30 years NEET AIPMT chapterwise solutions chemistry 1

CHAPTERWISE SOLUTIONS

CHEMISTRY
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Contents
1. Some Basic Concepts of Chemistry

........ 1

16.Solutions

���� 130

2.

........ 9

17.Electrochemistry

���� 139

18. Chemical Kinetics

���� 149

19. Surface Chemistry

���� 158

Structure of Atom

3. Classification of Elements and


Periodicity in Properties

������ 17

4. Chemical Bonding and

20. General Principles and Processes
������ 22



5. States of Matter : Gases and Liquids

������ 37

21. p-Block Elements (Group 15 to 18)

���� 164

6.Thermodynamics

������ 45

22. d-and f-Block Elements

���� 175

7.Equilibrium

������ 56

23. Coordination Compounds

���� 185

8. Redox Reactions

������ 72

24. Haloalkanes and Haloarenes

���� 197

9.Hydrogen

������ 75

25. Alcohols, Phenols and Ethers

���� 204

10. s-Block Elements

������ 78

11. p-Block Elements (Group 13 and 14)

������ 85



Molecular Structure

���� 161

26. Aldehydes, Ketones and


Carboxylic Acids

���� 214

27. Organic Compounds


12. Organic Chemistry – Some Basic

of Isolation of Elements

Containing Nitrogen

���� 231

������ 91

28.Biomolecules

���� 241

13.Hydrocarbons

���� 106

29.Polymers

���� 250

14. Environmental Chemistry

���� 121

30. Chemistry in Everyday Life

���� 255

15. Solid State

���� 123

31. Nuclear Chemistry

���� 258



Principles and Techniques

iv

Syllabus

ô

UNIT I: SOME BASIC CONCEPTS OF CHEMISTRY
ã 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,
de Broglie’s relationship, Heisenberg 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, ionization 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).
*For

details, refer to latest prospectus


v
UNIT VI: THERMODYNAMICS
• First law of thermodynamics-internal energy and enthalpy, heat capacity and specific heat, measurement
of DU and DH, 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 hydrogen carbonate, 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
hyperconjugation.
• Homolytic and heterolytic fission of a covalent bond: free radials, carbocations, carbanions; electrophiles
and nucleophiles, types of organic reactions.


vi
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.
UNIT XV: 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 XVI: 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’t Hoff factor.
UNIT XVII: 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 cell- electrolytic 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 XVIII: 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 XIX: 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; emulsions- types of emulsions.


vii
UNIT XX: 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 XXI: 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 (PCl3, PCl5) 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, 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: 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 XXII: 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 XXIII: 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 XXIV: 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 XXV: 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, electrophilic substitution reactions, uses of phenols.
• Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses.


viii
UNIT XXVI: 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 XXVII: 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 XXVIII: 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 XXIX: POLYMERS
• Classification- Natural and synthetic, methods of polymerization (addition and condensation),
copolymerization. Some important polymers: natural and synthetic like polyesters, bakelite; rubber,
Biodegradable and non-biodegradable polymers.

UNIT XXX: 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 of Chemistry

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1.

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Suppose the elements X and Y combine to
form two compounds XY 2 and X 3 Y 2 . When
0.1 mole of XY2 weighs 10 g and 0.05 mole of
X3Y2 weighs 9 g, the atomic weights of X and
Y are
(a) 40, 30
(b) 60, 40
(c) 20, 30
(d) 30, 20
(NEET-II 2016)

2.

What is the mass of the precipitate formed
when 50 mL of 16.9% solution of AgNO3 is
mixed with 50 mL of 5.8% NaCl solution?
(Ag = 107.8, N = 14, O = 16, Na = 23,
Cl = 35.5)
(a) 3.5 g
(b) 7 g
(c) 14 g (d) 28 g
(2015)

3.

If Avogadro number N A , is changed from
6.022 × 1023 mol–1 to 6.022 × 10 20 mol–1, this
would change
(a) the mass of one mole of carbon
(b) the ratio of chemical species to each other
in a balanced equation
(c) the ratio of elements to each other in a
compound
(d) the definition of mass in units of grams.
(2015)

(a) 8 : 16 : 1
(c) 16 : 1 : 2

(b) 16 : 8 : 1

(d) 8 : 1 : 2 (2014)

7.

When 22.4 litres of H2(g) is mixed with 11.2
litres of Cl2(g), each at S.T.P, the moles of HCl(g)
formed is equal to
(b) 2 mol of HCl(g)
(a) 1 mol of HCl(g)
(c) 0.5 mol of HCl(g)
(d) 1.5 mol of HCl(g).
(2014)

8.

1.0 g of magnesium is burnt with 0.56 g O2 in
a closed vessel. Which reactant is left in excess
and how much? (At. wt. Mg = 24, O = 16)
(a) Mg, 0.16 g
(b) O2, 0.16 g
(c) Mg, 0.44 g
(d) O2, 0.28 g
(2014)

9.

6.02 × 1020 molecules of urea are present in
100 mL of its solution. The concentration of
solution is
(a) 0.001 M

(b) 0.1 M
(c) 0.02 M
(d) 0.01 M
(NEET 2013)

4.

The number of water molecules is maximum in
(a) 1.8 gram of water
(b) 18 gram of water
(c) 18 moles of water
(d) 18 molecules of water.
(2015)

10. In an experiment it showed that 10 mL of 0.05 M
solution of chloride required 10 mL of 0.1 M
solution of AgNO 3, which of the following
will be the formula of the chloride (X stands
for the symbol of the element other than
chlorine)
(a) X2Cl2 (b) XCl2 (c) XCl4 (d) X2Cl
(Karnataka NEET 2013)

5.

A mixture of gases contains H2 and O2 gases
in the ratio of 1 : 4 (w/w). What is the molar
ratio of the two gases in the mixture?
(a) 16 : 1 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1
(2015, Cancelled)

11. Which has the maximum number of molecules
among the following?
(a) 44 g CO2
(b) 48 g O3
(c) 8 g H2
(d) 64 g SO2
(Mains 2011)

6.

Equal masses of H 2, 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 : O2 : methane
would be

12. The number of atoms in 0.1 mol of a triatomic
gas is (NA = 6.02 × 10 23 mol–1)
(b) 1.806 × 1023
(a) 6.026 × 1022
23
(c) 3.600 × 10
(d) 1.800 × 1022
(2010)


2

13. 25.3 g of sodium carbonate, Na 2 CO 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 ions, CO 32– are respectively
(Molar mass of Na2CO3 = 106 g mol–1)
(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
(2010)
14. 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
(a) 3 mol
(b) 4 mol
(c) 1 mol
(d) 2 mol (2009)
15. What volume of oxygen gas (O 2) measured
at 0°C and 1 atm, is needed to burn completely
1 L of propane gas (C 3H 8) measured under
the same conditions?
(a) 5 L
(b) 10 L (c) 7 L
(d) 6 L
(2008)
16. How many moles of lead (II) chloride will be
formed from a reaction between 6.5 g of PbO
and 3.2 g HCl?
(a) 0.011
(b) 0.029 (c) 0.044 (d) 0.333

(2008)
17. 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
(a) CHO
(b) CH4O
(c) CH3O
(d) CH2O (2008)
18. An element, X has the following isotopic
composition:
200

199
202
X : 90%
X : 8.0%
X : 2.0%
The weighted average atomic mass of the
naturally occurring element X is closest to
(a) 201 amu
(b) 202 amu
(c) 199 amu
(d) 200 amu (2007)
19. The maximum number of molecules is present in
(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.
(2004)

