steam
(b) Mn2 + and O2
(d) Mn4 + and MnO2
The reaction of aqueousKMnO4 with H2O2
in acidic medium is
3H2SO4 + 2KMnO4 + 5H2O2 →
5 O2 + 2MnSO4 + 8H2O + K2SO4
In the above reaction,KMnO4 oxidises
H2O2 to O2 and itself i.e. [MnO−4 ] gets
reduced toMn2 + ion as MnSO4 . Hence,
aqueous solution of KMnO4 with H2O2
yields Mn2 + and O2 in acidic conditions.
45 Identify the alloy containing a
non-metal as a constituent in it.
[CBSE AIPMT 2012]
(a) Invar
(c) Bell-metal
Ans. (b)
Alloy
Invar
Steel
Bell-metal
Bronze
(b) Steel
(d) Bronze
Constituents
Fe + Ni
Fe + C
Cu (80%) + Sn(20%)e
Cu(75.90%) + Sn(10.25)%
Among these alloys, only steel contains
carbon which is a non-metal.
46 Acidified K 2Cr 2O 7 solution turns
green when Na 2SO 3 is added to it.
This is due to the formation of
[CBSE AIPMT 2011]
from air
0
(b) HgCl2 , SnCl2
(d) FeCl3, Kl
(a) CrO2–
4
(c) CrSO4
Ans. (d)
(b) Cr2 (SO 3) 3
(d) Cr2 (SO 4 ) 3
K2 Cr2O7 is oxidising reagent.
K2 Cr2O7 + 3Na2SO3 + 4H2SO4 →
3Na2SO4 + K2SO4 + Cr2 (SO4 ) 3 + 4H2O
141
d- and f-Block Elements
47 Copper sulphate dissolves in
excess of KCN to give
[CBSE AIPMT 2006]
(a) CuCN
(c) [Cu(CN) 4] 2–
Ans. (b)
(b) [Cu(CN) 4] 3–
(d) Cu(CN)2
Copper sulphate when react with KCN
first give precipitate of cupric cyanide
which reduce into Cu2 CN2 and dissolve in
excess of KCN to give soluble
K3 [Cu(CN) 4 ] complex salt
[CuSO4 + 2KCN → Cu(CN)2 + K2SO4 ] × 2
Cupric
cyanide
2Cu(CN)2 → Cu2 (CN)2 + NC—CN
Cyanogen
Cu2 (CN)2 + 6KCN → 2K3 [Cu(CN) 4 ]
Soluble complex salt
2CuSO 4 + 10KCN → 2K 3 [Cu(CN) 4] + 2K 2SO 4
+ (CN)2
48 CuSO 4 when reacts with KCN
forms CuCN which is insoluble in
water. It is soluble in excess of KCN
due to the formation of the
complex
[CBSE AIPMT 2002]
(a) K2 [Cu(CN) 4]
(c) Cu(CN)2
Ans. (b)
(b) K 3 [Cu(CN) 4]
(d) Cu[KCu(CN) 4]
CuSO4 reacts with KCN to give a white
precipitate of cuprous cyanide and
cyanogen gas. The cuprous cyanide
dissolves in excess of KCN forming
K3 [Cu(CN) 4 ].
50 In which of the following
compounds, transition metal has
zero oxidation state?
[CBSE AIPMT 1999]
Oxidation state of Fe inFe(CO) 5 is zero
because CO is a neutral ligand and it
shows zero oxidation state.
51 When a substance A reacts with
water it produces a combustible
gas B and a solution of substance C
in water. When another substance
D reacts with this solution of C, it
also produces the same gas B on
warming but D can also produce
gas B on reaction with dilute
sulphuric acid at room
temperature. A imparts a deep
golden yellow colour to a
smokeless flame of Bunsen burner.
A, B, C and D, respectively are
[CBSE AIPMT 1998]
(a) Na, H2 , NaOH, Zn
(b) K, H2 , KOH, Al
(c) CaH2 , Ca(OH)2 , Sn
(d) CaC2 , C2H2 , Ca(OH)2 , Fe
Ans. (a)
A
2Cu(CN)2 → 2CuCN + CN  CN
Insoluble cyanogen
CuCN + 3KCN → K3 [Cu(CN) 4 ]
C
B
Zn + 2NaOH —→ Na2ZnO2 + H2 ↑
D
C
B
Zn + dil. H2SO4 → ZnSO4 + H2 ↑
D
B
Na, produces golden yellow colour with
smokeless flame of Bunsen burner
Soluble
49 In the silver plating of copper,
K[Ag(CN) 2] is used instead of
AgNO 3 . The reason is
[CBSE AIPMT 2002]
(a) a thin layer of Ag is formed on Cu
(b) more voltage is required
(c) Ag + ions are completely removed
from solution
(d) less availability of Ag + ions, as Cu
cannot displace Ag from [Ag(CN)2 ] –
ion
Ans. (d)
In the silver plating of copper,
K [Ag(CN)2 ] is used instead of AgNO3.The
reason is less availability of Ag + ions, as
Cu cannot displace Ag from [Ag(CN)2 ] –
ion.
52 K 2Cr 2O 7 on heating with
aqueous NaOH gives
[CBSE AIPMT 1997]
(a) CrO2–
4
(c) Cr2O2–
7
Ans. (a)
Ans. (a)
Cuprous compound contains Cu+ ion
which has small size, so the hydration is
maximum and hence, the system has
lower energy which result in stability of
the compounds.
54 Stainless steel contains iron and
[CBSE AIPMT 1995]
(a) Cr + Ni
(c) Zn + Pb
Ans. (a)
(b) Cr + Zn
(d) Fe + Cr + Ni
Stainless steel is resistant to rusting. It
contains 73% iron, 18% chromium and
8% nickel. Stainless steel is used in
utensils, cycle, cutlery and automobile
parts.
55 By pasing H2S gas in acidified
KMnO 4 solution, we get
[CBSE AIPMT 1995]
(a) S
(c) MnO2
Ans. (a)
(b) K2 S
(d) K2 SO 3
2KMnO4 + 5H2S + 3H2SO4 →
K2SO4 + 5S + 2MnSO4 + 8H2O
Only Na gives golden colour to bunsen
flame. So, A is Na
2Na + 2H2O → 2NaOH + H2 ↑
CuSO4 + 2KCN → K2SO4 + Cu(CN) 2
Unstable
(b) NH2 ⋅NH2
(d) CrO 5
(a) Fe(CO) 5
(c) NOClO 4
Ans. (a)
(c) diamagnetic nature
(d) insolubility in water
(b) Cr(OH) 3
(d) Cr(OH)2
K2 Cr2O7 + 2NaOH → K2 CrO4
+ Na2 CrO4 + H2O
Hence, CrO24– ion is obtained.
53 Cuprous compounds such as CuCl,
CuCN and CuSCN are the only salts
stable in H2O due to
[CBSE AIPMT 1996]
(a) high hydration energy of Cu+ ions
(b) their inherent tendency not to
disproportionate
56 When (NH4 ) 2 Cr 2O 7 is heated, the
gas evolved is [CBSE AIPMT 1994]
(a) N2
(c) O2
Ans. (a)
(b) NO2
(d) N2O
When ammonium dichromate is heated,
it form nitrogen gas, Cr2O3 and water.
∆
(NH4 )2 Cr2O7 → Cr2O3 + 4H2O + N2
57 The most durable metal plating on
iron to protect against corrosion is
[CBSE AIPMT 1994]
(a) nickel plating (b) tin plating
(c) copper plating (d) zinc plating
Ans. (d)
Zinc is commonly used for covering iron
surfaces. The process of covering iron
with zinc is called galvanisation. If some
scratches occur on the protective zinc
film on coated iron, even then iron will
not be rusted. This is due to the fact that
because of scratches, both zinc and iron
get exposed to oxidation but zinc
undergoes oxidation in preference to
iron, because the reduction potential of
zinc is less than the reduction potential
of iron. So, zinc coating is the best and
durable method for protection of iron.
142
NEET Chapterwise Topicwise Chemistry
58 Nitriding is the process of surface
hardening of steel by treating it in
an atmosphere of
[CBSE AIPMT 1989]
(a) NH3
(c) O 3
Ans. (a)
(b) N2
(d) H2 S
Nitriding is the process of heating of
steel in the presence of ammonia. In this
process the layer of iron nitride is
formed which prevent the rusting of
iron.
59 Photographic plates and films have
an essential ingredient of
(a) silver nitrate [CBSE AIPMT 1989]
(b) silver bromide
(c) sodium chloride
(d) oleic acid
Ans. (b)
The photographic plate of film consists
of a glass plate or thin strip of celluloid
which is coated with the thin layer of an
emulsion of silver bromide dispersed in
gelatin.
TOPIC 3
Inner Transition Elements
60 The incorrect statement among the
following is
[NEET 2021]
(a) actinoid contraction is greater for
element to element than lanthanoid
contraction.
(b) most of the trivalent lanthanoid ions
are colourless in the solid state.
(c) lanthanoids are good conductors of
heat and electricity.
(d) actinoids are highly reactive metals,
especially when finely divided.
Ans. (b)
The shielding effect of 5f-orbitals in
actinoids is poor than the shielding
effect of 4f-orbitals. So, the effective
nuclear charge on valence electrons is
more in actinoids. Hence, actinoid
contraction is greater than lanthanoid
contraction.
∴ Statement (a) is correct.
Trivalent lanthanoid ions are coloured in
the solid state due to presence of
f-electrons.
∴ Statement (b) is incorrect.
Lanthanoids are inner transition metals.
So, they are good conductors of heat
and electricity.
∴ Statement (c) is correct.
The surface area increases when
actinoids are finely divided which results
in exposure of more reactant molecules
to react. Hence, rate increases and so,
actinoids are highly reactive metals
when finely divided.
∴ Statement (d) is correct.
61 Zr (Z = 40) and Hf (Z = 72) have
similar atomic and ionic radii
because of
[NEET 2021]
(a) belonging to same group
(b) diagonal relationship
(c) lanthanoid contraction
(d) having similar chemical properties
Ans. (c)
Zr (Z = 40) → 1s 2 , 2s 2 2p6 3s 2 3p6
3d 10 , 4s 2 4p6 4d2 , 5s 2
Hf(Z = 72) → 1s 2 , 2s 2 2p6 , 3s 2 3p6 3d 10 ,
4s 2 4p6 4d 10 , 5s 2 5p6 4f 14 5d2 , 6s 2
Hf is a post lanthanoid element. Due to
presence of4f-orbitals which have poor
shielding effect, the effective nuclear
charge on valence shell electrons is
more which result in the decrease of the
size of Hf. This effect is known as
lanthanoid contraction.
62 The reason for greater range of
oxidation states in actinoids is
attributed to
[NEET 2017]
(a) the radioactive nature of actinoids
(b) actinoid contraction
(c) 5f, 6d and 7s levels having
comparable energies
(d) 4f and 5d levels being close in
energies
Ans. (c)
The reason for greater range of
oxidation states in actinoid is attributed
to the 5f, 6d and 7s levels having
comparable energies.
The 5f-orbitals extend into space beyond
the 7s and 6p-orbitals and participate in
bonding. This is in direct contrast to the
lanthanides where the4f-orbitals are
buried deep inside the atom, totally
shielded by outer orbitals and thus
unable to take part in bonding.
63 The electronic configurations of Eu
(Atomic no. 63), Gd (Atomic no. 64)
and Tb (Atomic no. 65) are
[NEET 2016, Phase I]
(a) [Xe] 4f 6 5d 16s 2 , [Xe] 4f 7 5d 16s 2 and
[Xe] 4f 9 6s 2
(b) [Xe] 4f 6 5d 16s 2 , [Xe] 4f 7 5d 16s 2 and
[Xe] 4f 8 5d 16s 2
(c) [Xe] 4f 7 6s 2 , [Xe] 4f 7 5d 16s 2 and
[Xe] 4f 9 6s 2
(d) [Xe] 4f 7 6s 2 , [Xe] 4f 8 6s 2 and
[Xe] 4f 8 5d 16s 2
Ans. (c)
Electronic configuration of
7
2
63Eu = [Xe] 54 4f 6s
Electronic configuration of
7
1
2
64 Gd = [Xe] 54 4f 5d 6s
Electronic configuration of
Tb = [Xe] 54 4f 9 6s 2
65
64 Which one of the following
statements related to lanthanons is
incorrect?
[NEET 2016, Phase II]
(a) Europium shows +2 oxidation state
(b) The basicity decreases as the ionic
radius decreases from Pr to Lu
(c) All the lanthanons are much more
reactive than aluminium
(d) Ce (+4) solutuion are widely used as
oxidising agent in volumetric
analysis
Ans. (c)
Eu(63) = 4f 7 ⋅ 5d 0 , 6s 2 , Eu2 + = 4f 7
In lanthanoids series, ionic radius
decreases and covalent character
increases, thus basicity decreases.
Lanthanons are less reactive than
aluminium due to high ionisation
potential. The reason for this high
ionisation potential is lanthanoid
contraction. Ce4+ is a good oxidising
agent, it is easily converted to Ce3+
65 Gadolinium belongs to 4f series. It’s
atomic number is 64. Which of the
following is the correct electronic
configuration of gadolinium?
[CBSE AIPMT 2015]
(a) [Xe]4 f 8 6d2
(b) [Xe]4 f 9 5 s 1
(c) [Xe]4f 7 5d1 6s 2 (d) [Xe]4 f 6 5 d2 6 s 2
Ans. (c)
64 Gd =
[Xe] 4f 7 5d 1 6s 2
66 Because of lanthanoid contraction,
which of the following pairs of
elements have nearly same atomic
radii? (Numbers in the parenthesis
are atomic numbers).
[CBSE AIPMT 2015]
(a) Ti (22) and Zr (40)
(b) Zr (40) and Nb (41)
(c) Zr (40) and Hf (72)
(d) Zr (40) and Ta (73)
Ans. (c)
Because of the lanthanoid contraction Zr
(atomic radii 160 pm) and Hf (atomic radii
158 pm) have nearly same atomic radii.
143
d- and f-Block Elements
Lanthanoids include the elements from
lanthanum La (Z = 57) to lutetium
Lu(Z = 71). zirconiumZr (40) belong to the
second transition series (4d) and Hf (72)
belongs to third transition series (5d).
Lanthanoid contraction is associated
with the intervention of the4f orbitals
which are filled before the 5d-series of
elements starts. The filling of4f orbitals
before 5d-orbitals results in regular
decrease in atomic radii which
compensates the expected increase in
atomic size with increasing atomic
number. As a result of this lanthanoid
contraction, the elements of second and
third transition series have almost
similar atomic radii.
67 Reason of lanthanoid contraction is
[CBSE AIPMT 2014]
(a) negligible screening effect of
f-orbitals
(b) increasing nuclear charge
(c) decreasing nuclear charge
(d) decreasing screening effect
Ans. (a)
(a) La3+ (Z = 57)
(c) Lu 3+ (Z = 71)
Ans. (b)
(b) Ti3+ (Z = 22)
(d) Sc 3+ (Z = 21)
Key Idea Colour is obtained as a
consequence of d-d (or f-f) transition, and
for d-d (or f-f ) transition, presence of
unpaired electrons is the necessary
condition.
Electronic configuration of
La3+ (Z = 57 ) = [Xe] 4f 0 5 d 0 6s 0
(no unpaired electron)
Ti3+ (Z = 22) = [Ar] 3 d 14s 0
(one unpaired electron)
3+
Lu (Z = 71) = [Xe]4f 14 5 d 0 6s 0
(no unpaired electron)
Sc3+ (Z = 21) = [Ar] 3d 0 4s 0
(no unpaired electron)
70 Identify the incorrect statement
among the following.
[CBSE AIPMT 2007]
Lanthanoid contraction is the regular
decrease in atomic and ionic radii of
lanthanides. This is due to the imperfect
shielding [or poor screening effect] of
f-orbitals due to their diffused shape
which unable to counterbalance the
effect to the increased nuclear charge.
Hence, the net result is a contraction in
size of lanthanoids.
68 Which of the following lanthanoid
ions is diamagnetic?
(At. no. Ce = 58, Sm = 62 , Eu = 63,
Yb = 70)
[NEET 2013]
(a) Ce2+ (b) Sm2+ (c) Eu2+
Ans. (d)
69 Which of the following ions will
exhibit colour in aqueous
solutions?
[CBSE AIPMT 2010]
(d) Yb2+
Lanthanoid ion with no unpaired electron
is diamagnetic in nature.
Ce58 = [Xe] 4f2 5d 0 6s 2
Ce2 + = [Xe] 4f2
(two unpaired electrons)
Sm62 = [Xe] 4f 6 5 d 0 6s 2
Sm2 + = [Xe] 4f 6
(six unpaired electrons)
Eu63 = [Xe] 4f 7 5 d 0 6s 2
Eu2 + = [Xe] 4f 7
(seven unpaired electrons)
Yb70 = [Xe] 4f 14 5 d 0 6s 2
Yb2 + = [Xe] 4f 14
(no unpaired electrons)
Because of the absence of unpaired
electrons, Yb2 + is diamagnetic.
(a) There is a decrease in the radii of the
atoms or ions as one proceeds from
La or Lu
(b) Lanthanide contraction is the
accumulation of successive
shrinkages
(c) As a result of lanthanide contraction,
the properties of4d series of the
transition elements have no
similarities with the 5d series of
elements
(d) Shielding power of4f electrons is
quite weak
Ans. (c)
The regular decrease in the radii of
lanthanide ions fromLa3+ to Lu3+ is
known as lanthanides contraction.
It is due to the greater effect of the
increased nuclear charge than that of
screening effect (shielding effect).
As a result of lanthanide contraction, the
atomic radii of element of4d and 5d
come closer, so the properties of4d and
5d-transition element shows the
similarities.
71 More number of oxidation states
are exhibited by the actinides than
by the lanthanides. The main
reason for this is
[CBSE AIPMT 2006, 2005]
(a) more energy difference between 5f
and 6d-orbitals than that between 4f
and 5d-orbitals
(b) lesser energy difference between 5f
and 6d-orbitals than that between 4f
and 5d-orbitals
(c) greater metallic character of the
lanthanides than that of the
corresponding actinides
(d) more active nature of the actinides
Ans. (b)
More number of oxidation states are
exhibited by the actinides than by the
corresponding lanthanides due to lesser
energy difference between 5f and 6d
orbitals than that between4f and
5d-orbitals.
72 Lanthanides are
[CBSE AIPMT 2004]
(a) 14 elements in the sixth period
(At. no. = 90 to 103 ) that are filling4f
sub-level
(b) 14 elements in the seventh period
(At. no. = 90 to 103) that are filling 5f
sub-level
(c) 14 elements in the sixth period
(At. no. = 58 to 71) that are filling4f
sub-level
(d) 14 elements in the seventh period
(At. no. = 58 to 71) that are filling4f
sub-level
Ans. (c)
Lanthanides are the 14 elements of IIIB
group and sixth period (at. no. = 58 to 71)
that are filling4f sub-shell of
antipenultimate shell from 1 to 14.
Actually, they are placed below the
periodic table in horizontal row as
lanthanide series.
73 The correct order of ionic radii of
Y 3 + , La 3 + , Eu 3 + and Lu 3+ is
[CBSE AIPMT 2003]
(At. no. Y = 39, La = 57, Eu = 63, Lu
= 71)
(a) Lu 3+ < Eu 3+ < La3+ < Y3+
(b) La3+ < Eu 3 + < Lu 3+ < Y3+
(c) Y3 + < La3+ < Eu 3+ < Lu 3+
(d) Y3+ < Lu 3+ < Eu 3+ < La3+
Ans. (d)
The correct order of ionic radii of
Y 3 + , La3 + , Eu3 + and Lu3 + , is
Y 3+ < Lu3+ < Eu3+ < La3 + because Eu and
Lu are the members of lanthanide series
(so they show lanthanide contraction)
and La is the representative element of
all elements of such series and Y 3+ ion
has lower radii as comparison toLa3+
because it lies immediately above it in
the periodic table.
144
NEET Chapterwise Topicwise Chemistry
74 General electronic configuration
of lanthanides are
[CBSE AIPMT 2002]
(a) (n − 2) f 1− 14 (n − 1) s 2 p 6d 0− 1ns 2
(b) (n − 2) f 10− 14 (n − 1) d 0− 1ns 2
(c) (n − 2) f 0− 14 (n − 1) d 10ns 2
(d) (n − 2) d 0− 1 (n − 1) f 1− 14 ns 2
Ans. (a)
In lanthanides (at. no. of elements 57
to 71) the electronic configuration of
outermost shells are
(n − 2)f 1 − 14 (n − 1) s 2 p6d 0 to 1ns 2 .
75 Which of the following
statements is not correct?
[CBSE AIPMT 2001]
(a) La(OH) 3 is less basic thanLi(OH) 3
(b) In lanthanide series, ionic radius
of Ln3+ ion decreases
(c) La is actually an element of
transition series rather lanthanide
(d) Atomic radius of Zr and Hf are
same because of lanthanide
contraction
Ans. (a)
La3 + ions larger thanLi3 + . So, it easily
gives OH− ion La (OH) 3 is more basic
than Li(OH) 3. In lanthanides the basic
character of hydroxides decreases as
the ionic radius decreases.
76 Which one of the following elements
shows maximum number of different
oxidation states in its compounds?
[CBSE AIPMT 1998]
(a) Eu
(b) La
Ans. (d)
(c) Gd
(d) Am
Oxidation states shown by elements are as
follows:
La = + 3
Eu and Gd = + 2 and + 3
Am = + 2 + 3 + 4 + 5
and
+6
Am shows maximum number of different
oxidation state due to its larger size and low
ionisation energy.
77 The lanthanide contraction is
responsible for the fact that
[CBSE AIPMT 1997]
(a)
(b)
(c)
(d)
Zr and Yt have about the same radius
Zr and Nb have similar oxidation state
Zr and Hf have about the same radius
Zr and Zn have the same oxidation state
Ans. (c)
The elements of second and third transition
series resembles more in properties than
the elements of first and second transition
series. It is due to lanthanide contraction.
So, due to lanthanide contraction Zr and Hf
have the same radius and also known as
twins.
78 Among the lanthanides, the one
obtained by synthetic method is
[CBSE AIPMT 1994]
(a) Lu
(c) Pr
Ans. (b)
(b) Pm
(d) Gd
Promethium (Pm) is the element which is
prepared only by synthetic methods. It is
not present in nature. It is the only
synthetic radioactive lanthanoid.
79 Actinides
[CBSE AIPMT 1994]
(a) are all synthetic elements
(b) include element 104
(c) have any short lived isotopes
(d) have variable valency
Ans. (d)
All actinides show different oxidation
states such as +2 +3 +4 +5 and +7.
However, +3 oxidation state is most
common among all the actinides. The
wide range of oxidation states of
actinides is attributed to the fact that
the 5f,6d and 7s energy levels are of
comparable energies. Therefore, all
these three sub-shells can participate.
19
Co-ordination Compounds
TOPIC 1
number of Co is 6 with octahedral
geometry.
Nomenclature, Isomerism
and Werner’s Theory
Cl
(a) hexadentate ligand with four O and
two N donor atoms
(b) unidentate ligand
(c) bidentate ligand with two N donor
atoms
(d) tridentate ligand with three N
donor atoms
Ans. (a)
Ethylene diaminetetraacetate (EDTA)
ion.
O
O—C—CH2
O
N ––CH2 —CH2—N
CH2
×× ×× ×× ××
Cl
Cl
01 Ethylene diaminetetraacetate
(EDTA) ion is
[NEET 2021]
O
3d
Ni(CO)4
CH2—C—O
CH2—C—O
C
EDTA is a hexadentate ligand with four ‘O’
and two ‘N’ donor atoms.
02 The type of isomerism shown by
the complex [COCl 2 (en) 2] is
[NEET 2018]
(a) ionisation isomerism
(b) coordination isomerism
(c) geometrical isomerism
(d) linkage isomerism
Ans. (c)
Isomers in which the atoms or ligands
occupy different positions around
central metal/ion are called geometrical
isomers. Complexes having coordination
number of central atom/ion 6 with
formula M(AA)2 B 2 exhibit geometrical
isomerism [where, AA is a bidentate
ligand]. In [CoCl2 (en)2 ], coordination
Co
en
CO CO CO CO
en
en
Co
en
Cl
cis-isomer
(Optically
active)
trans-isomer
(Optically
inactive)
sp3-hybridised
(Tetrahedral geometry)
There is no unpaired electron, hence,
Ni(CO) 4 is diamagnetic with tetrahedral
geometry.
CO
Thus, [CoCl2 (en)2 ] show geometrical
isomerism.
03 The geometry and magnetic
behaviour of the complex [Ni(CO) 4]
are
[NEET 2018]
(a) square planar geometry and
paramagnetic
(b) tetrahedral geometry and
diamagnetic
(c) square planar geometry and
diamagnetic
(d) tetrahedral geometry and
paramagnetic
Ans. (b)
Key Concept The complexes having
sp3-hybridisation are tetrahedral while
having dsp2 -hybridisation are square
planar. The magnetic behaviour of
complexes can be paramagnetic and
diamagnetic based on the presence and
absence of unpaired electrons,
respectively.
Electronic configuration of Ni (Z =28) is
[Ar] 18 3d 8 4s 2 . Due to presence of CO
(neutral ligand), oxidation state of Ni in
[Ni(CO) 4 ] is 0.
3d
Ni
CO
OC
CO
04 The correct order of the
stoichiometrics of AgCl formed
when AgNO 3 in excess is
treated with the complexes:
CoCl 3 ⋅ 6NH3 , CoCl 3 ⋅ 5NH3 ,
CoCl 3 ⋅ 4NH3 respectively is
(a) 1 AgCl, 3 AgCl, 2 AgCl
(b) 3 AgCl, 1 AgCl, 2 AgCl
(c) 3 AgCl, 2 AgCl, 1 AgCl
(d) 2 AgCl, 3 AgCl, 1 AgCl
Ans. (c)
According to Werner’s theory,
CoCl 3 ⋅6NH3 → [Co(NH3) 6 ] 3 + 3Cl −
CoCl 3 ⋅ 5NH3 → [Co(NH3) 5 Cl]2 + 2Cl −
CoCl 3 ⋅4NH3 → [Co(NH3) 4 Cl2 ] + Cl −
When AgNO3 in excess is treated with
these complexes then following
reactions takes place :
[Co(NH3) 6 ] 3+ 3Cl − + AgNO3 → 3AgCl
(Excess)
+ [Co(NH3) 6 ] 3+
[Co(NH3) 5 Cl] 2Cl + AgNO3 →
2+
Ni-atom
4s
−
(Excess)
4p
2AgCl + [Co(NH3) 5 Cl]2 +
[Co(NH3) 4 Cl2 ] Cl + AgNO3 → AgCl
+
Since, CO is a strong field ligand, it pair
up the unpaired electrons of Ni.
[NEET 2017]
−
(excess)
+ [Co(NH3) 4 Cl2 ] +
146
NEET Chapterwise Topicwise Chemistry
05 The correct increasing order of
trans-effect of the following species
is
[NEET 2016, Phase II]
−
NH3 > CN > Br > C 6H−5
CN− > C 6H−5 > Br − > NH3
Br − > CN− > NH3 > C 6H−5
CN− > Br − > C 6H−5 > NH3
(a)
(b)
(c)
(d)
Ans. (b)
−
[CBSE AIPMT 2015]
(a) 2
(b) 1
Ans. (c)
06 Jahn-Teller effect is not observed
in high spin complexes of
[NEET 2016, Phase II]
(c) d 4
(d) d 9
Key Idea Jahn-Teller distortion is
observed in those octahedral complexes
in which d-electrons are filled
unsymmetrically.
eg
eg
t2g
d4
t2g
d7
eg
eg
t22g
t2g
d8
d9
Except d 8 , all are unsymmetrically filled,
thus d 8 complex will not show
Jahn-Teller distortion.
07 Cobalt (III) chloride forms several
octahedral complexes with
ammonia. Which of the following
will not give test for chloride ions
with silver nitrate at 25°C?
[CBSE AIPMT 2015]
(a) CoCl3 ⋅ 3NH3
(c) CoCl3 ⋅5NH3
Ans. (a)
(c) 3
(d) 4
[Co(en)2 Cl2 ]Cl
Possible isomers are
Trans effect is the effect of a
coordinated group upon the rate of
substitution at the position trans to itself
in a square or octahedral complex.As the
rate of substitution of the trans ligand
increases, the intensity of trans effect
also increases. Thus, correct order is,
CN− > C6H–5 > Br – > NH3
(a) d 7
(b) d 8
Ans. (b)
08 Number of possible isomers for the
complex [Co(en) 2 Cl 2]Cl will be
(en = ethylenediamine)
(b) CoCl3 ⋅4NH3
(d) CoCl3 ⋅ 6NH3
[Co (NH3) 6 ]Cl 3 → [Co(NH3) 6 ] 3+ + 3Cl −
[Co (NH3) 3 Cl 3] → [Co(NH3) 3 Cl 3]
[Co(NH3) 4 Cl2 ] Cl → [Co(NH3) 4 Cl2 ] +
+ Cl −
2+
[Co(NH3) 5 Cl] Cl2 → [Co(NH3) 5 Cl]
+ 2Cl −
So, [Co (NH3) 3 Cl 3] does not ionise so
does not give test for chloride ions.
Cl
en
Cl
Cl
Co
en
en
cis
Optically active
Stereoisomers =2
Co
en
Cl
trans
Optically inactive
Stereoisomers =1
Hence, total number of stereoisomers
= 2 + 1 = 3.
09 The sum of coordination number
and oxidation number of the metal
M in the complex [M(en) 2 (C 2O 4 )]Cl
(where, en is ethylenediamine) is
[CBSE AIPMT 2015]
(a) 9
(b) 6
Ans. (b)
(c) 7
(d) Linkage isomerism
Ans. (b)
The complexes [Co(NH3) 6 ][Cr(CN) 6 ] and
[Cr(NH3) 6 ][Co(CN) 6 ] are the examples of
coordination isomerisms. This
isomerism occurs only in those
complexes in which both cation and
anion are complex. It occurs due to
exchange of ligands between cation and
anion.
12 The complex, [Pt(Py)(NH3 )BrCl] will
have how many geometrical
isomers?
[CBSE AIPMT 2011]
(a) 4
(b) 0
Ans. (d)
(c) 2
(d) 3
The complex is square planar and is of
the type [M(abcd)]. It has three
geometrical isomers.
Py
NH3 Py
Br
Pt
Pt
Cl
Br ;
Cl
NH3 ;
Py
NH3
(d) 8
Given complex compound is
[M(en)2 (C2O4 )]Cl
Let oxidation number of M is x.
∴
x −2−2= − 1
or
x=+3
Now, as coordination number is defined
as the total number of binding sites
attached to the metal. Hence, in the
given complex coordination number is 6.
10 The name of complex ion,
[Fe(CN) 6] 3− is [CBSE AIPMT 2015]
(a) hexacyanoiron (III) ion
(b) hexacyanitoferrate (III) ion
(c) tricyanoferrate (III) ion
(d) hexacyanidoferrate (III)ion
Ans. (d)
Key Concept When complex ion is an
anion, the name of the metal ends with
suffix -ate along with its oxidation
number in the complex ion.
[Fe(CN) 6 ] 3 − = Hexacyanoferrate (III) ion
11 The complex [Co(NH3) 6][Cr(CN) 6]
and [Cr(NH3 ) 6][Co(CN) 6] are the
examples of which type of
isomerism?
[CBSE AIPMT 2011]
(a) Ionisation isomerism
(b) Coordination isomerism
(c) Geometrical isomerism
Pt
Br
Cl
13 The existence of two different
coloured complexes with the
composition of [Co(NH3 ) 4 Cl 2] + is
due to
[CBSE AIPMT 2010]
(a) linkage isomerism
(b) geometrical isomerism
(c) coordination isomerism
(d) ionisation isomerism
Ans. (b)
Key Idea Complexes of [MA4 B2 ] type
exhibit geometrical isomerism.
The complex [Co(NH3) 4 Cl2 ] + is a [MA4 B2 ]
type complex and thus, fulfills the
conditions that are necessary to exhibit
geometrical isomerism. Hence, it has two
geometrical isomers of different colours
as :
The structure of the geometrical
isomers is as
Cl
H3N
Cl
Co
H3N
NH3
Cis-form
NH3
147
Co-ordination Compounds
Ans. (b)
Cl
H3N
NH3
Co
H3N
Cl
NH3
Trans-form
For linkage isomerism, presence of
ambidentate ligand is necessary.
For coordination isomerism, both the
cation and anion of the complex must be
complex ions.
For ionisation isomerism, an anion
different to the ligands must be present
outside the coordination sphere. All
these conditions are not satisfied by this
complex. Hence, it does not exhibit
other given isomerisms.
14 Which of the following does not
show optical isomerism?
(en = ethylenediamine)
[CBSE AIPMT 2009]
(a) [Co(en)2 Cl2] +
(b) [Co(NH3) 3Cl3] 0
(c) [Co(en)Cl2 (NH3)2] +
(d) [Co(en) 3] 3+
Ans. (b)
Optical isomerism is exhibited only by
those complexes in which plane of
symmetry are absent. Octahedral
complexes of the types [M(aa) 3],
[M(aa) x2 , y2 ] and [M(aa)2 x2 ] have absence
of plane of symmetry, thus exhibit
optical isomerism. Here, aa represents
bidentate ligand, x or y represents
monodentate ligand and M represents
central metal ion.
Hence, [Co(NH3) 3 Cl 3] 0 due to presence
of symmetry elements does not exhibit
optical isomerism.
or
Octahedral complexes of [M(AA)2 B2 ]
type, e.g. [Co(en)2 Cl2 ] + , [M(AA) B2C2 ]
type, e.g. [Co(en)Cl2
(NH3)2 ] and [M(AA) 3] type, e.g. [Co(en) 3] 3+
show optical isomensim, whereas
complexes of [ MA3 B3 ] type, e.g.
[Co(NH3)Cl2 ] 0 do not show optical
isomerism.
15 Which of the following will give a
pair of enantiomers?
(en =NH2CH2CH2NH2 )
[CBSE AIPMT 2007]
(a) [Cr(NH3) 6] [Co(CN) 6]
(b) [Co(en)2 Cl2]Cl
(c) [Pt(NH3) 4] [PtCl6]
(d) [Co(NH3) 4 Cl2]NO2
Enantiomorphs or Enantiomers A pair
of molecules related to each other as an
object and its mirror images are known
as enantiomorphs or enantiomers.
These are not superimposable on its
mirror image.
The example is [Co(en)2 Cl2 ] +
Dichlorobis (ethylene diamine) cobalt (III)
Mirror
en
en
Cl
Cl
Co
Co
Cl
Cl
en
en
17 Which one of the following is
expected to exhibit optical
isomerism?
(en = ethylenediamine)
[CBSE AIPMT 2005]
(a) Cis-[Pt(NH3)2 Cl2]
(b) Trans-[Co(en)2 Cl2] +
(c) Trans-[Pt(NH3)2 Cl2]
(d) Cis-[Co(en)2 Cl2] +
Ans. (d)
Cis- [Co(en)2 Cl2 ] + is able to show the
phenomenon of optical isomerism
because it can form a superimposable
mirror image.
Enantiomorphs
en
[CBSE AIPMT 2006]
The compound [Co(NH3) 4 (NO2 )2 ]Cl
exhibits linkage, ionisation and
geometrical isomerism. Hence,
(i) its linkage isomers are
[Co(NH3) 4 (NO2 )2 ] Cl and
[Co(NH3) 4 (ONO)2 ]Cl
(ii) its ionisation isomers are
[Co(NH3) 4 (NO2 )Cl]NO2 and
[Co(NH3) 4 (NO2 )2 ]Cl
(iii) its geometrical isomers are
+
NH3
H3N
en
Co
16 [Co(NH3 ) 4 (NO 2 ) 2]Cl exhibits
(a) linkage isomerism, geometrical
isomerism and optical isomerism
(b) linkage isomerism, ionisation
isomerism and optical isomerism
(c) linkage isomerism, ionisation
isomerism and geometrical
isomerism
(d) ionisation isomerism, geometrical
isomerism and optical isomerism
Ans. (c)
+
Cl
Mirror image
Cl
It gives super imposable structure.
but trans-form is optically inactive
Cl
Co
Transform en
en
Cl
Optically active form
18 Which of the following coordination
compounds would exhibit optical
isomerism?
[CBSE AIPMT 2004]
(a) Pentaamminenitrocobalt (III)
iodide
(b) Diamminedichloroplatinum (II)
(c) Trans-dicyanobis
(ethylenediamine) chromium (III)
chloride
(d) Tris-(ethylenediamine) cobalt (III)
bromide
Ans. (d)
Tris-(ethylenediamine) cobalt (III)
bromide [Co(en) 3]Br3 exhibits optical
isomerism :
NO2
Co
3+
en
H3N
NO2
en
NH3
Co
Cis-isomer
+
NO2
H3N
and
NH3
Co
en
Mirror
d-form
NH3
H3N
NO2
Trans-isomer
3+
en
Co
en
l-form
en
148
NEET Chapterwise Topicwise Chemistry
19 According to IUPAC nomenclature
sodium nitroprusside is named as
23 Which one of the following
complexes will have four isomers?
[CBSE AIPMT 2003]
[CBSE AIPMT 2000]
(a) sodium pentacyanonitrosyl ferrate
(II)
(b) sodium pentacyanonitrosyl ferrate
(III)
(c) sodium nitroferricyanide
(d) sodium nitroferrocyanide
Ans. (b)
(a) [Co(en) 3]Cl3
(b) [Co(en)2 Cl2]Cl
(c) [Co(PPh3)2 (NH3)Cl2]Cl
(d) [Co(PPh3) 3Cl]Cl2
Ans. (b)
20 Which one of the following
octahedral complexes will not show
geometrical isomerism? (A and B
are monodentate ligands)
[CBSE AIPMT 2003]
(a) [MA4B2]
(c) [MA2B4]
Ans. (b)
(b) [MA5B]
(d) [MA3B3]
[MA 5 B] due to absence of symmetry of
B cannot exist in the form of
cis-trans-isomer.
21 The hypothetical complex chloro
diaquatriammine cobalt (III)
chloride can be represented as
[CBSE AIPMT 2002]
(a) [CoCl(NH3) 3 (H2O)2]Cl2
(b) [Co(NH3) 3 (H2O)Cl3]
(c) [Co(NH2 ) 3 (H2O)2 Cl]
(d) [Co(NH3) 3 (H2O) 3]Cl3
Ans. (a)
Chlorodiaquatriammine cobalt (III)
chloride is [CoCl(NH3) 3 (H2O)2] Cl2 .
22 Which of the following will give
maximum number of isomers?
[CBSE AIPMT 2001]
(a) [Co(NH3) 4 Cl2]
(b) [Ni(en)(NH3) 4] 2+
(c) [Ni(C2O4 )(en)2] 2–
(d) [Cr(SCN)2 (NH3) 4] +
Ans. (d)
+
Cl
en
Cl
en
Co
en
+
26 IUPAC name of
[Pt(NH3 ) 3 (Br) (NO 2 ) Cl] Cl is
en
Cl
[CBSE AIPMT 1998]
Cis
(a) triamminebromochloronitroplatinum
(IV) chloride
(b) triamminebromonitrochloroplatinum
(IV) chloride
(c) triamminechlorobromonitroplatinum
(IV) chloride
(d) triamminenitrochlorobromoplatinum
(IV) chloride
(ii) Optical isomers
+
Cl
Cl
en
+
Cl
Cl
Co
en
Co
en
(b) 6
(d) 4
Given complex gives four isomers that
are as follow.
(i) [Cu(NH3) 4 ] [PtCl 4 ]
(ii) [Cu(NH3) 3 Cl] [Pt(NH3)Cl 3]
(iii) [Pt(NH3) 3 Cl] [Cu(NH3)Cl 3]
(iv) [Pt(NH3) 4 ] [CuCl 4 ]
Cl
Co
Trans
en
Optically active cis-[Co(en)2Cl2]Cl
24 A coordination complex compound
of cobalt has the molecular formula
containing five ammonia
molecules, one nitro group and two
chlorine atoms for one cobalt
atom. One mole of this compound
produces three mole ions in an
aqueous solution. On reacting this
solution with excess of
AgNO 3 s olution, we get two moles
of AgCl precipitate. The ionic
formula for this complex would be
Ans. (a)
The IUPAC name of
[Pt(NH3) 3 (Br)(NO2 )Cl]Cl is
triamminebromochloronitroplatinum (IV)
chloride.
(Oxidation number ofPt = + 4, and ligands
are arranged in alphabetical order)
27 The number of geometrical
isomers of the complex
[Co(NO 2 ) 3 (NH3 ) 3] is
[CBSE AIPMT 1997]
(a) 4
(c) 2
Ans. (c)
[CBSE AIPMT 1998]
(a) [Co(NH3) 5 (NO2 )]Cl2
(b) [Co(NH3) 5 Cl] [Cl(NO2 )]
(c) [Co(NH3) 4 (NO2 )Cl] [(NH3)Cl]
(d) [Co(NH3) 5] [(NO2 )2 Cl2]
Ans.
(b) 0
(d) 3
Geometrical isomers of the complex
[Co(NO2 ) 3 (NH3) 3] are two. These are
NO2
H3N
NO2
NO2 H3N
NO2
(a)
Co
Co
The complex gives three ions in aqueous
H
N
NH
NO2
H
N
3
3
3
solution.
∴The complex should be
NH3
NO2
[CO(NH3) 5 NO2 ]Cl2 .
Facial
isomers
Maridonal isomers
It will give three ions on dissociation as
follows :
28 The formula of dichlorobis (urea)
[Co(NH3) 5 NO2 ] Cl2 → [Co(NH3) 5 NO2 ]2 +
copper (II) is
[CBSE AIPMT 1997]
1 mol
+ 2Cl − (Counter ion)
+
[Cr(SCN)2 (NH3) 4 ] Oxidation number of
Cr is 6 shows linkage, geometrical and
optical isomerism.
[CBSE AIPMT 1998]
(a) 5
(c) 3
Ans. (d)
Complex [Co(en)2 Cl2 ]Cl will have four
different isomers.
(i) Geometrical isomers
IUPAC name of sodium nitroprusside
Na2 [Fe (CN) 5 NO] is sodium pentacyanoni
trosyl ferrate (III) because in it NO is
neutral ligand and the oxidation number
of Fe is +3. Which is calculated as
Na2 [Fe (CN) 5 NO]
2 × (+1) + x + 5 × (– 1) + 1 × 0 =0
[where x = oxidation state of Fe]
2+ x − 5=0
x −3=0 ⇒ x = +3
25 The total number of possible
isomers for the complex compound
[CuII (NH3 ) 4] [PtII Cl 4] are
2 mol
2Cl + 2AgNO3 → 2AgCl + 2NO–3
–
Silver ppt
(a) [Cu{O == C(NH2 )2 Cl}] Cl
(b) [CuCl2 {O == C(NH2 )2 }2]
(c) [Cu{O == C(NH2 )2 }] Cl2
(d) [CuCl2 {O == C(NH2 )2 H2 }]
149
Co-ordination Compounds
Ans. (b)
Ans. (d)
The formula of dichlorobis (urea) copper
(II) is [CuCl2 {O == C(NH2 )2 }2 ].
A. [Fe(CN)6 ]3− Oxidation number of
Fe = x + 6 (−1) = − 3
x − 6 = −3 ⇒ x = + 3
Electronic configuration of
Fe → [Ar] 3d 6 4s 2
Electronic configuration of
Fe3+ → [Ar] 3d 5 4s 0
29 The number of geometrical
isomers for [Pt(NH3 ) 2]Cl 2 is
[CBSE AIPMT 1995]
(a) 3
(c) 1
Ans. (d)
(b) 4
(d) 2
3d 5
Geometrical isomers for [Pt(NH3)2 Cl2 ] is
Cl
NH3
32 Urea reacts with water to form A
which will decompose to form B. B
when passed through Cu 2+ (aq),
deep blue colour solutionC is formed.
What is the formula of C from the
following?
[NEET (Sep.) 2020]
4s 0
CN− is a strong field ligand so, pairing
of electrons occurs.
3d 5
Pt
Complete reaction is
O
4s 0
3–
[Fe(CN) ] =
Cl
Cl
NH3
= 1.73 BM
is +3.
Water is weak field ligand, so pairing
of electron do not take place.
Cl
[CBSE AIPMT 1989]
(a) bleaching powder
(b) K 4 [Fe(CN) 6]
(c) hypo
(d) potash alum
Ans. (d)
(∆O<P)
d5
t2 g
Number of unpaired electrons = 5.
µ s = n(n + 2 ) = 5 × (5 + 2 )
C. [Fe(CN)6 ] 4− Oxidation state of Fe
is +2.
Electronic configuration of
Fe2 + = [Ar] 3d 6 4s 0
CN− is a strong field ligand, so pairing
of electrons take place.
eg
TOPIC 2
(DO>P)
Bonding in Coordination
Compounds
31 Match List-I with List-II. [NEET 2021]
List-I
List-II
A.
[Fe(CN) 6] 3 −
1.
B.
[Fe(H2O) 6] 3 +
2. 0 BM
C.
[Fe(CN) 6] 4 −
3. 4.90 BM
[Fe(H2O) 6]
5.92 BM
Choose the correct answer from
the options given below
A B C
(a) 4 2 1
D
3
A B
(b) 2 4
C D
3 1
(c) 1
2
(d) 4
2 3
3
4
d6
t2g
Number of unpaired electrons = 0
µ s = 0 (0 + 1) = 0 BM
D. [Fe(H2O)6 ]4− Oxidation state of Fe
is +2 (3d 6 ).
Water is weak field ligand, so pairing
of electron do not take place.
eg
4. 1.73 BM
1
[Cu(NH3)4] (aq)
B
C
Deep blue solution
Cu2+(aq)
Number of unpaired electrons = 4
µ s = 4(4 + 2) = 4 × 6
= 24 = 4.90 BM
Hence, correct match is
A → 4, → B → 1, C → 2, D → 3.
(a) SCN− < F − < CN− < C2O24−
(b) F − < SCN− < C2O24− < CN−
(c) CN− < C2O24− < SCN− < F −
(d) SCN− < F − < C2O24− < CN−
Ans. (d)
Increasing field strength of ligands to
form coordination compounds, we have
to follow the spectrochemical series.
I− < Br − < SCN− < Cl − < S2 − < F − < OH− <
C2O24 − < H2O < NCS − < EDTA4 − < NH3 < en <
CN− < CO
So, the order given in option (d) :
SCN− < F − < C2O24 − < CN − is correct.
34 The Crystal Field Stabilisation
Energy (CFSE) for [CoCl 6] 4− is
18000 cm −1 . The CFSE for
[CoCl 4] 2− will be
[NEET (Odisha) 2019]
(∆O<P)
d6
33 Which of the following is the
correct order of increasing field
strength of ligands to form
coordination compounds?
[NEET (Sep.) 2020]
= 5 × 7 = 35 = 5.92 BM
Double salts are additon or molecular
compounds which are formed by two
apparently saturated compounds but
they lose their identity when dissolved in
water. The most common example of
double salt is potash alum
K2SO4 ⋅ Al2 (SO4 ) 3 ⋅24H2O.
CO2(g) + NH3(g)
2+
eg
30 An example of a double salt is
D.
A
Ammonium
carbonate
Urea
B. [Fe(H2O)6 ]3+ Oxidation state of Fe
(Trans)
dark yellow
2 +
H2N—C—ONH4
H2N—C—NH2 + H2O
Number of unpaired electrons = 1
µ s = n(n + 1) = 1 (1 + 2) = 3
Pt
H3N
O
6
NH3
(Cis)
pale yellow
(a) [Cu(NH3) 4] 2+
(b) Cu(OH)2
(c) CuCO 3 ⋅Cu(OH)2
(d) CuSO 4
Ans. (a)
t2g
(a) 6000 cm −1
(c) 18000 cm −1
Ans. (d)
(b) 16000 cm −1
(d) 8000 cm −1
[CoCl 6 ] 4− is an octahedral while
[CoCl 4 ]2 − is a tetrahedral complex.
∆ octahedral = 18000 cm −1
150
NEET Chapterwise Topicwise Chemistry
We know that,
4
∆ tetrahedral = ∆ octahedral
9
4
= × 18000 cm −1 = 8000 cm −1
9
35 Aluminium chloride in acidified
aqueous solution forms a complex ‘
A’, in which hybridisation state of Al
is ‘B’. What are ‘A’ and ‘B’,
respectively? [NEET (Odisha) 2019]
(a) [Al(H2O) 6] 3+ , sp 3d2
(b) [Al(H2O) 4] 3+ , sp 3
(c) [Al(H2O) 4] 3+ ,dsp2
(d) [Al(H2O) 6] 3+ ,d2 sp 3
Ans. (a)
Aluminium chloride in acidified aqueous
solution forms an octahedral complex.
Aqueous solution is mostly water and
when the compound is dissolved in
acidified aqueous solution, the water
fills its vacancies and an octahedral
complex, ‘A’ which is [Al(H2O) 6 ] 3+ is
formed.
Al(13) = s ,22s 2 , 2p6 , 3s 2 , 3p1
Al 3+ =1s 2 , 2s 2 , 2p6 .
2s
37 What is the correct electronic
configuration of the central atom in
K 4 [Fe(CN) 6] based on crystal field
theory?
[NEET (National) 2019]
(a) t26g e g0 (b) e 3 t23
Ans.
(c) e 4 t22
dx2–y2 dz2
d-orbitals
free ion
eg
Average energy of the
d-orbitals in spherical
crystal field
Hence, the electronic configuration of
the central atom inK4 [Fe(CN) 6 ] is t26g e g0 .
(a) trinuclear
(c) tetranuclear
Ans. (b)
2p
[Al(H2O)6]3+ =
3p
3d
H2O H2O H2O H2O H2O H2O
sp3d2
Thus, the hybridisation state of Al in
[Al(H2O) 6 ] 3+ is sp3d2 (B).
36 Which of the following species is
not stable?
[NEET (Nationa) 2019]
(a) [GeCl6] 2−
(c) [SiCl6] 2−
Ans. (c)
t2g
Splitting of d-orbitals
in octahedral
crystal field
[NEET 2018]
2s
(b) [Sn(OH) 6] 2−
(d) [SiF6] 2−
[SiCl 6 ]2 − is not stable and does not exist
because
(i) six large chloride ions cannot be
accomdated around Si4 + due to
limitation of its size.
(ii) interaction between lone pair of
chloride ion andSi4 + is not very
strong.
On the other hand, due to presence of
d-orbital in Si, Ge and Sn they form
species like [SiF6 ]2 − , [GeCl 6 ]2 − and
[Sn(OH) 6 ]2 − . Hence, option (c) is correct.
(a) ruthenocene
(b) Grignard’s reagent
(c) ferrocene
(d) cobaltocene
Ans. (b)
The organometallic compounds having
sigma bond between carbon and metal
are sigma bonded organometallic. An
example of a sigma bonded
organometallic compound is Grignard’s
reagent.
R Mg  X
12
4 4
3
σ− bond
dxy dxz dyz
38 Iron carbonyl, Fe(CO) 5 is
2p
39 An example of a sigma bonded
organometallic compound is
[NEET 2017]
(a)
In K4 [Fe(CN) 6 ],
Fe2 + = [Ar] 3d 6 4s 0 .
CN− is a strong field ligand and as it
approaches the metal ion, the electrons
must pair up.
The splitting of thed-orbitals in two sets
orbitals in an octahedral complex,
K4 [Fe(CN) 6 ] may be represented as :
Al3+=
3s
(d) t24g e2g
The examples of dinuclear, trinuclear
complexes are Co2 (CO) 18 , Fe3 (CO) 12
respectively.
(b) mononuclear
(d) dinuclear
Key Concept Compounds of transition
metal with carbonyls (carbon monoxide)
are known as metal carbonyls. These are
classified into mononuclear, dinuclear,
trinuclear and so on based on the number
of central metal atoms/ions present in a
complex.
Complexes following EAN rule have EAN
of central metal/ion equal to nearest
inert gas configuration and hence, are
stable.
Effective atomic number (EAN) of the
metal in a complex is given by
EAN = Atomic number (Z)
− Oxidation number (O.N)
+ 2 (Coordination number)
= 26 − 0 + 2 (5 ) = 36
Thus, Fe(CO) 5 is a stable complex/ion.
Since, there is only one central metal
atom present in iron carbonyl,Fe(CO) 5 ,
thus it is mononuclear. The structure of
Fe(CO) 5 is shown below:
CO
Fe
CO
Ru
Fe
CO
Co
Ruthenocene Ferrocene Cobaltocene
40 Correct increasing order for the
wavelengths of absorption in the
visible region for the complexes of
CO 3+ is
[NEET 2017]
(a) [Co(en) 3] 3+ , [Co(NH3) 6] 3+ ,
[Co(H2O) 6] 3+
(b) [Co(H2O) 6] 3+ , [Co(en) 3] 3+ ,
[Co(NH3) 6] 3+
(c) [Co(H2O) 6] 3+ , [Co(NH3) 6] 3+ ,
[Co(en) 3] 3+
(d) [Co(NH3) 6] 3+ , [Co(en) 3] 3+ ,
[Co(H2O) 6] 3+
Ans. (a)
Key concept Wavelength (λ) of
absorption is inversely proportional to
CFSE (∆O value) of ligands attached with
the central metal ion
1
i.e.
λ∝
∆O
According to spectrochemical series.
I − < Br − < S2 − < SCN − < Cl − < F − < OH−
< C2O24− < O2 − < H2O < NSS − < NH3 < en <
NO2− < CN −
Weak field Increasing order of ∆
o
ligands
OC
OC
Whereas, ruthenocene, ferrocene and
cobaltocene are not sigma bonded
organometallic compound.
Strong
field
ligands
The CFSE of ligands attached with Co3+
ion is in the order
151
Co-ordination Compounds
en > NH3 > H2O (From spectrochemical
series)
Q Wavelength of absorbed light (λ) ∝
1
∆o
∴ For ligand the order of wavelength of
absorption in the visible region will be :
en < NH3 < H2O
or, [Co(en) 3] 3+ < [Co(NH3) 6 ] 3+ <
[Co(H2O) 6 ] 3+
41 Pick out the correct statement
with respect to [Mn(CN) 6] 3−
[NEET 2017]
3 2
(a) It is sp d hybridised and
octahedral
(b) It is sp 3d2 hybridised and
tetrahedral
(c) It is d2 sp 3 hybridised and
octahedral
(d) It is dsp2 hybridised and square
planar
Ans. (c)
[Mn(CN) 6 ] 3− is d2 sp3-hybridised and
octahedral complex. In [Mn(CN) 6 ] 3− , Mn
is in +3 oxidation state
Mn3+ = 3d 4 4s 0
4p
4s
3d
Orbitals of
=
3+
Mn ion
3d
4s
4p
[Mn (CN)6]3– =
d2sp3-hybridised
(Octahedral)
42 Which of the following has longest
C—O bond length? (Free C—O bond
length in CO is 1.128 Å.)
[NEET 2016, Phase I]
(a) [Co(CO) 4] −
(c) [Mn(CO) 6] +
Ans. (b)
(b) [Fe(CO) 4] 2 −
(d) Ni(CO) 4
As negative charge on metal carbonyl
complex increases, back π-bonding
increases and hence bond length of C—O
bond increases while bond length of
metal-carbon bond decreases.
Hence, [Fe(CO) 4 ]2 − has longest C—O
bond length among the given complexes.
The correct order of bond length of the
given complexes is
[Mn(CO) 6 ] + < [Ni(CO) 4 ] < [Co(CO) 4 ] −
< [Fe(CO) 4 ]2 −
43 Which of these statements about
[Co(CN) 6] 3– is true?
[CBSE AIPMT 2015]
(a) [Co(CN) 6] 3– has no unpaired
electrons and will be in a low-spin
configuration.
(b) [Co(CN) 6] 3– has four unpaired
electrons and will be in a low-spin
configuration.
(c) [Co(CN) 6] 3– has four unpaired
electrons and will be in a high-spin
configuration.
(d) [Co(CN) 6] 3– has no unpaired
electrons and will be in a high-spin
configuration.
Ans. (a)
[Co(CN) 6 ] 3−
Co3+ = 1s 2 2s 2 2p6 3s 2 3p6 3d 6
CN− is a strong field ligand and as it
approaches the metal ion, the electrons
must pair up.
The splitting of thed-orbitals into two
sets of orbitals in an octahedral
[Co(CN) 6 ] 3− may be represented as
3d
dx2–y2 dz2
eg
∇orbital
o
dxy dxz dxz
t2g
orbital
Here, for d 6 ions, three electrons first
enter orbitals with parallel spin put the
remaining may pair up int2 g orbital giving
rise to low spin complex (strong ligand)
field.
∴ [Co(CN) 6 ] 3− has no unpaired electron
and will be in a low spin configuration.
44 The hybridisation involved in
complex [Ni(CN) 4] 2− is (Atomic
number of Ni = 28)
[CBSE AIPMT 2015]
(a) dsp 2
(c) d 2 sp 2
Ans. (a)
(b) sp 3
(d) d2sp 3
[Ni(CN) 4 ]2 −
Let oxidation state of Ni in [Ni(CN) 4 ]2 − is
x.
∴ x −4= −2
or
x =2
Now, Ni2 + = [Ar] 3d 8 4s 0
3d
3s
4p
Q CN– is a strong field ligand. Hence, all
unpaired electrons are paired up.
3d
3s
CN–
4p
CN– CN– CN–
∴ Hybridisation of [Ni(CN) 4 ]2 − is dsp2
45 Among the following complexes,
the one which shows zero crystal
field stabilisation energy (CFSE) is
[CBSE AIPMT 2014]
3+
(a) [Mn(H2O) 6]
(c) [Co(H2O) 6] 2+
Ans. (b)
(b) [Fe(H2O) 6] 3+
(d) [Co(H2O) 6] 3+
The CFSE for octahedral complex is
given by
CFSE = [−0.4 t2 g e − + 0.6 e g e − ]
For Mn3+ , [3d 4 ] → t23g e g1
∴ CFSE = [(−0.4 × 3) + (0.6 × 1)] = −0.6
For Fe3+ , [3d 5 ] → t23g e2g
CFSE = [− (0.4 × 3) + (0.6 × 2)] = 0
For Co2 + , [3d 7 ] → t25g e2g
CFSE = [(−0.4 × 5) + (2 × 0.6)] = −0.8
For Co3+ , [3d 6 ] → t24g e2g
CFSE = [(−0.4 × 4) + (2 × 0.6)] = −0.4
46 A magnetic moment of 1.73 BM will
be shown by one among the
following
[CBSE AIPMT 2013]
(a) [Cu(NH3) 4] 2+
(c) TiCl4
Ans. (a)
(b) [Ni(CN) 4] 2−
(d) [CoCl6] 4−
Magnetic moment,µ is related with
number of unpaired electrons as
µ = n (n + 2) BM
(1.73)2 = n (n + 2)
On solving
n= 1
Thus, the complex/compound having
one unpaired electron exhibit a magnetic
moment of 1.73 BM.
(a) In [Cu(NH3) 4 ]2 +
Cu2 + = [Ar] 3d 9
(Although in the presence of strong field
ligand NH3, the unpaired electron
gets excited to higher energy level but it
still remains unpaired).
(b) In [Ni(CN) 4 ]2 −
Ni2 + = [Ar] 3d 8
But CN− being strong field ligand pair up
the unpaired electrons and hence in this
complex, number of unpaired electrons
= 0.
(c) In [TiCl 4 ]
Ti 4+ = [Ar]
no unpaired electron.
152
NEET Chapterwise Topicwise Chemistry
(c) In [Cr(NH3) 6 ] 3+ ,
Cr 3+ = [Ar] 3d 3
(d) In [CoCl 6 ] 4−
Co2 + = [Ar] 3d 7
[Cr(NH3)6]3+ =
3d
47 Which one of the following is an
outer orbital complex and exhibits
paramagnetic behaviour?
[CBSE AIPMT 2012]
2+
2+
(a) [Ni(NH3) 6]
(c) [Cr(NH3) 6] 3+
Ans. (a)
(b) [Zn(NH3) 6]
(d) [Co(NH3) 6] 3+
4p
{
d2sp3 hybridisation
Three
unpaired
electrons
3d
[Co(NH3)6]3+ =
3d
4p
4s
Because of the involvement of (n − 1) d
orbital in hybridisation, it is an inner
orbital complex. As all the electrons are
paired, it is a diamagnetic complex.
3d
Two
Unpaired
electrons
4s
d2sp3
⇒ Inner orbital complex
4s
[Ni(NH 3)6 ]2+ =
4d
4p
sp 3d 2 hybridisation
So, this is an outer orbital complex as it
involve 4d- orbitals for bonding, but
having paramagnetic character.
(b) In [Zn(NH3) 6 ]2 + :
Zn2 + = [Ar] 3d 10
48 Of the following complex ions
which is diamagnetic in nature?
[CBSE AIPMT 2011]
(a) [Ni(CN) 4] 2–
(c) [CoF6] 3–
Ans. (a)
(b) [CuCl4] 2−
(d) [NiCl4] 2–
Electronic configuration of Ni2 + in
[Ni(CN) 4 ]2 – is
Ni2+ = 3d8 4s0
3d
4s
4p
Ni2 + has dsp2 hybridisation, as CN− is a
strong field ligand.
3d
(a) In [Ni(CN) 4 ]2 − , Ni is present asNi2 + .
∴
Ni2 + = [Ar] 3d 8 4 s 0
[Ni(CN)4]2– =
3d
Because of the involvement of (n − 1)d,
i.e.3d-orbital in hybridisation, it is an
inner orbital complex. Its nature is
paramagnetic because of the presence
of three unpaired electrons.
(d) In [Co(NH3) 6 ] 3+
Co3+ = [Ar] 3d 6
3d
{
Outer orbital complex utilises
(n − 1) d-orbitals for bonding and exhibit
paramagnetic behaviour, only if there
present unpaired electrons.
(a) In [Ni(NH3) 6 ]2 + :
Ni2 + = [Ar] 3d 8 4 s 0
4s
{
It contains three unpaired electrons.
Thus, [Cu(NH3) 4 ]2 + is the complex that
exhibits a magnetic moment 1.73 BM.
Ans. (a)
Key Idea For the absorption of visible
light, presence of unpairedd-electrons is
the necessity.
[Zn(NH3)6]2+ =
××
××
×× ××
4s
4p
14444244443
dsp2 hybridisation
(Pairing occurs because CN− is a strong
field ligand).
Since, in [Ni(CN) 4 ]2 − , no unpaired
electron is present in d-orbitals so it
does not absorb visible light.
(b) In [Cr(NH3) 6 ] 3+ , Cr is present as Cr 3+ .
Cr 3+ = [Ar] 3d 34 s 0 (Three unpaired
electrons)
(c) In [Fe(H2O) 6 ] 3+ , Fe is present asFe2 + .
[Ar]3d 6 4 s 0 (Four unpaired electrons)
(d) In [Ni (H2O) 6 ]2 + , Ni is present asNi2 + .
Ni2 + = [Ar] 3d 8 4 s 0 (Two paired electrons)
The complex given in option (b), (c), (d)
have unpaired electrons, thus absorb
visible light.
NOTE In presence ofNH3 and H2O pairing
does not occur as they are strong field
ligand.
50 Crystal field stabilisation energy for
high spin d 4 octahedral complex is
[CBSE AIPMT 2010]
(a) −1.8 ∆ o
(c) −1.2 ∆ o
Ans. (d)
(b) − 1.6 ∆ o + P
(d) −0.6 ∆ o
Key Idea In case of high spin complex,
∆ o is small. than the pairing energy. That
means, the energy required to pair up the
fourth electron with the electrons of
lower energyd-orbitals would be higher
than that required to place the electrons
in the higher d-orbital. Thus, pairing does
not occur.
For high spind 4 octahedral complex,
3d
eg
CN–
4p
4d
{
4s
sp3d2 hybridisation
Thus, it is also an outer orbital complex as
it involve4d- orbitals for bonding but it is
diamagnetic as all the electrons are
paired.
CN– CN– CN–
2−
∴ [Ni(CN) 4 ] is diamagnetic (because of
the absence of unpaired electrons).
49 Which of the following complex
ions is not expected to absorb
visible light?
[CBSE AIPMT 2010]
(a) [Ni (CN) 4] 2−
(c) [Fe (H2O) 6] 2+
(b) [Cr (NH3) 6] 3+
(d) [Ni (H2O) 6] 2+
Degenerate
d-orbitals
0.6∆
0.4∆
∴Crystal field stabilisation energy
= (− 3 × 0.4 + 1 × 0.6) ∆ o
= ( − 1.2 + 0.6) ∆ o
= − 0.6 ∆ o
t2g
153
Co-ordination Compounds
51 Out of TiF62− , CoF63− , Cu 2Cl 2 and
NiCl 2−
4 (At. no. Z of Ti = 22, Co = 27,
Cu = 29,Ni= 28), the colourless
species are
[CBSE AIPMT 2009]
(a) TiF62 − and CoF63−
(b) Cu2Cl2 and NiCl2−
4
(c) TiF62 − and Cu2Cl2
(d) CoF63− and NiCl24−
Ans. (c)
In TiF62 − , Ti is present asTi4+ .
4+
0
0
Ti = [Ar] 3d 4s
Hence,TiF62 − is colourless due to the
absence of unpaired electrons.
In Cu2 Cl2 , Cu is present as Cu+ .
Cu+ = [Ar]
10
3d
4s
Due to absence of unpaired electrons,
Cu2 Cl2 is colourless.
52 Which of the following complex
ions is expected to absorb visible
light? (At. no. of
Zn = 30, Sc = 21, Ti = 22, Cr = 24)
[CBSE AIPMT 2009]
(a) [Sc(H2O) 3 (NH3) 3] 3+
(b) [Ti(en)2 (NH3)2] 4+
(c) [Cr(NH3) 6] 3+
(d) [Zn(NH3) 6] 2+
Ans. (c)
In [Cr(NH3) 6 ] 3+ , Cr is present as Cr 3+ .
Cr 3+ = [Ar] 3d 3, 4s 0
3d
4s
[Cr(NH3) 6 ] 3+ = [Ar] 3d 3
3d
4s
NH3
NH3 NH3 NH3
{
NH3 NH3
4p
Electron donated by NH3 ligand
hybridisation d 2sp3
Since, this complex has three unpaired
electrons, excitation of electrons is
possible and thus, it is expected that this
complex will absorb visible light.
53 In which of the following
coordination entities the
magnitude of ∆ o (CFSE in
octahedral field) will be maximum?
(At. no. of Co = 27)
[CBSE AIPMT 2008]
(a) [Co(H2O) 6] 3+
(c) [Co(CN) 6] 3–
(b) [Co(NH3) 6] 3+
(d) [Co(C2O 4 ) 3] 3−
Ans. (c)
As in all the given complex the central
metal atom is same and contains same
number of d electrons, thus CFSE is
decided by ligands. In case of strong
field ligand, CFSE is maximum. CN− is a
strong field ligand, Hence, in [Co(CN) 6 ] 3−
magnitude of CFSE i.e. ∆ 0 is maximum.
54 Which of the following complexes
exhibits the highest paramagnetic
behaviour?
where, gly = glycine, en =
ethylenediamine and bpy =
bipyridyl moities)
(At. no. of Ti = 22,C = 23,Fe = 26,
[CBSE AIPMT 2008]
Co = 27)
(a) [V(gly)2 (OH)2 (NH3)2] +
(b) [Fe(en)(py)(NH3)2] 2+
(c) [Co(ox)2 (OH)2] −
(d) [Ti(NH3) 6] 3+
Ans. (c)
The electronic configuration of
V(23) = [Ar] 4s 2 , 3d 3
Let in [V(gly)2 (OH)2 (NH3)2 ] + oxidation
state of V is x.
x + (−1) × 2 + (−1)2 + (0 × 2) = + 1
x=+5
V5 + = [Ar] 4s 0 , 3d 0
(no unpaired electrons)
The electronic configuration of
Fe(26) = [Ar] 4s 2 , 3d 6
Let the oxidation state of Fe in
[Fe(en)(ppy)(NH3)2 ]2 + is x.
[x + (0) + (0) + (0) × 2] = + 2
x = +2
Fe2 + = [Ar],3d 6
(Q 4 unpaired electron)
but, bpy, en and NH3 all are strong field
ligands, so pairing occurs and thus,Fe2 +
contains no unpaired electron.
The electronic configuration of
Co(27) = [Ar] 4s 2 , 3d 7
Let the oxidation state of Co in
[Co(ox)2 (OH)2 ] − is x
x + (−2) × 2 + (−1) × 2 = − 1
x=+5
Co5 + = [Ar], 3d 4 [4 unpaired electrons]
ox and OH are weak field ligands, thus
pairing of electron units does not occur.
The electronic configuration of
Ti(22) = [Ar] 4s 2 , 3d2
Oxidation state of Ti in [Ti(NH3) 6 ] 3+ is 3.
Ti3+ = [Ar] 3d 1
(one unpaired electron)
Hence, complex [Co(ox)2 (OH)2 ] − has
maximum number of unpaired electrons,
thus show maximum paramagnetism.
55 [Cr(H2O) 6]Cl 3 (at. no. of Cr = 24)
has a magnetic moment of 3.83
BM, the correct distribution of 3d
electrons in the chromium of the
complex is
[CBSE AIPMT 2006]
1
1
(a) 3 dxy
, 3 dyz
, 3 dz1 2
1
(b) 3 d 2 2 , 3 dz1 2 , 3 dxz
(x – y )
(c) 3 dxy , 3 d
(x 2 − y 2 )
1
, 3 dyz
1
1
1
(d) 3 d xy
, 3 d yz
, 3 d zx
Ans. (d)
Magnetic moment (µ ) = n(n + 2) BM
or
3 .83 = n (n + 2)
or
3 .83 × 3 .83 = n2 + 2n
14 .6689 = n2 + 2n
n ~− 3
Hence, number of unpaired electrons in
d-subshell of chromium (Cr= 24) = 3.
So, the configuration of chromium ion is
Cr 3+ = 1s 2 , 2 s 2 2p6 , 3s 2 3p6 3d 3
In [Cr(H2O) 6 ]Cl2 , oxidation state of Cr is +3.
Hence, in3d 3 the distribution of
electrons
1
1
1
3d xy
,3d yz
,3d zx
,3d 02 2 ,3d 02
x
−y
z
56 Which one of the following is an
inner orbital complex as well as
diamagnetic in behaviour?
(At. no. of Zn = 30, Cr = 24, Co =
27, Ni = 28)
[CBSE AIPMT 2005]
(a) [Zn(NH3) 6] 2+
(c) [Co(NH3) 6] 3+
Ans.
(b) [Cr(NH3) 6] 3+
(d) [Ni (NH3) 6] 2+
(c) In [Co(NH3 )6 ] 3+ , oxidation state of
Co = +3 and its coordination number is
six.
2
2
6
2
6
7
2
27 Co = 1s , 2s 2p , 3s 3p 3d , 4s
3+
2
2
6
2
6
6
Co = 1s , 2s 2p , 3s 3p 3d
d
s
p
d 2sp 3
[Co(NH3) 6 ] 3+ is an inner orbital complex
as well as diamagnetic in behaviour (due
to absence of unpaired electrons).
[Zn(NH3) 6 ] 2 + → sp3d2
hybridisation (outer orbital complex and
diamagnetic in nature).
[Cr(NH3) 6 ] 3+ → d2 sp3
hybridisation (inner orbital complex and
paramagnetic in nature).
154
NEET Chapterwise Topicwise Chemistry
57 Among [Ni(CO) 4], [Ni(CN) 4] 2– ,
[NiCl 4] 2– species, the hybridisation
states of the Ni atom are,
respectively (At. no. of Ni = 28)
[CBSE AIPMT 2004]
3
2,
2
(a) sp , dsp dsp
(c) sp 3, sp 3, dsp2
Ans. (b)
3
2
3
(b) sp , dsp , sp
(d) dsp2 , sp 3, sp 3
In Ni(CO) 4 , nickel is sp3 hybridised
because in it oxidation state of Ni is zero.
So, configuration of
2
2
6
2
6
8
2
28 Ni = 1s , 2 s 2p , 3 s 3p 3d , 4s
p
s
d
sp 3
(CO is a strong field ligand, hence pairing
of electrons will occur)
In [Ni(CN) 4 ]2 – , nickel is present asNi2 + ,
so its configuration
= 1s 2 ,2s 2 2p6 , 3s 2 3p6 3d 8
p
s
d
of weak field, ligand there will be no
pairing of electrons in 3d. So, it will
form high spin complex due to
presence of five unpaired electrons.
59 CN − is strong field ligand. This is
due to the fact that
[CBSE AIPMT 2004]
(a) it carries negative charge
(b) it is a pseudohalide
(c) it can accept electrons from metal
species
(d) it forms high spin complexes with
metal species
Ans. (b)
CN− is a strong field ligand because it is
an example of pseudohalide.
Pseudohalide ions are stronger
coordinating ligands and they have the
ability to form σ and π-bonds.
60 Among the following which is not
the π-bonded organometallic
compound?
[CBSE AIPMT 2003]
(a) K[PtCl3 (η2 − C2H4 )]
dsp 2
−
CN is strong field ligand, hence it makes
Ni2 + electrons to be paired up.
In [NiCl 4 ] 2 − , nickel is present as Ni2 + , so
its configuration
= 1s 2 , 2s 2 2p6 , 3 s 2 3p6 , 3d 8
p
s
d
sp 3
Cl − is a weak field ligand, hence inNi2 +
electrons are not paired.
[CBSE AIPMT 2004]
(a) 3
(b) 5
Ans. (b)
(c) 2
(d) 4
In [Mn(H2O) 6 ] 2 + , Mn is present asMn2 + or
Mn (II), so its electronic configuration
= 1s , 2 s 2 p , 3s 3p , 3d
2
2
6
2
6
5
In (CH3) 4 Sn (organometallic compounds
of tin) single bonds are present in form
of sigma bond. pi-bonded
organometallic compound includes
alkenes, alkynes and some other carbon
containing compounds having
pi-electrons in their molecular orbitals.
d
s
p
d
sp 3d 2
In [Mn(H2O) 6] 2+ the coordination
number of Mn is six, and in presence
CH3

