Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019) Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019) Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019) Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019) Preview Complete Companion For Jee Main 2020 Chemistry Vol 1 by A.K. Singhal, U.K. Singhal (2019)
F IF T H ED IT IO N Complete Companion for FIFTH EDITION JEE Main 2020 Complete Companion for CHEMISTRY Complete Companion for JEE Main AK Singhal I UK Singhal Complete coverage of JEE Main curriculum with emphasis on important concepts ‘info boxes’, and ‘concept notes’ to aid in quick last-minute revision Includes ample number of solved examples along with NCERT Exemplar Updated Practice Exercises as per recent patterns of JEE Main for result-oriented preparation Cover Image: Anton Khrupin anttoniart.shutterstock.com Key points highlighted within text along with features like – ‘Concepts at a glance’, Y L L D U E F VIS RE HIGHLIGHTS Coverage of key topics along with solved examples CHEMISTRY H I G H LI G H TS Provides varieties of solved example after each concept for better understanding CHEMISTRY 2020 The Complete Resource Book for JEE Main series has been designed to be an independent resource to enable faster and effective learning This series includes three separate books on Physics, Chemistry and Mathematics where the core objective of each book is to provide ‘effective preparation via modular and graded content’ Developed by highly experienced and qualified faculties, these books would act as trusted content for aspirants who are aiming to clear the JEE (Joint Entrance Examinations) and other key engineering entrance examinations JEE Main 2020 Includes practice problems with complete solutions Chapter-wise Previous 18 years’ AIEEE/JEE Main questions Fully-solved JEE Main 2019 questions (April & Jan) Includes online tests based on JEE Main pattern FREE Online Mock Tests MRP Inclusive of all Taxes `550.00 in.pearson.com 5e Spine: 22mm ISBN: 9789353435059 Title Sub Title Edition Singhal Singhal Size: 203x254mm AK Singhal Authors / Editors Name UK Singhal With CD Red Band Territory line URL Price mQuest About Pearson Pearson is the world’s learning company, with presence across 70 countries worldwide Our unique insights and world−class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world We believe learning opens up opportunities, creates fulfilling careers and hence better lives We hence collaborate with the best of minds to deliver you class−leading products, spread across the Higher Education and K12 spectrum Superior learning experience and improved outcomes are at the heart of everything we This product is the result of one such effort Your feedback plays a critical role in the evolution of our products and you can contact us at reachus@pearson.com We look forward to it This page is intentionally left blank Complete Companion for JEE Main 2020 CHEMISTRY Also useful for other Engineering Entrance Examinations AK Singhal UK Singhal This page is intentionally left blank Complete Companion for JEE Main 2020 CHEMISTRY Also useful for other Engineering Entrance Examinations AK Singhal UK Singhal Copyright © 2019 Pearson India Education Services Pvt Ltd Published by Pearson India Education Services Pvt Ltd, CIN: U72200TN2005PTC057128 No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent This eBook may or may not include all assets that were part of the print version The publisher reserves the right to remove any material in this eBook at any time ISBN 978-93-534-3505-9 eISBN: Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No 1, Block-C, Sector 16, Noida 201 301, Uttar Pradesh, India Registered Office: The HIVE, 3rd Floor, Metro Zone, No.44, Pillayar Koil Street, Jawaharlal Nehru Road, Anna Nagar, Chennai 600 040, Tamil Nadu, India Phone: 044-66540100 Website: in.pearson.com, Email: companysecretary.india@pearson.com Contents Prefacexi Acknowledgementsxii Chemistry Trend Analysis (2011 to 2019) xiii Chapter 1 Some Basic Concepts of Chemistry1.1 Matter 1.1 Physical Quantities and their Measurements1.2 Laws of Chemical Combinations 1.4 Avogadro’s Law 1.4 Mole 1.5 Mass 1.5 Equivalent Weight 1.6 Mole Fraction 1.7 Chemical Equation and Stoichiometry of Chemical Reactions 1.7 NCERT Exemplars 1.13 Practice Exercises 1.15 Chapter 2 Structure of Atoms 2.1 Atom and Atomic Theory Thomson Atomic Model Rutherford Atomic Model Electronic Structure of Atoms Spectrum Bohr’s Atomic Model De Broglie Equation and Dual Nature Theory Heisenberg’s Uncertainty Principle Schrodinger Wave Equation Quantum Numbers Aufbau Principle Pauli’s Exclusion Principle Hund’s Rule of Maximum Multiplicity NCERT Exemplars Practice Exercises 2.1 2.4 2.4 2.4 2.6 2.7 2.8 2.9 2.9 2.11 2.12 2.13 2.14 2.19 2.21 Chapter 3 Classification of Elements and Periodicity in Properties 3.1 Modern Periodic Law 3.1 Trends in Periodic Properties of Elements NCERT Exemplars Practice Exercises 3.4 3.12 3.14 Chapter 4 Chemical Bonding and Molecular Structure 4.1 Valency 4.1 Chemical Bond 4.1 Ionic or Kernel Bond 4.2 Covalent Bond 4.3 Coordinate or Semi-polar Bond 4.5 Hydrogen Bond 4.6 Metallic Bonding 4.7 Resonance 4.7 Hybridization 4.8 Molecular Orbital Theory NCERT Exemplars Practice Exercises 4.13 4.18 4.20 Chapter 5A States of Matter: Gaseous State5.1 General Properties 5.1 Gas Laws 5.1 Kinetic Theory of Gases 5.4 Ideal and Real Gases 5.6 Van Der Waals Equation 5.6 Critical Phenomenon and Liquefaction of Gases 5.7 NCERT Exemplars 5.15 Practice Exercises 5.17 viii Contents Chapter 5B States of Matter: Solid State 5.33 General Properties 5.33 Types of Solids 5.33 Bragg’s Equation 5.34 Unit Cell 5.35 Mathematical Analysis of Cubic System 5.37 Packing of Constituents in Crystals 5.40 Imperfections in Solids 5.42 Magnetic Properties of Solids 5.44 Electrical Properties of Solids 5.45 General Properties NCERT Exemplars Practice Exercises 5.45 5.52 5.55 Chapter 6 Thermodynamics and Thermochemistry Ostwald Dilution Law 8.2 Ionic Product of Water 8.3 Buffer Solution 8.4 Solubility and Solubility Product 8.5 Salt Hydrolysis 8.6 Acid