20. Which has maximum molecules?
(a) 7 g N2
(b) 2 g H2
(c) 16 g NO2
(d) 16 g O2 (2002)

21. Percentage of Se in peroxidase anhydrous
enzyme is 0.5% by weight (at. wt. = 78.4) then
minimum molecular weight of peroxidase
anhydrous enzyme is
(a) 1.568 × 104
(c) 15.68

(b) 1.568 × 103
(d) 2.136 × 104
(2001)
22. Molarity of liquid HCl, if density of solution is
1.17 g/cc is
(a) 36.5
(b) 18.25
(c) 32.05
(d) 42.10
(2001)
23. Specific volume of cylindrical virus particle is
6.02 × 10–2 cc/g whose radius and length are
7 Å and 10 Å respectively. If NA = 6.02 × 1023,
find molecular weight of virus.
(a) 15.4 kg/mol
(b) 1.54 × 104 kg/mol
4

(c) 3.08 × 10 kg/mol (d) 3.08 × 103 kg/mol
(2001)
24. In quantitative analysis of second group in
laboratory, H2S gas is passed in acidic medium
for precipitation. When Cu2+ and Cd2+ react
with KCN, then for product, true statement is
(a) K2[Cu(CN)4] more soluble
(b) K2[Cd(CN)4] less stable
(c) K3[Cu(CN)2] less stable
(2000)
(d) K2[Cd(CN)3] more stable.
25. Volume of CO 2 obtained by the complete
decomposition of 9.85 g of BaCO3 is
(a) 2.24 L
(b) 1.12 L
(c) 0.84 L
(d) 0.56 L (2000)
26. Oxidation numbers of A, B, C are +2, +5 and
–2 respectively. Possible formula of compound is
(b) A3(BC 4) 2
(a) A2(BC2) 2
(c) A2(BC3) 2
(d) A3(B 2C) 2
(2000)
27. The number of atoms in 4.25 g of NH 3 is
approximately
(b) 2 × 2023
(a) 4 × 1023
23
(c) 1 × 10

(d) 6 × 1023 (1999)
28. Given the numbers: 161 cm, 0.161 cm, 0.0161 cm.
The number of significant figures for the three
numbers is
(a) 3, 3 and 4 respectively
(b) 3, 4 and 4 respectively
(c) 3, 4 and 5 respectively
(d) 3, 3 and 3 respectively.
(1998)


Some Basic Concepts of Chemistry

29. Haemoglobin contains 0.334% of iron by
weight. The molecular weight of haemoglobin
is approximately 67200. The number of iron
atoms (Atomic weight of Fe is 56) present in
one molecule of haemoglobin is
(a) 4
(b) 6
(c) 3
(d) 2
(1998)
30. In the reaction,
4NH3(g) + 5O2(g) ® 4NO(g) + 6H2O(l)
when 1 mole of ammonia and 1 mole of O2 are
made to react to completion :
(a) All the oxygen will be consumed.
(b) 1.0 mole of NO will be produced.
(c) 1.0 mole of H2O is produced.

(d) All the ammonia will be consumed.
(1998)
31. Among the following which one is not
paramagnetic? [Atomic numbers; Be = 4,
Ne = 10, As = 33, Cl = 17]
(a) Ne2+
(b) Be+
(c) Cl–
(d) As +
(1998)
32. 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)
(a) 95.93
(b) 59.93 (c) 95.39 (d) 5.993
(1996)
33. The amount of zinc required to produce 224 mL
of H2 at STP on treatment with dilute H2SO4
will be
(a) 65 g
(b) 0.065 g (c) 0.65 g (d) 6.5 g
(1996)
34. The
that
(a)
(b)
(c)
(d)

dimensions of pressure are the same as

of
force per unit volume
energy per unit volume
force
energy.
(1995)

35. The number of moles of oxygen in one litre of
air containing 21% oxygen by volume, under
standard conditions, is
(a) 0.0093 mol
(b) 2.10 mol
(c) 0.186 mol
(d) 0.21 mol.
(1995)
36. The total number of valence electrons in 4.2 g
of N3– ion is (NA is the Avogadro’s number)
(a) 2.1 NA
(b) 4.2 NA
(c) 1.6 NA
(d) 3.2 NA (1994)

3

37. A 5 molar solution of H2SO4 is diluted from 1
litre to a volume of 10 litres, the normality of
the solution will be
(a) 1 N
(b) 0.1 N
(c) 5 N

(d) 0.5 N (1991)
38. The number of gram molecules of oxygen in
6.02 × 1024 CO molecules is
(a) 10 g molecules
(b) 5 g molecules
(c) 1 g molecules
(d) 0.5 g molecules.
(1990)
39. Boron has two stable isotopes, 10B(19%) and
11
B(81%). Calculate average at. wt. of boron in
the periodic table
(a) 10.8
(b) 10.2
(c) 11.2
(d) 10.0
(1990)
40. The molecular weight of O2 and SO2 are 32 and
64 respectively. At 15°C and 150 mmHg
pressure, one litre of O2 contains ‘N’ molecules.
The number of molecules in two litres of SO2
under the same conditions of temperature and
pressure will be
(a) N/2
(b) N
(c) 2 N
(d) 4 N
(1990)
41. A metal oxide has the formula Z2O3. 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
(a) 27.9
(b) 159.6
(c) 79.8
(d) 55.8
(1989)
42. Ratio of Cp and CV of a gas ‘X’ is 1.4. The
number of atoms of the gas ‘X’ present in 11.2
litres of it at NTP will be
(b) 1.2 × 1023
(a) 6.02 × 1023
23
(c) 3.01 × 10
(d) 2.01 × 1023
(1989)
43. What is the weight of oxygen required for the
complete combustion of 2.8 kg of ethylene?
(a) 2.8 kg (b) 6.4 kg (c) 9.6 kg (d) 96 kg
(1989)
44. The number of oxygen atoms in 4.4 g of CO2
is
(b) 6 × 1022
(a) 1.2 × 1023
23
(c) 6 × 10
(d) 12 × 1023
(1989)

4

45. At S.T.P. the density of CCl 4 vapour in g/L
will be nearest to
(a) 6.87
(b) 3.42
(c) 10.26 (d) 4.57
(1988)

47. 1 cc N 2O at NTP contains
(a)

46. One litre hard water contains 12.00 mg Mg2+.
Milli-equivalents of washing soda required to
remove its hardness is
(a) 1
(b) 12.16
(c) 1 × 10–3
(d) 12.16 × 10–3
(1988)

(b)
(c)

1.8
u 1022 atoms
224
6.02
u 1023 molecules

22400
1.32
u 1023 electrons
224

(d) All the above.

(1988)

Answer Key

1.
11.
21.
31.
41.

(a)
(c)
(a)
(c)
(d)

2.
12.
22.
32.
42.

(b)

(b)
(c)
(b)
(a)

3.
13.
23.
33.
43.

(a)
(b)
(a)
(c)
(c)

4.
14.
24.
34.
44.

(c)
(b)
(c)
(b)
(a)

5.

15.
25.
35.
45.

(d)
(a)
(b)
(a)
(a)

6.
16.
26.
36.
46.

(c)
(b)
(b)
(c)
(a)

7.
17.
27.
37.
47.

(a)

(c)
(d)
(a)
(d)

8.
18.
28.
38.

