Download Chemical Bonding Class11th by PS Sir IIT JEE PDF

TitleChemical Bonding Class11th by PS Sir IIT JEE
TagsChemical Bond Ion Ionic Bonding Chemical Polarity Covalent Bond
File Size826.1 KB
Total Pages44
Table of Contents
                            Chemical Bonding Class11th By PS Sir IIT JEE /PMT Course
	The Lewis Theory
	Criteria for Ionic Bond Formation
	Energy Change During the Formation of Ionic Bond
	Determination of Lattice Energy
	Characteristics of Ionic Compounds
	Valence Bond Theory (VBT)
	Sigma and Pi Bonding
	Co-ordinate Covalency
	Conditions for Formation of Covalent Bond
	Polar and Non-Polar Covalent Bonds
	Maximum Covalency
	Dipole Moment
	Calculation of Resultant Bond Moments
	Bond Length
		Important Features of Bond Length
	Bond Energy or Bond Strength
	Rules for writing Resonating Structures
	Important Characteristics of Hybridization?
	Rule for Determination of total Number of Hybrid Orbitals
	Regular Geometry of Molecules
		Number of electron pairs
		Arrangement of electrons
		Molecular geometry
	Irregular Geometry of Molecules and VSPER Theory
		Molecule Type
		No. of Bonding pairs
		No. of lone pair
		Arrangement of electrons pairs
		Shape (Geometry)
		Total N/2
		Shape of molecule or ion
	Introduction to Molecular Orbital Theory
	Bond Order
	Application of MOT to Homonuclear Diatomic Molecules
	M.O. of Some Diatomic Heteronuclei Molecules
		Total No. of Electrons
		Magnetic Behaviour
		Inert Pair Effect
Covalent Character in Ionic Compounds
	Fajan's Rule
	Percentage of Ionic Character
	Metallic Bonding
	Hydrogen Bonding
		Conditions for Hydrogen Bonding
		Types of Hydrogen Bonding
		Importance of Hydrogen Bonding in Biological Systems
		Effect of Hydrogen Bonding
Solved Problems of Chemical Bonding
Document Text Contents
Page 1

Chemical Bonding Class11th By
PS Sir IIT JEE /PMT Course


Chemical bond is an

attractive force which keeps tow atoms or ions together in a molecule.A

molecule is formed if it is more stable and has lower energy than the

individual atoms. Normally only electrons in the outermost shell of an atom

are involved in bond formation and in this process each atom attains a

stable electronic configuration of inert gas. Atoms may attain stable

electronic configuration in three different ways by losing or gaining

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electrons by sharing electrons. The attractive forces which hold various

constituents (atoms, ions etc) together in different chemical species are

called chemical bonds. Elements may be divided into three classes.

 Electropositive elements, whose atoms give up one or more electrons
easily, they have low ionization potentials.

 Electronegative elements, which can gain electrons. They have
higher value of electronegativity.

 Elements which have little tendency to loose or gain electrons.

Three different types of bond may be formed depending on electropositive
or electronegative character of atoms involved.

Electropositive element + Electronegative element = Ionic bond
(electrovalent bond)

Electronegative element + Electronegative element = Covalent bond

or less electro positive + Electronegative element = Covalent bond

Electropositive + Electropositive element = Metallic bond.

The Lewis Theory

The octet rule:- The Lewis theory gave the first explanation of a covalent

bond in terms of electrons that was generally accepted. If two electrons

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positive charge. Carbon dioxide is best represented by structures (I) and

The mixing or merging of dissimilar orbitals of similar energies to form new

orbitals is known as hybridization and the new orbitals formed are known

as hybrid orbitals.

Important Characteristics of Hybridization?
1. Orbitals belonging to the same atom or ion having similar energies

get hybridized.

2. Number of hybrid orbitals is equal to the no. of orbitals taking part in


3. The hybrid orbitals are always equivalent in energy and shape.

4. The hybrid orbitals form more stable bond than the pure atom


5. The reason hybridization takes place is to produce equivalent

orbitals which give maximum symmetry.

6. It is not known whether actually hybridization takes place or not. It is

a concept which explains the known behaviour of molecules.

7. The hybrid orbitals are directed in space in same preferred direction

to have some stable arrangement and giving suitable geometry to the


Depending upon the different combination of s and p orbitals, these types

of hybridization are known.

sp3 hybridization: In this case, one s and three p orbitals hybridise to form

four sp3 hybrid orbitals. These four sp3 hybrid orbitals are oriented in a

tetrahedral arrangement.

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sp2 hybridization: In this case one s and two p orbitals mix together to

form three sp2 hybrid orbitals and are oriented in a trigonal planar


The remaining p orbital if required form side ways overlapping with the

other unhybridized p orbital of other C atom and leads to formation of p2C

= CH2 bond as in H

sp hybridization: In this case, one s and one p orbital mix together to form

two sp hybrid orbitals and are oriented in a linear shape.

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“If the electronegativity of the peripheral atoms is more, then the bond

angle will be less”. For example if we consider NH3 and NF3, F – N – F bond

angle will be lower than H – N – H bond angle. This is because in NF3 the

bond pair is displaced more towards F and in NH3 it is displaced more

towards N. So accordingly the b.p. – b.p. interaction is less in NF3 and more

in NH3.

Prob 13. The bond angle of H2O is 104° while that that of F2O is 102°.

Solution:Both H2O and F2O have a lone pair of electrons. But fluorine being

highly electronegative, the bond pair of electrons are drawn more towards

F in F2O, whereas in H2O it is drawn towards O. So in F2O the bond pairs

being displaced away from the central atom, has very little tendency to

open up the angle. But in H2O this opening up is more as the bond pair

electrons are closer to each other. So bond Ð of F2O is less than H2O.

Prob 14. Predict the hybridization for the central atom in POCl3, OSF4,



Total No. of V.E. = 5+6+21/8 = 32/8 = 4

So, hybridization = sp3

OSF4 = 6+6+28/8 = 40/8 = 5

So, hybridization of s = dsp3

OIF5 = 6+7+35/8 = 48/8 = 6

So, hybridization of I = d2sp3

Prob 15. Out of the three molecules XeF4, SF4 and SiF4 one which has

tetrahedral structures is

(A) All of three

(B) Only SiF4

(C) Both SF4 and XeF4

(D) Only SF4 and XeF4


Hybridization of XeF4 = sp3d2, SF4 = sp3d, SiF4 = sp3

Hence (B) is correct.

Prob 16. Among the following compounds in which case central element

uses d-orbital to make bonds with attached atom

(A) BeF2 (B) XeF2 (C) SiF4 (D) BF3


In XeF2. Xe atom has sp3d hybridisation. Hence (B) is correct.

Prob 17.

When NH3 is treated with HCl, state of hybridisation on central nitrogen

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(A) Changes from sp3 to sp2

(B) Remains unchanged

(C) Changes from sp3 to sp3d

(D) Changes from sp3 to sp


On NH4+ state of hybridisation on central nitrogen atom is sp3 as in NH3.

Hence (B) is the correct answer.

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