and Bases NCERT Exemplars Practice Exercises 8.6 8.16 8.18 Chapter 9 Hydrogen 9.1 Rogue Element or Hydrogen 6.1 6.1 Thermodynamic Process 6.2 Thermodynamic Equilibrium 6.3 Heat 6.3 Work 6.3 6.3 First Law of Thermodynamics or Law of 9.1 Resemblance of Hydrogen with Alkali Metals (IA) 9.1 Resemblance with Halogens 9.2 Types of Hydrogen 9.3 Methods of Preparation of Dihydrogen 9.3 Hydrides Thermodynamics Internal Energy 9.6 Water 9.9 Hard and Soft Water 9.10 Heavy Water or Deuterium Oxide D2O9.12 Hydrogen Peroxide (Auxochrome) H2O29.13 NCERT Exemplars 9.18 Practice Exercises 9.20 Chapter 10 s-Block Elements 10.1 s-Block Elements 10.1 Conservation of Energy 6.4 Alkali Metals [Ia] 10.1 6.5 Entropy 6.5 Sodium (11Na23)10.6 Zeroth Law of Thermodynamics 6.6 Sodium Oxide (Na2O)10.8 Third Law of Thermodyanmics 6.6 Caustic Soda or Sodium Hydroxide (NaOH) Gibbs Free Energy (G) 6.6 Thermochemistry 6.7 Sodium Carbonate or Washing Soda Heat or Enthalpy of Reaction 6.7 Sources of Energy 6.10 (Na2CO3.10H2O)10.11 Conservation of Energy NCERT Exemplars Practice Exercises 6.11 6.17 6.19 Second Law of Thermodynamics Chapter 7 Chemical Equilibrium Types of Reactions 7.1 7.1 Equilibrium Practice Exercises 7.2 7.10 Chapter 8 Ionic Equilibrium 8.1 Sodium Peroxide (Na2O2)10.8 Sodium Chloride (NaCl) 10.9 10.11 Sodium Bicarbonate or Baking Soda (NaHCO3)10.13 Micro Cosmic Salt (Na(NH4) HPO4.4H2O)10.14 39 Potassium (19K )10.14 Potassium Super Oxide (KO2)10.15 Potassium Hydroxide (KOH) 10.15 Potassium Chloride (KCl) 10.15 Potassium Carbonate (K2CO3)10.16 Potassium Iodide (KI) Potassium Sulphate (K2SO4)10.17 Potassium Bicarbonates (KHCO3)10.17 Types of Substances 8.1 Biological Role of Sodium (Na) and Arrhenius Theory 8.1 Potassium (K) 10.16 10.17 Contents ix Chapter 11B p-Block Elements II Alkaline Earth Metals and their Compounds [IIa (ns )]10.17 Chemical Properties and Compounds 24 Magnesium (12Mg )10.23 Magnesia (MgO) 10.19 10.25 Magnesium Chloride (MgCl2)10.25 Magnesium Sulphate or Epsom Salt or Epsomite [MgSO4 7H2O]10.26 Magnesium Hydroxide Mg(OH)210.27 Magnesium Carbonate (MgCO3)10.27 Magnesium Bicarbonate Mg(HCO3)210.28 Calcium (20Ca )10.28 Calcium Oxide or Quick–Lime (Cao) 40 10.29 Calcium Hydroxide or Slaked Lime Ca(OH)210.29 Calcium Oxide or Marble or Lime Stone (CaCO3)10.30 Calcium Sulphate Dihydrate or Gypsum (CaSO4 2H2O)10.30 Plaster of Paris (CaSO4 ½ H2O) or [(CaSO4)2 H2O]10.31 Bleaching Powder CaOCl2 or Ca(OCl)Cl 10.31 Cement 10.32 10.33 10.35 10.37 Biological Role of Mg and Ca NCERT Exemplars Practice Exercises Chapter 11A p-Block Elements I 11.1 IIIa (ns2 np1)11.1 Boron 11.5 Compounds of Boron Borax or Tincal (Na2B4O7 .10H2O)11.8 Boric Acid or Ortho Boric Acid (H3BO3 [IV (ns2 np2) or Group 14] 11.31 11.31 Carbon 11.35 Compounds of Carbon 11.38 Carbon Dioxide O = C = O or CO211.40 Carbon Disulphide (CS2)11.42 Carbides 11.43 Fuels 11.44 11.45 Silicon and its Compounds Silicon Compounds of Silicon Carborundum (Silicon Carbide) (SiC) 11.46 Silicones 11.47 Silicates 11.48 Glass 11.50 TIN and its Compounds TIN (Sn) 11.51 Properties of TIN 11.53 Compounds of Tin Stannous Oxide (SnO) 11.54 Stannic Oxide (SnO2)11.55 Stannous Chloride (SnCl2)11.55 Stannic Chloride (SnCl4)11.56 Lead and its Compounds Lead or Plumbum (Pb) 11.56 11.57 Properties Of Lead Lead Mono Oxide or Litharge or Plumbous Oxide or Lead (II) Oxide (PbO) 11.59 Plumbic Oxide or Lead Dioxide or Lead (IV) Oxide (PbO2)11.59 Red Lead or Tri Lead Tetra-Oxide (Pb3O4)11.60 Lead (II) Sulphide (PbS) 11.60 Lead (II) Halides or Plumbous Halides (PbX2)11.60 Lead Chloride or Plumbous Chloride (PbCl2)11.60 Lead (IV) Halides or Plumbic Halides (PbX4)11.61 or B(OH)3 )11.9 Lead Tetrach loride or Plumbic Chloride (PbCl4)11.61 Hydrides of Boron Lead Acetate or Sugar of Lead (CH3COO)2Pb11.61 Diborane (B2H6 )11.10 Basic Lead Carbonate or White Lead Boron Nitride (BN) 2PbCO3.Pb(OH)211.62 NCERT Exemplars 11.63 Practice Exercises 11.65 11.10 11.12 27 Aluminium ( 9Al )11.13 Aluminium Chloride AlCl3 or Al2Cl611.16 Aluminium Oxide or Alumina Al2O311.17 Alums 11.18 Potash Alum (K2SO4Al2(SO4)3 24H2O)11.18 Ultra Marine (Na5Al3Si3S3O12)11.19 Practice Exercises 11.20 Chapter 12A Organic Chemistry I 12.1 Organic and Inorganic Compounds 12.1 Homologous Series 12.4 Hybridization 12.4 12.6 IUPAC System 2.20 Chapter 12 Orbital angular momentum depends on (a) l (b) n and l (c) n and m (d) m and s 13 Chlorine exists in two isotopic forms, Cl-37 and Cl35 but its atomic mass is 35.5 This indicates the ratio of Cl-37 and Cl-35 is approximately (a) 1:2 (b) 1:1 (c) 1:3 (d) 3:1 14 The pair of ions having same electronic configuration is (a) Cr 3+, Fe3+ (b) Fe3+, Mn2+ 3+ 3+ (c) Fe , Co (d) Sc3+, Cr 3+ 15 For the electrons of oxygen atom, which of the following statements is correct? (a) Zeff for an electron in a 2s orbital is the same as Zeff for an electron in a 2p orbital (b) An electron in the 2s orbital has the same energy as an electron in the 2p orbital (c) Zeff for an electron in 1s orbital is the same as Zeff for an electron in a 2s orbital (d) The two electrons present in the 2s orbital have spin quantum numbers ms but of opposite sign 16 If travelling at same speeds, which of the following matter waves have the shortest wavelength? (a) Electron (b) Alpha particle (He2+ ) (c) Neutron (d) Proton ANSWER K EYS 1. (c) 11. (c) 2. (b) 12. (a) 3. (d) 13. (c) 4. (d) 14. (b) 5. (b) 15. (d) 6. (a) 16. (b) 7. (d) 8. (d) 9. (c) 10. (b) HINTS AND EXPLANATIONS HINTS AND EXPLANATIONS Concept of circular path was proposed by Bohr but not derived from Rutherford’s scattering experiment 8 n= 3-1-1 = the density of electrons for 2s orbit first increases and then decreases and followed by increasing again 14 Fe3+, Mn2+ have 23 electrons Since both have same number of electrons so same electronic configuration Mass of electron (9.1 × 10 –31 kg) is very small when compared it with mass of neutron (1.674 × 10 –27 kg) 11 n2 = 32 = 16 for same value of frequency, larger the value of mass m, wavelength gets shorter Structure of Atoms 2.21 PRACTICE EXERCISES Single Option Correct Type Wavelength of spectral line emitted is