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Preview IIT JEE Main Complete Chemistry K L Kapoor MHE Mc Graw Hill Education by K L Kapoor TMH Preview IIT JEE Main Complete Chemistry K L Kapoor MHE Mc Graw Hill Education by K L Kapoor TMH Preview IIT JEE Main Complete Chemistry K L Kapoor MHE Mc Graw Hill Education by K L Kapoor TMH Preview IIT JEE Main Complete Chemistry K L Kapoor MHE Mc Graw Hill Education by K L Kapoor TMH Preview IIT JEE Main Complete Chemistry K L Kapoor MHE Mc Graw Hill Education by K L Kapoor TMH Preview IIT JEE Main Complete Chemistry K L Kapoor MHE Mc Graw Hill Education by K L Kapoor TMH
JEE MAIN COMPLETE CHEMISTRY JEE MAIN COMPLETE CHEMISTRY K.L Kapoor Formerly Associate Professor, Hindu College, University of Delhi, Delhi McGraw Hill Education (India) Private Limited CHENNAI McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur, Chennai-600116 Complete Chemistry—JEE Main Copyright © 2018, McGraw Hill Education (India) Private Limited No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited Price: `875/1 2 3 4 5 6 7 8 9 7085462 22 21 20 19 18 Printed and bound in India ISBN (13): 978-93-87572-55-3 ISBN (10): 93-87572-55-2 Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services If such services are required, the assistance of an appropriate professional should be sought Typeset at Sri Krishna Graphics, Delhi and printed at Cover Designer: Neeraj Dayal visit us at: www.mheducation.co.in To Our Readers How to Crack the JEE T o help students preparing for the JEE Main, there was need for a book which included a variety of Multiple Choice Questions (MCQs) designed on the basis of the prescribed syllabus for this examination This book is an attempt in this direction and will help students in developing a strong foundation and enough confidence to take the JEE Main The various topics of chemistry may be classified into three branches—Physical, Inorganic and Organic The book covers these three branches in 29 chapters Each chapter starts with the synopsis of theory followed by MCQs along with answers and hints and solutions to arrive at correct answers Wherever needed, the chapter is divided into sections to cover the subject in easily understandable portions to help in better grasping of the subject matter Each section/chapter ends with MCQs from the previous years' AIEEE and JEE Main This will help students in getting an idea about the types and levels of questions asked in this competitive examination The answers and solutions to these questions are provided separately, immediately after the questions The analyses of these papers (provided on the next page) reveals that in most of the cases, one question is asked from each chapter and the entire syllabus is covered in the examination paper This book provides extensive coverage of the theory as well as the associated MCQs The contents of each chapter are covered in various sections At the end of all the sections, more extensive MCQs based on the Entire Chapter along with their solutions are also included It will be beneficial for the students to adopt the present book as the reference book along with their main text book The MCQs included in this book should be attempted along with the class-room teaching of the subject matter A regular and periodical review of the theory and MCQs from this book will help students in gaining enough confidence to appear in the JEE Main and enable them to face the challenge of successfully clearing this examination From the analyses of previous years’ question papers, a pattern of predominant topics emerges on which students should pay more attention while preparing for the examination These are: Physical Chemistry—Entire portion Inorganic Chemistry has been thoroughly revised and updated—Chemical families—perioidic properties, structures of compounds containing Si, N, P, S, halogens and inert