Atul singhal the pearson guide to inorganic chemistry for the JEE advanced pearson education (2014)

Chapter ContentsOrbital overlap and covalent bond; Hybridization involving s, p and d orbitals only; Orbital energy diagrams for homonuclear diatomic species; Hydrogen bond; Polarity in

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This book is designed to help aspiring engineers understand the various

important aspects of ‘inorganic chemistry’ Each book in this series approaches

the subject in a very conceptual and coherent manner The illustrative approach

adopted in this series will help students to familiarize themselves with complex

concepts and their applications in a simple manner This book also includes a

wide variety of questions

SALIENT FEATURES

● 1750+ Multiple-Choice Questions for practice

● Hints and solutions provided for most of the questions

● Follows the latest pattern of JEE Advanced

● Provides extensive pedagogy to demystify the examination pattern

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The Pearson Guide to Inorganic Chemistry

for the

JEE Advanced

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Licensees of Pearson Education in South Asia

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 9789332520899

eISBN 9789332537095

Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India

Registered Office: 11 Community Centre, Panchsheel Park, New Delhi 110 017, India

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Dedicated to

My Grand Parents, Parents and Teachers

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This page is intentionally left blank.

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Hybridization 1.16

• Straight Objective Type Questions • Brainteasers Objective Type Questions • Multiple

Correct Answer Type Questions • Linked-Comprehensions Type Questions • Assertion and

Reasoning Questions • Matrix–Match Type Questions • The IIT–JEE Corner • Solved

Subjective Questions • Questions for Self-assessment • Integer Type Questions

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3 Preparation and Properties of Non-metals

4 Compounds of Lighter Metals–1

5 Compounds of p-block Elements–1

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Oxides of Phosphorous 5.24

6 Compounds of p-block Elements–2

7 Transition Elements and Co-ordination Chemistry

8 Metallurgy

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9 Compounds of Heavy Metals

Oxides 9.7

Halides 9.9

Sulphates 9.12

10 Principles of Qualitative Analysis

Characteristic Test of Anions (Acidic Radicals) 10.2

Subjective Questions • Questionsfor Self-assessment • Integer Type Questions

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Preface

The Pearson Guide to Inorganic Chemistry for the JEE Advanced is an invaluable book for all the students

preparing for the prestigious engineering entrance examination It provides class-tested course

mate-rial and problems that will supplement any kind of coaching or resource the students might be using

Because of its comprehensive and in-depth approach, it will be especially helpful for those students

who do not have enough time or money to take classroom courses

A careful scrutiny of previous years’ IIT papers and various other competitive examinations

dur-ing the last 10 to 12 years was made before writdur-ing this book It is strictly based on the latest IIT

syllabus (2014–15) recommended by the executive board It covers the subject in a structured way

and familiarizes students with the trends in these examinations Not many books in the market

can stand up to this material when it comes to the strict alignment with the prescribed syllabus

It is written in a lucid manner to assist students to understand the concepts without the help of

any guide

The objective of this book is to provide this vast subject in a structured and useful manner so as to

famil-iarize the candidates taking the current examinations with the current trends and types of

multiple-choice questions asked

The multiple-choice questions have been arranged in following categories: Straight Objective

Type Questions (single choice), Brainteasers Objective Type Questions (single choice), Multiple

Correct Answer Type Questions (more than one choice), Linked-Comprehension Type Questions,

Assertion and Reasoning Questions, Matrix-Match Type Questions, the IIT JEE Corner and

Integer Type

This book is written to pass on to another generation, my fascination with descriptive inorganic

chemistry Thus, the comments of the readers, both students and instructors, will be sincerely

appre-ciated Any suggestions for added or updated additional readings would also be welcome, students

can reach me directly at singhal.atul1974@gmail.com

Atul Singhal

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The contentment and ecstasy that accompany the successful completion of any work would remain

essentially incomplete if I fail to mention the people whose constant guidance and support has

encouraged me

I am grateful to all my reverend teachers, especially, the late J K Mishra, Dr D K Rastogi, the

late A K Rastogi and my honourable guide, Dr S K Agarwal Their knowledge and wisdom has

continued to assist me to present in this work

I am thankful to my colleagues and friends, Deepak Bhatia, Er Vikas Kaushik, Er A R Khan,

Vipul Agarwal, Er Ankit Arora, Er Wasim, Akhilesh Pathak, Akhil Mishra, Alok Gupta, Mr Anupam

Shrivastav, Mr Rajiv Jain, Mr Ashok Kumar, Mr Sandeep Singhal, Mr Chandan Kumar, Mr P S Rana,

Amitabh Bhatacharya, Ashutosh Tripathi, N C Joshi and Rajneesh Shukla

I am indebted to my father, B K Singhal, mother Usha Singhal, brothers, Amit Singhal and

Katar Singh, and sister, Ambika, who have been my motivation at every step Their never-ending

affection has provided me with moral support and encouragement while writing this book

Last but not the least, I wish to express my deepest gratitude to my wife Urmila and my little,

but witty beyond years, daughters Khushi and Shanvi who always supported me during my work

Atul Singhal

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Chapter ContentsOrbital overlap and covalent bond; Hybridization involving s, p and d orbitals

only; Orbital energy diagrams for homonuclear diatomic species; Hydrogen bond;

Polarity in molecules; Dipole moment (qualitative aspects only); VSEPR model and

shapes of molecules (linear, angular, triangular, square planar, pyramidal, square

pyramidal, trigonal bipyramidal, tetrahedral and octahedral) and various levels

of multiple-choice questions.