inversely pro portional to (a) energy (b) velocity (c) radius (d) quantum number The number of electrons, neutrons and protons in a species are equal to 10, and respectively The proper symbol of the species is (a) 16O8 (b) 18O8 (c) 18Ne10 (d) 16O8–2 Which set has same number of unpaired electrons? (a) Fe2+, Mn+2 (b) Fe3+, Mn+2 (c) Cr+3, Ni2+ (d) Zn2+ Cu2+ Which one of the following forms a colourless solu tion in aqueous medium? (a) V3+ (b) Cr3+ (c) Ti3+ (d) Sc3+ Atomic number Sc = 21, Ti = 22, V = 23, Cr =24 The incorrect configuration is (a) K = [Ar] 4s1 (b) Cr = [Ar] 3d5 4s1 (c) Cr = [Ar] 3d4 4s2 (d) Cu = [Ar] 3d10 4s1 Which of the following elements has least number of electrons in its M shell? (a) Mn (b) Ni (c) K (d) Sc In which orbital is the angular momentum of an elec tron zero? (a) 2p (b) 2s (c) 3d (d) 4f Chromium is represented configuration (a) [Ne] 3s2 3p6 3d1 4s2 (b) [Ne] 3s2 3p6 3d2 4s1 (c) [Ne] 3s2 3p6 3d5 4s1 (d) [Ne] 3s2 3p6 4s2 3d4 by the electronic 10 The correct set of quantum numbers is (a) n = 2, l = l, m = –2, s = (b) n = 2, l = –2, m = l, s = + ½ (c) n = 2, l= 2, m = –l, s = –½ (d) n = 2, l = l, m = 0, s = + ½ 11 The values of four quantum numbers of valence elec tron of an element are n = 4, l = 0, m = and s = + ½ The element is (a) Ti (b) K (c) Na (d) Sc 12 Set of isoelectronic species is (a) H2, CO2, CN–, O– (b) N, H2S, CO (c) N2, CO, CN–, O2+2 (d) Ca, Mg, Cl 13 Number of orbitals in L energy level (a) 1 (b) 2 (c) 3 (d) 4 14 Ratio of radii of second and first Bohr orbits of H atom is (a) 2 (b) 4 (c) 3 (d) 4 15 If a source of power 4Kw produces 1020 photon Per second, the radiation belongs to a part of the spectrum called (a) γ-Rays (b) U.V Rays (c) Microwaves (d) X-Rays 16 The outer most configuration of most electronegative element is (a) ns2 np5 (b) ns2 np6 (c) ns2 np4 (d) ns2 np6 17 Which one of the following expressions represent the electron probability function D? (a) 4πr dr ψ2 (b) 4πr2 dr ψ (c) 4πr2 dr ψ2 (d) 4πr dr ψ 18 If the electron of a hydrogen atom is present in the first orbit, the total energy of the electron is (a) –e2/2r (b) –e2/r 2 (c) –e /r (d) –e2/2r2 PRACTICE EXERCISES How many e– are present in p-sub-orbits in Zn-atom? (a) 2 (b) 6 (c) 12 (d) 18 2.22 Chapter 19 Quantum numbers of an atom can be defined on the basis of (a) Aufbau’s principle (b) Heisenberg’s uncertainity principle (c) Hund’s rule (d) Pauli’s exclusion principle 20 Which of the following is isoelectronic? (a) CO2, NO2 (b) NO2–, CO2 – (c) CN , CO (d) SO2, CO2 21 Which of the following not travel with speed of light? (a) De-Broglie waves (b) X-rays (c) Gamma rays (d) All of these 22 How many electrons can be accommodated in a p-orbital? (a) electrons (b) electrons (c) electrons (d) none of these PRACTICE EXERCISES 23 The quantum number ‘m’ of a free gaseous atom is associated with (a) the effective volume of the orbital (b) the shape of the orbital (c) the spatial orientation of the orbital (d) the energy of the orbital in the absence of a mag netic field 24 In ground state, an element has 13 electrons in its M shell The element is (a) copper (b) iron (c) nickel (d) chromium 25 Which of the following pair of ions have same para magnetic moment? (a) Cu2+, Ti3+ (b) Ti3+, Ni2+ 4+ 2+ (c) Ti , Cu (d) Mn2+, Cu2+ 26 The following quantum numbers are possible for how many orbital? n = 3, l = 2, m = +2 (a) 1 (b) 2 (c) 3 (d) 4 27 The number of nodal planes in a px orbital is (a) 1 (b) 2 (c) 3 (d) 0 28 How many d-electrons are present in Cr2+ ion? (a) 5 (b) 6 (c) 3 (d) 4 29 l = 3, then the values of magnetic quantum numbers are Å (a) ± l, ± 2, ± (b) 0, ± 1, ± 2, ± (c) –1, –2, –3 (d) 0, + l, + 2, + 30 Which one of the following pairs of ions have the same electronic configuration? (a) Sr3+, Cr3+ (b) Fe3+, Mn2+ 3+ 3+ (c) Fe , Co (d) Cr3+, Fe3+ 31 The radius of hydrogen atom is 0.53 Å The radius of Li2+ is of (a) 1.27 Å (b) 0.17 Å (c) 0.57 Å (d) 0.99 Å 32 The atomic number of an element is 35 What is the total number of electrons present in all the p-orbitals of the ground state atom of that element? (a) 17 (b) ll (c) 23 (d) 6 33 The atomic number of an element is 17 The number of orbitals electron pairs in its valence shell is (a) 3 (b) 4 (c) 6 (d) 8 34 The correct representation for d orbital is (a) (n – 1) d1–9 ns1 (b) (n – 1) d1–10 ns1–2 (c) (n–1)d1–5 (d) (n–l) d1–10 ns2 35 An element has electrons in its K shell, electrons in L shell, 13 electrons in M shell and one electron in N shell The element is (a) Cr (b) Fe (c) V (d) Ti 36 Effective magnetic moment of Sc3+ ion is (a) 0 (b) 1.73 (c) 2.83 (d) 3.87 37 Which of the following is not possible? (a) n = 2, l = l, m = (b) n = 2, l = 0, m = –l (c) n = 3, l = 0, m = (d) n = 3, l = 1, m = –1 38 The most probable radius (in pm) for finding the elec tron in He+ is (a) 105.8 (b) 52.9 (c) 26.5 (d) 0.0 Structure of Atoms 2.23 40 When potassium metal is exposed to violet light (a) there is no effect (b) ejection of electron takes place (c) the absorption of electrons takes place (d) ejection of some potassium atoms occurs 41 The atomic number of Ni and Cu are 28 and 29 respectively The electronic configuration 1s2 2s2 2p6 3s2 3p6 3d10 represents (a) Cu+ (b) Cu2+ 2+ (c) Ni (d) Ni 42 Rutherford’s experiment, which established the nu clear model of the atom, used a beam of (a) β-particles, which impinged on a metal foil and got absorbed (b) γ-rays, which impinged on a metal foil and ejected electrons (c) Helium atoms, which impinged on a metal foil and got scattered (d) Helium nuclei, which impinged on a metal foil and got scatterd 43 The quantum number + ½ and – ½ for the electron spin represent (a) rotation of the electron in clockwise and anti clockwise direction respectively (b) rotation of the electron in anti clockwise and clockwise direction respectively (c) magnetic moment of the electron pointing up and down respectively (d) two quantum mechanical spin states which have no classical analogues 44 The electronic configuration of an element is 1s2 2s2 2p6 3s2 3p6 3d5 4s1 This represents (a) excited state (b) ground state (c) cationic form (d) anionic form 45 The