Sn

CH3
Atoms, ions or molecules having
unpaired electrons are paramagnetic. In
[Cr(NH3) 6 ] 3+ , Cr is present as Cr (III).
Cr 3+ = 1s 2 , 2 s 2 2p6 , 3s 2 3p6 3d 3
In excited state
d
s
p
d 2sp3 hybridisation
Number of unpaired electrons = 3, so it is
paramagnetic while rest of the species
are diamagnetic.
63 Which of the following
organometallic compounds is σ and
[CBSE AIPMT 2001]
π-bonded?
[Co(CO) 5 NH3]2 + . In this complex,
Co-atom attached with NH3 through
σ-bonding and with CO through dative
π-bond.
64 Coordination number of Ni in
[Ni(C 2O 4 ) 3] 4– is
[CBSE AIPMT 2001]
CH3
61 The number of unpaired electrons
in the complex ion [CoF6] 3– is
[CBSE AIPMT 2003]
(At. no. of Co = 27)
(a) 3
(c) 4
Ans. (c)
(b) [Fe(CO) 5]
(d) [Cr(NH3) 6] 3+
(b) K[PtCl3 (η2 —C2H4 )]
(c) [Co(CO) 5 NH3] 2+
(d) Fe(CH3) 3
Ans. (c)
(c) Cr (η6 –C 6H6 )2
(d) (CH3) 4 Sn
Ans. (d)
H3C
[CBSE AIPMT 2002]
(a) [Cr(CO) 6]
(c) [Fe(CN) 6] 4–
Ans. (d)
(a) [Fe(η5 —C 5H5 )2]
(b) Fe(η5 − C 5H5 )2
58 Considering H2O as a weak field
ligand, the number of unpaired
electrons in [Mn(H2O) 6] 2+ will be
(At. no. of Mn = 25)
62 Atomic number of Cr and Fe are
respectively 24 and 26, which of
the following is paramagnetic with
the spin of electron?
(b) 2
(d) 0
In complex ion[CoF6 ]3 – , Co is present in
+3 oxidation state
2
2
6
2
6
7
2
27 Co = 1s , 2 s 2p , 3s 3p 3d , 4s
3+
2
2
6
2
6
6
Co = 1s , 2 s 2p , 3 s 3p 3d
Therefore, the number of unpaired
electrons in3d subshell of [CoF6 ] 3− is 4.
(a) 3
(c) 4
Ans. (b)
(b) 6
(d) 2
Coordination number of nickel in
[Ni(C2O4 ) 3] 4– is 6 because C2O24– is a
bidentate ligand.
65 Which statement is incorrect?
[CBSE AIPMT 2001]
(a) Ni(CO) 4 -tetrahedral, paramagnetic
(b) [Ni(CN) 4] 2− -square planar,
diamagnetic
(c) Ni(CO) 4 -tetrahedral, diamagnetic
(d) [Ni(Cl) 4] 2– tetrahedral,
paramagnetic
Ans. (a)
In Ni(CO) 4 , Ni has zero oxidation state. It
is sp3 hybridised.
155
Co-ordination Compounds
Ni(CO) 4 →
68 Shape of Fe(CO) 5 is
3d
[CBSE AIPMT 2000]
4s
4p
sp3 hybridisation
Hence, it has no unpaired electron so, it
shows the property of diamagnetism and
has tetrahedral structure.
NOTE The valence shell electronic
configuration of ground state Ni atom is
3d 8 4s 2 . The two electrons of 4s are
pushed into3d orbitals and get paired up
because of the presentce of strong
ligand (CO).
(a) octahedral
(b) square planar
(c) trigonal bipyramidal
(d) square pyramidal
Ans. (c)
In Fe(CO) 5 , the Fe-atom is indsp3
hybridised state, therefore the shape of
molecule is trigonal bipyramidal. The
hybridisation is as follows :
2
2
6
2
6
6
2
0
26 Fe = 1s , 2s 2p , 3s 3p 3d , 4s 4p
In presence of strong field ligand (CO),,
the electrons of4s are pushed in3d
orbital and get paired up.
In Fe(CO) 5 , the Fe-atom is
66 Which of the following will exhibit
maximum ionic conductivity?
3d
[CBSE AIPMT 2001]
(a) K 4 [Fe(CN) 6]
(c) [Cu(NH3) 4]Cl2
Ans. (a)
(b) [Co(NH3) 6]Cl3
(d) [Ni(CO) 4]
4s
4p
dsp3 hybridised
Ionic conductivity depends upon the
number of ions produced in aqueous
solution. K4 [Fe(CN) 6 ] produces
maximum number of ions, i.e. 5.
4K+ + [Fe(CN) 6 ] 4 −
144424443
CO
CO
[Co(NH3) 6 ]Cl 3 produces 3, [Cu(NH3) 4 ] Cl2
produces 3 and [Ni(Co) 4 ] gives zero ions.
67 In the separation of Cu 2+ and Cd 2+
of IInd group in qualitative analysis
of cations, tetrammine copper (II)
sulphate and tetrammine cadmium
(II) sulphate react with KCN to form
the corresponding cyano
complexes, which one of the
following pairs of the complexes
and their relative stability enables
the separation of Cu 2+ and Cd 2+ ?
[CBSE AIPMT 2000]
(a) K 3 [Cu(CN) 4] : less stable and
K2 [Cd(CN) 4] : more stable
(b) K 3 [Cu(CN) 4] : more stable and
K2 [Cd(CN) 4] : less stable
(c) K2 [Cu(CN) 4] : less stable and
K2 [Cd(CN) 4] : more stable
(d) K2 [Cu(CN) 4] : more stable and
K2 [Cd(CN) 4] : less stable
Ans. (b)
K3 [Cu(CN) 4 ] is more stable while in
K2 [Cd(CN) 4 ] is less stable.
Here, Cu in + 1 oxidation state.
Fe
CO
Total 5 ions
CO
71 Which of the following ligands is
expected to be bidentate?
[CBSE AIPMT 1994]
(a) CH3NH2
(b) CH3C ≡≡N
(c) Br
(d) C2O2–
4
Ans. (d)
C2O24– is a bidentate ligand because it has
two donor atoms (sites) and can
coordinate to the central ion at two
positions.
–
M +
–
CO
69 The coordination number and
oxidation state of Cr in
K 3 [Cr(C 2O 4 ) 3] are respectively
[CBSE AIPMT 1995]
(a) 3 and +3
(c) 6 and +3
Ans. (c)
atoms. These compounds contain both
σ and π-bonded complexes.
σ-bond between metal and carbon atom
is formed when a vacant hybrid orbital of
the metal atom overlaps with an orbital
on C-atom of carbon monooxide
containing a low pair of electron.
Formation of π-bond is occurs when a
filled orbital of the metal atom overlaps
with a vacant antibonding π- orbital of
C-atom of Co.
(b) 3 and 0
(d) 4 and +2
Coordination number of Cr is 6 (oxalate is
bidentate ligand) and oxidation state of
Cr in K3 [Cr(C2O4 ) 3] is calculated below.
3 (1) + x + 3(−2) = 0
3 + x + (−6) = 0
x =6−3
x = +3
70 In metal carbonyl having general
formula M (CO) x where, M = metal,
x = 4 and the metal is bonded to
[CBSE AIPMT 1995]
(a) carbon and oxygen
(b) C ≡≡O
(c) oxygen
(d) carbon
Ans. (d)
In metal carbonyl M(CO) 4 , metal is
bonded to the ligand CO through carbon
C
+ +
M
C
O
O
72 Which one of the following
statements is not correct?
[CBSE AIPMT 1994]
(a) Mercury (II) iodide dissolves in
excess of potassium iodide
solution
(b) Tin (IV) chloride is made by
dissolving tin solution in
concentrated hydrochloric acid
(c) Zinc dissolves in sodium hydroxide
solution
(d) Carbon monoxide reduces iron (III)
oxide to iron
Ans. (b)
SnCl 4 is obtained by passing chlorine
over tin.
So Sn (IV) chloride is made by dissolving
tin solution in concentrated solution
statement is incorrect and answer is (b).
73 The complex ion [Co(NH3 ) 6] 3+ is
formed by sp 3d 2 hybridisation.
Hence, the ion should possess
[CBSE AIPMT 1990]
(a) octahedral geometry
(b) tetrahedral geometry
(c) square planar geometry
(d) tetragonal geometry
156
Ans. (a)
Since the hybridisation of central metal
tin [Co(NH3) 6 ] 3+ complex ion is sp3d2 and
coordination number of Co3+ is 6. So, its
geometry is octahedral.
TOPIC 3
Importance of
Co-ordination Compounds
74 Which of the following complexes
is used to be as an anticancer
agent?
[CBSE AIPMT 2014]
(a) Mer- [Co(NH3) 3Cl3]
(b) Cis- [PtCl2 (NH3)2]
(c) Cis- K2 [PtCl2Br2]
(d) Na2CoCl4
Ans. (b)
Cis-platin is known as anticancer agent.
The formula of cis-platin is cis[PtCl2 (NH3)2 ]. Here, the word cis refers
to cis geometrical isomer of
[PtCl2 (NH3)2 ]. It is used as an antitumour
agent.
NEET Chapterwise Topicwise Chemistry
75 Which of the following does not
have a metal-carbon bond?
(a) Al(OC2H5 ) 3
[CBSE AIPMT 2004]
(b) C2H5MgBr
(c) K[Pt(C2H4 )Cl3]
(d) Ni(CO) 4
Ans. (a)
Al(OC2H5 ) 3 does not have metal-carbon
bond, (i.e. it is not an example of
organometallic compound)
Structure
(a)
CH2
Cl
(b)
Cl
Cl
Pt
Cl
(c)
CH2
Cl
Pt
Cl
H3N
Cl
Cl
Pt
H3N
Cl
H H
H H
H—C—C—O—Al
H H
O—C—C—H
(d)
H3N
Cl
H H
H H
O—C—C—H
H H
76 Which of the following is
considered to be an anticancer
species?
[CBSE AIPMT 2004]
Cl
Pt
NH3
Ans. (c)
Cis-platin is the isomer of [Pt(NH3)2 Cl2 ]
which is used as an anticancer drug for
treating several types of malignant
tumours.
Cl
Cl
Pt
Cl
Cl
20
Purification and
Characterisation
of Organic Compounds
TOPIC 1
O
Methods of Purification
O
Intramolecular H-bonding
in o-nitrophenol
[NEET 2017, CBSE AIPMT 99, 94]
Steam distillation is used to purify the
substances which
(i) are volatile in steam but are
immiscible with water.
(ii) possess sufficiently high vapour
pressure at the boiling point of water.
(iii) contain non- volatile impurities.
The process of steam distillation can also
be used to separate a mixture of two
organic compounds one of which is
steam volatile while the other is not.
In ortho and para-nitrophenol,
ortho-nitrophenol has intramolecular
H-bonding. So, it has lower boiling point.
Intermolecular H-Bonding more strong
then intramolecular H-bonding. Whereas
para-nitrophenol has intermolecular
H-bonding. So, it has higher boiling point.
Due to difference in boiling points ortho
and para-nitrophenol can be separated
from each other by distillation.
O
N
01 The most suitable method of
separation of 1 : 1 mixture of ortho
and para-nitrophenols is
(a) sublimation
(b) chromatography
(c) crystallisation
(d) steam distillation
Ans. (d)
H-bonding
H
---HO—
H-bonding
O
—N

O---HO
—N
O
O---
Intermolecular H-bonding in p-nitrophenol
02 Which of the statements is not
true?
[CBSE AIPMT 2012]
(a) On passing H2 S through acidified
K2Cr2O 7 solution, a milky colour
is observed
(b) Na2Cr2O 7 is preferred over
K2Cr2O 7 in volumetric analysis
(c) K2Cr2O 7 solution in acidic
medium is orange
(d) K2Cr2O 7 solution becomes yellow
on increasing the pH beyond 7
Ans. (b)
Being hygroscopic, sodium
dichromate,Na2 Cr2O7 cannot be used
in volumetric analysis.
All other given statements are true.
03 The best method for the separation
of naphthalene and benzoic acid
from their mixture is
(a) chromatography [CBSE AIPMT 2005]
(b) crystallisation
(c) distillation
(d) sublimation
Ans. (d)
The best method for the separation of
naphthalene and benzoic acid from their
mixture is sublimation because it is
applicable for those organic compounds
which pass directly from solid to vapour
state on heating and vice versa on cooling.
In these compounds naphthalene is
volatile and benzoic acid is non-volatile
due to the formation of dimer via hydrogen
bonding (intermolecular).
04 Camphor is often used in molecular
mass determination because
[CBSE AIPMT 2004]
(a) it is readily available
(b) it has a very high cryoscopic constant
(c) it is volatile
(d) it is solvent for organic substances
Ans. (c)
Camphor is used in molecular mass
determination due to its volatile nature.
The method is called Rast’s camphor
method. Camphor acts as a solid solvent
which is volatile, hence can be removed
easily.
158
NEET Chapterwise Topicwise Chemistry
05 In steam distillation of toluene, the
pressure of toluene in vapour is
[CBSE AIPMT 2001]
(a) equal to the pressure of barometer
(b) less than the pressure of
barometer
(c) equal to vapour pressure of
toluene in simple distillation
(d) more than vapour pressure of
toluene in simple distillation
Ans. (b)
In steam distillation of toluene, the
pressure of toluene in vapour is less than
pressure of barometer, because it is
carried out when a solid or liquid is
insoluble in water and is volatile with
steam but the impurities are
non-volatile.
06 Which of the following techniques
is most suitable for purification of
cyclohexanone from a mixture
containing benzoic acid, isoamyl
alcohol, cyclohexane and
cyclohexanone? [CBSE AIPMT 1997]
(a) Crystallisation (b) IR spectroscopy
(c) Sublimation
(d) Evaporation
Ans. (b)
IR spectroscopy is used for the
purification of cyclohexanone from a
mixture of benzoic acid, isoamyl alcohol,
cyclohexane and cyclohexanone
because in IR spectroscopy each
functional group appears at a certain
peak. IR spectroscopy exploits the fact
that molecules absorb specific
frequencies that are characteristic of
their structure.
07 A is a lighter phenol and B is an
aromatic carboxylic acid. Separation
of a mixture of A and B can be
carried out easily by using a
solution of
[CBSE AIPMT 1992]
(a) sodium hydroxide
(b) sodium sulphate
(c) calcium chloride
(d) sodium bicarbonate
Ans. (d)
Carboxylic acids are soluble in sodium
bicarbonate but phenol are not dissolve
in it, so they are separated because
carboxylic acid react withNaHCO3 and
form sodium carboxylate.
R — COOH + NaHCO3
→ R — COO−Na+ + H2 CO3
08 Prussian blue is formed when
[CBSE AIPMT 1989]
(a) ferrous sulphate reacts with FeCl3
(b) ferric sulphate reacts with
Na4 [Fe(CN) 6]
(c) ferrous ammonium sulphate
reacts with FeCl3
(d) ammonium sulphate reacts with
FeCl3
Ans. (b)
When the sodium fusion extract is added
with FeCl 3 and then the resulting solution
is acidified with dilute hydrochloric acid,
the appearance of Prussian blue
colouration confirms the presence of
nitrogen in the organic compound.
Na + C + N → NaCN
FeSO4 + 2NaCN → Fe(CN) 2 + Na2SO4
Fe(CN)2 + 4NaCN → Na4 [Fe(CN) 6 ]
3Na4 [Fe(CN) 6 ] + 4FeCl 3 →
Fe4 [Fe(CN) 6 ] 3 + 12NaCl
Prussian blue
TOPIC 2
Qualitative Analysis
09 The Lassaigne’s extract is boiled
with conc. HNO 3 while testing for
halogens. By doing so it
[CBSE AIPMT 2011]
(a) helps in the precipitation of AgCl
(b) increases the solubility product of
AgCl
(c) increases the concentration of
NO –3 ions
(d) decomposes Na2 S and NaCN, if
formed
Ans. (d)
Na2S and NaCN, if present in the extract,
will be decomposed toH2S and HCN by
HNO3 .
NaCN + HNO3 → NaNO3 + HCN
Na2S + 2HNO3 → 2NaNO3 + H2S
These will escape from the solution and
will not interfere with the test for
halogens.
10 Lassaigne’s test for the detection
of nitrogen fails in
[CBSE AIPMT 1994]
(a) NH2CONHNH2 ⋅HCl
(b) NH2NH2 ⋅HCl
(c) NH2CONH2
(d) C 6H5NHNH2 ⋅HCl
Ans. (b)
Lassaigne’s test is given by only those
compounds which contain both carbon
and nitrogen. When compounds
containing C and N heated with sodium,
then it form NaCN which is easily
detected byFeCl 3.
Or
Some compounds live hydrazine
(NH2 ⋅ NH2 ) although contain nitrogen but
they do not respond Lassaigne’s test
because they do not have any carbon
and hence, NaCN is not formed.
11 In sodium fusion test of organic
compounds, the nitrogen of the
organic compound is converted
into
[CBSE AIPMT 1991]
(a) sodamide
(c) sodium nitrite
Ans. (b)
(b) sodium cyanide
(d) sodium nitrate
When the nitrogen containing compound
is heated with sodium, then nitrogen and
carbon of organic compound converted
into sodium cyanide.
Na +
C2
+3
N
→ NaCN
1
From organic compound
12 Lassaigne’s test is used to detect
[CBSE AIPMT 1990]
(a) nitrogen
(c) chlorine
Ans. (d)
(b) sulphur
(d) All of these
The detection of chlorine, sulphur and
nitrogen in organic compounds is done
by Lassaigne’s test.
TOPIC 3
Quantitative Analysis
13 In Duma’s method for estimation of
nitrogen, 0.25g of an organic
compound gave 40 mL of nitrogen
collected at 300 K temperature and
725 mm pressure. If the aqueous
tension at 300 K is 25 mm, the
percentage of nitrogen in the
compound is
[CBSE AIPMT 2015]
(a) 17.36 (b) 18.20 (c) 16.76 (d) 15.76
Ans. (c)
Mass of the substance taken = 0.25 g
Volume of nitrogen collected = 40 mL
Atmospheric pressure = 725 mm
159
Purification and Characterisation of Organic Compounds
Room temperature = 300 K
Aqueous tension at 300 K = 25 mm
Actual pressure of the gas
= (725 − 25) mmHg = 700 mm
To convert the volume at experimental
conditions to volume at STP.
Experimental value
At STP
p1 = 700 mm
p2 = 760 mm
V1 = 40 ml
V2 = ?
T1 = 300 k
T2 = 273 k
Substituting these values in the gase eq.
p2V2 p1V1
=
T2
T1
760 × V2
700 × 40
273
300
700 × 40 273
×
V2 =
300
760
= 33.53 mL
To convert volume at STP into mass
22400 ml of nitrogen at STP weigh = 28 g
∴ 33.53 ml of nitrogen at STP will
28 × 33.53
weigh =
22400 × 0.25
To calculate percentage of nitrogen
28 × 33.53
=
× 100
22400 × 0.25
we get,
= 16.76 %
=
14 In Duma’s method of estimation of
nitrogen 0.35g of an organic
compound gave 55 ml of nitrogen
collected at 300K temperature and
175 mm pressure. The percentage
composition of nitrogen in the
compound would be (Aqueous
tension at 300 K = 15mm]
[CBSE AIPMT 2011]
(a) 16.45
(c) 14.45
Ans. (a)
(b) 17.45
(d) 15.45
According to combined gas equation,
p1V1 p2V2
=
T1
T2
Where, p2 = pressure of N2 at STP
= 760 mm
T2 = Temperature ofN2 at STP = 273 K
V2 = ?
Volume of N2 at STP (By gas equation)
 ρ − ρ1 
273
= V2

 V1 ×
760
 t + 273 
Where, p1 = ρ − ρ1
ρ = 715 mm (pressure at whichN2
collected)
ρ1 = aqueous tension of water = 15 mm
T1 = t + 273 = 300 K
V1 = 55 mL = volume of moist nitrogen in
nitrometer
(715 − 15) × 55 273
×
∴ V2 =
300
760
mL
= 46098
.
% of nitrogen in given compound
V
28
=
× 2 × 100
22400 W
.
28
46098
=
×
× 100
22400
0.35
= 1645
. %
15 Kjeldahl’s method is used in the
estimation of [CBSE AIPMT 1990]
(a) nitrogen
(c) sulphur
Ans. (a)
(b) halogens
(d) oxygen
Kjeldahl’s method is simpler and more
convenient than Duma’s method. This
method is largely used for the estimation
of nitrogen in food stuff, drugs,
fertilisers and many other organic
compounds. However, this method
cannot be used for organic compounds
containing nitrogen in the ring such as
pyridine, quinoline and organic
compounds containing nitro (—NO2 ) and
diazo (—N == N—) groups.
21
Some Basic Principles of
Organic Chemistry
TOPIC 1
Nomenclature
01 The IUPAC name of the compound
[NEET 2017]
(a) 3-keto-2-methylhex-4-enal
(b) 5-formylhex-2-en-3-one
(c) 5-methyl-4-oxohex-2-en-5-al
(d) 3-keto-2-methylhex-5-enal
Ans. (a)
O
H
1
2
3
4
is
(a) 3-ethyl-4-ethenylheptane
(b) 3-ethyl-4-propylhex-5-ene
(c) 3-(1-ethyl propyl) hex-1-ene
(d) 4-ethyl-3-propylhex-1-ene
Ans. (d)
1
2
3
1-bromo prop-2-ene
6
— CHO group gets higher priority over >
C == O and
C == C group in numbering of
principal carbon chain.
IUPAC name =
3-keto-2-methylhex-4-enal.
02 The structure of isobutyl group in
an organic compound is [NEET 2013]
CH3
04 The correct IUPAC name of the
compound
[CBSE AIPMT 2011]
(a) Br CH2 CH == CH2
5
(a)
03 Which nomenclature is not
according to IUPAC system?
[CBSE AIPMT 2012]
O
C
iso-butyl group in an organic compound
is
CH3

CH3  C  CH2 

H
(‘yl’ suffix is used to represent one —H
less than the parent hydrocarbon.)
CH CH2 
CH3

(b) CH 3CH 2 CCH 2CH  CH 3


CH3
Br
4-bromo-2, 4-dimethylhexane
(c) CH3CHCHCH2CH3

CH3
2-methyl-3-phenylpentane
(d) CH3C—CH2CH2CH2COOH

CH3
(b) CH3 CH CH2  CH3

(c) CH3 CH2 CH2 CH2 
CH3

(d) CH3 C 

CH3
Ans. (a)
‘Iso’ mean’s one Me group is present in
side chain. Hence, the structure of

O
5-oxohexanoic acid
Ans. (a)
In IUPAC system of nomenclature,
preference is given to multiple bond than
halogen substituent, so the correct
name of
3
2
4
5
6
4-ethyl-3-propylhex-1-ene
Priority order→ = > −
05 The IUPAC name of the compound
having the formula
CH≡≡C—CH==CH2 is
[CBSE AIPMT 2009]
(a) 3-butene-1-yne
(b) 1-butyn-3-ene
(c) but-1-yne-3-ene
(d) 1-butene-3-yne
Ans. (d)
In IUPAC nomenclature, double bond is
given more preference than triple bond.
4
3
2
1
CH ≡≡ C  CH == CH2
1-butene-3-yne
1
Br  CH2  CH==CH2 is
3-bromoprop-1-ene
Priority is given to double bond than
halogen.
06 The state of hybridisation of
C 2 , C 3 , C 5 and C 6 of the
hydrocarbon,
161
Some Basic Principles of Organic Chemistry
CH3

CH3 C  CH == CH
7
5
4
6

CH3
CH3

CH C ≡≡ CH
3
2
respectively. Due to higher s-character
electron attracting tendency, i.e.
electronegativity increases.
09 The IUPAC name of is
Cl
1
[CBSE AIPMT 2009]
(a) sp, sp 3, sp2 and sp 3
(b) sp 3, sp2 , sp2 and sp
(c) sp, sp2 , sp2 and sp 3
(d) sp, sp2 , sp 3 and sp2
Ans. (a)
If number of σ bonds = 2; hybridisation is
sp.
If number of σ bonds = 3; hybridisation is
sp2 ,
If number of σ bonds = 4; hybridisation is
sp3.
CH3 H H CH3
σ
σ σ σ σ
σ
σ σ
C—C—C
CH
σ5
3
2
1
4
σ
σ
(a) 3, 4-dimethylpentanoyl chloride
(b) 1-chloro-1-oxo-2,
3-dimethylpentane
(c) 2-ethyl-3-methylbutanoyl chloride
(d) 2, 3-dimethyl pentanoyl chloride
Ans. (d)
IUPAC name of the given compound is
CH3
4

2CH
CH2
Cl
5
3
1
CH3
CH
C

CH3
O
2, 3-dimethyl pentanoyl chloride
H
3s-bonds 4s-bonds 2s-bonds
(sp)
(sp3)
(sp2)
4s-bonds
(sp3)
07 In the hydrocarbon
CH3 — CH==CH — CH2 — C ≡≡ CH
6
5
4
3
2
1
(a) CH3 CH  CH  CH3


CH3
OH
(b) sp, sp 3, sp2
(d) sp 3, sp2 , sp
CH3CH
6
5
CHCH2C
4
3
2
CH
sp3
sp2
sp2
sp3
sp
sp
3-methyl-2-butanol
08 The correct order regarding the
electronegativity of hybrid orbitals
of carbon is
[CBSE AIPMT 2006]
(b) sp > sp2 > sp 3
(d) sp < sp2 < sp 3
The correct order regarding the
electronegativity of hybrid orbitals of
carbon is sp > sp2 > sp3 because in sp, sp2
and sp3 hybrid orbitals, s-orbital
character is 50%, 33.3% and 25%
5
4
7
6
8
2-ethyl-3-methylbut-1-ene
(d) CH3  CH2  CH2
CH3

 CH  CH  CH2CH3

CH2CH3
3-methyl-4-ethyl heptane
Ans. (d)
CH3
5
4
2
1
6
3
CH 3CH 2CH 2CHCHCH 2 CH 3

CH2—CH3
Correct IUPAC name is 4-ethyl-3-methyl
heptane.
11 Name of the compound given
below
[CBSE AIPMT 2003]
CH3
4-ethyl-3-methyl octane
12 IUPAC name of the following is
[CBSE AIPMT 2002]
CH2 ==CH — CH2 — CH2 — C ≡≡ CH
(a) 1, 5-hexenyne (b) 1-hexene-5-yne
(c) 1-hexyne-5-ene (d) 1, 5-hexynene
Ans. (b)
The double bond gets more priority over
triple bond. So, the IUPAC name of
compound is
2
1
CH3
CH3
4
3
5
6
CH2 == CH— CH2 — CH2 — C ≡≡ CH
hex-1-en-5-yne or 1-hexene-5-yne
13 The incorrect IUPAC name is
[CBSE AIPMT 2001]
(b) CH3 CH  CH  CH3
|
|
CH3 CH2 —CH3
2, 3-dimethyl pentane
(c) CH3 —C ≡≡ CCH(CH3)2
4-methyl-2-pentyne
(d) CH3 CH CH CH3
|
|
Br
Cl
2-bromo-3-chloro butane
Ans. (a)
1
2
3
4
CH3  C  CH  CH3
||
|
O
CH3
Correct name is 3-methyl-2-butanone.
14 IUPAC name of the compound
[CBSE AIPMT 1998]
Cl
C ==C
H 3C
CH3
H3C
CH3
2-methyl-3-butanone
(b) CH3 C ≡≡C  CH(CH3)2
1
Thus, the hybridisation of C1, C3 and C5
are sp, sp3, sp2 respectively.
3
(a) CH3  C CH  CH3
 |
O CH3
(c) CH3 CH2 C  CH  CH3
|
||
CH2 CH3
[CBSE AIPMT 2008]
(a) sp > sp2 < sp 3
(c) sp < sp2 > sp 3
Ans. (b)
10 Names of some compounds are
given. Which one is not correct in
IUPAC system? [CBSE AIPMT 2005]
4-methyl-2-pentyne
the state of hybridisation of
carbons 1, 3 and 5 are in the
following sequence
(a) sp2 , sp, sp 3
(c) sp, sp2 , sp 3
Ans. (b)
H3C 1 2
[CBSE AIPMT 2006]
σ
H3C—C—–C
CH3
The IUPAC name of the given compound is
CH3
O
is in the following sequence
(a) 2, 3-diethylheptane
(b) 5-ethyl-6-methyloctane
(c) 4-ethyl-3-methyloctane
(d) 3-methyl-4-ethyloctane
Ans. (c)
CH2CH3
is
I
(a) trans-3- iodo-4-chloro-3-pentene
(b) cis-2-chloro-3-iodo-2-pentene
162
NEET Chapterwise Topicwise Chemistry
(c) trans-2-chloro-3-iodo-2-pentene
(d) cis-3-iodo-4-chloro-3-pentene
Ans. (c)
Cl
CH2 — CH3
C == C
H3C
[CBSE AIPMT 1992]
I
In this compound groups preferential
order is
 Cl >  CH3 and I >  CH2  CH3
Hence, more preferential order
containing groups are attached at
opposite sides. So, it is E (trans)-isomer.
Thus, its name is
trans-2-chloro-3-iodo-2-pentene.
15 The IUPAC name of
CH3 —CH— CH— CH2


CH3 CH3
3
5
4
Its IUPAC name is 2, 3, 6-trimethyl
heptane.
16 For the compound
H3C — CH — CH — CH2 — CH3


CH3 CH2CH3
which of the following IUPAC
names is correct?
[CBSE AIPMT 1994]
(a) 2-methyl-3-ethyl pentane
(b) 3-ethyl-2-methyl pentane
(c) 2-ethyl-3-methyl pentane
(d) 3-methyl-2-ethyl pentane
Ans. (b)
4
3
CH3CHCH
OH
1
2
CH 2
(a)
2
1
CH3—CH2—CH—CH—CH3
C2H5 CH3
3-ethyl-2-methyl pentane
CH 3
CCHO
CH 3
CH3
(c)
CH 3
CH3

(a) CH3 — CH —CH2CH3
(b) CH3 —C ==CH — CH3

CH3
(c) CH3 — CH2 —C ==CH2

CH3
(d) CH3 —CH — CH == CH2

CH3
NO 2
CH 2
(b)
18 2-methyl-2-butene will be
represented as [CBSE AIPMT 1992]
CH3 — CH — CH — CH2 — CH2


CH3 CH3
6
7
— CH— CH3

CH3
5
3
4
5
20 Identify the compound that will
react with Hinsberg’s reagent to
give a solid which dissolves in
alkali.
[NEET 2021]
CH 3
NH
CH 2
NH 2
CH 2
(d)
(a) 1,3-isopropyl-3-methyl propane
(b) 2,3,6-trimethyl heptane
(c) 2,5,6-trimethyl heptane
(d) 2,6,3-trimethyl heptane
Ans. (b)
2
(a) 4-hydroxy-1-methyl pentanal
(b) 4-hydroxy-2-methyl pent-2-en-1-al
(c) 2-hydroxy-4-methyl pent-3-en-5-al
(d) 2-hydroxy-3-methyl pent-2-en-5-al
Ans. (b)
TOPIC 2
Isomerism
4-hydroxy -2-methyl pent -2-en-1-al
—CH2 — CH — CH3 is

CH3 [CBSE AIPMT 1996]
1
17 The IUPAC name of
CH3 — CH — CH==C — CHO


OH
CH3
CH 3
CH 2
N
CH 3
CH3
Ans. (c)
Hinsberg test is used to test the amine
compounds. Only 1° and 2° amines give
product after reacting with Hinsberg
reagent (benzene sulphonyl chloride). 3°
amines do not react.
The product formed can be
distinguished by dissolving in alkali. If it
dissolves, then amine is 1° otherwise the
reactant is secondary amine.
CH3  CH2NH2 + C6H5SO2 Cl →
Ethyl Amine
Hinsberg reagent
C6H5SO2NHCH2 CH3 + HCl
Alkali KOH
C6H5SO2NHCH2 CH3 →
Ans. (b)
The structure of 2-methyl-2-butene is
CH3
CH3—C
1
Soluble in alkali medium
CH—CH3
2
3
4
19 Which of the following possesses a
sp-carbon in its structure?
[CBSE AIPMT 1989]
(a) CH2 == CCl — CH ==CH2
(b) CCl2 == CCl2
(c) CH2 ==C ==CH2
(d) CH2 ==CH — CH ==CH2
Ans. (c)
In structure CH2 == C == CH2 , the middle
carbon is attached with two π-bonds, so
it have sp hybridisation.
CH2==C==CH2
sp2
sp
Also, called allenes.
s
C6H5SO2 N CH2 CH3K⊕
sp2
21 With respect to the conformers of
ethane, which of the following
statements is true?
(a) Bond angle remains same but
bond length changes [NEET 2017]
(b) Bond angle changes but bond
length remains same
(c) Both bond angle and bond length
change
(d) Both bond angles and bond length
remain same
Ans. (d)
Isomers which are possible by rotation
about single bonds without cleavage of
any bond are called conformers. In
ethane, an infinite number of
conformations are possible. There are
two extreme forms, the staggered
conformation, which is the most stable
163
Some Basic Principles of Organic Chemistry
and the eclipsed conformation which is
least stable. Among the conformers of
ethane, bond angle and bond length
remains same while their energy,
stability and dihedral angle are different.
22 The correct statement regarding a
carbonyl compound with a
hydrogen atom on its alpha-carbon,
is
[NEET 2016, Phase I]
(a) a carbonyl compound with a
hydrogen atom on its
alpha-carbon rapidly equilibrates
with its corresponding enol and
this process is known as
aldehyde-ketone equilibration
(b) a carbonyl compound with a
hydrogen atom on its
alpha-carbon rapidly equilibrates
with its corresponding enol and
this process is known as
carbonylation
(c) a carbonyl compound with a
hydrogen atom on its
alpha-carbon rapidly equilibrates
with its corresponding enol and
this process is known as keto-enol
tautomerism
(d) a carbonyl compound with a
hydrogen atom on its
alpha-carbon never equilibrates
with its corresponding enol
Ans. (c)
In keto-enol tautomerism, a carbonyl
compound with a hydrogen atom on its
alpha-carbon rapidly equilibrates with its
corresponding enol.
OH
O