gases, d-block elements and coordination chemistry Organic Chemistry—Stereoisomerism, SN1 and S22 Reactions, Reactions involving rearrangement, Chemistry of typical reactions shown by phenols, aldehydes and ketones and amines, relative acidity/basicity of phenolic, Carboxylic acids and amines, polymers, carbohydrates, stereochemistry involved in halogenation of alkenes and dehalogenation of halogenated compounds to give alkene, reactions involving Grignard reagent and diazonium salt K.L Kapoor Trend Analysis Physical Chemistry Chapter 10 Some Basic Concepts States of Matter Gaseous State Liquid State Solid State Atomic Structure Chemical Bonding Solutions Chemical Thermodynamics Chemical and Ionic Equilibria Chemical Equilibria Ionic Equilibria Redox Reactions and Electrochemistry Redox Reactions Electrolysis Conduction Galvanic Cells Chemical Kinetics Surface Chemistry Total 2014 (offline) I 2014 (online) II III 1 2015 2016 (offline) (offline) IV — 1 — 1 — 1 — — — 1 — 1 1 1 1 — — — 2 — — — — 1 1 — 1 — 13 — — — — 13 — — — — 14 1 — — — 12 — — — — 14 — — — — — 12 2017 (offline) 1 1 1 1 2 1 1 11 1 15 Inorganic Chemistry Chapter 11 12 13 14 Chemical Families–Periodic properties General Principles and Processes of Isolation of Metals Hydrogen s-Block Elements 2014 (offline) — I — 2014 (online) II III IV 1 1 — — — — — — — — — — 1 1 — — — 2015 2016 (offline) (offline) 2017 (offline) 1 (Contd ) viii Trend Analysis Chapter 15 16 17 18 19 Study of p-Block Elements (Groups 13, 14 and 15) Study of p-Block Elements (Groups 16, 17 and 18) d– and f– Block Elements Coordination Chemistry and Organometallics Nuclear Chemistry Total 2014 2014 (online) 2015 1 2 2016 (offline) 2017 — 1 — 2 — 1 2 — — — — — — 10 Organic Chemistry Chapter 2014 (offline) I 20 21 22 23 24 25 26 27 28 29 Purification and Characterization of Organic Compounds Some Basic Principles Hydrocarbons Alkanes Alkenes Alkynes Benzene Organic Compounds Containing Halogens Organic Compounds Containing Oxygen Alcohols Phenols Ethers Aldehydes & Ketones Carboxylic Acids Organic Compounds Containing Nitrogen Synthetic & Natural Polymers Biomolecules & Biological Processes Chemistry in Action Principles Related to Practical Chemistry Total 2014 (online) II III 2015 2016 2017 (offline) (offline) (offline) IV — — — — — 1 — — — — — — — — — 1 — 1 — — — — — — — 2 — — — 1 — — — — — — — — 1 — — — — — 11 11 2 1 — — 1 — — — — — 1 1 1 — 1 1 — 1 1 — — 2 1 11 — 10 — 10 — 9 11 About JEE Main 1. Introduction and Scheme of Examination The Joint Entrance Examination from the year 2013 for admission to the undergraduate programmes in Engineering is being held in two parts, JEE-Main and JEE-Advanced Only the top 1,50,000 candidates (including all categories) based on performance in JEE Main will qualify to appear in the JEE Advanced examination Admissions to IITs will be based only on category-wise All India Rank (AIR) in JEE Advanced, subject to condition that such candidates are in the top 20 percentile categories Admission to NITs will be based on 40% weightage for performance in Class XII board marks (normalized) and the remainder 60% weightage would be given to performance in JEE Main and a combined All India Rank (AIR) would be decided accordingly In case any State opts to admit students in the engineering Colleges affiliated to state Universities where States require separate merit list to be provided based on relative weightages adopted by the states, then the merit list shall be prepared with such relative weightages as may be indicated by States 2. Eligibility Criteria and List of Qualifying Examinations for JEE(Main) Exam The minimum academic qualification for appearing in JEE(Main) is that the candidate must have passed in final examination of 10+2 (Class XII) or its equivalent referred to as the qualifying examination (see below) However, admission criteria in the concerned