CHEMICAL BOND

Chemical bond is the force of attraction that

binds two atoms together A chemical bond

bal-ances the force of attraction and force of

repul-sion at a particular distance

A chemical bond is formed to:

attain the octet state

minimize energy

gain stability

decrease reactivity

When two atoms come close to each other,

forces of attraction and repulsion operate

between them The distance at which the

attrac-tive forces overcome repulsive forces is called

bond distance The potential energy for the

sys-tem is lowest, hence the bond is formed

Following are the six types of chemical bonds

They are listed in the decreasing order of their respective bond strengths

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Metallic bond, hydrogen bond and van der

Waals bond are interactions

Octet Rule

It was introduced by Lewis and Kossel

According to this rule, each atom tries to

obtain the octet state, that is, a state with

eight valence electrons

Exceptions to the octet rule

Transition metal ions like Cr3+, Mn2+

and Fe2+

Pseudo inert gas configuration cations

like Zn2+ and Cd2+

Contraction of octet state

The central atom is electron deficient or

does not have an octet state For example,

BeX2

4

BX36

AIX36

Ge (CH3)36e−

Expansion of octet state

The central atom has more than 8

electrons due to empty d-orbitals For

example, PCl5, SF6, OsF8, ICl3, etc

IONIC OR KERNEL BOND

An ionic bond is formed by the complete fer of valence electrons from a metal to a non-metal This was first studied by Kossel

Maximum number of electrons transferred

by a metal to non-metal is three, as in the case with AlF3 (Al metal transfers three elec-trons to F)

During electron transfer, the outermost orbit

of metal is destroyed The remaining portion

is called core or kernel, hence this bond is also called kernel bond

Nature of ionic bond is electrostatic or coloumbic force of attraction

Metal must have low ionization energy

Non-metals must have high electron affinity

Ions must have high lattice energy

Cation should be large with low tivity

Anion must be small with high tivity

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electronega-Born–Haber Cycle

The formation of an ionic compound in terms

of energy can be shown by Born–Haber cycle

It is also used to find lattice energy, ionization

energy and electron affinity

For the formation of an ionic solid, energy must

be released during its formation, that is, ΔH

must be negative

−E − U > S + ½ D + I

Properties of Ionic Compounds

1 Ionic compounds have solid crystalline

structures (flat surfaces), with definite

geometry, due to strong electrostatic force

of attraction as constituents are arranged in

a definite pattern

2 These compounds are hard in nature

Hardness ∝ Electrostatic force of

attraction

∝ Charge on ion

∝ 1 Ionic radius

3 Ionic compounds have high value of ing point, melting point and density due

boil-to strong electrostatic force of attraction

Boiling point, melting point ∝ Electrostatic force of attraction

Volatile nature

∝ 1

Electrostatic force of attraction

4 Ionic compounds show isomorphism, that is, they have same crystalline structure For exam-ple, all alums, NaF and MgO, ZnSO

4⋅7H2O and FeSO

Lattice Energy

Lattice energy is the energy needed to break an ionic solid molecule into its con-stitutent ions It is denoted by U

U ∝ Charge on ion ∝ 1 Size of ion

Hence, lattice energy for the following compounds increases in the order shown under:

NaCl < MgCl2 < AlCl3 < SiCl4

As charge on a metal atom increases, its size decreases

In case of univalent and bivalent ionic pounds, lattice energy decreases as follows:

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Bi-bi > Uni-bi or Bi-uni > Uni-uni

For example,

MgO > MgCl2 > NaCl

Some orders of lattice energy

(i) LiX > NaX > KX > RbX > CsX

(ii) LiF > LiCl > LiI

(iii) AgF > AgCl > AgI

(iv) BeO > MgO > CaO > SrO

9 Ionic compounds are soluble in polar

sol-vents like water due to high dielectric

con-stant of these solvents The force of attraction

between ions is destroyed and hence they

dissolve in the solvent

Facts Related to Solubility

If ΔH (hydration) > Lattice energy then

the ionic compound

is soluble

If ΔH (hydration) < Lattice energy then

the ionic compound

is insoluble

If ΔH (hydration) = Lattice energy then

the compound is at equilibrium state

Some Solubility Orders

a LiX < NaX < KX < RbX < CsX

b LiOH < NaOH < KOH < RbOH <

CsOH

c BeX2 < MgX2 < CaX2 < BaX2

d Be(OH)2 < Mg(OH)2 < Ca(OH)2 <

Ba(OH)2

e BeSO4 > MgSO4 > CaSO4 > SrSO4 >

BaSO4

f AIF3 > AlCl3 > AlBr3 > AlI3

Crystals of high ionic charges are less

soluble For example, compounds of

CO3–2, SO4–2, PO4–3 are less soluble

Compounds Ba+2, Pb+2 are insoluble as lattice energy > ΔHhy

Compounds of Ag (salts) are insoluble as lattice energy > ΔHhy

Presence of common ions decrease bility For example, solubility of AgCl decreases in presence of AgNO3 or KCl, due to the presence of common ions, that is, Ag+ and Cl− respectively

solu-Note: The concept of ionic bond is not a

part of IIT-JEE syllabus but has been cussed for the better understanding of the chapter

dis-COVALENT BOND

A covalent bond is formed by the equal ing of electrons between two similar or different atoms

If atoms are same or their electronegativity

is same, the covalent bond between them is non-polar For example, X – X, O = O, N ≡ N