(a)
(d)
(d)
(b)

9.
19.
29.
39.

(d)
(a)
(a)
(a)

10.
20.
30.
40.

(b)
(b)
(a)
(c)


5

Some Basic Concepts of Chemistry

1.

(a) : Let atomic weight of element X is x and
that of element Y is y.

O

For XY2,



X

.PM XU




⇒ x + 2y =
Y  Z



For X3Y2,

O

...(i)

X

7.

.PM XU


⇒ Y   Z
...(ii)


Y   Z

On solving equations (i) and (ii), we get y = 30
x + 2(30) = 100 ⇒ x = 100 – 60 = 40


2. (b) : 16.9% solution of AgNO3 means 16.9 g of
AgNO3 in 100 mL of solution.
16.9 g of AgNO3 in 100 mL solution ≡ 8.45 g of
AgNO3 in 50 mL solution.
Similarly, 5.8% of NaCl in 100 mL solution

º 2.9 g of NaCl in 50 mL solution.
The reaction can be represented as :
AgCl + NaNO3
AgNO3 + NaCl
Initial
8.45/170 2.9/58.5
mole
= 0.049 = 0.049
Final moles
0
0

0

0

0.049

0.049

\ Mass of AgCl precipitated = 0.049 × 143.3
= 7.02 » 7 g
(a) : Mass of 1 mol (6.022 × 1023 atoms) of carbon

3.
= 12 g
If Avogadro number is changed to 6.022 × 1020 atoms
then mass of 1 mol of carbon
12 u 6.022 u 10 20
6.022 u 10 23

12 u 10 3 g

6.023 u 10 23
u 1.8
4. (c) : 1.8 gram of water =
18
= 6.023 × 1022 molecules
18 gram of water = 6.023 × 1023 molecules
18 moles of water = 18 × 6.023 × 1023 molecules

5.

(d) : Number of moles of H2 =

Number of moles of O2 =
Hence, molar ratio =

4
32

Mass
Mol. mass
w
w
w
nH 2
; n
; n
2 O 2 32 CH 4 16

w w w
So, the ratio is
:
:
or 16 :1: 2.
2 32 16

So, no. of moles =

1
2

1 4
:
=4:1
2 32

6. (c) : According to Avogadro’s hypothesis,
ratio of the volumes of gases will be equal to the
ratio of their no. of moles.

(a) : 1 mole ≡ 22.4 litres at S.T.P.

22.4
1 mol ; nCl 2
22.4
Reaction is as,
H2(g) +

11.2

22.4

nH 2

Initial
Final

1 mol
(1 – 0.5)
= 0.5 mol

0.5 mol

Cl2(g)

2HCl(g)

0.5 mol
(0.5 – 0.5)
= 0 mol

0
2 × 0.5
1 mol

Here, Cl2 is limiting reagent. So, 1 mole of HCl(g) is
formed.
1
= 0.0416 moles
8. (a) : nMg

24
0.56
nO 2
= 0.0175 moles
32
The balanced equation is
1O
MgO
Mg
+
2 2
Initial
Final

0.0416 moles

0.0175 moles

0

2 × 0.0175
(0.0416 – 2 × 0.0175)
0
= 0.0066 moles (O 2 is limiting reagent.)

∴ Mass of Mg left in excess = 0.0066 × 24 = 0.16 g
9.

(d) : Moles of urea =

Concentration of solution =

×
×

= 0.001
× 1000 = 0.01 M

10. (b) : Millimoles of solution of chloride
= 0.05 × 10 = 0.5
Millimoles of AgNO3 solution = 10 × 0.1 = 1
So, the millimoles of AgNO3 are double than the
chloride solution.
∴ XCl2 + 2AgNO3 → 2AgCl + X(NO3)2
11. (c) : 8 g H2 has 4 moles while the others has
1 mole each.
12. (b) : No. of atoms = NA × No. of moles × 3
= 6.023 × 1023 × 0.1 × 3 = 1.806 × 1023
13. (b) : Given that molar mass of Na2CO3 = 106 g
25.3 × 1000
∴ Molarity of solution =
106 × 250
= 0.9547 M = 0.955 M
Na2CO3 → 2Na+ + CO2–
3


6

[Na+] = 2[Na2CO3] = 2 × 0.955 = 1.910 M

[CO2–
3 ] = [Na2CO3] = 0.955 M

2 g H2 = 6.023 × 1023
23
0.5 g H2 = 6.023 × 10 × 0.5 = 1.505 × 10 23
2
32 g O2 = 6.023 × 1023

14. (b) : H2 + 1/2O2 → H2O
2g
1 mol

16 g
0.5

18 g
1 mol

10 g of H2 = 5 mol and 64 g of O2 = 2 mol
∴ In this reaction, oxygen is the limiting reagent so
amount of H2O produced depends on that of O2.
Since 0.5 mol of O2 gives 1 mol H2O
∴ 2 mol of O2 will give 4 mol H2O
15. (a) :
According to the above equation
1 vol. or 1 litre of propane requires to 5 vol.
or 5 litre of O2 to burn completely.
16. (b) : PbO

n mol

+

2HCl → PbCl2 + H2O

2n mol

n mol

6.5
3.2
mol
mol
224
36.5
0.029 mol 0.087 mol
Formation of moles of lead (II) chloride depends
upon the no. of moles of PbO which acts as a limiting
factor here. So, no. of moles of PbCl2 formed will be
equal to the no. of moles of PbO i.e. 0.029.
17. (c) :
Element

%

Atomic
mass

mole

ratio

simple
ratio

C

38.71

12

3.22
38.71
=1
= 3.22
12
3.22

H

9.67

1

9.67
9.67
=3
= 9.67
1
3.22

O

51.62

16

3.22
51.62
= 3.22
=1
16
3.22

Hence empirical formula of the compound would be
CH3O.
18. (d) : Average isotopic mass of X

200 × 90 + 199 × 8 + 202 × 2
=
90 + 8 + 2
=

18000 + 1592 + 404
= 199.96 a.m.u. ≈ 200 a.m.u.
100

19. (a) : At STP, 22.4 L H2 = 6.023 × 1023 molecules
15 L H2 =
5 L N2 =

6.023 × 10 23 × 15
= 4.033 × 10 23
22.4
6.023 × 10 23 × 5
= 1.344 × 10 23
22.4

10 g of O2 =

6.023 × 10 23 × 10
= 1.882 × 10 23
32

20. (b) : 1 mole of any element contain 6.023 × 1023
number of molecules.
1 g mole of O2 = 32 g O2
⇒ 16 g of O2 = 0.5 g mole O2
1 g mole of N2 = 28 g N2
⇒ 7 g N2 = 0.25 g mole N2
1 g mole of H2 = 2 g H2
⇒ 2 g H2 = 1.0 g mole H2
1 g mole NO2 = 14 + 16 × 2 = 46
⇒ 16 g of NO2 = 0.35 mole NO2
2 g H2 (1 g mole H2) contain maximum molecules.
21. (a) : In peroxidase anhydrous enzyme 0.5% Se
is present means, 0.5 g Se is present in 100 g of
enzyme. In a molecule of enzyme one Se atom must
be present. Hence 78.4 g Se will be present in

