Preview Arihant Chemistry JEE Main Chapterwise Solutions 20192002 Solved Papers by Arihant Prakashan Series (2020)

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All the 16 Question Papers

of JEE Main Online 2019 (Jan & Apr Attempt)

ARIHANT PRAKASHAN (Series), MEERUT

CHAPTERWISE

SOLUTIONS

2 0 1 9 - 2 0 0 2

JEE Main

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JEE Main is a gateway examination for candidates expecting to seek admission in Bachelor in Engineering (BE), Bachelor of Technology (B.Tech) and Bachelor of Architecture (B.Arch) at Indian Institutes of Information Technology (IIITs), National Institutes of Technology (NITs), Delhi Technological University and other Centrally Funded Technical Institutes (CFTIs)

JEE Main is also an examination which is like screening examination for JEE Advanced (The gateway examination to India's most reputed Technical Institutes, Indian Institutes of Technology— IITs).

We hope this book would be highly beneficial for the students We would be grateful if any discrepancy or mistake in the questions or answers is brought to our notice so that these could be rectified in subsequent editions.

To make the students well-versed with the pattern as well as the level of the questions asked in the exam, this book contains

Chapterwise Solutions of Questions asked in Last 18 Years’ Examinations of JEE Main (formerly known as AIEEE) Solutions to

all the questions have been kept very detailed and accurate Along with the indication of level of the exam, this book will also teach you

to solve the questions objectively in the examination.

To give the students a complete practice, along with Chapterwise Solutions, this book contains 5 Practice Sets, based exactly on JEE Main Syllabus and Pattern By practicing these sets, students can attain efficiency in Time Management during the examination

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PREFACE

Solutions 57-72 6.

Analytical Chemistry and Chemistry in Everyday Life 308-314 20.

Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 239-276

UNIT 1 Some Basic Concepts in hemistry

Matter and its nature, Dalton's atomic theory;

Concept of atom, molecule, element and

compound; Physical quantities and their

measurements in Chemistry, precision and

accuracy, significant figures, S.I Units,

dimensional analysis; Laws of chemical

combination; Atomic and molecular masses,

mole concept, molar mass, percentage

composition, empirical and molecular

formulae; Chemical equations and

stoichiometry

UNIT 2 States of Matter

Gaseous State Measurable properties of gases;

Gas laws - Boyle's law, Charle's law, Graham's

law of diffusion, Avogadro's law, Dalton's law of

partial pressure; Concept of Absolute scale of

temperature; Ideal gas equation, Kinetic theory

of gases (only postulates); Concept of average,

root mean square and most probable velocities;

Real gases, deviation from Ideal behaviour,

compressibility factor, van der Waals’

Classification of matter into solid, liquid and

gaseous states

Equation, liquefaction of gases, critical

constants

Liquid State Properties of liquids - vapour

pressure, viscosity and surface tension and

effect of temperature on them (qualitative

treatment only)

Solid State Classification of solids: molecular, ionic, covalent and metallic solids, amorphous and crystalline solids (elementary idea); Bragg's Law and its applications, Unit cell and lattices, packing in solids (fcc, bcc and hcp lattices), voids, calculations involving unit cell parameters, imperfection in solids; electrical, magnetic and dielectric properties

UNIT 3 Atomic Structure

Discovery of sub-atomic particles (electron, proton and neutron); Thomson and Rutherford atomic models and their limitations; Nature of electromagnetic radiation, photoelectric effect; spectrum of hydrogen atom, Bohr model of hydrogen atom - its postulates, derivation of the relations for energy of the electron and radii of the different orbits, limitations of Bohr's model; dual nature of matter, de-Broglie's relationship, Heisenberg uncertainty principle

Elementary ideas of quantum mechanics, quantum mechanical model of atom, its important features,ψ and ψ2, concept of atomic orbitals as one electron wave functions; Variation

of and ψ ψ2 with r for 1s and 2s orbitals; various quantum numbers (principal, angular

momentum and magnetic quantum numbers) and their significance; shapes of s, p and d - orbitals, electron spin and spin quantum number; rules for filling electrons in orbitals – aufbau principle, Pauli's exclusion principle and Hund's rule, electronic configuration of elements, extra stability of half-filled and completely filled orbitals

SECTION- A (Physical Chemistry)

SYLLABUS

Electrochemical cells - Electrolytic and Galvanic cells, different types of electrodes, electrode potentials including standard electrode potential, half - cell and cell reactions, emf of a Galvanic cell and its

UNIT 5 Chemical Thermodynamics

UNIT 7 Equilibrium

Covalent Bonding Concept of electronegativity,

Fajan's rule, dipole moment; Valence Shell

Electron Pair Repulsion (VSEPR) theory and

shapes of simple molecules

Fundamentals of thermodynamics System and

surroundings, extensive and intensive

properties, state functions, types of processes

UNIT 8 Redox Reactions and Electrochemistry

First law of thermodynamics Concept of work,

heat internal energy and enthalpy, heat capacity,

molar heat capacity, Hess's law of constant heat

summation; Enthalpies of bond dissociation,

combustion, formation, atomization,

sublimation, phase transition, hydration,

ionization and solution

Elementary idea of metallic bonding Hydrogen

bonding and its applications

Molecular Orbital Theory Its important features,

LCAOs, types of molecular orbitals (bonding,

antibonding), sigma and pi-bonds, molecular

orbital electronic configurations of homonuclear

diatomic molecules, concept of bond order,

bond length and bond energy

Kossel Lewis approach to chemical bond

formation, concept of ionic and covalent bonds

Equilibria involving chemical processes Law of chemical equilibrium, equilibrium constants(K and K) and their significance, significance of G Δand G in chemical equilibria, factors affecting oΔ

equilibrium concentration, pressure, temperature, effect of catalyst; Le -Chatelier’s principle

UNIT 6 Solutions

Quantum mechanical approach to covalent

bonding Valence bond theory - Its important

features, concept of hybridization involving s, p

and d orbitals; Resonance

Second law of thermodynamics Spontaneity of

processes; ΔS of the universe and G of the Δ

system as criteria for spontaneity, G o

Δ(Standard Gibb's energy change) and

equilibrium constant

Meaning of equilibrium, concept of dynamic equilibrium

Ionic equilibrium Weak and strong electrolytes,

ionization of electrolytes, various concepts of acids and bases (Arrhenius, Bronsted - Lowry and Lewis) and their ionization, acid-base equilibria (including multistage ionization) and ionization constants, ionization of water, pH scale, common ion effect, hydrolysis of salts and pH of their solutions, solubility

of sparingly soluble salts and solubility products, buffer solutions

Equilibria involving physical processes Solid

-liquid, liquid - gas and solid - gas equilibria, Henry’s law, general characteristics of equilibrium involving physical processes

Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for assigning oxidation number, balancing of redox reactions

Colligative properties of dilute solutions - relative lowering of vapour pressure, depression of freezing point, elevation of boiling point and osmotic pressure; Determination of molecular mass using colligative properties; Abnormal value of molar mass, van’t Hoff factor and its significance

Eectrolytic and metallic conduction, conductance

in electrolytic solutions, specific and molar conductivities and their variation with

concentration: Kohlrausch's law and its

applications.