R —C—CH2 R ′ r R —C—CHR ′
Enol
Ketone
(containing α-hydrogen)
H
H
H H
H
(a) 18 sigma bonds and 2 pi-bonds
(b) 16 sigma bonds and 1 pi-bond
(c) 9 sigma bonds and 2 pi-bonds
(d) 9 sigma bonds and 1 pi-bond
Ans. (a)
H
all carbons are sp2 -hybridised.
24 Which among the given molecules
can exhibit tautomerism?
[NEET 2016, Phase II]
O
I
O
Ph
Ph
O
II
(a) III Only
(c) Both I and II
Ans. (a)
H
III
(b) Both I and III
(d) Both II and III
CH3
CN
C
C
(d)
CN
Ans. (a)
The molecule in which all the carbon
atoms are sp2 -hybridised will be
coplanar. Thus, in option (a) i.e.
C
H OC2H5
H
H
σ
Hσ
H
σ
C s σ C σ O
p-bond
σ
σ C
C
H σ p-bond σ Hσ Hσ
O
O
C σ C
σ
σ
σ
σ
σH
H
OH
27 Which of the following compounds
will undergoes racemisation when
solution of KOH hydrolysis?
H
III
O
C
C
H
O
H
Enol form
Keto form
[CBSE AIPMT 2014]
So only, III molecule will show
tautomerism. Thus, correct option is (a).
25 Two possible stereo-structures of
CH3CHOH⋅ COOH, which are
optically active, are called
[CBSE AIPMT 2015]
(a) diastereomers (b) atropisomers
(c) enantiomers
(d) mesomers
Ans. (c)
COOH
COOH
OH
H
HO
CH3
H
Both are enantiomers.
H3C
C
OH
H
C
C
O
OC2 H5
H3C
C
O
H2
C
CH2  Cl
(a)
(b)
CH3  CH2  CH2  Cl
CH3
(c) CH3  CH  CH 2  Cl
CH3
(d)
H
C
Cl
C2H5
Ans. (c)
CH3
[CBSE AIPMT 2015]
CH3
H3C
O
26 The enolic form of ethyl
acetoacetate as below has
(c)
The enolic form of ethyl acetoacetate
has 16 single bonds i.e. 16 σ-bonds and 2
double bonds i.e. 2 σ-bonds and 2
π-bonds.
Hence, the given structure has 18
σ-bonds and 2 π-bonds.
In tautomerism,α −hydrogen must be
present in the molecule. Thus, molecule
II will not show tautomerism. As at
bridge, double bond is highly unstable.
Thus, molecule I will also not show
tautomerism.
For III molecule,
[NEET 2016, Phase II]
(b)
H H
H
23 In which of the following
molecules, all atoms are coplanar?
(a)
H
Option (c) is most probable answer. Out
of compound (c) or (d), the (d) results in
formation of racemic product due to
chirality. Compound (c) although not
chiral but can form racemic product.
After removal of Cl − gives a carbocation
which in turn undergo racemisation after
rearrangement.
28 The order of stability of the
following tautomeric compound is
C
O
OC2 H5
[NEET 2013]
O
OH


CH2 == C  CH2 C CH3
I
º
164
NEET Chapterwise Topicwise Chemistry
O
O


CH3 C CH2 C CH3
II
OH
O


CH3 C == CH C CH3
III
(a) I > II > III
(c) II > I> III
Ans. (b)
º
º
(b) III > II > I
(d) II > III > I
30 Which of the following compounds
will exhibit cis-trans (geometrical)
isomerism?
[CBSE AIPMT 2009]
(a) 2-butene
(c) 2-butyne
Ans. (a)
(b) Butanol
(d) 2-butenol
H
CH3CH ==CHCH3 ⇒
H
C ==C
H3C
2 - butene
(Stabilised by conjugation
and H− bonding)
III
O
O


> CH3  C  CH2  C  CH3
II
OH
O


> CH2 == C  CH2  C  CH3
I
Less stable as (== ) bond
is not in conjugation with carbonyl group
29 Which of the following acid does
not exhibit optical isomerism?
[CBSE AIPMT 2012]
(a) Maleic acid
(c) Lactic acid
Ans. (a)
(b) α-amino acid
(d) Tartaric acid
Only those compounds exhibit optical
isomerism, which have chiral centre
and/or absence of symmetrical
elements. (Chiral carbon is the carbon in
which all the four valencies are satisfied
by four different groups.)
*
R—CH—COOH
CHCOOH
(a) 
(b)

NH2
CHCOOH
Maleic acid
(no chiral centre,
so, optically
inactive
a-amino acids
(one chiral centre,
so, optically active)
*
CH(OH)COOH
*
(c) CH3CH(OH)COOH (d) ∗
CH(OH)COOH
Lactic acid
(one chiral centre,
so, optically
active)
Tartaric acid
(two chiral centre,
so, optically active)
Thus, maleic acid does not exhibit
optical isomerism.
NOTE If R == H, theα-amino acid is
achiral.
H3C
CH3

(d) H C  Cl

C2H5
Ans. (c)
CH3 — CHCl — CH2 — CH3, Priority order
is — Cl > — C2H5 > — CH3 > — H
2
CH3
Cis -form
The enols of β-dicarbonyl compounds
are more stable because of conjugation
and intramolecular H–bonding. Thus, the
order of stability is
OH
O


H3C  C == CH  C  CH3
C2H5

(c) Cl C  CH3

H
H
No
interchange
(a) H3CCCl
1
C == C
H
C2H5
3
2
H
C2H5
3
CH3
H3CCCl
Trans-form
1
CH3CH2 CH2 CH2OH ⇒no geometrical
H
2
Butanol
isomers
CH3C ≡≡ CCH3 ⇒no geometrical isomers
CH3CH2 C(OH) ==CH2 ⇒no geometrical
isomers.
31 How many stereoisomers does this
molecule have? [CBSE AIPMT 2008]
CH3CH== CHCH2CHBrCH3
(a) 4
(c) 8
Ans. (a)
(b) 6
(d) 2
C2H5
4
3
HCCH3
(b)
Interchange
3
1 Cl
4
2
C2H5
3
1
(c) ClCCH3
4
C* = asymmetric carbon
Number of optical isomers = 2n
where, n = number of asymmetric
carbon atoms = 21 = 2
Number of geometrical isomers = 2n
where, n = number of double bonds
= 21 = 2
Hence, total number of stereoisomers =
Total optical isomers + Total geometrical
isomers
= 2 + 2= 4
32 CH3  CHCl  CH2  CH3 has a
chiral centre. Which one of the
following represents its
R-configuration?
[CBSE AIPMT 2007]
C2H5

(a) H3C  C  Cl

H
C2H5

(b) H  C CH3

Cl
H
No
interchange
2
H
C2H5
1
3
ClCCH3
R
4H
3
Br
2
ClCC2H5 S
*
CH3C== CCH 2CCH
3
H
CH3
1
H
H
S
CH3
4
1
(d) HCCl
Interchange
2C
2H5
Cl
1
2
3
H5C2CCH3 S
H
4
33 If there is no rotation of plane
polarised light by a compound in a
specific solvent, though to be
chiral, it may mean that
[CBSE AIPMT 2007]
(a) the compound is certainly a chiral
(b) the compound is certainly meso
(c) there is no compound in the
solvent
(d) the compound may be a racemic
mixture
Ans. (b)
The compounds in which asymmetric
carbon atom is present, are called
optically active, they rotate the plane
polarised light but the compounds which
165
Some Basic Principles of Organic Chemistry
do not show optical activity inspite of the
presence of chiral carbon atoms are
called meso-compounds. The absence
of optical activity in these compounds is
due to the presence of plane of
symmetry in the molecule.
34 Which of the following is not chiral?
[CBSE AIPMT 2006]
(a) 2-butanol
(b) 2, 3-dibromo pentane
(c) 3-bromo pentane
(d) 2-hydroxy propanoic acid
Ans. (c)
36 Which one of the following pairs
represents stereoisomerism?
[CBSE AIPMT 2005]
(a) Chain isomerism and rotational
isomerism
(b) Structural isomerism and
geometrical isomerism
(c) Linkage isomerism and
geometrical isomerism
(d) Optical isomerism and geometrical
isomerism
Ans. (d)
Pair of optical isomerism and
geometrical isomerism are able to
exhibit the phenomenon of
stereoisomerism because both type of
isomers differ only in their orientation in
space.
(a) 2-butanol
H
Chiral ‘C’
|
CH3CCH2CH3
|
OH
(b) 2,3-dibromo pentane
H H
 
CH3CCCH2CH3
 
Br Br
37 The molecular formula of diphenyl
methane
CH 2
(a) 6
(c) 8
Ans. (b)
Not chiral ‘C’
(d) 2-hydroxy propanoic acid
H

CH 3 CH 2 CH 2 CCOOH

OH
Chiral ‘C’
Hence, 3-bromopentane is not a chiral
molecule due to absence of chiral
C-atom.
CH2 
(diphenyl methane)
C13H12
Monochloro derivatives
Cl
[CBSE AIPMT 2005]
Br
C
(a) R
(b) S
Ans. (a)
H is
Cl
(c) Z
(d) E
4
H3C
H
Cl
CH2
(ii) Cl—
Cl
1 Br
C
CH2
(i)
[CBSE AIPMT 2003]
CH3
CH3
OH
OH
H
OH
and
H
H
H
(a)
HO
CH3
CH3
CH3
CH3
H
H
H
HO
and
OH
HO
HO
(b)
H
CH3
CH3
CH3
CH3
H
OH
HO
OH
and
H
H
H
(c)
HO
CH3
CH3
CH3
CH3
H
H
OH
HO
and
H
HO
H
(d)
HO
CH3
CH3
Ans. (c)
Enantiomers are mirror images of each
other but they are not superimposable
on each other.
CH3
H
e.g.
HO
CH3
OH
HO
and
H
H
CH3
H
OH
CH3
39 Geometrical isomers differ in
[CBSE AIPMT 2002]
(a) position of functional group
(b) position of atoms
(c) spatial arrangement of atoms
(d) length of carbon chain
Ans. (c)
Geometrical isomers The isomers
having same molecular formula but
differ in the position of atoms or groups
in space due to hindered rotation about a
double bond.
s
••
Br
3
(b) 4
(d) 7
The molecular formula of diphenyl
methane shows four isomers in form of
mono chloro derivatives.
35 The chirality of the compound
H3C
is C13H12
How many structural isomers are
possible when one of the
hydrogen is replaced by a chlorine
atom?
[CBSE AIPMT 2004]
Chiral ‘C’
(c) 3-bromo pentane
H

CH3CH2CCH2CH3

Br
38 Which of the following pair of
compounds are enantiomers ?
(iii)
CH2
(iv)
CH2
|
Cl
4
3
H3C
H
[R]
C
Cl
40 CH2 —C —CH3 and CH2==C — CH3


• •
O
• O•
•• s
are
[CBSE AIPMT 2002]
(a) resonating structures
(b) tautomers
(c) geometrical isomers
(d) optical isomers
166
NEET Chapterwise Topicwise Chemistry
Ans. (a)
–
CH2CCH3
||
O
CH2==C CH3
|
O –
Structure II
Structure I
Structure I and structure II are the
resonating forms because the position
of atoms remains the same and only
redistribution of electrons take place.
41 CH3 — CH2— CH CH3 obtained
|
Cl
by chlorination of n-butane, will be
[CBSE AIPMT 2001]
(a) meso form
(b) racemic mixture
(c) d-form
(d) l-form
Ans. (b)
Chlorination of n-butane takes place by
free radical mechanism as follows:
•
•
Step I Cl2 → Cl + Cl
•
Step II CH3CH2 CH2 CH3 + Cl →
•
CH3 C HCH2 CH3 + HCl
•
Step III CH3 C HCH2 CH3 + Cl2 →
CH3
|
H—C—Cl
|
C2H 5
+
CH3
|
Cl—C—H
|
C2H 5
Racemic mixture
42 A compound of molecular formula
C 7 H16 shows optical isomerism,
compound will be
[CBSE AIPMT 2001]
(a) 2, 3-dimethyl pentane
(b) 2, 2-dimethyl butane
(c) 2-methyl hexane
(d) None of the above
Ans. (a)
(b)
2,3-dimethylpentane
Chiral carbon
CH3  CH  *CH  CH2  CH3


CH3 CH3
shows the property of optical isomerism
due to presence of an asymmetric
*
C-atom.
43 The (R)- and (S)- enantiomers of an
optically active compound differ in
[CBSE AIPMT 2000]
(a) their solubility in a chiral solvent
(b) their reactivity with a chiral
reagent
(c) their optical rotation of plane
polarised light
(d) their melting points
Ans. (c)
R and S forms of an optically active
compound differ in their behaviour
towards plane polarised light. The
species which rotate the plane polarised
light towards right is called R-form
(rectus form) or d-form (dextro-form) and
the species which rotate the plane
polarised light towards left is known as
S-form (sinister-form) or l-form
(laevo-form).
44 But-2-ene exhibits
cis-trans-isomerism due to
[CBSE AIPMT 2000]
(a) rotation around C2 —C 3 double
bond
(b) rotation around C 3 —C 4 sigma bond
(c) rotation around C 1 —C2 bond
(d) restricted rotation around C ==C
bond
Ans. (d)
Due to presence of>C == C <in
but-2-ene, it shows restricted rotation.
Hence, give two types of arrangements
around the space of >C == C< as cis and
trans-forms.
45 Which of the following compounds
is not chiral ?
[CBSE AIPMT 1998]
(a) DCH2CH2CH2Cl
(c) CH3CHDCH2Cl
Ans. (a)
(b) CH3CH2CHDCl
(d) CH3CHClCH2D
Chiral carbon is that carbon whose all
the four valencies are satisfied by four
different groups.
Due to absence of asymmetric (chiral) Catom
D — CH2 — CH2 — CH2 Cl
molecule is not a chiral molecule.
46 Tautomerism will be exhibited by
[CBSE AIPMT 1997]
(a) (CH3) 3 CNO
(c) R3CNO2
Ans. (d)
O
R—CH2—N
O
Nitro form
OH
N
R—CH
Azo form
O
47 Which of the following will not
show cis-trans-isomerism?
[CBSE AIPMT 1996]
(a) CH3 — CH == CH — CH3
(b) CH3 — CH2 — CH ==CH — CH2 — CH3
(c) CH3 — CH ==CH — CH2 — CH3

CH3
(d) CH3 — CH — CH ==CH — CH2 — CH3

Ans. (c) CH3
CH3 — C == CH — CH2 — CH3 will not

CH3
show cis-trans isomerism because
double bonded carbon atom have two
same groups (—CH3, methyl group).
48 Which of the following will exhibit
chirality?
[CBSE AIPMT 1996]
(a) 2-methyl hexane
(b) 3-methyl hexane
(c) Neopentane
(d) Isopentane
Ans. (b)
3-methyl hexane contains chiral carbon
atom. Here it exhibits chirality.
H

CH3— CH2 — C *— CH2 CH2 CH3

CH3
49 The number of possible isomers of
the compound with molecular
formula C 7 H8O is [CBSE AIPMT 1995]
(a) 3
(b) 5
Ans. (b)
(c) 7
(d) 9
The possible isomers of C7H8O are
OCH3
CH2OH
OH
CH3
Anisole
Benzyl alcohol
OH
o-cresol
OH
(b) (CH3)2 NH
(d) RCH2NO2
It is due to the presence ofα-hydrogen
atom in NO2 and NO compounds.
CH3
m-cresol
CH3
p-cresol
167
Some Basic Principles of Organic Chemistry
50 The process of separation of a
racemic modification into d and
l-enantiomers is called
[CBSE AIPMT 1994]
(a) resolution
(b) dehydration
(c) revolution
(d) dehydrohalogenation
Ans. (a)
The process of separation of a racemic
mixture into d- or l-forms (enantiomers)
is called resolution. The racemic mixture
of enantiomers is resolved by treating
with an enantiomers of some chiral
compound. The products are
diastereomers which can be separated
by usual methods such as
recrystallisation, chromatography, etc.
51 The most important chemical
method to resolve a racemic
mixture makes use of the
formation of
[CBSE AIPMT 1994]
(a) a meso-compound
(b) enantiomers
(c) diastereomers
(d) racemates
Ans. (c)
Diastereomers have different physical
properties such as melting point, boiling
point, solubilities in a given solvent,
densities, refractive index, etc. Because
of the differences in the boiling points
and the solubilities, they can be
separated by fractional distillation and
fractional crystallisation.
52 Isomers of a substance must have
the same
[CBSE AIPMT 1991]
(a) structural formula
(b) physical properties
(c) chemical properties
(d) molecular formula
Ans. (d)
Two or more compounds having the
same molecular formula but different
physical, chemical properties and
structural formula are called isomers.
53 Which one of the following can
exhibit cis-trans-isomerism?
[CBSE AIPMT 1989]
(a) CH3  CHCl  COOH
(b) H  C ≡≡C  Cl
(c) Cl  CH==CHCl
(d) ClCH2  CH2Cl
Ans. (c)
TOPIC 3
The main conditions for geometrical
isomerism are
(i) The molecule must have a double
bond and the double bond should
have restricted rotation
(ii) The two atoms or groups attached to
the same carbon atom must be
different.
These two conditions are only obeyed by
Cl — CH==CH — Cl, so it shows
geometrical isomerism.
Bond Fission and
Electric Displacement
54 Which of the following is an
optically active compound?
[CBSE AIPMT 1988]
(a) 1-butanol
(b) 1-propanol
(c) 2-chlorobutane
(d) 4-hydroxybutanal
Ans. (c)
2-chlorobutane has the structure
Cl
 Chiral carbon
CH3 — C* — C2H5

H
It is optically active due to the presence
of asymmetric carbon atom (C*).
55 How many chain isomers could be
obtained from the alkane, C 6H14 ?
[CBSE AIPMT 1988]
(a) Four
(b) Five
(c) Six
(d) Seven
Ans. (b)
Hexane (C6H14 ) has five chain isomers
and these are
(i) CH3  CH2  CH2  CH2
 CH2  CH3
(ii) CH3  CH2  CH2  CH  CH3

CH3
CH3

(iii) CH3  CH2  CH  CH2  CH3
(iv) CH3  CH  CH  CH3


CH3 CH3
CH3

(v) CH3  CH2  C  CH3

CH3
56 The compound that is most
difficult to protonate is
[NEET (National) 2019]
O
(a) H3C
(c) Ph
O
H (b) H3C
O
O
(d) H
H
CH3
H
Ans. (c)
In CH3OH and CH3OCH3,  CH3 group
shows + I-effect due to which oxygen
acquires partial negative charge in both
the compounds.
δ–
δδ–
O
O
H3C
H
CH3
(+I-effect)
(+I-effect)
CH3
(+I-effect)
On the other hand inPh OH, Phshows
−I effect due to which oxygen acquire
partial positive charge.
δ+
O
Ph
H
Along with, the lone pair of oxygen also
show conjugation with benzene ring
which further increases the positive
charge on oxygen as:
δ+
O—H
O—H
δ–
δ–
δ–
Thus, further decreasing the process of
protonation.
Therefore, incoming proton will not be
able to attack easily on partial positive ‘O’
atom. Hence, phenol is most difficult to
protonate. Thus, option (c) is correct.
57 Which of the following is correct
with respect to −I effect of the
substituents? (R = alkyl) [NEET 2018]
(a) NH2 > OR > F
(b) NR2 < OR < F
(c) NH2 < OR < F
(d) NH2 > OR > F
Ans. (b, c)
–I effect is related to the ability of
substituent for the electron attraction
capacity from the attached carbon
atom.
168
NEET Chapterwise Topicwise Chemistry
i.e. It is based on electronegativity of an
atom. This effect increases with
increase in the electronegativity of an
atom.
From above we can conclude that
options (b) and (c) are correct.
−NR2 < − OR < − F (−I effect)
−NH2 < −OR < −F (−I effect)
Also, options (a) and (d) shows the order
of +I effect.
— NH2 > −OR > −F (+I effect)
—NR2 > −OR > −F (+I effect)
58 In pyrrole
3
4
5
2
N1
H
the electron density is maximum
on
[NEET 2016, Phase II]
(a) 2 and 3
(c) 2 and 4
Ans. (d)
C
CH3
Ph
2
N 1
r
N
H
I
H
II
s
Ph
CH
CH3
(I)
C
Ph
+
2°-allyl cation
CH3
(III)
Ans. (c)
Hyperconjugation occurs through the
H-atoms present on the carbon atom
next to the double bond i.e.α-hydrogen
atoms.
There is noα-H in the structures I and II.
So, hyperconjugation occurs in structure
III only i.e.
C
H
H
H
H
H
III
H

H3C  C  CH == CH2 ←→
H

+
H3C  C == CH  CH2
1°-allyl cation
H
r
N
H

H3C  C == CH  CH2  Cl ←→
(II)
CH3
s
3
5
[CBSE AIPMT 2015]
CH3
(b) 3 and 4
(d) 2 and 5
The conjugation (delocalisation) of
electrons in pyrrole can be visualised as
4
60 Consider the following compounds
C
α-hydrogen
H
C
CH2
C
C
CH3
C
H
H
Hence, it undergoes nucleophilic
reaction readily.
62 The total number of π- bond
electrons in the following structure
is
[CBSE AIPMT 2015]
H H
H3C
H
CH3
H3C
H CH3
H2C
(a) 4
(c) 12
Ans. (b)
(b) 8
(d) 16
In a given structure there are4π-bonds.
Hence, total number of π-electrons are
8.
H H
CH3
H
π-bond
d
b
π-
CH3
α-hydrogen
H2C
on
πbo
CH3
nd
π-bond
H
CH3
s
r
N
s
H
IV
r
N
H
V
As resonating structures III and IV are
more stable than II and V. Thus,
maximum electron density will be found
on carbon 2 and 5.
59 The pair of electron in the given
carbanion, CH3C ≡C − , is present in
which orbitals?
[NEET 2016, Phase I]
(a) sp 3
(c) sp
Ans. (c)
(b) sp2
(d) 2p
Number of σ -electrons
2
2 + 2 (negative ion)
=
2
Hybridisation =
= 2 = sp
Hence, in the carbanion, CH3C ≡ C! , pair
of electron as (–)ve charge is present in
sp-hybridised-orbital.
61 Which of the following is the most
correct electron displacement for a
nucleophilic reaction to take place?
[CBSE AIPMT 2015]
H2
H
(a) H3C →– C == C  C  Cl
H
H2
H
(b) H3C →– C == C  C  Cl
H
H2
H
(c) H3C →– C == C  C  Cl
H
H2
H
(d) H 3C →– C == C  C  Cl
H
Ans. (d)
Allylic and benzylic halides show high
reactivity towards S N 1 reaction.
Further, due to greater stabilisation of
allyl and benzyl carbocations
intermediates by resonance, primary
allylic and primary benzylic halides
show higher reactivity in S N 1 reactions
than other simple primary halides.
63 The correct order of decreasing
acidic strength of trichloroacetic
acid (A), trifluoroacetic acid (B),
acetic acid (C) and formic acid (D) is
[CBSE AIPMT 2012]
(a) B > A > D > C
(c) A > B > C > D
Ans. (a)
(b) B > D > C > A
(d) A > C > B > D
If an electron withdrawing group
(−I-showing group) is present, e.g.  CF3
has more (−I-effect) withdrawing power
than  CCl 3, it makes the removal of
proton more easy by stabilising the
remaining carboxylate ion and thus,
makes the acid more acidic.
The order of acidity of given compounds is
F
Cl
F CCOOH
F
> Cl CCOOH
Trifluoroacetic
acid
(B)
Cl
Trichloroacetic
acid
(A)
> HCOOH > CH3COOH
Acetic
Formic
acid
acid
(D)
(C)
169
Some Basic Principles of Organic Chemistry
64 The correct order of increasing
bond length of
C  H, C  O, C  C and C == Cis
[CBSE AIPMT 2011]
(a) C—C < C == C < C — O < C — H
(b) C—O < C—H < C—C < C == C
(c) C—H <C— O <C—C <C ==C
(d) C—H <C == C <C—O <C—C
Ans. (d)
C — H :0.109 nm
C ==C : 0.134 nm
C — O :0.143 nm
C — C :0.154 nm
∴ Bond length order is
C — H < C == C < C — O < C — C
65 Which one of the following
compounds has the most acidic
nature?
[CBSE AIPMT 2010]
CH2OH
(a)
H
+
H C
(a) (i) > (ii) > (iii) > (iv)
(b) (ii) > (iii) > (iv) > (i)
(c) (iv) > (ii) > (iii) > (i)
(d) (i) > (iii) > (ii) > (iv)
Ans. (a)
The carbanion with more s-character is
more stable. Thus, the order of stability
is
s
s
s
RC ≡≡ C > C6Hs
> R2 C==CH > R 3C — CH2
5
67 Base strength of
s
(i) H3CC H2
s
(ii) H2C== CH
(iii) H—C ≡≡ C s
(b)
OH
(c)
+
[CBSE AIPMT 2008]
s
(b) (iii) > (ii) > (i)
(d) (i) > (ii) > (iii)
s
s
CH3 C H2 > H2 C == C H > H — C ≡≡ C
OH
68 Which amongst the following is the
most stable carbocation?
CH
(d)
[CBSE AIPMT 2005]
+
Ans. (b)
Key Idea Presence of electron
withdrawing substituent increases the
acidity while electron releasing
substituent, decreases the acidity.
Phenyl is an electron withdrawing
substituent while CH3 is an electron
releasing substituent. Moreover,
phenoxide ion is more resonance
stabilised as compared to benzyloxide
ion, thus releases proton more easily.
That’s why phenol is a strong acid among
the given compounds.
66 The stability of carbanions in the
following
s
s
(i) RC
C
(ii)
s
(iii) R 2C==CH
s
(iv) R 3C —CH2
is in the order of
[CBSE AIPMT 2008]
(a) CH3 — C — H
|
CH3
+
CH3

(b) CH3  C +

CH3
+
(c) CH3
(d) CH3CH2
Ans. (b)
The most stable carbocation is t-alkyl
carbocation. The order of stability of
alkyl carbocation is
ter-alkyl>sec-alkyl> pri-alkyl > CH+3
carbocation. This stability order is
described with the help of
hyperconjugation and inductive effect.
On the basis of hyperconjugation,
+
(CH3)2 CH shows six resonating
structures due to the presence of six
α- C  H bonds,
H
|
+
HC——CH
H
CH3
H CH3
CH3
|
CH3 — C+ shows nine resonating
|
CH3
structures due to the presence of nine
α- C  H bonds.
Greater theα H-atom greater will be the
hyper conjugation resonating structure
and therefore, greater will be the
stability.
+
Weaker the acid, stronger is its
conjugate base. Among alkane, alkene
and alkyne, alkynes are most acidic and
alkanes are least acidic, so the order of
base strength is
alkane > alkene > alkyne or
OH
HC C H, etc.
H+
CH3
C H
CH3 does not show the property of
is in the order of
(a) (ii) > (i) > (iii)
(c) (i) > (iii) > (ii)
Ans. (d)
H
H+
HC C H
|
|
H CH3
resonance while CH3 — CH2 shows three
resonating structures due to presence
of three α- C H bonds. Hence, larger
number of resonating structures are
possible in (b), so it is most stable. The
above order of stability is also explained
with the help of (+) I-effect of —CH3
group. More the number of — CH3 group
more will be tendency to displace the
electrons towards positively charged
carbon of carbocation. Thus, positive
charge is decreased or compensated
and stability of carbocation is increased.
69 In HS − , I − , RNH2 , NH3 order of
proton accepting tendency will be
[CBSE AIPMT 2001]
(a) I– > NH3 > RNH2 > HS–
(b) NH3 > RNH2 > HS– > I–
(c) RNH2 > NH3 > HS– > I–
(d) HS– > RNH2 > NH3 > I–
Ans. (c)
Basic strength ∝ rate of accepting a
proton.
••
In R  N H2 ,N — has lone pair of
electron which increases the intensity
due to electron releasing R-group and
increases the tendency to donate lone
pair of electrons toH+ . Secondly as the
size of the ion increases, there is less
attraction for H+ and form weaker bond
with H-atom and less basic. The order of
the given series is
RNH2 > NH3 > HS – > I–
70 Which one of the following orders
is correct regarding the –I-effect of
the substituents ?
[CBSE AIPMT 1998]
(a) —NR2 < —OR > —F
(b) —NR2 > — OR > —F
(c) —NR2 < —OR < —F
(d) —NR2 > —OR < —F
170
NEET Chapterwise Topicwise Chemistry
NO2
Ans. (c)
The electronegativity follows the order
N < O < F.So, due to electronegative
character the order of −I-effect is
 NR2 < − OR < − F
71 Correct the increasing order of
acidity is as
[CBSE AIPMT 1994]
(a) H2O,C2H2 ,H2CO 3, phenol
(b) C2H2 ,H2O,H2CO 3, phenol
(c) Phenol, C2H2 ,H2CO 3,H2O
(d) C2H2 ,H2O, phenol and H2CO 3
Ans. (d)
C2H2 have acidic nature but it is less
acidic than water. Phenol is more acidic
than water due to resonance
stabilisation of phenoxide ion.H2 CO3 is
most acidic due to resonating
stabilisation of carbonate ion (CO23– ).
Thus, the correct order is
C2H2 < H2O < phenol < H2 CO33
NO2
H
(c)
r
(d) Y
r
H
Y
Ans. (a)
NO2 group is an electron withdrawing
group and exhibit −I effect. This effect
increases with decrease in distance of
positive charge present on C-atom and
hence lesser is the stability of
carbocation.
In option (a), the positive charge is at
maximum distance toNO2 group, so −I
effect due to NO2 group will be minimum
and stability will be maximum.
The above given reaction is known as
Riemer-Tiemann reaction. In this
reaction, electrophile involved is
dichlorocarbene (•• CCl2 ) which is formed
in theIst step of mechanism. It is given as
follows :
Mechanism
Step I Generation of electrophile
CHCl 3 + OH− r – CCl 3 + H2O
–
CCl 3 →
• CCl + Cl −
•
2
(Electrophile)
Dichlorocarbene
Step II Reaction of etectrophile with
phenoxide
O–
O
H
NO2
–
+ CCl2
CCl2
H
O–
Y
CHCl2
r
In option (b) and (d) the positive charges
is at minimum distance toNO2 group
hence the stability will be minimum.
TOPIC 4
Reaction
Intermediates
o-dichloromethyl
phenoxide
NO2
NO2
H
Step III Hydrolysis
r
r
O–
Y
O–
CHCl2
72 The most stable carbocation,
among the following is
[NEET (Odisha) 2019]
⊕
(a) (CH3) 3C CH CH3
Y
OH
2 OH–
H
Also, in option (c) the distance of positive
charge to NO2 group is more than ortho
but less than para, so it will be less stable
as compared to option (a).
NO2
⊕
OH
CH
O–
OH
CHO
–H2O
CHO
H+
(b) CH3 CH2 CH CH2 CH3
⊕
Salicylaldehyde
(c) CH3 CH CH2 CH2 CH3
r
⊕
(d) CH3 CH2 CH2
75 The correct statement regarding
electrophile is
[NEET 2017]
Y H
Ans. (c)
+
CH3 CH CH2 CH2 CH3 is the most stable
74 In the reaction
carbocation among the given
carbocations. It is because the number
of α-H atom attached to carbocation is
has maximum number of hyperconjugating
structure hence, it is most stable.
CHO
+ CHCl3+NaOH
+
maximum in CH3 CH CH2 CH2 CH3. Thus, it
O–Na+
OH
the electrophile involved is
[NEET 2018]
–
(a) dichloromethyl anion (C HCl2 )
73 Which of the following
carbocations is expected to be
most stable?
[NEET 2018]
NO2
NO2
s
O
OH
r
(a)
+
(b) formyl cation (CHO)
+
(c) dichloromethyl cation (CHCl2 )
(d) dichlorocarbene (•• C Cl2 )
Ans. (d)
(b)
H
Y
+
Na+
CHO
+ CHCl3+NaOH
Y
H
(a) Electrophile is a negatively charge
species and can form a bond by
accepting a pair of electrons from
a nucleophile
(b) Electrophile is a negatively
charged species and can form a
bond by accepting a pair of
electrons from another
electrophile
(c) Electrophiles are generally neutral
species and can form a bond by
accepting a pair of electrons from
a nucleophile
(d) Electrophile can be either neutral
or positively charged species and
can form a bond by accepting a
pair of electrons from a
nucleophile
171
Some Basic Principles of Organic Chemistry
Cl
Ans. (d)
An electrophile is defined as electron
deficient species which attacks on
electron rich areas. Being electron
deficient, the electrophiles behaves as
Lewis acids.
e.g.
+
+
+
+
+
H , Cl , Br , NO2 ,NO etc.
The electrophiles can be seen in the
form of neutral molecules also
••
•• CR , N R,BF etc.
2
3
e.g.
Thus, we can say that electrophile can
be either neutral or positively charged
species and can form a bond by
accepting a pair of electrons from a
nucleophile.
76 Which of the following statements
is not correct for a nucleophile?
[CBSE AIPMT 2015]
(a) Nucleophile is a Lewis acid
(b) Ammonia is a nucleophile
(c) Nucleophiles attack low electrons
density sites
(d) Nucleophiles are not electron
seeking
Ans. (a)
Nucleophiles are electron rich species.
Hence, act as a Lewis base but not Lewis
acid.
CH2 is
77 The radical
aromatic because it has [NEET 2013]
(a) 6p-orbitals and 6 unpaired
electrons
(b) 7p-2orbitals and 6 unpaired
electrons
(c) 7p-orbitals and 7 unpaired
electrons
(d) 6p-orbitals and 7 unpaired
electrons
Ans. (a)
H
Because of the
C
H
C
C
H
C
C
C
presence of
6p-orbitals and 6
H unpaired electrons, it
is aromatic in nature
H as these unpaired
electrons delocalise
in p-orbitals.
H
(c)
(d)
Ans. (b)
The group showing electron-donating
effect (such as —NH2 , —OH) should
stabilise the intermediate ions, i.e.
makes the ring more reactive towards
electrophilic substitution than benzene
and are called activating group while
electron withdrawing group (such as —Cl,
NO2 ) increases the positive charge on
ring, thus deactivates the ring. Hence,
phenol is more readily attacked by an
electrophile and is most reactive
towards an electroptile .
79 The order of decreasing reactivity
towards an electrophilic reagent,
for the following
(i) Benzene
(ii) Toluene
(iii) Chlorobenzene
(iv) Phenol
would be
[CBSE AIPMT 2007]
(a) (i) > (ii) > (iii) > (iv)
(b) (ii) > (iv) > (i) > (iii)
(c) (iv) > (iii) > (ii) > (i)
(d) (iv) > (ii) > (i) > (iii)
Ans. (d)
Phenol >Toluene >Benzene
(iv)
(ii)
(i)
> Chlorobenzene
(iii)
80 Which one of the following
compounds will be most easily
attacked by an electrophile?
Cl
(a)
OH
CH2OH
(b)
(b)
OH
CH3
(c)
—Cl-atom shows + R-effect that
o/p-directive influence but deactivate
the benzene ring. While —OH, —CH3
groups also shows o/p-influence but
activate the benzene ring. But in these
—OH group activates more than — CH3.
Hence, order of electrophilic
substitution is
OH
CH3
Cl
>
>
>
81 Which one of the following
compounds is resistant to
nucleophilic attack by hydroxyl
ions?
[CBSE AIPMT 1998]
(a) Methyl acetate
(b) Acetonitrile
(c) Dimethyl ether
(d) Acetamide
Ans. (c)
Dimethyl ether does not show
nucleophilic attack due to absence of
multiple bond.
Other compounds have multiple bonded
C-atom and bears partial positive
charges, therefore they undergo
nucleophilic attack byOH− ions.
82 The reaction is described as
Benzene having any activating group, i.e.
OH, R, etc, undergoes electrophilic
substitution very easily as compared to
benzene itself. Thus, toluene (C6H5 CH3),
phenol (C6H5OH) undergo elecrophilic
substitution very readily than benzene.
Chlorine with + E and + M-effect
deactivates the ring due to strong
−I-effect. So, it is difficult to carry out
the substitution in chlorobenzene than
in benzene, so the correct order is
[CBSE AIPMT 99, 1998]
78 Which one of the following is most
reactive towards electrophilic
attack?
[CBSE AIPMT 2008]
(a)
Ans. (d)
NO2
(d)
[CBSE AIPMT 1997]
H
CH3(CH2)5
H3C
CBr
OH –
H
(CH2)5CH3
HO C
CH3
(a) SE2
(c) SN2
Ans. (c)
(b) SN1
(d) S N0
When the attack of nucleophile takes
place on the opposite side of the leaving
group in the substrate molecule, a
transition state is obtained. It is partially
bonded to both the attacking
nucleophile and the leaving group and
results in the formation of product. Such
reactions are called asS N 2 (bimolecular
nucleophilic substitution reaction)
because in such reactions, rate depends
on the concentration of both, the
substrate and the nucleophile.
172
NEET Chapterwise Topicwise Chemistry
TOPIC 5
Types of Organic
Reactions
It is an example of electrophilic addition
reaction.
Br2
(c) CH4 
→ CH3Br + HBr
hν
83 Which of the following compound
is most reactive in electrophilic
aromatic substitution?
[NEET (Oct.) 2020]
Cl
NO2
(a)
(b)
(Organic peroxide)
CH3CH2 CH2Br
(Major)
It is a free radical addition reaction.
Hence, option (c) is correct.
85 Among the following, the reaction
that proceeds through an
electrophilic substitution, is
OH
(c)
It is a free radical substitution reaction,
which have free radical intermediates
like Br •, CH3•, H• etc.
HBr / (C 6H 5 COO) 2
(d) CH3  CH == CH2  
→
(d)
[NEET (National) 2019]
Ans. (d)
Electron Donating Group (EDG : + R > +
hyperconjugation) increases electron
density of benzene nucleus (which acts
as nucleophile) and make it more
susceptible for electrophilic attack in
electrophilic aromatic substitution
(ArSE2) reaction.
Lone pair of electrons on the central
atom of a group (substituent) show + R or
+ M effect with the benzene ring, likeOH (Power of +R effect)
Cl
+ Cl2
(a)
O
H +R
N
O –R
87 Which of the following organic
compounds has same hybridisation
as its combustion (CO 2 ) product?
[CBSE AIPMT 2014]
(a) Ethane
(c) Ethene
Ans. (b)
Hybridisation of carbon
= sp3 − number ofπ - bonds
In CO2 , (O==C==O) hybridisation of
carbon 1π 1π
2π
+ Cl2
(b)
Cl
Cl
Cl
Cl
Cl
—CH2OH + HCl
(c)
(H  C ≡≡≡ C  H) hybridisation of carbon
= sp3− 2 = sp
UV light
Cl
88 Which one is most reactive
towards nucleophilic addition
reaction?
[CBSE AIPMT 2014]
CHO
Heat
(a)
O
—CH2Cl
+ H2O
+
—N2Cl–
(d)
C
(b)
Cu2Cl2
+R
So, with the options, order of reactivity
in ArSE2 reaction will be
(d) > (a) > (c) > (b)
Ans. (a)
[(NEET (Oct.) 2020]
(a) Benzene with Br2 / AlCl3
(b) Acetylene with HBr
(c) Methane with Br2 /hv
(d) Propene with HBr/(C 6H5COO)2
Ans. (c)
Br / AlCl
2
(a) C6H5 H  
3→ C6H5 Br
−HBr
It is an electrophilic
aromatic substitution (ArSE2) reaction.
HBr
(b) HC ≡≡ CH → CH2 == CH Br
HBr
→ CH3  CHBr2
(c)
Halogenation of benzene is an example
of electrophilic substitution reaction.
+ Cl2
84 Which of the following is a free
radical substitution reaction?
CH3
CHO
—Cl + N2
–R
(b) Ethyne
(d) Ethanol
= sp3− 2 = sp
In ethyne or acetylene, C2H2 ,
—Cl + HCl
whereas NO2 group shows − R effect
and makes the benzene nucleus least
reactive in ArSE2 reaction.
O
AlCl3
Ans. (b)
In S N 1 reaction, there is slight more
inversion product than retention product
because S N 1 reaction also depend on
shielding effect of the front side of the
reacting carbon.
AlCl3
CH3
—Cl + HCl
CHO
Chlorobenzene
Reaction given in option (b) is an example
of addition reaction, reaction in option
(c) is an example of nucleophilic
substitution and option (d) is an example
of substitution reaction.
86 In an SN 1 reaction on chiral centres
there is
[CBSE AIPMT 2015]
(a) 100% racemisation
(b) inversion more than retention
leading to partial racemisation
(c) 100% retention
(d) 100% inversion
(d)
NO2
Ans. (d)
Reactivity of carbonyl compounds
towards nucleophilic addition reactions
depends on the presence of substituted
group.
Electron withdrawing (−I, − M) groups
increase reactivity towards nucleophilic
addition reactions. Thus, correct order is
173
Some Basic Principles of Organic Chemistry
CHO
CHO
>
90 Which one of the following is most
reactive towards electrophilic
reagent?
[CBSE AIPMT 2011]
>
CH3
(a)
OH
NO2 (–I, –M)
O
CHO
C
CH3
CH3
(b)
NHCOCH3
CH3
(c)
>
CH2OH
(iv) Double bond character
CX bond acquire some double bond
character due to resonance. Presence of
electron withdrawing groups like NO2
at ortho and para-positions facilitate the
nucleophilic displacement of X of aryl
halide. Among alkyl halides, 3° halides
are more reactive as compared to 2°
halides due to the formation of more
stable carbocation. Hence, the order of
reactivity of CX bond towards
nucleophile is as
X
X
NO2 < (CH3) CH—X
2
CH3
CH3 (+I)
<
(d)
89 Among the following compound
one that is most reactive towards
electrophilic nitration is
OCH3
IV
I
Ans. (a)
< (CH3)3C—X
NO2
II
CH3
III
[CBSE AIPMT 2012]
(a) benzoic acid
(c) toluene
Ans. (c)
(b) nitrobenzene
(d) benzene
Presence of electron releasing groups
like  R, OH, etc. increases the
electron density at o/p-position and
thus, makes the benzene ring more
reactive (at o/p-positions) towards an
electrophile. On the other hand, electron
withdrawing groups like COOH, NO2 ,
etc. if present, reduces electron density
and thus, reduces the activity of
benzene nucleus towards an
electrophile. Thus, the order of the given
compounds towards electrophilic
nitration is
Electron
withdrawing group (EWG)
NO2
COOH
<
Nitrobenzene
Benzoic
acid
Electron releasing
group (ERG)
CH3
<
<
Benzene
Toluene
Thus, toluene is most reactive towards
electrophilic nitration.
OH
(o-cresol)
Due to + M-effect of —OH group and
hyperconjugation of — CH3 group, the
benzene of o-cresol is highly reactive
towards electrophilic substitution.
91 The correct order of increasing
reactivity of C X bond towards
nucleophile in the following
compounds is [CBSE AIPMT 2010]
X
I
(CH3)3C—X
III
(a) I < II < IV < III
(c) IV < III < I < II
Ans. (a)
X
92 Which of the following reactions is an
example of nucleophilic substitution
reaction?
[CBSE AIPMT 2009]
(a) RX + KOH → ROH + K X
(b) 2 RX + 2 Na → R — R + 2 NaX
(c) RX + H2 → RH + HX
(d) R X + Mg → RMg X
Ans. (a)
RX +
KOH → K+ + OH−
OH
→ R — OH+ X –
–
Nucleophile
NO2
NO2
II
(CH)2CH—X
IV
(b) II < III < I < IV
(d) III < II < I < IV
Key Idea Alkyl halides are more reactive
towards nucleophilic substitution.
Reactivity depends upon the stability of
carbocation intermediate formed.
Among the given halides, aryl halide
(C6H5 X) is least reactive towards
nucleophile, due to the four possible
reason
(i) Reasonance effect
(ii) Hybridisation
(iii) Unstability of phenyl cation
−
OH is a stronger nucleophile than
halogen. So it easily replace the weaker
nucleophile.
Nucleophiles are either negative charge
or lone pair of electrons bearing species,
••
e.g. OH− , NH3, etc.
93 Which of the following is least
reactive in a nucleophilic
substitution reaction?
[CBSE AIPMT 2004]
(a) (CH3) 3C — Cl
(c) CH3CH2Cl
Ans. (b)
(b) CH2 == CHCl
(d) CH2 == CHCH2Cl
Chlorine of vinyl chloride (CH2 == CHCl) is
non-reactive (less reactive) towards
nucleophile (in nucleophilic substitution
reaction) because it shows the following
resonating structure due to + M-effect
of — Cl-atom.
••
δ–
δ+
CH2 == CH — Cl •• ↔ CH2 — CH== Cl
I
••
II
174
NEET Chapterwise Topicwise Chemistry
In structure II, Cl-atom have positive
charge and partial double bond
character with C of vinyl group, so it is
more tightly attracted towards the
nucleus and does not get replaced by
nucleophile inS N reaction.
94 Which one of the following is a free
radical substitution reaction ?
[CBSE AIPMT 2003]
CH2Cl
(a)
+ AgNO2
CH2NO2
CH2
(b)
CH2Cl
+ Cl2
+ Cl
Benzylic free radical
is resonating stable.
(III)
Cl + Cl
Cl2 (termination)
95 The correct order of reactivity
towards the electrophilic
substitution of the compounds
aniline (I), benzene (II) and
nitrobenzene (III) is
Ans. (b)
In —CH3, —OCH3 and CF3, CH3 and —OCH3
are electron donating group. Hence, they
activate the benzene nucleus. In these,
order of activation is —OCH3 > — CH3
while —CF3 group deactivates the
benzene nucleus. So, it shows lower rate
of electrophilic substitution on benzene
ring. Thus, order of electrophilic
substitution is
CF3
<
<
(b) CH3CHO + HCN
CH3CH(OH)CN
CH3
(c)
Boiling
+ Cl2
CH2Cl
+ CH3Cl
(d)
Anhy. AlCl3
CH3
(b) I > II > III
(d) II > III > I
Ans. (c)
(I)
>
I
II
III
96 Among the following compounds
the decreasing order of reactivity
towards electrophilic substitution
is
[CBSE AIPMT 2000]
Cl — Cl
2Cl
Free radical (Initiation)
CH3
(II) (a)
NO2
>
At higher temperature the reaction of
toluene with chlorine is an example of
free radical substitution.
<
In aniline —NH2 group is attached with
benzene ring. —NH2 group shows
+M-effect. So, it activates the benzene
ring. Hence, rate of electrophilic
substitution is increased due to increase
in the electron density at o/p-position. In
case of nitrobenzene, (—NO2 ) −M-effect
deactivates the benzene ring. So in
nitrobenzene, rate of electrophilic
substitution is lower than benzene.
Hence, order ofSE reaction is
NH2
OCH3
CH3
[CBSE AIPMT 2003]
(a) II < III > I
(c) III > II > I
Ans. (b)
+ Cl
CH2
97 Among the following compounds
(I–III), the correct order of reaction
with electrophile is
NO2
OCH3
I
II
III
[CBSE AIPMT 1997]
(a) II > III > I
(b) III < I < II
(c) I > II > III
(d) I > III > II
Ans. (c)
Due to +I as well as + R-effect of —OCH3
group, it activates the benzene ring.
While —NO2 deactivates the benzene
ring due to its −I-effect and also
decrease the reaction rate, as well as
−R-effect.
So, order of SE reaction is
OCH3
+ HCl
(Propagation)
CH3
I
II
(a) II > I > III > IV
(c) IV > I > II > III
OCH3
CF3
III
IV
(b) III > I > II > IV
(d) I > II > III > IV
NO2
>
I
>
II
III
22
Hydrocarbons
TOPIC 1
Alkanes
Ans. (b)
01 Which of the following alkane
cannot be made in good yield by
Wurtz reaction?
[NEET 2020]
(a) 2, 3-dimethylbutane
(b) n-heptane
(c) n-butane
(d) n-hexane
Ans. (b)
In Wurtz reaction, an alkyl halide (R  X)
give a symmetrical alkane (R  R) of
even number of carbon atom with 100%
yield.
2Na/ether
2R  X →
R  R ( 100%)
−2NaX
An unsymmetrical alkane (R  R ′) with
odd number of carbon atom can be
obtained with 33% yield only, if we take a
mixture of two different alkyl halides,
R  X and R ′ X.
R  X + R ′ X → R  R + R ′ − R ′
1442443
Key Idea Alkanes which contain all
equivalent hydrogen atoms forms only
one monochloro derivative on
halogenation.
CH3

Neo-pentane,H3C  C  CH3 contains all

CH3
equivalent hydrogen atoms. So, it will
give only one monochloro derivative on
halogenation.
03 Hydrocarbon (A) reacts with
bromine by substitution to form an
alkyl bromide which by Wurtz
reaction is converted to gaseous
hydrocarbon containing less than
four carbon atoms. A is [NEET 2018]
(a) CH3  CH3
(c) CH ≡≡CH
Ans. (d)
The given reaction takes place as follows
67%
+ R − R′
123
33%
So, here n-heptane (C7H16 ) cannot be
obtained with good yield.
Hydrocarbons in options (a), (c) and (d)
have even number of carbon atom
(symmetrical, alkanes).
02 The alkane that gives only one
mono-chloro product on
chlorination with Cl 2 in presence of
diffused sunlight is
[NEET (Odisha) 2019]
(a) 2, 2-dimethylbutane
(b) neopentane
(c) n-pentane
(d) isopentane
(b) CH2 == CH2
(d) CH4
Br2 / hν
Na / dry ether
CH4 → CH3Br →
(A )
Step 1
04 The correct statement regarding
the comparison of staggered and
eclipsed conformations of ethane,
is
[NEET 2016, Phase I]
(a) The eclipsed conformation of
ethane is more stable than
staggered conformation, because
eclipsed conformation has no
torsional strain
(b) The eclipsed conformation of
ethane is more stable than
staggered conformation even
though the eclipsed conformation
has torsional strain
(c) The staggered conformation of
ethane is more stable than
eclipsed conformation, because
staggered conformation has no
torsional strain
(d) The staggered conformation of
ethane is less stable than
eclipsed conformation, because
staggered conformation has
torsional strain
Ans. (c)
Wurtz reaction
(Step 2)
CH3  CH3
(A )
Step I Alkyl halide is formed by free
radical halogenation of alkane in the
presence of UV-light.
Step II The formed alkyl halide reacts
with sodium in presence of dry ether to
form alkane containing double number
of carbon atoms present in alkyl halide.
This reaction is known as Wurtz
reaction.
From the above mechanism, it is
concluded that option (d) is correct as in
all other cases the hydrocarbon formed
in step 2 will contain more than four
carbon atoms.
Due to the absence of torsional strain
staggered conformation of ethane is
more stable than eclipsed conformation
of it.
H
H
HH
H
H
H
H
H
H
H
H
Staggered conformation Eclipsed conformation
05 The compound that will react most
readily with gaseous bromine has
the formula
[NEET 2016, Phase II]
(a) C 3H6
(c) C 4H10
(b) C2H2
(d) C2H4
176
NEET Chapterwise Topicwise Chemistry
Ans. (c)
In gaseous state, Br2 forms free radicals
and saturated hydrocarbons are more
prone to have free radical substitutions.
As C4H10 reacts most readily with
gaseous bromine via free radical
mechanism as shown below:
C4H10 + Br2 → C4H9Br + HBr
Therefore, option (c) is correct.
06 Liquid hydrocarbons can be
converted to a mixture of gaseous
hydrocarbons by
[CBSE AIPMT 2010]
(a) oxidation
(b) cracking
(c) distillation under reduced pressure
(d) hydrolysis
Ans. (b)
Key Idea Lower hydrocarbons exist in
gaseous state while higher ones are in
liquid state or solid state.
On cracking or pyrolysis, the
hydrocarbon with higher molecular mass
gives a mixture of hydrocarbons having
lower molecular mass. Hence, we can
say that by cracking a liquid hydrocarbon
can be converted into a mixture of
gaseous hydrocarbons.
07 In the following the most stable
conformation of n-butane is
[CBSE AIPMT 2010]
CH3
CH3
CH3
H
(a)
H
H
H
H
(b)
H
H
H
CH3
CH3
CH3
(c)
H
H
CH3
H
(d)
H
H
H
H
H
H3C
Ans. (b)
Key Idea The conformation in which the
heavier groups are present at maximum
possible distances, so that the forces of
repulsion get weak, is more stable.
Among the given conformations of
n-butane, the conformation shown in
option (b), i.e. anti conformation is most
stable as in it the bulkier group, i.e.
CH3 group are present at maximum
possible distance and get lower energy.
08 Which of the following reactions is
expected to readily give a
hydrocarbon product in good
yields ?
[CBSE AIPMT 1997]
Electrolytic
oxidation
(a) RCOOK →
H
H
H
H
CH3
CH3
(Structure III)
Gauche or
Skew form
I
2
(b) RCOOAg →
Cl2
(c) CH3CH3 →
hν
C2H 5 OH
(d) (CH3)2 CCl →
Ans. (a)
Electrolysis of sodium or potassium salt
of carboxylic acid gives good yield of
hydrocarbon
Staggered conformation has minimum
repulsion, so it is the most stable. The
order of stability is
staggered > gauche > eclipsed
Energy order eclipsed> gauche >
staggered
Electrolytic
2RCOOK → 2 RCOO–
oxidation
11 The distance between two
adjacent carbon atoms is largest in
Anode
+ 2K
+
Cathode
At anode
•
2 RCOO → 2 RCOO+2 e
–
•
2 RCOO → R  R +2CO2
09 In the commercial gasolines, the
type of hydrocarbons which is more
desirable, is
[CBSE AIPMT 1997]
(a) branched hydrocarbon
(b) straight chain hydrocarbon
(c) linear, unsaturated hydrocarbon
(d) toluene
Ans. (a)
On increasing the number of branches,
knocking is decreased and octane
number is increased. So, branched chain
hydrocarbons have less knocking and is
more desirable.
10 The most stable configuration of
n-butane will be [CBSE AIPMT 1997]
(a) chlorine free radical
(b) hydrogen chloride
(c) methyl radical
(d) chloromethyl radical
Ans. (a)
hν
CH4 + Cl2 → CH3Cl
Chain initiation step involves the fission
of Cl2 molecule into chlorine free radical.
Cl2 → Cl • + Cl •
hν
[CBSE AIPMT 1994]
3
H
H
H
H
H
12 In the free radical chlorination of
methane, the chain initiating
step involves the
formation of
[CBSE AIPMT 1994]
13 Which of the following compounds
has the lowest boiling point?
(b) eclipsed
(d) staggered-anti
Conformations of n-butane are as under
CH3
CH
H
(b) ethene
(d) ethyne
The C—C bond length is maximum for
single bond, butane have largest C—C
bond length because it contains
carbon-carbon single bond.
At cathode
2K+ + 2 e – → 2K
2K + H2O → 2KOH +H2 ↑
(a) skew boat
(c) gauche
Ans. (d)
[CBSE AIPMT 1994]
(a) benzene
(c) butane
Ans. (c)
–
H
CH3
(Structure I)
Staggered or
anti-form
H
CH3
(Structure II)
Eclipsed form
(a) CH3CH2CH2CH2CH3
(b) CH3CH ==CH —CH2CH3
(c) CH3CH ==CH —CH ==CH2
(d) CH3CH2CH2CH3
Ans. (d)
Boiling points of alkanes increase as the
number of carbon atom increases or
molecular mass increases. Alkenes have
high boiling point, so CH3CH2 CH2 CH3 has
the lowest boiling point.
177
Hydrocarbons
14 Reactivity of hydrogen atoms
attached to different carbon atoms
in alkanes has the order
CH3
|
(d) Cl  CH2  CH2  CH
|
CH3
CH2 CH2 CH3
(c)
[CBSE AIPMT 1993]
(a) tertiary > primary > secondary
(b) primary > secondary > tertiary
(c) Both (a) and (b)
(d) tertiary > secondary > primary
Ans. (d)
Tertiary alkanes are more reactive as
they form tertiary free radical which is
more stable. On the other hand,
primary alkanes are less reactive
because they form 1° free radicals
which are less stable.
CH3
CH3
H
H