100
× 78.4 = 1.568 × 104
0.5
22. (c) : Density = 1.17 g/cc.
⇒ 1 cc. solution contains 1.17 g of HCl
1.17 × 1000
= 32.05
∴ Molarity =
36.5 × 1
23. (a) : Specific volume (vol. of 1 g) cylindrical
virus particle = 6.02 × 10–2 cc/g
Radius of virus, r = 7 Å = 7 × 10–8 cm
Volume of virus = πr2l
22
× (7 × 10 −8 ) 2 × 10 × 10 −8 = 154 × 10–23 cc
=
7
Volume
wt. of one virus particle = Specific volume

154 × 10 −23
g
6 .0 2 × 1 0 − 2
∴ Molecular wt. of virus = wt. of NA particle

⇒

154 × 10 −23
× 6.02 × 10 −23 g/mol.
−2

6.02 × 10
= 15400 g/mol = 15.4 kg/mol

=

24. (c) : K3[Cu(CN)2] = 3(+1) + x + 2(–1) = 0
⇒ x = –1
As the oxidation no. of ‘Cu’ is –1 (–ve), so this
complex is unstable and is not formed.
25. (b) : BaCO3 → BaO + CO2
197 ⋅ 34 g → 22 ⋅ 4 L at N.T.P.
22 ⋅ 4
× 9 ⋅ 85 = 1 ⋅ 118 L
9 ⋅ 85 g →
197 ⋅ 34


7

Some Basic Concepts of Chemistry

⇒ 9 ⋅ 85 g BaCO3 will produce 1.118 L CO2 at N.T.P.
on the complete decomposition.
26. (b) : In A3(BC4)2, (+2) × 3 + 2[+5 + 4(-2)]
⇒ + 6 + 10 − 16 = 0
Hence in the compound A3(BC4)2, the oxidation no.
of ‘A’, ‘B’ and ‘C’ are +2, +5 and −2 respectively.
27. (d) : No. of molecules in 4.25 g NH 3

4.25

× 6.023 × 1023 = 2.5 × 6.023 × 1022
17
Number of atoms in 4.25 g NH3
= 4 × 2.5 × 6.023 × 10 22 = 6.023 × 1023
=

28. (d) : Zeros placed left to the number are never
significant, therefore the no. of significant figures
for the numbers.
161 cm = 0.161 cm and 0.0161 cm are same, i.e. 3
29. (a) : Quantity of iron in one molecule
67200
=
× 0.334 = 224.45 amu
100
No. of iron atoms in one molecule of haemoglobin
224.45
= 56 = 4
30. (a) : 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
4 mole + 5 mole → 4 mole + 6 mole
5
⇒ 1 mole of NH3 requires = = 1.25 mole of oxygen
4
4
while 1 mole of O2 requires =
= 0.8 mole of NH3.
5
As there is 1 mole of NH3 and 1 mole of O2, so all
oxygen will be consumed.
31. (c) : Ne2+(8) ⇒ 1s22s2 2 p x2 2 p1y 2 p1z

Be+(3) ⇒ 1s22s1
Cl–(18) ⇒ 1s22s22p63s23p6

As + (32) ⇒ 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p1x 4 p1y
Cl– is not paramagnetic , as it has no unpaired electron.

32. (b) : Weight of gas = 0.24 g, Volume of gas =
45 mL = 0.045 litre and density of H2 = 0.089.
We know that weight of 45 mL of H2 =
Density × Volume = 0.089 × 0.045 = 4.005 × 10−3 g
Therefore vapour density
Weight of certain volume of substance
=
Weight of same volume of hydrogen
0.24
= 59.93
=
4.005 × 10−3
33. (c) : Zn + H2SO4 → ZnSO4 + H2
(65 g)

(22400 mL)

Since 65 g of zinc reacts to liberate 22400 mL of H2
at STP, therefore amount of zinc needed to produce
224 mL of H2 at STP

65
× 224 = 0.65 g
22400

Force
34. (b) : Pressure =
Area
=

MLT −2
L2
= ML−1T–2
and dimensions of energy per unit volume
Energy
ML2T −2
=
=
= ML−1T−2
Volume
L3

Therefore dimensions of pressure =

35. (a) : Volume of oxygen in one litre of air
21
× 1000 = 210 mL
=
100
210
Therefore no. of mol =
= 0.0093 mol
22400
36. (c) : Each nitrogen atom has 5 valence
electrons, therefore total number of electrons in N3−

ion is 16. Since the molecular mass of N3 is 42,
therefore total number of electrons in 4.2 g of N3− ion
4.2
× 16 × N A = 1.6 NA
=
42
37. (a) : 5M H2SO4 = 10N H2SO4
N1V1 = N2V2 ị 10 ì 1 = N2 ì 10 ị N2 = 1N
38. (b) : Avogadro’s No., NA = 6.02 × 1023 molecules.
\ 6.02 × 1024 CO molecules = 10 moles CO
= 10 g atoms of O = 5 g molecules of O2
39. (a) : Average atomic mass
19 u 10  81 u 11
10.81
100
40. (c) : If 1L of one gas contains N molecules,
2 L of any gas under the same conditions will contain
2 N molecules.
41. (d) : Z2O3 + 3H2 ® 2Z + 3H2O
Valency of metal in Z2O3 = 3
0.1596 g of Z2O3 react with 6 mg of H2.
[1 mg = 0.001 g = 10–3g]

0.1596
= 26.6 g of Z2O3
0.006
\ Eq. wt. of Z2O3 = 26.6
Now, Eq. wt. of Z + Eq. wt. of O = Eq. wt. of Z + 8 = 26.6
Þ Eq. wt. of Z = 26.6 – 8 = 18.6
\ At. wt. of Z = 18.6 × 3 = 55.8

Atomic wt.
ê
ô Eq. wt = Valency of metal ằ
ơ
ẳ
\ 1 g of H2 react with

42. (a) : Here, Cp/CV = 1.4, which shows that the
gas is diatomic.
22.4 L at NTP = 6.02 × 1023 molecules


8

\ 11.2 L at NTP = 3.01 × 1023 molecules
Since gas is diatomic.
\ 11.2 L at NTP = 6.02 × 1023 atom
43.

46. (a) : Mg2+ + Na2CO3 → MgCO3 + 2Na+
1g eq.

(c) : C2H4 + 3O2 ® 2CO2 + 2H2O
28 g

96 g

96 g
2.8 kg C2H4 = 28 g u 2.8 kg

96
u 2.8 u 103 g = 9.6 × 103 g = 9.6 kg
28
44. (a) : 1 mol of CO2 = 44 g of CO2
\ 4·4 g CO2 = 0.1 mol CO2 = 6 × 1022 molecules

[Since, 1 mole CO2 = 6 × 1023 molecules]

= 2 × 6 × 1022 atoms of O = 1.2 × 1023 atoms of O
45. (a) : 1 mol CCl4 vapour = 12 + 4 × 35.5
= 154 g = 22.4 L
154
g L–1
\ Density of CCl4 vapour
22.4
= 6.875 g L

1g eq.

1g eq. of Mg2+ = 12g of Mg2+ = 12000 mg
Now, 1000 millieq. of Na2CO3 = 12000 mg of Mg2+
\
1 millieq. of Na2CO3 = 12 mg of Mg2+
47. (d) : As we know,
22400 cc of N2O contain 6.02 × 1023 molecules

–1

6.02 u 1023
molecules

22400
Since in N2O molecule there are 3 atoms

\

1 cc of N2O contain

3 u 6.02 u 1023
atoms
22400
22
1.8 u 10
atoms
224
No. of electrons in a molecule of N2O = 7 + 7 + 8 = 22

\

1 cc N2O

6.02 u 1023
u 22 electrons
22400

Hence, no. of electrons
=

1.32
u 1023 electrons
224

9

Structure of Atom

$IBQUFS

R…ƒ†h…†ƒpÁ€qÁ4…€x

"

1.