Different methods for expressing

concentration of solution - molality, molarity,

mole fraction, percentage (by volume and mass

both), vapour pressure of solutions and Raoult's

Law - Ideal and non-ideal solutions, vapour

Ionic Bonding Formation of ionic bonds, factors

affecting the formation of ionic bonds;

calculation of lattice enthalpy

SECTION- B (Inorganic Chemistry)

General introduction, electronic configuration and general trends in physical and chemical properties

of elements, anomalous properties of the first element of each group, diagonal relationships

UNIT 11 Classification of Elements and

Periodicity in Properties

Periodic Law and Present Form of the Periodic

Table, s, p, d and Block Elements, Periodic f

Trends in Properties of Elementsatomic and Ionic

Radii, Ionization Enthalpy, Electron Gain

Enthalpy, Valence, Oxidation States and

Chemical Reactivity

UNIT 13 Hydrogen

Group 1 and 2 Elements

Modes of occurrence of elements in nature,

minerals, ores; steps involved in the extraction of

metals - concentration, reduction (chemical and

electrolytic methods) and refining with special

reference to the extraction of Al, Cu, Zn and Fe;

Thermodynamic and electrochemical principles

involved in the extraction of metals

Position of hydrogen in periodic table, isotopes,

preparation, properties and uses of hydrogen;

physical and chemical properties of water and

heavy water; Structure, preparation, reactions

and uses of hydrogen peroxide; Classification of

hydrides ionic, covalent and interstitial;

Hydrogen as a fuel

UNIT 12 General Principles and

Processes of Isolation of Metals

UNIT 14 s - Block Elements

(Alkali and Alkaline Earth Metals)

Preparation and properties of some important compounds - sodium carbonate, sodium chloride, sodium hydroxide and sodium hydrogen carbonate; Industrial uses of lime, limestone, Plaster of Paris and cement; Biological significance of Na, K, Mg and Ca

Group wise study of the – block elements p

UNIT 15 p - Block Elements

Group 13 to Group 18 Elements

General Introduction Electronic configuration and general trends in physical and chemical properties

of elements across the periods and down the groups; unique behaviour of the first element in each group

Group 13 Preparation, properties and uses of boron

and aluminium; structure, properties and uses of borax, boric acid, diborane, boron trifluoride, aluminium chloride and alums

and Gibbs’ energy change; Dry cell and lead

accumulator; Fuel cells; Corrosion and its

prevention

Colloidal state distinction among true solutions, colloids and suspensions, classification of colloids - lyophilic, lyophobic; multi molecular, macromole-cular and associated colloids (micelles), preparation and properties of colloids Tyndall effect, Brownian movement,

electrophoresis, dialysis, coagulation and flocculation; Emulsions and their characteristics

Rate of a chemical reaction, factors affecting the rate

of reactions concentration, temperature, pressure

and catalyst; elementary and complex reactions,

order and molecularity of reactions, rate law, rate

constant and its units, differential and integral forms

of zero and first order reactions, their characteristics

and half - lives, effect of temperature on rate of

reactions - Arrhenius theory, activation energy and

its calculation, collision theory of bimolecular

gaseous reactions (no derivation)

their characteristics, factors affecting adsorption of gases on solids- Freundlich and Langmuir adsorption isotherms, adsorption from solutions

UNIT 9 Chemical Kinetics Catalysis Homogeneous and heterogeneous,

activity and selectivity of solid catalysts, enzyme catalysis and its mechanism

Group 17 Preparation, properties and uses of

chlorine and hydrochloric acid; Trends in the acidic

nature of hydrogen halides; Structures of

Interhalogen compounds and oxides and oxoacids

of halogens

Transition Elements General introduction, electronic

configuration, occurrence and characteristics,

general trends in properties of the first row

transition elements - physical properties, ionization

enthalpy, oxidation states,

atomic radii, colour, catalytic behaviour, magnetic

properties, complex formation, interstitial

compounds, alloy formation; Preparation,

properties and uses of K Cr O and KMnO 2 2 7 4

Group 16 Preparation, properties, structures and

uses of dioxygen and ozone; Allotropic forms of

sulphur; Preparation, properties, structures and uses

of sulphur dioxide, sulphuric acid (including its

industrial preparation); Structures of oxoacids of

sulphur

Group 18 Occurrence and uses of noble gases;

Structures of fluorides and oxides of xenon

UNIT 16 d – and f – Block Elements

carbon, silicon tetrachloride, silicates, zeolites and

silicones

Group 15 Properties and uses of nitrogen and

phosphorus; Allotrophic forms of phosphorus;

Preparation, properties, structure and uses of

ammonia nitric acid, phosphine and phosphorus

halides,(PCl , PCl ); Structures of oxides and oxoacids 3 5

of nitrogen and phosphorus

Inner Transition Elements Lanthanoids Electronic

configuration, oxidation states, chemical reactivity

and lanthanoid contraction Actinoids Electronic

configuration and oxidation states

Werner's theory; ligands, coordination number, denticity, chelation; IUPAC nomenclature of mononuclear coordination compounds, isomerism; Bonding Valence bond approach and basic ideas of Crystal field theory, colour and magnetic properties; importance of coordination compounds

(in qualitative analysis, extraction of metals and in biological systems)

Unit 18 Environmental Chemistry

Particulate pollutants Smoke, dust, smog, fumes, mist; their sources, harmful effects and

Stratospheric pollution Formation and breakdown of ozone, depletion of ozone layer - its mechanism and effects

Environmental pollution Atmospheric, water and soil

Water pollution Major pollutants such as, pathogens, organic wastes and chemical pollutants their harmful effects and prevention.Soil pollution Major pollutants such as: Pesticides (insecticides, herbicides and fungicides), their harmful effects and prevention

Strategies to control environmental pollution

Nomenclature (Trivial and IUPAC)

Electronic displacement in a covalent bond Inductive

effect, electromeric effect, resonance and

hyperconjugation

UNIT 19 Purification & Characterisation

of Organic Compounds

Calculations of empirical formulae and molecular

formulae; Numerical problems in organic quantitative

analysis

Alcohols, Phenols and Ethers

Purification Crystallization, sublimation, distillation,

differential extraction and chromatography principles

and their applications

Qualitative analysis Detection of nitrogen, sulphur,

phosphorus and halogens

UNIT 20 Some Basic Principles

of Organic Chemistry

Quantitative analysis (basic principles only)

Estimation of carbon, hydrogen, nitrogen, halogens,

sulphur, phosphorus

Tetravalency of carbon; Shapes of simple molecules

hybridization ( and ); Classification of organic s p

compounds based on functional groups:

—C=C—,—C=C— and those containing halogens,

oxygen, nitrogen and sulphur, Homologous series;

Isomerism - structural and stereoisomerism

Covalent bond fission Homolytic and heterolytic free

radicals, carbocations and carbanions; stability of

carbocations and free radicals, electrophiles and

nucleophiles

Common types of organic reactions Substitution,

addition, elimination and rearrangement

UNIT 21 Hydrocarbons

Alkanes Conformations: Sawhorse and Newman

projections (of ethane); Mechanism of halogenation of

alkanes

Alkenes Geometrical isomerism; Mechanism of

electrophilic addition: addition of hydrogen, halogens,

water, hydrogen halides (Markownikoff's and peroxide effect); Ozonolysis, oxidation, and polymerization

Aromatic hydrocarbons Nomenclature, benzene structure and aromaticity; Mechanism of electrophilic substitution: halogenation, nitration, Friedel – Craft's alkylation and acylation, directive influence of functional group in mono-substituted benzene

Classification, isomerism, IUPAC nomenclature, general

methods of preparation, properties and reactions

Alkynes acidic character; addition of hydrogen, halogens, water and hydrogen halides;

UNIT 23 Organic Compounds Containing Oxygen

General methods of preparation, properties, reactions and uses

Alcohols Identification of primary, secondary and tertiary alcohols; mechanism of dehydration.Phenols Acidic nature, electrophilic substitution reactions: halogenation, nitration and

sulphonation, Reimer - Tiemann reaction

Ethers Structure

Aldehyde and Ketones Nature of carbonyl group;Nucleophilic addition to >C=O group, relative reactivities of aldehydes and ketones; Important reactions such as - Nucleophilic addition reactions (addition of HCN, NH and its derivatives), Grignard