•
•
•
CH3 —C > CH3 —C > CH3 — C > H — C•




CH3
H
H
H
CH
CH
Ans. (b)
CH3
The given road map problem is
Alkene (A)
(d)
HCl
Ans. (b)
Let us identity the ozonolysis [(i) O2 /THF,
(ii) Zn/H2O] products from each of the
following :
Addition
reaction
B
(Major product)
O
O
oo
—CH2CH== CH2
group of these arose from the π-bonds
of alkene. Thus, most probable alkene
will be
H3C
—CH2CH==O+ CH2==O
(a)
(c)
oo
H3C
H3C
oo
H3C
H3C
—CH==CH CH3
C===CHCH3
2-methyl but-2-ene
(A)
(d)
6
7
3
8
2
1
10
9
The structure can be identified with the
help of branches at 2 and 6 positions and
a double bond at 4 position.
16 An alkene on ozonolysis gives
methanal as one of the product. Its
structure is
[NEET (Sep.) 2020]
CH2
CH2
CH3
CH2
CH
CH2
(a)
(b)
H3C
C == O+O ==CHCH3
Ethanal
Propanone
Ans. (a)
4
O3
Zn/H2O
H3C
—CH==O + CH3  CH==O
5
CH3
The reaction taking place will be
O
Structure of 2,6-dimethyl-dec-4-ene is
H
C == C
Alkene
CH2CH2CH3
(d)
O ==C
CH3
CH2CH2CH3
O
(b)
H
C== O
H3C
(Methanal)
(c)
Ethanal
The products of first reaction, i.e.
ozonolysis can give an idea about the
probable alkene as C —
—O
CH2CH2CH3
[NEET 2021]
C = O + O = CH—CH3
H3C
Propanone
—CH2CH2CH3
(a)
(b)
15 The correct structure of
2,6-dimethyl-dec-4-ene is
H3C
oo
Stability decreases
TOPIC 2
Alkenes
O3
Zn/H2O
HCl
We get, methanal from the hydrocarbon
in option (b) only.
H3 C
17 An alkene ‘A’ on reaction with O 3
and Zn -H2O gives propanone and
ethanal in equimolar ratio. Addition
of HCl to alkene ‘A’ gives ‘B’ as the
major product. The structure of
product ‘B’ is [NEET (National) 2019]
H3 C
Addition
reaction
C—CH2—CH3 +
Cl
2-chloro-2-methylbutane
(B)
(major product)
H3 C
H3 C
Cl
CH—CH—CH3
CH2 Cl
|
(a) H3C  CH2  CH  CH3
2-chloro-3-methylbutane
(minor product)
CH3
|
(b) H3C  CH2  C  CH3
|
Cl
CH3
|
(c) H3C  CH  CH
|
|
Cl
CH3
18 Which of the following molecules
represents the order of
hybridisation sp 2 , sp 2 , sp, sp from
left to right atoms?
[NEET 2018]
(a)
(b)
(c)
(d)
CH2 == CH  CH == CH2
CH2 == CH C ≡≡ CH
CH ≡≡ C C ≡≡ CH
CH3 CH == CH  CH3
178
NEET Chapterwise Topicwise Chemistry
Ans. (b)
Key Idea While judging the hybridisation
in the given type of organic molecules
always look for the number of σ and π
bonds formed by C-atom involved.
The hybridisation of the given molecules
are.
sp
2
sp
2
sp
2
sp
2
H2
(d) H3C  C  CH2Br
Ans. (c)
H2C
CH2
C
H2
sp 2
sp
sp
sp
sp
3
sp
sp
2
2
sp
Br
H
[CBSE AIPMT 2008]
CH3
C
CH
CH3
CH3
CH3
(c) CH3 — CH — CH — CH3


Br CH3
CH3 CH3
C
CH
CH3 CH3
C
CH
20 Which of the following compounds
shall not produce propene by
reaction with HBr followed by
elimination or direct only
elimination reaction?
[NEET 2016, Phase II]
CH2
C
H2
H2
(b) H3C  C  CH2OH
(c) H2 C == C ==O
(a) CH3 — CH — CH— CH2Br

CH3
Br

(b) CH3 — C  CH2 CH3

CH3
CH3
(a) H2C
q
O

H3C  C Br
22 Which of the following compounds
with molecular formula, C 5H10
yields acetone on ozonolysis?
[CBSE AIPMT 2007]
(a) 2-methyl-2-butene
(b) 3-methyl-1-butene
(c) Cyclopentane
(d) 2-methyl-1-butene
Ans. (a)
2-methyl-2-butene (molecular formula
C5H10 ) yields acetone on ozonolysis.
23 Which one of the following alkenes
will react faster with H2 under
catalytic hydrogenation conditions ?
(R =alkyl substituent)
[CBSE AIPMT 2005]
R
H
(a)
A (predominantly) is
+
CH3
addition
21 H3C —CH—CH== CH2 + HBr → A

CH3
CH3
D
–H
Nucleophilic
CH3  CH == CH2
Thus, option (c) is correct.
Ans. (b)
CH3
CH2 == C== O → H2 C== C OH
Direct elimination
(c) (CH3)2 C ==CH — CH2 — CH2
CH2

HBr
CH3  CH2  CH2  Br →
(d) (CH3)2 CH — CH2 — CH == CH2
1-2-methyl
shift
CH2
Elimination
(b) (CH3) 3C —CH == CH2
CH3
CH
→ H3C  CH == CH2
(a) (CH3)2 CH —CH — CH == CH3

CH3
CH3
H3C
CH3  CH2  CH2 Br
19 2,3-dimethyl-2-butene can be
prepared by heating which of the
following compounds with a strong
acid?
[NEET 2016, Phase I]
CH
CH2
Electrophilic addition
3
Therefore, the correct option is (b).
C
CH2
CH3  CH2  CH2  OH →
(iv) CH3  CH == CH  CH3
CH3
CH3
HBr
sp
sp
HBr
Electrophilic
addition
Elimination
(iii) CH ≡≡ C  C ≡≡ CH
Br

CH3 — C — CH2 CH3

CH3
Br
sp
(ii) CH2 == CH  C ≡≡ CH
−
3° carbocation
(more stable)
The said reactions can be visualised as
(i) CH2 == CH  CH == CH2
sp 2
⊕
H-shift
Br
→
CH3 — C — CH2 — CH3 →

CH3
R
H
R
R
(c)
H
H
H
 ⊕
CH3  C  CH  CH3

CH3
2° carbocation
(less stable)
R
R
R
R
H
Ans. (a)
1
Heat of hydrogenation of alkene
Greater the number of alkyl groups
attached to the double bonded carbon
atoms, more stable is the alkene. Hence,
given alkene follow the following order
of stability.
R
R
>
=
H3C — CH — CH == CH2 + HBr →
−Br −

CH3
R
Stability of alkene
R
Ans. (b)
R
(d)
∝
(d) CH3 — CH — CH— CH3


CH3 Br
R
(b)
R
R
R
R
R
R
>
=
H
R
R
R
H
H
=
H
H
R
R
>
=
H
R
=
H
=
R
H
Hence, faster hydrogenation occurs in
H
R
=
R
H
179
Hydrocarbons
24 Reaction of HBr with propene in
the presence of peroxide gives
[CBSE AIPMT 2004]
(a) iso-propyl bromide
(b) 3-bromo propane
(c) allyl bromide
(d) n-propyl bromide
Ans.
(d) Reaction of HBr with propene in the
presence of peroxide gives n-propyl
bromide. This addition reaction is an
example of anti-Markownikoff’s addition
reaction.
(i.e. it is completed in form of free radical
addition)
25 The compound,
CH3

CH3 —C == CH—CH3 on reaction
with NaIO 4 in the presence of
KMnO 4 gives [CBSE AIPMT 2003]
(a) CH3COCH3 + CH3CHO
(b) CH3CHO + CO2
(c) CH3COCH3
(d) CH3COCH3 + CH3COOH
Ans. (d)
H3C
NaIO
in presence
of KMnO4
H3C
Peroxide
H3C
CH3 — CH2 —CH2Br
Benzoyl peroxide
C ==O + CH3COOH
H3C
n− propyl bromide
Mechanism of this reaction is
represented asfollows :
Step I Formation of free radical of
peroxide by means of decomposition.
∆
C6H5 — C—O O— C — C6H5 →


O
O
26 Which alkene on ozonolysis gives
CH3CH2CHO and CH3 C CH3 ?

O
2C6H5 — COO
Step II Benzoate free radical forms
bromine free radical with HBr.
(b) CH3CH2 CH ==CHCH2 CH3
(c) CH3CH2 CH==CHCH3
•
C6H5 COO + H  Br → C6H5 COOH + Br
Br
1° free radical
(less stable)
•
2° free radical
(more stable)
CH3
CH3—CH2—CH == C
•
+ O3
CH2OH
R
→

CH2OH
O
product of this step.
Step IV More stable free radical accept
hydrogen free radical from benzoic acid
and give final product of reaction. i.e.
n-propyl bromide.
C
CH3—CH2—CH
O
•
CH3 — CH— CH2Br + C6H5 COOH →
•
CH3 — CH2 — CH2Br + C6H5 COO
CH3
CH3
Ozonide
CH3—CH2—CHO + CH3COCH3
+ ZnO + H2O
O
Ans. (b)
CH2OH
HOCl
CH2 == CH2 →

CH2 Cl
‘ M’
CH2OH
R
→ 
NaHCO 3 ( aq )
CH2OH
Glycol
CH3
Hence, CH3 — C H — CH2Br is the major
C6H5 COO + C6H5 COO → (C6H5 COO)2
Hypochlorous
acid
CH2 == CH2 →M
(d) CH2  CH2 and heat
When O3 reacts with alkene, it forms
ozonide, which on reaction with Zn and
acid or H2 / Ni gives aldehydes and/or
ketones. These products helps in
locating the position of a double bond as
CH3  C H  CH2Br
•
28 In a reaction,
(c) CH3CH2OH and HCl
Ans. (a)
•
CH3 — CH==CH2 +Br → CH3 — CH— CH2 +
|
•
Stability of an alkene depends upon the
heat of hydrogenation of an alkene.
Therefore, lower the heat of
hydrogenation of an alkene, higher will
be stability.
Order of stability is
trans-but-2-ene > cis-but-2-ene >
but-1-ene.
Heat of hydrogenation (kJ/mol) are 115.5,
119.6 and 126.8 respectively.
(a) CH3CH2 Cl and NaOH
(b) CH2 Cl ⋅ CH2OH and aq. NaHCO3
(d) CH3 — C==CHCH3

CH3
Step III Bromine free radical attacks on
C== C of propene to form intermediate
free radical.
n-propyl bromide
1
Heat of hydrogenation of alkene
M = molecules, R = reagent. M and R
are
[CBSE AIPMT 1997]
CH3
Benzoate free radical
Step V Benzoate free radicals are
changed into benzoyl peroxide for the
termination of free radical chain.
Stability of alkene
CH3
(a) CH3CH2 CH==C
•
(b) III > II > I
(d) I > II > III
[CBSE AIPMT 2001]
•
•
[CBSE AIPMT 2000]
(a) II > I > III
(c) III > I > II
Ans. (b)
∝
4
C == CH— CH3 + 3[O]  
→
CH3 — CH == CH2 + HBr →
the decreasing order of stability is
O
(Reduction)
Zn/H2O
27 Among the following alkenes,
1-butene cis-2-butene
I
III
trans-2-butene
II
CH2OH
So, M is 
and R is NaHCO3.
CH2 Cl
29 In the presence of platinum
catalyst, hydrocarbon A adds
hydrogen to form n-hexane. When
hydrogen bromide is added to A
instead of hydrogen only a single
bromo compound is formed. Which
of the following is A?
[CBSE AIPMT 1996]
(a) CH3 —CH2 —CH ==CH —CH2 —CH3
(b) CH3 —CH2 —CH2 —CH ==CH —CH3
180
NEET Chapterwise Topicwise Chemistry
(c) CH3 —CH ==CH — CH2 —CH2 —CH3
(d) CH2 ==CH —CH2 —CH2 —CH2 —CH3
Ans. (a)
CH3 — CH2 — CH == CH— CH2 — CH3
(A)
HBr
Pt
H2
CH3CH2 CH2 CH2 CH2 CH3
n-hexane
CH3CH2 CH CH2 CH2 CH3

Br
3-bromohexane
(c) R —C == O

CH3
(d) R — CH — CH2


OH OH
But-2-yne
1
Here, mole of B2H6 react with alkene by
2
Markownikoff’s addition and form
trialkylborone called Hydroboration,
H2O2 / OH− gives oxidation. So,
trialkyborone oxidise in alcohols and
reaction is also called
Hydroboration-oxidation.
••
CH3 O H


CH3 — C —– C — CH3 + H+ →
Protonation


CH3 H
2° carbocation
CH3

→ CH3 — C —– C — CH3
shift


CH3 H
⊕
Methyl
3°carbocation
( R — CH2 — CH2 ) 3 — B
‘ B’
→ 3R — CH2 — CH2 —OH
31 The alkene R—CH== CH2 reacts
readily with B 2H6 and formed the
product B which on oxidation with
alkaline H2O 2 produces
[CBSE AIPMT 1995]
(a) R —CH2 CHO
(b) R —CH2 —CH2 — OH
H
H
cis-2-butene
(syn addition)
34 The number of sigma (σ) and pi (π)
bonds in pent-2-en-4-yne is
(a) 8σ-bonds and 5π-bonds
(b) 11σ-bonds and 2π-bonds
(c) 13σ-bonds and no π-bonds
(d) 10σ-bonds and 3π-bonds
Ans. (d)
Single bond = 1σ, Double bond = 1σ + 1 π,
Triple bond = 1σ + 2 π.
The structure of pent-2-en-4-yne is
TOPIC 3
Alkynes
32 How many (i) sp 2 hybridised carbon
atoms and (ii) π bonds are present
in the following compound?
—C—C—COOCH3
[NEET (Oct.) 2020]
(a) 7, 5
(c) 7, 6
Ans. (c)
sp2
(b) 8, 6
(d) 8, 5
O
sp2
sp2
sp2
sp2
C
C
C
sp2
OCH3
sp2
Number of sp2 -carbon atom = 7
π-bonds = 6
Hence, option (c) is correct.
33 The most suitable reagent for the
following conversion, is
[NEET (National) 2019]
CH3

−H +
→ CH3 — C == C — CH3
rearrangement

CH3
CH3
C=
=C
[NEET (National) 2019]
H2 O2 /H +
⊕
CH3 OH2


Dehydration
CH3 — C —– C — CH3 →
–H2 O


CH3 H
CH3

⊕
CH3  C   C  CH3


H
CH3
H2, Pd/C
Quinoline
H3C
BH
2 6
3R —CH == CH2 →
[CBSE AIPMT 1995]
2,3-dimethyl-2-butene
3,3-dimethyl-2-butene
2,3-dimethyl-1-butene
cis and trans-isomers of
2,3-dimethyl-2- butene
Ans. (a)
—
H3C—C=
=C—CH3
Ans. (b)
30 When 3,3-dimethyl-2-butanol is
heated with conc. H2SO4 , the major
product obtained is
(a)
(b)
(c)
(d)
alkene thus formed has
cis-configuration.
H3C—C≡≡C—CH3
H3C
H
CH3
H
cis-2-butene
(a) H2 , Pd/C, quinoline
(b) Zn/HCl
(c) Hg2+ /H+ , H2O
(d) Na/liquid NH3
Ans. (a)
Hydrogenation of alkynes in the
presence of Pd/C, quinoline proceeds
through syn addition of hydrogen and the
H
H
| 1σ 1σ | 1σ 1σ 1σ
H  C  C ==C 1σC ≡≡C H
1σ | 1σ | 1π
2π
1σ
H H
1σ
∴ The number of sigma (σ) bonds are 10
and pi ( π) bonds are 3.
35 Which one is the correct order of
acidity?
[NEET 2017]
(a) CH2 == CH2 > CH3  CH == CH2
> CH3  C ≡≡ CH > CH ≡≡ CH
(b) CH ≡≡ CH > CH3 C ≡≡ CH >
CH2 == CH2 >CH3 CH3
(c) CH ≡≡ CH > CH2 == CH2 > CH3
 C ≡≡ CH > CH3 CH3
(d) CH3  CH3 > CH2 == CH2 > CH3
 C ≡≡ CH > CH ≡≡ CH
Ans. (b)
Greater the s-character of C-atom in
hydrocarbons, greater the
electronegativity of that carbon and thus
greater the acidic nature of the H
attached to electronegative carbon.
CH ≡≡ CH CH2 == CH2 CH3  CH3
Hybridisation :
sp
sp 2
sp 3
s-character :
50%
33%
25%
Max.
Electronegativity: ←

Max.
Acidic character of ←

terminal H
Thus, CH ≡≡ CH > CH3C ≡≡ CH
> CH2 == CH2 > CH3  CH3
181
Hydrocarbons
36 In the reaction,
(i) NaNH /liq.NH
3
2
HC ≡CH →
(ii) CH3 CH2Br
(i) NaNH /liq.NH
2
3
X →
Y
(ii) CH3 CH2Br
X and Y are
[NEET 2016, Phase I]
(a) X = 2 -butyne; Y = 3 -hexyne
(b) X = 2 -butyne; Y = 2 -hexyne
(c) X = 1 -butyne; Y = 2 -hexyne
(d) X = 1 -butyne; Y = 3 -hexyne
Ans. (d)
Since,NaNH2 /liq.NH3 behaves as a base,
so it abstracts proton from acetylene to
form acetylide anion followed by
alkylation to give compound (X) i.e.
1-butyne. (X) further reacts with
NaNH2 /liq.NH3 followed by alkylation with
ethyl bromide yields 3-hexyne (Y).
Hydration of alkyne,
Hg 2 + /H+
CH ≡≡ CH + H2O → CH2 == CH

OH
I

(b) CH3 — CH2 — CH2 —C — H

Cl
I

(c) CH3 — CH2 — CH — CH2Cl
I

(d) CH3CH2  C  CH3

Cl
Ans. (d)
Unsaturated
alcohol
(unstable)
 Tautomerisation
↓
CH3  CHO
41 The cylindrical shape of an alkyne
is due to
[CBSE AIPMT 1997]
Followed by Markownikoff’s rule.
CH3 — CH2 — C ≡≡ CH + HCl →
HI
CH3CH2 — C == CH2 →

Cl
I

CH3 — CH2 — C — CH3

Cl
H  C ≡≡ CH (1)
NaNH2/Liq.NH3
Ans. (d)
(a) CH3 — CH—CH2CH2I

Cl
H — C ≡≡ Cs
(2) CH3CH2—Br
alkylation
HBr + H — C ≡≡ C — CH2CH3
X
1-butyne
2-chloro-2 -iodobutane
(1) NaNH2/liq.NH3
(2) CH3CH2—Br
alkylation
s
C ≡≡ C — CH2CH3
H3CH2C — C ≡≡ C — CH2CH3 + HBr
(Y)
3-hexyne
39 Products of the following reaction
[CBSE AIPMT 2005]
(i) O
3
CH3C ≡≡ C ⋅ CH2CH3 →
(ii) Hydrolysis
… are
37 Which of the following reagents will
be able to distinguish between
1-butyne and 2-butyne?
[CBSE AIPMT 2012]
(a) NaNH2
(c) O2
Ans. (a)
(b) HCl
(d) Br2
(a) CH3CHO + CH3CH2CHO
(b) CH3COOH + CH3COCH3
(c) CH3COOH + HOOC ⋅CH2CH3
(d) CH3COOH + CO2
Ans. (c)
O
NaNH
2
CH3CH2 C ≡≡ CH →
CH3CH2 C ≡≡ CNa
1
+ H2 ↑
2
NaNH 2
CH3  C ≡≡ C H → No reaction
38 Predict the product C obtained in
the following reaction of butyne-1.
[CBSE AIPMT 2007]
CH3CH2  C ≡≡ CH + HCl →
HI
B →
C
O
CH3C

O
2H2O
CCH2CH3

O
CH3COOH + CH3CH2COOH
Acetic acid
Propanoic acid
40 When acetylene is passed through
dil ⋅ H2SO4 in presence of HgSO 4 ,
the compound formed is
[CBSE AIPMT 1999]
(a) ether
(b) ketone
(c) acetic acid
(d) acetaldehyde
Ans. (d)
In alkynes C ≡≡ C is present, out of these
three bonds one sigma and two π-bonds
are present. Sigma bond is formed by
sp-hybrid orbitals whereas π-bonds are
formed by unhybridised orbitals.
Hence, it shows cylindrical shape.
42 A compound is treated with NaNH2
to give sodium salt. Identify the
compound.
[CBSE AIPMT 1993]
(a) C2H2 (b) C 6H6 (c) C2H6
Ans. (a)
(d) C2H4
Sodamide is strong base. Therefore it
attracts the more acidic hydrogen and
gives sodium salt
H— C ≡≡ C —H + NaNH2 →
1
H— C ≡≡ C–Na+ + H2
2
Sodium ethynide
3
CH3  C ≡≡ C  CH2  CH3 →
NaNH2 is used to distinguish between
1-butyne and 2-butyne.
(a) three sigma C—C bonds
(b) three π C—C bonds
(c) two sigma C—C and one π C—C
bonds
(d) one sigma C—C and two π C—C
bonds
43 Reduction of 2-butyne with sodium
in liquid ammonia gives
predominantly [CBSE AIPMT 1993]
(a) cis-2-butene
(c) no reaction
Ans. (b)
(b) trans-2-butene
(d) n-butane
In presence of Na and liquidNH3 liq ⋅NH3.
Trans addition of hydrogen on alkene
occurs.
Na, liq ⋅NH
3
CH3 — C ≡≡ C — CH3 →
Birch reduction
but -2-yne
(2-butyne)
H3C
H
H
C == C
CH3
Trans -but-2-ene
182
NEET Chapterwise Topicwise Chemistry
Reagent
44 R—CH2 —CCl 2 —R →
R —C ≡≡ C —R. The reagent is
[CBSE AIPMT 1993]
(a) Na
(b) HCl in H2O
(c) KOH in C2H5OH (d) Zn in alcohol
Ans. (c)
Dehydrohalogenation reaction,
Cl

C 2 H 5 OH + KOH
RCH2 — C — R →
–KCl, –H 2 O

Cl
KOH + C H OH
2 5
RCH== C — R →
–KCl, –H2 O

Cl
R — C ≡≡ C — R
In presence of ethanolic KOH, substrate
gives elimination reaction.
45 Select the true statement about
benzene amongst the following.
[CBSE AIPMT 1992]
(a) Because of unsaturation benzene
easily undergoes addition
(b) There are two types of C—C bonds
in benzene molecule
(c) There is cyclic delocalisation of
pi-bonds in benzene
(d) Monosubstitution of benzene gives
three isomeric products
Ans. (c)
According to the orbital concept, each
carbon atom in benzene is
sp2 -hybridised and one p-orbital of each
carbon remains unhybridised. The
π-electron charge in benzene is not
confined to space between two carbon
atoms as in ethylene, but is spread over
a greater area. This is known as the
delocalisation of the electron charge.
46 Which one of the following has the
shortest carbon-carbon bond
length?
[CBSE AIPMT 1992]
(a) Benzene
(c) Ethyne
Ans. (c)
(b) Ethene
(d) Ethane
Ethyne (acetylene) have shortest C—C
bond length because it have C ≡≡ C i.e.
triple bond. The bond length follows the
following order
C ≡≡ C < C == C < C  C
47 The shortest C—C bond distance is
found in
[CBSE AIPMT 1991]
(a) diamond
benzene
(b) ethane (c)
(d) acetylene
Ans. (d)
Acetylene have shortest C—C bond
length because it have C ≡≡ C triple bond.
The bond length follows the following
order
C — C > C ≡≡ C > C ≡≡ C
48 Which is the most suitable reagent
among the following to distinguish
compound (III) from rest of the
compounds?
[CBSE AIPMT 1989]
I. CH3 —C ≡≡ C —CH3
II. CH3 —CH2 —CH2 —CH3
III. CH3 — CH2 — C ≡≡ CH
IV. CH3 — CH== CH2
(a) Br2 / CCl4
(b) Br2 / CH3COOH
(c) Alk. KMnO 4
(d) Ammoniacal AgNO 3
Ans. (d)
It may be noted that only the terminal
alkynes react with ammoniacal silver
nitrate. Therefore, this reaction can be
used to distinguish between 1-alkynes
and others such as alkane, alkenes and
non-terminal alkynes.
CH3 — CH2 — C ≡≡ CH + 2[Ag(NH3) 2 ]NO3
→ CH3CH2 C ≡≡ CAg ↓
White ppt.
towards the carbon atom and help in the
release of H+ ions.
For a example ethyne react with sodium
metal and release theH+ ions.
CH≡≡ CH + Na → HC ≡≡ C−Na+
TOPIC 4
Aromatic
Hydrocarbons
50 In the following reaction,
Red hot iron tube
H3C C ≡≡ CH →
A,
873 K
the number of sigma (σ) bonds
present in the product A is
[NEET (Odisha) 2019]
(a) 21
(c) 24
Ans. (a)
(b) 9
(d) 18
Key Idea Single bond = 1σ, Double bond
= 1σ, 1 π; Triple bond = 1σ and 2π.
Alkynes form aromatic compounds when
their vapours are passed over red hot
copper or iron tubes.
Red hot iron tube
3CH3—C≡≡CH
873 K
CH3
49 Acetylenic hydrogens are acidic
because
[CBSE AIPMT 1989]
(a) sigma electron density of C—H
bond in acetylene is nearer to
carbon which has 50% s-character
(b) acetylene has only one hydrogen
on each carbon
(c) acetylene contains least number
of hydrogens among the possible
hydrocarbons having two carbons
(d) acetylene belong to the class of
alkynes with molecular formula
C nH2 n −2
Ans. (a)
Acidic character of alkynes can be
explained on the basis of
sp-hybridisation state of the carbon
atom in alkynes. We know that an
electron in s orbital is more tightly held
than in a p orbital because s-electrons
are more closer to the nucleus. In
sp-hybridisation, s-character is more
(50%) as compared to sp2 -(33%) or
sp3-(25%) hybrid orbitals. Due to very
large s-character, the electrons in
sp-hybrid orbitals are held tightly by the
nucleus and are quite electronegative.
Consequently, the electron pair of
F — C ≡≡ bond gets displaced more
H3C
CH3
Mesitylene
(A)
The number of sigma (σ) bonds present
in the product (A) is 21.
H
σ|
σ
H—
σ C—H
σ
H
H
σ
H
σ|
C
σ|
H
σ
σ
σ
σ
π
π
σ
σ
H
π
σ
H
σ
σ
σ
σ
H
|σ
C
σ|
H
σ
H
51 In the given reaction
HF
0°C
+
P
the product P is
[NEET 2016, Phase I]
F
(a)
(b)
F
183
Hydrocarbons
Ans. (c)
53 Given,
(c)
(d)
H3 C
Key Idea It is an example of Friedel-Craft
reaction. First π-electrons of
cyclohexene attack at H + ion of HF and
form carbocation. This carbocation
further reacts with benzene and forms
addition product.
H
CH3
(II)
H2C
CH2
CH2
(III)
The enthalpy of hydrogenation of
these compounds will be in the
order as
[CBSE AIPMT 2015]
r
Electrophile
Thus, the correct option is (c).
52 In the reaction with HCl, an alkene
reacts in accordance with the
Markownikoff’s rule, to give a
product 1-chloro-1methylcyclohexane. The possible
alkane is
[CBSE AIPMT 2015]
CH2
(a) I > II > III
(c) II > III > I
Ans. (b)
(b) III > II > I
(d) II > I > III
The enthalpy of hydrogenation of given
compounds is inversely proportional to
stability of alkene.
H2C
CH2
H3C
CH2
Less stable
CH2
2
–
+
CH2
H
CH3
II
Ans. (c)
For structure A,
CH3
H3C
+
CH3
H3C
– CH3
+
H2C
Cl
CH3
Cl–
CH3
CH3
For structure B,
The rearrangement of carbocation occur
because 3°-carbocation is more stable
than 2°-carbocation.
CH3
3
H3C
(d)
H+
CH2
+
CH2
III
CH3
CH2
–
–
–
(b)
(c) (a) and (b)
+
+
H2C
CH3
(a)
CH2
CH3
(I)
Ans. (c)
HF
CH3
H3C
Stable
H3C
CH3
= Already aromatic
compound, more stable.
CH3
+
H
Rearrangement
CH3
H3C
–
Cl
Cl
54 Which of the following compounds
will not undergoes Friedel-Craft’s
reaction easily?
[NEET 2013]
(a) Cumene
(c) Nitrobenzene
55 Some meta-directing substituents
in aromatic substitution are given.
Which one is most deactivating?
[NEET 2013]
(a) C ≡≡ N
(c) COOH
Ans. (d)
(b) SO 3H
(d) NO2
The deactivating tendency of given
groups follows the order
O

NO2 >  SO3H >  C ≡≡ N >  C OH
Thus, NO2 is the most deactivating
group.
56 The reaction of toluene with Cl 2 in
the presence of FeCl 3 gives ‘X’ and
reaction in presence of light gives
‘Y ’. Thus, ‘X’ and ‘Y’ are
[CBSE AIPMT 2010]
(a) X = benzal chloride,
Y = o-chlorotoluene
(b) X = m-chlorotoluene,
Y = p -chlorotoluene
(c) X = o and p-chlorotoluene,
Y = trichloromethyl benzene
(d) X = benzyl chloride,
Y = m-chlorotoluene
Ans. (c)
Key Idea In the presence of halogen
carrier, electrophilic substitution occurs
while in the presence of sunlight,
substitution, occurs at the side chain.
CH3
CH3
Cl
Cl2
I
Hence, correct order is III > II > I.
2º-carbocation
3º-carbocation
CH3
Nitro group being electron withdrawing,
deactivates the benzene nucleus to such
an extent . that an electrophile cannot
attack on benzene ring easily due to
deactivation of benzene ring. Hence
becomes incapable to give
Friedel-Craft’s reaction.
NOTE Nitrobenzene because of its
unreactivity towards Friedel-Craft’s
reaction is used as a solvent for this
reaction.
(b) Xylene
(d) Toluene
FeCl3
o-chlorotoluene
CH3
+
Cl
p-chlorotoluene
184
NEET Chapterwise Topicwise Chemistry
FeCl 3 + Cl2 → FeCl −4 + Cl +
Electrophile attacking species
(Q—CH3 is an o/p-directing group.)
In presence ofhν, reaction is free radical
substitution reaction.
CH3
CH2Cl
(c) H2C ==CH2 + C 6H6
(d) H3C —CH3 + C 6H6
Ans. (c)
From the reaction of benzene with
ethylene in the presence of anhy. AlCl 3,
ethylbenzene isproduced.
CH2 — CH 3
Cl2
Cl2
hν
hν
+ CH2 ==CH 2
CHCl2
CCl3
Cl2
Anhy. AlCl3
95°C
60 In Friedel-Craft’s synthesis of
toluene, the reactants in addition
to anhydrous AlCl 3 are
[CBSE AIPMT 2000]
hν
Trichloromethyl
benzene
57 Nitrobenzene can be prepared
from benzene by using a mixture of
conc. HNO 3 and conc.H2SO4 . In
the mixture, nitric acid acts as a/an
[CBSE AIPMT 2009]
(a) reducing agent (b) acid
(c) base
(d) catalyst
Ans. (c)
(a) C 6H5Cl + CH4
(c) C 6H6 + CH4
Ans. (d)
(b) C 6H5Cl + CH3Cl
(d) C 6H6 + CH3Cl
+
CH3Cl + AlCl 3 →
C H3
Electrophilie
NO2+
H CH3
+
Slow
+ CH3
Hence, in this reactionHNO3 acts as a
base andH2SO4 as an acid.
58 Benzene reacts with CH3Cl in the
presence of anhy. AlCl 3 to form
[CBSE AIPMT 2009]
(a) toluene
(b) chlorobenzene
(c) benzylchloride (d) xylene
Ans. (a)
CH3
Anhy. AlCl3
Benzene
Toluene
This reaction is known as Friedel-Craft’s
alkylation of benzene.
59 Using anhy. AlCl 3 as catalyst, which
one of the following reactions
produce ethylbenzene (PhEt)?
[CBSE AIPMT 2004]
(a) H3C —CH2OH + C 6H6
(b) CH3 —CH ==CH2 + C 6H6
+
Intermediate
carbocation
+ H2O
Nitronium
ion
Attacking species
(electrophile)
+ CH3Cl
+ AlCl –4
Step II
Acid
H2NO+3 →
(a) Greater stability [CBSE AIPMT 1998]
(b) Delocalisation of π-electrons
(c) Electrophilic additions
(d) Resonance
Ans. (c)
Arenes gives an electrophilic
substitution reactions. They do not give
electrophilic addition reaction in normal
state due to resonance stabilisation.
63 Which is the correct symbol
relating the hetero Kekule
structure of benzene?
[CBSE AIPMT 1993]
(a)
Friedel-Craft’s alkylation of benzene
(Ar—H)
Mechanism of this reaction is
represented as follows :
Step I
Conc. H2SO4 and conc. HNO3 react in the
following manner
HNO3 +H2SO4 → H2NO+3 + HSO–4
Base
62 Which one of these, is not
compatible with arenes?
º
(b) → (c) ≡≡
(d) ←→
Ans. (d)
Kekule in 1865 proposed a ring
structure for benzene in which the
position of the three double bonds are
not fixed. He suggested that the
double bond keeps on changing their
position and this is called resonance.
According to Kekule, benzene is a
resonance hybrid of two structures (a)
and (b) and the hybrid structure may
be represented as (c). The (a) and (b)
are called resonating structures and
are represented by putting ( ←→)
double headed arrow between them.
Step III
CH3
H CH3
+
–
+ AlCl4
(a)
Fast
+ AlCl3 + HCl
Thus, C6H6 and CH3Cl are required in
addition to AlCl 3.
61 In Friedel-Craft’s alkylation, besides
AlCl 3 the other reactants are
[CBSE AIPMT 1999]
(a) C 6H6 + NH2
(c) C 6H6 + CH3Cl
Ans. (c)
(b) C 6H6 + CH4
(d) C 6H6 + CH3COCl
Friedel-Craft’s alkylation When
benzene reacts with alkyl halide in
presence of anhy. AlCl3 , toluene is
obtained. It is called Friedel-Craft’s
alkylation reaction. Friedel-Craft’s
reaction is type of an electrophilic
substitution reaction.
Anhy. AlCl 3
C6H6 + CH3Cl →
C6 H5 CH3 + HCl
Toluene
(b)
(c)
64 Benzene reacts with n-propyl
chloride in the presence of
anhydrous AlCl 3 to give
[CBSE AIPMT 1993]
(a) 3-propyl-1-chlorobenzene
(b) n-propyl benzene
(c) no reaction
(d) iso-propyl benzene
Ans. (d)
Electrophilic substitution reaction of
benzene ring yields iso-propyl benzene.
+ Cl—CH2—CH2—CH3 Anhy. AlCl3
n-propyl chloride
Benzene
H3C—CH—CH3
+ HCl
Iso-propyl benzene
(Cumene)
23
Organic Compounds
Containing Halogens
Strong base
TOPIC 1
Haloalkane
–HBr
CH3CH2—CH2—CH==CH2
Pent-1-ene
(Hofmann product)
[Minor]
+ CH3CH2—CH==CH—CH3
+?
01 CH3CH2COO − Na + NaOH


→
Pent-2-ene (cis+trans)
(Zaitsev product)
[Major]
Heat
CH3CH3 + Na 2CO 3
Consider the above reaction and
identify the missing
reagent/chemical.
[NEET 2021]
(a) B2H6
(c) CaO
Ans. (c)
(b) Red phosphorus
(d) DIBAL-H
In this reaction, removal of carbon
dioxide takes place. So, this is a
decarboxylation reaction. A
decarboxylation reaction takes place
with soda lime (NaOH + CaO).
So, missing reagent is CaO.
02 Elimination reaction of
2-bromo-pentane to form
pent-2-ene is
1. β-elimination reaction.
2. Follows Zaitsev rule.
3. Dehydrohalogenation reaction.
4. Dehydration reaction.
In the above reaction,β-elimination
takes placevia E2 mechanism
(anti-elimination).
According to Zaitsev rule, β-carbon
carrying lesser number of H-atoms get
involved in the elimination to give a more
substituted alkene (pent-2-ene) as the
major product.
Here, loss of HBr, i.e.
dehydrobromination
(dehydrohalogenation) takes place.
So, 1, 2 and 3 are correct combination.
03 For the following reactions,
(b) (2), (3), (4)
(d) (1), (2), (3)
Br
β
β′
CH3CH2CHCHCH2
α
H
Zaitsev or
Saytzeff elimination
centre for β-H
(two β-H atoms)
H
Hoffmann elimination
centre for β′-H
(three β′-H atoms)
(i) CH3CH2 CH2Br + KOH→
CH3CH == CH2 + KBr + H2O
Elimination reaction
(ii)
H3C
CH3
+ KOH
Br
Substitution reaction
H3C
CH3
+ KBr
OH
Br
(iii)
+ Br2
Addition reaction
Br
[NEET 2016, Phase I]
(i) CH3CH2CH2Br + KOH→
CH3CH == CH2 + KBr + H2O
(ii) H3C CH3
+ KOH
Br
H3C CH3
+ KBr
Br
(iii)
04 In a SN2 substitution reaction of the
DMF
type R —Br + Cl– →
R —Cl + Br − ,
Which one of the following has the
highest relative rate?
[CBSE AIPMT 2008]
OH
[NEET (Sep.) 2020]
(a) (1), (3), (4)
(c) (1), (2), (4)
Ans. (d)
(c) (i) is substitution, (ii) and (iii) are
addition reactions
(d) (i) and (ii) are elimination reactions
and (iii) is addition reaction
Ans. (a)
+ Br2
Br
Which of the following statements
is correct?
(a) (i) is elimination reaction, (ii) is
substitution and (iii) is addition
reaction
(b) (i) is elimination, (ii) and (iii) are
substitution reactions
(a) CH3 — CH2 — CH2Br
(b) CH3 — CH — CH2Br

CH3
CH3

(c) CH3 —C — CH2Br

CH3
(d) CH3CH2Br
186
NEET Chapterwise Topicwise Chemistry
Ans. (d)
05 Which of the following undergoes
nucleophilic substitution exclusively
bySN 1 mechanism?
[CBSE AIPMT 2005]
(a) Benzyl chloride
(b) Ethyl chloride
(c) Chlorobenzene
(d) Isopropyl chloride
Ans. (a)
+
C6H5  CH2 + Nu
Fast
→
C6H5 CH2Nu
06 Chloropicrin is obtained by the
reaction of
[CBSE AIPMT 2004]
(a) steam on carbon tetrachloride
(b) nitric acid on chlorobenzene
(c) chlorine on picric acid
(d) nitric acid on chloroform
Ans. (d)
Chloroform on reaction with nitric acid
gives chloropicrin (nitro chloroform). Its
reaction is shown below
CHCl 3 + HNO3 → C(NO2 )Cl 3 + H2O
Chloroform
Nitro chloroform
(chloropicrin)
07 Which of the following is
responsible for depletion of the
ozone layer in the upper strata of
the atmosphere?
(a) Polyhalogens
(c) Fullerenes
Ans. (d)
+
C6H5 CH2 > CH3  CH  CH3
2° carbocation
+
> CH3  CH2
1° carbocation
and in step (ii) nucleophile is attracted
towards the carbonium ion in form of fast
step to give final product.
Hence, in benzyl chloride, ethyl chloride
and isopropyl chloride order ofS N 1
reaction is benzyl chloride > isopropyl
chloride > ethyl chloride.
In chlorobenzene, mechanism of S N 1
reaction differ to aliphatic alkyl halide.
The aryl halides are much less reactive
as compared to alkyl halides, towards
nucleophilic reagents in either S N 1 or
S N 2 reaction.
The carbon-halogen bond in the aryl
halide is quite strong and only forcing
conditions can break up this bond.
Hence,
Step (i)
[C6H5 CH2 Cl]
(First order kinetics)
(b) Ferrocenes
(d) Freons
Cl• + O3 → ClO• +O2
ClO + O3 →
•
NaNH 2 ( Sodamide)
CH3 — CH2 — CHCl2 →
∆
NaNH
2
CH3 — CH== CHCl →
∆
CH3 — C ≡≡ CH
Final product
Ni/H
2
NaCN
09 CH3CH2Cl →
X →
Y
Acetic anhydride
→
Z
In above reaction sequence, Z is
[CBSE AIPMT 2002]
(a) CH3CH2CH2NHCOCH3
(b) CH3CH2CH2NH2
(c) CH3CH2CH2CONHCH3
(d) CH3CH2CH2CONHCOCH3
Ans. (a)
CH3CH2 Cl NaCN

→ CH3 — CH2 — CN
X
Ni/H2
Acetic
→ CH3 — CH2 CH2NH2 →
Y
anhydride
CH3 — CH2 — CH2 — NHCOCH3
Z
Freons or chlorofluoro carbons are
responsible for depletion of the ozone
layer in the upper strata of the
atmosphere. They are used as
propellants, aerosol spray caps,
refrigerants, fire fighting reagents, etc.
They are stable and chemically inert
compounds. They absorb UV-radiation
and break down liberating free atomic
chlorine which causes decomposition of
ozone through free radical reaction. This
results in the depletion of the ozone layer.
They form free radical of chlorine in
presence of UV-radiation. Such free
radical decomposesO3 as follows :
Cl
•
+ 2O2
Chlorine free
radical
08 When CH3CH2CHCl 2 is treated with
NaNH2 , the product formed is
[CBSE AIPMT 2002]
(a) CH3 — CH == CH2
(b) CH3 — C ≡≡ CH
NH2
(c) CH3CH2CH
NH2
+
Slow C H CH + Cl –
C6H5 CH2 — Cl →
2
6 5
Rate ∝
–
[CBSE AIPMT 2004]
AliphaticS N 1 reaction is carried out in
two steps. In form of slow step
Step (i) carbonium (carbocation) ion is
formed and its formation is based upon
the stability
Stability order of carbocation benzylic
carbocation (resonating stable)
+
Ans. (b)
Step (ii)
Aprotic solvents like DMF increases the
reactivity of nucleophile and favoursS N 2
reaction.
The relative reactivity of alkyl halides
towards S N 2 reactions is as follows
CH3 X >Primary > Secondary >
Tertiary
However, if the primary alkyl halide or
the nucleophile/base is sterically
hindered the nucleophile will have
difficulty to getting the back side of the
α-carbon as a result of this, the
elimination product will be predominant.
Here, CH3CH2Br is the least hindered,
hence it has the highest relative rate
towards S N 2 reaction.
Cl
(d) CH3CH2CH
NH2
10 Reactivity order of halides for
dehydrohalogenation is
[CBSE AIPMT 2002]
(a) R—F > R—Cl > R—Br > R— I
(b) R—I > R—Br > R—Cl > R—F
(c) R—I> R—Cl> R—Br > R—F
(d) R—F> R—I> R—Br> R—Cl
Ans. (b)
F, Cl, Br and I are the elements of VII A
group. In a group atomic radii increases
from top to bottom and the bond
dissociation energy decreases as
R—F > R— Cl > R—Br > R—I
So, during dehydrohalogenation R—I
bond breaks more easily than R—I bond.
Hence, order of reactivity will be
R—I > R—Br > R—Cl > R—F
11 An organic compound A(C 4 H9 Cl) on
reaction with Na/diethyl ether gives a
hydrocarbon which on
monochlorination gives only one
chloro derivative, then A is
[CBSE AIPMT 2001]
(a) t-butyl chloride
(b) s-butyl chloride
(c) iso-butyl chloride
(d) n-butyl chloride
187
Organic Compounds Containing Halogens
Ans. (a)
Alkyl halides reacts with Na in presence
of dry ether to form alkanes. This
reaction is known as Wurtz reaction.
R  X + 2Na + X  R
Dry ether
→ R  R + 2NaX
In the given question t-butyl chloride
C4H9 Cl is A. It reacts with Na metal in dry
ether to form a hydrocarbon that on
chlorination gives only one monochloro
derivative.
CH3
CH3
|
|
CH3 — C — Cl + 2Na + Cl — C — CH3
|
|
CH3
CH3
Dry ether
→
–2NaCl
CH3 CH3


—CH3
CH3  C —— C —


CH3 CH3
2, 2, 3, 3—tetramethnyl-butane
12 2-bromopentane is heated with
potassium ethoxide in ethanol. The
major product obtained is
[CBSE AIPMT 1998]
(a) 2-ethoxypentane
(b) pentene-1
(c) trans-pentene-2
(d) cis-pentene-2
Ans. (c)
–
C 2 H5 O K +
CH3 — CH— CH2 — CH2 — CH3 →
|
Br
14 When chlorine is passed through
propene at 400°C, which of the
following is formed?
(a) PVC
[CBSE AIPMT 1993]
(b) Allyl chloride
(c) Nickel chloride
(d) 1,2-dichloro ethane
Ans. (b)
When chlorine gas is reacted with
propene at high temperature (400°C),
then substitution occurs in place of
addition reaction. Hence, allyl chloride is
formed
CH3 — CH == CH2 + Cl2 → CH2 —
(Allylic substitution)

Cl
H
C == C
H
CH2 — CH3
Trans-pentene-2
13 The alkyl halide is converted into an
alcohol by
[CBSE AIPMT 1997]
(a) addition
(b) substitution
(c) dehydrohalogenation
(d) elimination
Ans. (b)
RCl + NaOH(aq) —→ ROH + NaCl
It is an example of nucleophilic
substitution reaction.
Cl
MgCl
+ Mg
Dry ether
Chlorobenzene
CH == CH2 + HCl
15 Industrial preparation of
chloroform employs acetone and
Phenyl magnesium
chloride
C2H5OH
+ C2H5OMgCl
Alcohols contains
active hydrogen
Benzene
[CBSE AIPMT 1993]
(a) phosgene
(b) calcium hypochlorite
(c) chlorine gas
(d) sodium chloride
Ans. (b)
18 HBr reacts fastest with
[CBSE AIPMT 1992]
The industrial preparation of chloroform
involves the following steps :
(i) CaOCl2 + H2O → Ca(OH) 2 + Cl2
(ii) CH3COCH3 + 3Cl2 → CCl 3COCH3
+3HCl
O H CCl 3 — COCH3
(iii) Ca
+
→
O H CCl 3 — COCH3
CH3COO
2CHCl 3 +
+
CH3
(a) phenol
(b) benzene
(c) ethyl benzene
(d) phenyl ether
Ans. (b)
400 ° C
2-bromopentane
–
(–Br )
CH3 — CH— CH2 — CH2 — CH3 
→
+
–H
17 Chlorobenzene reacts with Mg in
dry ether to give a compound (A)
which further reacts with ethanol
to yield
[CBSE AIPMT 1993]
Ca
CH3COO
16 When hydrochloric acid gas is
treated with propene in presence
of benzoyl peroxide, it gives
(a) 2-methyl propan-1-ol
(b) 2-methyl propan-2-ol
(c) propan-2-ol
(d) propan-1-ol
Ans. (b)
2-methylpropan-2-ol gives 3°
carbocation, so it reacts with HBr at
faster speed.
CH3

CH3  C  CH3 + H+ →

OH
2-methylpropan-2-ol
CH3
CH3


CH3  C  CH3 → CH3  C  CH3
–H2 O
⊕

3 ° carbocation
OH2
+
CH3
[CBSE AIPMT 1993]
Br − CH  
→
C  CH3
3
(a) 2-chloropropane
(b) alkyl chloride
(c) no reaction
(d) n-propyl chloride
Ans. (a)

Br
Kharasch effect or peroxide effect is
only observed in case of addition of HBr
to unsymmetrical alkenes, so the
addition of HCl with propene takes place
as usual by Markownikoff’s rule
Benzoyl
peroxide
CH3 CH == CH2 + HCl →
Cl

CH3  CH  CH3
2-chloropropane
19 In compound ‘X’ all the bond angles
are exactly 109°28′, ‘X’ is
[CBSE AIPMT 1991]
(a) chloromethane
(b) carbon tetrachloride
(c) iodoform
(d) chloroform
Ans. (b)
Carbon tetrachloride (CCl 4 ) have sp3
hybridisation and symmetrical
188
NEET Chapterwise Topicwise Chemistry
structure, so it have all the bond angle
of 109° 28′.
Cl
109°28'
C
Cl
Cl
sp3 hybridised
Cl
Tetrahedral
structure
20 The Cl—C—Cl angle in 1,1,2,2-tetrachloroethene and
tetrachloromethane will be about
(d)
CH3
CBrCH2CH3
CH3
CH3
CH3
3-methylbutene
CH3
CHCH2CH2Br +
1-bromo-3-methyl
butane (Major)
CH3
Cl
Br
2-bromo, 3-methyl
butane (Minor)
[NEET (Oct.) 2020]
[CBSE AIPMT 1988]
(a) CH2 == CH CH2Cl
(b) (CH3) 3CCl
CH2CH2Cl
(c)
CH3
CH3
CH3
The product ‘C’ is
(C6H5CO)2O2
CHCH2CH2Br
CHCH2CH2OCOC6H5
Given,
Br2 /Fe
Zn/HCl
The reaction in the above road map can
be explained by the following steps.
Step I Toluene (A) undergoes side chain
halogenation with excess of chlorine to
give benzotrichloride (A).
CCl3
Ans. (c)
An SN1 reaction proceeds through
formation of a stable carbocation as
an intermediate. Here,
(a) CH2
?
CH
CH2Cl
–Cl–
r
Step II In compound (A), the substituent
CCl 3 is an electron withdrawing group, so
the electrophile will attack at m-position.
Thus, benzotrichloride reacts with
bromine in presence of Fe- catalyst to
give m-bromobenzotrichloride (B).
CCl3
CH2
CH
− Cl−
CH2CH2Cl
CCl3
Br2/Fe
⊕
(b) (CH3) 3 CCl → CH3  C  CH3

CH3
tert-butyl carbocation (stable)
(9 hyper conjugation)
(c)
(A)
(Benzotrichloride)
C7H8
(Toluene)
(d)
(A)
Br
(B)
m-bromobenzotrichloride
Step III m-bromobenzotrichloride
undergoes reduction with Zn in presence
of HCl to give m-bromotoluene (C).
–Cl–
CCl3
–I
r
CH2CH2
1°-carbocation (less stable due
to –I effect of the phenyl group)
CH3
Reduction
Zn/HCl
CHCHCH3
Br
[NEET 2018]
(a) 3-bromo-2, 4, 6-trichlorotoluene
(b) o-bromotoluene
(c) m-bromotoluene
(d) p-bromotoluene
Ans. (c)
Allyl-1°-carbocation
(stable)
22 The major product of the following
chemical reaction is
[NEET 2021]
CH3
Zn / HCl
D
CH2
CH2+HBr
Br2 / Fe
CH2Cl
TOPIC 2
Haloarene
(b)
3Cl2 / ∆
C 7 H8 → A → B → C
3Cl2
+HCl
CH3
24 The compound C 7 H8 undergoes
the following reactions :
CH3
(phosgene)
CH3
So, option (c) will not undergoS N 1
reaction withOH− .
(c)
Chloroform is slowly oxidised into a
poisonous compound called phosgene in
the presence of air or light. This
compound is also called carbonyl
chloride (COCl2 )
1
Air
COCl2
CHCl 3 + O2 →
light
2
Carbonyl chloride
(a)
Benzyl-1°-carbocation
(stable)
3Cl2 / ∆
(a) phosphonyl chloride
(b) thionyl chloride
(c) carbon dioxide and phosphine
(d) carbonyl chloride
Ans. (d)
CHCH
CH2
C7H8 → A → B → C
23 Which of the following will not
undergo SN 1 reaction with OH ?
21 Phosgene is a common name for
CH3
CHCHCH3
CH3
C == C
CH3
(C6H5CO)2O2
CHCH CH2+HBr
CH3
1,1,2,2-tetrachloro ethene
–Cl–
r
Addition of HBr to an alkene in presence
of a peroxide (benzoyl peroxide
[(C6H5 CO)2 O2 ] gives an antiMarkownikoff’s product.
Anti-Markownikoff’s rule states that
hydrogen is added to a more substituted
carbon atom of an unsymmetrical
alkene.
(a) 120° and 109 °28′
(b) 90° and 109.5°
(c) 109.5° and 90°
(d) 109.5° and 120°
Ans. (a)
Cl
Cl
Carbon sp2 hybridised
CH2Cl
Ans. (a)
[CBSE AIPMT 1988]
Cl
(d)
(B)
Br
Br
(C)
m-bromotoluene
189
Organic Compounds Containing Halogens
25 Which of the following can be used
as the halide component for
Friedel-Crafts reaction?
[NEET 2016, Phase II]
(a) Chlorobenzene
(b) Bromobenzene
(c) Chloroethene
(d) Isopropyl chloride
Ans. (d)
2°
Key Idea In chlorobenzene,
bromobenzene and chloroethene, lone
pair of halogen is delocalised with
π-bonds so it attains double bond
character. Thus, these are not suitable as
a halide component for Friedel-Crafts
reaction.
+ CH3
CH
Cl
Anhd. AlCl3
CH3
CH3
C
H
CH3
Other halides, i.e. chloro and
bromobenzene along with chloroethene
have carbon halogen bond as
C
X
26 In which of the following
compounds, the C—Cl bond
ionisation shall give most stable
carbonium ion? [CBSE AIPMT 2015]
(a)
H3C
CH
Cl
(b)
H3C
H
(c)
H3C
Cl
C
H3C
CH
CH3
H
CH
carbocation containing
6 α-hydrogen showing six
hyperconjugative structure along
with two +I group.
(c) H
CH
Cl
O2NH2C
The stability of carbocation follow the
order 3° > 2° > 1° > methyl. More the
number of alkyl group attached with the
carbon atom carrying the positive
charge greater would be the tendency to
stabilise positive charge via inductive
effect and hence more stable is that
carbocation.
CH3
CH3
+