Which one is the wrong statement?
(a) The uncertainty principle is

Δ& × ΔU ≥

6.

The number of d-electrons in Fe2+ (Z = 26) is
not equal to the number of electrons in which
one of the following?
(a) d-electrons in Fe (Z = 26)
(b) p-electrons in Ne (Z = 10)
(c) s-electrons in Mg (Z = 12)
(d) p-electrons in Cl (Z = 17)
(2015, Cancelled)

7.

The angular momentum of electron in ‘d’ orbital
is equal to

I
π

(b) Half filled and fully filled orbitals have
greater stability due to greater exchange
energ y, greater symmetry and more
balanced arrangement.
(c) The energy of 2s-orbital is less than the
energy of 2p-orbital in case of hydrogen
like atoms.
(d) de-Broglie’s wavelength is given by

I
λ=
, where m = mass of the particle,
NW
v = group velocity of the particle.
(NEET 2017)

(a)

2 3=

(c)

6=

(b) 0 =
(d)

2=
(2015, Cancelled)

8.

What is the maximum number of orbitals that
can be identified with the following quantum
numbers?
n = 3, l = 1, ml = 0
(a) 1
(b) 2
(c) 3
(d) 4
(2014)

9.

Calculate the energy in joule corresponding
to light of wavelength 45 nm. (Planck’s
constant, h = 6.63 × 10–34 J s, speed of light,
c = 3 × 108 m s–1)
(b) 6.67 × 1011
(a) 6.67 × 1015
–15
(c) 4.42 × 10

(d) 4.42 × 10–18
(2014)

2.

How many electrons can fit in the orbital for
which n = 3 and l = 1?
(a) 2
(b) 6
(c) 10
(d) 14
(NEET-II 2016)

3.

Which of the following pairs of d-orbitals will
have electron density along the axes?
(a) dz2, dxz
(b) dxz, dyz
(c) dz2, dx2 – y2
(d) dxy, dx2 – y2
(NEET-II 2016)

4.

Two electrons occupying the same orbital are
distinguished by
(a) azimuthal quantum number
(b) spin quantum number
(c) principal quantum number

(d) magnetic quantum number.
(NEET-I 2016)

10. Be2+ is isoelectronic with which of the following
ions?
(a) H+
(b) Li+
(c) Na+ (d) Mg2+
(2014)

Which is the correct order of increasing energy
of the listed orbitals in the atom of titanium?
(At. no. Z = 22)
(a) 4s 3s 3p 3d
(b) 3s 3p 3d 4s
(c) 3s 3p 4s 3d
(d) 3s 4s 3p 3d
(2015)

11. 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) 4
(b) 2
(c) 10
(d) 6
(NEET 2013)

5.

10

⎛Z ⎞
12. Based on equation E = – 2.178 × 10–18 J ⎜ ⎟ ,
⎝n ⎠
certain conclusions are written. Which of them
is not correct?
(a) Equation can be used to calculate the
change in energy when the electron
changes orbit.
(b) For n = 1, the electron has a more negative
energy than it does for n = 6 which means
that the electron is more loosely bound in
the smallest allowed orbit.
(c) The negative sign in equation simply
means that the energy of electron bound
to the nucleus is lower than it would be if
the electrons were at the infinite distance
from the nucleus.
(d) Larger the value of n, the larger is the
orbit radius.
(NEET 2013)

18. The orbital angular momentum of a p-electron
is given as
(a)

h

S (b)

h

S

(c)

h

h

(d)
S
S
(Mains 2012)

19. The total number of atomic orbitals in fourth
energy level of an atom is
(a) 8
(b) 16
(c) 32
(d) 4
(2011)
20. The energies E1 and E2 of two radiations are
25 eV and 50 eV respectively. The relation
between their wavelengths i.e., λ1 and λ2 will
be
(a) λ1 = λ2

(b) λ1 = 2λ2
(c) λ1 = 4λ2

(d) O

O

(2011)

–34

13. The value of Planck’s constant is 6.63 × 10 J s.
The speed of light is 3 × 1017 nm s–1. Which
value is closest to the wavelength in nanometer
of a quantum of light with frequency of
6 × 1015 s–1?
(a) 50
(b) 75
(c) 10
(d) 25
(NEET 2013)
14. The outer electronic configuration of Gd
(At. No. 64) is
(a) 4f 5 5d4 6s1
(b) 4f 7 5d1 6s2
3
5
2
(c) 4f  5d 6s
(d) 4f 4 5d5 6s1

(Karnataka NEET 2013)
15. According to law of photochemical equivalence
the energy absorbed (in ergs/mole) is given
as (h = 6.62 × 10–27 ergs, c = 3 × 1010 cm s–1,
NA = 6.02 × 10–23 mol–1)
(a)

u
O

(b)

u
O

(c)

u
O

(d)

u
O

(Karnataka NEET 2013)

16. Maximum number of electrons in a subshell
with l = 3 and n = 4 is
(a) 14

(b) 16
(c) 10
(d) 12
(2012)
17. The correct set of four quantum numbers for
the valence electron of rubidium atom
(Z = 37) is
(a) 5, 1, 1, +1/2
(b) 6, 0, 0, +1/2
(c) 5, 0, 0, +1/2
(d) 5, 1, 0, +1/2
(2012)

21. If n = 6, the correct sequence for filling of
electrons will be
(a) ns → (n – 2)f → (n – 1)d → np
(b) ns → (n – 1)d → (n – 2)f → np
(c) ns → (n – 2)f → np → (n – 1)d
(d) ns → np(n – 1)d → (n – 2)f
(2011)
22. According to the Bohr theory, which of the
following transitions in the hydrogen atom
will give rise to the least energetic photon?
(a) n = 6 to n = 1
(b) n = 5 to n = 4
(c) n = 6 to n = 5
(d) n = 5 to n = 3
(Mains 2011)
23. A 0.66 kg ball is moving with a speed of 100 m/s.
The associated wavelength will be

(h = 6.6 × 10–34 J s)
(a) 6.6 × 10–32 m
(b) 6.6 × 10–34 m
–35
(c) 1.0 × 10 m
(d) 1.0 × 10–32 m
(Mains 2010)
24. Maximum number of electrons in a subshell
of an atom is determined by the following
(a) 2l + 1
(b) 4l – 2
(c) 2n2
(d) 4l + 2
(2009)
25. Which of the following is not permissible
arrangement of electrons in an atom?
(a) n = 5, l = 3, m = 0, s = +1/2
(b) n = 3, l = 2, m = –3, s = –1/2
(c) n = 3, l = 2, m = –2, s = –1/2
(d) n = 4, l = 0, m = 0, s = –1/2
(2009)
26. If uncertainty in position and momentum are
equal, then uncertainty in velocity is


11

Structure of Atom

(a)

1 h
m π

(b)

h
π

(c)

1 h
2m π

(d)

h
2π

(2008)

27. 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)
(b) 1 × 1011 cm s–1
(a) 1 × 105 cm s–1
9
–1

(c) 1 × 10 cm s
(d) 1 × 106 cm s–1
(Prelims 2008)
28. 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?
(a) (i), (ii), (iii) and (iv)