If atoms are different or have different value

of electronegativity, the covalent bond formed between them is polar For example, +δ −δ +δ, +δ −δ

H − O − H, H − X The number of electrons shared or covalent bonds represent covalency

One atom can share maximum three electrons with another atom For example, in ammonia the covalency of nitrogen atom is three

Properties of Covalent Compounds

1 Covalent compounds mostly occur in uid and gaseous state but if molecular weight of the compound is high they may occur in solid state too

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For example, F2, Cl2 Br2 I2

‘g’ ‘g’ ‘l’ ‘s’

Glucose SugarMolecular wt 180 342

(less solid) (more solid)

2 Solubility of these compounds follows the

concepts ‘like dissolves like’, that is,

non-polar solute dissolves in non-non-polar solvent

For example, CCl4 dissolves in organic

solvents Similarly, polar solutes dissolves

in polar solvent For example, alcohol and

ammonia dissolve in water

3 Covalent compounds have lower boiling

point and melting point values than those

of ionic compounds This is because

cova-lent bond is a weak van der Waals force in

nature

For example, KOH > HX

ionic force van der

of attarction Waals forces

Boiling point and melting point

H-bonding weight decreases

4 Covalent compounds are non-conductors

due to absence of free ions, but graphite is

a conductor, as the free electrons are

avail-able in its hexagonal sheet like structure

In case of diamond, the structure is

tetra-hedral hence free electrons are not

avail-able Therefore, it is not a conductor

5 Covalent bond is directional, hence these

compounds can show structural and space

isomerisms

6 The reactions involving covalent bond are

slow as these need higher activation energy

COORDINATE OR DATIVE SEMI-POLAR BOND

Coordinate bond is a special type of bond which

is formed by donation of electron pair from the donor to the receiver, that is, it involves par-tial transfer or unequal sharing of electrons It is denoted as ( ) from donor to receiver

Donor or Receiver Lewis base Lewis acidCoordinate bond is intermediate between ionic and covalent bonds, but more closely resembles

a covalent bond The properties of coordinate compounds are more close to covalent com-pounds For example,

HNHO

HBF

FF

HOH

H, NH4+, K4[Fe(CN)6], N2O+

ClCl

Fig 1.2

MODERN CONCEPT OF COVALENT BOND

The nature of covalent bond is explained on the basis of Heitler–London’s Valence bond theory,

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Pauling and Slater’s overlapping theory and

Hund, Mullikan’s theory

Valence Bond Theory or

Heitler–London Theory

Orbital concept of covalent bond was

intro-duced by Heitler and London According to this

concept, “A covalent bond is formed due to the

half-filled atomic orbitals having electrons with

opposite spin to each other”

Features

1 The atoms should have unpaired electrons

to form a covalent bond

2 A covalent bond is formed by the pairing

of electrons

3 The maximum electron density lies

between the bonded atoms

4 There is a tendency to form close shells of

the atom, though the octet is not attained

in BeCl2, BF3, etc., and is exceeded in PCl5,

SF6, IF7, etc

Due to overlapping the potential energy of

system decreases

The internuclear distance with maximum

overlapping and greater decrease of potential

energy is known as bond length

Limitations

1 It cannot explain the formation of odd

electron molecule such as ClO3, NO, H2+,

etc In such molecules, electron pairing

does not take place

2 It cannot explain the formation of

coordinate bond where only one atom

donates a lone pair of electrons to the other

atom, molecule or ion

3 It cannot explain the formation of π-bond

4 It cannot explain the stereochemistry of

the molecules and ions

Pauling and Slater’s Theory

It deals with the directional nature of the bond formed and is simply an extension of Heitler–

London theory

1 Greater the overlapping, stronger will be the bond formed It means bond strength depends upon the overlapping and is directly proportional to the extent of overlapping

2 A spherically symmetrical orbit, say, s-orbital will not show any preference in direction whereas non-spherical orbitals, say, p- or d-orbitals will tend to form a bond in the direction of maximum electron density with the orbital

REMEMBER

The overlapping of the orbitals of only those electrons which take part in the bond formation will occur and not with the electrons of other atoms

The wave function of an electron of s-orbital is spherically symmetrical, therefore such an electron exhibits no directional preference in bond forma-tion The two orbitals having similar energy level, the one which is more directionally concentrated, will form a stronger bond

Strongness of overlapping ∝ mode of overlapping

For example, linear overlapping is stronger than lateral overlapping

Strongness of overlapping ∝ directional nature of orbital

For example,p-p > s-p > s-s > p-pLinear or axial Lateral overlapping overlapping

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Types of Overlapping

1 s-s Overlapping: Overlapping between

s-s electrons of two similar or dissimilar

atoms is called s-s overlapping and forms a

single covalent bond

Fig 1.3 Formation of hydrogen molecule by

s-s overlapping.