(a) CH3  C  Cl → CH3  C
— Cl

 3°
CH3
CH3
This carbocation is more stable due
to nine α-hydrogen and (nine
C6H5  CH  CH2  CH3

Br
Addition product
Electrophilic addition reaction takes
place via more stable carbocation.
28 What products are formed when
the following compound is treated
with Br 2 in the presence of FeBr 3 ?
[CBSE AIPMT 2014]
+
–
–Cl
CH2
CH3
Benzyl carbocation
It has slightly lesser stability as
compared to 3°-alkyl carbocation due to
presence of three electron donating
alkyl group in3°-alkyl carbocation.
Although the stabilities of 3° and benzyl
carbonium ion are almost same and
cannot be compared in solution but
whenever a comparison is made
between Resonance (the cause of
stability in benzyl carbonium ion) and No
bond resonance (the cause of stability in
3° carbonium ion) then the former is
always preferred hence here in this
question benzyl carbonium ion is more
stable than 3° carbonium ion.
(d)
H
O2NH2C
C
H
Cl
CH3
CH3
CH3
Br
and
(a)
CH3
CH3
Br
CH3
CH3
Br
Br
and
(b)
CH3
CH3
–
–Cl
CH3
CH3
Br
–
O
+
N
C
H2
O
+
and
(c)
CH2
CH3
CH3
Br
1° carbocation less stable than all present here.
CH3
and
CH3
[CBSE AIPMT 2015]
(a) C6H5 C HCH2 CH3
CH3
Br
Br

Br
Ans. (c)
CH3 is a o/p-directing group, thus
electrophilic substitution reaction of
toluene
(b) C 6H5CH2 C HCH3

Br
CH3
CH3
(c) C 6H5CH2CH2CH2Br
CH
CH3
(d)
with HBr produces
Ans. (c)
tert-butyl
carbocation
Cl
+
Fast
C6H5  CH  CH2  CH3 + Br − →
27 The reaction of C 6 H5 CH==CHCH3
Cl
(d)
hyperconjugative structures) three +I
groups.
H
CH3
CH3

+
(b)
C  Cl →
CH
2°
− Cl −
CH3
CH3
Br
CHCH3
Br2/FeBr3
Electrophilic
CH3 substitution
(d)
CH3
CH3
Br
Ans. (a)
+
Slow
C6H5 CH == CHCH3 + H+ →
+
C6H5  CH  CH2  CH3
Stable carbocation
CH3
Br
190
NEET Chapterwise Topicwise Chemistry
CH3
CH3
Br2/FeBr3
H
Br
Cl3C
Electrophilic
CH3 substitution
H
—Cl
H
—Cl
MgBr
O+
C
Trichloroacetaldehyde
CH3
Not possible
due to steric
hinderance
Chlorobenzene
Conc. H2SO4
29 Trichloroacetaldehyde, CCl 3CHO
reacts with chlorobenzene in the
presence of sulphuric acid and
produces
[CBSE AIPMT 2009]
H
Cl3C
C
–H2O
C
Dichlorodiphenyl
trichloroethane (DDT)
CH2Cl
MgBr
30
Cl
P
+
(ii) H3O
C
Cl
In the above reaction product ‘P ’
is
[CBSE AIPMT 2002]
(c) Cl
C
Cl
(a)
CH
CCl3
Ans. (d)
(b)
OH
Cl
(d) Cl
COOH
CHO
H
OH
Cl
(c)
Ans. (b)
O
||
C—O—H
+ Mg(OH)Br
31 Replacement Cl of chlorobenzene
to give phenol requires drastic
conditions but chlorine of 2,
4-dinitro chlorobenzene is readily
replaced. This is because
[CBSE AIPMT 1997]
(i) CO2
(b) Cl
H3O
P
—Cl
Cl
+
CO2
—Cl
Cl
(a) Cl
O
|
C—O—MgBr
O
||
(d) C6H5—C—C6H5
(a) NO2 makes the ring electron rich at
ortho and para-positions
(b) NO2 withdraw electrons from
meta-position
(c) NO2 donates electrons at
meta-position
(d) NO2 withdraw electrons from ortho/
para-positions
Ans. (d)
—NO2 group is electron withdrawing
group, so it deactivates the benzene ring.
Due to electron withdrawing nature of
—NO2 group, it develops positive charge
at o/pposition. This cause easier for the
removal of —Cl-atom.
24
Organic Compounds
Containing Oxygen
TOPIC 1
Alcohols and Phenols
01 The major product formed in
dehydrohalogenation reaction of
2-bromo pentane is pent-2-ene.
This product formation is based on
Ans. (a)
Ans. (a)
EDG (+ R, + I) decreases acidity and EWG
(− R, − I) increases acidity of phenol.
(a)
OH
(b)
02 Which of the following substituted
phenols is the strongest acid?
[NEET (Oct.) 2020]
OH
[NEET (National) 2019]
CH
OH
(c, d)
(b)
CH3
(R = CH3, C2H5)
OH
OH
(d)
C2H5
A
H+
H2O
OH
R
O
+ H3C
+I from para position
decreases its acidity
CH3
(a) H3CCOOH
03 CH3CH2CH== CH2  −→ Z
OCH3
(c)
CH3
B2H6
NO2
CH3
H2 O,H2 O2 ,OH
what is Z?
OH
04 The structure of intermediate A in
the following reaction, is
OCH3
Thus, option (a) is correct.
(a)
–
It is hydroboration-oxidation (HBO)
reaction of an alkene which undergoes
hydration to give an alcohol.
Here, anti-Markownikoff’s addition of H2O
takes place.
OH
O2
Minor (pent-1-ene)
(i) B2H6
(ii) H2O, H2O2, OH
(Z )
Butan-1-ol
+
OH
CH2
CH3CH2—CH—CH2
–R from para position
makes its a stronger acid
Major (pent-2-ene)
2-bromopentane
CH
Least substituted
sp2 carbon
H
+R from para-position
deceases its acid remarkably
–HBr
OH
NO2
Saytzeff’s rule states that more
substituted alkene is formed in a
dehydrohalogenation reaction.
In dehydrohalogenation of
2-bromopentane, pent-2-ene is formed
as major product which is a more
substituted alkene.
Br
δ–
H
Anti-Markownikoff's
addition of H2O (overall)
[NEET 2021]
(a) Saytzeff’s rule (b) Hund’s rule
(c) Hoffmann rule (d) Huckel’s rule
Ans. (a)
CH3CH2
δ+
[NEET (Oct.) 2020]
(a) CH3CH2CH2CH2OH
(b) CH3 CH2 CHCH3

OH
(c) CH3CH2CH2CHO (d) CH3CH2CH2CH3
CH3
(b)
OOCH
CH3
CH3
192
NEET Chapterwise Topicwise Chemistry
CH2OOH
which is a mild oxidising agent forms
aromatic aldehydes.
HC
CH3
CH2OH
(c)
CH3
Benzaldehyde
In the remaining options benzoic acid is
formed as follows:
CH
O
CHO
PCC
(Pyridinium
chlorochromate)
CH3
CH2OH
COOH
K2Cr2O7
(d)
Benzoic acid
Ans. (a)
COCH3
The given reaction is of cumene process
for phenol production and intermediate (
A) is cumene hydroperoxide. In the
process, cumene (isopropylbenzene) is
oxidised in the presence of air to
cumene hydroperoxide. Which is then
converted to phenol and acetone by
treating with dilute acid. Acetone, a
by-product of this reaction is also
obtained in large quantities by this
method. The reaction takes place as
follows :
(i) NaOCl
(ii) H3O+
COOH
+ CHCl3
KMnO4/H+
CH3
Benzoic acid
O2
06 The hydrolysis reaction that takes
place at the slowest rate, among
the following is
Cumene
hydroperoxide
(A)
[NEET (Odisha) 2019]
OH
(a)
H+
Cl
+ CH3COCH3
H2O
Acetone
K2Cr2O7
aq. NaOH
(b) H3C CH2 Cl →
H3C CH2 OH
aq. NaOH
(d)
(i) NaOCl
(b)
CH2Cl
aq.NaOH
(ii) H3O+
CH2OH
CH2OH
PCC
(c)
(Pyridinium
chlorochromate)
CH2OH
+
(d)
ONa
(c) H2 C == CH  CH2 Cl →
COCH3
KMnO4/H
Ans. (c)
Primary aromatic alcohols on reaction
with pyridinium chlorochromate (PCC)
Cu/573 K
OH
H3C
C=
=O+H2
H3C
Propanone
(Ketone)
08 The compound A on treatment with
Na gives B, and with PCl 5 gives C. B
and C react together to give diethyl
ether. A, B and C are in the order
[NEET 2018]
(a) C2H5Cl, C2H6 , C2H5OH
(b) C2H5OH, C2H5Cl, C2H5ONa
(c) C2H5OH, C2H6 , C2H5Cl
Ans. (d)
CH3
CH2OH
(a)
H
C
(d) C2H5OH, C2H5ONa, C2H5 Cl
aq.NaOH
CH3
Phenol
05 The reaction that does not give
benzoic acid as the major product
is
[NEET (Odisha) 2019]
H3C
Thus, option (c) is correct.
Thus, option (c) is correct.
Cumene
Ans. (c)
When vapours of alcohols are passed
over heated copper at 573 K, primary and
secondary alcohols undergo
dehydrogenation to give aldehydes and
ketones, respectively. While tertiary
alcohols undergo dehydration to give
alkenes.
COOH
H3CCOOH
CH
(c) a ketone
(d) an alkene
Propan-2-ol
(2°)
CH3
CH3
(a) a carboxylic acid
(b) an aldehyde
H3C
Benzoic acid
CH2OH
07 When vapours of a secondary
alcohol is passed over heated
copper at 573 K, the product
formed is
[NEET (Odisha) 2019]
According to given question and options
(A) must be C2H5OH, as it reacts with Na
to give C2H5ONa. The reaction sequence
is as follows.
Na
(i) C2H5OH →
Ethanol
(A )
PCl5
there is a partial double bond character
between sp3-hybridised C atom next to
an aromatic ring and Cl. It is most
difficult to break this bond and hence it
undergoes hydrolysis reaction with
slowest rate.
Sodium ethoxide
(B)

↓
C2H5 Cl
Ethyl chloride
(C )
S N2
−
+
(ii) C2H5 O N a + C2H5 Cl →
C2H5  O  C2H5 + NaCl
Ans. (a)
Cl is a benzylic halide. Thus,
− +
C2H5 ONa
Diethyl ether
The above reaction is known as
Williamson’s ether synthesis. It involves
nucleophilic attack of alkoxide ion on
alkyl halide according toS N 2
mechanism.
C2H5O− + CH3  CH2 Cl →
Slow
Nucleophile
(Alkoxide ion)
Substrate
(Alkyl halide)
193
Organic Compounds Containing Oxygen
Fast
+
C2H5 O… CH2 … Cl − →

CH3
OH
OH
O2N
(c)
NO2
Ans. (d)
OH
CHO
C2H5 O  C2H5 + Cl −
09 Compound A, C 8H10 O, is found to
react with NaOI (produced by
reacting Y with NaOH) and yields a
yellow precipitate with
characteristic smell.
A and Y are respectively.
[NEET 2018]
(a)
CH
CH3 and I2
OH
(b)
CH2
(c) H3C
CH2
OH and I2
CH2
OH and I2
CH3
+ CHCl3
NO2
NO2
Iodoform reaction with sodium
hypoiodite is used for the detection of
CH3CO group. Also compounds
containing CH3CH(OH) group shows
positive iodoform test as it produces
CH3CO group on oxidation.
Since, among the compounds,
CH3CH(OH) group is given only in the
substrate of option (a) hence, it is
correct. The reaction of compound A
with NaOI is given as follows :
2NaOH + I2 → NaOI + NaI + H2O
CH
CH3
NaOI
OH
Thinking process This problem is based
on the acidic character of phenol.
Electron -withdrawing group ato and
p-position w.r.t. —OH group of phenol,
increase the acidic strength.
Picric acid (2, 4, 6-trinitrophenol) is
extremely more acidic than given
compounds because its pKa value is
close to zero also due to the presence of
three strong electron withdrawing group
( NO2 group) at ortho and
para-positions, picric is more acidic
compound.
11 Which of the following reagents
would distinguish cis-cyclopenta-1,
2-diol from the trans-isomer?
(a) Ozone
(b) MnO2
(c) Aluminium isopropoxide
(d) Acetone
Ans. (d)
cis-cyclopenta-1,2-diol when reacts
with acetone, forms cyclic ketal
whereas trans-isomer of cyclopenta-1,
2-diol can not form cyclic ketal.
OH
CH3
+ O == C
cis – cyclopenta
-1, 2-diol
CHO
Minor
This is Reimer-Tiemann reaction. So
finally —CHO group is introduced
13 Which of the following reaction(s)
can be used for the preparation of
alkyl halides?
[CBSE AIPMT 2015]
anh.ZnCl
I. CH3CH2OH+ HCl  2→
II. CH3CH2OH+ HCl →
III. (CH3 ) 3 COH+ HCl →
anh.ZnCl
IV. (CH3 ) 2 CHOH+ HCl  2→
(a) I, III and IV
(c) Only IV
Ans. (a)
14 Which of the following is not the
product of dehydration of
?
OH
CH3
O
C
C
CH3
Cyclic ketal
OH
Acetophenone
(a)
(b)
(c)
(d)
CH3
+ O == C
-
CONa++ CHI3
O
[CBSE AIPMT 2015]
CH3
But,
O
(b) I and II
(d) III and IV
In (I) and (IV) due to the presence of
Lucas reagent (HCl + anh. ZnCl2 )
alcohols give alkyl halides while in (III)
alkyl halide is formed due toS N 1
reaction.
Acetone
O
I2/NaOH
+
CH3
OH
OH
(A)
Major
Phenol Chloroform
[NEET 2016, Phase II]
Ans. (a)
Dil./NaOH
Ans. (d)
OH and I2
(d) CH3
OH
(d)
Iodoform
(yellow ppt.)
CH3
OH
Trans-cyclopenta -1, 2-diol
No reaction
Sodium benzoate
Ans. (b)
10 Which one is the most acidic
compound?
[NEET 2017]
OH
OH
(a)
(b)
CH3
12 Reaction of phenol with chloroform
in the presence of dilute sodium
hydroxide finally introduces, which
one of the following functional
group?
[CBSE AIPMT 2015]
(a)  CH2Cl
(c)  CHCl2
(b) COOH
(d) CHO
Key Concept When
intermediate carbocation is
stable, no rearrangement takes place in
carbocation.
194
NEET Chapterwise Topicwise Chemistry
CH3
∆
–H2O
OH
+
18 Which one is a nucleophilic
substitution reaction among the
following?
[CBSE AIPMT 2011]

(a) H3C—C—CH—CH3
 
OH CH3
CH3
+
(a) RCHO + R′ MgX → R — CH— R′

OH
CH3

(b) CH3 CH2 CH  CH2 Br

15 Which of the following will not be
soluble in sodium hydrogen
carbonate?
[CBSE AIPMT 2014]
(a) 2,4,6-trinitrophenol
(b) Benzoic acid
(c) o-nitrophenol
(d) Benzenesulphonic acid
Ans. (c)
(b) CH2—C—CH2—CH3


OH
CH3
CH3
(c) H3C—C—CH—CH3
 
CH3OH
CH3
(c) CH3CHO + HCN → CH3 —CH(OH)CN

H+
(d) CH3 CH == CH2 + H2O →
(d) H3C—C—CH2—CH2
OH
NO2
NaHCO3
1No reaction
O-nitrophenol is insoluble in sodium
hydrogen carbonate. While
2,4,6-trinitrophenol, benzoic acid and
benzene sulphonic acid are soluble in
NaHCO3.
Infact, Acid + NaHCO3 → Salt + H2 CO3
This reaction is possible in forward
direction if acid is more acidic than
H2 CO3. o-nitrophenol is less acidic than
H2 CO3. Hence, it is not soluble in sodium
hydrogen carbonate.
16 In the following sequence of
reactions,
KCN
H 3O +
LiAlH 4
CH3 Br → A → B → C


CH3
OH
Ans. (a)
CH3
|
H3CCCH== CH2
|
CH3
CH3
H+
−KBr
‘B’
H
A
Ethanoic
acid
B
CH3
|
CH3  C  CH  CH3
+
|
CH3
H3CCCCH3
Ether
H
3° carbocation
(more stable)
Ethanol
C
In the presence ofLiAlH4 carboxylic acid
reduce in alcohols directly.
17 In the following reaction,
CH3

H 2O/H +
H3C C  CH==CH2 →

CH3
A
B
+
Major product
Minor product
[CBSE AIPMT 2012]
H 2O

+
CH3 H
 
–H+
H3CCCCH3
+
H
 
O CH3
H
R  CH  R ′

CH3
OH

(b) CH3  CH2  CH  CH2Br + NH3
CH3CH(OH)CN
(d) CH3  CH == CH2 + H2O
H+
→ CH3  CH  CH3
(Electrophilic

addition)
OH
19 In the following reactions,
 
LiAlH
(Nucleophilic
addition)
(Nucleophilic
addition)
CH3 CH3
4
→
CH3CH2OH
(a) RCHO+ R ′MgX →
(c) CH3CHO + HCN →
2° carbocation
(less stable)
1, 2-Me shift
Ans. (b)
(Nucleophilic
substitution)
H
H O+
CH3 CH  CH3

OH
CH3

→ CH3  CH2  CH  CH2NH2
CH3
|
CH3  C  CH  CH3 →
|
|
CH3 O+
(b) methane
(d) ethyl alcohol
Ethane Complete
nitrile hydrolysis
(ii) –H+
Minor product
[CBSE AIPMT 2012]
KCN
3
CH3Br →
CH3CN →
CH3COOH
(i) H2O
H3C CCHCH3
| |
CH3 OH
the end product C is
(a) acetone
(c) acetaldehyde
Ans. (d)
CH3
| +
H3CCCHCH3
|
CH3
H+
2° carbocation
ether
Methyl
bromide
+ NH3 → CH3 CH2
CH3

 CH  CH2 — NH2

CH3 CH3


H3CCCHCH3

OH
2,3-dimethyl butan-2-ol
(major product)
‘ A’
[CBSE AIPMT 2011]
CH3

H + /Heat
I. CH3—CH—CH—CH3 →

OH
A + B
Major
product
Minor
product
HBr, dark
II. A →
in absence of peroxide
C
 Major 
 product 


+
D
 Major 
 product 


the major products A and C are
respectively
195
Organic Compounds Containing Oxygen
CH3

(a) CH3 C == CH CH3 and
CH3

CH3 —C — CH2 —CH3

Br
CH3

(b) CH 3 C ==CH CH3 and
CH3

CH3— CH— CH —CH3

Br
CH3

(c) CH 2 C ==CH2 CH3 and
CH3

CH3—C— CH2 —CH3

Br
CH3

(d) CH2 = = C— CH2 —CH3 and
CH3

CH2 — CH— CH2 —CH3

Br
Ans. (a)
CH3

(I) CH3  CH  CH  CH3

OH
CH3

+
⊕
H /∆
→ CH3  C H  C  CH3
Protonation
and dehydration
2° carbocation
CH3

1, 2-hydride
→ CH3  C  CH2  CH3
shift
+
More stable 3°carbonium ion
CH3

H
→ CH3  C == CH  CH3
+
Rearrangement
Major A
CH3

+ CH3  CH  CH == CH2
Minor B
A part is major because more
substituted alkenes are more stable.
CH3

(II) CH3  C == CH  CH3
Major A
CH3

→ CH3  C  CH2  CH3
In the absence

of peroxide
Br
HBr (dark)
Major C
CH3

+ CH3  C  CH  CH3
 
H Br
Minor D
20 Given are cyclohexanol (I), acetic
acid (II), 2, 4, 6-trinitrophenol (III)
and phenol (IV). In these, the order
of decreasing acidic character will
be
[CBSE AIPMT 2010]
(a) III > II > IV > I
(c) II > III > IV > I
Ans. (a)
(a) Reimer-Tiemann reaction
(b) Cannizaro reaction
(c) Wurtz reaction
(d) Friedel-Crafts’s acylation
Ans. (b)
(a) Reimer-Tiemann reaction,
(Here, a new C — C bond is formed.)
Riemer-Tiemann reaction is an
electrophilic substitution reaction.
OH
+CHCl3 + NaOH
OH
CHO
(b) II > III > I > IV
(d) III > IV > II > I
Key Idea Higher the tendency to give a
proton, higher is the acidic character and
tendency to lose a proton depends upon
the stability of intermediate, i.e.
carbanion formed.
2, 4, 6-trinitrophenol after the loss of a
proton gives 2,4,6-trinitrophenoxide ion
which is stabilised by resonance,
–I-effect and –M-effect, thus is most
acidic among the given compounds.
Phenol after losing a proton form
phenoxide ion which is also stabilised by
resonance, − M and – I effects but is less
stabilised as compared to
2, 4, 6-trinitrophenoxide ions. Thus, it is
less acidic as compared to 2, 4,
6-trinitrophenol. (CH3COOH) after losing
O 

 ion
a proton gives acetate  CH3C



O− 
+ NaCl + H2O
(b) Cannizaro reaction,
(disproportionation
reaction)
In this reaction, 1-molecule of HCHO
convert in methanol and another
molecule convert in salt.
Conc. NaOH
2 HCHO → CH3OH
+ HCOO−N+ a
(No new C—C bond is formed in this
reaction.)
(c) Wurtz reaction,
R  X + 2Na + R ′ X + dry Na
Ether
→ R  R
Here, R and R ′ must be equal
otherwise mixture of alkanes will
form
(One new C—C bond is formed).
(d) Friedel-Craft’s acylation,
Carboxylate ion
which is stabilised by only resonance.
However, it is more resonance stabilised
as compared to a phenoxide ion, thus
more acidic as compared to phenol.
2, 4, 6-trinitrophenol, however, is more
acidic than acetic acid due to the
presence of three electron withdrawing
—NO2 groups. Cyclohexanol gives an
anion that is least stable among the
given, thus, it is least acidic.
Hence, the correct order of acidic
strength is
2, 4, 6-trinitrophenol > acetic acid >
phenol > cyclohexanol
III > II > IV > I
21 Which of the following reactions
will not result in the formation of
carbon-carbon bonds?
[CBSE AIPMT 2010]
+CH3COCl
Anhy. AlCl3
COCH3
(New C—C bond is formed)
Thus, among the given reactions,
only Cannizaro reaction does not
involve the formation of a new C—C
bond.
22 Consider the following reaction,
PBr
Alc. KOH
3
Ethanol →
X →
Y
(i) H SO , room temperature
2
4
→
Z
(ii) H2 O, heat
The product Z is
[CBSE AIPMT 2009]
196
NEET Chapterwise Topicwise Chemistry
(a) CH2 ==CH2
(c) CH3CH2OSO 3H
Ans. (d)
(b) CH3CH2OCH2CH3
(d) CH3CH2OH
OH
COOH
CH3
O
Alk. KMnO4
C2H5OH → C2H5Br
Ethanol
Alc. KOH
CH2 == CH2
→
–K Br
–H 2 O
β- elimination
H O/ ∆
H SO
Benzoic acid
Z
Toluene
Y
25 Ethylene oxide when treated with
Grignard reagent yields
2
4
2
→
CH3  CH2OSO3H →
–H2 SO4
[CBSE AIPMT 2006]
CH3CH2OH
Ethanol
23 H2COH⋅ CH2OH on heating with
periodic acid gives
[CBSE AIPMT 2009]
(a) 2CO2
CHO
(c) 
CHO
(b) 2HCOOH
H
(d) 2
C==O
H
CH2CH2 + RMgX
KOH
–NaCl,
–H2O
RCH2CH2OH
Primary alcohol
Hydrolysis
–Mg(OH)X
Salicylaldehyde
(a) C nH2 nO2
(c) C nH2 n + 1O
Ans. (d)
Ans. (d)
→
Periodic acid
(Oxidising agent)
Ethylene glycol
2
H
H
C==O + HIO3
Zn -dust
CH Cl
3
→ X →
Y
Anhy. AlCl 3
(b) C nH2 nO
(d) C nH2 n + 2O
Alkanols are the derivatives of alkanes
which are derived by the replacement of
—H of alkanes with —OH (hydroxyl
groups).
–H C H
CnH2 n + 2 →
n 2 n + 1 OH
24 Consider the following reaction,
Phenol
Alk.
KMnO4
→
or CnH2 n + 2O
Z
[CBSE AIPMT 2005]
(b) benzaldehyde
(d) benzene
Phenol
reduction
of Phenol Benzene
X
OH
(a) ClCH2—CH2OH (b)
NO2
OH
OH
Friedel-Craft's
reaction
(c)
(d)
CH3
CH3Cl
anhy. AlCl3
O
II
III
O
O
IV
V
s
s
28 The enzyme which hydrolysis
triglycerides into fatty acids and
glycerol is called
(a) maltase
(c) zymase
Ans. (b)
CH2OOCR
CHOOCR + 3H2O
(b) lipase
(d) pepsin
CH2OH
Lipase
hydrolysis
CH2OOCR
Triglycerides
CHOH
CH2OH
Glycerol
+ RCOOH
Fatty acid
[CBSE AIPMT 2004]
27 Which one of the following
compounds is most acidic ?
[CBSE AIPMT 2009]
Zn-dust
–ZnO
O
29 Which of the following will not form
a yellow precipitate on heating with
an alkaline solution of iodine?
+OH Alkanols
Alkanes
The product Z is
(a) toluene
(c) benzoic acid
Ans. (c)
OH
s
[CBSE AIPMT 2004]
26 The general molecular formula, which
represents the homologous series of
alkanols is
[CBSE AIPMT 2006]
HIO4
O
RCH2CH2OMg
CHO
+
Phenoxide ion is stabilised due to
following resonating structures :
s
Ethylene oxide
OH
Phenol
Phenoxide ion
I
Grignard’s
reagent
O
Phenol
s
(a) secondary alcohol
(b) tertiary alcohol
(c) cyclopropyl alcohol
(d) primary alcohol
Ans. (d)
Ethylene oxide on treatment with
Grignard reagent give additive product
which undergo hydrolysis to give primary
alcohol as final product
OH
+ CHCl3
+ H+
h
PBr3
CH2OH

CH2OH
–
Ans. (c)
Phenols are much more acidic than
alcohols, due to the stabilisation of
phenoxide ion by resonance
(a) CH3CH(OH)CH3
(b) CH3CH2CH(OH)CH3
(c) CH3OH
(d) CH3CH2OH
Ans. (c)
An organic compound form yellow
precipitate of iodoform with I2 in
presence of alkali, if it has
CH3CO — group directly or it has
H

CH3  C  group.

OH
NaOH
(a) CH3CH(OH)CH3 + I2 → CH3COCH3
+ 2HI + 3NaI + CH3COO−N+ a + 3H2O
197
Organic Compounds Containing Oxygen
OH
CH3COCH3 + 3I2 + 4NaOH → CHI3 ↓
Yellow ppt
+ 3NaI + CH3COO−N+ a + 3H2O
(b) CH2  CH2 CH(OH) CH3 + I2 →
CH3  CH2  C  CH3 + 2HI

O
Ethyl methyl ketone
It gives iodoform test
CH3CH2  C  CH3 + 3I2 + 4NaOH →

O
CHI3 ↓ + 3NaI + CH3CH2 COONa + 3H2O
34 Which of the following is correct?
CHO
[CBSE AIPMT 2001]
Major product
(Salicylaldehyde)
OH
+
H
Cl
Cl
C
Cl
30 The —OH group of an alcohol or the
—COOH group of a carboxylic acid
can be replaced by —Cl using
[CBSE AIPMT 2004]
(a) phosphorus pentachloride
(b) hypochlorous acid
(c) chlorine
(d) hydrochloric acid
Ans. (a)
The —OH group of alcohol or the —COOH
group of a carboxylic acid is replaced by
—Cl using phosphorus penta chloride (i.e.
PCl 5 )
ROH + PCl 5 → RCl + POCl 3 + HCl
+H2O
Ans. (b)
+Cl–
Diastase is an example of enzyme which
is used for the conversion of starch into
maltose by hydrolysis
C
Cl
Cl
Dichlorocarbene
(attacking species)
Yellow ppt
(c) CH3OH + I2 → HCHO + 2HI
It does not have methyl ketonic
group, so it does not give yellow ppt.
with I2 in presence of alkali.
(d) CH3CH2OH+ I2 → CH3  C  H + 2HI

O
CH3  C  H+ 3I2 + 4NaOH → CHI3
Yellow ppt.

O
+ HCOONa + 3NaI + 2H2O
Due to the presence of —COCH3
group, it gives Haloform test.
(a) Cycloheptane is an aromatic
compound
(b) Diastase is an enzyme
(c) Acetophenone is an ether
(d) All of the above
+ 3NaCl + 2H2O
[CBSE AIPMT 2002]
(a) PCl5
(b) reduction
(c) oxidation with potassium
dichromate
(d) ozonolysis
Ans. (c)
[O]
CH3 CH2  CH2OH →
K Cr O /H SO
2
4
CH3 — CH2 — CHO
Propionaldehyde
[O]
CH3 CH  CH3 →
K2 Cr2 O 7 /H2 SO 4
|
OH
(a) Reduction of any aldehyde gives
secondary alcohol
(b) Reaction of vegetable oil withH2SO4
gives glycerine
(c) Alcoholic iodine with NaOH gives
iodoform
(d) Sucrose on reaction with NaCl gives
invert sugar
C2H5OH + 4I2 + 6NaOH → CHI3 ↓
Iodoform
+ HCOO−N+ a + 5NaI + 5H2O
Iodoform is a pale yellow solid.
36 The correct acidic order of following
is
[CBSE AIPMT 2001]
(a) I > II > III
(c) II > III > I
Ans.
CH3  C  CH3

O
(b) III > I > II
(d) I > III > II
OH
Iso-propyl alcohol
2° alcohol
Alcohol
OH
(I)
Acid
31 When phenol is treated with CHCl 3
and NaOH, the product formed is
[CBSE AIPMT 2002]
(a) benzaldehyde (b) salicylaldehyde
(c) salicylic acid
(d) benzoic acid
Ans. (b)
Reimer-Tiemann reaction When
phenol is treated with chloroform and
NaOH, salicylaldehyde is obtained.
OH
33 In preparation of alkene from
alcohol using Al 2O 3 , which is
effective factor?
[CBSE AIPMT 2001]
(a) Porosity of Al2O3
(b) Temperature
(c) Concentration
(d) Surface area of Al2O3
Ans. (b)
Temperature is the effective factor for
dehydration of alcohol by
Al2O3(dehydrating reagent).
Al2 O
3
R — CH2 — CH2OH →
350 ° - 380 ° C
+ CHCl3 + 3NaOH
R — CH==CH2 + H2O
While at 220-250°C, it forms ether.
OH
(III)
(II)
Acetone
RCOOH + PCl 5 → RCOCl + POCl 3 + HCl
(maltose)
Ans. (c)
n-propyl alcohol and iso-propyl alcohol
gives different products on oxidation
with K2 Cr2O7
n -propyl alcohol
1° alcohol
enzyme
(starch)
35 Which one of the following is
correct?
[CBSE AIPMT 2001]
32 n-propyl alcohol and iso-propyl
alcohol can be chemically
distinguished by which reagent?
2 2 7
Diastase
2(C6H10O5 ) n + nH2O →
nC12H22O11
NO2
CH3
(a) I > II > III
(c) II > III > I
Ans. (b)
(b) III > I > II
(d) I > III > II
The acidic behaviour of phenols may be
explained on the basis of two following
reasons.
(a) Due to resonance (which is not
possible in alcohols), the oxygen
atom of the — OHgroup acquires a
positive charge which helps in the
release of a proton.
+
OH
–
OH
198
NEET Chapterwise Topicwise Chemistry
+
+
OH
OH
–
Resonance
representation
of phenol
–
δ +OH
OH
δ-
+1
H O/H+
2
R  CH2  CH2 OH ←
Resonance stabilisation of phenoxide ion
O
–
O
O
–
δ-
–
Mg(OH) X +
Ans. (c)
d-
–
(b) In the dissociation of phenol to
phenoxide ion and a proton the
equilibrium lies mainly towards the
right hand side as the resulting
phenoxide ion is more stabilised by
resonance as compared to phenol.
O
R CH2 CH2 OMgX
(c) phenoxide ion is stabilised through
delocalisation
(d) phenoxide ion is less stable than
ethoxide
O
O
–
–
(a) NO2 group at p-position behaves in a
different way from that at o-position
(b) intramolecular hydrogen bonding
exists in p-nitrophenol
(c) there is intermolecular hydrogen
bonding in p-nitrophenol
(d) p-nitrophenol has a higher molecular
weight than o-nitrophenol
Ans. (c)
C2H5  O —H
O
Stable
ºCH
–
2 5
O +H
–
+
Unstable
ethoxide ion
(due to absence of resonance)
Phenoxide ion is more stable than
ethoxide ion due to resonance.
Therefore, the ionisation constant of
phenol is higher than ethanol.
O
O
40 The boiling point of p-nitrophenol is
higher than that of o-nitrophenol
because
[CBSE AIPMT 1994]
The boiling point of p-nitrophenol is
higher than that of o-nitrophenol
because p-nitrophenol have
intermolecular hydrogen bonding
whereas o-nitrophenol have
intramolecular H-bonding as given
below.
O
H
O
–
Resonance
representation
of phenoxide ion
–
O
[CBSE AIPMT 2000]
O

(a) CH3  C O O H
δ-
–
O
δ-
δδ-
–
o-nitrophenol easily give
steam volatile
HO
(b) H3BO 3
(c) B2H6 /NaOH H2O2
(d) H2 SO4 / H2O
Ans. (c)
+
Ar — OH
ArO + H
The acidic strength of phenols depends
on the nature of substituents present in
the benzene nucleus.
Electron withdrawing groups like NO2 ,
—CN, — CHO, — COOH, etc, when
present at the ortho and para-positions
with respect to phenolic group increases
the acidity of phenol due to greater
stabilisation of phenoxide ion. While the
presence of electron releasing group like
—NH2 , — CH3, etc, decrease the acidity
of phenols. This explains the following
order of acidity
p-nitrophenol > phenol > p-cresol.
º
N
+ O
38 Propan-1-ol may be prepared by
reaction of propene with
37 The ionisation constant of phenol is
higher than that of ethanol because
[CBSE AIPMT 2000]
(a) phenoxide ion is bulkier than
ethoxide
(b) phenoxide ion is stronger base than
ethoxide
+
O
—N
O---HO
O
+
N
O
Hydroboration-oxidation reaction
B H
2 6
6CH3CH == CH2 →
2(CH3CH2 CH2 ) 3 B
Propene
Tri-n-propylborane
H2 O2 / OH
–
6CH3CH2 CH2OH
Propan-1-ol
39 Reaction of H2 C CH2 with RMg X
O
leads to formation of
[CBSE AIPMT 1998]
(a) RCHOHR
(b) RCHOHCH3
R
(d)
CHCH2OH
(c) RCH2CH2OH
R
Ans. (c)
δ−
+δ
CH2  CH2 + R MgX →
O
Ethylene oxide
41 What is formed when a primary
alcohol undergoes catalytic
dehydrogenation?
[CBSE AIPMT 1993]
(a) Aldehyde
(c) Alkene
Ans. (a)
(b) Ketone
(d) Acid
Aldehydes can be prepared by the
dehydrogenation of primary alcohols. It
is carried out by passing the vapour of
primary alcohol over reduced copper at
573 K.
H

Cu , 573 K
R — C— O — H →

H
(1° alcohol)
R — C == O + H2

H
Aldehyde
199
Organic Compounds Containing Oxygen
1° alcohol gives aldehyde by catalytic
dehydrogenation
e.g.
Cu + 573 K
CH3  CH  O → CH3CHO


H
H
Ans. (d)
Phenol reacts with bromine water
(aqueous solution to give a precipitate of
2,4,6-tribromophenol) due to polar
solvent.
OH
OH
Br
42 Which one of the following on
oxidation gives a ketone?
[CBSE AIPMT 1993]
(a) Primary alcohol
(b) Secondary alcohol
(c) Tertiary alcohol (d) All of these
Ans. (b)
Ketones can be prepared by the
oxidation of secondary alcohols by using
oxidising agent such asK2 Cr2O7 / H2SO4
H3C
CHOH + [O] →
H3C
2° alcohol
H3C
C == O + H2O
H3C
Dimethyl ketone
43 Increasing order of acidic strength
among p-methoxy phenol (I),
p-methyl phenol (II) and
p-nitrophenol (III) is
[CBSE AIPMT 1992]
(a) III, I, II (b) II, I, III (c) III, II, I (d) I, II, III
Ans. (d)
OH
OH
OH
,
Br
→ CH3  CH2  CH2 OH
phenol
Br
Propan-1-ol
2,4,6-tribromophenol
(White ppt)
45 Methanol is industrially prepared by
[CBSE AIPMT 1992]
(a) oxidation of CH4 by steam at 900°C
(b) reduction of HCHO using LiAlH4
(c) reaction of HCHO with a solution of
NaOH
(d) reduction of CO usingH2 and
ZnO − Cr2O3
Ans. (d)
Commercially methanol is prepared from
water gas which is a mixture of carbon
monoxide and hydrogen. In this method,
CO gas is mixed with its half volume of
hydrogen and is passed over heated
Cr2O3 − ZnO catalyst at 673 K under high
pressure.
Cr2 O -ZnO
3
CO + 2H2 →
CH3OH
673 K, high pressure
[CBSE AIPMT 1992]
I
CH3
II
NO2
III
Nitro group is an electron withdrawing
group, so increases the acidic character
of phenol. Whereas —CH3 and —OCH3
both are electron releasing groups, so it
decrease the acidic character of phenol.
But —CH3 group is less electron
donating or releasing, so p-methyl
phenol is slightly more acidic as compare
to p-methoxy phenol and p-nitro phenol
is most acidic. So, the order of acidic
character is
p-methoxy phenol < p-methyl phenol
< p-nitro phenol.
44 When phenol is treated with excess
of bromine water, it gives
[CBSE AIPMT 1992]
(a) m-bromophenol
(b) o- and p-bromophenols
(c) 2,4-dibromophenol
(d) 2,4,6-tribromophenol
(CH3  CH2  CH2 ) 3B
H2O2 /OH–
be primary alcohols?
OCH3
B H
2 6
3CH3 CH == CH2 →
+ 3HBr
+ 3Br2(aq)
46 How many isomers of C 5 H11 OH will
,
(a) KMnO 4 (alkaline)
(b) Osmium tetroxide (OsO 4 / CH2Cl2 )
(c) B2H6 and alk H2O2
(d) O 3 / Zn
Ans. (c)
(a) 5
(b) 4
Ans. (b)
(c) 2
(d) 3
The primary alcohols isomers of C5H11OH
are
(i) CH3  CH2  CH2  CH2  CH2 OH
(ii) CH3  CH  CH2  CH2 OH

CH3
(iii) CH3  CH2  CH  CH2  OH

CH3 CH3

(iv) CH3  C  CH2 OH

CH3
47 Propene, CH3  CH == CH2 can be
converted into 1-propanol by
oxidation. Indicate which set of
reagents amongst the following is
ideal to affect the above
conversion?
[CBSE AIPMT 1989]
Here, half mol of (B2H6 ) diborane react
with propane by Markownikoff's addition
it gives tripropyl borane called
hydroboration. In presence ofH2O2 in
basic medium tripropyl borane gives
alcohol. Remember that product is
Anti-Markownikoff’s rule that is
1-propanol. Reaction is called
hydroboration oxidation.
48 The compound which reacts
fastest with Lucas reagent is (at
room temperature)
[CBSE AIPMT 1989]
(a) butan-1-ol
(b) butan-2-ol
(c) 2-methyl propan-1-ol
(d) 2-methyl propan-2-ol
Ans. (d)
In Lucas test when Lucas reagent is
treated with 1°, 2° and 3° alcohols, then
turbidity appears, if turbidity is appeared
immediately, then alcohol is tertiary.
2-methyl propan-2-ol is a tertiary
alcohol. Hence, it reacts fastest with
Lucas reagent.
49 Which chloro derivative of benzene
among the following would undergo
hydrolysis most readily with aq.
NaOH to furnish the corresponding
hydroxy derivative?
[CBSE AIPMT 1989]
NO2
(a)
O2N—
—Cl
NO2
(b) O2N—
(c)
Me2N—
(d) C6H5—Cl
—Cl
—Cl
200
NEET Chapterwise Topicwise Chemistry
Ans. (a)
OH
Cl
NO2
O2N
NO2
NO2
H2O
Warm
dil. HCl
NO2
NO2
Picric acid
or 2, 4, 6-trinitrochlorobenzene
50 When phenol is heated with CHCl 3
and alcoholic KOH, salicylaldehyde is
produced. This reaction is known as
[CBSE AIPMT 1989, 88]
(a) Rosenmund’s reaction
(b) Reimer-Tiemann reaction
(c) Friedel-Craft’s reaction
(d) Sommelet reaction
Ans. (b)
When phenol is heated with chloroform
(CHCl 3) and alcoholic KOH,
salicylaldehyde is formed. This reaction
is known as Reimer-Tiemann reaction.
of functional group are known as
metamers and this phenomenon is
known as metamerism.
(a) C5H12 contains no functional group.
So, it cannot show metamerism.
(b) C3H8O has ether functional group in
which only one arrangement is
possible. So, it does not show
metamerism.
CH3 — O — CH2 — CH3
(c) C3H6O has carbonyl functional group
in which following two arrangements
are possible.
O

CH3 — C— CH3, CH3CH2 CHO
So, it shows functional group
isomerism and does not show
metamerism.
(d) C4H10O has ether functional group in
which following two arrangements
are possible. So, it shows
metamerism.
CH3 O  CH2  CH2  CH3
CH3  CH2  O  CH2  CH3
+ CHCl3
Phenol
Ethoxy ethane
53 Anisole on cleavage with HI gives:
[NEET (Sep.) 2020]
OH
I
I
CHO
(a)
Salicylaldehyde
51 Lucas reagent is [CBSE AIPMT 1988]
(a) conc. HCl and anhy. ZnCl2
(b) conc. HNO 3 and anhy. ZnCl2
(c) conc. HCl and hydrous ZnCl2
(d) conc. HNO 3 and hydrous ZnCl2
Ans. (a)
The equimolar mixture of concentrated
hydrochloric acid and anhydrousZnCl2 is
called Lucas reagent. Lucas reagent is
used to distinguish between 1°, 2° and 3°
alcohols.
TOPIC 2
Ethers
52 The compound which shows
metamerism is
[NEET 2021]
(a) C 5H12 (b) C 3H8O (c) C 3H6O (d) C 4H10O
Ans. (d)
Metamerism compound which have
same molecular formula but different
number of carbon atoms on either sides
(d) H3C  CH2  CH2 OH and
HO C(CH3) 3
Ans. (a)
Ethers are readily cleaved by heating in
presence of halogen acids to form
alcohol and an alkyl halide. In case of
unsymmetrical ethers, halogen goes
preferentially with smaller alkyl group or
more stable carbocation.
If excess of acid is used then only alkyl
halide is formed because alcohol formed
reacts further with halogen acid to form
corresponding alkyl halide.
+ CH3OH (b)
+ C2H5I
OH
I
+ CH3I
+ C2H5OH (d)
(c)
Ans. (d)
H
OCH3
H—I
–I
OCH3 I
∆
CH3CH2CH2OH + (CH3)3 I
‘D’
∆ Excess
HI
CH3CH2CH2I
‘C’
No SN2-attack
of I to
the aryl carbon
is possible
55 The heating of phenyl-methyl
ethers with HI produces. [NEET 2017]
(a) ethyl chlorides (b) iodobenzene
(c) phenol
(d) benzene
Ans. (c)
Thinking Process This problem is based
on the resonance stabilisation.
In anisol, methyl phenyl oxonium ion is
formed by protonation of ether. The
bond between O — CH3 is weaker than
the bond between O — C6H5 , because the
carbon of phenyl group is sp2 -hybridised
and there is a partial double bond
character. Thus, the reaction yields
phenol and alkyl halide.
H + CH3
OH
OCH3
O
SN2
attack
(Anisole)
HI
CH3CH2CH2OC(CH3)3
Methoxy propane
OH
KOH
–NaCl,
–H2O
(b) H3C CH2 CH2 OH and
I C(CH3) 3
(c) H3C CH2 CH2 I and
HO C(CH3) 3
H
I
I
–
+ CH3I
OH + CH3I
Phenol
Anisol
Option (d)
So, even with excess of HI anisole will
give always phenol and methyl iodide (as
in option-d)
54 The major products C and D formed
in the following reactions
respectively are
56 The reaction
OH
Excess HI
→ C + D
∆
(a) H3C CH2 CH2 I and I C(CH3) 3
OsNa+
MeI
O
Me
can be classified as
[NEET (Odisha) 2019]
H3C CH2 CH2 OC(CH3 ) 3
NaH
[NEET 2016, Phase I]
(a)
(b)
(c)
(d)
Alcohol formation reaction
Dehydration reaction
Williamson alcohol synthesis reaction
Williamson ether synthesis reaction
201
Organic Compounds Containing Oxygen
Ans. (d)
(c) CH3 (CH2 ) 4  O  CH3
The formation of ether from alcohol in
the presence of base followed by
alkylation is known as Williamson ether
synthesis reaction.
OH
NaH
–H2
OsNa+
(d) CH3CH2  CH(CH3) O CH2CH3
Ans. (a)
HBr / H2 O2
CH3CH2 CH == CH2 →
Anti - Markownikoff' s rule
CH3CH2 CH2 CH2Br →
Alkylation MeI
Bromo-butanane (1° product)
O
NaI +
Me
C 2 H ONa
5
CH3CH2 CH2 CH2Br →
SN 2 reaction
57 The reaction,
CH3

CH3 C ONa + CH3CH2Cl

CH3
CH3

→
CH3 C O − CH2 − CH3
−NaCl

CH3
is called
[CBSE AIPMT 2015]
(a) Williamson synthesis
(b) Williamson continuous etherification
process
(c) Etard reaction
(d) Gatterman-Koch reaction
(Williamson’s synthesis)
CH3CH2 CH2 CH2OC2H5
Ethoxy-butane
59 Among the following sets of
reactants which one produces
anisole?
[CBSE AIPMT 2014]
(a) CH3CHO, RMgX
(b) C 6H5OH, NaOH, CH3 I
(c) C 6H5OH, neutral FeCl3
(d) C 6H5 —CH3, CH3COCl, AlCl3
Ans. (b)
Sodium alkoxide
R O  R ′+NaX
Ether
CH3

CH3  C ONa + CH3CH2 Cl →
−NaCl

CH3
CH3

CH3  C O  CH2  CH3

CH3
58 Identity Z in the sequence of
reactions,
[CBSE AIPMT 2014]
HBr /H2O 2
CH3CH2CH == CH2 → Y
C 2H ONa
5
Z
→
(a) CH3  (CH2 ) 3 O CH2CH3
(b) (CH3)2 CH2  O  CH2CH3
⋅⋅
⋅⋅
CH3
 +
→ H3C C  O CH3
 
CH3 H
CH3

→ H3C  C+ +CH3OH

CH3
CH3

→ CH3  C  I + CH3OH
Methanol

CH3
+I È
O-N+a
OH
+ NaOH
Ans. (a)
Alkyl
halide
one of the product. The reaction
proceeds as
CH3

H3C  C  O  CH3 + H+

CH3
3°carbocation
Williamson’s synthesis
The reaction of alkyl halides with sodium
alkoxide or sodium phenoxide to form
ethers is called Williamson synthesis.
Here, in this reaction alkyl halide should
be primary and alkoxide, should be
bulkier as shown below,
R  X + R ′  ONa →
product with hot conc. HI. The order of
stability of carbocation is
3° > 2° > 1°
CH3

Thus, CH3  C  OCH3 gives CH3OH as

CH3
OCH3
CH3I
61 The reaction
CH3

H3C CH  CH2 O CH2 CH3
Heated
SN2
Anisole
60 Among the following ethers which
one will produce methyl alcohol on
treatment with hot concentrated
HI?
[CBSE AIPMT 2013]
(a) CH3 CH2 CH2 CH2 O  CH3
(b) CH3 CH2 C H  O CH3

CH3
CH3

(c) CH3 C  O CH3

CH3
(d) CH3 CH  CH2 O CH3

CH3
Ans. (c)
The ether which gives more stable
carbocation, forms CH3OH as one of the
+ HI → --Which of the following
compounds will be formed?
[CBSE AIPMT 2007]
CH3

(a) H3C CH  CH2 I + CH3CH2OH
(b) CH3 CH  CH3 + CH3CH2OH

CH3
(c) CH3 CH  CH2OH + CH3CH3

CH3
CH3

(d) H3C CH  CH2OH + CH3 CH2 I
Ans. (d)
When conc. HI or HBr reacts with ether,
the corresponding alcohol and alkyl
iodide is formed. When there is a case of
mixed ethers the halogen atom attaches
to the smaller alkyl group, due to steric
effect.
202
NEET Chapterwise Topicwise Chemistry
CH3

CH3  CH  CH2 O  CH2  CH3 + HI
CH3

→ CH3  CH  CH2 OH + CH3CH2I
∆
62 The major organic product in the
reaction, CH3OCH(CH3 ) 2 + HI →
Product, is/are [CBSE AIPMT 2006]
(a) CH3OH + (CH3)2 CHI
(b) ICH2OCH(CH3)2
(c) CH3OC (CH3)2
|
I
(d) CH3I + (CH3)2 CHOH
Ans. (d)
373 K
CH3I
CH3 O  CH(CH3)2 + HI →
Unsymmetrical ether
+ (CH3)2 CHOH
In case of unsymmetrical ether, the alkyl
halide is always formed from smaller alkyl
group. This happens, becauseI− ion being
larger in size approaches smaller alkyl
group to avoid steric hindrance.
63 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 [CBSE AIPMT 1993]
(a) H-bonding in ethanol
(b) H-bonding in dimethyl ether
(c) —CH3 group in ethanol
(d) —CH3 group in dimethyl ether
Ans. (a)
Alcohols have higher boiling points as
compared to other organic compounds
of similar molecular masses such as
ethers. This is due to the presence of
intermolecular hydrogen bonding in
alcohols which is absent in ethers.
Because of hydrogen bonding in
alcohols, these exist as associated
molecules rather than discrete
molecules. Consequently, a large
amount of energy is required to break
these bonds and therefore, their boiling
points are high.
R
R
R



- - - O H- - - O H- - - O H- - H-bonding
64 Which one is formed when sodium
phenoxide is heated with ethyl
iodide?
[CBSE AIPMT 1988]
(a) Phenetole
(b) Ethyl phenyl alcohol
(c) Phenol
(d) None of the above
Ans. (a)
When sodium phenoxide (C 6H5O −N+ a)
is heated with ethyl iodide (C2H5I) it
form ethyl phenyl ether which is also
called phenetole. This reaction is
called Williamson’s synthesis
66 Match List-I with List-II.
List-I
List-II
1. Hell-VolhardZelinsky
reaction
CO, HCl
Anhyd. AICI3/CuCl
A.
O

B. R C CH 3 + NaOX →
2. GattermannKoch
reaction
C. R CH 2 OH + R ′ COOH 3. Haloform
Conc. H 2SO 4
reaction

→
-+
—ONa + IC2H5
D. R CH 2 COOH
OC2H5 + NaI
4. Esterification
(i) X /Red P
 2 →
(ii) H 2O
Phenetole
[NEET 2021]
Choose the correct answer from
the options given below.
TOPIC 3
A
(a) 4
(c) 1
Aldehydes and Ketones
65 The product formed in the
following chemical reaction is
[NEET 2021]
CH2  C  OCH3
CH3
NaBH4
C2H5OH
CH3
OH
CH2CH2OH
(b)
Aldehyde
CH3
H
CH2CCH3
(c)
CH3
CH2COCH3
Benzaldehyde
O

R  C Os Na⊕ + CHX3
Acid
O

R  CH2 O  C  R ′
CH3
Ester
Ans. (d)
NaBH4 is a weak reducing agent. It can
reduce aldehyde/ketone to alcohol but
cannot reduce ester group.
O
CH2COCH3
CH3
CHO
Alcohol
(d)
O
C D
1 4
4 1
C. Esterification Carboxylic acid reacts
with an alcohol in acidic medium.
R  CH2 OH + R ′ COOH
OH
O
OH
B
2
3
B. Haloform reaction Treatment of
carbonyl compound having atleast one
methyl group attached to the C==O with
X2 / NaOHor NaOX.
O