(b) (ii), (iv) and (v)
(c) (i) and (iii)
(d) (ii), (iii) and (iv).
(2007)
29. The orientation of an atomic orbital is governed
by
(a) principal quantum number
(b) azimuthal quantum number
(c) spin quantum number
(d) magnetic quantum number.
(2006)

32. The frequency of radiation emitted when the
electron falls from n = 4 to n = 1 in a hydrogen
atom will be (Given ionization energy of
H = 2.18 × 10–18 J atom–1 and h = 6.625 × 10–34 J s)
(b) 1.03 × 1015 s–1
(a) 1.54 × 1015 s–1
15 –1
(c) 3.08 × 10 s
(d) 2.00 × 1015 s–1
(2004)
33. The value of Planck’s constant is 6.63 × 10–34 J s.
The velocity of light is 3.0 × 108 m s–1. Which
value is closest to the wavelength in
nanometers of a quantum of light with
frequency of 8 × 1015 s –1 ?
(b) 5 × 10–18
(a) 2 × 10–25
1

(c) 4 × 10
(d) 3 × 107
(2003)
34. In hydrogen atom, energy of first excited state
is –3.4 eV. Then find out K.E. of same orbit
of hydrogen atom
(a) +3.4 eV
(b) +6.8 eV
(c) –13.6 eV
(d) +13.6 eV (2002)
35. Main axis of a diatomic molecule is z, molecular
orbital px and p y overlap to form which of the
following orbitals.
(a) π molecular orbital
(b) σ molecular orbital
(c) δ molecular orbital
(d) No bond will form.
(2001)
36. The following quantum numbers are possible
for how many orbitals : n = 3, l = 2, m = +2 ?
(a) 1
(b) 2
(c) 3
(d) 4
(2001)
37. For given energy, E = 3.03 × 10 –19 Joules
corresponding wavelength is
(h = 6.626 × 10 –34 J sec, c = 3 × 10 8 m/sec)
(a) 65.6 nm
(b) 6.56 nm

(c) 3.4 nm
(d) 656 nm
(2000)

30. Given : The mass of electron is 9.11 × 10–31 kg,
Planck constant is 6.626 × 10 –34 J s, the
uncertainty involved in the measurement of
velocity within a distance of 0.1 Å is
(a) 5.79 × 105 m s–1 (b) 5.79 × 106 m s–1
(c) 5.79 × 107 m s–1 (d) 5.79 × 108 m s–1
(2006)

38. Isoelectronic species are
(a) CO, CN–, NO+, C22–
(b) CO–, CN, NO, C2–
(c) CO+, CN+, NO–, C2
(d) CO, CN, NO, C2

31. The energy of second Bohr orbit of the
hydrogen atom is –328 kJ mol–1; hence the
energy of fourth Bohr orbit would be
(b) –82 kJ mol–1
(a) – 41 kJ mol–1
–1
(c) –164 kJ mol
(d) –1312 kJ mol–1
(2005)

39. The uncertainty in momentum of an electron
is 1 × 10 –5 kg m/s. The uncertainty in its

position will be (h = 6.62 × 10 –34 kg m2/s)
(b) 1.05 × 10–26 m
(a) 5.27 × 10–30 m
–28
(d) 5.25 × 10–28 m
(c) 1.05 × 10 m
(1999)

(2000)


12

40. Who modified Bohr’s theory by introducing
elliptical orbits for electron path?
(a) Rutherford
(b) Thomson
(c) Hund
(d) Sommerfield
(1999)

48. The radius of hydrogen atom in the ground
state is 0.53 Å. The radius of Li2+ ion (atomic
number = 3) in a similar state is
(a) 0.53 Å
(b) 1.06 Å
(c) 0.17 Å
(d) 0.265 Å (1995)

41. The de Broglie wavelength of a particle with

mass 1 g and velocity 100 m/s is
(b) 6.63 × 10–34 m
(a) 6.63 × 10–35 m
–33
(d) 6.65 × 10–35 m
(c) 6.63 × 10 m
(1999)

49. For which of the following sets of four
quantum numbers, an electron will have the
highest energy?
n
l
m
s
(a) 3
2
1 +1/2
(b) 4
2
– 1 +1/2
(c) 4
1
0 –1/2
(d) 5
0
0 –1/2
(1994)

42. The Bohr orbit radius for the hydrogen atom

(n = 1) is approximately 0.530 Å. The radius
for the first excited state (n = 2) orbit is (in Å)
(a) 4.77 (b) 1.06 (c) 0.13 (d) 2.12
(1998)
43. The position of both, an electron and a helium
atom is known within 1.0 nm. Further the
momentum of the electron is known within
5.0 × 10–26 kg m s–1. The minimum uncertainty
in the measurement of the momentum of the
helium atom is
(a) 8.0 × 10–26 kg m s–1
(b) 80 kg m s–1
(c) 50 kg m s–1
(d) 5.0 × 10–26 kg m s–1
(1998)
44. The ion that is isoelectronic with CO is
(a) CN– (b) N2+
(c) O2–
(d) N–2
(1997)
45. What will be the longest wavelength line in
Balmer series of spectrum?
(a) 546 nm
(b) 656 nm
(c) 566 nm
(d) 556 nm
(1996)
46. In a Bohr’s model of an atom, when an electron
jumps from n = 1 to n = 3, how much energy
will be emitted or absorbed?

(a) 2.389 × 10–12 ergs
(b) 0.239 × 10–10 ergs
(c) 2.15 × 10–11 ergs
(1996)
(d) 0.1936 × 10–10 ergs
47. Uncertainty in position of an electron (mass
= 9.1 × 10–28 g) moving with a velocity of
3 × 104 cm/s accurate upto 0.001% will be
(Use h/(4π) in uncertainty expression where
h = 6.626 × 10–27 erg second)
(a) 5.76 cm
(b) 7.68 cm
(c) 1.93 cm
(d) 3.84 cm
(1995)

50. Which one of the following is not isoelectronic
with O2–?
(a) Tl+
(b) Na+ (c) N3–
(d) F–
(1994)
51. Electronic configuration of calcium atom can
be written as
(a) [Ne] 4p2
(b) [Ar] 4s2
2
(c) [Ne] 4s
(d) [Kr] 4p2 (1992)
52. The energy of an electron in the nth Bohr orbit

of hydrogen atom is
(a)
(c)

13.6
n

4

13.6
n2

eV

(b)

eV

(d)

13.6
n3

eV

13.6
eV
n

(1992)

53. In a given atom no two electrons can have
the same values for all the four quantum
numbers. This is called
(a) Hund’s Rule
(b) Aufbau principle
(c) Uncertainty principle
(d) Pauli’s Exclusion principle.
(1991)
54. For azimuthal quantum number l = 3, the
maximum number of electrons will be
(a) 2
(b) 6
(c) 0
(d) 14
(1991)
55. The order of filling of electrons in the orbitals
of an atom will be
(a) 3d, 4s, 4p, 4d, 5s
(b) 4s, 3d, 4p, 5s, 4d
(c) 5s, 4p, 3d, 4d, 5s
(d) 3d, 4p, 4s, 4d, 5s
(1991)