angular momentum (L) of an electron in a Bohr orbit is given as (a) L = nh/2π (b) √[l(l + 1) h/2π (c) L = mg/2π (d) L = h/4π 46 Which one of the following is the standard for atomic mass? (a) 1H1 (b) C12 (c) 6C4 (d) O16 47 Which of the following relates to photons both as wave motion and as a stream of particles? (a) interference (b) E = mc2 (c) diffraction (d) E = hυ 48 Electromagnetic radiation with maximum wavelength is (a) radiowave (b) X-ray (c) infraredt (d) ultraviolet 49 When the e– of a H-atom moves from n = to n = the number of spectral lines emitted is (a) 3 (b) 6 (c) 9 (d) 15 50 Which set represents the quantum Numbers of 19th electron in Cr–atom? (a) 4, 0, 0, ½ (b) 4, 1, 0, ½ (c) 3, 2, 2, ½ (d) 3, 2, –2, ½ 51 The orbital diagram in which both the Pauli’s exclu sion principle and Hund’s rule are violated is ↑ (a) ↑ ↑↑ ↑ ↑ ↑ ↑ ↑ (b) ↑ ↑ ↑ ↑ ↑ ↑ ↑ (c) ↑ ↑ ↑ ↑ (d) ↑ ↑ ↑ ↑ 52 The maximum number of 3d electrons having s = + ½ are (a) 10 (b) 5 (c) 14 (d) 7 53 The fourth electron of Be atom will have which of the following quantum numbers? (a) 1, 0, 0, ½ (b) 2, 0, 0, – ½ (c) 2, l, 0, + ½ (d) 1, 1, 1, + ½ 54 The electron density between 1s and 2s orbital is (a) high (b) low (c) zero (d) none of these 55 The Angular momentum of an e– can’t be (a) h π 2h (b) 2π (c) 2.5h 2π 0.5h (d) π PRACTICE EXERCISES 39 In which of the following pairs are both the ions coloured in aqueous solution? (a) Sc3+, Co2+ (b) Ni2+, Cu+ 2+ 3+ (c) Ni , Ti (d) Sc3+, Ti3+ (At no Sc = 21, Ti = 22, Ni = 28,Cu = 29, Co = 27) 2.24 Chapter 56 If the wavelength of an electromagnetic radiation is 2000 Å What is the energy in ergs? (a) 9.92 × 10–19 (b) 9.94 × 10–12 –12 (c) 4.97 × 10 (d) 4.97 × 10–19 57 The de Broglie wavelength of the electron in the ground state of hydrogen atom is [K.E.= 13.6 eV]; 1eV = 1.602 × 10–19 J (a) 33.28 nm (b) 3.328 nm (c) 0.3328 nm (d) 0.0332 nm 58 The value of Planck constant is 6.63 × 10–34 Js he velocity of light is 3.0 × 108 m s–1 which value is closest to the wavelength in nanometres of a quantum of light with frequency of × 1015 s–1? (a) 4 × 101 (b) × l07 –25 (c) 2 × 10 (d) × 10–18 59 The total number of electrons present in all the s orbitals, all the p orbitals and all the d orbitals of cesium ion are respectively (a) 12, 20, 22 (b) 8, 22, 24 (c) 10, 24, 20 (d) 8, 26, 10 PRACTICE EXERCISES 60 The frequency of radiation emitted when the electron falls from n = to n = in a hydrogen atom will be (Given ionization energy of H = 2.18 × 1018 J atom–1 and h = 6.625 × 10–34 Js) (a) 1.54 × 10l5 s–1 (b) 1.03 × 10l5 s–1 (c) 3.08 × 10l5 s–1 (d) 2.00 × l0l5 s–1 61 Among the following series of transition metal ions, the one where all metal ion have 3d2 electronic configuration is (a) Ti3+, V2+, Cr3+, Mn4+ (b) Ti+, V4+, Cr6+, Mn7+ (c) Ti4+, V3+, Cr2+, Mn3+ (d) Ti2+, V3+, Cr4+, Mn5+ (At wt Ti = 22, V = 23, Cr = 24, Mn = 25 ) 62 The relationship between energy E, of the radiation with a wavelength 8000 Å and the energy of the radiation with a wavelength 16000 Å is (a) E1 = 2E2 (b) E1 = 4E2 (c) E1 = 6E2 (d) E1 = E2 63 The atomic numbers of elements X, Y, Z are 19.21 and 25 respectively The number of electrons present in the ‘M’ shells of these elements follow the order (a) Z > Y > X (b) X > Y > Z (c) Z > X > Y (d) Y > Z > X 64 An electron is moving in Bohr’s fourth orbit, its de-Broglie wavelength is X What is the circumference of the fourth orbit? (a) 2λ (b) 2/λ (c) 3λ (d) 4λ 65 The correct order of number of unpaired electrons in the ion Cu2+ Ni2+, Fe3+ and Cr3+ is (a) Cu2+ > Ni2+ > Cr3+ > Fe3+ (b) Ni2+ > Cu2+ > Fe3+ > Cr3+ (c) Fe3+ > Cr3+ > Ni2+ > Cu2+ (d) Fe3+ > Cr3+ > Cu2+ > Ni2+ 66 Find the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25 (a) 6.9 B.M (b) 5.9 B.M (c) 4.9 B.M (d) 3.0 B.M 67 The magnetic moment of Cu2+ ion is (a) 2.6 (b) 2.76 (c) 1.73 (d) 0 68 Given: the mass of electron is 9.11 × 10–31 kg Planck constant is 6.626 × 10–34 Js, the uncertainty involved in the measurement of velocity within a distance of 0.1 Å is (a) 5.79 × 108 ms–1 (b) 5.79 × 105 ms–1 (c) 5.79 × 106 ms–1 (d) 5.79 × 107 ms–1 69 The energy ratio of a photon of wavelength 3000Å and 6000Å is (a) : (b) : (c) : (d) : 70 The energy of second Bohr orbit of the hydrogen atom is –328 kJ mol–1, hence the energy of fourth bohr orbit would be (a) –164 kJ mol–1 (b) –41 kJ mol–1 –1 (c) –82 kJ mol (d) –1312 kJ mol–1 71 The de Broglie wavelength associated with a particle of mass 10–6 kg moving with a velocity of 10 ms–1 is (a) 6.63 × 10–7 m (b) 6.63 × 10–16 m –21 (c) 6.63 × 10 m (d) 6.63 × 10–29 m 72 In hydrogen atom, energy of first excited state is –3.4 eV The kinetic energy of the same orbit of hydrogen atom would be (a) + 3.4 eV (b) + 6.8 eV (c) –13.6eV (d) +13.6eV 73 The velocity of an electron in the second shell of hydrogen atom is (a) 10.94 × 106 ms–1 (b) 18.88 × 106 ms–1 (c) 1.888 × 106 ms–1 (d) 1.094 × 106 ms–1 Structure of Atoms 2.25 74 Electron energy of a photon is given as: ΔE/atom = 3.03 × 10–19 J atom–1 then, the wavelength of the photon is (a) 6.56 nm (b) 65.6 nm (c) 656 nm (d) 0.656 nm Given, h (Planck constant) = 6.63 × 10–34 Js c (velocity of light) = 3.00 × 108 ms–1 (a) 1, 2, (c) 3, 75 If the nitrogen atom has electronic configuration 1s7, it would have energy lower than that of the normal ground state configuration 1s2 2s2 2p3 , because the electrons would be closer to the nucleus Yet 1s7 is not observed because it violates (a) Heisenberg uncertainty principle (b) Hund’s rule (c) Pauli’s exclusion principle (d) Bohr postulates of stationary orbits h (a) + 12 2π h (c) 2π 77 The radius of which of the following orbits is same as that of the first Bohr’s orbit of hydrogen atom? (a) He+ (n = 2) (b) Li2+ (n = 2) 2+ (c) Li (n = 3) (d) Be3+ (n = 2) 78 The wavelength associated with a golf ball weighing 200 g and moving at a speed of m/h is of the order (a) 10–10 m (b) 10–20 m –30 (c) 10 m (d) 10–40 m 79 Ground state electronic configuration of nitrogen atom can be represented by (1) (2) 82 The orbital angular momentum of an electron in 2s orbital is (b) zero h (d) √2 2π 83 The mass of an electron is m Its charge is e and it is