institution/university will be followed as prescribed by concerned university/institution and as per the guidelines & criteria prescribed by AICTE Qualifying Examinations List of Qualifying Examinations (i) The +2 level examination in the 10+2 pattern of examination of any recognized Central/State Board of Secondary Examination, such as Central Board of Secondary Education, New Delhi, and Council for Indian School Certificate Examination, New Delhi (ii) Intermediate or two-year Pre-University Examination conducted by a recognized Board/University (iii) Final Examination of the two-year course of the Joint Services Wing of the National Defence Academy (iv) Any Public School/Board/University Examination in India or in foreign countries recognized by the Association of Indian Universities as equivalent to 10+2 system (v) H.S.C Vocational Examination (vi) A pass grade in the Senior Secondary School Examination conducted by the National Open School with a minimum of five subjects (vii) or 4-year diploma recognized by AICTE or a State Board of Technical Education x About JEE Main 3. Pattern of Examination Subject combination for each paper and type of questions in each paper are given below: Paper Paper Subjects Physics, Chemistry & Mathematics Mathematics – Part I Aptitude Test – Part II & Drawing Test – Part III Type of Questions Objective type questions with equal weightage to Physics, Chemistry & Mathematics Objective type questions Objective type questions questions to test Drawing Aptitude Duration Hours Hours Requirement of papers for different courses is given in the table below: Course B.E/B.TECH B.ARCH/B PLANNING Papers Paper – Paper – Scoring and Negative Marking There will be objective type questions with four options having single correct answer For each incorrect response, one fourth (1/4) of the total marks allotted to the question would be deducted No deduction from the total score will, however, be made if no response is indicated for an item in the answer sheet Atomic Structure 13 19 25 31 (a) (d) (b) (d) 14 20 26 32 (d) (c) (c) (b) 15 21 27 33 (b) (d) (c) (c) 16 22 28 34 (b) (b) (a) (c) 17 23 29 35 (a) (b) (a) (b) 3.37 18 (a) 24 (a) 30 (c) HINTS AND SOLUTIONS (2s)2 (2p)6 (3s)2 (3p)6 (3d)6 (4s)2 With the removal of 4s electrons, Fe2+ is formed Hence, 3d orbitals retain electrons All the three species in choice c contain 10 electrons h (6.63 ¥ 10-34 J s) = = 1.1 ¥ 10-33 m l = mu (60 ¥ 10-3 kg)(10 m s -1 ) For a 4f orbital, we have Principal quantum number, n = 4; Azimuthal quantum number, l = Magnetic quantum number, m = any one value from +3, +2, +1, –1, –2, –3 For an electron in this orbital, spin quantum number, s = either +1/2 or –1/2 (1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)5 (4s)1 24Cr is l = corresponds to p orbital The number of electrons in p orbitals is + = 12 l = corresponds to d orbital The number of electrons in d orbitals is Ê 1ˆ 1 We have DE = R• Á - ˜ = R• ÊÁ - ˆ˜ = R• ; Ë1 Ë n1 n2 ¯ • ¯ l= = 0.91 Ơ 10-7 m = 91nm Rã The isoelectronic species have the same number of electrons Each of the species K+, Ca2+, Sc3+ and Cl– has the same number (=18) of electrons n + l possess the same electronic energy 11 The number of electrons of each species in the choice a is 50 The number of electrons of each species in the choice b is 14 The number of electrons of each species in the choice c are: SO32–(42), CO32–(32) and NO3– (42) The number of electrons of each species in the choice d is 32 12 For a single electron species, the energy of electron depends only on the principal quantum number Hence, 3s, 3p and 3d orbitals will have the same energy 13 In Bohr’s theory, the angular momentum of electron is given by L = n(h/2p), where n is the quantum number For 5th orbit, n = 14 p = mv = (9.1 ¥ 10–31 kg) (300 m s–1) = 27.3 ¥ 10–29 kg m s–1 Dp = (0.001%)p = (10–5) (27.3 ¥ 10–29 kg m s–1) = 27.3 ¥ 