13

Structure of Atom

56. The electronic configuration of Cu (atomic

number 29) is
(a) 1s2, 2s22p6, 3s23p6, 4s23d9
(b) 1s2, 2s22p6, 3s23p63d10, 4s1
(c) 1s2, 2s22p6, 3s23p6, 4s24p6, 5s25p1
(d) 1s2, 2s22p6, 3s23p6, 4s24p63d3
(1991)
57. The total number of electrons that can be
accommodated in all the orbitals having
principal quantum number 2 and azimuthal
quantum number 1 are
(a) 2
(b) 4
(c) 6
(d) 8
(1990)
58. An ion has 18 electrons in the outermost shell,
it is
(a) Cu+
(b) Th 4+
+
(c) Cs
(d) K+
(1990)
59. Which of the following statements do not form
a part of Bohr’s model of hydrogen atom?
(a) Energy of the electrons in the orbits are
quantized.
(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 the electrons
in the orbit cannot be determined
simultaneously.
(1989)
60. Number of unpaired electrons in N2+ is/are
(a) 2
(b) 0
(c) 1
(d) 3
(1989)
61. The maximum number of electrons in a subshell
is given by the expression
(a) 4l – 2
(b) 4l + 2
(c) 2l + 2
(d) 2n 2
(1989)
62. The spectrum of He is expected to be similar
to that
(a) H
(b) Li+
(c) Na
(d) He+
(1988)
63. The number of spherical nodes in 3p orbitals
are/is
(a) one
(b) three

(c) none
(d) two
(1988)
64. If r is the radius of the first orbit, the radius
of nth orbit of H-atom is given by
(a) r0n 2
(b) r0 n
(c) r0/n
(d) r02n 2
(1988)

Answer Key

1.
11.
21.
31.
41.
51.
61.

(c)
(b)
(a)
(b)
(c)
(b)
(b)

2.

12.
22.
32.
42.
52.
62.

(a)
(b)
(c)
(c)
(d)
(c)
(b)

3.
13.
23.
33.
43.
53.
63.

(c)
(a)
(c)
(c)
(d)
(d)
(a)

4.
14.
24.
34.
44.
54.
64.

(b)
(b)
(d)
(a)
(a)
(d)
(a)

5.
15.
25.
35.
45.
55.

(c)
(a)
(b)
(a)
(b)
(b)

6.
16.
26.
36.
46.
56.

(d)
(a)
(c)
(a)
(d)
(b)

7.
17.
27.
37.
47.
57.

(c)
(c)
(c)
(d)
(c)
(c)

8.

18.
28.
38.
48.
58.

(a)
(a)
(b)
(a)
(c)
(a)

9.
19.
29.
39.
49.
59.

(d)
(b)
(d)
(a)
(b)
(d)

10.
20.
30.

40.
50.
60.

(b)
(b)
(b)
(d)
(a)
(c)


14

1. (c) : In case of hydrogen like atoms, energy
depends on the principal quantum number only.
Hence, 2s-orbital will have energy equal to 2p-orbital.
2. (a) : For n = 3 and l = 1, the subshell is 3p and a
particular 3p orbital can accommodate only 2
electrons.
3. (c) : dx2 – y2 and dz2 orbitals have electron density
along the axes while dxy, dyz and dxz orbitals have
electron density inbetween the axes.
4. (b) : For the two electrons occupying the same
orbital values of n, l and ml are same but ms is


different, i.e.,  BOE  



5. (c) : Ti(22) : 1s22s22p63s23p64s23d2
∴ Order of increasing energy is 3s, 3p, 4s, 3d
6. (d) : Number of d-electrons in Fe2+ = 6
Number of p-electrons in Cl = 11
7. (c) : Angular momentum
For d orbital, l = 2
Angular momentum

l (l  1) =

2(2  1) =

6=

8. (a) : Only one orbital, 3pz has following set of
quantum numbers, n = 3, l = 1 and ml = 0.
hc
9. (d) : E
[Given, λ = 45 nm = 45 × 10–9 m]

O

On putting the given values in the equation, we get
6.63 u 10 34 u 3 u 108
4.42 u 10 18 J
E
45 u 10 9
10. (b) :
Species
No. of electrons

Be2+
2
H+
0
Li+
2
Na+
10
Mg 2+
10
11. (b) : The orbital associated with n = 3, l = 1 is 3p.
One orbital (with m = –1) of 3p-subshell can
accomodate maximum 2 electrons.
12. (b) : The electron is more tightly bound in the
smallest allowed orbit.
13. (a) : c = υλ
λ=

×
c
=
υ
×

= 50 nm

14. (b) : The electronic configuration of Gd 64 is
[Xe]4f 75d16s2.
15. (a) : We know that, E

hcNA

O



u
O

u u
O

u

u
u
O

u
Ž›œ –˜• 

16. (a) : l = 3 and n = 4 represent 4f. So, total number
of electrons in a subshell = 2(2l + 1) = 2(2 × 3 + 1) = 14
electrons. Hence f-subshell can contain maximum
14 electrons.
17. (c) : Rb(37) : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1
For 5s, n = 5, l = 0, m = 0, s = +1/2 or –1/2
18. (a) : Orbital angular momentum
h

l l

(m)

S
For p-electrons; l = 1


Thus, m

h

h

h

S
S
S
19. (b) : Total number of atomic orbitals in any
energy level is given by n2.
hc

20. (b) : E
E
E

or

hc u

O
O
O

O
O

hc

˜›

Š— E

O
O

O
O

hc

O

;

Ÿ O

O

21. (a)
22. (c) : We know that
⎡1
1 ⎤
ΔE ∝ ⎢ 2 − 2 ⎥ , where n2 > n1
⎣⎢ n1 n2 ⎦⎥

∴ n = 6 to n = 5 will give least energetic photon.
23. (c) : According to de-Broglie equation, O

h
mv

Given, h = 6.6 × 10–34 J s ; m = 0.66 kg ; v = 100 m s–1

O

u



–

u 
u
24. (d) : For a given shell, l,
the number of subshells, ml = (2l + 1)
Since each subshell can accommodate 2 electrons
of opposite spin, so maximum number of electrons
in a subshell = 2(2l + 1) = 4l + 2.

25. (b) : In an atom, for any value of n, the values of
l = 0 to (n – 1).
For a given value of l, the values of ml = –l to 0 to +l
and the value of s = +1/2 or –1/2.
In option (b), l = 2 and ml = –3
This is not possible, as values of ml which are
possible for l = 2 are –2, –1, 0, +1 and +2 only.
∴


15

Structure of Atom

26. (c) : From Heisenberg uncertainty principle
h
h
Δp ⋅ Δx ≥
or mΔv × Δx ≥
4π
4π
h
2
or (mΔv ) ≥
( Δx = Δp )
4π
1 h
or Δv ≥
2m π
27. (c) : Uncertainty in momentum

(mΔv) = 1 × 10–18 g cm s–1
Uncertainty in velocity,
1 × 10−18
( Δv) =
= 1.1 × 109 cm s −1
9 × 10−28
28. (b) : (i) represents an electron in 3s orbital.
(ii) is not possible as value of l varies from
0, 1, ... (n – 1).
(iii) represents an electron in 4f orbital.
(iv) is not possible as value of m varies from
– l ... +l.
(v) is not possible as value of m varies from
–l ... +l, it can never be greater than l.
29. (d) : Principal quantum number represents the
name, size and energy of the shell to which the
electron belongs.
Azimuthal quantum number describes the spatial
distribution of electron cloud and angular
momentum. Magnetic quantum number describes
the orientation or distribution of electron cloud. Spin
quantum number represents the direction of electron
spin around its own axis.