accelerated from rest through a potential difference V The velocity acquired by the electron will be (a) √V/m (b) √eV/m (c) √2e V/m (d) none 84 An atom A has the electronic configuration of 1s2 2s2 2p1 Atom B has the electronic configuration of 1s2 2s2 2p1 The empirical formula of the compound obtained from the reaction of A and B is (a) AB (b) AB3 (c) A3B3 (d) A2 B6 85 A 600 W mercury lamp emits monochromatic radiation of wavelength 331.3 nm How many photons are emitted from the lamp per second? (h = 6.626 × 10–34 Js; velocity of light = × 108 ms–1 (a) × 1019 (b) × 1020 21 (c) × 10 (d) × 1023 86 The shortest wavelength in hydrogen spectrum of Lyman series when RH = 109678 cm–1 is (a) 1002.7 Å (b) 1215.67 Å (c) 1127.30 Å (d) 911.7 Å 87 Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon? (a) 1s (b) 2s (c) 2p (d) 3s (3) (4) (a) only (c) l, 81 For a d-electron, the orbital angular momentum is (a) √6h (b) √2h (c) h (d) 2h (b) 1, (d) 2, 80 Which of the following statement(s) are correct? (1) the electronic configuration of Cr is [Ar] 3d5 4s1 (atomic number of Cr = 24) (2) the magnetic quantum number may have a nega tive value (3) in silver atom, 23 electrons have a spin of one type and 24 of the opposite type (atomic number of Ag = 47) (4) the oxidation state of nitrogen in HN3 is –3 88 Correct set of four quantum numbers for the valence (outermost) electron of rubidium (Z = 37) is (a) 5, 0, , ± ½ (b) 6, 0, , + ½ (c) 5, 1, l, ± ½ (d) 5, l, 0, ± ½ 89 The increasing order (lowest first) for the values of e/m (charge/mass) for electron (e), proton (p), neutron (n) and alpha particle (a) is (a) n, p, a , e (b) n, p, e, a (c) n, a, p, e (d) e, p, n, a PRACTICE EXERCISES 76 Radial nodes present in 3s and 2p orbitals are respectively (a) 0, (b) 2, (c) 2, (d) 1,2 (b) 2, 3, (d) 1, 2, 2.26 Chapter 90 The ionization energy of hydrogen atom is 13.6 eV What will be the ionization energy of He+? (a) 13.6 eV (b) 54.4 eV (c) l22.4 eV (d) zero 91 If S, be the specific charge (e/m) of cathode rays and S2 be that of positive rays then which is true? (a) S1 = S2 (b) S1 < S2 (c) S1 > S2 (d) None of these 92 Predict the total spin in Ni2+ ion (a) ±5/2 (b) ±3/2 (c) ± l/2 (d) ± l 93 The orbital diagram in which Aufbau principle is violated is ↑ ↑ (a) ↑ ↑ ↑ ↑ (b) ↑ ↑ ↑ ↑ ↑ (c) ↑ ↑ ↑ ↑ ↑ ↑ ↑ (d) ↑ ↑ ↑ ↑ 94 Wave function vs distance from nucleus graph of an orbital is given below: + Ψ - The number of nodal sphere of this orbital is (a) 1 (c) 3 PRACTICE EXERCISES r (b) 2 (d) 4 95 For the electronic transition from n = → n = 1, which of the following will produce shortest wavelength? (a) Li2+ ion (b) D atom (c) He+ ion (d) H atom 96 Which of the following curves may represent the speed of the electron in a hydrogen atom as a func tion of the principal quantum number n? b v a d c n (a) d (c) b (b) c (d) a 97 For n = the correct set of l, m are (a) l = 2, m = –2,–1, +1, +2 (b) l = m = –2,–1, +1, +2 (c) l = m = –1, 0, +1 (d) l = o m = –1, 0, +1 98 Probability of finding the electron ψ2 of s orbital doesn’t depend upon (a) azimuthal quantum number (b) energy of s orbital (c) principal quantum number (d) distance from nucleus (r) 99 The charge cloud of a single electron in a 2px atomic orbital has two lobes of electron density This means (a) there is a high probability of locating the electron in a 2px atomic orbital at values of x > (b) there is a great probability of finding a p electron right at the nucleus (c) there is a high probability of locating it values of x < but no probability at alloy locating if any where in the yz plane along which x = (d) both (a) and (c) 100 The wavelength of the de Broglie wave of the elec tron revolving in the fifth orbit of the hydrogen atom is (r0 is the Bohr’s radius = 0.529 Å) (a) 20 r0 (b) (10 π) r0 (c) 5 π r0 (d) 15 π r0 101 A monoenergetic electron beam with a de Broglie wavelength of x Å is accelerated till its wavelength is halved By what factor is its kinetic energy changed? (a) 8 (b) 6 (c) 4 (d) 3 102 The de Broglie wavelength associated with a ball of mass kg having a kinetic energy 0.5 J is (a) 6.626 × 10–34 m (b) 13.2 × 10–34 m (c) 10.38 × 10–21 m (d) 6.626 Å 103 The size of a microscopic particle is one micron and its mass is × 10–13 gm If its position may be measured to within 0.1% of its size, the uncertainty in velocity, in cm s–1, is approximately (a) 10–6/3 π (b) 10–7/2 π –5 (c) 10 /4 π (d) 10–7/4 π 104 The electrons, identified by quantum numbers n and l (i) n = l = (ii) n = 4, l = (iii) n = 3, l = (iv) n = 3, l = can be placed in order of increasing energy from the lowest to highest as Structure of Atoms 2.27 (a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii) (c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii) 105 What is the wavelength of the radiation emitted pro duced in a line in the Lyman series when an electron falls from fourth stationary state in hydrogen atom? (RH = 1.1 × 107 m–1) (a) 96.97 nm (b) 969.7 nm (c) 9.697 nm (d) none 106 Rearrange the following (I to IV) in the order of in creasing masses and choose the correct answer from (a), (b), (c), (d) (atomic masses: N =14, O = 16, Cu = 63) I. molecule of oxygen II. atom of nitrogen III. 1 × 1010 g molecular weight of oxygen IV. 1 × 10–18 g atomic weight of copper (a) II < I < IV < III (b) IV < III < II < I (c) II < III < I < IV (d) III < IV < I < II 107 The ionization energy of He+ is 19.6 × 10–18 J atom–1 Calculate the energy of the first stationary state of Li2+ (a) 19.6 × 10–18 J atom–1 (b) 4.41 × 10–18 J atom–1 (c) 19.6 × 10–19 J atom–1 (d) 4.41 × 10–17 J atom–1 108 The masses of photons corresponding to the first lines of the Lyman and the Balmer series of the atomic spectrum of hydrogen are in the ratio of (a) : (b) 27 : (c) : (d) : 27 109 An X-ray tube is operated at 50,000 volts The short-wavelength limit of the X-rays produced is (a) 0.1245 Å (b) 0.3485 Å (c) 0.2485 Å (d) 0.03456 Å 110 The ratio of the difference between 2nd and 3rd Bohr’s orbit energy to that between 3rd and 4th orbit energy is (a) 7/20 (b) 20/7 (c) 27/9 (d) 9/27 111 An electron in a hydrogen atom in its ground state absorbs 1: 