R  C  CH3 + NaOX →
O
OH
A
(b) 3
(d) 2
CO, HCl
Anhy. AlCl3/CuCl
Benzene
CH2COCH3
(a)
D
3
2
A. Gattermann-Koch reaction Benzene
or its derivatives are treated with CO and
HCl in presence of anhydrous AlCl 3/ CuCl.
?
H
OH
C
2
3
Ans. (d)
O
O
B
1
4
NaBH4
C2H5OH
O
OH
CH2COCH3
CH3
D. Hell-Volhard Zelinsky reaction
Treatment of carboxylic acid having
α-hydrogen withPX3 or X2 / Red P.
O

(i) X 2 / Red P
R  CH2  C OH →
(ii) H 2 O
Carboxylic acid
O

R  CH  C OH

X
α-halocarboxylic acid
Hence, correct match is
A→ 2, B→ 3, C→ 4, D→ 1.
203
Organic Compounds Containing Oxygen
O
(i) PhCCH2
[NEET 2021]
(i) C2H5MgBr, Dry ether
product
Acetone →
(ii) H O, H+
2
OH
O
PhCCH2
C
Tautomerism
O —H
Oδ
O
H2O
OMgBr
CH3CCH3
Acetone
Dry ether
CH3CCH3
Benzoyl chloride
(A)
It is Rosenmund reaction, in which an
acid chloride gets converted into an
aldehyde.
O
–H2O
Ph CH CH C Ph
H
PhCH==CH COPh
Aldol
69 Identify compound X in the
following sequence of reactions.
C2H5
CH3
71 Predict the correct intermediate
and product in the following
reaction.
[NEET 2017]
H2 O, H2 SO4
H3C C ≡≡ CH  →
HgSO4
CHO
Intermediate → Product
OH
Cl2/hv
CH3—C—CH3
(ii) H2O, H+
CH2
CH2Cl
(Product)
IUPAC name of product is
2-methylbutan-2-ol.
(b)
CCl2
(b) A =H3C  C == CH2 ;

OH
Cl
(c)
B =H3C  C == CH2

SO4
(d)
[NEET (Sept.) 2020]
(a) Cannizzaro’s reaction
(b) Cross Cannizzaro’s reaction
(c) Cross aldol condensation
(d) Aldol condensation
Ans. (c)
Ans. (b)
An 1, 1-gem-dihalide on hot hydrolysis
(H2O/373 K) can produce an aldehyde.
Cl
CH3
Dilute NaOH is the reagent for aldol
condensation. Dilute NaOH process
enolate ion from acetophenone which
attacks benzaldehyde to give aldol.
CH
Cl
2Cl2
hν
PhCH==O
373 K
Acetophenone
O
PhCHCH2CPh
O
–H2O
70 Identify compound (A) in the
following reaction.
[NEET (Oct.) 2020]
CHO
PhCH==CHCPh
(Cis + trans)
It is cross aldol condensation or
Claisen-Schmidt reaction or Claisen
reaction.
(c) A =H3C  C  CH3;

O
B =H3C  C ≡≡ CH
(d) A =H3C  C == CH2 ;

OH
B =H3C  C  CH3

O
Ans. (d)
H2O
+ CH3CPh
OH
Dil. NaOH
CHO
(X)
O
Benzaldehyde
(a) A =H3C  C == CH2 ;

SO4
B=H3C  C  CH3

O
CHCl2
(a)
68 Reaction between benzaldehyde
and acetophenone in presence of
dilute NaOH is known as
( A)
H2O
X 373 K
[NEET (Sept.) 2020]
CH3
CHO
Benzaldehyde
–OH s
– +
(i) C2H5MgBr
H2
Pd/BaSO4
(Lindlar catalyst)
H2O
+ CH2  C  Ph
δ
O
Cl
O
OH
Ans. (d)
Ph C==CH2
–OH
–
(ii) Ph CH
+
Acetone on reaction with Grignard
reagent and on further hydrolysis gives
2-methyl butan-2-ol as follows
O
H
PhC==CH2
(a) 2-methyl propan 2-ol
(b) pentan-2-ol
(c) pentan-3-ol
(d) 2-methyl butan-2-ol
Ans. (a)
Mechanism
Enolate ion
67 What is the IUPAC name of the
organic compound formed in the
following chemical reaction?
A
H2/Pd/BaSO4
(a) Benzoyl chloride (b) Toluene
(c) Acetophenone (d) Benzoic acid
( B)
204
NEET Chapterwise Topicwise Chemistry
OH

Therefore, A = CH3  C == CH2
O

B = CH3  C  CH3
Identify A, X , Y and Z
72 Of the following which is the
product formed when
cyclohexanone undergoes aldol
condensation followed by heating?
75 The correct structure of the
product A formed in the reaction
[NEET 2016, Phase II]
O
H2 (gas, l atmosphere)
A is
Pd/carbon, ethanol
OH
OH
OH
O
Ans. (c)
[NEET 2017]
O
[NEET 2017]
(a) A-methoxymethane, X-ethanoic acid,
Y-acetate ion, Z-hydrazine
(b) A-methoxymethane, X-ethanol,
Y-ethanoic acid, Z-semicarbazide
(c) A-ethanal, X-Acetaldelyde,
Y-but-2-enal, Z-semicarbazone
(d) A-ethanol, X-acetaldehyde,
Y-butanone, Z-hydrazone
Aldehydes gives silver mirror test so, ‘X’
may be alcohol which is oxidised by Cu
gives aldehydes.
Therefore,
A is acetaldehyde (CH3CHO)
(a)
OH
Cu/573 K
C 2H5OH →
CH3CHO
oxidation
(b)
Acetaldehydol
(A )
[Ag(NH3 )2]
→
OH /D
Tollen's reagent
O
r
(d)
Ans. (b)
In presence of Pd-catalyst, selective
reduction of α,β-unsaturated carbonyl is
observed as hydrogenation takes place
of carbon-carbon double bond only.
O
O
Silver
mirror observed
H2 (gas 1, atmosphere)
Pd/ Carbon, Ethanol
NH
2
(c)
(c)
(b)
(a)
O
OH
C
O
2
O
NH
O
Ans. (b)
Aldehydes and ketones containingα H
atoms undergo aldol condensation in
presence of dilute alkali as catalyst and
gives α, β unsaturated compound with
the elimination ofH2O molecule.
H
H
–
2
OH/∆
CH3 — CH == CH—C—H
But-2-en-1-al
Aldol
(Y)
condensation
NH
(d)
O
OH
O
OH
–
O
O
H2O
–
– OH
O
O
∆ –H2O
Semicarbazone
(Z)
74 The product formed by the reaction
of an aldehyde with a primary amine
is
R  C  H + R′  NH2
O
73 Consider the reactions,
+
Cu
A
573 K
[Ag(NH3)2]
–OH ∆
–OH, ∆
O
NH2
Z
NH
C
Primary
amine
r
OH2

R  C  N  R′


H
H
(α, β- unsaturated
compound)
X
(C2H6O)
s
O
H

+
R  C  N  R′


H
H
O
Aldehyde
Dehydration
Y
NH2
Silver mirror observed
H+
OH

R  C  N  R′


H
H
–H2O
r
R  C== N  R′
 
H H
O with methyl lithium
gives which of the following
species?
[CBSE AIPMT 2015]
(a) Cyclopentanonyl anion
CH3 — CH == NH—NH—C—NH2
(a) Ketone
[NEET 2016, Phase I]
(b) Carboxylic acid
(c) Aromatic acid
(d) Schiff base
Ans. (d)
–
H
O
76 Treatment of cyclopentanone
(b) Cyclopentanonyl cation
(c) Cyclopentanonyl radical
(d) Cyclopentanonyl biradical
Ans. (a)
α
–
R  C==N  R′

H
Schiff's base
+
O + CH3 Li
Cyclopentanone
Methyl lithium
+
–
Li
O
+ CH4
Lithium
cyclopentanoyl anion
(Intermediate)
Here, CH3Li abstract is an active proton
from cyclo pentanone forming methane
leaving behind an intermediate lithium
cyclopentanoyl anion.
77 A single compound of the structure is
CH3
OHC
–H+
H
CH3
C
C
C H C
O
H2
H2
obtainable from ozonolysis of
which of the following cyclic
compounds? [CBSE AIPMT 2015]
205
Organic Compounds Containing Oxygen
H3C
H3C
(a)
CH3 (b)
H3C
H3C
Ans. (a)
Ans. (*)
In keto-enol tautomerism keto form
should haveα-hydrogen (structure I and II).
Reaction of carbonyl compounds with
ammonia derivatives give addition
product followed by the elimination
reaction. Slightly acidic medium
generate a nucleophilic centre for the
attack of weak base like ammonia
derivatives.
CH3
CH3
CH3
(c)
CH3
O
CH3
OH
CH3
CH3
α CH3
CH3
Ans. (a)
II.
1
O3
CH3 Zn,H
2
CH3
2
OHC1
H3C
2
C2
CH2
H
O
H γ H
CH3
C
CH2
O
Zn,H2
O
H
CH3
C
1
H3C
CH2
CH
CH
CH3
O3
OHC
1
Zn,H2
2
CHO
2
CH3
CH3
1
C
CH2
CH2
2
CH3
Zn,H2
H3C
CH2
CH3
C
1
CH2
CH2
O
C
2
CH
O
(II)
CH3
O
CH3
79 An organic compound X having
molecular formula C 5H10 O yields
phenyl hydrazone and gives
negative response to the iodoform
test and Tollen's test. It produces
n-pentane on reduction. X could be
Air
V2O5
CH3
O
(III)
CH3
[CBSE AIPMT 2015]
CHC
O
CHC
O
Maleic anhydride
82 Reaction by which benzaldehyde
cannot be prepared? [NEET 2013]
CH3
(a)
+ CrO2Cl2 and CS2
followed by H3O+
COCl
(b)
+ H2 in presence of
Pd-BaSO4
Since, the compound X yields phenyl
hydrazone and gives negative response
to the iodoform test and Tollen’s test , it
must contain a C == O group but is
neither a methyl ketone nor an aldehyde.
The structure of X could be
O

CH3  CH2  C  CH2  CH3
having molecular formula C5H10O.
Which of the given compounds
can exhibit tautomerism?
(a) I and II
(b) I and III
(c) II and III
(d) I, II and III
CH3
[CBSE AIPMT 2015]
(a) benzoic anhydride
(b) maleic anhydride
(c) benzoic acid
(d) benzaldehyde
Ans. (b)
+ CO + HCl in presence of
anhy. AlCl3
(c)
COOH
+ Zn / Hg and conc. HCl
(d)
Ans. (d)
3-pentanone
CH3
CH3
O
(I)
CH3
(b) 2-pentanone
(c) 3-pentanone
(d) n-amyl alcohol
Ans. (c)
78 Given,
CH3
α
in the presence of air produces
(a) pentanal
CHO
2
O
CH3
O3
β
[CBSE AIPMT 2015]
CH3
1
III.
OH
81 The oxidation of benzene by V2O 5
O
OH
Here, γ-H participates in tautomerism.
O3
1
H3C
CH3
I.
(d)
CH3
CH3
80 Reaction of a carbonyl compound
with one of the following reagents
involves nucleophilic addition
followed by the elimination of
water. The reagents is
[CBSE AIPMT 2015]
(a) a Grignard reagent
(b) hydrazine in presence of feebly
acidic solution
(c) hydrocyanic acid
(d) sodium hydrogen sulphite
(a)
CH3
2CrO2Cl2
CS2
CH3.2CrO2Cl2
Brown
(addition product)
CHO
H2O
Benzaldehyde
This reaction is known as Etard
reaction.
206
NEET Chapterwise Topicwise Chemistry
O
C
(b)

(a) CH 3CH 2CH 2CCH 3
Pd/BaSO4, S
(Boiling xylene)
O
Benzoyl
chloride
CHO
+ HCl
This reaction is called Rosenmund
reaction.
(c)
+ CO + HCl
Anhy. AlCl3
CHO
+ HCl
The above reaction is known as
Gattermann-Koch aldehyde synthesis.
COOH
(d)
Ans. (c)
O
Cl
+ Zn/Hg
Conc. HCl

(b) CH 3 CH 2 CH 2 C CH 2 CH 2 CH 3
CHO
OH
(c) (CH3)2C
(d)
(CH3)2C
(a) Benedict test
(b) iodoform test
(c) Tollen’s reagent test
(d) Fehling solution test
Ans. (b)
CH3CHO and C6H5 CH2 CHO both being
aliphatic aldehydes react with Tollen’s
reagent, Fehling solution and Benedict
solution. So, these reagents cannot be
used to distinguish them.
CH3CHO due to the presence of
O 


 
 CH — C —  group reacts with NaOH
 3





and I2 to give yellow crystals of iodoform
while C6H5 CH2 CHO does not react with it.
CH3CHO + 3I2 + 4NaOH → CHI3
+ HCOONa + 3NaI + 3H2O
C6H5 CH2 CHO + I2 + NaOH →
No reaction
Thus, CH3CHO and C6H5 CH2 CHO can be
distinguished by iodoform test.
84 Acetone is treated with excess of
ethanol in the presence of
hydrochloric acid. The product
obtained is
[CBSE AIPMT 2012]
Cl
(CH3)2C ==O+ C2H5OH

Cl
OH
OC2H5
OC2H5
(CH3)2C
OC2H5
Ketal
Cl
Potassium-3-chloro
benzoate
Hemiketal
NOTE Formation of hemiketal is a
nucleophilic addition reaction.
85 Predict the products in the given
reaction,
[CBSE AIPMT 2012]
COO–
2

ethanol
C2H5OH
(excess)
HCl
or
3
HCl
(CH3)2C
–+
COOK
1
When carbonyl compounds are treated
with alcohol, they form hemiacetal
(hemiketal and acetal/ketal.)
+
1 Cl
3-hydroxy methyl
chlorobenzene
OC2H5
Acetone
(2-propanone)
2
50% KOH
Ans. (d)
Thus, from the reactants given in option
(d) benzaldehyde is not obtained.
[CBSE AIPMT 2012]
3½
OC2H5
+ H2O
distinguished chemically by
CH2OH

OC2H5
COCl
83 CH3CHO and C 6H5CH2CHO can be
When benzaldehyde is treated with 50%
alkali, it undergoes oxidation to give an
acid salt as well as reduction to give an
alcohol. This reaction is called
Cannizaro’s reaction.
86 Clemmensen reduction of a ketone
is carried out in the presence of
which of the following?
[CBSE AIPMT 2011]
(a) Zn-Hg with HCl
(b) LIAlH4
(c) H2 and Pt as catalyst
(d) Glycol with KOH
Ans. (a)
The reducing agent used in Clemmensen
reduction is Zn-Hg and HCl.
CHO

Zn-Hg /HCl
C == O →
50% KOH
CH2
Cl
–
CH2COO
CH2OH
(a)

Cl
+
87 Acetophenone when reacted with a
base, C 2H5ONa, yields a stable
compound which has the structure
[CBSE AIPMT 2008]

Cl
CH2OH
OH
(a)
C
CHC
O
CH3
(b)

OH
+

OH
(b)
CH3 CH3
(c)

Cl
+
(c)

Cl
CC
OH OH
COO–
CH2OH
(d)
CHCH
OH
(d)

OH
+

OH
O
CH3
COO–
CH2OH
CHCH2C
OH
Ans. (a)
Aldehydes and ketones withα-hydrogen
atom, when reacted with a base yields
207
Organic Compounds Containing Oxygen
aldol which on heating loses water
molecule to giveα, β-unsaturated
aldehydes or ketones. This reaction is
called aldol condensation reaction.
C2H5ONa
C2H5O– + Na+
i
Base
C
O
–
+ C2H5O
CH3
Base
(Abstract the acid hydrogen)
O
s
CH2—C
C2H5OH +
(Attacking species)
(Nu)
CH3
—C
Ans. (b)
Condensation between two molecules of
an aldehyde or a ketone having atleast
one α-hydrogen atom in presence of a
base to form a β-hydroxy aldehyde or
β-hydroxy ketone is known as aldol
condensation. Aldol condensation are
divided into two parts one is self aldol
condensation and another is cross-aldol
condensation, when both molecules are
same then it is called self aldol and vice
versa.
H
H


O
O OH −
+H C C
H  C C
H
H


H
H
3
α -hydrogen
O
s
O + CH2C
CH3
(c) an alpha-hydroxy aldehyde or ketone
(d) an alpha, beta unsaturated ester
H H
 
→ CH3  C  C  C
α β
OH H
O
O
CH3
+
H
O
C−CHC
—C
90 Which one of the following on
treatment with 50% aqueous
sodium hydroxide yields the
corresponding alcohol and acid?
[CBSE AIPMT 2007]
OH H
CH3
H
O
CHC
–H2O
(a) C 6H5CH2CHO
(b) C 6H5CHO
(c) CH3CH2CH2CHO
O

(d) CH3  C  CH3
Ans. (b)
88 A strong base can abstract an
α-hydrogen from
[CBSE AIPMT 2008]
(a) alkene
(c) ketone
Ans. (c)
(b) amine
(d) alkane
Since the carbonyl carbon is electron
deficient, so most susceptible to attack
by nucleophilic reagents or base. A base
increases the acidity of hydrogen atom
attached to theα-C of the ketones or
aldehydes. That’s why α-hydrogen is
easily abstracted from ketones by a
base, e.g. in aldol condensation reaction,
α-hydrogen atom of aldehyde or ketone
is abstracted by a strong base.
89 The product formed in aldol
condensation is [CBSE AIPMT 2007]
(a) a beta-hydroxy acid
(b) a beta-hydroxy aldehyde or a
beta-hydroxy ketone
Clemmensen reduction Aldehydes and
ketones are reduced to the
corresponding alkanes by means of
amalgamated zinc and HCl.
Zn-Hg /HCl
C==O +4[H] →
Aldehydes which do not have any
α-hydrogen atom when heated with a
concentrated solution of NaOH undergo
a simultaneous oxidation and reduction
(disproportionation) forming a salt of
carboxylic acid and alcohol. This
reaction is called Cannizaro reaction.
2C6H5 CHO + NaOH → C6H5 CH2OH
Benzyl alcohol
− +
+ C6H5 COONa
Sodium benzoate
91 Reduction of aldehydes and
ketones into hydrocarbons using
zinc amalgam and conc. HCl is
called
[CBSE AIPMT 2007]
(a) Clemmensen reduction
(b) Cope reduction
(c) Dow reduction
(d) Wolff-Kishner reduction
CH2 + H2O
92 Nucleophilic addition reaction will
be most favoured in
[CBSE AIPMT 2006]
O

(a) CH3  CH2  CH2 C  CH3
(b) (CH3)2 C ==O
(c) CH3CH2CHO
(d) CH3CHO
Ans. (d)
Nu
Slow
C==O
Nucleophile
3-hydroxy butanal or
b-hydroxy aldehyde
CCH2C
–
O
Ans. (a)
Carbonyl
compound
O–
C
Nu
Tetrahedral
intermediate
H+
Fast
OH
C
Adduct Nu
The carbonyl compounds undergo
nucleophilic addition reaction because
oxygen is more electronegative than
carbon. As such, it withdraws shared
π-electron pair towards itself and gets
partial negative charge, therefore
carbon get partial positive charge and
becomes susceptible to nucleophilic
attack.
Aldehydes are more reactive than
ketones towards nucleophiles. This can
be explained on the basis of inductive
effect as well as steric effect. The
addition of nucleophiles is based upon
the positive charge present on carbon
atom of
C==O group. In aldehyde
C==O group is present with at least
one alkyl group (except formaldehyde)
which has +I-effect (electron donating
effect) and which decreases the positive
charge of carbon, thereby making the
attack to nucleophile difficult. The
nucleophilic attack becomes more
difficult in ketones having minimum of
two alkyl groups.
Hence, by means of attachment of alkyl
groups (due to +I-effect) rate of
nucleophilic addition decreases. That
means e − density at C-atom decreases,
nucleophilic addition reaction increases.
Order of +I-effect in alkyl group
208
NEET Chapterwise Topicwise Chemistry
R
R
<
CH—< R
C
R
R
Order of nucleophilic addition in given
carbonyl compound is
CH3CHO > CH3  CH2  CHO > (CH3) 2 CO >
O

CH3  CH2  CH2  C  CH3
 CH3 < R  CH2−
93 A carbonyl compound reacts with
hydrogen cyanide to form
cyanohydrin which on hydrolysis
forms a racemic mixture of
α-hydroxy acid. The carbonyl
compound is [CBSE AIPMT 2006]
(a) acetaldehyde (b) acetone
(c) diethyl ketone (d) formaldehyde
Ans. (a)
OH
C==O + HCN
C
CN
OH
H2O
C
COOH
(It is α-hydroxy acid)
In this reaction, by the complete
hydrolysis of cyanide gives acid and
partial hydrolysis gives amide.
If it is racemic mixture, therefore such
C-atom must be asymmertic carbon
atom.
CH3
C==O + HCN
CH3
H
H
Acetaldehyde
H2O
OH
C
CH3
|
HCOH +
|
COOH
d-
CH3
C==O + HCN
CH3
|
HO—C—H
|
COOH
lOH
CH3
C
CH3
Acetone
H2O
CH3
CN
OH
C
CH3
COOH
It is not optically active
racemic mixture is nor formed
C2H5
C==O + HCN
C2H5
OH
C
C2H5
C2H5
Diethyl ketone
H2O
C2H5
C2H5
OH
H
C ==O + HCN
C
H
CN
OH
H
H
Formaldehyde
H2O
C
94 The major organic product formed
from the following reaction
[CBSE AIPMT 2005]
O
(i) CH3NH2
... is
(ii) LiAlH4
(iii) H2O
CH3
(a)
CN
OH
C
COOH
It is not optically active
(c) A = RR′ C
CN
, B = CH3
COOH
(d) A = RR′C
CN
,
OH
B =LiAlH4
Ans. (d)
In the presence ofLiAlH4 , cyanide group
gives amine by reduction.
OH
R
R
HCN
C ==O →
C
KCN
R′
R′
CN
A
Cyanide
(b)
OH
NHCH3
ONHCH3
(c)
(a) A = RR′CH2CN,
B = NaOH
OH
(b) A = RR′ C
, B = CH3
COOH
COOH
H
It is not optically active
C
R′
CH2NH2
Amine
OH
OH
OH
R
Ans. (b)
O
OH
R
LiAlH (B )
4
→
NHCH3
(d)
Hence, A is
CH3NH2
and B is
C
R′
CN
LiAlH4
Ketone
NCH3
Schiff base
NHCH3
[H]
(i) LiAlH4
(ii) H2O
or
NHCH3
2° amine
95 Which one of the following can be
oxidised to the corresponding
carbonyl compound ?
[CBSE AIPMT 2004]
CN
Racemic mixture
CH3
H
(a) 2-hydroxy propane
(b) Ortho-nitro phenol
(c) Phenol
(d) 2-methyl-2-hydroxy propane
Ans. (a)
2-hydroxy propane or secondary alcohol
is oxidised into propanone
(corresponding because in 2-hydroxy
propane, secondary alcoholic group is
present and it is oxidised into ketone).
[O]
CH3  CH  CH3 → CH3  C  CH3
|

OH
O
2-hydroxy propane
(2° alcohol)
Propanone
(Ketone)
96 A and B in the following reactions
are
[CBSE AIPMT 2003]
HCN/
R — C — R ′ →
A →
KCN

O
OH
R —C
CH2NH2
R′
B
97 In this reaction,
CH3CHO + HCN→ CH3CH(OH)CN
H⋅ OH
→ CH3CH(OH)COOH
an asymmetric centre is
generated. The acid obtained
would be
[CBSE AIPMT 2003]
(a) 50% D + 50% L-isomer
(b) 20% D + 80% L-isomer
(c) D-isomer
(d) L-isomer
Ans. (a)
Lactic acid obtained in the given
reaction is an optically active compound
due to the presence of chiral C-atom. It
exits as d and l-forms whose
ratio is 1:1.
CH3
C ==O + HCN→
H
CH3
CH3


HO  C  H
H  C  OH +


CN
CN
50% L-isomer
50% D-isomer
Cyanohydrine
98 Polarisation of electrons in acrolein
may be written as
[CBSE AIPMT 2000]
δ+
δ−
(a) CH 2 ==CH —CH==O
δ+
δ+
(b) CH 2 ==CH —CH==O
209
Organic Compounds Containing Oxygen
δ+
δ+
CH3  COCH2  CH2 CH3 + 3I2 + 4NaOH →
CHI3 ↓ + CH3CH2 CH2 COONa
+ 3NaI +3H2O
Iodoform
(c) CH 2==CH —CH==O
δ–
δ+
(d) CH2 ==CH — CH ==O
(yellow ppt.)
Ans. (a)
CH3 CHO + 3I2 +4NaOH —→
CHI3 ↓ + HCOONa + 3NaI + 3H2O
In CH2 ==CH— CHO due to – M-effect of
—CHO group, polarisation of electron
takes place as follows :
+
CH2==CHC==O
|
H
Iodoform
(yellow ppt)
(CH3) 3 COH 8 [O]
→ CH3COOH + 2CO2 + 3H2O
2
C2H5OH →
CH3CHO
CH3CHO+ 3I2 + 4NaOH → CHI3 ↓
Iodoform
99 During reduction of aldehydes with
hydrazine and potassium
hydroxide, the first is the formation
of
[CBSE AIPMT 2000]
(a) R—CH==N—NH2
(b) R—C ≡≡N
(c) R—C —NH2

O
(d) R—CH==NH
Ans. (a)
R
H
3-pentanone
→ No reaction.
KOH
Aldehyde
R
H
Hydrazine
C ==N —NH2 + H2O
Aldehyde hydrazone
100 Aldol condensation will not take
place in
[CBSE AIPMT 1999]
(a) HCHO
(c) CH3COCH3
Ans. (a)
+ HCOONa + 3NaI+ 3H2O
But due to absence of CH3  CH

OH
or CH3  C == O group in 3-pentanone, it

does not give iodoform.
O

CH3  CH2  C  CH2  CH3 + I2 + NaOH
C ==O + N H2 —NH2 →
(b) CH3CHO
(d) CH3CH2CHO
102 1-phenyl ethanol can be prepared
by the reaction of benzaldehyde
with
[CBSE AIPMT 1997]
(a)
(b)
(c)
(d)
methyl bromide
ethyl iodide and magnesium
methyl iodide and magnesium
methyl bromide and aluminium
bromide
Ans. (c)
1-phenyl ethanol is prepared by reacting
benzaldehyde with methyl magnesium
iodide (mixture of methyl iodide and
magnesium as )
Dry ether
Aldol condensation in aldehydes is due
to presence ofα-hydrogen atoms. Those
aldehydes which does not have
α-hydrogen atom like HCHO, does not
give aldol condensation reaction.
101 Iodoform test is not given by
[CBSE AIPMT 1998]
(a) 2-pentanone
(b) ethanol
(c) ethanal
(d) 3-pentanone
Ans. (d)
The compounds which contain either
CH3  CO  group or CH3 —CH group
give positive

OH
iodoform test. In 2-pentanone,
(CH3CH2 CH2 COCH3), CH3CHO and C2H5OH,
required groups are present, thus they
give iodoform as follows
CH3I+ Mg → CH3MgI
CH3  MgI +
Ans. (c)
By oxidation of tertiary alcohol with
stronger oxidising agents, ketones may
be formed along with carboxylic acid.
4 [O]
→
CH3COCH3 + CO2 + 2H2O
I
–
CH2—CH= CO
|
H
(c) oxidation of tertiary alcohol
(d) reaction of acid halide with alcohols
C6H5
104 (CH3 ) 3 C  CHO does not undergo
aldol condensation due to
[CBSE AIPMT 1996]
(a) three electron donating methyl
groups
(b) cleavage taking place between
—C—CHO bond
(c) absence of alpha hydrogen atom in
the molecule
(d) bulky (CH3) 3 C — group
Ans. (c)
CH3

CH3 — C — CHO does not undergo aldol

CH3
condensation because it does not
contain α-hydrogen atom.
105 Acetone reacts with iodine (I 2 ) to
form iodoform in the presence of
[CBSE AIPMT 1995]
(a) CaCO 3
(c) KOH
Ans. (b)
(b) NaOH
(d) MgCO 3
O

CH3  C  CH3 + 3I2 + 4NaOH →
Acetone
− +
C == O →
CHI3 + CH3COONa + 3NaI + 3H2O
Iodoform
H
C6H5
CH — O—MgI
CH3
HO
–Mg(OH)I
2
HOHC
→
C6H5
CH3
(1-phenyl ethanol)
103 Ketones[R —C—R 1 ], where
||
O
R = R 1 = alkyl group, can be
obtained in one step by
[CBSE AIPMT 1997]
(a) hydrolysis of esters
(b) oxidation of primary alcohol
106 Which of the following compounds
will undergo self aldol
condensation in the presence of
cold dilute alkali?
[CBSE AIPMT 1994]
(a) CH2 ==CH  CHO
(b) CH ≡≡C—CHO
(c) C 6H5CHO
(d) CH3  CH2CHO
Ans. (d)
Only those aldehyde undergoes aldol
condensation which haveα-hydrogen, so
CH3CH2 CHO give this reaction because it
contains α-hydrogen atom. Aldol
condensation proceed in presence of
strong base. Aldol condensation are
divided into two parts one is self aldol
210
NEET Chapterwise Topicwise Chemistry
condensation and another is cross-aldol
condensation. When both molecules are
same called as self aldol and vice versa.
107 Aldehydes and ketones will not
form crystalline derivatives with
[CBSE AIPMT 1994]
(a) sodium bisulphite
(b) phenyl hydrazine
(c) semicarbazide hydrochloride
(d) dihydrogen sodium phosphate
Ans. (d)
Dihydrogen sodium phosphate
(NaH2PO 4 ) does not react with
aldehydes and ketones because
NaH2PO 4 does not have any lone pair
of electron on phosphorus atom, so it
cannot act as a nucleophile.
108 Benzaldehyde reacts with ethanolic
KCN to give
[CBSE AIPMT 1994]
(a) C 6H5CHOHCN
(b) C 6H5CHOHCOC 6H5
(c) C 6H5CHOHCOOH
(d) C 6H5CHOHCHOHC 6H5
Ans. (b)
H
C
C
KCN
C2H5OH
OH O
HCC
Benzoin
(hydroxy ketone)
This reaction is also called benzoin
condensation. Benzoin is chiral and it
exists as a pair of enantiomer, i.e.
R-benzoin and S-benzoin.
109 Pinacolone is
[CBSE AIPMT 1994]
(a) 2,3-dimethyl-2,3-butanediol
(b) 3,3-dimethyl-2-butanone
(c) 1-phenyl-2-propanone
(d) 1,1-diphenyl-2-ethanediol
Ans. (b)
The structure of pinacolone is
CH3 O
CH3CCCH3
4
3
2
CH3
110 (CH3 ) 2 C==CHCOCH3 can be
oxidised to (CH3 ) 2 C==CHCOOH by
Consequently, the carbonyl carbon is
positively charged while the oxygen is
negatively charged. The positively
charged carbon is easily attacked by a
nucleophilic reagent (Nu− ).
[CBSE AIPMT 1993]
(a) chromic acid
(c) Cu at 300°C
Ans. (b)
(b) NaOI
(d) KMnO 4
Haloform reaction,
O

(CH3)2 C==CHCOCH3 contains CH3 — C —
unit so it can be oxidised to
(CH3)2 C==CH— COOH by NaOI.
O

NaOI
CH3  C == CH  C  CH3 →

CH3
CH3  C == CH  COOH + CHI3

CH3
1
(a) CH3COCH3
(c) CH3CH2CH2OH
Ans. (a)
(b) CCl3CH2CHO
(d) CH3CH2CHO
Ketones are not easily oxidised.
However, under drastic conditions or
with powerful oxidising agents such as
conc. HNO3, KMnO4 / H2SO4 or
K2 Cr2O7 / H2SO4 , cleavage of
carbon-carbon bond takes place giving a
mixture of carboxylic acids having less
number of carbon atoms than the
original ketone.
O

[O]
CH3  C  CH3 → HCOOH
Acetone
conc. HNO3
Formic acid
+ CH3COOH
Acetic acid
112 Acetaldehyde reacts with
[CBSE AIPMT 1991]
(a)
(b)
(c)
(d)
C==O
δ−
δ+
C ------- O
or
O–
Nu
Slow
C == O + Nu– →
C
[CBSE AIPMT 1992]
O
+
oxygen atom. The actual structure may
be represented as
111 In which of the following the
number of carbon atoms does not
remain same when carboxylic acid
is obtained by oxidation?
H
O
So, its IUPAC name is
3,3-dimethyl-2-butanone
(Colourless liquid) (Camphor odour)
only electrophiles
only nucleophiles
only free radicals
both electrophiles and nucleophiles
Ans. (b)
The carbonyl group is highly reactive
polar group. It is polarised due to the
higher electronegativity of oxygen in
comparison to carbon. As a result, the
electrons present between carbon and
oxygen are more attracted towards
Intermediate
Aldehyde
or ketone
sp2 hybridisation planar structure.
+
H
→
C
fast
OH
Nu
Addition product
sp3 hybridisation tetrahedral structure.
113 The reagents which can be used to
distinguish acetophenone from
benzophenone is (are)
[CBSE AIPMT 1990]
(a) 2,4-dinitrophenyl hydrazine
(b) aqueous solution of NaHSO3
(c) Benedict reagent
(d) I2 and Na2 CO3
Ans. (d)
The structures of acetophenone and
benzophenone are
O
CH3C
O
Acetophenone
C
Benzophenone
When acetophenone containing
O

CH3  C  unit, is treated withI2 and
Na2 CO3 it forms yellow precipitate of
CHI3 whereas benzophenone does not
give this test.Acetophenons gives
iodoform test due to presence of
COCH3 group.
HCl
114 3CH3COCH3 →
(A)
–H 2O
(CH3 ) 2 C==CH—CO—CH==C(CH3 ) 2
(B)
This polymer (B) is obtained when
acetone is saturated with HCl gas,
B can be
[CBSE AIPMT 1989]
(a) phorone
(b) formose
(c) diacetone alcohol
(d) mesityl oxide
211
Organic Compounds Containing Oxygen
Ans. (a)
O

C == O + H2 CH  C  CH H2
CH3
CH3
CH3 Dry HCl
→
–2H2O
CH3
+ O == C
O

C == CH  C  CH == C
CH3
CH3
CH3
CH3
2,6-dimethylhepta-2,5-diene-4-one
(Phorone) (B)
Phorone is self condensation product of
acetone. It can also be obtained from
certain camphor compound. Phorone is
combustible when exposed to heat or
flame.
CH3
, the compound
115
H3C
CH3
CH2 O
TOPIC 4
CH2 , the
CH2 O
Carboxylic Acids
shown polymer is obtained when a
carbonyl compound is allowed to
119 Which of the following acid will
stand. It is a white solid. The
form an (i) anhydride on heating
polymer is
[CBSE AIPMT 1989]
and (ii) acid imide on strong heating
(a) trioxane
with ammonia? [NEET (Oct.) 2020]
116 O
(b) para-formaldehyde
(c) formose
(d) meta-aldehyde
Ans. (a)
When formaldehyde is allowed to
stand at room temperature, it slowly
undergoes, polymerisation and forms
a white solid called
meta-formaldehyde or trioxane.
CH2
CH2
O
O
O
O
Allowed to
CH2
O
CH2
stand
H2C
O
CH2
meta-formaldehyde
describes a condensation polymer
(Trioxane)
which can be obtained in two
ways, either treating 3 molecules 117 Formalin is an aqueous solution of
of acetone (CH3COCH3 ) with conc.
[CBSE AIPMT 1988]
H2SO 4 or passing propyne
(a) fluorescein
(b) formic acid
(CH3 —C ≡≡ CH) through a red hot
(c) formaldehyde (d) furfuraldehyde
tube, the polymer is
Ans. (c)
[CBSE AIPMT 1989]
(a) phorone
(b) deacetonyl alcohol
(c) mesityl oxide (d) mesitylene
Ans. (d)
When acetone is treated withH2SO4 ,
three molecules get condensed to give
mesitylene,
CH3
O
C
CH H2
H2
CH
CH3
C
C
O H2 HC
–3H2O
CH3
HC
C
CH3
HC
CH3
CH
C
C
C
CH
C—CH3
CH
Red hot tube
CH3
CH3
C
HC
H3C
C
C
H
(b)
COOH
COOH
COOH
OH
(c)
(d)
COOH
COOH
Ans. (a)
An α, β-dicarboxylic acid with same-side
(syn) orientation of — COOH group is able
to form anhydride (cyclic) and imide
(cyclic).
Among isomeric benzene dicarboxylic
acids, only benzene-1,2-dicarboxylic acid
(phthalic acid) will respond to the below
reactions.
O
O
C
C
OH
O
C
Heating
–H2O
O
C
H
O
O
Phthalic anhydride
O
C
COOH Strong
heating
NH3
COOH –H3O
Phthalic acid
O H2 HN
C
O
CH
C
Mesitylene
(a) methane
(b) methyl alcohol
(c) ethyl formate
(d) acetylene
Ans. (b)
–H2O
O
C
NH
When α-hydrogen is absent in carbonyl
group, those compound gives cannizaro
reaction.
This reaction show disproportionation.
The oxidation product is salt of
carboxylic acid and reduced product is
alcohol.
Mesitylene
CH
(a)
[CBSE AIPMT 1988]
CH3
C
CH3
118 If formaldehyde is heated with
KOH, then we get
H2SO4 (conc.)
O
The 40% solution of formaldehyde in
water is sold in market under the name
of formalin. Formaldehyde in the form of
formalin (40% formaldehyde, 8%
methanol and 52% water) is used for
preserving biological specimens.
Formaline solution also used as a
disinfectants and commonly used in
hardeness and nail varnish.
COOH
COOH
CH3
KOH (conc. )
HCHO + HCHO →
CH3OH
C
O
Phthalimide
120 The major product of the following
reaction is : [NEET (National) 2019]
COOH
+ NH3
Methyl alcohol
+ HCOO–K+
COOH
Strong heating
212
NEET Chapterwise Topicwise Chemistry
O
COOH
(a)
(b)
NH
NH2
O
The hydrogen bonds are not
completely broken in the vapour state.
In fact mostly carboxylic acids exist as
dimer in the vapour state or aprotic
solvent.
–
δ
O
COOH
NH2
(c)
O
NH2
CONH2
Ans. (a)
Carboxylic acids react with ammonia to
give ammonium salt which on further
heating at high temperature give
amides. Further, on strong heating,
ammonia is removed from phthalamide
and phthalimide is formed.
The reaction takes place as follows :
+ NH3
Phthalic
acid
Ammonium
phthalate
∆ –2H2O
O
CONH2
Strong heating
– NH3
C
Phthalamide
CONH2
O
Phthalimide
121 Carboxylic acids have higher boiling
points than aldehydes, ketones and
even alcohols of comparable
molecular mass. It is due to their
COOH
Ans. (d)
Carboxylic acids have higher boiling
points than aldehyde, ketones and
even alcohols of comparable
molecular mass because of the extent
of intermolecular-hydrogen bonding
with water, due to which they exist as
associated molecules.
C
H
O
R
C
δ–
O
δ+
C
H
O
δ–
H
δ+
O δ–
124 In a set of reactions, ethyl benzene
yielded a product D.
[CBSE AIPMT 2010]
CH2CH3
KMnO4
is
O
KOH
III
(a) I > II > III
(c) III > II > I
Ans. (b)
C
(b) II > III > I
(d) II > I > III
D would be
COOH
COOH
>
>
C2H5OH
H+
D
III
Less I-effect
(Moderately
acidic)
(a)
Br
Br
(b)
Br
CH2COOC2H5
COOH
COOH
(c)
OC2H5
I
No I-effect
(Least acidic)
123 Which one of the following esters
gets hydrolysed most easily under
alkaline conditions?
[CBSE AIPMT 2015]
OCOCH3
(a)
O2N
Br2
FeCl3
CH2 CHCOOC2H5
Key Idea Order of strengths of the given
carboxylic acids can be determined by the
concept of I-effect.
The oxygen atom present in the ring
shows I-effect. As the distance between
oxygen and COOH group increases,
–I-effect of oxygen decreases.
Thus, corresponding carboxylic acid will
show less acidic nature.
The correct order of strengths of the
carboxylic acids is
II
More-effect
(Most acidic)
B
OCOCH3
(b)
OCOCH3
COOC2H5
(d)
Br
Ans. (d)
Alkaline KMnO4 converts complete
carbon chain (that is directly attached to
benzene nucleus) to COOH group. Br2
in the presence of halogen carrier
causes bromination by electrophilic
substitution reaction and ethyl alcohol in
acidic medium results in esterification.
CH2CH3
COOH
(c)
δ+
H
O
O
II
H3CO
O
O2N
COOH
O
H
OCOCH3
COOH
I
[NEET 2018]
(a) more extensive association of
carboxylic acid via van der Waals’
force of attraction
(b) formation of carboxylate ion
(c) formation of intramolecular
H-bonding
(d) formation of intermolecular
H-bonding
R
δ
[NEET 2016, Phase II]
– +
COONH4
R
122 The correct order of strengths of
the carboxylic acids
COONH4
COOH
NH
δ
– +
COOH
C
C
R
(d)
δ+
O
H
H-bonds
C
H+
O–
Ans. (a)
Electron withdrawing group attach to
the benzene ring increases the reactivity
towards nucleophilic sustitution
reaction. Since, NO2 group is strong
electron withdrawing group. Hence, in
basic medium ester containing NO2
group will hydrolysed most easily.
R
OCOCH3
Br2/FeCl3
KMnO4
KOH
(d)
Cl
(m-directing)
B
213
Organic Compounds Containing Oxygen
COOC2H5
COOH
C2H5 OH
H+
Cl – < RCOO– < R ′O– > NH2–
Thus, order of reactivity is
RCOCl > (RCO)2 O > RCOOR ′
Acyl chloride
Acid anhydride
Ester
Br
Br
>RCONH2
Amide
D
C
125 Propionic acid with Br 2 —P yields a
dibromo product. Its structure
would be
[CBSE AIPMT 2009]
(a) CH2Br CHBr COOH
Br

(b) H  C CH2COOH

Br
(c) CH2Br  CH2  COBr
Br

(d) CH3  C  COOH

Br
Ans. (d)
Br
H


Br2 /P
CH3  C  COOH → CH3  C  COOH
−2HBr


Br
H
Propionic acid
Carbonylic acids reacts with Cl2 or Br2 in
presence of red P to give exclusively
α-chloro or α-bromo acids.
This reaction is called
Hell-Volhard-Zelinsky (HVZ) reduction.
This reaction is example of
α-H-substitution.
126 The relative reactivities of acyl
compounds towards nucleophilic
substitution are in the order of
[CBSE AIPMT 2008]
(a) acyl chloride > acid anhydride > ester
> amide
(b) ester > acyl chloride > amide > acid
anhydride
(c) acid anhydride > amide > ester > acyl
chloride
(d) acyl chloride > ester > acid anhydride
> amide
Ans. (a)
In acyl compounds (i.e. acyl chloride,
acid anhydride, ester and amide) RCO—
group is same, thus reactivity depends
upon the nature of group Z (i.e. Cl − ,
RCOO− , R ′O − ,NH2– , etc.)
If group Z is a weak base, then it is a
strong leaving agent and its reactivity
towards nucleophilic substitution is
high.
The order of basic nature of Z groups is
127 Which of the following represents
the correct order of acidity in the
given compounds?
The structure of D would be
[CBSE AIPMT 2006]
(a) CH3CH2CH2NH2 (b) CH3CH2CONH2
(c) CH3CH2NHCH3 (d) CH3CH2NH2
Ans. (d)
For the reaction,
SOCl2
CH3CH2 COOH →
CH3 CH2 COCl
–
[CBSE AIPMT 2007]
(a) FCH2 COOH > CH3COOH > BrCH2 COOH
> ClCH2 COOH
(b) BrCH2 COOH > ClCH2 COOH >
FCH2 COOH > CH3COOH
(c) FCH2 COOH > ClCH2 COOH > BrCH2 COOH
> CH3COOH
(d) CH3COOH > BrCH2 COOH > ClCH2 COOH
< FCH2 COOH
Ans. (c)
The acidity of halogenated acid
increases with increase in
electronegativity of the halogen present.
The electronegativity of halogen
decreases in order asF > Ce > Br.
Therefore correct order of given
compounds is
FCH2 COOH > ClCH2 COOH > BrCH2 COOH
> CH3COOH
128 Self condensation of two moles of
ethyl acetate in the presence of
sodium ethoxide yields
SO2
–HCl
A
NH
4KOH + Br
– HCl
(–2KBr, –K 2 CO3 ,
–2H2 O)
3
2
→
CH3CH2 CONH2 →
C
CH3CH2NH2
D
(Ethyl amine)
Hence, it is also called Hofmann
bromamide degradation reaction.
Hence, compound ‘D’ is
CH3  CH2  NH2 .
Hofmann bromamide reaction degrade
the one C in amine product from amide
130 In a set of reactions, acetic acid
yielded a product D.
CH3COOH
SOCl2
On condensation, two moles of ethyl
acetate in the presence of sodium
ethoxide, gives ethyl acetoacetate
(ester). This condensation is an example
of Claisen condensation because it is
possible in those ester which have
α-hydrogen atom.
O

CH3  C O  C2H5 + H  CH2 COOC2H5
O

NaOC2H5
→ CH3C CH2 COOC2H5 + C2H5OH
Ethyl acetoacetate
(ester)
Benzene
anhy. AlCl3
C
HOH
B
D
The structure of D would be
[CBSE AIPMT 2005]
OH
|
CCOOH
|
CH3
(a)
COOH
|
CH2 CCH3
|
OH
(b)
OH
|
CH2 CCH3
|
CN
(c)
CN
|
CCH3
|
OH
(d)
Ans. (a)
129 In a set of reactions propionic acid
yielded a compound D.
SOCl
A
HCN
[CBSE AIPMT 2006]
(a) ethyl butyrate (b) acetoacetic ester
(c) methyl acetoacetate
(d) ethyl propionate
Ans. (b)
B
CH3COOH
SOCl2
NH
2
3
CH3CH2COOH →
B →
C
KOH
→ D
Br2
Benzene
Anhy. AlCl3
B
CH3COCl
A
COCH3
214
NEET Chapterwise Topicwise Chemistry
OH
H2 O
R  COOR ′ → R  COOH + R ′OH
C—CH3
HCN
CN
OH
(i) C2H5MgBr
C6H6
Anhy. AlCl3
C
LiAlH4 / ether
(ii) Ether hydrolysis
B
C6H5
C2H5
C
C—CH3
2HOH
Ans. (d)
COCH3
CH3
OH
C
COOH
OH
D
C—COOH
or
CH3
131 Which one of the following orders
of acidic strength is correct ?
[CBSE AIPMT 2003]
(a) RCOOH > HOH > HC ≡≡ CH > ROH
(b) RCOOH > HC ≡≡ CH > HOH > ROH
(c) RCOOH >ROH > HOH > HC ≡≡ CH
(d) RCOOH > HOH > ROH > HC ≡≡ CH
133 In the following reaction, product P
is
Carboxylic acid is stronger than alcohol
and water because after removal of
proton, carboxylate ion is stabilised by
resonance. Hence, correct order of acid
strength is
RCOOH > HOH > ROH > HC ≡≡ CH
Which is based upon the rate of donation
of proton or strength of base, thus order
of basic strength is
–
–
—
—
—
—
—
—
C ≡≡ CH > R — O > OH– > RCOO–
O–
O
C
—
—C
O
O–
Resonating structures of carboxylate ion
132 In a set of the given reactions,
acetic acid yielded a product C.
C H
6 6
B
CH3COOH + PCl 5 → A →
anhy. AlCl 3
C H MgBr
2 5
→
C
Ether
Product C would be
[CBSE AIPMT 2002]
(a) RCH2OH
(c) RCHO
Ans. (c)
(b) RCOOH
(d) RCH3
The given reaction is Rosenmund
reaction
R —C— Cl → R — C—H + HCl
Pd -BaSO4


O
O
134 Benzoic acid may be converted into
ethyl benzoate by reaction with
[CBSE AIPMT 2000]
A
(a) CH3(CH2)3 COC2H5 + HCHCOOC2H5
|
CH2CH2CH3
O
Claisen
No reaction, because for Claisen
condensation an ester with
α-hydrogen atoms is required.
Ethyl benzoate is prepared by reacting
benzoic acid and ethanol in the presence
of dry HCl. This reaction is known as
esterification reaction.
C6H5 COOH + C2H5OH
Benzoic acid
Ethanol
CH3CH2CH2CH2CCHCOOC2H5

O CH2CH2CH3+C2H5OH
(b) C6H5 COOC2H5 + C6H5 COOC2H5
(a) sodium ethoxide
(b) ethyl chloride
(c) dry HCl, C2H5OH
(d) ethanol
Ans. (c)
Dry
C6H5 COOC2H5
3
HCl
Ethyl benzoate
+ H2O
This reaction proceed with equilibrium.
Therefore, H2O continuously removed
from reaction for preparation of ester
product.
[CBSE AIPMT 2000]
CH3COOH+ PCl 5 → CH3COCl
The ester which containsα-hydrogen
atom undergoes Claisen-self
condensation :
Condensation
135 Reduction by LiAlH4 of hydrolysed
product of an ester gives
(c) CH3CH(OH)C2H5
(d) CH3COC 6H5
Ans. (b)
[CBSE AIPMT 1998]
(a) CH3CH2CH2CH2COOC2H5
(b) C 6H5COOC2H5
(c) C 6H5CH2COOC2H5
(d) C 6H11CH2COOC2H5
Ans. (b)
( P)
[CBSE AIPMT 2003]
(a) CH3CH(OH)C 6H5
C2H5

(b) CH3  C(OH) C 6H5
136 Which one of the following esters
cannot undergo Claisen
self-condensation ?
H
2
R —C —Cl →
P
Pd -BaSO 4

O
H2
Ans. (d)
R — CH2OH + R ′OH
According to the above equation, it is
clear that reduction of hydrolysed
product of ester byLiAlH4 gives two
alcohols.
(a) two acids
(b) two aldehydes
(c) one molecule of alcohol and
another of carboxylic acid
(d) two alcohols
(c) C6H5CH2CO OC2H5 + H CHCOOC2H5

C6H5
Claisen
Condensation
C6H5CH2COCHCOOC2H5

C6H5
(d) C6H11CH2CO OC2H5 + HCHCOOC2H5

C6H11
Claisen
Condensation
C6H11CH2COCHCOOC2H5 + C2H5OH

C6H11
137 An ester (A) with molecular formula
C 9 H10 O 2 was treated with excess of
CH3MgBr and the complex so
formed was treated with H2SO 4 to
give an olefin (B). Ozonolysis of (B)
gave a ketone with molecular
formula C 8H8O which shows
positive iodoform test. The
structure of (A) is [CBSE AIPMT 1998]
215
Organic Compounds Containing Oxygen
(a) C 6H5COOC2H5
(b) C 6H5COOC 6H5
(c) H3CCOOC 6H5
(d) p -H3COC 6H4COCH3
Ans. (a)
C6H5COOC2H5
A
139 An ester is boiled with KOH. The
product is cooled and acidified with
conc. HCl. A white crystalline acid
separates. The ester is
[CBSE AIPMT 1994]
OMgBr

C6H5COC2H5

CH3
CH3MgBr
–(C2H5O)MgBr
OMgBr

C6H5CCH3

CH3
CH3MgBr
(a) methyl acetate (b) ethyl acetate
(c) ethyl formate (d) ethyl benzoate
Ans. (d)
Boiling
H+ /H O
2
C6H5 COOK →
Benzoic acid
(white precipitate)
C6H5C==CH2

CH3
B
Conc. H2SO4
–H2O
O3/H2O
C6H5CCH2

CH3
(B)
C6H5CCH3
||
O
(C8H8O)
I
2
C6H5CCH3 NaOH CHI3 + C6H5COONa
|
Iodoform
O
138 Consider the following
transformations
CaCO3
Heat
I2
→ C The molecular formula of
NaOH
C is
[CBSE AIPMT 1996]
OH

(a) CH3  C  CH3 (b) ICH2 COCH3

I
(c) CHI3
Ans. (c)
(d) CH3I
CH COO
2CH3COOH → 3
CH3COO
A
CaCO 3
140 Schotten-Baumann reaction is a
reaction of phenols with
[CBSE AIPMT 1994]
(a) benzoyl chloride and NaOH
(b) acetyl chloride and NaOH
(c) salicylic acid and conc. H2 SO 4
(d) acetyl chloride and conc. H2 SO 4
Ans. (a)
Schotten-Baumann reaction
.
pyridine
Aq NaOH
C6H5 COCl + C6H5OH →
C6H5 COOC6H5 + HCl
141 The preparation of ethyl
acetoacetate involves
CH3COOH → A → B
[CBSE AIPMT 1994]
(a) Wittig reaction
(b) Cannizaro’s reaction
(c) Reformatsky reaction
(d) Claisen condensation
Ans. (d)
Claisen condensation
O
–

C2 H5 ONa+
CH3 C OC2H5 + HCH2 COOC2H5 →
O

CH3 C CH2 COOC2H5 + C2H5OH
Ca
Ethyl acetoacetate
Heat
Na2 CO3 + H2O
→
– CaCO 3
COOH

COOH
Glycerol
→ == HCOOH + CO2 ↑
Heat 373K
Formic acid
Oxalic acid
143 Among acetic acid, phenol and
n-hexanol which one of the
following compound will react with
+ C2H5OH
NaHCO 3 solution to give sodium
C6H5 COOH ↓
salt and CO 2 ? [CBSE AIPMT 1993, 99]
H2O
OH

C6H5CCH3

CH3
Ans. (c)
KOH
C6H5 COOC2H5 →
C6H5 COOK+
Ethyl benzoate
O
||
C6H5CCH3
(c) glycerol is heated with oxalic acid at
373 K
(d) acetaldehyde is oxidised withK2 Cr2O7
and H2SO4
142 Formic acid is obtained when
(a) Acetic acid
(b) n-hexanol
(c) Acetic acid and phenol
(d) Phenol
Ans. (a)
+
CH3COOH + NaHCO3 → CH3COON a
Acetic acid
Sodium
carbonate
+ H2O + CO2 ↑
144 Sodium formate on heating yields.
[CBSE AIPMT 1993]
(a) Oxalic acid and H2
(b) Sodium oxalate and H2
(c) CO2 and NaOH
(d) Sodium oxalate
Ans. (b)
+
COONa
Heat
→
+ H2 ↑