10–34 kg m s–1 Dx = Ê 6.626 ¥ 10-34 J s ˆ Ê hˆ ÁË ˜¯ = ˜¯ = 0.0192 m Dp 4p ¥ 3.14 (27.3 ¥ 10-34 kg m s -1 ) ÁË 15 19K+, 17Cl–, 20Ca2+ and 21Sc3+ have the same number of 18 electrons 16 Larger the value of n + l, higher the energy of atomic orbital 17 The energy difference between the two orbits of hydrogen atom is given by Ê 1ˆ DE = N A R• hc Á - ˜ Ë n1 n2 ¯ For ionization, n1 = and n2 = ã Hence, For n =1 ặ n = 2, the energy difference is NARãhc = 1.312 Ơ 106 J mol–1 3.38 Complete Chemistry—JEE Main 3 ( N A Rã hc) = Ơ 1.312 Ơ 106 J mol-1 = 9.84 ¥ 105 J mol-1 4 18 According to the uncertainty principle Dp Dx ≥ (h/4p) i.e (m Du) Dx ≥ (h/4p) DE = Hence, Dx ≥ 19 We have (6.6 ¥ 10-34 kg m s -1 ) h = = 1.92 ¥ 10-3 m -31 -1 4p m( Du ) 4(22 / 7)(9.1 ¥ 10 kg){(600 m s )(0.005 / 100)} (6.63 ¥ 10-34 J s) h h = = = 4.0 ¥ 10-10 m = 0.40 nm p mu (1.67 ¥ 10-27 kg)(1.0 ¥ 103 m s -1 ) l= 20 (IE)He+ = Z12 Rã hc = 19.6 Ơ 10-18 J Rã hc = This gives 19.6 ¥ 10-18 J 19.6 ¥ 10-18 J = Z12 (9)(19.6 ¥ 10-18 J/ 4) = -4.41 ¥ 10-17 J n12 (12 ) (355 nm)(680 nm) 1 = + or l = = 743 nm 21 The conversion of energy required that (680 - 355) nm 355 nm 680 nm l ( E1 ) Li2 + = - Z 22 R• hc =- 22 The expression of emission of special transitions is For the given transition in He+, we have Ê 1ˆ DE = Z R• hc Á - ˜ Ë n2 n1 ¯ Ê 1ˆ DE = (2) R• hc Á - ˜ = R• hc Ë2 ¯ In hydrogen atom, this transition will correspond to the transition n2 = ă n1 = 2, since Ê1 1ˆ DE = (1)2 R• hc Á - ˜ = R• hc Ë1 ¯ 23 Energy of an electron in an atom increases with increase in the value of n + l For the same value of n + l, the energy is larger for the larger value of n Hence, we have Choice (i); n+l=4+1=5 Choice (iii); n+l=3+2=5 Choice (ii); n+l=4+0=4 Choice (iv); n+l=3+1=4 Hence, the correct order of increasing energy is (iv) < (ii) < (iii) < (i) 1 24 DE = E2 - E1 = (-2.178 ¥ 10-18 J) ÊÁ - ˆ˜ = 1.6335 ¥ 10-18 J Ë 1¯ l= hc (6.62 ¥ 10-34 J)(3 ¥ 108 m s -1 ) = = 1.214 ¥ 10-7 m -18 DE (1.6335 ¥ 10 J) l = 0, m = and ms = +1/2 26 The energy of electron in Bohr orbits is E=- Z2 R• n2 (where R• is Rydberg constant) E = –R• = –13.6 eV 9 R = - R• = - (13.6 eV) • n n n When n = 2, E = –30.6 eV(given choice) For Li2+ (Z = 3), we have 27 E0 = hc/l0 and E=- E = hc/l Ê 1 ˆ hc(l0 - l ) ; mu = E - E0 = hc Á - ˜ = Ë l l0 ¯ ll0 1/ È 2hc (l0 - l ) ˘ u=Í ll0 ˙˚ Ỵ m Thus it has n = 5, Atomic Structure 3.39 Ê -18 Ê 1ˆ 1ˆ -18 -18 28 DE = -(2.0 ¥ 10-18 J) Á - ˜ = -(2.0 ¥ 10 J) ÁË - ˜¯ = (2.0 ¥ 10 J)(3 / 4) = 1.5 ¥ 10 J Ë n2 n1 ¯ l= 29 l = hc (6.625 ¥ 10-34 J s)(3 ¥ 108 m s -1 ) = = 1.325 ¥ 10 –7 m DE (1.5 ¥ 10-18 J) h (6.626 ¥ 10-34 J s) = = 10-33 m mu (6.63 ¥ 10-3 kg)(100 m s -1 ) 30 E = hn = (6.63 ¥ 10-34 J s)(2.46 ¥ 1015 s -1 ) = 1.64 ¥ 10-18 J 31 Ionization energy per gaseous atom is Ê 495.5 ¥ 103 J mol-1 ˆ E (8.228 ¥ 10-19 J) -19 = ¥ = = E=Á 228 10 J; n = 1.24 ¥ 1015 s -1 h (6.626 ¥ 10-34 J s) Ë 6.022 ¥ 1023 mol-1 ˜¯ 32 The guiding principle is Energy increases with increase in the value of n + l For the same value of n + l, smaller the value of n, lesser the energy For the given orbitals, we have ns np (n – 1)d (n – 2)f lỈ n+l n n+1 n+1 n+1 ns Ỉ (n – 2)f Ỉ (n – 1)d Ỉ np 33 Lesser the mass of particle, larger its de Broglie wavelength (l = h/mu) Hence, the order of wavelength is Visible photon > Thermal electron > Thermal neutron 34 The energy of orbits in hydrogen atom is given by the expression 13.6 eV En = n2 For n = 2, E2 = –3.4 eV 35 The number of orbitals in the principal quantum number n is n2 Hence, answer is 25 Chemical Bonding and Molecular Structure SECTION Bond Formation and VSEPR Theory The formation of a chemical bond between two atoms implies that the system consisting of these two atoms at stable internuclear distance is energetically more stable than the two isolated atoms A general study on the reactivity of different elements revealed that noble gases have little tendency to combine