50 times as much energy as the minimum required for it to escape from the atom What is the wavelength of the emitted electron? (a) 4.7 Å (b) 4.70 pm (c) 6.3 Å (d) 8.4 Å 112 If the radius of the first Bohr orbit is ‘a’, then de Broglie wavelength of electron in 3rd orbit is nearly (a) 2π a (b) 6π a (c) 3a (d) a/3 113 If the shortest wavelength of H atom in Lyman series is ‘a’, then longest wavelength in Balmer series of He+ is (a) a/4 (b) 5a/9 (c) 4a/9 (d) 9a/5 114 In hydrogen atom, an orbit has a diameter of about 16.92 A What is the maximum number of electrons that can be accommodated? (a) 32 (b) 16 (c) 48 (d) 72 115 Energy levels A, B, C of a certain atom corresponds to increasing values of energy, i.e., EA < EB < Ec If X1, X2 and X3 are the wavelengths of radiations cor responding to the transitions C to B, B to A and C to A respectively, which of the following statement is correct? C B X1 X2 X3 (a) X1 + X2 + X3 = (c) X32 = X12 + X22 A (b) X3 = X1 + X2 X X (d) X3 = X 1+ X2 116 A 1000 watt radio transmitter operates at a frequency of 880 kc/sec How many photons per sec does it emit? [h = 6.626 × 10–34 Js] (a) 2.51 × 1030 (b) 2.27 × l028 30 (c) 1.72 × 10 (d) 1.77 × 1027 117 How many moles of electrons weigh one kilogram? (mass of electron = 9.108 × 10–31 kg, Avogadro number = 6.023 × 1023) (a) 6.023 × 1023 6.023 (c) _ × 1054 9.108 (b) 1/9.108 × 1031 (d) × 108 9.108 × 6.023 118 Calculate the wavelength and energy of the radiation emitted for the electronic transition from infinity (∞) to stationary state first of the hydrogen atom (RH = 1.09678 × 107 m–1, h = 6.6256 × 10–34 Js) (a) 2.18 × 10–21 kJ (c) 1.18 × 10–23 kJ (b) 3.18 × 10–22 kJ (d) 2.18 × l0–31 kJ PRACTICE EXERCISES 2.28 Chapter Previous Years' Questions 119 Energy of H-atom in the ground state is –3.6 eV, hence energy in the second excited state is [2002] (a) –6.8 eV (b) –3.4eV (c) –1.51 eV (d) –4.53 eV 120 Uncertainty in position of a particle of 25 g in space is 10–5 m Hence uncertainty in velocity (ms–1) is (Planck constant h = 6.6 × 10–34 Js) [2002] (a) 2.1 × 10–28 (b) 2.1 × 10–34 (c) 0.5 × 10–34 (d) 5.0 × 10–24 121 Which of the following ions has the maximum magnetic moment? [2002] 2+ 2+ (a) Mn (b) Fe (c) Ti2+ (d) Cr2+ 122 In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?[2003] (a) 3 (b) (c) 4 (d) 2 PRACTICE EXERCISES 123 The de Broglie wavelength of a tennins ball of mass 60 g moving with a velocity of 10 metres per second is approximately [2003] –31 (Planck constant h = 6.63 × 10 Js) (a) 10–33 m (b) 10–31m (c) 10–16 m (d) 10–25 m 124 The orbital angular momentum for an electron re volving in an orbit is given by /(l + 1) h/2π This momentum for an s electron will be given by [2003] (a) + ½.h/2 π (b) zero (c) h/2 π (d) √2.h/2 π 125 Which one of the following grouping represents a collection of isoelectronic species? (At numbers Cs-55, Br-35) [2003] + 2+ 2+ (a) Na , Ca , Mg (b) N3–, F–, Na+ (c) Be, Al3+, Cl– (d) Ca2+, Cs+, Br 126 The number of d electrons retained in Fe2+ (At number of Fe = 26) ions is [2003] (a) 3 (b) 4 (c) 5 (d) 6 127 Which of the following sets of quantum numbers is correct for an electron in 4f orbital?[2004] (a) n = 4, l = 3, m = +4, s = + ½ (b) n = 4, l = 4, m = –4, s = –½ (c) n = 4, l = 3, m = + l, s = + ½ (d) n = 3, l = 2, m = –2, s = + ½ 128 Consider the ground state of Cr atom (Z = 24) The numbers of electrons with the azimuthal quantum numbers, l =1 and are, respectively[2004] (a) 12 and (b) 12 and (c) 16 and (d) 16 and 129 The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to station ary state 1, would be (Rydberg constant = 1.097 × 107 m–1) [2004] (a) 91nm (b) 192nm (c) 406nm (d) 9.1 × 10–8 nm 130 Which one of the following sets of ions represents the collection of isoelectronic species? [2004] 2+ 3+ – + 2+ (a) K+ Ca , Sc , Cl (b) Na , Ca , Sc3+, F– + – 2+ 3+ (c) K , Cl , Mg , Sc (d) Na+ Mg2+, Al3+, Cl– 131 In a multielectron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic field and electric fields? [2005] (a) n= l, l = 0, m = (b) n = 2, l = 0, m = (c) n = 3, l = l,m=l (d) n = 3, l =2, m=l (e) n = 3, l = 2, m = (a) b and c (b) d and e (c) c and d (d) a and b 132 Which of the following statements in relation to the hydrogen atom is correct? [2005] (a) 3s and 3p orbitals are of lower energy than 3d orbital (b) 3p orbital is lower in energy than 3d orbital (c) 3s orbital is lower in energy than 3p orbital (d) 3s, 3p and 3d orbitals all have the same energy 133 Of the following sets which one does not contain iso electronic species? [2005] Structure of Atoms 2.29 (b) CN–, N2, C22– (d) BO33–, CO32–, NO3– 134 According to Bohr theory, the angular momentum of an electron in 5th orbit is [2006] (a) 25h/π (b) 1.0 h/π (c) 10 h/π (d) 2.5 h/π 135 Which one of the following sets of ions represents a collection of isoelectronic species? [2006] (a) K+, Cl–, Ca2+, Sc3+ (b) Ba2+, Sr2+, K+, Ca2+ (c) N3–, O2–, F–, S2– (d) Li+, Na+, Mg2+ Ca2+ 136 Which of the following sets of quantum numbers represents the highest energy of an atom? [2007] (a) n = 3, l = 2, m= l,s = + 1/2 (b) n = 4, l = 0, m = 0, s = +1/2 (c) n = 3, l = 0, m = 0, s = + 1/2 (d) n = 3, l = l,m= l,s = + 1/2 137 Which one of the following constitutes a group of the isoelectronic species? [2008] 2– – (a) C2 , O2 , CO, NO – (b) NO+, C2– , CN , N2 – 2– (c) CN , N2, O2 , C2 2– – + (d) N2, O2 , NO , CO 138 In an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0.005% Certainity with which the position of the electron can be located is (h = 6.6 × 10–34 kg m2 s–1, mass of electron, em = 9.1 × 10–31 kg): [2009] –3 –3 (a) 5.10 × 10 m (b) 1.92 × 10 m (c) 3.84 × 10–3 m (d) 1.52 × 10–3 m 139 Calculate the wavelength (in monometer) associated with a proton moving at 1.0 × 103 ms–1 (Mass of proton = 1.67 × 10–27 kg and h = 6.63 × 10–34 Js): [2009] (a) 0.40 nm (b) 2.5 nm (c) 14.0 nm (d) 0.032 nm 140 Ionization energy of He+ is 19.6 × 10–18 J atom–1 The energy of the first stationary state (n = 1) of Li2+ is [2010] –16 –1 (a) 4.41 × 10 J atom (b) –4.41 × 10–17 J atom–1 (c) –2.2 × 10–15 J atom–1 (d) 8.82 × 10–17 J atom–1 141 The energy required to break one mole