2 s-p Overlapping: Overlapping between

s and p electrons is called s-p

Strong bond can be formed only when

hydrogen electrons approach in the

direc-tion of X, Y and Z axis at right angles to

each other

s — p

Fig 1.4

3 p-p Overlapping: p-p overlapping is

formed by the overlapping of the p-orbitals

of the atoms In case of chlorine molecule,

it is formed by the overlapping of the 3pz orbitals of two chlorine atoms

It is determined by X-ray diffraction and spectroscopic methods

In case of ionic compounds, it is the sum of ionic radius of cation and anion, while in case

of covalent compounds, it is sum of their covalent radius

Factors affecting bond length

Bond length ∝ Size of atom For example,

HF < HCI < HBr < HI (Atomic size)

Since F < CI < Br < I Bond length ∝ _ 1

Bond order or multiplicity For example, C – C > C = C > C ≡ C

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Resonance and hyperconjugation also

change bond length

For example, in benzene, C – C bond length is

1.39 Å, that is, in between C – C and C=C

Bond Energy

It is the energy needed to break one mole of

bond of a particular type, so as to separate

them into gaseous atoms It is also called

bond dissociation energy

Bond energy can also be defined as the energy

released during the formation of one mole of

a particular bond

Factors affecting bond energy

Bond energy ∝ Bond order or multiplicity

It is the angle between the lines representing

the directions of the bonds or the orbitals

hav-ing bondhav-ing pair of electrons

Factors affecting bond angle

Bond angle ∝ Bond order ∝ s%

∝ 1 Bond length For example,

Size of terminal atom For example, I2O > Br2O > Cl2O > OF2 Bond angle

∝ _ 1 Size of central atom/electronegativity Normally, bond angle decreases when we move down the group, as electronegativity decreases

For example, NH3 > PH3 > AsH3 > BiH3

H2O > H2S > H2Se > H2Te

BF3 > PF3 > ClF3 Bond angle ∝ Electronegativity of terminal atom

For example, PF3 > PCl3 < PBr3 < PI3

PF3 has more bond angle than PCl3 due to pπ-dπ bonding

REMEMBER

PF3 has greater bond angle when compared

to PH3 due to resonance in PF3, where a ble bond character develops NH3 has more bond angle value than NF3 as F-atom pulls the bpe– away from N-atom in NF3

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dou-POLORIZATION AND FAJAN’S

RULE

When cation and anion are close to each other,

the shape of anion is distorted by the cation This

is known as polarization Due to this, covalent

nature develops in an ionic molecule

Polarization ∝ Covalent nature ∝ 1

Polarization or covalent nature is explained

by the following rules:

Charge on Cation Polarization, covalent

nature or polarizing power of a cation ∝

charge on cation That is greater the charge

on cation, greater will be its polarizing

power and more will be covalent nature

For example,

SiCl4 > AlCl3 > MgCl2 > NaCl

Size of Cation When the charge is same and

the anion is common, consider that the

cova-lent nature ∝ _ 1

Size of cation That is, smaller cation has more polarizing

power

For example,

LiCl > NaCl > KCl > RbCl > CsCl

Max covalent Max ionic

Least ionic Least covalent

Li+ < Na+ < K+ < Rb+ < Cs+

Smallest Largest

Size of Anion This property is taken into

account when the charges are same and

the cation is common

Polarization or covalent nature ∝ size

of anion Hence, larger anions are more polarized

For example, LiF < LiCl < LiBr > LiI Since, F− < Cl− < Br− < I−

Larger the size of anion, easier will be its polarization

A cation with 18 valence electrons has more polarizing power than a cation with 8 valence electrons

For example, Group IB > Group IA

Cu+ Na+

Αg+ K+ Group IIB > Group IIA

Zn+2 Mg+2 For example, ZnO > MgO

inten-Dipole Moment

–q

r+q

Fig 1.7

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Dipole moment is used to measure the polarity

in a molecule It is denoted by μ

Mathemati-cally, it is given as

μ = q × r coulomb metre

μ = e × d esu cm

1 debye = 1 × 10−18 esu cm

It is represented by ( ) from

electro-positive to electronegative species or less

electronegative to more electronegative

spe-cies For example, AX3

AX

μ = 0Non-polar

ence For example, HF > HCI > HBr > HI

Dipole moment ∝ Number of lone pair of

electrons

For example, HF > H2O > NH3

Fluorine has 3 lone pair, oxygen has 2 lone

pair, and ammonia has 1 lone pair of electron

F

μnet= 0 F

Xe

F F

μ net = 0 C

OHH

μ(net) = 1.82D

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Ammonia has more dipole moment than NF3

as in ammonia μ (net) is in the direction of lone pair electrons i.e., it is additive while in NF3 μ (net) is opposite to lone pair i.e., substractive

Dipole moment of a cis-alkene is more than trans-alkene In trans-alkenes, it is zero due

to symmetry in most of the cases

Dipole Moment of Some Common Molecules

Exception: Unsymmetric alkenes with odd

number of carbon atoms have some value of

dipole moment

For example, trans-2-pentene

C=CH

CH2CH3H

Specific cases of dipole moment

→ CH3Cl > CH2Cl2 > CHCl3 > CCl4

Highly polar Non-polar

→ CH3Cl > CH3F > CH3Br > CH3I

Uses

To find geometry of a complex/molecule etc

To find ionic character or nature in a covalent species

Ionic nature % = μobserved

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2.87 10

1004.26 10

= 67.4%

2 The dipole moment of KCl is 3.336 × 10−29

coulomb meter which indicates that it is

a highly polar molecule The interatomic

distance between K+ and Cl− in this

mol-ecule is 2.6 × 10−10 m Calculate the dipole

moment of KCl molecule if there were

opposite charges of one fundamental unit

located at each nucleus Calculate the

per-centage ionic character of KCl

Sigma bond is formed by axial or headtohead or

linear overlapping between two s – s or s – p or

2 The minimum and maximum number of

σ bonds between two bonded atoms is 1

3 Stability ∝ Number of sigma bonds

Fig 1.9

1 It is a weak or less stable bond, and fore more reactive, due to less effective overlapping

2 Minimum and maximum number of π bonds between two bonded atoms are 0 and 2, respectively

3 Stability ∝ 1

Number of π bonds .