+
CO
ONa
Sodium formate
Hydrogen
HCOONa
+
HCOONa
Sodium
oxalate
145 Benzoic acid gives benzene on
being heated with X and phenol
gives benzene on being heated with
Y. Therefore, X and Y are respectively
[CBSE AIPMT 1992]
(a) sodalime and copper
(b) Zn dust and NaOH
(c) Zn dust and sodalime
(d) sodalime and zinc dust
Ans. (d)
COOH
[CBSE AIPMT 1994]
I +NaOH
2
+ CHI3 ←
CH3COCH3
Acetone
C
B
+ CH3COONa
(a) calcium acetate is heated with conc.
H2SO4
(b) calcium formate is heated with
calcium acetate
NaOH + CaO
Decarboxylation
Benzoic acid
Benzene
216
NEET Chapterwise Topicwise Chemistry
OH
Phenol
Here x = NaOH + CaO (soda line)
y = Zn dust
146 The compound formed when
malonic acid is heated with urea, is
[CBSE AIPMT 1989]
(a) cinnamic acid (b) butyric acid
(c) barbituric acid (d) crotonic acid
Ans. (c)
COOH H2N
+
COOH H2N
Malonic acid
Heat
−H2 O
C == O →
Urea
CO — NH
CO — NH
C == O
Barbituric acid
Zn
Dust
CH2
CH2
147 Among the following the strongest
acid is
[CBSE AIPMT 1988]
(a) CH3COOH
(c) CH2ClCOOH
Ans. (c)
(b) CH2ClCH2COOH
(d) CH3CH2COOH
Inductance effect distance depending
factor. It decreases rapidly with
distance. Therefore, as the distance of
Cl-atom increases the acidic character
decreases.
148 Which of the following represent
the correct decreasing order of
acidic strength of following?
[CBSE AIPMT 1988]
(i)
(ii)
(iii)
(iv)
Methanoic acid
Ethanoic acid
Propanoic acid
Butanoic acid
(a) (i) > (ii) > (iii) > (iv) (b) (ii) > (iii) > (iv) > (i)
(c) (i) > (iv) > (iii) > (ii) (d) (iv) > (i) > (iii) > (ii)
Ans. (a)
The correct order of acidic strength is
methanoic acid > ethanoic acid >
propanoic acid > butanoic acid because
the +I–effect of alkyl group increases in
the order.
CH3 < C2H5 < C3H7 < C4H9
− I - effect (EWG)
Acidic Nature ∝
+I - effect (ERG)
−I-effect increases hence, acidic nature
increases.
25
Organic Compounds
Containing Nitrogen
TOPIC 1
Aliphatic Amines
01 Reaction of propanamide with
ethanolic sodium hydroxide and
bromine will give [NEET (Oct.) 2020]
(a) ethylamine
(c) propylamine
Ans. (a)
(b) methylamine
(d) aniline
Ans. (a)
O
CH3CH2
C
NH2

CH3  C  CH2 CH2 CH3,

CH3
1° amine
CH3

CH3  C  CH  NH2
 
CH3 CH3
1° amine
NH2

(c) CH3 C C H2CH2CH3

CH3
CH3

(d) CH3 C  CH  NH2
 
CH3 CH3
NH2
Propanamide
(No. of C-atoms =3)
CH3CH2
NaOH-C2H5OH
Br2
NH2 + Na2CO3+NaBr
Ethylamine (1°-amine)
(No. of C-atoms =2)
It is Hofmann bromamide reaction
through which an acid amide degrades
into an one carbon less primary (1°)
amine.
Here, the carbon atom of the amide
 O

 

group   C NH2  gets decarboxylated






in the form of carbonate salt (Na2 CO3).
02 The amine that reacts will
Hinsberg’s reagent to give an alkali
insoluble product is
Secondary amines on reaction with
Hinsberg’s reagent (benzene sulphonyl
chloride) forms N,N-dialkyl benzene
sulphonamide.
The product formed is not acidic as no
H-atom is attached to N thus it is
insoluble in alkali.
The reaction of 2º amine given in option
(a) takes place as follows:
O
—S—Cl+CH3—CH—NH—CH—CH3
CH3 CH3
O
2º amine
O
CH3
—SNCHCH3 + HCl
O H3C— CH
CH3
(a) CH3  CH NH  CH CH3


CH3
CH3
On the other hand, 1° amines react with
Hinsberg’s reagent to yield N-alkyl
benzene sulphonamide which is soluble
in alkali and 3° amines do not react with
C6H5SO2 Cl. The remaining options
contain 1° and 3° amines.
CH2CH3

(b) CH3 CH2 N CH2C H3
CH2 CH3

CH3CH2  N  CH2 CH3,
3° amine
[NEET (Odisha) 2019]
Thus, option (a) is correct.
03 The correct order of the basic
strength of methyl substituted
amines in aqueous solution is
[NEET (National) 2019]
(a) (CH3) 3N > CH3NH2 > (CH3)2 NH
(b) (CH3) 3N > (CH3)2 NH > CH3NH2
(c) CH3NH2 > (CH3)2 NH > (CH3) 3N
(d) (CH3)2 NH > CH3NH2 > (CH3) 3N
Ans. (d)
Basic strength of methyl substituted
amines in aqueous solution depends upon
the ease of formation of cation by
accepting a proton from the acid.
H
RN +H
H
+
+
RNH
H
H
The basic strength can be decided by
both inductive effect and solvation
effect of alkyl group. In aqueous phase,
+
the substitutedNH4 cations get
stabilised not only by electron releasing
effect of alkyl group (+ I) but also by
solvation with water molecules. Greater
the size of ion, lesser will be solvation
and less stabilised is the ion. The order is
as follows :
218
NEET Chapterwise Topicwise Chemistry
Inductive effect
(CH3) 3N > (CH3)2 (NH) > (CH3)NH2
H
+
H3C—N—H----OH2 >
H3C
H----OH2
+
N
H----OH2
H3C
H
(2°)
OH2
H3C
(1°)
>
+
N—H----OH2
H3C
Step IV
O
 ••
CH3  C N •• intermolecular
→ CH3NCO
alkyl migration
∆
StepV CH3NCO + 2OH− →
CH3NH2 + CO23−
CH3
(3°)
Strong +I-effect and hydrogen bonding
favours higher basic strength of 2º
amine. Thus, the correct order of basic
strength in aqueous media will be
(CH3)2 NH > CH3NH2 > (CH3) 3N
04 Which of the following reactions is
appropriate for converting acetamide
to methanamine?
[NEET 2017]
(a) Carbylamine reaction
(b) Hofmann hypobromamide reaction
(c) Stephens reaction
(d) Gabriels phthalimide synthesis
Ans. (b)
The conversion of amide with no
substituent on nitrogen to an amine
containing one carbon less by the action
of alkaline hypobromide or bromine in
presence of NaOH. It involves the
migration of alkyl or aryl group with its
electron pair to electron deficient N
from adjacent carbon. The reaction
involves the intermediates of
isocyanate.
O

∆
CH3  C  NH2 + Br2 + NaOH →
Acetamide
CH3NH2
O

s
Step III CH3  C N  Br →
O
 ••
s
CH3  C N •• + Br
+ NaBr + Na2 CO3 + H2O
05 Which one of the following nitrocompounds does not react with
nitrous acid? [NEET 2016, Phase II]
H2
C
(a) H3C
C NO2
H2
H3C
(b)
H2
C
CH NO2
H3C
CH3
H3C
(c) H3C C
H3C
(d) H3C
NO2
C
H
NO2
O
Ans. (c)
Key Idea 1° and 2° nitro compounds
react withHNO2 while 3°-nitro compound
does not.
The reactions of given compounds with
HNO2 are as follow
HON == O
CH3CH2 CH2NO2 →
1° -nitro
compound
CH3  CH2  C NO2

N OH
H2
C
H3C
H3C
HO
N
1°-nitro compound
Mathanamine
CH3
O

Step I CH3  C NH2 + Br2
CH
N
OH
C
NO2
CH3
O

Step II CH3  C N Br + OH−

H
O

→ CH3  C N Br + H2O
s
CH3
C
CH
O
1°-nitro compound
NO2
HO
CH3
N
CH
O
C
CH3 N
H3C
H3C
H3C
C
NO2
HO
N
The given reaction followsS N 2
mechanism andS N 2 reactions are
favoured in polar aprotic medium like
DMSO, DMF... etc.
DMF
CH3CH2 CH2Br + NaCN →
CH3CH2 CH2 CN + NaBr
So, the correct option is (c).
07 Method by which aniline cannot be
prepared is
[CBSE AIPMT 2015]
(a) hydrolysis phenyl isocyanide with
acidic solution
(b) degradation of benzamide with
bromine in alkaline solution
(c) reduction of nitrobenzene with
H2 / Pd in ethanol
(d) potassium salt of phthalimide
treated with chlorobenzene followed
by the hydrolysis with aqueous NaOH
solution
Ans. (d)
Due to resonance in chlorobenzene C—Cl
bond acquires double bond character
hence, C—Cl bond is inert towards
nucleophile (phthalimide ion). Therefore
aniline cannot be prepared.
O
3°-nitro compound
Thus, option (c) is incorrect.
08 The number of structural isomers
possible from the molecular
formula C 3H9 N is
[CBSE AIPMT 2015]
(a) 4
(c) 2
Ans. (d)
CH3
O

→ CH3  C N Br

H
(a) ethanol
[NEET 2016, Phase II]
(b) methanol
(c) N, N′-dimethylformamide (DMF)
(d) water
Ans. (c)
O
NO2
CH
06 Consider the reaction
CH3CH2CH2Br + NaCN →
CH3CH2CH2CN + NaBr
This reaction will be the fastest in
(b) 5
(d) 3
Structural isomers of C3H9N are
NO2
OH
No reaction
CH3CH2 CH2NH2 ,CH3  CH  CH3

NH2
1444442444443
1° -amine
CH3  CH2 NH  CH3, CH3 N  CH3

2 ° -amine
CH3
3° - amine
219
Organic Compounds Containing Nitrogen
09 Acetamide is treated with the
following reagents separately.
Which one of these would yield
methyl amine? [CBSE AIPMT 2010]
(a) NaOH/ Br2
(b) Sodalime
(c) Hot conc. H2 SO 4
(d) PCl5
Ans. (a)
−
••
••
Methyl amine
Acetamide
+ NaBr + Na2 CO3 + H2O
10 Which one of the following on
reduction with LiAlH 4 yields a
secondary amine?
[CBSE AIPMT 2007]
(a) Methyl isocyanide
(b) Acetamide
(c) Methyl cyanide
••
K+
(iii) R  C  N Br → R  C  N + KBr

 • •
O
O
(iv) On rearrangement
R  C— N
 ••
O
Key Idea The reagent which can convert
CONH2 group into —NH2 group is used
for this reaction.
Among the given reagents only NaOH/
Br2 converts CONH2 group to NH2
group, thus it is used for converting
acetamide to methyl amine. This
reaction is called Hoffmann bromamide
reaction, in which primary amides on
treatment withBr2 / NaOH form primary
amines.
CH3CONH2 + NaOH+ Br2 → CH3NH2
−
••
–
OH
(ii) RCONHBr →
R  CO N Br + H2O
••
º R  N == C == O
••
Ans. (a)
LiAlH
4
CH3 N ≡≡ C + 4[H] →
CH3NHCH3
Dimethylamine
On catalytic reduction or with lithium
aluminium hydride (LiAlH4 ) or with
nascent hydrogen, alkyl isocyanide yield
2° amine whereas cyanide gives 1° amine
on reduction.
11 Intermediates formed during
reaction of RCNH2 with Br 2 and

O
KOH are
[CBSE AIPMT 2001]
(a) RCONHBr and RNCO
(b) RNHCOBr and RNCO
(c) RNHBr and RCONHBr
(d) RCONBr2
Ans. (a)
The reaction,
RCONH2 + Br2 + KOH → RNH2
is known as Hofmann bromamide
reaction. The mechanism of this
reaction is given as :
(i) RCONH2 + Br2 → RCONHBr + HBr
Boiling H O
2
HCl
CH3CN + 2H →
X →
Y,
SnCl2
the term Y is
[CBSE AIPMT 1999]
(a) acetone
(c) acetaldehyde
Ans. (c)
(b) ethanamine
(d) dimethyl amine
HCl
SnCl 2
CH3CN + 2H → CH3  CH == NH
X
Imide
(v) R  N == C == O + 2KOH → RNH2
H2 O
+ K2 CO3
CH3  CH ==O ←
Boil
Y
12 Amides can be converted into
amines by a reaction named after
[CBSE AIPMT 1999]
(a) Perkin
(c) Hofmann
Ans. (c)
(b) Claisen
(d) Kekule
Amides can be converted into amines by
Hofmann’s bromamide reaction. This
reaction is named after Hofmann. The
reaction is as follow.
 CONH2 + Br2 (l ) + 4KOH → NH2
+2KBr + K2 CO3 +2H2O
13 The decomposition of organic
compounds, in the presence of
oxygen and without the development
of odoriferous substances, is called
[CBSE AIPMT 1999]
(d) Nitroethane
15 In the reaction,
(a) decay
(c) nitrification
Ans. (a)
(b) N2 -fixation
(d) denitrification
Decomposition of organic compounds
in the presence of oxygen is generally
called decay. The remaining three
reactions takes place in the presence of
bacteria.
14 Phenyl isocyanides are prepared
from which of the following
reaction?
[CBSE AIPMT 1999]
Acetaldehyde
So, Y is acetaldehyde.
16 The compound obtained by heating
a mixture of primary amine and
chloroform with ethanolic
potassium hydroxide (KOH) is
[CBSE AIPMT 1997]
(a) an alkyl isocyanide
(b) an alkyl halide
(c) an amide
(d) an amide and nitro compound
Ans. (a)
RNH2 + CHCl 3 + 3KOH (alc.) →
RNC
Alkyl isocyanide
This reaction is known as
carbylamine test.(only 1° amine gives
this reaction)
17 Consider the following sequence of
reactions
Reduction
[CBSE AIPMT 1996]
(a) CH3CH2CN
(c) CH3NC
Ans. (d)
(b) CH3NO2
(d) CH3CN
Reduction
CH3C ≡≡ N →
Both aliphatic and aromatic primary
amines react with chloroform and alc.
KOH to give isocyanides or carbylamines
and the reaction is known as
carbylamine reaction.
C6H5NH2 + CHCl 3 + 3KOH → C6H5NC
∴ A is CH3CN.
Phenyl
isocyanide
+ 3KCl +3H2O
HNO 2
Compound [A] → [B] →
CH3CH2OH
The compound [A] is
(a) Rosenmund’s reaction
(b) Carbylamine reaction
(c) Reimer-Tiemann reaction
(d) Wurtz reaction
Ans. (b)
(alc. )
+ 3KCl + 3H2O
LiAlH4
A
HNO
2
CH3  CH2 NH2 →
CH3CH2OH
B
1° amine (Ethanamine)
18 Which is formed when acetonitrile
is hydrolysed partially with cold
conc. HCl?
[CBSE AIPMT 1995]
(a) Acetic acid
(b) Acetamide
(c) Methyl cyanide
(d) Acetic anhydride
220
NEET Chapterwise Topicwise Chemistry
Ans. (b)
22 Mark the correct statement.
The partial hydrolysis of alkyl cyanides
with cold conc. HCl or H2SO4 gives
amides.
Conc. HCl
CH3  C ≡≡ N →
CH3CONH2
–
Alkylcyanides
H2 O/OH
Acetamide
19 Acetamide and ethyl amine can be
distinguished by reacting with
[CBSE AIPMT 1994]
(a) aq. HCl and heat
(b) aq. NaOH and heat
(c) acidified KMnO 4
(d) bromine water
Ans. (b)
When acetamide is heated with aq.
NaOH it forms NH3 gas but ethylamine
cannot form NH3.
NaOH ∆
CH3CONH2 + H2O → CH3COONa + NH3
NaOH ∆
CH3CH2NH2 + H2O →
No reaction
[CBSE AIPMT 1989]
(a) Methyl amine is slightly acidic
(b) Methyl amine is less basic than ammonia
(c) Methyl amine is a stronger base than
NH3
(d) Methyl amine forms salts with alkalies
any primary amine and chloroform
chloroform and silver powder
a primary amine and an alkyl halide
a mono alkyl amine and
trichloromethane
+
NH3
NH2
H+
Ans. (a)
Methyl amine is a stronger base thanNH3.
This is due to the reason that alkyl
groups are electron releasing groups
(+I-effect). As a result of which it
increase the electron density on the
nitrogen atom and therefore, they can
donate electron pair more easily than
ammonia.
Aniline
Anilinium ion
Since, anilinium ion so formed is meta
directing, thus besides ortho and para
derivatives, significant amount of meta
derivative is also formed.
NH2
NH2
HNO3,
H2SO4
NO2
Aniline
m-nitroaniline
(47%)
p-nitroaniline
(51%)
NH
NO2
Aromatic Amines
2
23 Which of the following amine will
give the carbylamine test?
NO2
+
[NEET (Sep.) 2020]
N(CH3)2
NHCH3
(a)
o-nitroaniline
(2%)
25 Identify A and predict the type of
reactions.
[NEET 2017]
(b)
OCH3
NH2
NHC2H5
NaNH2
Ans. (a)
(c)
Aliphatic and aromatic primary amines
when warmed with chloroform and an
alcoholic solution of KOH, form
isocyanide or carbylamine which has
very unpleasant smell.
Warm
CH3CH2NH2 + CHCl 3 + 3KOH →
CH3CH2NC + 3KCl + 3H2O
21 Indicate which nitrogen compound
amongst the following would
undergo Hofmann reaction?
[CBSE AIPMT 1989]
(a) RCONHCH3
(b) RCOONH4
(c) RCONH2
(d) RCONHOH
Ans. (a)
When amides react with bromine in the
presence of caustic alkali to form a
primary amine carrying one carbon atom
less than the parent amide, then the
reaction is known as Hofmann
bromamide reaction.
Heat
RCONH2 + Br2 + 4KOH → RNH2 + K2 CO3
+ 2KBr + 2H2O
NH2
+
288K
TOPIC 2
20 For carbylamine reaction, we need
hot alc. KOH and [CBSE AIPMT 1992]
(a)
(b)
(c)
(d)
Ans. (d)
In strongly acidic medium, aniline is
protonated to form the anilinium ion.
(d)
A
Br
OCH3
Ans. (d)
Carbylamine test is responded by
primary or 1°-amine (aliphatic or
aromatic) only.
NH2
CHCl 3
OCH3
R NH2  → R NC + KCl + H2O
KOH/ ∆
CHCl 3
Ar NH2  → Ar NC + KCl + H2O
KOH/ ∆
Carbyl amine
Options
(a) Ph NH  CH3; 2° amine
(b) Ph N(CH3)2 ; 3° amine
(c) Ph NH  C2H5 ; 2° amine
(d) Ph NH2 ; 1° amine (aniline)
It responds to carbylamine test.
24 Nitration of aniline in strong acidic
medium also gives m-nitroaniline
because
[NEET 2018]
(a) in absence of substituents nitro
group always goes to m-position
(b) in electrophilic substitution reactions
amino group is meta directive
(c) in spite of substituents nitro group
always goes to only m-position
(d) in acidic (strong) medium aniline is
present as anilinium ion
and substitution
reaction
(a)
NH2
and elimination
addition reaction
(b)
OCH3
Br
(c)
and cine substitution
reaction
OCH3
and cine substitution
reaction
(d)
Ans. (a)
OCH3
OCH3
OCH3
NaNH2
Br
NH2
+
NH
2
cine
direct
substitution substitution
221
Organic Compounds Containing Nitrogen
If nucleophile occupies same position of
the leaving group, product is called direct
substitution product.
If nucleophile occupies adjacent position
of the leaving group, product is called
cine substitution product. Intermediate
formed in this reaction is benzyne.
OCH3
OCH3
s
H + NH2
Br
Benzyne
OCH3
OCH3
s
s
+ NH2
Attack of nucleophile
at the original position
(from where Br– leaves)
+ HNH2
NH2
OCH3
NH2
Direct substitution
product
OCH3
OCH3
NH2
s
+ NH2
s
H—NH2
Attack of
nucleophile at
the adjacent
carbon
OCH3
NH2
Cine substitution
product
In III, CH3 group is an electron
donating ando / p directing group which
increase the electron density on
benzene ring at ortho or para position
while in II, NO2 group is an electron
withdrawing group which decrease the
electron density on benzene ring. Hence,
the III is more basic than II.
In I, there is no substituent attached, due
to which I is more basic than II and less
basic than III.
Therefore, the correct order of basic
strength of above compounds is II < I < III.
27 The correct statement regarding
the basicity of arylamines is
[NEET 2016, Phase I]
(a) Arylamines are generally more basic
than alkylamines because the
nitrogen lone-pair electrons are not
delocalized by interaction with the
aromatic ring π-electron system
(b) Arylamines are generally more basic
than alkylamines because of aryl
group
(c) Arylamines are generally more basic
than alkylamines, because the
nitrogen atom in arylamines is
sp-hybridized
(d) Arylamines are generally less basic
than alkylamines because the
nitrogen lone-pair electrons are
delocalized by interaction with the
aromatic ring π-electron system.
Ans. (d)
NH2
(I)
NH2
NO2
(II)
(a) II < III < I
(c) III < II < I
Ans. (d)
NH2
CH3
(III)
(b) III < I < II
(d) II < I < III
Thinking process This type of problem
can be solved by application of electronwithdrawing and electron donating group.
NH2
NH2
NH2
(I)
NO2
(II)
CH3
(III)
R — NH2
Aryl amine
(less basic)
NO2
(b)
CN
(c)
CONH2
(d)
Ans. (b)
The complete road map of the reaction
can be seen as
NO2
+
NH2
–
N2Cl
HNO2
Sn/HCl
Reduction
(A)
(B)
Aniline
Nitro benzene
Benzene diazonium
chloride unstable
N
N
OH
PhOH
Red colour dye
29 The following reaction,
NH2
NaOH
+ Cl
O
H
N
O
NH2
26 The correct increasing order of
basic strength for the following
compounds is
[NEET 2017]
NH2
(a)
Alkyl amine
(more basic)
Due to delocalisation of lone pair of
electrons of N-atom to the benzene ring,
it losses its basicity and becomes less
basic than alkyl amine.
On the other hand, alkyl amine has free
lone pair of electron as well as +I-effect
of alkyl group increases electron density
on N-atom enhancing its basic nature.
28 A given nitrogen-containing
aromatic compound A reacts with
Sn/HCl, followed by HNO 2 to give
an unstable compound B. B, on
treatment with phenol, forms a
beautiful coloured compound C
with the molecular formula
C 12H10 N 2O. The structure of
compound A is [NEET 2016, Phase II]
Ans. (d)
Schotten-Baumann reaction is a
method to synthesise amides from
amines and acid chlorides.
30 Nitrobenzene on reaction with
conc. HNO 3 /H2SO4 at 80-100°C
forms which one of the following
products?
[NEET 2013]
(a) 1, 2-dinitrobenzene
(b) 1, 3-dinitrobenzene
(c) 1, 4-dinitrobenzene
(d) 1, 2, 4-trinitrobenzene
Ans. (b)
NO2 group being electron withdrawing
that’s why it reduces the electron
density at ortho and para-positions.
Hence, as compare to ortho and para
the meta-position is electron rich on
which the electrophile (nitronium ion)
can easily attacks during nitration.
222
NEET Chapterwise Topicwise Chemistry
(a)
HNO2
C6H5NH2 → C6H5 N2+ Cl
(b)
Aryl amine
NH2
Electrophile
(nitronium ion)
NO2
NH2
COOH
HNO 3 + H2 SO 4 → H2NO 3+ + HSO 4−

↓
H2O + ↓NO2+
Br
SO2NH2
CONH2
(c)
(d)
NO2
Br
+ NO2+
COOH
COCl
m-dinitrobenzene
or 1, 3-dinitrobenzene
NH3
SOCl2
31 What is the product obtained in the
following reaction?
Br
B
CONH2
at 0-5°C temperature
NaNO2 + HCl → HNO2 + NaCl
Thus, HNO2 does not convert aryl amines
into phenol.
34 The correct order of increasing
reactivity of C X bond towards
nucleophile in the following
compounds is [CBSE AIPMT 2010]
Br
Ans. (b)
NO2
0 - 5° C
(273-278K)
X
X
Br
NO2
,
NH2
(CH3)3CX (CH)2CHX
IV
III
[CBSE AIPMT 2011]
Zn
NH4Cl
.............?
C
N
(a)
–
O
[CBSE AIPMT 2010]
(a) Alkyl amines are stronger bases than
aryl amines
(b) Alkyl amines react with nitrous acid
to produce alcohols
(c) Aryl amines react with nitrous acid to
produce phenols
(d) Alkyl amines are stronger bases than
ammonia
NH2
(c)
NHOH
(d)
Ans. (d)
Reduction of nitrobenzene with Zn/
NH4Cl (neutral medium) gives phenyl
hydroxyamine.
NO2
Zn/NH4Cl
(Neutral medium)
NH—OH
+H2O
32 In a set of reactions,
m-bromobenzoic acid gave a
product D. Identify the product D.
[CBSE AIPMT 2011]
COOH
Ans. (c)
Key Idea (i) Presence of electron
withdrawing substituent decreases the
basicity while the presence of electron
releasing substituent like,
 CH3,  C2H5 , etc, increases the
acidity.
(ii) HNO2 converts NH2 group of
aliphatic amine into OHwhile that of
aromatic amines into N == NCl.
Since, phenyl group is a electron
withdrawing group, it decreases the
basicity. Alkyl group, on the other hand,
being electron releasing, increases the
basicity. Thus, alkyl amines are more
basic as compared to aryl amines as well
as ammonia.
HNO2
R  NH2 → R OH
Alkyl
amine
SOCl2
Br
D
Br
33 Which of the following statements
about primary amines is false ?
N===N
+
+4[H]
Br
+Br2
The conversion of ‘C’ to ‘D’ is an
example of Hofmann bromamide
degradation reaction.
N
(b)
I
NaOH
NO2
B
NH3
C
NaOH
Br2
D
Thus, HNO2 (nitrous acid) converts alkyl
amines to alcohols.
But aryl amines react with nitrous acid to
form diazonium salt.
NO2
II
(a) I < II < IV < III
(c) IV < III < I < II
Ans. (a)
(b) II < III < I < IV
(d) III < II < I < IV
Key Idea Alkyl halides are more reactive
towards nucleophilic substitution.
Reactivity depends upon the stability of
carbocation intermediate formed.
Among the given halides, aryl halide
(C6H5 X ) is least reactive towards
nucleophile as in it the CX bond acquire
some double bond character due to
resonance. Presence of electron
withdrawing groups like NO2 at ortho
and para-positions facilitate the
nucleophilic displacement of X of aryl
halide. Among alkyl halides, 3° halides are
more reactive as compared to 2° halides
due to the formation of more stable
carbocation. Hence, the order of
reactivity of CX bond towards
nucleophile is as follows:
X
X
NO2
< (CH3)2 CH—X
IV
<
I
NO2
II
< (CH3)3C—X
III
35 Which of the following is more basic
than aniline?
[CBSE AIPMT 2006]
(a) Diphenylamine (b) Triphenylamine
(c) p-nitroaniline (d) Benzylamine
Ans. (d)
••
Benzylamine, C6H5 CH2 —NH2 is more
basic than aniline because benzyl group
(C6H5 CH2 —) is electron donating group
due to +I-effect. So, it is able to increase
223
Organic Compounds Containing Nitrogen
the electron density of N of NH2
group. Thus, due to higher electron
density, rate of donation of free pair of
electron is increased, i.e. basic
character is higher. Phenyl and nitro
group are electron withdrawing groups,
so they decreases the electron density
NHCOCH3
CH3
CH3
on N of —NH2 group. Hence, they are less
basic than aniline.
B
C
Br
H2O/H+
NO2
(a) aniline
(b) nitrosobenzene
(c) N-phenyl hydroxylamine
(d) p-hydroxyaniline
Ans. (a)
+ CH3MgBr
O
CCH3
(b)
OCH3
+H2O
OCH3
CHO
COOH
Aniline
(c)
(d)
OCH3
NH2
OCH3
Ans. (b)
Ac2O
A
Br2
CH3COOH
B
H2O
H+
CH3
|
C==N.MgBr
C
C—
—N
CH3
δ–
would be
NHCOCH3
Br
(a) 1, 3, 5-tribromobenzene
(b) p-bromofluorobenzene
(c) p-bromoaniline
(d) 2, 4, 6-tribromofluorobenzene
Ans. (d)
NH2 group is greatly activating group.
Hence, reaction takes place rapidly.
δ+
NH2
δ–
OCH3
OCH3
+
2H3O
–NH3
–Mg(OH)Br
δ–
δ–
O
C—CH3
CH3
CH3
NH2
COCH3
COCH3
(c)
40 Aniline is reacted with bromine water
and the resulting product is treated
with an aqueous solution of sodium
nitrite in presence of
dilute hydrochloric acid. The
compound so formed is converted
into a tetrafluoroborate which is
subsequently heated. The final
product is
[CBSE AIPMT 1998]
δ+
Br
(b)
(a)
N-methylaniline
+ CH3MgBr
[CBSE AIPMT 2003]
NH2
It is an o,p-directing group.
H3C—C==O
NH2
P
OCH3
NH2
Br
Or
Br
Br
+ 3Br2
OCH3
CH3
CH3
NH2
NHCOCH3
Br2
Ac2O
CH3COOH
CH3
CH3
A
39 An organic compound A on
reduction gives compound B which
on reaction with chloroform and
potassium hydroxide forms C. The
compound C on catalytic reduction
gives N-methylaniline. The
compound A is [CBSE AIPMT 2000]
(a) nitrobenzene
(b) nitromethane
(c) methylamine
(d) aniline
NaNO2
dil. HCl
(water)
(d)
Ans. (a)
Aryl isocyanide
C
Catalytic
4H reduction
OCH3
(a)
37 The final product C, obtained in this
reaction
KOH
Aniline
B
P
[CBSE AIPMT 2002]
NH2
C
H—N—CH3
OH
|
CHCH3
Electrolytic
N
CHCl3
Nitrobenzene
A
Product P in the above reaction is
NO2
NH2
C—
—N
38
36 Electrolytic reduction of
nitrobenzene in weakly acidic
medium gives [CBSE AIPMT 2005]
Reduction in weakly
acidic medium
Ans. (d)
Br
Red P
HI
••
+ 4H
NH2
Br
+
–
N2BF4
N2Cl
Br
Br
NaBF4
Br
Br
(–NaCl)
Br
Br
F
Br
∆
Br
–N2, –BF3
Br
2,4,6-tribromofluorobenzene
224
NEET Chapterwise Topicwise Chemistry
41 A reagent suitable for the
determination of N-terminal
residue of a peptide is
Ans. (a)
Ans. (d)
—CHO +H2N—C6H5
Aniline
[CBSE AIPMT 1996]
(a) p-toluene sulphonyl chloride
(b) 2,4-dinitrophenyl hydrazine
(c) carboxypeptidase
(d) 2,4-dinitrofluorobenzene
Ans. (d)
2,4-dinitrofluorobenzene is called
Sanger’s reagent. When this reagent
reacts with amino group of peptide
chain, it form 2,4-dinitrophenyl
derivatives which on hydrolysis form
DNP derivatives of amino acids.
42 Aniline is an activated system for
electrophilic substitution. The
compound formed on heating
aniline with acetic anhydride is
[CBSE AIPMT 1996]
NH2
(a)
COCH3
(b)
Benzaldehyde
—CH
N—C6H5 + H2O
Benzal aniline
(Schiff’s base, anils)
44 What is the decreasing order of
basicity of 1°, 2° and 3° ethyl
amines and ammonia?
[CBSE AIPMT 1994]
(a) NH3>C2H5NH2 > (C2H5 )2 NH> (C2H5 ) 3N
(b) (C2H5 ) 3N> (C2H5 )2 NH >C2H5NH2 > NH3
(c) (C2H5 )2 NH> C2H5NH2> (C2H5 ) 3N> NH3
(d) (C2H5 )2 NH> (C2H5 ) 3N> C2H5NH2> NH3
Ans. (d)
As the number of alkyl groups increases,
the electron density on nitrogen atom
also increases, so the basic character
increases but 3° amines are less basic
than 2° amines due to steric hindrance
of 3° amines, so the correct order of
basicity is
Ammonia < 1° < 3° < 2°
NH3 < C2H5NH2 < (C2H5 ) 3N < (C2H5 )2 NH
TOPIC 3
COCH3
45 The intermediate compound ‘X’ in
the following chemical reaction is
(d)
Ans. (d)
[NEET 2021]
NH2
CH3
H3O–
CS2
X
+CrO2Cl2
O
NHCOCH3
CH3CO
O
CH3CO
C
Aniline
Acetanilide
43 When aniline reacts with oil of
bitter almonds (C 6H5CHO)
condensation takes place and
benzal derivative is formed. This is
known as
[CBSE AIPMT 1995]
(a) Millon’s base
(b) Schiff’s reagent
(c) Schiff’s base
(d) Benedict’s reagent
CH(OCrOHCl2)2
(a)
CH(OCOCH3)2
(b)
CH
(c)
CH
(d)
CH(OCrOHCl2)2
(X)
H3O
CHO
+
Benzaldehyde
46 Consider the nitration of benzene
using mixed conc. H2 SO 4 and
HNO 3 . If a large amount of KHSO 4
is added to the mixture, the rate of
nitration will be [NEET 2016, Phase I]
(a) slower
(b) unchanged
(c) doubled
(d) faster
Ans. (d)
In the nitration of benzene in the
presence of conc.H2SO4 and HNO3,
nitrobenzene is formed.
HNO3 + H2SO4 r NO2+ +
HSO−4 +
Electrophile
Diazonium Salts
NHCOCH3
CH3
CS
+ CrO2Cl2 2
Toluene
NH2
(c)
This is Etard reaction in which reaction
of toluene with chromyl chloride in CCl 4
followed by hydrolysis gives
benzaldehyde. Toluene reacts with
chromyl chloride to form a precipitate
called the Etard complex.
Cl
Cl
Cl
H
H
H2O
Nucleophile
If large amount of KHSO 4 is added to this
mixture, moreHSO−4 ion furnishes and
hence the concentration ofNO2+ , i.e.
electrophile decreases.
As concentration of electrophile
decreases, rate of electrophilic
aromatic reaction also decreases.
47 Which of the following will be most
stable diazonium salt RN +2 X − ?
[CBSE AIPMT 2014]
(a) CH3N2+ X −
(b) C 6H5N2+ X −
(c) CH3CH2N2+ X −
(d) C 6H5CH2N2+ X −
Ans. (b)
Diazonium salt containing aryl group
directly linked to the nitrogen atom is
most stable due to resonance
stabilisation between the benzene
nucleus and N-atom. Diazonium ion
act as a electrophile.
225
Organic Compounds Containing Nitrogen
+
N
+
N
N
N
+
–
N
N
N==NCl
–
H3C
CH3
CH3
N
NH
+ NaNO2
+ HCl or HNO2
+
+
N-methylaniline
(2°amine)
+
+
N
+
N
N
–
N
N N
H
X
CH3
—N
—N==N—
N-nitroso-N-methylaniline
(nitroso compound)
(yellow oily liquid)
CH3
[Resonance structure of benzene diazonium ion]
Coupling product
Y
48 Aniline in a set of the following
reactions yielded a coloured
product Y.
[CBSE AIPMT 2010]
50 In a reaction of aniline a coloured
products C was obtained.
49 Predict the product,
[CBSE AIPMT 2009]
NHCH3
NH2
NH2
+ NaNO2+HCl
X
N, N-dimethylaniline
CH3
—N==N
CH3
N N
(a)
CH3
CH3
(b) HN
—NH
Cold
N
(a)
NCH2
N
CH3
—NH
NH2
N==N
CH3
C
[CBSE AIPMT 2008]
NNO2
(b)
NHCH3
(c) H3C
CH3
B
The structure of C would be
O
CH3
—N
CH3
NaNO2
HCl
Product
CH3
—N
A
Y
The structure of Y would be
(a)
O
N,N-dimethylaniline
+
NaNO2/HCl
(273-278 K)
CH3
NHCH3
CH3
NO
(c)
+
CH3
N
(b)
N
CH3
 N==N 
(d) HN
NO
—NH
OH
Ans. (a)
Key Idea NaNO2 /HCl causes
diazotisation of —NH2 group and the
diazonium chloride gives a coupling
product with active aryl nucleus.
NH2
N==NCl
NaNO2 /HCl
(273-278-K)
X
(d)
NHNH
(c)
N
CH3
CH3
NCH3
—N==N 
(d)
N
CH3
CH3
Ans. (a)
Both aliphatic and aromatic secondary
amines reacts with
NaNO2 + HCl or (HNO2 ) to form
N-nitrosamines which are insoluble in
dilute mineral acids and separate out as
neutral yellow oily compounds.
Ans. (d)
NH2
N
NaNO2
+ –
NCl
HCl
273-278 K
A
Aniline
B
Benzene diazonium chloride
226
NEET Chapterwise Topicwise Chemistry
H
N
Ans. (d)
Ch3
CH3
NH2
N
Cold
N
CuCN
(Sandmeyer's
reaction)
51 Aniline in a set of reactions yielded
a product
[CBSE AIPMT 2005]
NH2
NaNO2
HCl
A
CuCN
B
H2
Ni
HNO2
The structure of the product D
would be
(a) C 6H5CH2NH2
(b) C 6H5NHCH2CH3
(c) C 6H5NHOH
(d) C 6H5CH2OH
H2 /Ni
(Reduction)
B
(c) (CH3)2N
N==N 
(d) (CH3)2N
N==N 
—CH3
Ans. (c)
CH2NH2
CH2OH
HNO2
Benzyl amine
NH2
CN
Aniline on diazotisation in cold (at0° to
5° C) gives benzene diazonium chloride.
0-5°C
NH2 + NaNH2 + 2HCl Diazotisation
C
D
NHCH3
A
Cyanobenzene
C
N==N 
N==N 
(b) CH3
Benzene diazonium
chloride
CH3
CH3
C
+
NaNO2
HCl
(Diazotisation)
p-(N,N-dimethyl) amine azobenzene
(azodye)
N
(a) CH3NH
+ –
N2Cl
D
Benzyl alcohol
52 Aniline when diazotised in cold and
then treated with dimethyl aniline,
gives a coloured product. Its
structure would be
[CBSE AIPMT 2004]
+ _
N==N—Cl + NaCl + 2H 2 O
Benzene diazonium
chloride
This benzene diazonium chloride on
coupling with dimethyl aniline gives a
coloured product, i.e. p-(N,N-dimethyl)
amino azobenzene (azo dye).
26
Polymers
TOPIC 1
Isoprene shows 1, 4-addition with
themselves to give cis-1, 4-polyisoprene
or natural rubber.
CH3
CH3
Classification of
Polymers
01 Which of the following statement is
correct about bakelite?
----CH2==CCH==CH2 CH2==C—CH==CH2---1
2
1′
3 4
[NEET (Oct.) 2020]
(a) It is a cross linked polymer.
(b) It is an addition polymer.
(c) It is a branched chain polymer.
(d) It is a linear polymer.
Ans. (a)
Bakelite is a cross linked condensation
thermosetting copolymer of phenol and
formaldehyde.
OH
O
+ CH2
CH2
Laderer Manasse
reaction
CH2
CH2
CH2
OH
H3C
H
H3C
C==C
CH2
1
2 3
CH2
CH2
2′ 3′
1′
CH2
OH
Bakelite or Phenol
formaldehyde resin
02 Which of the following is a natural
polymer?
[NEET (Sep.) 2020]
(a) Poly (Butadiene-styrene)
(b) Polybutadiene
(c) Poly (Butadiene-acrylonitrile)
(d) Cis-1, 4-polyisoprene
Ans. (d)
CH2
4′
Natural rubber
All polymers mentioned in options (a), (b)
and (c) are synthetic 1,4-addition
polymers. Where (a) and (c) are buna-S
and buna-N respectively.
(a) melamine
(b) nylon-6, 6
(c) polyacrylonitrile (d) buna-N
Ans. (c)
Natural polymer (soft) is an addition
homopolymer of isoprene which is a
conjugated diene.
homopolymer of monomer CH2 == CHCN
(vinyl cyanide). It is used in making
synthetic fibres and synthetic wool.
Thus, it is a substitute for wool in making
commercial fibres.
04 The biodegradable polymer is
[NEET (National) 2019]
(a) nylon-2-nylon-6 (b) nylon-6
(c) buna-S
(d) nylon-6,6
Ans. (a)
Nylon-2-nylon-6
O


( HN  CH2 CONH(CH2 ) 5 C 
)n

( CH2  CH == CH  CH2  CH 
)n
are non-biodegradable polymers.
Hence, option (a) is correct.
H
C==C
Polyacrylonitrile or orlon or acrilan,
CN

( CH2  CH 
) n is an addition
CH2
CH2
4′
[NEET (Odisha) 2019]
OH
CH2
3′
03 The polymer that is used as a
substitute for wool in making
commercial fibres is
–
H+or OH (catalyat)
OH
2′
1, 4-addition
polymerisation
glycine (H2N  CH2  COOH) and amino
caproic acid (H2N(CH2 ) 5 COOH).
The remaining polymers, i.e. nylon-6,6,
( NH(CH2 ) 6 NHCO(CH2 ) 4 CO)
n ,
nylon-6- ( CO(CH2 ) 5 NH 
) n and buna-S
C6H5
is a
biodegradable polymer. It is an
alternating polyamide copolymer of
05 Regarding cross-linked or network
polymers, which of the following
statements is incorrect? [NEET 2018]
(a) Examples are bakelite and melamine
(b) They are formed from bi- and
tri-functional monomers
(c) They contain covalent bonds
between various linear polymer chains
(d) They contain strong covalent bonds
in their polymer chains
Ans. (d)
Cross-linked or network polymers are
formed from bi-functional and
tri-functional monomers and contain
strong covalent bonds between various
linear polymer chains. These are hard,
rigid and brittle due to cross-links
e.g. bakelite, melamine etc. Thus, option
(d) is incorrect.
06 Natural rubber has
[NEET 2016, Phase I]
(a) All trans-configuration
(b) Alternate cis- and transconfiguration
(c) Random cis- and trans-configuration
(d) All cis-configuration
Ans. (d)
The repeating unit in natural rubber has
the cis-configurations with chain
extensions on the same side of the
ethylene double bond, which is essential
for elasticity. If the configuration is
228
NEET Chapterwise Topicwise Chemistry
trans, the polymer is either a hard plastic
or a substance like gutta-percha.
CH3
C == C
H2C
H
CH2
CH2
CH2
C == C
H
H3C
H3C
C == C
CH2
H
CH2
07 Which one of the following is an
example of a thermosetting
polymer?
[CBSE AIPMT 2014]
(CH2  C ==CH  CH2
)n

Cl
(b) 
( CH2  CH 
)n

Cl
H
H O

 
(c) (N  (CH2 ) 6  N  C
(a)
12 Of the following which one is
classified as polyester polymer?
[CBSE AIPMT 2011]
(d) CH2 === CH  CH === CH2
Ans. (c)
Neoprene is synthetic rubber and is a
polymer of chloroprene which is
chemically 2-chlorobuta- 1,3-diene.
n CH2
C
CH
CH2
Cl
CH 2 C
CH
CH 2
Cl
n
Neoprene
09 Nylon is an example of
OH
CH2
CH2
Nylon-66 is a fibre not a elastomer. As in
it the forces of attraction are H-bonding
. All other given statements are true.
(c) CH2 === C  CH === CH2

Cl
Chloroprene
O

 (CH2 ) 4  C)
n
OH
Ans. (b)
(b) CH2 ===C  CH === CH2

CH3
(d)
(a) polyester
(c) polyamide
Ans. (d)
(b) polysaccharide
(d) polythene
The general structure of any nylon
polymer is
O
R
n
Ans. (d)
[NEET 2013]
C
N
R
n
H
Amide linkage
OH
OH
CH2
Because of the presence of amide
linkage, nylon belongs to polyamides.
CH2
n
Novolac, a condensation polymer of
phenol and formaldehyde is a
thermosetting polymer.
Neoprene rubber
(a) 
[ CH2  C == CH  CH2 
]n and PVC

Cl
(b) 
[ CH2  CH 
]n

Cl
are thermoplastic polymers while
nylon-66
O

(c) 
[ NH  (CH2 ) 6 NH  C  (CH2 ) 4
O

C 
] n is a polyamide which is
commonly known as fibre.
08 Which is the monomer of neoprene
in the following?
[NEET 2013]
(a) CH2 ==CH C ==CH
10 Which one of the following is not a
condensation polymer?
[CBSE AIPMT 2012]
(a) Melamine
(c) Dacron
Ans. (d)
(b) Glyptal
(d) Neoprene
Condensation polymers are obtained by
bifunctional molecules (monomers) with
the elimination of smaller molecules
whereas additional polymers are
obtained from multiple bond containing
monomers. Neoprene is a polymer of
chloroprene (CH2 ==C(Cl)  CH==CH2 )
so it is an addition polymer, not a
condensation polymer.
11 Which of the following statements
is false?
[CBSE AIPMT 2012]
(a) Artificial silk is derived from cellulose
(b) Nylon-66 is an example of elastomer
(c) The repeat unit in natural rubber is
isoprene
(d) Both starch and cellulose are
polymers of glucose
(a) Bakelite
(b) Melamine
(c) Nylon-66
(d) Terylene
Ans. (d)
Terylene (or dacron) is a polyester
because it contains ester groups and
formed by the monomer units
terephthalic acid and ethylene glycol
CH2 — OH




CH2 — OH


13 Structures of some common
polymers are given. Which one is
not correctly presented?
[CBSE AIPMT 2009]
(a) Teflon 
( CF2 —CF2 
)n
(b) Neoprene
 CH2 C ===CH CH2 CH2 







Cl

n
(c) Terylene
COOCH2CH2O
)n
—OC
(
(d) Nylon 66
—
[ NH(CH2 ) 6 NHCO(CH2 ) 4 — CO —]n
Ans. (b)
Neoprene is a polymer of chloroprene
(2-chloro-1,3-butadiene) and also
called homopolymer addition
polymer).
nCH2
CCH
CH2
Polymerisation
Cl
Chloroprene
CH2C
CHCH2
Cl
Neoprene
(synthetic rubber)
n
14 [NH(CH2 ) 6 NHCO(CH2 ) 4 CO]n
is a
[CBSE AIPMT 2006]
(a) copolymer
(b) addition polymer
(c) thermo-setting polymer
(d) homopolymer
229
Polymers
Ans. (a)

[ NH(CH2 ) 6 NHCO(CH2 ) 4 CO 
] n is a
copolymer. Polymers whose repeating
structural units are derived form two
or more types of monomer units are
called copolymer
n H2N(CH2 ) 6 NH2 + nHOOC(CH2 ) 4 COOH
Polymerisation
→
–nH2O


—
 NH(CH2 ) 6 NHCO(CH2 ) 4 CO 


(a) butadiene
(b) ethyne
(c) styrene
(d) isoprene
Ans. (d)
Polyisoprene is the natural rubber
which is the polymer of isoprene.
[CBSE AIPMT 2002]
(b) fructose
(d) sucrose
Polyisoprene (natural rubber) n
(polymer)
18 In elastomer, the intermolecular
forces are
[CBSE AIPMT 1996]
16 Which one of the following is not
correctly matched?
[CBSE AIPMT 2001]





n
(a) strong
(c) nil
Ans. (b)
(b) weak
(d) None of these
In elastomers, the polymer chains are
held together by weak van der Waals’
forces, e.g. natural rubber.
TOPIC 2
NH(CH 2) 6NH
CO(CH 2) 4CO 
Methods of
Polymerisation
n
(c) Terylene
O
—OCH2CH2C
C
n


CH3




 CH2  C 
(d) PMMA 





COOCH
3 n

(a)
O
C—
n
Terylene (Dacron)
Hence, the structure of terylene given in
question is incorrect.
(b)
n
Ans. (d)
Nylon-6, 6 polymer is formed as
HOOC(CH2)4COOH + H2N(CH2)6NH2
Adipic acid
(c)
H2 H
C C
NH3
66
H2 H
C C
CH2
O
O
C (CH2)4CNH (CH2)6NH
Nylon-6,6
n
Thus, option (d) is correct.
20 Biodegradable polymer which can be
produced from glycine and
aminocapric acid is
[CBSE AIPMT 2015]
(a) nylon 2-nylon 6
(b) PHBV
(c) buna-N
(d) nylon-6, 6
Ans. (d)
Nylon-2-nylon-6
It is an alternating polyamide of glycine
(containing two carbon atoms) and
amino caproic acid or 6-aminohexanoic
acid (containing six carbon atoms).
nH2N  CH2  COOH
Amino caproic acid


NH CH2  C NH( CH2 )5  C




O
O





 n
21 Caprolactum is used for the
manufacture of
[CBSE AIPMT 2015]
66
(a) nylon-6
(c) terylene
Ans. (a)
O
H2 H
C C
NH2
Hexamethylene
diamine
It is a biodegradable step-growth
copolymer.
H2 H
C C
NH2
NH
Nylon - 2 - nylon - 6
CH3
H2 H
C C
(CH2)6
2
+ nH2N  (CH2 ) 5  COOH →
H2 H
C C
NH2
Ans. (c)
Terylene is formed by the condensation
of dimethyl terephthalate and glycol.
Its structure is
O
O
||
||
OCH2CH2OC
H2 H
C C
H
N
Glycine
19 Which one of the following
structures represents nylon 6, 6
polymer?
[NEET 2016, Phase II]
O
C
H2
H2
C C
Polymerisation
→




—

 CH2 — CH == C  CH2 —
|




CH3
Cellulose is a polymer of glucose, i.e.
C6H12O6 .
(a) Neoprene



 CH2  C == CH — CH2
|


Cl
(b) Nylon-66
C
(d)
Isoprene
(monomer)
Polymerisation
15 Cellulose is a polymer of
(a) glucose
(c) ribose
Ans. (a)
[CBSE AIPMT 1999]




n—

 CH2== CH — C ==CH2 —
|




CH3
n
O
17 Natural rubber is a polymer of
Cl
6
H2 H
C C
NH
6
COOH
(b) teflon
(d) nylon-6 6
O
H
N
533K
N2
n
Nylon-6
230
NEET Chapterwise Topicwise Chemistry
22 Which of the following organic
compounds polymerises to form
the polyester dacron?
[CBSE AIPMT 2014]
(a) Propylene and para
HO  (C6H4 ) OH
(b) Benzoic acid and ethanol
(c) Terephthalic acid and ethylene glycol
(d) Benzoic acid and para
HO  (C6H4 ) OH
Ans. (c)
Dacron, commonly known as terylene, is
obtained by heating a mixture of
terephthalic acid and ethylene glycol at
420-460 K in the presence of zinc
acetate and antimony trioxide as a
catalyst.
COOH
nHOOC
Terephthalic acid
+ nHOCH2CH2OH
Ethylene glycol
∆
– HCl
[CBSE AIPMT 2008]
(a) In vulcanisation, the formation of
sulphur bridges between different
chains make rubber harder and
stronger
(b) Natural rubber has the
trans-configuration at every double
bond
(c) Buna-S is a copolymer of butadiene and
styrene
(d) Natural rubber is a 1,4-polymer of
isoprene
Ans. (b)
Natural rubber is cis-1, 4-polyisoprene
and has all cis configurations about the
double bond as shown below. It is
prepared from latex which is obtained in
cis form called Havia Rubber latex is
obtained from rubber tree (Havea
brasiliensis).
H 3C
C
H 2C

[ O  CH2  CH2  O—CO
Polyesteric bond
24 Which one of the following
statements is not true?
(a) Starch
(b) Nucleic acid
(c) Polystyrene
(d) Protein
Ans. (c)
Chain growth polymerisation requires an
initiator (such as organic peroxides) to
produce a free radical to which the
monomers are added in a chain fashion.
Initiators are added in a very small
quantities and are decomposed by heat,
light or oxidation-reduction reaction to
produce reactive species, e.g. free
radical.
Polystyrene is an example of chain
growth polymer because in it styrene
molecules are associated in the form of
monomer
CH 2CH 3
CH 2== C H 2
AlCl3
H
CH 2
C
CH 2
Dacron
C
H 3C
C
CH 2
CO ] n
CH==CH2
CH 2
C
H 3C
H
H
C
CH 2
Fe2O 3 /C r 2O 3
650°C
—— CHCH2
25 The monomer of the polymer
23 Which of the following structures
represents neoprene polymer?
[CBSE AIPMT 2010]
(a) (CH2 C == CH CH2 )
n

Cl
CN
|
(b) —( CH2 CH—
)n
Cl
|
(c) —
( CH2  CH—)n
[CBSE AIPMT 2005]
(a) H2C == C
CH3
CH3
The monomer of polymer
CH3
CH3

+
is
— CH2 — C — CH2 — C

CH3
CH3
Ans. (a)
Neoprene (synthetic rubber) is a
polymer of chloroprene, i.e.
2-chloro-1, 3-butadiene.
Polymerisation
CH2 == C
Cl H
CH3
because
CH3
2-chloro-1,3-butadiene
(chloroprene)
2-methylpropene shows cationic
polymerisation.
CH2—C==C—CH2
Cl
H
Neoprene
(synthetic rubber)
n
( C 6 H 5CO )2O 2
Polystyrene
n
27 Which one of the following
monomers gives the polymer
neoprene on polymerisation?
[CBSE AIPMT 2003]
(b) (CH3)2 C == C(CH3)2
(c) CH3CH ==CH⋅CH3
(d) CH3CH == CH2
Ans. (a)
(d) —( CH CH2—) n
|
C 6H5
nH2C==C—C==CH2
CH3
CH3