3reagent; oxidation; reduction (Wolff Kishner and Clemmensen) acidity of -hydrogen, aldol αcondensation, Cannizzaro reaction, Haloform reaction; Chemical tests to distinguish between aldehydes and Ketones Carboxylic Acids Acidic strength and factors affecting it

General methods of preparation, properties, reactions

and uses

Amines Nomenclature, classification, structure basic

character and identification of primary, secondary and

tertiary amines and their basic character

Diazonium Salts Importance in synthetic organic

Cl , Br , I (Insoluble salts excluded).- -

-UNIT 27 Chemistry in Everyday Life

Carbohydrates Classification: aldoses and ketoses;

monosaccharides (glucose and fructose), constituent

monosaccharides of oligosacchorides (sucrose, lactose,

maltose) and polysaccharides (starch, cellulose,

glycogen)

General introduction and importance of biomolecules

— Detection of extra elements (N, S, halogens) in organic compounds; Detection of the following functional groups: hydroxyl (alcoholic and phenolic), carbonyl (aldehyde and ketone), carboxyl and amino groups in organic compounds

Cleansing agents Soaps and detergents, cleansing action

General introduction and classification of polymers,

general methods of polymerization-addition and

condensation, copolymerization; Natural and synthetic

rubber and vulcanization; some important polymers

with emphasis on their monomers and uses -

polythene, nylon, polyester and bakelite

Nucleic Acids Chemical constitution of DNA and RNA

Biological functions of Nucleic acids

— Organic compounds Acetanilide, p-nitroacetan ilide, aniline yellow, iodoform

UNIT 28 Principles Related to Practical Chemistry

Chemicals in medicines Analgesics, tranquilizers,

antiseptics, disinfectants, antimicrobials, antifertility

Proteins Elementary Idea of -amino acids, peptide α

bond, polypeptides; proteins: primary, secondary,

tertiary and quaternary structure (qualitative idea

only), denaturation of proteins, enzymes

UNIT 26 Biomolecules

Vitamins Classification and functions

Chemicals in food Preservatives, artificial sweetening agents - common examples

— Chemistry involved in the titrimetric excercises - Acids bases and the use of indicators, oxali acid vs KMnO , Mohr's salt vs KMnO 4 4

1 Enthalpy of solution of CuSO4

2 Enthalpy of neutralization of strong acid and strong base

3 Preparation of lyophilic and lyophobic sols

— Chemical principles involved in the qualitative salt analysis

1. In order to oxidise a mixture of one mole of

each of FeC O2 4, Fe (C O )2 2 4 3, FeSO4 and

Fe (SO )2 4 3 in acidic medium, the number of

moles of KMnO4required is

[JEE Main 2019, 8 April Shift-I]

Exp. (a)

The oxidation of a mixture of one mole of each of

FeC O , Fe (C O )2 4 2 2 4 3FeSO4and

Fe2(SO )4 3in acidic medium with KMnO4is as follows :

FeC O2 4+KMnO4 →Fe3++CO2+Mn2+ …(i)

Fe (C O )2 2 4 3+KMnO4→ Fe3++CO2+ Mn2+…(ii)

FeSO4+KMnO4 →Fe3++SO4−+Mn2+…(iii)

Change in oxidation number of Mn is 5 Change in

oxidation number of Fe in (i), (ii) and (iii) are

+ +3, 6,+1, respectively

neqKMnO4 =neq[FeC O2 4+Fe (C O )2 2 4 3+FeSO ]4

n× = × + × + ×5 1 3 1 6 1 1

2. 100 mL of a water sample contains 0.81 g of

calcium bicarbonate and 0.73 g of magnesium

bicarbonate The hardness of this water

sample expressed in terms of equivalents of

CaCO3is (molar mass of calcium bicarbonate

is 162 g mol−1and magnesium bicarbonate is

146 g mol−1) [JEE Main 2019, 8 April Shift-I]

100 10

6

=10 000, ppm

3. 0.27 g of a long chain fatty acid was dissolved in

100 cm3of hexane 10 mL of this solution wasadded dropwise to the surface of water in around watch glass Hexane evaporates and amonolayer is formed The distance from edge

to centre of the watch glass is 10 cm What isthe height of the monolayer?

[Density of fatty acid=0 9 g cm−3;π =3]

[JEE Main 2019, 8 April Shift-II]

(a) 10−6m (b) 10−4m (c) 10−8m (d) 10−2m

Exp. (a)

100 mL (cm3) of hexane contains 0.27 g of fatty acid

In 10 mL solution, mass of the fatty acid,

Density of fatty acid, d =0 9 g cm−3

∴Volume of the fatty acid over the watch glass,

V m d

= =0 027 =

0 9

0.03 cm

3

Let, height of the cylindrical monolayer=h cm

Q Volume of the cylinder=Volume of fatty acid

cmcm

3 2

= ×1 10− 4cm = ×1 10− 6m

4. For a reaction,

N2( )g +3H2( )g →2NH3( )g , identify

dihydrogen (H2)as a limiting reagent in the

following reaction mixtures

[JEE Main 2019, 9 April Shift-I]

Key IdeaThe reactant which is present in the lesser

amount, i.e which limits the amount of product formed

is called limiting reagent

When 56 g of N2+10 gof H2is taken as a

combination then dihydrogen (H )2 act as a limiting

reagent in the reaction

12 g of H2gas is required for 56 g of N2 gas but

only 10 g of H2gas is present in option (a)

Hence, H2gas is the limiting reagent

For 14 g of N2+4 gof H2

As we know 28 g of N2reacts with 6 g of H2

14 g of N2reacts with 6

28×14g of H2 ⇒3 g of H2.For 28 g of N2+6 gof H2, i.e 28 g of N2reacts with 6 g

of H2(by equation I)

5. At 300 K and 1 atmospheric pressure,

10 mL of a hydrocarbon required 55 mL of O2for complete combustion and 40 mL of CO2isformed The formula of the hydrocarbon is

[JEE Main 2019, 10 April Shift-I]

(a) C H Cl4 7 (b) C H4 6(c) C H4 10 (d) C H4 8

4 15⇒y=15× 4 =

10 6

So, the hydrocarbon (C Hx y) is C H4 6

6. The minimum amount of O2( )g consumed per

gram of reactant is for the reaction (Givenatomic mass : Fe=56, O=16, Mg=24, P=31,

C=12, H=1) [JEE Main 2019, 10 April Shift-II]

Exp. (b)

(a) C H ( ) + 5O ( )3 8

44g

2 160g

g g →3CO ( )2g +4H O2 ( )l

⇒1g of reactant=160

44g of O2consumed

=3.64 g(b) P ( ) + 5O ( )4

124g

2 160g

s g →P O4 10( )s

⇒1g of reactant160

124g of O2consumed=129 g(c) 4Fe( ) + 3O ( )

So, minimum amount of O2is consumed per gram of

reactant (Fe) in reaction (c)

7. 5 moles of AB2weight125 10× −3kg and 10 moles

of A B2 2weight 300 10× −3kg The molar mass of

A M( A)and molar mass of B M( B)in kg mol−1are

[JEE Main 2019, 12 April Shift-I]

Key IdeaTo find the mass of A and B in the given

question, mole concept is used

Number of moles ( )n w

M

= given mass ( )molecular mass ( )Compound Mass of A (g) Mass of B (g)