4 Reactivity ∝ Number of π bonds

5 In case of a π bond, free rotation is not possible

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6 It does not determine the shape of a

mol-ecule but shortens the bond length

(C − C > C = C > C ≡ C)

π-bonds 1 2

Strength of s- and π-Bonds

The strength of a bond depends upon the extent

of overlapping of half-filled atomic orbitals

The extent of overlapping between two atoms

is always greater when there is end-to-end

overlapping of orbitals Therefore, a σ-bond is

always stronger than π-bond

REMEMBER

It is the number of unpaired s- or p-electrons

present in its atom in the ground state Thus,

covalency of hydrogen atom is 1

To find sigma and pi bonds in a

HH

It has 9σ and 1π bond, and 2 lone pairs

NC

It has 9σ and 9π bond, and 4 lone pairs

O and N Here, the hydrogen atom is lently bonded to any of these

The nature of a hydrogen bond is either dipole–dipole type, ion–dipole type or dipole–

induced dipole type HCl has no H-bonding as chlorine is large in size

H-bond strength for the following order is

10 kcal per mole, 7 kcal per mole and 2 kcal mole, respectively HF > H2O > NH3 Hydrogen bonding is of the following two types:

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Intermolecular H-bonding

Intermolecular H-bonding is formed between

two or more different molecules of the same or

different types For example, HF, H2O, NH3,

R – OH, R – COOH, etc

HH

HHH

Fig 1.10

Facts Related to Intermolecular

Hydrogen Bonding

One water molecule can form hydrogen

bonding with four other water molecules

Due to hydrogen bonding in water, the water

molecules are closely packed, hence water has

less volume but more density than ice where

an open cage like structure is observed

Water has maximum density at 4°C as above

4°C some hydrogen bonds are broken

lead-ing to a decrease in the density

Two ice cubes when pressed against each

other, form one block due to hydrogen

bonding

Effects of Intermolecular

H-bonding

Increase in boiling point, melting point,

solubility, thermal stability, viscosity, surface

tension and occurence liquid state is observed

as molecules get associated more closely due

to intermolecular H-bonding

HF is a liquid and has a higher boiling point than other HX molecules which are gases at room temperature (Here X = halogens)

Alcohols are highly soluble in water in any proportion and have higher boiling points than others which are very less soluble in water

Glycerol is highly viscous with a high boiling point

Glycerol > Glycol > C2 H5OHB.P., Viscosity decrease Acids have higher boiling point and solubil-ity than their corresponding acid derivatives

In DNA and RNA, the complementary strands are held together by intermolecular H-bonding between the nitrogenous bases of the two strands

Nucleic acid and proteins are held together

by hydrogen bonds

KHF2 or HF2− exists due to hydrogen ing, but formation of other HX2− (for exam-ple HCl2) is not possible, due to absence of hydrogen bonding because of large sizes of the halogen atoms

The extent of hydrogen bonding in water is higher than H2O, hence it has a higher boil-ing point than HF

H2O > HF > NH3 Acids can dimerize due to intermolecular hydrogen bonding For example, acetic acid dimerizes in benzene

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H

C

H O

F

C O

O N

H O

Fig 1.12

Other examples include

pyridine-2-carbon-aldoxime and o-hydroxybenzoic acid, chloral

hydrate HSO5−, acetoacetic ester etc

Effects of Intramolecular

H-bonding

Due to this bonding the boiling point and

acidic nature of the molecule decrease but its

volatile nature increases

O-nitrophenol has a lower boiling point and

reduced acidic nature, but is more volatile

than p-nitrophenol A mixture of both these

componds can be separated by steam

distilla-tion method

Resonance

When all the properties of a molecule cannot

be explained by a single structural formula,

then such molecules are represented by many

structural formulas that are canonical

struc-tures or contributing or resonating structural

representating a single compound

It is observed due to the delocalization of π

electrons

Facts About Canonical Structures

Canonical structures for a given molecule

have the same arrangement of atoms

Position and arrangement of atoms are same

in canonical structures, they only differ in

the distribution of electrons

Canonical structures are depicted by the

symbol (↔) betweem them

Canonical structures should be planar or

nearly planar

Total number of paired and unpaired

elec-trons are also same in canonical structures

O

N O–

O _O¯

O

_ O

O _

S O

O_

O O

S O_

Resonance changes bond length, for ple, in benzene C − C = 1.39 Å, which is an intermediate value between (C – C) = 1.54

exam-Å, (C = C) = 1.34 Å

Resonance Energy

Resonance energy = Energy of most stable canonical structure – Resonance hybrid energy

Resonance energy ∝ Number of canonical structure Resonance energy ∝ Stability

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Resonance energy ∝ _ 1

Reactivity Resonance energy = Expected heat of hydro-

genation − Actual heat

of hydrogenation

Due to high resonance energy, benzene is

quite stable and undergoes electrophilic

sub-stitution reactions It does not undergo

addi-tion reacaddi-tions, although it has double bonds

(due to delocalization of π electrons or

reso-nance)