of C1 – C1 bonds in C12 is 242 kJ mol–1 the longest wavelength of light capable of breaking a single C1 – C1 bond is (c = × 108 ms–1 and NA = 6.02 × 1023 mol–1 [2010] (a) 594 nm (b) 640 nm (c) 700 nm (d) 494 nm 142 A gas adsorbs a photon of 355 nm and emits at two wavelengths If one of the emissions is at 680 nm, the other is at: [2011] (a) 743 nm (b) 376 nm (c) 453 nm (d) 581 nm 143 The electrons identified by quantum numbers n and l [2012] (1) n = 4, l = (2) n = 4, l = (3) n = 3, l = (4) n = 3, l = can be placed in order of increasing energy as (a) < < < (b) < < < (c) < < < (d) < < < 144 Energy of an electron is given by, [2013] Z2 E = –2.178 × 10–18 n Wavelength of light required to excite an electron in an hydrogen atom from level n = to n = will be (h = 6.62 × 10–34 Js and c = 3.0 × 108 ms–1) (a) 6.500 × 10–7 m (b) 8.500 × 10–7 m –7 (c) 1.214 × 10 m (d) 2.816 × 10–7 m 145 The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is:[2014] 1 (a) 5, 1, 1, (b) 5, 0, 1, 2 1 (c) 5, 0, 0, (d) 5, 1, 0, 2 146 Which of the following is the energy of a possible excited state of hydrogen? [2015] (a) +13.6 eV (b) –6.8 eV (c) –3.4 eV (d) +6.8 eV 147 A stream of electrons from a heated filament was passed between two charged plates kept at a potential difference V esu If e and m are charge and mass of an electron, respectively, then the value of h/λ (where λ is wavelength associated with electron wave) is given by: [2016] PRACTICE EXERCISES (a) PO43–, SO42–, ClO4– (c) SO32–, CO32–, NO3– 2.30 Chapter (a) 2meV (b) meV (c) (d) meV 2meV (a) 1.65Å (c) 0.529Å 148 The radius of the second Bohr orbit for hydrogen atom is: (Plank’s constant, h = 6.6262 × 10−34 Js; mass of electron = 9.1091 × 10−31 kg; charge of electron, e = 1.60210 × 10−19 C; permittivity of vaccum, ∈0 = 8.854185 × 10−12 kg−1 m−3 A2) [2017] (b) 4.76Å (d) 2.12Å 149 The isotopes of hydrogen are: (a) Tritium and protium only (b) Deuterium and tritium only (c) Protium and deuterum only (d) Protium, deuterium and tritium [2019] ANSWER K EYS Single Option Correct Type 1. (c) 2. (a) 11. (b) 12. (c) 21. (a) 22. (b) 31. (b) 32. (a) 41. (a) 42. (d) 51. (a) 52. (b) 61. (d) 62. (a) 71. (d) 72. (a) 81. (a) 82. (b) 91. (c) 92. (d) 101. (c) 102. (a) 111. (a) 112. (b) 3. (d) 13. (d) 23. (c) 33. (a) 43. (d) 53. (b) 63. (a) 73. (d) 83. (c) 93. (b) 103. (d) 113. (d) 4. (b) 14. (b) 24. (d) 34. (b) 44. (b) 54. (c) 64. (d) 74. (c) 84. (a) 94. (a) 104. (a) 114. (a) 5. (d) 6. (c) 7. (c) 15. (d) 16. (a) 17. (c) 25. (a) 26. (b) 27. (a) 35. (a) 36. (a) 37. (b) 45. (a) 46. (b) 47. (d) 55. (c) 56. (b) 57. (c) 65. (c) 66. (b) 67. (c) 75. (c) 76. (b) 77. (d) 85. (c) 86. (d) 87. (a) 95. (a) 96. (b) 97. (c) 105. (a) 106. (a) 107. (d) 115. (d) 116. (c) 117. (d) 8. (b) 9. (c) 10. (d) 18. (a) 19. (d) 20. (c) 28. (d) 29. (b) 30. (b) 38. (c) 39. (c) 40. (b) 48. (a) 49. (b) 50. (a) 58. (a) 59. (c) 60. (c) 68. (c) 69. (b) 70. (c) 78. (c) 79. (c) 80. (a) 88. (a) 89. (c) 90. (b) 98. (a) 99. (d) 100. (b) 108. (b) 109. (c) 110. (b) 118 (a) 122. (b) 132. (d) 142. (a) 123. (a) 124. (b) 125. (b) 133. (c) 134. (d) 135. (a) 143. (a) 144. (c) 145. (c) 126. (d) 127. (c) 128. (b) 136. (a) 137. (b) 138. (b) 146. (c) 147. (c) 148. (d) PRACTICE EXERCISES Previous Years’ Questions 19. (c) 120. (a) 129. (a) 130. (a) 139. (a) 140. (b) 149. (d) 121. (a) 131. (b) 141. (d) Structure of Atoms 2.31 HINTS AND EXPLANATIONS Single Option Correct Type 14 As rn α n2 So r2 = r1 15 Power (P) = λ= nhc λ nhc 10 20 × 6.62 × 10 −34 × × 108 = × 10 p = 4.965 × 10–9 m = 49.65 A° 16 As halogens are most electronegative so the configu ration is ns2 np5 20 As both have 14 electrons 24 As 24Cr = 2, 8, 13, i.e., M shell has 13 electrons 25 As n = 30 Both Fe3+ and Mn2+ have [Ar] 3d5 configuration × I2 n2 0.53 _ 31 rn = 0.53 × Z = 3 = 0.17 Å 32 36 37 39 (Z = 35) : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5 No of electron in all the p orbitals = × + = 17 As it has no unpaired electrons As values of m is from –1 to +1 including zero As Ni2+, Ti3+ has 2, unpaired electrons respectively so both are coloured 41 It represent the configuration of Cu+ 43 These represent two quantum mechanical spin states which have no classical analogues 49 No of lines = 50 Cr 24 n(n – 1) = 4(4 − 1) =6 1s2 2s2 2p6 3s2 3p6 4s1 3d5 The 19th e– is in 4s1 hence n, l, m, s are 4,0, 55 Angular momentom = n 2.5 respectively h as 2.5 is not an intiger so 2π h is not possible 2π 56 E = hc/λ = 2000 × 10–8 = 9.94 × 10–12 ergs 64 As circumference = nλ = 4λ 65 As they have 5, 3, 2, number of unpaired electrons respectively 66 It is Mn2+ having five unpaired electrons so its magnetic moment is µ = √5 (5 + 2) = 5.9 B.M 67 As it has only one unpaired electron so magnetic moment is √l (l + 2) = 1.73 B.M 68 Δv = h/4π m × Δx 6.626 x 10–34 = x 3.14 x 9.1 x 10–31 x 0.1 x 10–10 = 5.8 × 106 ms–1 69 E1/E2 = λ2/λ1 = 6000/3000 = : 70 As En ∞ 1/n2 71 Use the relation; λ = h/mv Z 73 = 2.188 × 106 × n = 2.188 × 106 × 2 = 1.094 × 106 ms–1 74 Use the relation ΔE = hc/λ 75 As maximum number of electrons in any orbit, sub-orbit or orbital is decided by Pauli’s law 6.626 × 10–34 _ 78 λ = h/mv = 200 × 10–3 × 5 = 6.626 × _ 10–32 81 As mvr = √ 2l(l + 1) h Here l = so mvr = –h 82 Here l = so mvr = 84 Both A and B have the valencies So, the formula is AB 85 Power = 6.626 × 10 × × 10 –27 ×108 _ 58 v = c/λ = 8 × 1015 = 37.5 × 10–9 m = 37.5 nm ≈ × 101 nm 59 Cs+ : 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 No of electrons in s orbitals = × = 10 No of electrons in p orbitals = × = 24 No of electrons in d orbitals = 10 × = 20 63 As Z has 13 electrons, Y has electrons and X has e lectrons in M shell 600 = Energy Time = nhc λ×t n × 6.626 × 10 −34 × × 108 331.3 × 10 −9 × HINTS AND EXPLANATIONS Both Mn+2 and Fe3+ have un-paired electrons As Sc3+ has no unpaired electron so it is colour less 12 All have 14 electrons 2.32 Chapter ⇒ n = 600 × 331.3 × 10 −9 6.626 × 10 −34 × × 108 0.1 _ = × 10 ∆ x = 100 × 10–4 cm = 10–7 cm 21 _ 86 The shortest wavelength in hydrogen spectrum of Lyman series cab be given : as follows λ = RH n = RH = 105 n1= , n2 = Now 1/ λ = RH [l/n12 – 1/n22] = 1.1 × 10–7 [l/l2 – l/42] After solving λ = 96.9 nm 109678 1 07 I.E of He+ = E × 22 (as Z for He = 2) I.E of Li2+ = E × 32 (as Z for Li = 3) = 911.7 × 10 −10 m = 911.7 Å Hence, 87 It is the 1s level, the ground state, where the H-atom can only absorb a photon and go to higher excited states = × 19.6 × 10–18 = 4.41 × 10–17 J atom–1 08 ΔE = mc2 For Lymann series, m1c2 α (l/l2 – 1/22) For Balmer series, m2c2 α (1/22 – 1/32) m1/m2 = × 36/4 × m, : m2 = 27 : 90 I.E of He+ = 13.6 eV × Z2 = 13.6 eV × = 54.4 eV 91 Mass of positively charged ions in positive rays is more than mass of electrons 109 Energy of electrons striking the anti-cathode HINTS AND EXPLANATIONS = 50000 × 1.6 × 10–19 J = 8.0 × 10–15 J 92 No of unpaired electrons in Ni2+ is two Total spin = ẵ ì No of unpaired electrons I.E (He+) _ = I.E (Li2+) Therefore, I.E (Li2+) = × I.E (He+) 88 Rb has the configuration, 1s2 2s2 2p2 3s2 3p6 3d10 4s2 4p6 5s1 ; so n = 5, l = 0, m = and s = ½ 10–7 _ = 4 π × 1.1 × 10–7 cm s–1 = 4 π cm s–1 ⇒ λ = 9.117 × 10 −6 cm 6.626 × 10–27 _ ∆v = 4 π × × 10–13 × 10–7 cm s–1 c 6.626 × 10–34 × × 108 93 s-subshell should be filled first as it possesses lower energy level than p-subshell h λ = λ 19.88 × 10 −26 = J λ 95 Li2+ ion has shortest wavelength 100 Radius of the 5th orbit = 52r0 = 25 r0 Circumference = π (25 r0) = 50 πr0 = λ, n λ.= 2π rn λ = (10 π) rn h √ 2mE 101 λ = λE½ = constant So λ1 √E1 = λ2 √E2 λl/ λ1 = √(E2/E1) = Therefore, E2/E1 = 102 λ = h/√2m E 6.626 × 10–34 _ = 6.626 × 10–34 m √ × × 0.5 = 103 Δp Δx = h / 2π m Δv Δx = h/2π h _ _ ∆ v = 4 π × m ∆x m = × 10–13 g 19.88 × 10–26 λ = × 10–15 m = 2.485 × 10–11 m = 0.2485 Å 110 As ΔE3 – = –1312[(1/n22) – (1/n12) = –1312 (1/9 – 1/4) = –1312 × –5 36 Similarly, ΔE4 – = –1312 (1/16 – 1/9) –1312 × –7 = 16 × 9 ∆E 3–2 20 _ ∆E = 7 4–3 111 As 13.6 eV is needed for ionization, 20.4 eV (13.6 × 1.5) must have been absorbed Of this, 6.8 eV is converted to kinetic energy h _ √(2 m K.E.) 6.626 × 10–34 Js _ = √(2 × 9.1 × 10–31 × 6.8 × 1.6 × 10–19 J) λ = = 4.70 × 10–10 m = 4.70 Å 112 rn = r1 × n2 r = 32a = 9a Structure of Atoms 2.33 3h _ 2π mvr3 = 3h h = π 9a πa h.6 π a h _ _ λ = mv = = π a h mv = 113 For shortest λ of Lyman series of H n2 = ∝ 1 _ _ λ = RH I2 – ∝2 L For longest λ of Balmer series of He+ n2 = _ λ = Z2 RH 22 – 32 hυ = [6.626 × 10–34] × 880 × 103 J = 583.1 × 10–30 J Power transmitted = 1000 watts = 1000 J/s Number of photons emitted per sec 1000 = 583 × 10 = 1.72 × 1030 –30 B λ _ n = (N – shell) Number of electrons = 2n2 = × 42 = 32 116 Frequency = 880 kc/sec = 880 × 103 cycles/sec h = 6.626 × 10–34Js _ × 36 So, λ B = × 4 A Hence, λB = λA /5 = 9a/5 117 As mass of one electron = 9.108 electrons = 1/9.108 × 10–31 = = 6.92 14 r = 2 = 8.46 Å rn = r1 n2 r n × 10–31 kg So, kg of 1031 _ mol 9.108 × 6.022 × 1023 31–23 10 mol = 9.108 × 6.022 108 _ mol 9.108 × 6.022 8.46 Å = 16 0.529 Å n2 = r = Previous Years’ Questions λ _ n1 _ n2 = RH 2 – 2 λ = 1.09678 × 107 λ = 9.11 × 10–8 m E = hv = h × c/λ (c = × 108 m sec–1) − ∝2 6.6256 × 10–34 × 108 Value of l < n, l s p d f g value of m: – l, 0, + l value of s: + ½ or – ½ thus for 4f: n = 4, l = 3, m = any value between –3 to + 128 E.C of Cr (Z = 24) is 1s2 2s2 2p6 3s2 3p6 3d5 4s1 so electrons with l = (p), are 12 with l = (d), are _ _ = = 2.18 × 10–18 J 9.11 × 10–8 129 λ = vH = RH n 2 – n 2 = 2.18 × 10–21 kj = 1.097 × 107 12 – ∝ 2 13.6 En = n2 ev For second excited state n = 3, 13.6 E3 = – 9 = –1.51 eV h _ 120 ∆x ∆v ≥ π m 2+ 121 Here, Mn has unpaired electrons so it has maximum magnetic moment 122 The lines at the red end suggest Balmer series These are obtained for the jumps n = from n = and second line from n = and third line from n = 5, that is, 123 λ = h/mv 6.626 × 10–34 λ = 60 × 10–3 × 10 = 10–33 m (nearly) 124 As for s orbital, l = 125 As N , F and Na has 10 electrons –3 – + 127 Any sub-orbit is represented as nl such that n is the principle quantum number (in the form of values) and l is the azimuthal quantum number (its name) _ λ = 1.097 × 107 m = 9.11 × 10–8 m –9 = 91.1 × 10 m = 91.1 nm (1 nm = 10–9 m) 130 As these have 18 electrons 131 Orbitals having same (n + l) value in the absence of electric and magnetic field will have same energy 132 They not have same energy as they have different n + l values and for hydrogen the energy order of orbital is s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f 133 Number of electron in SO32– =16 + × + = 42 No of e– in CO = + × + = 32 No of e– in NO = + × + = 32 –3 – These are not isoelectronic species as number of electrons are not same 134 As mvr = nh/2π = 5h/2 π = 2.5h/π 135 As all of these have 18 electrons HINTS AND EXPLANATIONS 119 n1 = 1, n2 = ∝ 2.34 Chapter 137 NO+, C2– , CN– and N2 all are isoelectronic as all have four2 teen electrons h 138 ∆x ⋅ m ⋅ ∆v = 143 The order is decided by sum of (n + l) so < < < (3p) (4s) (3d) (4p) 4π h ∆x = 4π m∆v 0.005 ∆v = 600 × = 0.03 100 6.626 × 10 ∆x = E − E1 = 144 +2.178 × 10 −8 − 1 4 −34 = × 3.14 × 9.1 × 10 −31 × 0.03 −3 = 1.92 × 10 m h = 139 −19.6 × 10 kZ n2 12 H I N T S AND A N DEXPLANATIONS SOLUTIONS HINTS 2+ E1 (Li ) = − 4.9 × 10 −18 × 12 = – 44.1 × 10–18 J atom–1 = – 4.41 × 10 –17 141 E = = = J atom 6.023 × 10 23 = hc J molecule = 1 + = 743 nm ; −13.6 22 = 3.4 e.v h mv h 2m(KE) h λ = 2meV Radius of second Bohr orbit = 0.53 ´ ( 2) 149 Isotopes of hydrogen is: Proteium Deuterium Tritium = 494.7 mm e.v/atom = 2.12 Å 6.626 × × 6.023 × 10 −3 = 0.4947 × 10 −6 m 142 × 10 −8 148 Radius of nth Bohr orbit in H-atom = 0.53 6.626 × 10 −34 × × 108 242 × 10 h2 = ∴ –1 242 × 10 2.178 In Ist excitation state n=2 147 λ = K= 4.9 × 10 −18 −13.6 En = En = − k × 22 = 6.6 × 146 for H-atom 140 Ionization energy of He+ = E∞ – E1 = – E1 E1 (He+) = –19.6 × 10–18 J atm-1 −18 6.6 × 10 −34 × × 108 = 1.212 × 10 −7 m mv 1.67 × 10 −27 × 10 = 0.40 nm En = ∞ = 12.12 × 10 −8 6.63 × 10 −34 = = hc (1H1) (1H2) (1H3) 355 = 680 + n2 Å Z ... 2PbCO3.Pb(OH)2 11 .62 NCERT Exemplars 11 .63 Practice Exercises 11 .65 11 .10 11 .12 27 Aluminium ( 9Al ) 11 .13 Aluminium Chloride AlCl3 or Al2Cl6 11 .16 Aluminium Oxide or Alumina Al2O3 11 .17 Alums 11 .18 ... 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