On solving the equation, we obtain

M A= ×5 10− 3 and M B=10×10− 3

So, the molar mass of A M( A) is

5×10−3kgmol− 1and B M( B)is10×10−3kg mol− 1

8. 25 g of an unknown hydrocarbon uponburning produces 88 g of CO2and 9 g of H O2 This unknown hydrocarbon contains

[JEE Main 2019, 12 April Shift-II]

(a) 20 g of carbon and 5 g of hydrogen(b) 22 g of carbon and 3 g of hydrogen(c) 24 g of carbon and 1 g of hydrogen(d) 18 g of carbon and 7 g of hydrogen

Exp. (c)

Hydrocarbon containing C and H upon burningproduces CO2and water vapour respectively Theequation is represented as

C Hx y+(x+ y/ )4O2 → xCO2+( / )y 2 H O2Mass of carbon=12×

44 mass of CO2

=12×

44 88 g=24 gMass of hydrogen= 2 ×

1780 g of C H57 110O6produced=1980 g of H O.2445g of C H57 110O6produced=1980×

1780 445 g of H O2

=495 of H O

10.In the reaction of oxalate with permanganate

in acidic medium, the number of electrons

involved in producing one molecule of CO2is

So, number of electrons involved in producing

10 molecules of CO2is 10 Thus, number of electrons

involved in producing 1 molecule of CO2is 1

11.A 10 mg effervescent tablet containing sodium

bicarbonate and oxalic acid releases 0.25 mL of

CO2 at T=298.15 K and p=1 bar If molar

volume of CO2is 25.0 L under such condition,

what is the percentage of sodium bicarbonate

2 mol 1 mol 2 mol

⇒ In the reaction, number of mole of CO2

3 3

=8 56 %

12. The hardness of a water sample (in terms ofequivalents of CaCO3) containing

10−3M CaSO4is(Molar mass of CaSO4=136g mol−1)

(a) 100 ppm (b) 10 ppm(c) 50 ppm (d) 90 ppm

[ JEE Main 2019, 12 Jan Shift-I]

Exp. (a)

Hardness of water sample can be calculated in terms

of ppm concentration of CaCO3.Given, molarity=10− 3Mi.e 1000 mL of solution contains10−3mole ofCaCO3

∴Hardness of water=ppm of CaCO3

13. Decomposition of X exhibits a rate constant of

0.05µg/year How many years are required forthe decomposition of 5µg of X into 2.5µg?

[ JEE Main 2019, 12 Jan (Shift-I)]

Exp. (d)

Given, rate constant (k)=0 05 µg/year

Thus, from the unit of k, it is clear that the reaction is

zero order Now, we know thathalf-life (t1 2/ )for zero order reaction= a

=50 yearsThus, 50 years are required for the decomposition of

5µg of X into 2.5µg

14. The ratio of mass per cent of C and H of anorganic compound (C H Ox y z) is 6 : 1 If onemolecule of the above compound (C H Ox y z)contains half as much oxygen as required to burnone molecule of compound C Hx y completely

to CO2 and H O2 The empirical formula ofcompound C H Ox y zis [ JEE Main 2018]

(a) C H O3 6 3 (b) C H O2 4(c) C H O3 4 2 (d) C H O2 4 3

Exp. (d)

We can calculate the simplest whole number ratio of C

and H from the data given, as

Element Relative

mass

Molar mass

Relative mole

Simplest whole no.

in the terms of x and y as

1

x

y = (given and molar mass of C=12, H=1)

Now, after calculating this ratio look for condition 2

given in the question i.e quantity of oxygen is half

of the quantity required to burn one molecule of

compound C Hx ycompletely to CO2and H O2 We

can calculate number of oxygen atoms from this as

consider the equation

Here we consider x and y as simplest ratios for C

and H so now putting the values of x and y in the

15.The most abundant elements by mass in the

body of a healthy human adult are Oxygen

(61.4%); Carbon (22.9%), Hydrogen (10.0 %);

and Nitrogen (2.6%) The weight which a 75 kg

person would gain if all1H atoms are replaced

by2H atoms is [ JEE Main 2017 (Offline)]

Total weight of person=75 kgMass due to1H=75×10=

100 7 5 kg

1Hatoms are replaced by2H atoms,Mass due to2H = (7.5×2)kg

∴Mass gain by person=7 5 kg

16. 1 g of a carbonate (M2CO ) on treatment with3excess HCl produces 0.01186 mole of CO2 The

molar mass of M2CO in g mol3 −1is

[ JEE Main 2017 (Offline)]

(a) 1186 (b) 84.3(c) 118.6 (d) 11.86

Exp. (b)

M CO2 3 2HCl 2M Cl H O CO

1 g

2 2 0.01186 mole

of the hydrocarbon is [ JEE Main 2016 (Offline)]

(a) C H3 8 (b) C H4 8(c) C H4 10 (d) C H3 6

75

15 ⇒ x+ y=

4 5

18.The molecular formula of a commercial resin

used for exchanging ions in water softening is

C H SO Na8 7 3 (mol wt = 206) What would be

the maximum uptake of Ca2 +ions by the resin

when expressed in mole per gram resin?

Exp. (d)

We know the molecular weight of C H SO Na8 7 3

=12× + × +8 1 7 32+16× +3 23=206

We have to find, mole per gram of resin

∴1 g of C H SO Na8 7 3 has number of mole

= Weight of given resin

Molecular weight of resin= 1

206molNow, reaction looks like

2C H SO Na + Ca8 7 3 2+→(C H SO ) Ca + 2Na8 7 3 2 +

Q 2 moles of C H SO Na8 7 3 combines with 1 mol Ca2 +

∴1 mole of C H SO Na8 7 3 will combine with1

× mol Ca2+= 1

412mol Ca

2 +

19.3 g of activated charcoal was added to 50 mL of

acetic acid solution (0.06 N) in a flask After an

hour it was filtered and the strength of the

filtrate was found to be 0.042 N The amount of

acetic acid adsorbed (per gram of charcoal) is

∴ Initial m moles of CH COOH3 =0 06 ×50=3

Final m moles of CH COOH3 =0 042 ×50=2 1

∴ mmoles of CH COOH3 adsorbed

= −3 2 1 =0 9 m molHence, mass of CH COOH3 absorbed per gram of

charcoal

=0 9×603

(Q molar mass of CH COOH3 =60 gmol−1)

=54=

3 18 mg.

20. The ratio of masses of oxygen and nitrogen in aparticular gaseous mixture is 1 4: The ratio ofnumber of their molecule is [ JEE Main 2014]

(a) 1 4: (b) 7 32:(c) 1 8: (d) 3 16:

( )( )Thus, ratio of moles of O2and N2is given by

n n

2 2 2 2

= 

 ×  

14

28

32 = 732

(a) 0.875 M (b) 1.00 M(c) 1.75 M (d) 0.0975 M

=0 875 M

22. A gaseous hydrocarbon gives upon combustion,0.72 g of water and 3.08 g of CO2 The empiricalformula of the hydrocarbon is [ JEE Main 2013]

Exp. (d)

Total mass of solution=1000 g water + 120 g urea

=1120 gDensity of solution=1.15 g / mL

Thus, volume of solution= mass

density= 1120 g1.15 g / mL

= 973.91 mL =0 974 LMoles of solute=120=

60 2 molMolarity= moles of solute

volume (L) of solution= 2 mol

0.974 L

=2.05 mol L− 1

24.The mass of potassium dichromate crystals

required to oxidise 750 cm3of 0 6 M Mohr’s salt

solution is (molar mass=392) [AIEEE 2011]