Benzene has 36 kcal/mole of resonance

energy

Resonance energy of CO2 is 154.9 kJ

In tautomerism, arrangement of atoms is

different for its different arrangements but

in resonance, the arrangement of atoms is the

same

Stability of Different Canonical

Structures

1 A non-polar structure is always more

sta-ble than a polar structure In the following

example, the structures are arranged in a

decreasing order of stability

>

⊕ ⊕

>

In the last two structures the charges are

apart so they are less stable

2 Greater the number of covalent bonds

greater will be the stability Therefore,

3 The canonical structure in which positive

charge in an electro+positive atom and

negative charge on the electro−negative

atom is more stable Therefore,

−O

4 The canonical structure in which each atom has an octet state is more stable Therefore,

NO2, CO3−2

2 Heterovalent resonance Here, the canonical structures have different number

of bonds and charges For example, buta-

1, 3-diene, vinyl cyanide

Resonance and Bond Order

=

Total no of bonds or Total no of order between two atomsBond order

Total no of major canonical

structuresExample, In SO3, B.O.= 4/3 = 1.33

In ClO4−, B.O.= 7/4 = 1.75

HYBRIDIZATION

Pauling and Slater introduced this concept to explain the shape of molecules which could not be explained by the valence bond theory

It is the intermixing or re-distribution of energy among two or more half-filled, fully filled, incompletely filled or empty orbitals of comparable energy, to form same number of hybrid orbitals Hybrids have identical ener-gies and similar shapes

Trang 30

Facts About Hybridization

Number of atomic orbitals taking part in

hybridization is equal to number of hybrids

formed

Electrons do not undergo hybridization

A hybrid bond is always a sigma bond

A hybrid bond is always stronger than a

non-hybrid bond

Hybridization occurs at the time of bond

formation

Hybridization ∝ overlapping (for enough

overlapping, orbitals must be at an

approp-priate distance from each other, that is,

nei-ther very close nor very far)

Hybridization increases stability and

decreases reactivity and energy of a molecule

Hybridization occurs in the central atom in

Hybridization does not occur in isolated

atoms but in bonded atoms

Types of Hybridization

1 sp hybridization: Here, one s and one p

orbitals form two sp hybrid orbitals after

intermixing

Shape of molecule is linear and bond angle

is 180° For example, X – M – X (Mg, Be,

Zn, Hg)

H – C ≡ C – H

sp sp

Some other examples are CO2, and CS2

2 sp 2 hybridization: Here, one s and two

p orbitals intermix to form three new sp2

hybrid orbitals

Shape of these species is trigonal or

coplanar and the bond angle is 120°

For example,

X

XXB

3 sp 3 hybridization: Here, one s and three

p orbitals intermix to give four new sp3

hybrid orbitals

Shape of the species is tetrahedral and bond angle is 109° 28′ For example, C

4

−, NH

3, PH

3, H

2O, H

2S

HCHHH

4 dsp 2 hybridization: Here, one s, two p

and one d orbitals (dx2−y2) intermix to give four new dsp2 hybrid orbitals

Shape of the species is square planar and bond angle is 90° For example, [Ni(CN)4]−2, [Cu(CN)4]2−

2−

CNM

CNNC

NC

5 sp 3 d hybridization: Here, one s, three p

and one d-orbital (dz2) intermix to form five new sp3d hybrid orbitals

Trang 31

XX

6 sp 3 d 2 hybridization: Here, one s, three p

and two d-orbital (dz2 and dx2− y2) intermix

to form six new sp3d2 hybrid orbitals

Shape of the species is octahedral and

bond angle is 90° For example, SF6, XeF4

S

FF

FFFF

7 sp 3 d 3 hybridization: Here one s, three p

and three d orbital (dxy, dyz, dxz) intermix to

form seven new sp3d3 hybrid orbitals

Shape of the species is

pentago-nal bipyramidal and bond angle is between 72° to 90°

For example, IF7, XeF6

F

FF

Xe

:

Rules to Find the Type

of Hybridization

For covalent compounds and ions: 1 Count

the total number of valence electrons and (±)

charge, to find a particular value For example, in the PO4−3 number of valence elctrons is 5 + 4 × 6 + 3 = 32 For NH+

4, this number is 5 + 4 − 1 = 8

2 Now divide the total value of electrons to get the quotient X (number of bond pair electrons)

(a) If total value of electrons is between 2

P = ½ (V + M − C + A) Here, P = total numbers of pairs of elec-

trons around the central atom which gives the present hybrid-ization of the central atom as calculated above

A = Charge on anion

C = Charge on cation

M = Number of monovalent

atoms

Trang 32

V = Number of electrons in the

valence shell of the central atom

To find lone pair of electrons

Lone pair = P − N

Here,

P = Total numbers of pairs of electrons

around the central atom which gives hybridization as above

N = Number of atoms surrounding the

central atom or number of bond pairs

Hybridization in complexes:

Coordina-tion number of ligands is used to find the

2O, F, Cl, Br, I CO, CN, NH

3

For example, in [Fe(CN6)]−3 the

coordina-tion number is 6 and ligand is strong, hence the

hybridization is d2sp3

Similary, in [Fe(H2O)6]+3 the tion number is 6 and ligand is weak, hence the hybridization is sp3d2

coordina-Valence Shell Electron Pair Repulsion Theory (VSEPR)

Valence shell electron pair repulsion theory was introduced by Nyholm and Gillispie to predict the shape of polyatomic molecules and ions

According to this theory, besides tion, the nature of electrons around the cen-tral atom also decide the shape of molecule

There may be two types of electrons around the central atom, that is, bond pair or lone pair of electrons

These electrons undergo electron−electron repulsion and the decreasing order of elec-tronic repulsion follows lp − lp > lp − bp >

bp − bp

Due to this electronic repulsion, the shape

of the molecule becomes distorted and the bond angle changes

Distortion in shape ∝ e− − e− repulsionDistortion in shape or change in bond angle ∝ electronic repulsion