with other elements This leads to the ns)2 np)6 Kössel and Lewis Theory of Chemical Combination According to this theory, atoms can combine either by transfer of outer-shell electrons, known as valence electrons, eight electrons) in their respective valence shells formation of ionic bond between the two involved atoms Representation of a Bond by Lewis Structure In Lewis structure, a bond between the two atoms is shown by Lewis electron-dot symbols in which valence electrons are shown by dots around the letter symbol of the atom The dots are placed as follows Place a single dot on the four sides of the letter symbol followed by the second dot till all the valence-electrons have been accounted for Illustrations: Lithium Nitrogen Li N Beryllium Oxygen Be O Boron Fluorine Carbon B F Neon C Ne Formation of Covalent Bond(s) A covalent bond involves mutual sharing of valence electrons between two atoms The sharing of two, four and six electrons leads to the formation of a single, double and triple bond, respectively A covalent bond is formed if the atoms have lesser number of valence electrons as compared to the nearby noble formation of covalent bond is 4.2 Complete Chemistry—JEE Main Electronegative element + Electronegative element ỉỈ Covalent bond Exception to the octect rule is the hydrogen atom which can accommodate only two electrons which corresponds to 2) atom Illustrations Formation of Single Bond(s) Cl2 Cl + Cl Cl Cl or Cl Cl2 2H + O H O H or H O H NH3 3H + N H N H or H N H Cl H H 4H CH4 C + H H H C H or H C H H H Formation of Double Bond(s) O2 O + O CO2 O + C 2H O + C O O O H + C H C H C O C H H or O O or O C O or H C C H H H Formation of Triple Bond(s) N2 C 2H N N + N N H C H + C Exceptions to the Octect Rule or C H N or N H C C H The octect rule is generally obeyed by the elements of second and third periods with the following exceptions The Incomplete Octect BeCl2 Cl Be + + Cl Cl Be Cl electrons AlCl3 Cl + Al Cl Cl Al Cl electrons Odd-Electron Molecules All atoms of a compound containing odd number of electrons will not satisfy octect rule as even number of electrons are required for pairing of electrons Chemical Bonding and Molecular Structure NO N O + N N O O + 4.3 electrons O NO2 + N O + N + O O N O – + O N O electrons The Expanded Octect Elements of third period and beyond can accommodate more than electrons due to the availability of vacant d orbitals Cl PCl5 Cl F Cl P + F + + Cl P Cl Cl F F F F + F Coordinate Covalent (or Dative) Bond Cl P F If a pair of electrons shared between two atoms comes exclusively from ) Once a coordinate bond is formed, it behaves like a covalent bond H H3N—BF3 H N F + H B F F H H F N B H F F Writing a Lewis Structure The structure of a molecule or ion may be written by following the steps listed below Calculate the total numbers of valence electrons of the atoms in the molecule For an anion, add the number of negative charges and for a cation, subtract the number of positive charges Write the skeleton structure of the molecule or ion connecting every bonded pair of atoms by a single bond, i.e a pair of electron dots If there are fewer than eight electrons on the central atom, move one or two pairs of electrons from a surrounding Illustration Lewis structure of COCl2 Step Valence electrons are + + ¥ = 24 4.4 Complete Chemistry—JEE Main Carbon being the most electropositive atoms occupies the central position to which other atoms are bounded O Cl C Cl Step Assign electrons each to surrounding atoms O Cl C Cl Step Step There were 24 valence electrons and all of them have been distributed However, the central C atom has only electrons In order that this atom also has electrons, move one pair of electrons from O to the bond connecting C atom, thus forming a double bond O O O Cl C Cl Cl C Cl or Cl C Cl Formal Charge and Lewis Structure The formal charge on an atom is the difference