(a) 0.49 g (b) 0.45 g

(c) 22.05 g (d) 2.2 g

Exp. (c)

Mohr’s salt is FeSO4⋅(NH ) SO4 2 4⋅6H O2

Only oxidisable part Fe2+is

25.The molality of a urea solution in which

0.0100 g of urea, [( NH ) CO]2 2 is added to

0.3000 dm3of water at STP is [AIEEE 2011]

(a) 5.55×10−4M (b) 33.3M

(c) 3.33×10−2M (d) 0.555 M

Exp. (a)

Molality= moles of the solute

mass of the solvent in kg

Moles of urea (nurea)

=299.9 g=0.2999 kgMolality=

×

0.01000.2999

60 =5.55×10− 4mol kg− 1

26. Amount of oxalic acid present in a solution can

be determined by its titration with KMnO4solution in the presence of H SO2 4 The titration

gives unsatisfactory result when carried out inthe presence of HCl because HCl [AIEEE 2008]

(a) gets oxidised by oxalic acid to chlorine(b) furnishes H+ ions in addition to those fromoxalic acid

(c) 67.2 L H ( )2 g at STP is produced for every mole

Al that reacts(d)11.2 L H ( )2 g at STP is produced for every mole

HCl (aq consumed)

Exp. (d)

2Al( )s +6HCl(aq)→2Al3+(aq)+6Cl−(aq)+ 3H2( )g

From the equation, it is clear that,

6 mol of HCl produces 3 mol of H2

or 1 mole of HCl 22.4

6

=3× L of H2=11.2 L of H2

28. How many moles of magnesium phosphate,

Mg (PO )3 4 2 will contain 0.25 mole of oxygen

Hence, 0.25 moles of O-atom are contained by

=1×

8 0.25 mol Mg3(PO4 2) =3.125×10− 2

29.If we consider that 1/6, in place of 1/12, mass of

carbon atom is taken to be the relative atomic

mass unit, the mass of one mole of a substance

30.What volume of hydrogen gas, at 273 K and

1 atm pressure will be consumed in obtaining

21.6 g of elemental boron (atomic mass= 10.8)

from the reduction of boron trichloride by

(atomic mass=56 g mol−1) is [AIEEE 2002]

(a) twice that of 70 g N (b) half that of 20 g H

(c) Both (a) and (b) (d) None of these

Exp. (c)

560 g of FeNumber of moles= 560 g−

56 g mol 1 =10 molFor 70 g of N

14 gN=1 mol of N-atom

70 gN=5 mol of N-atomFor 20 g of H

1 gH=1 mol of H-atom

20 g H≡20 mol of H -atom

32. In an organic compound of molar mass

108 g mol–1C, H and N atoms are present in

9 : 1 : 3.5 by weight Molecular formula can be

[AIEEE 2002]

(a) C H N6 8 2 (b) C H N7 10(c) C H N5 6 3 (d) C H N4 18 3

Exp. (a)

Molar mass 108 g mol−1

Total part by weight= + +9 1 3.5=13.5Weight of carbon= 9 ×

13.5 108=72 gNumber of carbon atoms=72=

12 6Weight of hydrogen= 1 ×

13.5 108=8 gNumber of hydrogen atoms=8=

Weight of nitrogen= 3.5 ×

13.5 108=28 gNumber of nitrogen atom=28=

14 2Hence, molecular formula=C H N6 8 2

1. Element ‘B ’ forms ccp structure and ‘A ’

occupies half of the octahedral voids, while

oxygen atoms occupy all the tetrahedral voids

The structure of bimetallic oxide is

[JEE Main 2019, 8 April Shift-I]

Number of tetrahedral voids = 2N

Number of octahedral voids = N

∴Number of ‘A’ in the crystal=N= =

2

4

Number of oxygen (O) atoms=2N= × =2 4 8

∴The structure of bimetallic oxide=A B2 4O8

=AB2O4

2. Consider the bcc unit cells of the solids 1 and 2

with the position of atoms as shown below

The radius of atom B is twice that of atom A.

The unit cell edge length is 50% more is solid 2

than in 1 What is the approximate packing

efficiency in solid 2?

(a) 65% (b) 90% (c) 75% (d) 45%

[JEE Main 2019, 8 April Shift-II]

Exp. (b)Key IdeaPacking efficiency

=Volume occupied by sphere×

r A

a2=2 3r A

Packing efficiency= × + ×

43

43

3 3 2

πr z πr z a

πr πr a

43

4

2 3

3 3 3

πr π r r

πr πr

A B

Solid 2

3. Consider the van der Waals’ constants,

a and b, for the following gases.

Critical temperature is the temperature of a

gas above which it cannot be liquefied what

ever high the pressure may be The kinetic

energy of gas molecules above this

temperature is sufficient enough to overcome

the attractive forces It is represented as T c

For Xe, T c = ×

× × =

8 4.1

27 8.314 5.0 0.02

The value of T cis highest for Kr (Krypton)

4. At a given temperature T , gases Ne, Ar, Xe

and Kr are found to deviate from ideal gas

behaviour Their equation of state is given

as, p RT

=

− atT

Here, b is the van der Waals’ constant.

Which gas will exhibit steepest increase in

the plot of Z (compression factor)vs p?

[JEE Main 2019, 9 April Shift-II]

Exp. (a)

Noble gases such as Ne, Ar, Xe and Kr found

to deviate from ideal gas behaviour

Xe gas will exhibit steepest increase in plot of

Z vs p Equation of state is given as:

pb RT

The plot of z vs p is found to be

The gas with high value of b will be steepest as slope is directly proportional to b b is the van der

Waals’ constant and is equal to four times the actualvolume of the gas molecules Xe gas possess thelargest atomic volume among the given noble gases(Ne, Kr, Ar) Hence, it gives the steepest increase in

the plot of Z (compression factor) vsp.

5. 10 mL of 1 mM surfactant solution forms amonolayer covering 0.24 cm2on a polar substrate

If the polar head is approximated as a cube, what

is its edge length? [JEE Main 2019, 9 April Shift-II]

Now, number of molecules

=Number of moles×Avogadro’s number

p

6. Consider the following table.

Gas a/(k Pa dm 6 mol−1 ) b/(dm mol ) 3 −1

a and b are van der Waals’ constants The

correct statement about the gases is

[JEE Main 2019, 10 April Shift-I]

(a) gas C will occupy lesser volume than gas A; gas

B will be lesser compressible than gas D

(b) gas C will occupy more volume than gas A; gas

B will be more compressible than gas D

(c) gas C will occupy more volume than gas A; gas

B will be lesser compressible than gas D

(d) gas C will occupy more volume than gas A; gas

B will be lesser compressible than gas D

The constant ‘a’ measures the intermolecular

force of attraction of gas molecules and the

constant ‘b’ measures the volume correction by

gas molecules after a perfectly inelastic binary

collision of gas molecules

For gas A and gas C given value of ‘b’ is

0.05196 dm3mol− 1 Here,

a∝intermolecular force of attraction

∝compressibility∝real nature

volume occupied

Value of a/(kPa dm6mol−1) for gas A (642.32) >

gas C (431.91) So, gas C will occupy more

volume than gas A Similarly, for a given value of

asay 155.21 kPa dm6mol−1for gas B and gas D

1

b∝intermolecular force of attraction

∝compressibility∝real nature

volume accupied

b/(dm3mol− 1) for gas B (0.04136) < Gas D

(0.4382)

So, gas B will be more compressible than gas D.