Geometry of Some Molecules and Ions

sp 2 hybridization

A:

BB

Trigonal shape due

to bond pair of e¯

Angular or bent shape due to lone pair of e¯

Trang 33

For example, BX3, BH3, SO3

BX

S:

OO

Here, S atom has two bond pairs and

one lone pair of electron, so lp − bp type

of repulsion distorts the shape, that is, it

bends and changes the bond angle and the

shape becomes angular Same holds true for

SnCl2 and PbCl2

sp 3 hybridization

1 When the central atom has four bond pairs

of electrons, the shape will be normal with

normal bond angle of 109˚ 28' which The

Shape tetrahedral For example, CH4, CCl4,

SiCl4, NH4+, BX4

HCH

2 When the central atom has 3 bond pairs

and 1 lone pair of electron, there will be

lp − bp type of repulsion, which distorts

the shape and changes the bond angle,

that is, the shape becomes pyramidal and

the bond angles are less than 109˚ 28' For

In ammonia, the bond angle is 107˚

3 When the central atom has 2 lone pair

and 2 bond pair of electron, there will be

lp − lp type of electronic repulsion, hence

the shape will be distorted and it will be

angular or bent For example, H2O, H2S,

OF2, SCl2, SeCl2

AB

• • • •B

4 When the central atom has 3 lone pairs and 1 bond pair of electrons, there will be

lp − lp type of electronic repulsion hence, shape is highly distorted and it becomes linear For example, I − Cl, HCl

sp 3 d hybridization

1 When the central atom has 5 bond pair of electrons, the shape will be normal with nor-mal bond angle, that is, the shape becomes trigonal bipyramidal and bond angle of 90° and 120° As only bp − bp type of electronic repulsion occurs, hence there is no distor-tion in shape and no change in bond angle

For example, PCl5, AsF5 and PF5

ClCl

2 When the central atom has 4 bond pair and

1 lone pair of electrons, the shape will be distorted and it will possess a see-saw like structure For example, SeCl4, TeCl4, SF4

BAB

B

or

3 When the central atom has 3 bond pairs and 2 lone pair of electrons, the shape will

be distorted and it will be a T-shape like structure For example, ClF3, BrF3

Trang 34

BB

:

Here, lone pair of electrons occupy

equito-rial position to minimize e − e− repulsion

3 lone pair of electrons, the shape will be

distorted and the shape will be linear For

example, XeF2, I3−, ICl2−

1 When the central atom has 6 bond pairs of

elec-trons, the shape will be normal with normal

bond angles that is, octahedral (90˚) As only

bp − bp type of electronic repulsion occurs, so

there is no distortion in shape or change in the

bond angle For example, SF

6, Te Cl

6

A

SF

FFF

distorted and it will be square pyramidal

For example, BrF5, IF5

A

distorted and it will be square planar For

no change in bond angle For example, IF7

I

F F

FFF

2 When central atom has 6 bond pair and 1 lone pair of electrons, the shape will be dis-torted and the shape will be distorted pen-tagonal bipyramidal For example, XeF6

Xe :F

FCaped octahedron

FF

MOLECULAR ORBITAL THEORY

Molecular orbital theory was given by Hund and Mulliken

It is based on Linear Combination of Atomic Orbitals (LCAO) model

Atomic orbitals undergo linear combination

to form same number of molecular orbitals, if they fulfill the following conditions:

1 Atomic orbitals must have comparable energies

2 Atomic orbitals must overlap linearly for enough and effective overlapping

3 Atomic orbitals must have same try along with the major molecular axis,

Trang 35

symme-Hybridization and Shapes of Some Simple Molecules

Number

of Bonds

Number of Lone Pairs

Number of Charge Clouds

Molecular Geometry and Shape

Trigonal planar

Bent

H C H

bipyramidal

Trigonal bipyramidal

F

F F

F S

Cl

Cl P Cl Cl

Trang 36

Number

of Bonds

Number of Lone Pairs

Number of Charge Clouds

Molecular Geometry and Shape

F

Cl Cl

Cl F

F F F

F S F

2–

Octahedral

Square Pyramidal

Square Planar

for example, if Z axis is the main

molec-ular axis, then only pz − pz orbitals will

overlap and not px or py

Molecular orbitals are formed due to

con-structive and decon-structive interference of

atomic orbitals

Constructive interaction of orbitals between

orbital lobes having same wave function ψ

produces bonding molecular orbitals like

σ, π and Δ These are HoMOs (Highest

Occupied Molecular Orbitals)

Amplitude = a

Bonding MO Node

Constructive interaction

a a

+

+ +

2a

2a 2a 2a Amplitude = 2a a

Fig 1.13

Destructive interaction between orbitals having different sign of c produces anti-bonding molecular orbitals or LuMOs (Lowest Unoccupied Molecular Orbitals) For example, σ*, π*, d*