'

30. (b) : Δx ⋅ mΔv = h / 4π
0.1 × 10 −10 × 9.11 × 10−31 × Δv =

∴ Δv =

6.626 × 10 −34
4 × 3.143

6.626 ×10−34

0.1× 10−10 × 9.11× 10−31 × 4 × 3.143
= 5.79 × 106 m s–1
2

⎛Z⎞
31. (b) : E n = – K ⎜ ⎟
⎝n⎠
Z = 1 for hydrogen ; n = 2
–K × 1
E2 =
⇒ E2 = –328 kJ mol–1 ; K = 4 × 328
4
– K ×1
1
E4 =
⇒ E4= – 4 × 328 ×
= – 82 kJ mol–1
16
16
32. (c) : E = hυ or υ = E/h

−21.76 × 10 − 19
J atm − 1
n2

1
1
ΔE = − 21.76 × 10 − 19 2 − 2 = 20.40 × 10–19 J atm–1
4
1
20.40 × 10 − 19
υ=
= 3.079 × 10 15 s − 1
6.626 × 10 − 34

For H atom, E =

(

)

33. (c) : Applying (υ) = c/λ,
8
λ = c = 3 × 1015 = 37.5 × 10 −9 m = 37.5 nm ≈4 × 101 nm
υ 8 × 10

2

1 2 ⎛ πe2 ⎞
34. (a) : Kinetic energy = 2 mv = ⎜ nh ⎟ × 2m
⎝
⎠

2 Se 2
ằ

nh ẳ

'v

ê
ô
ơ

2

e2
2 2 me 4
Total energy En = − 2 2 = − ⎜ nh ⎟ × 2m = –K.E.
n h
⎝
⎠
∴ Kinetic energy = –En
Energy of first excited state is –3.4 eV
∴ Kinetic energy of same orbit (n = 2) will
be +3.4 eV.
35. (a) : For π overlap, the lobes of the atomic orbitals
are perpendicular to the line joining the nuclei.







¯

¯

¯

px

py

πoverlap

Hence, only sidewise overlapping takes place.
36. (a) : n = 3, l = 2, m = +2
It symbolises one of the five d-orbitals (3d).
m = +2 +1 0 –1 –2

hc ⇒
6 ⋅ 6 × 10−34 × 3 × 108
λ=
λ
3 ⋅ 03 × 10−19
= 656 nm
38. (a) : Species having same no. of electrons are
called isoelectronics.
The no. of electrons in CO = CN– = NO+ = C22– = 14.
So these are isoelectronics.
h
39. (a) : Δx × Δp =
4π
(Heisenberg uncertainty principle)

6.62 × 10−34
= 5.27 × 10–30 m
⇒ Δx =
4 × 3.14 × 10−5
40. (d) : Sommerfield modified Bohr’s theory
considering that in addition to circular orbits
electrons also move in elliptical orbits.
37. (d) : E =

41. (c) : λ =

h
6.63 × 10 −27 erg sec
=
mv
1g × 10 4 cm/s

= 6.63 × 10–31 cm = 6.63 × 10–33 m
42. (d) : for nth orbit of ‘H’ atom, rn = n2 × r1
⇒ radius of 2nd Bohr’s orbit.
r2 = 4 × r1 = 4 × 0.530 = 2.120 Å


16

43. (d) : According to uncertainty principle the
product of uncertainty in position and uncertainty
in momentum is constant for a particle.
h
i.e., Δx × Δp =

4π
As, Δx =1.0 nm for both electron and helium atom,
so Δp is also same for both the particles.
Thus uncertainty in momentum of the helium atom
is also 5.0 × 10–26 kg m s–1.
44. (a) : Since both CO and CN– have 14 electrons,
therefore these are isoelectronic (i.e. having same
number of electrons).
45. (b) : The longest wavelength means the lowest
energy. We know that relation for wavelength
1
1 Ã
Đ 1
RH ă 2  2 á
O
n2 ạ
â n1
here, n1 = 2, n2 = 3
RH (Rydberg constant = 109677 cm–1
⎛ 1
1
1 ⎞
⎟ = 15233
= 109677 ⎜
−
⎜ (2 )2 (3)2 ⎟
λ
⎝
⎠
1

or, λ =
= 6.56 × 10–5 cm = 6.56 × 10–7 m = 656 nm
15233
46. (d) : Energy of an atom when n = 1
E1 = −

1312
= – 1312 kJ mol-1
(1)2

Similarly energy when n = 3, (E3) = −

1312

(3)
= – 145.7 kJ mol-1
The energy absorbed when an electron jumps from
n = 1 to n = 3
E3 – E1 = – 145.7 – (– 1312) = 1166.3 kJ mol–1
2

1166.3
= 193.6 × 10–23 kJ
6.023 × 1023
= 193.6 × 10–20 J [1 Joule = 107 ergs]
=

⇒ 193.6 × 10–13 ergs = 0.1936 × 10-10 ergs
47. (c) : Mass of an electron (m) = 9.1 × 10–28 g
Velocity of electron (v) = 3 × 104 cm/s

0.001
Accuracy = 0.001% =
and Planck’s constant
100
-27
(h) = 6.626 × 10 erg-second.
We know that actual velocity of the electron
0.001
= 0.3 cm/s
(Δv) = 3 × 104 ×
100

Therefore, uncertainty in the position of the electron
h
6.626 × 10−27
=
= 1.93 cm
(Δx) =
4πmΔv 4π × 9.1 × 10−28 × 0.3
48. (c) : Due to ground state, state of hydrogen
atom (n) = 1
Radius of hydrogen atom (r) = 0.53 Å
Atomic no. of Li (Z) = 3
n2
(1) 2
0.53 u
0.17Å
Now, radius of Li2+ ion r u
Z
3

49. (b) : Energy of electron depends on the value of
(n + l). The subshell are 3d, 4d, 4p and 5s, 4d has
highest energy.
50. (a) : The number of electrons in O2–, N3–, F– and
Na+ is 10 each, but number of electrons in Tl+ is 80.
51. (b) : Atomic No. of Ca = 20
∴ Electronic configuration of Ca = [Ar]4s2
52. (c) : Energy of an electron in nth Bohr orbit of
13.6
eV
hydrogen atom
n2
53. (d) : This is a Pauli’s Exclusion principle.
54. (d) : l = 3 means f-subshell
Maximum no. of electrons in f-subshell = 14

(

)

f-subshell =
55. (b) : As per Aufbau Principle.
The principle states : In the ground state of the
atoms, the orbitals are filled in order of their
increasing energies.
56. (b) : Electronic configuration of Cu
= 1s22s22p63s23p63d104s1
57. (c) : n = 2, l = 1
It means 2p-orbitals
Total no. of electrons that can be accomodated in

2p orbitals = 6
58. (a) : Electronic configuration of Cu+ = [Ar]3d10
59. (d) : It is uncertainty principle and not Bohr’s
postulate.
60. (c) : N2+ = 1s22s22p1x
∴ No. of unpaired electrons = 1
61. (b) : No. of orbitals in a subshell = 2l + 1
⇒ No. of electrons = 2(2l + 1) = 4l + 2
62. (b) : Both He and Li+ contain 2 electrons each.
63. (a) : No. of radial nodes in 3p-orbital = n – l – 1
=3–1–1=1
64. (a) : Radius of nth orbit of H-atom = r0n2
where r0 = radius of the first orbit