between the valence electrons in an isolated atom and the number of electrons assigned to that atoms in a Lewis structure The equation for computing formal charge is bonding) electrons in a Lewis structure The sum of the formal charges of atoms in a Lewis structure is equal to the charge on the molecular species Illustration COCl2 molecule Lewis structure Cl C Cl O Atom Valence electrons in a free atom Electrons in Lewis structure Nonbonding bonding Cl + ¥ 2) = O 4 + ¥ 4) = C Formal Charge + ¥ 8) = Utility of Formal Charge Computing formal charge of atoms in a molecule or ion helps deciding a plausible Lewis structure of the species The guiding principles are as follows Amongst the several Lewis structures, the species having the lowest magnitude of formal charge is the preferred structure Amongst Lewis structures having similar distribution of formal charges, the one having negative formal charges on the more electronegative atoms is the preferred structure Polar Covalent Bond Each atom in a molecule has its own ability to attract the bonded pair of electrons This ability is known as electronegativity 2, O2, F2, Cl2, etc.) is shared equally by both of electrons is closer to the atom having larger electronegativity Consequently, this atom acquires a partial negative charge while the other atom acquires equal partial positive charge Because of the charge separation, the covalent bond between these two atoms is said to be a polar covalent bond The polarization of bonded pair of electrons between two atoms is expressed in terms of physical m m dq r where d q is the partial charge separation between two atoms and r is the distance between the two atoms Dipole Moment Chemical Bonding and Molecular Structure 4.5 Representation of Dipole Moment Dipole moment is a vector quantity, i.e it has magnitude as well as direction In chemistry, dipole moment is indicated by the crossed arrow as shown in the following.† positive end negative end that is, it is directed from positive end to the negative end dq Unit of Dipole Moment –18 –18 –18 r) = C m esu cm This value of dipole moment is known as debye Ï Ê 1.6 ¥ 10-19 C ˆ ¸ ˝ ) Ì(1esu ) Á Ë 4.8 ¥ 10-10 esu ˜¯ ˛ Ó –2 m) = 3.33 ¥ Cm Dipole Moment of a Polyatomic Molecule Each bond in a molecule has a dipole moment, known as bond moment The dipole moment of a molecule is obtained by the vector addition of these bond moments Illustration The bond moment of O H2O molecule will be H bond is 1.52 D The bond angle of H2 H mH2O = 2mOH O Nonpolar Polyatomic Molecule H The dipole moment of a nonpolar polyatomic molecule is zero inspite of the fact that the bond moments of the molecule is not zero This is due to the fact that the individual bond moments in the molecule is symmetrically placed so that their vector additions is zero Illustration F Cl F Be F B F O C Cl F C O Cl Cl Comparison of Dipole Moments of NH3 and NF3 Both NH3 and NF3 have pyramdial shapes with lone pair of electrons on nitrogen atom N N H F F F H In NH3, orbital dipole acts in the same direction as the sum of bond vectors of the three N H bond bonds F bond bonds In NF3, orbital dipole acts in the opposite direction to the sum of bond vectors of the three N These facts make the dipole moment of NH3 m = 1.57 D) larger than that of NF3 m H Per Cent Ionic Character of a Polar Band Per cent ionic character = The per cent ionic character of a polar band A m AB ¥ 100 mionic where mionic = e rAB † In actual practise, the dipole moment is represented as a simple arrow pointing from negative end to the positive end, that is negative end positive end In chemistry, the crossed arrow is used which indicates the direction of the shift in electronegativity in the molecule 4.6 Complete Chemistry—JEE Main Illustration will be The bond moment of O H bond is 1.52 D If bond length O Per cent ionic character = (1.52 D) (3.33 ¥ 10-30 C m / D) mOH ¥ 100 ¥ 100 = (1.6 ¥ 10-19 C) (95 ¥ 10-12 m) e rOH Concept of Resonance distribution of electrons over a given skeleton of atoms in a molecule None of the individual structures adequately explains the characteristics of the molecule However, these can be explained if the actual structure of the molecule is considered as the superposition of individual structures This phenomenon is known as resonance and the individual ) inserted between the resonating structures Illustrations O O O O O O written as O bond lengths equal in size 2– O C O O O O O written as C O C 2– O O 2– C O 2– O O Formation of an Ionic Bond leads to the formation of positive and negative ions, respectively The electrostatic attraction between the positive and negative ions results in the formation of an ionic bond between the involved ions Illustrations + Na F Na+ + Cl + Mg 5 + – Cl Mg2+ + F F – Energies Involved in the Formation of One Molecule of Sodium Chloride The formation of Na+Cl– – Ei = 8.24 ¥ J ặ Na+ ặ Cl Eea = 5.78 Ơ J + – Ỉ Na+Cl– PE where PE is the potential energy in the formation of ionic bond This is evaluated by the expression QQ PE = (4pe )r ¥ where Q1 = –Q2 we have PE = C and r = rNa+ + rCl– (1.60 ¥ 10-19 C)(-1.60 ¥ 10-19 C) = - 8.34 ¥ 10-19 J (4)(3.14)(8.854 ¥ 10-12 C2 N -1 m -2 )(276 ¥ 10-12 m) Q1 and Q2 as point charges, Chemical Bonding and Molecular Structure Ỉ Na+Cl– DE = Ei + DEea we have ¥ 4.7 DE =? J = –5.88 ¥ Comment The negative value of DE indicates that the formation of an isolated ionic bond Na+Cl– Essential Requirement for the formation of an Ionic Bond Ei Ei Hence Electropositive element + Electronegative element For DE Ionic Bond Formation of mol of Solid Ionic Compound from Constituent Elements Taking an example of sodium chloride, we have the following steps in the formation of solid compound –1 Ỉ DH1 –1 Ỉ DH2 Cl2 – –1 Ỉ Na+ DH3 – Ỉ Cl– –1 DH4 + – + – Ỉ Na Cl DH5 = –788 kJ mol–1 Cl2 Æ Na+ DH = –313 kJ mol–1 DH is negative, the formation of solid NaCl is energetically favourable From the values of DH’s listed above, it Ỉ Na+ – the ionic solid compound into gaseous constituent ions Larger the value of lattice energy, more stable the ionic compound listed above for the formation of solid ionic compound In this cycle, DH DH1, DH2, DH3 and DH4 from the value of DH gives the value of DH5 The lattice energy is negative of the value of DH5 Fajan Rules An ionic compound has partial covalent character and vice versa The partial covalency in an ionic compound my be explained qualitatively with the help of Fajan rules described in the following High Charge and Small Size of the Cation cationic electronic charge to penetrate partially into the anionic electronic cloud resulting into the partial covalent bond character to the ionic bond High Charge and Large Size of the Anion The electronic cloud of such an anion is most easily polarized by For two cations of the same size and charge, the cations of electronic n – 1)dx ns n – 1)s2 n – 1)p6 ns nucleus by the electronic cloud of transition metal ions as compared to that in the alkali and alkaline metal ions Hg2+ ion has larger polarizing effect than Ca2+ ion Lithium salts have more covalent character than the alkali salts I– ion is more easily polarized than Cl– ion by Ag+ ion Illustrations The Valence Shell Electron Pair Repulsion (VSEPR) Theory 4.8 Complete Chemistry—JEE Main The number of electron pairs in the valence shell of the central atom of a molecule decides the shape of the A multiple bond is treated as if it is a single electron pair The repulsive interaction of electron pairs decrease in the order Table Molecule Shapes of some molecules on the basis of VSEPR model Basic shape † Number of valence electrons around the central atom electron pairs bounding pairs BeCl2 2 Linear BCl3 3 Triangular planar CH4 4 NH3 H 2O 2 PF5 5 5 12 6 12 CIF3 IF5 lone pairs Tetrahedron