+
is
CH2  C  CH2  C

CH3
CH3
26 Which one of the following is a
chain growth polymer ?
[CBSE AIPMT 2004]
Cl
|
(a) CH2 == C CH == CH2
(b) CF2 ==CF2
(c) CH2 == CHCl
(d) CCl2 == CCl2
Ans. (a)
Neoprene is an addition polymer of
chloroprene or chloro-1,3-butadiene
(monomer).
Cl

Polymerisation
nCH2 == CH  C == CH2 →
Chloroprene

Cl



 CH2  CH ===C  CH2







n
231
Polymers
28 Acrilan is a hard, horny and a high
melting material. Which of the
following represents its structure?
[CBSE AIPMT 2003]
 CH2 CH 



|
(a) 
COOC2H5 



n
Ans. (b)
Ans. (b)
F2 C == CF2 is a monomer of well known
plastic teflon, a material inert to almost
all chemicals.
The monomer units of nylon-66 are
obtained by the reaction of
hexamethylene diamine and adipic
acid.
[
n
F2C== CF 2 
]
Polymerisation
H2O2, ∆
Monomer
(Tetra-flouroethene)
 —CH2 — CH —


|
(b) 



Cl

n
34 Bakelite is prepared by the reaction
between
[CBSE AIPMT 1995]
(a) urea and formaldehyde
(b) ethylene glycol
(c) phenol and formaldehyde
(d) tetramethylene glycol
Ans. (c)
Bakelite is a polymer obtained by the
condensation reaction between
phenol and formaldehyde. It is a
condensation polymer and basic unit
of Bakelite is Novolac.
(Teflon polymer)
CH2 — CH —


|
(c) 
CN 


n
CH3
|
(d) — CH2 — C 
|
COOCH3
F F
| |
— C— C—
| |
F F n
31 Terylene is a condensation polymer
of ethylene glycol and
[CBSE AIPMT 1999]
(a) benzoic acid
(c) salicylic acid
Ans. (d)
(b) phthalic acid
(d) terephthalic acid
Terylene is a condensation polymer of
ethylene glycol and terephthalic acid. It
is also called polyester
n
35 The reagent ‘R’ in the given
sequence of chemical reaction is
[NEET 2021]
Ans. (c)
Acrilan (or acrylonitrile) is monomer unit
of polyacrylonitrile (PAN). Its structure is
 CH2 — CH — 


|


CN


n

CH3


|
29 Monomer of  — C — CH2
|

CH3




 is
—


n
[CBSE AIPMT 2002]
(a) 2-methylpropene
(b) styrene
(c) propylene
(d) ethene
Ans. (a)

 CH3

|

 C— CH2—
is
Monomer of —

|


n
 CH3
2-methylpropene
or isobutene, H3C  C == CH2
|
CH3
30 CF2 == CF2 is a monomer of
[CBSE AIPMT 2000]
(a) buna-S
(b) teflon
(c) glyptal
(d) nylon-6
n[HOCH2CH2OH]+ n HOOC
Ethylene glycol
N2CI–
Br
Br
Terephthalic acid
Condensation
Polymerisation
Br
NaNO2, HCl
0-5°C
Br
Br
—OCH2CH2OC
||
O
Esteric
bond
+
NH2
COOH
Terylene
Br
C—
||
O n
Br
R
Br
Br
(a) H2O
(c) HI
Ans. (d)
32 Which one of the following is used
to make ‘non-stick’ cookware ?
[CBSE AIPMT 1997]
(a) PVC
(b) Polystyrene
(c) Polyethylene terephthalate
(d) Polytetrafluoro ethylene
Ans. (d)
Polytetrafluoro ethylene (C2F4 ) n or
teflon is used to make non-stick
cookware, because it is a tough
material, resistant to heat and also
the bad conductor of electricity.
(b) CH3CH2OH
(d) CuCN/KCN
Mild reducing agents like alcohol are
used to reduce diazonium salts to arene.
Alcohol (ethanol is oxidised to aldehyde
(ethanal).
N2Cl
NH2
Br
Br
NaNO2+HCl
Br
Br
0-5°C
Br
Br
2, 4, 6 tribromoaniline
33 Nylon-66 is a polyamide obtained
by the reaction of
CH3CH2OH
Br
Br
[CBSE AIPMT 1996]
(a) COOH(CH2 ) 4 COOH +H2NC 6H4NH2- (p)
(b) COOH(CH2 ) 4 COOH + NH2 (CH2 ) 6 NH2
(c) COOH(CH2 ) 6 COOH + NH2 (CH2 ) 4 NH2
(d) COOHC 6H4COOH- (p) + NH2 (CH2 ) 6 NH2
Br
1, 3, 5.-tribromobenzene
∴ Reagent used is CH3CH2OH (ethanol).
27
Biomolecules, Chemistry
in Everyday Life and
Environmental Chemistry
TOPIC 1
Biomolecules
Ans. (a)
01 Given below are two statements.
Statement I Aspirin and
paracetamol belong to the class of
narcotic analgesics.
Statement II Morphine and heroin
are non-narcotic analgesics.
In the light of the above statements,
choose the correct answer from the
options given below.
[NEET (Oct.) 2021]
(a) Both Statement I and Statement Il
are true.
(b) Both Statement I and Statement lI
are false.
(c) Statement I is true but Statement Il
is false.
(d) Statement I is false but Statement I
is true.
Ans. (b)
Narcotic analgesics They are the
analgesics drugs that are used to reduce
pain. They are obtained from opium
poppy, so they are also called opiates.
They are addictive in nature. Morphine
and heroin are narcotic analgesics.
Non-narcotic analgesics They are
non-addictive in nature and are not
obtained from opium poppy. Aspirin and
paracetamol are non-narcotic analgesics.
So, both statements I and II are false.
02 The reaction of concentrated
sulphuric acid with carbohydrates
(C 12H22O 11 ) is an example of
[NEET (Oct.) 2020]
(a) dehydration
(c) reduction
(b) oxidation
(d) sulphonation
Conc.H2SO4 removes water, i.e.
dehydrates carbohydrates into carbon
(black residue) or charred sugar.
Conc.H 2 SO 4
→ 12 C
C12H22O11 or C12 (H2O) 11  
− 11 H 2 O
(Black )
03 Deficiency of which vitamin causes
osteomalacia? [NEET (Oct.) 2020]
(a) Vitamin A
(b) Vitamin D
(c) Vitamin K
(d) Vitamin E
Ans. (b)
[NEET (Oct.) 2020]
(a) It is an aldohexose.
(b) It contains five hydroxyl groups.
(c) It is a reducing sugar.
(d) It is an aldopentose.
Ans. (d)
Glucose (C6H12O6 ) can be simply shown as,
⇒Shows reducing property.
So, glucose is an aldohexose (made of
six carbon atoms) and it is a reducing
sugar. Containing five OHgroups.
Hence, option (d) is incorrect statement.
one
1°– OH group
(b) Tyrosine
(d) Serine
(a) Alanine ⇒ CH3CH
NH2
COOH
(b) Tyrosine ⇒
—CH2—CH
NH2
COOH
(c) Lysine ⇒ H2N(CH2)4CH
04 Which of the following statements
is not true about glucose?
CH2
(a) Alanine
(c) Lysine
Ans. (c)
HO
Deficiency of
Vitamin A causes—Xerophthalmia
Vitamin D causes—Osteomalacia and
Rickets
Vitamin E causes—Fragility of RBCs
Vitamin K causes—Increase in blood
clotting time.
Hence, option (b) is correct.
HO
05 Which of the following is a basic
amino acid?
[NEET (Sep.) 2020]
(CHOH)4
Four
2°– OH group
CH
O
aldehyde
group
(d) Serine ⇒ HOCH2CH
NH2
COOH
NH2
COOH
Since, lysine contains more number of
—NH2 groups as compared to —COOH
groups, hence it is a basic amino acid.
06 The non-essential amino acid
among the following is
[NEET (National) 2019]
(a) leucine
(c) lysine
Ans. (b)
(b) alanine
(d) valine
The amino acids that can be synthesised
in our body and hence are not essentially
required in our diet are called
non-essential amino acids. e.g. glycine,
alanine, serine, proline, cysteine,
glutamine, tyrosine, aspartic acid,
glutamic acid, asparagine. Rest given
options are essential amino acids, i.e.
those can’t be synthesised in our body
hence essentially required in our diet.
Their other examples are isoleucine,
phenylalanine, methionine, tryptophan,
threonine, arginine and histidine.
Hence, option (b) is correct.
233
Biomolecules, Chemistry in Everyday Life and Environmental Chemistry
CH2OH
07 Which of the following compounds can form a Zwitter
ion?
[NEET 2018]
(a) Benzoic acid
(c) Aniline
Ans. (d)
Zwitter ion
−
OH
H+
−
H  C H  COO ← H  C H  COO


NH3
NH2
→ H C H  COOH

NH3
+
+
Anion in basic
medium
Cation in acidic
medium
Thus, amino acid bears a positive charge in acidic solution (low
pH) and a negative charge in basic solution (high pH). The pH at
which the amino acid has no net charge is called isoelectric point.
The isoelectric point of glycine is 5.97.
08 The difference between amylose and amylopectin is
[NEET 2018]
OH
H
CH2OH
O
H H
H
4
1
OH
H
H
OH
H H
O
5
CH2OH
O
4
1
OH
3
H
O
O
H
4
1
OH
H
H
OH
O—
2
H
OH
a-link
OH
H
H
4
1
a-Link
OH
Branch at C6
OH
H
O
6
CH2OH
H H
—O
CH2
O
H
4
1
OH
H
H H
O
5
H
4
1
OH
3
H
CH2OH
O
OH
H
O
H
H H
O
4
1
OH
H
H
OH
O—
2
H
a-Link
OH
a-Link
Amylopectin
09 Which of the following statements is not correct?
[NEET 2017]
(a) Insulin maintains sugar level in the blood of a human body
(b) Ovalbumin is a simple food reserve in egg white
(c) Blood proteins thrombin and fibrinogen are involved in blood
clotting
(d) Denaturation makes the proteins more active
Ans. (d)
Deprotonation of protein occur when it is subjected to physical
change like change in temperature or chemical change like
change in pH, the hydrogen bonds are disturbed. As a result,
globules unfolds and helix get uncoiled and protein losses its
biological activity. Hence, the denaturation of protein makes the
protein inactive.
10 In a protein molecule various amino acids are linked
together by
[NEET 2016, Phase I]
(a) β-glycosidic bond
(c) dative bond
Ans. (b)
(b) peptide bond
(d) α-glycosidic bond
Two amino acids in a protein are linked by a peptide bond.
CH3
Alanine
H2N  CH2  CO  NH  CH  COOH
Peptide linkage
CH3
Glycylalanine (Gly-Ala)
H H
H
O
–H2O
Starch contains two components amylose and amylopectin.
Chemically, amylose is a long unbranched chain with 200-1000
α-D- (+ )-glucose units held by C1-C4 glycosidic linkage.
6
O
H2N  CH2  CO OH + H2 N  CH COOH
Ans. (c)
—O
1
5
H H
e.g. glycylalanine is formed when carboxyl group of glycine
combines with the amino group of alanine.
amylopectin have 1 → 4 α-linkage and 1 → 6 β-linkage
amylose have 1 → 4 α-linkage and 1 → 6 β-linkage
amylopectin have 1 → 4 α-linkage and 1 → 6 α-linkage
amylose is made up of glucose and galactose
CH2OH
4
H
Zwitter ion is a cation in acidic medium and migrates to cathode
on passing electric current. It is an anion in basic medium and
migrates to anode on passing electric current.
(a)
(b)
(c)
(d)
H
2
Key Concept Ion containing positive as well as negative charge is
called Zwitter ion.
Among the given options, only glycine (H2N  CH2  COOH) is an
amino acid which contains both acidic (acquiring negative
charge) and basic group (acquiring positive charge).
Glycine can form a Zwitter ion. It is because glycine behave like
salts rather than simple amines or carboxylic acids. In aqueous
solution, the carboxyl group can lose a proton and amino group
can accept a proton giving rise to a dipolar ion known as Zwitter
ion.
H

H  C  COO−

NH3
+
−
H H
—O
(b) Acetanilide
(d) Glycine
CH2OH
O
a-link
Amylose
Amylopectin is a branched chain polymer ofα -D-glucose units in
which chain is formed by C1-C4 glycosidic linkage where
branching occurs by C1-C6 glycosidic linkage.
11 The correct statement regarding RNA and DNA,
respectively is
[NEET 2016, Phase I]
(a) The sugar component in RNA is ribose and the sugar
component in DNA is2′-deoxyribose
(b) The sugar component in RNA is arabinose and the sugar
component in DNA is ribose
(c) The sugar component in RNA is2′-deoxyribose and the sugar
component in DNA is arabinose
(d) The sugar component in RNA is arabinose and the sugar
component in DNA is2′-deoxyribose
234
NEET Chapterwise Topicwise Chemistry
Ans. (a)
In DNA, two helically twisted strands
connected together by steps. Each
strand consists of alternating molecules
of deoxyribose at 2′-position and
phosphate groups.
On the other hand, in RNA, the pentose
sugar has an identical structure with
deoxyribose sugar except that there is
an —OH group instead of —H on carbon
atom 2′.
Hence, it is only called ribose.
12 Which one given below is a
non-reducing sugar?
[NEET 2016, Phase I]
(a) Lactose
(c) Sucrose
Ans. (c)
(b) Glucose
(d) Maltose
Sucrose is non-reducing sugar because
O

reducing part of glucose (  C H) and
fructose (> C == O) are involved in
glycosidic linkage.
14 The correct corresponding order of
names of four aldoses with
configuration given below
HO
O
HO
HO
HO
H
H
OH
CH2OH
CHO
H
H
H
HO
OH
H
CH2OH
CH2OH
respectively, is [NEET 2016, Phase II]
(a) L-erythrose, L-threose, L-erythrose,
D-threose
(b) D-threose, D-erythrose, L-threose,
L-erythrose
(c) L-erythrose, L-threose, D-erythrose,
D-threose
(d) D-erythrose, D-threose, L-erythrose,
L-threose
Ans.
CHO
CHO
H
OH
HO
H
OH
H
H
H
H
OH
CH2OH
CH2OH
D-erythrose
OH
H
OH
OH
CH2OH
CHO
CH2OH
H
CHO
CHO
H
H
H
O
HO
H
H
CHO
HO
H
H
HO
H
HO
O
HOH2C
OH
H
CH2OH
H
OH
Sucrose
While, lactose, glucose and maltose are
reducing sugars.
(a) amino acids → proteins → DNA
(b) DNA → carbohydrates→ proteins
(c) DNA → RNA → proteins
(d) DNA → RNA → carbohydrates
Ans.
15 D-(+)-glucose reacts with hydroxyl
amine and yields an oxime. The
structure of the oxime would be
[CBSE AIPMT 2014]
CH
(a)
(c) The central dogma of molecular
genetics states that
Transcription
DNA → RNA
Translation
→ Protein
Thus, option (c) is correct.
L-threose
Thus, the correct option is (d).
13 The central dogma of molecular
genetics states that the genetic
information flows from
[NEET 2016, Phase II]
CH2OH
L-erythrose
OH CH2OH
(c)
NOH
HO
C
H
HO
C
H
H
H
C
OH
OH
H
C
OH
C
OH
HO
C
H
HO
C
H
C
(b)
CH2OH
CH2OH
CH
CH
NOH
HO
C
H
H
C
OH
HO
C
H
H
CH
NOH
H
C
OH
CH2OH
H
C
OH
HO
C
H
+ NH2OH
H
C
OH
H
C
OH
H
C
OH
HO
C
H
H
C
OH
H
C
OH
– H2O
CH2OH
Glucoxime
CH2OH
D-(+)-glucose
16 Which of the following hormones is
produced under the condition of
stress which stimulates
glycogenolysis in the liver of human
beings?
[CBSE AIPMT 2014]
(a) Thyroxin
(c) Adrenaline
Ans. (c)
(b) Insulin
(d) Estradiol
Adrenaline hormones increases pulse
rate and controls blood pressure. It
releases glucose from liver glycogen and
fatty acids from fats in emergency.
D-threose
CHO
Glycosidic
linkage
Ans. (d)
D-(+)-glucose contains aldehydic group
which reacts with hydroxyl amine
(NH2OH) to yield an oxime. The complete
reaction is
CH NOH
CH NOH
(d)
NOH
H
C
OH
HO
C
H
H
C
OH
H
C
OH
CH2OH
17 Which one of the following sets of
monosaccharides forms sucrose?
[CBSE AIPMT 2012]
(a) α-D-galactopyranose and
α-D-glucopyranose
(b) α-D-glucopyranose and
β-D-fructofuranose
(c) β-D-glucopyranose and
α-D-fructofuranose
(d) α -D-glucopyranose and
β-D-fructopyranose
Ans. (b)
Sucrose is composed of
α-D-glucopyranose and a
β-D-fructofuranose units which are
joined byα, β-glycosidic linkage between
C-1 of the glucose unit and C-2 of the
fructose unit
CH2OH
H
OH
H
H
HO
1
O
H
HOH2C
O
Glycosidic
linkage
HO
H
O
2
H
OH
3
OH
Structure of sucros
CH2OH
H
235
Biomolecules, Chemistry in Everyday Life and Environmental Chemistry
18 Which one of the following
statements is not true regarding (+)
lactose?
[CBSE AIPMT 2011]
(a) (+) lactose is a β-glycoside formed by
the union of a molecule of D-(+)glucose and a molecule of D-(+)galactose
(b) (+) lactose is a reducing sugar and
does not exhibit mutarotation
(c) (+) lactose, C12H22O11 contains
8 O  H groups
(d) On hydrolysis (+) lactose gives equal
amount of D-(+)- glucose and D-(+)galactose
Ans. (b)
CH2OH
HO
H
O
Glycosidic linkage
H
H
OH
H
H
Lactose
H
OH
H
OH
[CBSE AIPMT 2009]
(a) Insulin
(c) Adrenaline
Ans. (d)
(b) Testosterone
(d) Thyroxine
HO
H
O
CCOOH
I
OH
I
H
CH2OH
19 Which one of the following is
employed as antihistamine?
[CBSE AIPMT 2011]
(a) Diphenyl hydramine
(b) Norethindrone
(c) Omeprazole
(d) Chloramphenicol
Ans. (a)
Diphenylhydramine (benadryl) is used as
an antihistamine.
20 Which of the following does not
exihibit the phenomena of
mutarotation? [CBSE AIPMT 2010]
(a) (+) Sucrose
(c) (+) Maltose
Ans. (a)
21 Which of the following hormones
contains iodine?
Thyroxine is 3, 5,3′, 5′-tetra
iodothyronine. It is secreted by follicular
cells of thyroid glands.
Its structure is given as
H
I
O
H
OH
undergo mutarotation in aqueous
solution.
Among the given carbohydrates, only
sucrose is a non-reducing sugar as in it
the hemiacetal and hemiketal groups of
glucose and fructose are linked together
through O-atom and thus, not free. Due
to the absence of free hemiacetal or
hemiketal group, sucrose does not
exhibit mutarotation.
(b) (+) Lactose
(d) (−) Fructose
Key Idea Reducing sugars that exist in
hemiacetal and hemiketal forms, exhibit
the phenomenon of mutarotation in
aqueous solution. During mutarotation,
the ring open upto give the open chain
form which then reclose either in the
inverted position or in the original
position giving an equilibrium mixture of
two anomers with a small amount of
open chain form. Thus, all reducing
monosaccharides and disaccharides
NH2
O
I
Thyroxine stimulates the consumption
of oxygen and thus, the metabolism of all
cells or tissues in the body.
22 The segment of DNA which acts as
the instrumental manual for the
synthesis of the protein is
[CBSE AIPMT 2009]
(a) nucleotide
(c) gene
Ans. (c)
(b) ribose
(d) nucleoside
The segment of DNA which acts as the
instrumental manual for the synthesis of
the protein is gene. Every protein in a
cell has a corresponding gene.
23 Which one of the following is an
amine hormone?
[CBSE AIPMT 2008]
(a) Thyroxin
(c) Insulin
Ans. (a)
(b) Oxypurin
(d) Progesterone
(c) chiral phosphate ester units
(d) D-sugar component
Ans. (d)
RNA and DNA molecules have ribose and
deoxyribose sugar respectively. Both are
chiral, their chirality is due to D-ribose or
deoxyribose sugar components.
25 Which one of the following vitamins
is water-soluble?
[CBSE AIPMT 2007]
(a) Vitamin-B
(c) Vitamin-K
Ans. (a)
(b) Vitamin-E
(d) Vitamin-A
Vitamins are classified as
(i) Fat soluble vitamin-A, D, E, K
(ii) Water soluble vitamin-B complex, H
and C
So, vitamin-B is water soluble.
26 Which one of the following is a
peptide hormone?
[CBSE AIPMT 2006]
(a) Glucagon
(b) Testosterone
(c) Thyroxin
(d) Adrenaline
Ans. (a)
Glucagon is a peptide hormone because
in it peptide linkage is present.
27 The human body does not produce
[CBSE AIPMT 2006]
(a) DNA
(c) hormones
Ans. (b)
(b) vitamins
(d) enzymes
The organic compounds other than
carbohydrates, proteins which maintain
normal growth and nutrition in the human
body (but not produced in human body) are
called vitamins.
28 During the process of digestion,
the proteins present in food
materials are hydrolysed to amino
acids. The two enzymes involved in
the process
Enzyme (A)
Proteins →
Thyroxin is an amine hormone which is
secreted by thyroid gland.
24 RNA and DNA are chiral molecules,
their chirality is due to
[CBSE AIPMT 2007]
(a) L-sugar component
(b) chiral bases
Enzyme (B)
→ Amino acids,
are respectively
[CBSE AIPMT 2006]
(a) amylase and maltase
(b) diastase and lipase
(c) pepsin and trypsin
(d) invertase and zymase
236
NEET Chapterwise Topicwise Chemistry
Ans. (c)
In the process of digestion the proteins
present in food material are hydrolysed
to amino acid. In this process two
enzymes pepsin and trypsin are involved
as follows:
Pepsin
Proteins → Polypeptide
(Enzyme A)
|
|
Trypsin
Amino acid ←—
—
(Enzyme B )
29 Which functional group
participates in disulphide bond
formation in proteins?
[CBSE AIPMT 2005]
(a) Thiolactone
(c) Thioether
Ans. (b)
(b) Thiol
(d) Thioester
Disulphide bond may be reduced to thiol
by means of reagents, i.e. NaBH4 which
shows the presence of thiol group in
disulphide bond formation.
HO
*
36 Which one of the following
structures represents the peptide
chain?
[CBSE AIPMT 2004]
H
C

HC*OH

HOC* H

H  C* OH

HC*

CH2OH
O
C*
Chiral (Asymmetric)
carbon atom
33 The correct statement in respect
of protein haemoglobin is that it
[CBSE AIPMT 2004]
(a) functions as a catalyst for biological
reactions
(b) maintains blood sugar level
(c) act as an oxygen carrier in the blood
(d) forms antibodies and offers
resistance to diseases
Ans. (c)
30 The cell membranes are mainly
composed of [CBSE AIPMT 2005]
(a) carbohydrates (b) proteins
(c) phospholipids (d) fats
Ans.
(c) The cell membranes are mainly
composed of phospholipids.
31 The helical structure of protein is
stabilised by
[CBSE AIPMT 2004]
(a) dipeptide bonds
(b) hydrogen bonds
(c) ether bonds
(d) peptide bonds
Ans. (b)
The helical structure of protein is
stabilised by hydrogen bonds between
amide group of the same peptide chain.
These bonds are formed by —NH—group of
one unit and oxygen of carbonyl group of
the other unit. It takes 3.6 amino acid to
complete one turn of the helix to enable.
Such H-bonding and a 13 memberring is
formed by H-bonding. This H-bonding is
responsible for holding helix in a position.
32 Number of chiral carbon atoms in
β-D-(+)-glucose is
[CBSE AIPMT 2004]
(a) five
(c) three
Ans. (a)
(b) six
(d) four
The number of chiral carbon atoms in
β-D-(+) glucose are five
Haemoglobin acts as oxygen carrier in
the blood because fourFe2 + ions of each
haemoglobin can bind with four
molecules of O2 and form
oxyhaemoglobin
4Hb + 4O2 →
Hb4O8
Oxy -haemoglobin
34 The hormone that helps in the
conversion of glucose to glycogen
is
[CBSE AIPMT 2004]
(a) cortisone
(b) bile acids
(c) adrenaline
(d) insulin
Ans. (d)
Insulin hormone helps in the conversion
of glucose into glycogen by the liver and
skeletal muscle. Insulin is secreted by
pancreas that lower blood glucose level.
35 A sequence of how many
nucleotides in messenger RNA
makes a codon for an amino acid?
[CBSE AIPMT 2004]
(a) Three
(c) One
Ans. (a)
(b) Four
(d) Two
A sequence of three nucleotides in
messenger RNA makes a codon for an
amino acid because four bases in
messenger RNA adenine, cytosine,
guanine and uracil have been shown to
act in the form of triplet.
H
|

(a)  N  C  N  C  NH
|
||

O
O H
||
 C NH 
H
H


  
(b)  N  C  C  C  C  N
||   
O
  
C  C  C 
  
H
H
H





(c) N  N  C  C  N  C  C 
||
 ||

O
O
H
H




N CC N C
||


O
O
H
   


(d)  C  N  C  C  C  N C

 
 
H
H


 
C N CCC
||  

O
Ans. (c)
The peptide linkage ( NH  CO ) is
formed by the condensation of amino
acids molecules
HNH C H  COH
|
||
R
O
+ H ⋅NH C H  C OH →


R
O
HN CH  C NH CH  C  Hence,




R
R
O
O
following structure represents the
peptide chain
H
H
H
O
 
 
  
N  C  C N  C  C N  C  C 
 
 

O
O
37 Chargaff’s rule states that in an
organism
[CBSE AIPMT 2003]
(a) amount of adenine (A) is equal to that
of cytosine (C) and the amount of
thymine (T) is equal to that of
guanine (G)
237
Biomolecules, Chemistry in Everyday Life and Environmental Chemistry
(b) amounts of all bases are equal
(c) amount of adenine (A) is equal to that
of thymine (T) and the amount of
guanine (G) is equal to that of
cytosine (C)
(d) amount of adenine (A) is equal to that
of guanine (G) and the amount of
thymine (T) is equal to that of
cytosine (C)
Ans. (c)
[CBSE AIPMT 2002]
(a) α-carbon of α-amino acid is
asymmetric
(b) All proteins are found in L-form
(c) Human body can synthesise all proteins
they need
(d) At pH = 7 both amino and carboxylic
groups exist in ionised form
Ans. (b)
Chargaff ’s rule states that amount of
adenine (A) is equal to that of the amount
of thymine (T) and the amount of guanine
(G) is equal to that of the amount of
cytosine (C).
38 Glycolysis is
Ans. (a)
41 Which is not true statement?
[CBSE AIPMT 2003]
(a) oxidation of glucose to pyruvate
(b) conversion of glucose to haem
(c) oxidation of glucose to glutamate
(d) conversion of pyruvate to citrate
Ans. (a)
Glycolysis is the first stage in the
oxidation of glucose. It is an anaerobic
process and involves the degradation of
glucose into two molecules of pyruvate
with the generation of two molecules of
ATP.
All proteins are not found in L-form but
they may be present in D or L- form.
42 Enzymes are made up of
[CBSE AIPMT 2002]
(a) edible proteins
(b) proteins with specific structure
(c) nitrogen containing carbohydrates
(d) carbohydrates
Ans. (b)
Enzymes are made up of proteins with
specific structure and acts as a catalyst
for biochemical reactions.
43 Which statement is incorrect about
O
|| • •
peptide bond C NH  ?
[CBSE AIPMT 2001]
(a) one carboxylic acid residue and two
phosphate groups
(b) three phoshate groups
(c) three carboxylic acid residues
(d) two carboxylic acid residues and one
phosphate groups
(a) C—N bond length in proteins is longer
than usual bond length of C—N bond
(b) Spectroscopic analysis show planar
structure of  C NH  group
||
O
(c) C—N bond length in proteins is
smaller than usual bond length of
C—N bond
(d) None of the above
Ans. (d)
Ans. (c)
Phospholipids are esters of glycerol with
two carboxylic acid residue and one
phosphate group.
Hence, phospholipids may be regarded
as derivative of glycerol in which two
hydroxyl groups are esterified with fatty
acid, while third is esterified with
phosphoric acid.
Peptide bond is formed by the reaction
of —COOH group of one amino acid with
the —NH2 group of another amino acid
and represented as
39 Phospholipids are esters of
glycerol with [CBSE AIPMT 2003]
40 Vitamin-B 12 contains
[CBSE AIPMT 2003]
(a) Zn (II)
(b) Ca (II)
(c) Fe (II)
(d) Co (III)
Ans. (d)
The molecular formula of vitamin-B12 is
C63H88 N14O14PCo and the chemical name
is cyanocobalamine. So, cobalt is
present in vitamin-B12 .
O

CNH
O

C
–
+
NH
As partial double bond character found
between C—N bond, the bond length of
C—N in protein should be smaller than
usual C—N bond.
44 Which of the following is correct
statement?
[CBSE AIPMT 2001]
(a) Starch is a polymer ofα-glucose
(b) Amylose is a component of cellulose
(c) Proteins are composed of only one
type of amino acid
(d) In cyclic structure of fructose, there
are four carbons and one oxygen
atom
Starch is also known as amylum which
occurs in all green plants. A molecule of
starch (C6H10O5 ) n is built of a large
number of α-glucose ring joined through
oxygen atom.
45 Which of the following is correct
about H-bonding in nucleotide?
[CBSE AIPMT 2001]
(a) A–T, G–C
(c) G–T, A–C
Ans. (a)
(b) A–G, T– C
(d) A–A, T–T
The structure of DNA molecule is a
double helical structure. In this structure
double helix are made up of two right
handed helical polynucleotide chains
which are held together by H-bonds. In
these helixes the adenine (A) base is
linked with thymine (T) by two H-bonds
and guanine (G) is linked with cytocine
(C) by three H-bonds as A==T, and G ≡≡ C.
46 Which one of the following gives
positive Fehling’s solution test?
[CBSE AIPMT 2001]
(a) Sucrose
(c) Fats
Ans. (b)
(b) Glucose
(d) Protein
Glucose reduces Fehling solution to give
red ppt. of Cu2O.
COOH
CHO


(CHOH) 4 + 2CuO → (CHOH) 4 + Cu2O ↓
Fehling


Red ppt.
CH2OH solution
CH2OH
Glucose
47 The hormone which controls the
processes like burning of fats,
proteins and carbohydrates to
liberate energy in the body is
[CBSE AIPMT 2000]
(a) cortisone
(c) adrenaline
Ans. (b)
(b) thyroxine
(d) insulin
Thyroxine is a hormone secreted by
thyroid gland. This hormone controls
various biochemical reactions involving
burning of proteins, carbohydrates, fats
to release energy.
It is an iodinated derivatirve of amino
acid tyrosine.
48 α-D-(+)-glucose and β-D-(+)-glucose
are
[CBSE AIPMT 2000]
(a) anomers
(b) epimers
(c) enantiomers
(d) geometrical isomers
238
NEET Chapterwise Topicwise Chemistry
Ans. (a)
52 Aspirin is an acetylation product of
Those diastereomers which differ only in
configuration at C-1 are known as
anomers
H—C—OH
HO—C—H
|
|
H—C—OH
H—C—OH
|
|
O
HO—C—H
HO—C—H
|
|
H —C —OH
H—C—OH
|
|
H—C
H—C
|
|
CH 2OH
CH 2OH
α-D-(+)-glucose
O
β-D-(+)-glucose
[CBSE AIPMT 1998]
(a) o -hydroxybenzoic acid
(b) o -hydroxybenzene
(c) m -hydroxybenzoic acid
(d) p -dihydroxybenzene
Ans. (a)
OCOCH3
COOH
Structure of Aspirin is
It is prepared by the reaction of acetic
anhydride with salicyclic acid in the
presence of a catalyst (H2SO4 )
OH
COOH
+ (CH3CO)2O
Anomers
49 Which one of the following has
magnesium? [CBSE AIPMT 2000]
(a) Vitamin-B 12
(c) Haemocyanin
anhydrase
Ans. (b)
(b) Chlorophyll
(d) Carbonic
Formula of chlorophyll is C55H72MgN2O6 .
So Mg is present in chlorophyll. It is the
green colouring matter of leaves and
green stems.
Vitamin-B12 contains cobalt,
haemocyanin contains copper and
carbonic anhydrase contain zinc.
50 Which of the following is the
sweetest sugar? [CBSE AIPMT 1999]
(a) Sucrose
(c) Fructose
Ans. (c)
C
CH3 —C—OH
53 Glucose molecule reacts with ‘ X ’
number of molecules of phenyl
hydrazine to yield osazone. The
value of ‘ X ’ is
[CBSE AIPMT 1998]
(a) four (b) one
Ans. (d)
(c) two
CHOH
C
glucose phenyl hydrazone
NH
CH
O
–C6H5NH2
C6H5NH·NH2
H2O
CH2OH
imino ketone
CH
C
NH
N·NHC6H5
(CHOH)3
CH2OH
Enzymes are globular proteins which
catalyse biochemical reaction in the
living systems.
56 Which one of the following
chemical units is certainly to be
found in enzyme?
[CBSE AIPMT 1997]
OH
|
O
(b)
N
N
H
|
(c) NC
||
O
NH
O
NHC6H5
C
O
H
(d)
(CHOH)3
CH20H
hydrogen bonded
intermediate
C3H5NH·NH2
–NH3
CH
C
O
| O
OH
HO
CH2OH
glucose
(CHOH)3
[CBSE AIPMT 1997]
(a) transport oxygen
(b) provide immunity
(c) catalyse biochemical reactions
(d) provide energy
Ans. (c)
N·NHC6H5
(CHOH)3
–H2O
CH2OH
C
55 The function of enzymes in the
living system is to
OH
H2N·NHC6H5
(CHOH)3
CH
Haemoglobin is a globular protein of four
sub-units, it contains 94% globin.
H
O
[CBSE AIPMT 1997]
(a) an enzyme
(b) a globular protein
(c) a vitamin
(d) a carbohydrate
Ans. (b)
(d) three
CH
C
54 Haemoglobin is
(a)
H
51 In DNA, the complementary bases
are
[CBSE AIPMT 1998, 2008]
DNA has a double helical structure. These
helix contains polynucleotide chains and
these chains are held together by
hydrogen bonds. In these polynucleotide
chain of DNA, adenine has thymine
opposite to it and guanine has cytosine
opposite to it.
COOH
+ HCl
Acetyl salicylic acid
(Aspirin)
Fructose is the sweetest sugar and
highly soluble in water and slightly
soluble in alcohol. It is insoluble in ether
fructose in laevorotatory hence, called
laevulose.
Ans. (a)
OCOCH3
Salicylic
acid
(b) Glucose
(d) Maltose
(a) adenine and thymine, guanine and
cytosine
(b) uracil and adenine, cytosine and
guanine
(c) adenine and guanine, thymine and
cytosine
(d) adenine and thymine, guanine and
uracil
H2SO4
Thus, three phenyl hydrazine molecules
and one molecule of glucose is required
to form osazone.
O O
R
O
R
O
O
R
Ans. (c)
N·NH·C6H5
N·NH·C6H5
(CHOH)3
CH2OH
glycosazone
(yellow crystalline solid)
Peptide bonds are present in enzyme
because enzymes are made up of
proteins, and proteins are the polymer

 H

 |
 — N— C — 
|| 

O 

239
Biomolecules, Chemistry in Everyday Life and Environmental Chemistry
57 Sucrose in water is dextrorotatory,
[α] D = + 66.4° when boiled with dil.
HCl, the solution becomes
leavorotatory, [α] D = − 39.9° . In this
process the sucrose breaks into
[CBSE AIPMT 1996]
(a) L-glucose + D-fructose
(b) L-glucose + L-fructose
(c) D-glucose + D-fructose
(d) D-glucose + L-fructose
Ans. (c)
The aqueous solution of sucrose is
dextrorotatory having [α] D = + 66.4° .On
hydrolysis with dilute acids or enzyme
invertase, cane sugar (sucrose) gives
equimolar mixture of D-(+)-glucose and
D-(–)-fructose
HCl
C12H22O11 + H2O →
Cane sugar
C6H12O6
D -glucose
[α] D = + 52.5
[α] D = 66.4°
+
C6H12O6
D-fructose
[α] D = – 92.4°
So, sucrose is dextrorotatory but after
hydrolysis gives dextrorotatory glucose
and laevorotatory fructose.
D-(–)-fructose has a greater specific
rotation than D-(+)-glucose. Therefore,
the resultant solution is laevorotatory in
nature with specific rotation of –39.9°.
58 In reference to biological role, Ca 2+
ions are important in
[CBSE AIPMT 1996]
(a) triggering the contraction of
muscles
(b) generating the right electrode
potential across cell membrane
(c) hydrolysis of ATP
(d) defence mechanism
Ans. (b)
The Ca2 + ion generates the right
electrical potential across cell
membrane.
59 Which of the following protein
destroy the antigen when it enters
in body cell?
[CBSE AIPMT 1995]
(a) Antibodies
combined with biopolymers (known as
nucleic acids). Nucleic acids are
biologically important polymers which
are present in all living cells. DNA is the
polymer of nucleotides.
(b) Insulin
(c) Chromoprotein
(d) Phosphoprotein
Ans. (a)
Antibodies are the proteins which
protect the body against toxic
substances and infections. When an
antigen enter in the body cells, the
antibodies present in the body destroyed
the antigen.
60 The α-D-glucose and β-D-glucose
differ from each other due to
difference in carbon atom with
respect to its
[CBSE AIPMT 1995]
(a) conformation
(b) configuration
(c) number of OH-groups
(d) size of hemiacetal ring
Ans. (b)
The isomer having the hydroxyl group
(—OH) on the right is calledα-D-glucose
and one having the hydroxyl group (—OH)
on the left is calledβ-D-glucose. Such
pairs of optical isomers which differ in
the configuration only around C1 atom
are called anomers. Thus α-D-glucose
and β-D-glucose are anomers.
61 Chemically considering digestion is
basically
[CBSE AIPMT 1994]
(a) anabolism
(b) hydrogenation
(c) hydrolysis
(d) dehydrogenation
Ans. (c)
Digestion is basically hydrolysis reaction
in which large molecules are hydrolysed
to give smaller molecules. For example
when we eat proteins, it will hydrolyse
and form amino acids.
62 An example of biopolymer is
[CBSE AIPMT 1994]
(a) teflon
(c) nylon-66
Ans. (d)
(b) neoprene
(d) DNA
All living cells contains nucleoproteins,
i.e. substances made up of proteins
63 Diazo coupling is useful to prepare
some
[CBSE AIPMT 1994]
(a) pesticides
(c) proteins
Ans. (b)
(b) dyes
(d) vitamins
Diazo coupling is the reaction in which
phenol, or aniline reacts with
benzenediazonium salt to produce dyes
with (—N==N —) azo groups.
+
–
—N2Cl + H—
—N
–HCl
—NH2
N—
—NH2
azo dye
64 The couplings between base units
of DNA is through
[CBSE AIPMT 1992]
(a) hydrogen bonding
(b) electrostatic bonding
(c) covalent bonding
(d) van der Waals’ forces
Ans. (a)
In DNA the two strands are held together by
hydrogen bonds. For example guanine is
bonded to cytosine and adenine to thymine
by hydrogen bonding.
65 On hydrolysis of starch, we finally
get
[CBSE AIPMT 1991]
(a) glucose
(b) fructose
(c) Both (a) and (b) (d) sucrose
Ans. (a)
Starch is hydrolysed with dilute acids or
enzymes and break down to molecules
of variable complexity and finally gives
D-glucose.
(C6H10O5 ) n → (C6H10O5 ) n ′ →
Starch
Diastase
C12H22O11 → C6H12O6
Maltose
D-glucose
240
NEET Chapterwise Topicwise Chemistry
TOPIC 2
Chemistry in
Everyday LIfe
67 Which of the following will not
undergo SN 1 reaction with OH ?
[NEET (Oct.) 2020]
(a) CH2 ==CH CH2Cl
(b) (CH3) 3CCl
66 Which one of the following
polymers is prepared by addition
polymerisation?
[NEET 2021]
(a) Teflon
(c) Novolac
Ans. (a)
(b) Nylon-66
(d) Dacron
CH2Cl
(c)
(d)
F
F
F
F
F
Tetrafluoro
ethene
F
F n
An S N 1 reaction proceeds through
formation of a stable carbocation as an
intermediate. Here,
(a) CH2
Nylon-6, 6 Condensation polymerisation.
–Cl–
CH2Cl
CH2
(c)
Hexamethylene
diamine
4
(
(
Novolac Condensation polymerisation.
OH
OH
1°-carbocation (less stable due
to –I effect of the phenyl group)
(d)
CH2OH
Phenol
r
Benzyl-1°-carbocation (stable)
OH
CH2
n
Novolac
So, option (c) will not undergoS N 1
reaction withOH− .
68 Which of the following is not true
about chloramphenicol?
[NEET (Oct.) 2020]
Dacron Condensation polymerisation.
(a) It inhibits the growth of only
gram-positive bacteria
(b) It is a broad spectrum antibiotic
(c) It is not bactericidal
(d) It is bacteriostatic
COOH
OH
–H2O
+
–Cl–
CH2
H+
+ HCHO
OH–
Formaldehyde
CH2
HO
CH2Cl
Glycol
COOH
Ans. (a)
Terephthalic acid
O
O
C
n
O( CH2
OC
2
(
Dacron
È
cationic detergent.
(c) and (d) are anionic detergents.
Sodium dodecyl benzene sulphonate :
È
CH2CH2
Nylon-6, 6
Sodium stearate
Cetyltrimethyl ammonium bromide
Sodium dodecyl benzene sulphonate
Sodium lauryl sulphate
⊕
Sodium lauryl sulphate:
r
(
NHCH2NH
n
6
(a)
(b)
(c)
(d)
È
–I
( C( CH 

2 ( C
69 Which of the following is a cationic
detergent?
[NEET (Sep.) 2020]
p- CH3 (CH2 ) 11  C6H4 SO 3 N a
–Cl–
O
O
–H2O
CH2CH2Cl
Cl
(Chloramphenicol)
⊕
tert-butyl carbocation (stable)
( 6 NH2
+ H2N( CH2
Adipic acid
Cl
NHCOCH
CH3  (CH2 ) 15 N(CH3) 3 Br is a
(9 hyper
conjugations)
4
CH
(a) Solution stearate, C17H35 COONa is a
soap.
(b) Cetyltrimethyl ammonium bromide,
− Cl
(b) (CH3) 3 CCl →
CH3  C  CH3

CH3
( COH
( CH2
O
CH
OH
Ans. (b)
⊕
−
O
HO
r
CH2
CH
Allyl-1°-carbocation
(stable)
n
Teflon
CH
CH2
Narrow-spectrum antibiotics are
effective against gram-positive or
gram-negative bacteria.
Bactericidals show cidal or killing effect
on microbes.
Ans. (c)
Addition polymerisation Monomers are
added one after other in addition
polymerisation.
Teflon Addition polymerisation.
F
CH2CH2Cl
O2N
OH
Chloramphenicol is a bacteriostatic
(inhibitory) broad spectrum antibiotic,
which shows inhibitory effect on wide
range of gram-positive and
gram-negative bacteria (microbes).
⊕
CH3  (CH2 ) 10  CH2 OSO 3 N a
70 Among the following, the narrow
spectrum antibiotic is
[NEET (National) 2019]
(a) ampicillin
(b) amoxycillin
(c) chloramphenicol
(d) penicillin G
Ans. (d)
Key Idea Antibiotics which are effective
mainly against either gram positive or
gram negative bacteria are called narrow
spectrum antibiotics.
Penicillin G has a narrow spectrum.
Ampicillin and amoxycillin are synthetic
modifications of penicillin.
These have broad spectrum. Also,
chloramphenicol is a broad spectrum
antibiotic.
Hence, option (d) is correct.
71 The artificial sweetner stable at
cooking temperature and does not
provide calories is
[NEET (Odisha) 2019]
(a) saccharin
(c) sucralose
(b) aspartame
(d) alitame
241
Biomolecules, Chemistry in Everyday Life and Environmental Chemistry
Ans. (c)
Sucralose is a trichloro derivative of
sucrose and is about 650 times sweeter
than cane sugar. It is a zero calorie sugar
and stable at cooking temperature.
H
OH
O
H
Cl
HO
OH
OH
O
H
OH
CH2Cl
H
ClH2C
O
Sucralose
Cl
OH
S
Cl
HO
H
72 Mixture of chloroxylenol and
terpineol acts as
[NEET 2017]
(a) analgesic
(c) antipyretic
Ans. (b)
(a) softener
(b) dryer
(c) buffering agent (d) antiseptic
Ans. (d)
Bithional is added to soap to impart
antiseptic properties. It reduces odours
produced by bacterial decomposition of
organic matter on the skin .
OH
H
74 Bithional is generally added to the
soaps as an additive to function as
a/an
[CBSE AIPMT 2015]
(b) antiseptic
(d) antibiotic
(a) The medicines which are used to
reduce pain are known as analgesics.
For example paracetamol, ibuprofen,
morphine, etc.
(b) The chemicals which either prevent
the growth of microorganisms or kill
them but are not harmful to the living
tissues are known as antiseptics.
e.g. savlon, dettol. Dettol is one of
the most commonly used
antiseptics. It is a mixture of
chloroxylenol and α-terpineol.
(c) The chemical substances which are
used to bring down body temperature
during fever are called antipyretics,
e.g. aspirin, novalgin, etc.
(d) The chemicals which are obtained
from microorganisms such as
bacteria, fungi, etc., or by synthetic
methods and used to inhibit the
growth or even kill the
microorganisms are called
antibiotics, e.g. penicillin,
chloramphenicol, etc.
73 Which of the following is an
analgesic?
[NEET 2016, Phase I]
(a) Penicillin
(b) Streptomycin
(c) Chloromycetin (d) Novalgin
Ans. (d)
Novalgin (Dipyrone) is a non-narcotic
analgesic used as pain reliever.
Penicillin is an antibiotic used for
curing rheumatic fever.
Streptomycin is an antibiotic drug.
Chloromycetin is an antibiotic drug.
Cl
Cl
IUPAC Name 2,2sulfanediylbis
(4, 6-dichlorophenol)
75 Artificial sweetener which is stable
under cold conditions only is
[CBSE AIPMT 2014]
(a) saccharine
(c) aspartame
Ans. (c)
(b) sucralose
(d) alitame
Aspartame is the only artificial
sweetener which is stable at lower
temperature and decomposes at higher
temperature. It is also called Nutra
sweet. It’s relative sweetness value is
180.
76 Antiseptics and disinfectants either
kill or prevent growth of
microorganisms. Identify which of
the following is not true. [NEET 2013]
(a) A 0.2% solution of phenol is an
antiseptic while 1% solution acts as a
disinfectant
(b) Chlorine and iodine are used as
strong disinfectants
(c) Dilute solutions of boric acid and
hydrogen, peroxide are strong
antiseptics
(d) Disinfectants harm the living tissues
Ans. (c)
Antiseptics and disinfectants both either
kill or prevent the growth of
microorganisms. The main point of
difference between these two is that the
former (antispetics) are used for living
beings whereas disinfectants are not
safe for living tissues. These are actually
used for inanimate objects like floors,
tiles, etc.
A substance like phenol in its lower
concentration (0.2%) behaves as
antiseptic, whereas in higher
concentration (1%) as disinfectant.
Chlorine and iodine are strong
disinfectants whereas dilute solutions of
boric acid and hydrogen peroxide are
mild antiseptics.
77 Which one of the following is
employed as a tranquilizer drug?
[CBSE AIPMT 2010]
(a) Promethazine
(c) Naproxen
Ans. (b)
(b) Valium
(d) Mifepristone
Tranquilizer are the chemicals that
reduce anxiety and mental tensional.
Tranquilizer is the strain reliever also
used for mild and essential component
of sleeping pills. Thus, they are
sometimes called psychotherapeutic
drugs. Equanil, valium and serotonin and
barbiturates (hypnotic) are some
commonly used tranquilizers.
78 Which one of the following is
employed as a tranquiliser?
[CBSE AIPMT 2009]
(a) Equanil
(b) Naproxen
(c) Tetracycline (d) Chlorpheninamine
Ans. (a)
The drugs which are used to reduce
anxiety and for the treatment of mental
diseases, are called tranquilisers. These
drugs are also known as
pyschotherapeutic drugs. Luminal,
seconal and equanil are some commonly
used and are the example of tranquiliser.
79 Which of the following can possibly
be used as analgesic without
causing addiction and mood
modification? [CBSE AIPMT 1997]
(a) Morphine
(b) Diazepam
(c) Tetrahydrocational
(d) N-acetyl-para-aminophenol
Ans. (d)
N-acetyl-para-aminophenol or
paracetamol is used as analgesic as well
as antipyretic.
80 Commonly used antiseptic ‘Dettol’
is a mixture of [CBSE AIPMT 1996]
(a) o-chlorophenoxylenol + terpineol
(b) o-cresol + terpineol
(c) phenol + terpineol
(d) chloroxylenol + terpineol
Ans. (d)
Dettol is an antiseptic. It is a mixture of
chloroxylenol and terpineol in a suitable
solvent.
242
NEET Chapterwise Topicwise Chemistry
TOPIC 3
Environmental
Chemistry
81 The RBC deficiency is deficiency
disease of
[NEET 2021]
(a) vitamin-B 12
(c) vitamin-B 1
(b) vitamin-B 6
(d) vitamin-B2
Ans. (a)
VitaminB12 deficiency disease-RBC
deficiency (anaemia).
VitaminB6 deficiency disease–Dermatitis,
epilepsy.
VitaminB1 deficiency disease – Beri-beri
VitaminB2 deficiency disease –
Ariboflavinesis.
82 Which of the following statement is
not true about acid rain?
[NEET (Oct.) 2020]
(a) It is due to reaction ofSO2 , NO2 and
CO2 with rain water.
(b) Causes no damage to monuments
like Taj Mahal.
(c) It is harmful for plants.
(d) Its pH is less than 5.6
Ans. (b)
Air pollutants likeSO2 , NO2 and CO2 get
dissolved in rain water to produce acid
rain which constitutes mainlyH2SO4 ,
HNO3 and H2 CO3.
2SO2 (g) + O2 (g) + 2H2O(l ) → 2H2SO4 (aq)
4NO2 (g) + O2 (g) + 2H2O(g) → 4HNO3 (aq)
CO2 (g) + H2O(l ) → H2 CO3 (aq)
Due to the presence of these acids, pH
of rain water drops below 5.6.
Acid rain damages marbles of Taj Mahal.
CaCO3 (s ) + H2SO4 (aq) → CaSO4 (aq)
Marble
+ H2O(l ) + CO2 (g)
Thus, Taj Mahal gets disfigured,
discoloured and lustreless.
83 Which of the following statements
is not correct about carbon
monoxide?
[NEET (Sep.) 2020]
(a) It reduces oxygen carrying ability of
blood.
(b) The carboxyhaemoglobin
(haemoglobin bound to CO) is less
stable than oxyhaemoglobin.
(c) It is produced due to incomplete
combustion.
(d) It forms carboxyhaemoglobin.
Ans. (b)
Statements (a), (c) and (d) are correct
about carbon monoxide (CO).
Statement (b) is not correct about CO,
because carboxyhaemoglobin
(haemoglobin bound to CO) is about 300
times more stable than oxyhaemoglobin
complex.
84 Among the following, the one that
is not a green house gas is
[NEET (National) 2019]
(a) methane
(b) ozone
(c) sulphur dioxide (d) nitrous oxide
Ans. (c)
Sulphur dioxide (SO2 ) is not a green
house gas. Carbon dioxide (CO2 ),
methane (CH4 ), water vapour, nitrous
oxide (N2O), CFCs and ozone (O3) are the
green house gases. These gases are
responsible for global warming.
85 The liquified gas that is used in dry
cleaning along with a suitable
detergent is [NEET (Odisha) 2019]
(a) water gas
(c) NO2
Ans. (d)
(b) petroleum gas
(d) CO2
Solvents used to dry clean clothes are
usually chlorinated compounds which
are carcinogenic. Suitable detergents
which work in liquid carbon dioxide have
been discovered to replace the
chlorinated compounds. Thus, CO2 is the
liquified gas that is used in dry cleaning
along with a suitable detergent.
86 Which of the following is a sink for
CO ?
[NEET 2017]
(a) Haemoglobin
(b) Microorganisms present in the soil
(c) Oceans
(d) Plants
Ans. (b)
Microorganisms present in the soil act as
biggest source and sink. A sink is a
natural or artificial reservoir that
accumulates and stores some chemical
compound for an indefinite period. Thus
(b) is correct option.
87 Which one of the following is not a
common component of
photochemical smog?
(a) Ozone
[CBSE AIPMT 2014]
(b) Acrolein
(c) Peroxyacetyl nitrate
(d) Chlorofluorocarbons
Ans. (d)
Among the given chlorofluorocarbons
are the compounds that are responsible
for the ozone depletion which degrade
ozone into moleculer oxygen. It is not a
component of photochemical smog
while other are component of smog.
88 Which one of the following
statements regarding
photochemical smog is not correct?
[CBSE AIPMT 2012]
(a) Carbon monoxide does not play any
role in photochemical smog
formation
(b) Photochemical smog is an oxidising
agent in character
(c) Photochemical smog is formed
through photochemical reaction
involving solar energy
(d) Photochemical smog does not cause
irritation in eyes and throat
Ans. (d)
Photochemical smog is formed in warm
and sunny climate during day time by the
action of sunlight on primary pollutants. It
contains nitrogen oxides, ozone, PAN,
etc., which are oxidising in nature. So,
photochemical smog is an oxidising agent
in character. It causes irritation in eyes
and throat.
89 Green chemistry means such
reactions which [CBSE AIPMT 2008]
(a) produce colour during reactions
(b) reduce the use and production of
hazardous chemicals
(c) are related to the depletion of ozone
layer
(d) study the reactions in plants
Ans. (b)
Green chemistry means, the production
of chemicals of our daily needs by using
such reactions and chemical processes
which neither use toxic chemicals, nor
emit such chemicals into atmosphere.
Thus, green chemistry is an alternative
tool for reducing pollution.