7. Points I, II and III in the following plot

respectively correspond to (vmp : mostprobable velocity)

[JEE Main 2019, 10 April Shift-II]

8. An element has a face-centred cubic (fcc)

structure with a cell edge of a The distance

between the centres of two nearest tetrahedral

voids in the lattice is

[JEE Main 2019, 12 April Shift-I]

In fcc unit cell, two tetrahedral voids are formed on

each of the four non-parallel body diagonals of the

cube at a distance of 3a/ from every corner4

along the body diagonal

The angle between body diagonal and an edge

is cos ( /− 1 1 3 So, the projection of the line on)

9. The ratio of number of atoms present in a

simple cubic, body centered cubic and face

centered cubic structure are, respectively

[JEE Main 2019, 12 April Shift-II]

The ratio of number of atoms present in simple

cubic, body centred cubic and face centered

cubic structure are 1 : 2 : 4 respectively

10.At 100°C, copper (Cu) has FCC unit cell

structure with cell edge length of x Å What is

the approximate density of Cu (in g cm−3) atthis temperature?

For fcc, rank of the unit cell ( )Z =4

Mass of one Cu-atom, M=63 55 u

Avogadro’s number, N A =6 023 10 × 23atom

Edge length, a= xÅ= ×x 10− 8cmdensity ( )d Z M

(a) Hexagonal (b) Monoclinic(c) Tetragonal (d) Triclinic

[JEE Main 2019, 10 Jan Shift-I]

by the other atoms ?

(a) hcp lattice- A,2

3tetrahedral voids-B

(b) hcp lattice-A,1

3tetrahedral voids-B

One of the body diagonal

of cubic unit cell

Position of 2 TVs

on one of the body diagonal

of the unit cell

√ 3

—4a

1

—4 AB1

Total effective number of atoms in hcp unit lattice=

Number of octahedral voids in hcp=6

∴Number of tetrahedral voids (TV) in hcp

= ×2 Number of atoms in hcp lattice

3tetrahedral voids, B=hcp lattice

13. A solid having density of 9 10× 3kgm− 3 forms

face centred cubic crystals of edge length

200 2 pm What is the molar mass of the solid?

[Avogadro constant= ×6 1023mol−1,π =3]

(a) 0.03050 kg mol−1 (b) 0.4320 kg mol−1

NA =Avogadro’s constant= ×6 1023mol− 1

a=Edge length of the unit cell

(Edge length is represented by ‘a’)

(a) 0.134 a (b) 0.027 a (c) 0.047 a (d) 0.067 a

[JEE Main 2019, 11 Jan Shift-II]

pressures of the gases for equal number ofmoles are

On substituting in equation (i), we get

16.An open vessel at 27ºC is heated until two fifth

of the air (assumed as an ideal gas) in it has

escaped from the vessel Assuming that the

volume of the vessel remains constant, the

temperature at which the vessel has been

Volume of vessel=constant

Pressure in vessel=constant

Volume of air reduced by 2

5 so the remainingvolume of air is 3

5.

Let atT1the volume of air inside the vessel is nso at

T2the volume of air will be 3

17.Which type of ‘defect’ has the presence of

cations in the interstitial sites? [ JEE Main 2018]

(a) Schottky defect

18.A metal crystallises in a face centred cubic

structure If the edge length of its unit cell is ‘a’,

the closest approach between two atoms inmetallic crystal will be [ JEE Main 2017 (Offline)]

bulbs is then raised to T2 The final pressure p f

is [ JEE Main 2016 (Offline)]

i

+ –– ++ –– ++ –– ++ –– +

+ –– ++ ––+ –– ++ –– +

+ –– ++ –– ++ –– ++ –– +

+ –– ++ –– ++ –– ++ –– +

+

Originalvacant site

of cationCation ininterstitial site

After mixing, number of moles in left chamber

p V RT

p V RT

p V RT

20.Sodium metal crystallises in a body centred

cubic lattice with a unit cell edge of 4.29 Å The

radius of sodium atom is approximately

with unit cell edge, a=4.29Å

We have the formula for radius,

4 4.29 Å=1.86 Å

21.The intermolecular interaction that isdependent on the inverse cube of distancebetween the molecules is [ JEE Main 2015]

(a) ion-ion interaction (b) ion-dipole interaction(c) London force (d) hydrogen bond

22.If Z is a compressibility factor, van der Waals’

equation at low pressure can be written as

However, V is still large enough in comparison to

‘b’, hence, ‘b’ can be neglected Thus, van der

Waals’ equation becomes

C

a

a D

2r

r

r

23.CsCl crystallises in body centered cubic lattice.

If ‘a’ its edge length, then which of the following

expressions is correct? [JEE Main 2014]

In CsCl, Cl−lie at the corners of simple cube and

Cs+at the body centre Therefore, along the body

diagonal, Cs+and Cl−touch each other

2 a=rCs + +rCl −

24.Which of the following exists as covalent

crystals in the solid state? [JEE Main 2013]

(c) Sulphur (d) Phosphorus

Exp. (b)

Silicon exists as covalent crystal in solid state

(Network like structure, like diamond)

25.Experimentally, it was found that a metal oxide

has formula M0.98O Metal M, present as M2+

and M3+ in its oxide Fraction of the metal

which exists as M3+ would be [JEE Main 2013]

26.For gaseous state, if most probable speed is

denoted by C *, average speed by C and mean square speed by C, then for a large number of

molecules, the ratios of these speeds are

(a) C* : C C: =1 225 1 128 1. : . : [JEE Main 2013]

Note As no option correspond to mean square speed,

it is understood as misprint It should be root mean square speed.

27.The compressibility factor for a real gas at highpressure is [AIEEE 2012]

pb RT

= =1+ 

28.Lithium forms body-centred cubic structure.

The length of the side of its unit cell is 351 pm

Atomic radius of the lithium will be[AIEEE 2012]

(a) 75 pm (b) 300 pm (c) 240 pm (d) 152 pm

Exp. (d)

In body-centred cubic structure,

Edge length= =a 351 pm, Radius= =r ?

Then, r= 3a= ×

4

3 3514

=152 pm

29.a and b are van der Waals’ constants for gases.

Chlorine is more easily liquefied than ethane

(a) a and b for Cl2>a and b for C H2 6

(b) a and b for Cl2<a and b for C H2 6

(c) a for Cl2>a forC H2 6but b forCl2>b forC H2 6

(d) a for Cl2>a forC H2 6but b forCl2<b forC H2 6

Exp. (d)

van der Waals’ constant a is due to force of

attraction and b due to finite size of molecules.

Thus, greater the value a and smaller the value of

b, larger the liquefaction

Thus, a(Cl )2 >a(C H )2 6

and b(Cl2)<b(C H )2 6

30.In a face-centred cubic lattice, atom A occupies

the corner positions and atom B occupies the

face centred positions If one atom of B is

missing from one of the face centred points,

the formula of the compound is [AIEEE 2011]

31.The molecular velocity of any gas is

(a) inversely proportional to the square root oftemperature

(b) inversely proportional to absolutetemperature

(c) directly proportional to square of temperature(d) directly proportional to square root of

In all cases, molecular velocity∝ T

32.Copper crystallises in fcc lattice with a unit celledge of 361 pm The radius of copper atom is

[AIEEE 2011]

(a) 181 pm (b) 108 pm (c) 128 pm (d) 157 pm

Exp. (c)

Copper crystallises in fcc lattice

If, r=radius, a=edge length

then r= a =

2 2

361

2 2pm=127.633 pm≈128 pm

33.When r, p and M represent rate of diffusion,

pressure and molecular mass, respectively,then the ratio of the rates of diffusion( / )r A r B of

two gases A and B, is given as [AIEEE 2011]

(a) (p A/p B) (1 2/ M A/M B)(b) (p A/p B)(M B/M A)1 2/

(c) (p A/p B)1 2/ (M B/M A)(d) (p A/p B)(M A/M B)1 2/

Exp. (b)

Rate of diffusion, r∝p r

For gas B, r p

M

B B B

Dividing Eq (i) by Eq (ii), we get

r r

p p

M M

A B A B B A

M M

A B B A

1 2 /

34.If10− 4dm3of water is introduced into a1.0 dm3

flask at 300 K, then how many moles of water

are in the vapour phase when equilibrium is

established?