Trigonal bypyramid Octahedron A few examples of molecules containing lone pair electrons along with their geometry are described in the following N O O H H O H H H † The actual shapes of molecules containing lone pairs is a little distorted from the basic shape This is due to the fact that lone pair-bonded pair repulsion is larger than bonded pair-bounded repulsion Chemical Bonding and Molecular Structure F F F Cl F F F F Note: Lone pair equatorial position as it involves only two lp-bp repulsion It is not occupied in axial position as it will involve three lp-bp repulsions Note: Lone pair occupy equatorial positions so as to have lesser lp-bp repulsions Note: F F F Xe F F F F O F F I F F F F Xe F F I F I Cl I F F F F O Xe Xe F F O F F MULTIPLE CHOICE QUESTIONS ON SECTION Identify the correct choice in the following questions Which of the following statements regarding Hg2+ 2+ ion has more polarizing effect on an anion as compared to Ca2+ ion 2+ ion has less polarizing effect on an anion as compared to Ca2+ ion 2+ ion and Ca2+ have equal polarizing effect on an anion 2+ ion and Ca2+have no polarizing effect on an anion The increasing order of covalency in silver halides is 2+ The increasing order of melting point of carbonates of alkaline-earth metals is 3 3 3 3 3 3 Which of the following resonating structures of N2O is feasible in its linear structure? N N O N N O N N O + – 2– + + Which of the following statements regarding BrF3 molecule is correct? Br 4.9 N N O – + 4.10 Complete Chemistry—JEE Main Which of the following pairs of species are not isostructural? + + – + , IF 4, PF 4 and NH 2, I Which of the following order regarding bond angle O N O in the given species is correct? + – – + + – 2 NO 2 2 2 Which of the following facts regarding Fajan’s rules is not correct? – + – is not correct? I is correct? F bond has the same length Cl 11 The structure of ICl4 is Cl Cl Cl Cl Cl I I Cl Cl Cl 12 Which of the following statements is correct? 2 has nonlinear structures 2 has linear structures and CO2 has linear structures and CO2 has non-linear structures 13 The shape of XeF4 is Cl Cl Cl Cl I I Cl Cl Cl 14 Which one of the following compounds of xenon has trigonal bipryramid structure? 2F 15 The number of pairs of electrons around I in IF3, IF5 and IF7 ,respectively, are Cl 3F 16 Which of the following species has pyramidal shape? 2– 3 – 3 17 Which of the following species has linear structure? – + 2 18 Which of the following statements regarding bond angles in the given molecules is correct? 3 and OF2 3 and OF2 3 and OF2 3 and OF2 3 3 3 3 3 3 Chemical Bonding and Molecular Structure s – bond and p 4.11 s – bond, p – bond and coordinate bond ANSWERS HINTS AND SOLUTIONS Hg2+ ion belongs to transition metal and Ca2+ ion to alkaline-earth metal Thus Hg2+ ion has more polarizing effect on an anion As the size of halide is increased, it is more polarizable by Ag+ and hence larger is the covalency of silver halide 2+) of Group of periodic table, the polarizing power on CO2– is decreased The MCO3 becomes more ionic, hence their melting point increases F There are 28 valence electrons These are distributed as F Br F F Hence, its structure is pentagonal bipyramid with two lone pairs occupying equatorial positions Br Due to lone pair-bonding pair, the angle F XeF4 has pairs of electrons around Xe while PF+4 has pairs of electrons around P + O N – O O NO2+ N O NO2 NO–2 The molecule NO2 bonding electrons move away so as to reduce the repulsion between them F is Cl F F O N O NO–2 Br F F ... JEE MAIN COMPLETE CHEMISTRY K. L Kapoor Formerly Associate Professor, Hindu College, University of Delhi, Delhi McGraw Hill Education (India) Private Limited CHENNAI McGraw Hill Education. .. 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... York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto Published by McGraw Hill Education