Amplitude = 0Amplitude = a

+ +

Fig 1.14

Trang 37

Facts Related to

HoMOs and LuMOs

Energy: LuMOs > HoMOs

Wavelength: LuMOs < HoMOs

LuMOs have nodal planes while HoMOs

may or may not have nodal planes

Electrons contribute force of attraction in

HoMOs while they contribute repulsion in

LuMOs

The shape of the molecular orbitals formed

depend upon shape of atomic orbital from

which they are formed

Like atomic orbitals, molecular orbitals also

follow

1 Pauli exclusion principle—Any molecular

orbital can have a maximum of two electrons

with opposite spin

2 Hund’s rule—In degenerate molecular

orbital, before pairing, each molecular orbital

must have one electron

3 Aufbau principle—Electrons are filled from

mole cular orbital of lower energy to higher

Nodal plane σ∗1s or σ∗2s

σ1s or σ2s

Overlapping ψA + ψ

B Addition

ψ

A + ψB Subtraction

σ2pz σ∗2pz

Molecular Orbitals

Atomic Orbitals

Trang 38

2 Molecular orbital energy level diagram for

diatomic homonuclear molecules such as O2,

AtomicOrbitals

(i) Molecules with N2 configuration or 14 e−

σ 1s is the lowest energy molecular orbital while

σ*2pz is the highest energy molecular orbital

Due to intermixing of 2s and 2p orbitals in cases where the number of elecruons is more than 16, σ2pz is taken after σ*2s here

Bond order = n b− na

2 Here nb = Number of bonding molecular

orbital electrons

na = Number of anti-bonding

molecu-lar orbital electrons Bond order ∝ Bond dissociation energy ∝ Bond angle

∝ 1

Bond length Higher the bond order, higher will be stabil-ity and shorter will be the bond length

If unpaired electrons (n = 1, 2) are present in

Trang 39

Some Orders Related to

Molecular Orbital Theory

O22+ O2+ O2 O2− O2−2

Bond 3 2.5 2 1.5 1

order

Decreasing order of bond order, bond angle,

bond dissociation energy

Increasing order of bond length

CO, NO+, CN−, N2 (14 e−) all have bond

order 3 and are diamagnetic

NO, CN, N2−, N2+ all have a bond order

equal to 2.5 and are paramagnetic as n=1

H2, Li2, B2 all have a bond order equal to one

and are diamagnetic except B2 (Paramagnetic)

H2−, H2+, He2+ all have a bond equal to order

½ and are paramagnetic

All molecules with fractional bond order are

paramagnetic

Molecules with whole number bond order

are mostly diamagnetic, except O2, B2, N22−

REMEMBER

When z-axis is the major

molecu-lar axis, s molecumolecu-lar orbital will be

one nodal plane each

π*2px and π*2py have two nodal planes

each

SOME IMPORTANT GUIDELINES

Formyl Charge: It is equal to V − N − 1/2 B, where

V = Total number of valence electrons in the free atom

N = Total number of lone pair of electrons

B = Total number of shared electrons that is, bonded electrons

For example, in case of NH4+

NH3

H2H1

H+

4

Here, F.C on N-atom = 5 − 0 − ½ × 8 = + 1 F.C on 1, 2, 3 H-atoms = 1 − 0 − ½ × 2 = 0 F.C on H+ (4 H-atom) = 0 − 0 − ½ × 2 = −1

CO32−, NO3− involve only pπ − pπ bonding

SO42−, PO43−, ClO4− involve pπ − dπ bonding

In a hypervalent, for example, SO42−, PO4−

3, ClO4−, at least one atom has more than

8 electrons

Molecular solids have low heat of fusion

Breaking of covalent bonds occurs during melting of SiO2

Banana bonding is shown by boron hydride,

that is, diborane B2H6

Band Theory of Metallic Bonds It is

based on molecular orbital theory The est occupied energy band is called the valence band while the lowest occupied energy band

high-is called as conduction band The difference

in the energy between top of valence bond and bottom of conduction bond is called

energy gap For example,

1 When energy gap is very very small,

conduc-tion occurs (as in metals)

Trang 40

2 When energy gap is small, less conduction

occurs (as in semi-conductor)

3 When energy gap is large, no conduction

occurs (as in insulators)

(does not have zero dipole moment,)

C2 molecule has no σ bond but has a π bond

Bond energy: Cl2 > F2 > Br2 > I2

Hobs of CO is greater than μexpected due to

pres-ence of a coordinate bond

AgX, BaSO4, PbSO4, are nearly insoluble or

insoluble in H2O as HHyd > Lattice energy

In benzyne, the triple bond (≡) is partial due

to less effective overlapping In it all C atoms are sp2 hybridized

Bond length of CO > CO+ as bond order of

CO is 3 and of CO+ is 3.5

Bond energy order2s − 2s < 2s − 2p < 2.p − 2p

− p < s − s < s − p < p − p { ness of lateral ← axial overlaping → overlaping}

NO− >NO> NO2+ > NO+ (Bond length)

NO+ >NO2+ = NO > NO− (Bond order)

Straight Objective Type Questions

(Single Choice)

1 KF combines with HF to form KHF

2 The compound contains the species

(a) K+, F− and H+ (b) K+, F− and HF

(c) K+ and [HF2]− (d) [KHF]+ and F2

2 The bond order in O2− ion is

(c) 2.5 (d) 1.5

3 Which of the following molecules has the

smallest bond angle?

4 The number of sigma and pi bonds

pres-ent in tetracyanoethylene [(CN)2C =

C(CN)2] molecule are, respectively

(a) 5 σ and 9 π (b) 5 σ and 8π (c) 9 σ and 9 π (d) 9 σ and 7 π

5 Unusually high boiling point of water is the result of

(a) intermolecular hydrogen bonding

(b) both intra and inter molecular hydrogen bonding

(c) high specific heat

(d) intramolecular hydrogen bonding

6 Dissolution of ionic solid in water is accompanied by release of energy repre-sented by ΔHsolution This implies that (a) ΔHlattice > ΔHhydration

(b) ΔHlattice = ΔHhydration

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