(Given, vapour pressure of H O2 at 300 K is 3170

Pa; R 8.314 JK= −1mol )−1 [AIEEE 2010]

3

8.134

=1.27 10 mol× − 3

35.The edge length of a face centered cubic cell of

an ionic substance is 508 pm If the radius of

the cation is 110 pm, the radius of the anion is

36.Percentage of free space in cubic close packed

structure and in body centred packed structure

are respectively [AIEEE 2010]

(a) 30% and 26% (b) 26% and 32%

37.Copper crystallises in fcc with a unit cell length

of 361 pm What is the radius of copper atom?

Exp. (a)

Suppose atoms of element Y in ccp=100Number of tetrahedral voids= ×2 100Number of atoms of element

39.Total volume of atoms present in a

face-centred cubic unit cell of a metal is (r is

atomic radius) [AIEEE 2006]

(a) 203

(a) The area under the distribution curve remainsthe same as under the lower temperature(b) The distribution becomes broader(c) The fraction of the molecules with the mostprobable speed increases

(d) The most probable speed increases

Exp. (c)

Distribution of molecules ( )Nwith velocity ( )u at two

temperatures T1and T T2(2>T1)is shown below

At both temperatures, distribution of molecules

with increase in velocity first increases, reaches to

a maximum value and then decreases

41.An ionic compound has a unit cell consisting

of A ions at the corners of a cube and B ions on

the centres of the faces of the cube The

empirical formula for this compound would be

[AIEEE 2005]

(a) A B3 (b) AB3 (c) A B2 (d) AB

Exp. (b)

Unit cell consists of A ions at the corners.

Thus, number of ions of the type A= =8

42.As the temperature is raised from 20°C to 40°C,

the average kinetic energy of neon atoms

changes by a factor of which of the following?

43.In van der Waals’ equation of state of the gas

law, the constant ‘b’ is a measure of

(a) intermolecular repulsions [AIEEE 2004]

(b) intermolecular attraction

(c) volume occupied by the molecules

(d) intermolecular collisions per unit volume

Exp. (b)

When equal number of cations and anions (suchthat charges are equal) are missing(1Na , 1Cl / 1 Fe , 2Cl )+ − 2+ − then, it is a case ofSchottky defect

45.According to the kinetic theory of gases, in anideal gas, between two successive collisions agas molecule travels

(a) in a circular path [AIEEE 2003]

(b) in a wavy path(c) in a straight line path(d) with an accelerated velocity

Exp. (c)

In between two successive collisions, no force isacting on the gas molecules Resultantly theytravel with uniform velocity during this interval andhence, moves along a straight line

46.How many unit cells are present in a cubeshaped ideal crystal of NaCl of mass 1.00 g?[Atomic mass : Na = 23, Cl = 35.5 ]

∴Number of unit cells in 1g= 1

×

6.02 1058.5 4

23

=2.57 10× 21

47.Based on kinetic theory of gases following laws

can be proved [AIEEE 2002]

(a) Boyle’s law (b) Charles’ law

(c) Avogadro’s law (d) All of these

12

2= 

  

23

32

N

48.For an ideal gas, number of moles per litre in

terms of its pressure p , temperatureT and gas constant R is [AIEEE 2002]

1

49.Number of atoms in the unit cell of Na (bcctype crystal) and Mg (fcc type crystal) are,respectively [AIEEE 2002]

1. The quantum number of four electrons are

given below:

I n = 4 l = 2 ml= − 2 ms = − 1

2 , , ,

II n = 3 l = 2 ml= 1 ms = + 1

2 , , ,

III n = 4 l = 1 ml= 0 ms = + 1

2 , , ,

IV n = 3 l = 1 ml= 1 ms = − 1

2 , , ,

The correct order of their increasing

energies will be [JEE Main 2019, 8 April Shift-I]

(a) IV < III < II < I (b) I < II < 1III < IV

(c) IV < II < III < I (d) I < III < II < IV

Exp. (c)

Smaller the value of (n+l), smaller the energy If

two or more sub-orbits have same values of

(n+l), sub-orbits with lower values of n has lower

energy The (n+l)values of the given options are

Among II and III, n=3 has lower value of energy

Thus, the correct order of their increasing

energies will be

IV < II < III < I

2. If p is the momentum of the fastest electron

ejected from a metal surface after the

irradiation of light having wavelength λ ,

then for 1.5 p momentum of the

photoelectron, the wavelength of the light should be

(Assume kinetic energy of ejected photoelectron to be very high in comparison

to work function)[JEE Main 2019, 8 April Shift-II]

2 2

λ energy of incident light.

E0=threshold energy or work functions,1

2

12

p m

49

3. For any given series of spectral lines of

atomic hydrogen, let ∆ν ν = max− νminbe

the difference in maximum and minimum

frequencies in cm−1 The ratio

Let∆ν ν= max −νminbe the difference in

maximum and minimum frequencies in cm− 1

For Lyman series,

Lyman Balmer

is 9 : 4

4. Which one of the following about an

electron occupying the 1s-orbital in a

hydrogen atom is incorrect? (The Bohr

radius is represented by a0)

[JEE Main 2019, 9 April Shift-II]

(a) The electron can be found at a distance 2a0

from the nucleus

(b) The magnitude of the potential energy isdouble that of its kinetic energy on anaverage

(c) The probability density of finding theelectron is maximum at the nucleus.(d) The total energy of the electron is maximum

when it is at a distance a0from the nucleus

Exp. (d)

Statement (d) is incorrect For 1s-orbital radial

probability density (R2)against r is given as:

For 1s-orbital, probability density decreases

sharply as we move away from the nucleus.The radial distribution curves obtained by plotting

radial probability functions vs r for 1s-orbital is

The graph initially increases and then decreases Itreaches a maximum at a distance very close to thenucleus and then decreases The maximum in thecurve corresponds to the distance at which theprobability of finding the electron in maximum

5. The graph between | | ψ 2and r (radial

distance) is shown below This represents

[JEE Main 2019, 10 April Shift-I]

(a) 1s-orbital (b) 2p-orbital

(c) 3s-orbital (d) 2s-orbital

Exp. (d)

The graphs between| |ψ2and r are radial density

plots having (n l− −1 number of radial nodes For)

1s, 2s, 3s and 2 p-orbitals these are respectively.

Thus, the given graph between | |ψ2 and r

represents 2s-orbital.

6. The ratio of the shortest wavelength of two

spectral series of hydrogen spectrum is

found to be about 9 The spectral series are

[JEE Main 2019, 10 April Shift-II]

(a) Lyman and Paschen (b) Brackett and Pfund

(c) Paschen and Pfund (d) Balmer and Brackett

For shortest wavelength, i.e highest energy

spectral line, n2will be ( )∞.For the given spectral series, ratio of the shortestwavelength of two spectral series can becalculated as follows :

1

1 0

19

125

161

1625

125

91

925

116

41

14

2 2

2 2

NoteLyman= L (n1=1),Balmer=B (n1=2)Paschen=P(n1=3),Brackett Bk= (n1=4)Pfund=Pf (n1=5)

7. The electrons are more likely to be found

[JEE Main 2019, 12 April Shift-I]

(a) in the region a and c (b) in the region a and b (c) only in the region a (d) only in the region c

Exp. (a)

The electrons are more likely to be found in the

region a and c At b, wave function becomes zero

and is called radial nodal surface or simply node

| |Ψ 2

r

For 1 orbital number of radial node = 1–0–1=0s-

| |Ψ 2

r

For 2 orbital number of radial node = 2–0–1=1s-

| |Ψ 2

r

For 2 orbital number of radial node = 2–1–1=0p-

| |Ψ 2

r

For 3 orbital number of radial node = 3–0–1=2s-

The graph between wavefunction ( )ψand distance

(r) from the nucleus helps in determining the shape

of orbital

8. For emission line of atomic hydrogen from

ni= 8 to nf = n , the plot of wave number ( ) ν

against 12

n



   will be (The Rydberg constant,

RHis in wave number unit)

(a) non linear

(b) linear with slope−RH

(c) linear with slope RH

(d) linear with intercept−RH

[ JEE Main 2019, 9 Jan Shift-I]

[Qit is the case of emission]

H 1 164

n will belinear with slope (+RH)

9. Which of the following combination of

statements is true regarding the

interpretation of the atomic orbitals?

I An electron in an orbital of high

angular momentum stays away from

the nucleus than an electron in the

orbital of lower angular momentum.

II For a given value of the principal

quantum number, the size of the orbit

is inversely proportional to the

azimuthal quantum number.

III According to wave mechanics, the ground state angular momentum is equal to h

2 π .

IV The plot of ψ vs r for various azimuthal quantum numbers, shows peak

shifting towards higher r value.

(a) I, III (b) II, III (c) I, II (d) I, IV

[ JEE Main 2019, 9 Jan Shift-II]

(IV) The given plot is

10. Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface? [ JEE Main 2019, 10 Jan Shift-I]

K.E ofs

es

Intensity oflight0

K.E ofs

es

Frequency oflight0

K.E ofs

es

Frequency oflight0

b

c

Y

Exp. (d)

For photoelectric effect,

(i) KE=E−E0

where,

KE=Kinetic energy of ejected electrons

E=Energy of incident light=hν

E0=Threshold energy=hν0

ν =Frequency of incident light

ν0=Threshold frequency

Slope= ±1, intercept= −E0

So, option (a) is correct

(ii) KE of ejected electrons does not depend on the

intensity of incident light

So, option (b) is correct

(iii) When, number of ejected electrons is plotted with

frequency of light, we get

So, option (c) is also correct

(iv) KE=hν−hν0

Slope= +h, intercept= −hν0

So, option (d) is not correct

11. The 71st electron of an element X with an

atomic number of 71 enters into the orbital

(a) 4f (b) 6p (c) 5d (d) 6s

[ JEE Main 2019, 10 Jan Shift-II]

Exp. (c)

In the lanthanoid series, atomic number of

fourteen 4f-block elements ranges from 58 (Ce) to

71 (Lu)

Ytterbium, Yb(Z=70 has electronic configuration): [Xe]4 6f14 s So, the 71nth electron of lutetium,2

Lu(Z=71 should enter into 5d orbital and its)

(here, Lu is ‘X’) electronic configuration will be :

[Xe]4 5 6f14 d s It happens so, because f-block1 2

elements have general electronic configuration,(n−2)f1 14 − (n−1)d1 10 − ns2 Therefore, option (c) iscorrect

12. The ground state energy of hydrogen atom is

− 136 eV The energy of second excited state

of He+ion in eV is

(a)−54.4 (b)−3.4 (c)−6.04 (d)−27.2

[ JEE Main 2019, 10 Jan Shift-II]

Exp. (c)

The ground state energy of H-atom is+13 6 eV

For second excited state, n= + =2 1 3

n

3

2 2

13 6(He )+ = − × eV [Qfor He+, Z 2]=

= −13 6 2×

3

2 2

eV= −6 04 eV

13. Heat treatment of muscular pain involves radiation of wavelength of about 900 nm Which spectral line of H-atom is suitable for this purpose? [RH= × 1 10 cm5 –1,

0

Number

ofe ’–s

(n)Frequency of light

KE

0

1 2

3,

14. The de-Broglie wavelength ( ) λ associated

with a photoelectron varies with the

frequency ( ) ν of the incident radiation as,

15. What is the work function of the metal, if the

light of wavelength 4000 Å generates

photoelectron of velocity6 10 × 5ms from it?−1

Work function of metal ( )φ =hν0

where,ν0=threshold frequency

we get1

16. If the de-Broglie wavelength of the electron

in nth Bohr orbit in a hydrogenic atom is equal to 15 π a0(a0is Bohr radius), then the

Also, we know that radius( )r of an atom is given by

r a n Z

a a

= 91091 10 × − 31 kg ; charge of electron ( ) e = 160210 10 × −19C; permitivity of vacuum

(∈ =) × −

0 8 854185 10 12kg m A−1 −3 2)

[JEE Main 2017 (Offline)]

(a) 1.65 Å (b) 4.76 Å (c) 0.529 Å (d) 2.12 Å

18. A stream of electrons from a heated filament

was passed between two charged plates kept

at a potential differenceV esu If e and m are

charge and mass of an electron, respectively,

then the value ofh/ λ (where, λ is wavelength

associated with electron wave) is given by

[JEE Main 2016 (Offline)]

(a) 2 meV (b) meV (c) 2 meV (d) meV

Exp. (c)

mentioned hence, we have to find out

relation between h

λand energy For this, we

shall use de-Broglie wavelength and kinetic

energy term in eV.

de-Broglie wavelength for an electron ( )λ = h

19. Which of the following is the energy of a

possible excited state of hydrogen?

[JEE Main 2015]

(a)+13.6 eV (b)−6.8 eV

(c)−3.4 eV (d)+6.8 eV

Exp. (c)

Since, at n=1, the population of electrons is

maximum i.e at ground state So, maximum excitation will take place from n=1to n=2

Hence, n=2 is the possible excited state.Now, we have the formula for energy of H-atom

( )H 13.6 eV

2 2

(a) 5 0 0 1

2

2, , ,+

(c) 5 1 1 1

2

2, , ,+

J Wavelength of light required to excite an electron in an

hydrogen atom from level n = 1 to n = 2 will

( h 6.62 10 = × −34 J s and c = 3.0 10 ms × 8 −1)

(a) 1.214 10× − 7m (b) 2.816 10× − 7m(c) 6.500 10× − 7m (d) 8.500 10× − 7m

Higher the value of (n+l),higher is the energy If

(n+l)are same, then suborbit with lower value of

n has lower energy Thus,

3p<4s<3d<4p

(4) <(2) < (3) < (1)

23. A gas absorbs photon of 355 nm and emits at

two wavelengths If one of the emission is at

680 nm, the other is at [AIEEE 2011]

1680

24. The frequency of light emitted for the

transition n = 4 to n = 2 of He+is equal to the transition in H atom corresponding to which of the following? [AIEEE 2011]

14

13

14

2 2

E

E1

E2

1. The ion that has sp d3 2hybridisation

for the central atom, is

[JEE Main 2019, 8 April Shift-II]

(a) [ICl ]2− (b) [BrF ]2−

(c) [ICl ]4 − (d) [IF ]6 −

Exp. (c)

Key Idea The hybridisation for a central

atom in a species can be calculated

C=No of cationic (positive) charge

A=No of anionic (negative) charge

The hybridisation of given species are as

So order of bond length C2 O N O

(BO 3) = (BO 2) =2 2 (BO 1)2