Chemistry. CHEMICAL BONDING – SESSION I Session Opener

Chemistry

CHEMICAL BONDING – SESSION I

Session Opener

Session Objectives

1. Introduction

2. Octet rule

3. Different types of bonding

4. Lewis theory

5. VSEPR theory and shape of molecules

Session Objectives

Force of attraction holdinggroup(s) of atoms

What is Chemical bonding?

Chemical bonds

Na+ Cl-

Better stability against

chemical reagents

But why bonds are formed ??

Atoms

two electrons in the valenceshell (1s2)

Octet rule

noble gas configuration attain betterstability.

Na

2 8 1

Very reactive

Na

2 8

+

Ne

Cl

2 8 7

Very reactive

Cl

2 8 8

Ar

-

SFF

F

F

FF*

***

*

* .....

.

In SF6, ‘S’ has twelve electron in itsvalence shell, leads to minimisation of energy.

Other examples are: PCl5, BF3

Limitation of octet rule

Questions

The molecule that deviates from octet rule is

(a) NaCl (b) BeCl2

(c) MgO

(d) NH3

Illustrative Problem

The no. of valence electrons in different central atoms is:

Hence, the answer is (b).

Solution

Na+ 8 Be+2 2 Mg+2 8

N atom in NH3 (covalent compounds) 8

Bonding

Ionic

Covalent

Co-ordinate or dative

Metallic

Pi bond Sigma bond

Formation of ionic bond

Covalent bond

courtesy:www.lbw.cuny.edu

Formed by mutual

sharing of electrons

Covalent bonds

1 1

H2C CH2

Double bond

HC CHTriple bond

1 2

non-polar covalent bond between two carbon atomspolar covalent bond between carbon and hydrogen atoms.

Formation of covalent bond

Question

Covalent bonds are called directional while ionic bonds are called non-directional -explain

Solution:

Illustrative Problem

p and d-orbitals generate directional covalent bond.

electrostatic force of attraction.

Ionic bond

overlap of atomic orbitals

covalent bond

Strength of these sigma bonds is in the order:

sigma bond forms due to end-to-end or head-on overlap

p-p+

s-s

+

+

Types of covalent bonds

s-p

p-p > s-p >s-s

This is formed by lateral or sideways overlap which is possible for p or d-orbitals.

Sigma bond is stronger than pi bond due to greater extent of overlap.

+ or

Types of covalent bonds

Difference between sigma and pi bonds

Stronger as compared to bond

Weaker as compared to bond

H C C H

Formed by head-on overlapping of s-s or s-p or p-p or any hybrid orbital

Formed by side ways overlapping of unhybridised p-orbital

First bond between any two atoms is always sigma

Rest are bonds

In plane of molecule

Perpendicular to plane of molecule

Ionic compound(NaCl)

Covalent compound (CHCl3)

MP/BP

Very high Volatile liquid H20 solubility

Highly solubleAlmost insoluble

Benzene solubility

InsolubleHighly Insoluble

Directional nature

Non-directional

Except s-soverlap all are directional

Difference between ionic and covalent compound.

Question

How many sigma and pi-bondsare present in a benzene molecule?

Solution:

The structure of benzene molecule is H

H H

H

H

H

no. of pi bonds are 3 [C=C]

no. of sigma bonds are 12 [C-C and C-H]

Illustrative Problem

•Single atom donating lone pair•Shared by two atoms involved

Coordinate covalent bond

O

H

H+H :

H3O+

H

N

H

H+H :

[NH4]+

Question

NH3 and BF3 form an adduct readily-explain.

N- atom in NH3 have one lonepair and BF3

is electron deficient. They form an adductthrough coordinate bond, so BF3 can complete its octet.

Solution:

The adduct.N B

F

H

H

H

F

F:

Illustrative Problem

Metals lose their valence electrons to form cation in the pool of electrons. This is the Electron Sea Model for metallic bond.

Formation of metallic bond

Excellent electricaland thermal conductivity

Regular close packed structures

Characteristics of Metallic bond

Question

Which of the following has othertype of bonding with covalentbonding?

(a) CCl4 (b)AlI3

(c) NH4Cl (d) HCl

Illustrative Problem

Solution:

covalent bonding is between N and three H-atomsCo-ordinate bond is present between N and one H atom ionic bond is there between NH4

+ and Cl– ions.

Hence, the answer is (c).

Important aspects

i) Central atom.

[Exception: NH3, H2O more electronegative central atoms.]

ii) Formal charge on ‘each atom’= (valence electron in atom) – (no. of bonds) – (no. of unshared electrons)

Lewis theory

lesselectronegativ

e atom

iii) Multiple bonds complete the octet of atoms.

n1=4+4 =8;

n2=2x4+8x1 =16;

n3=n2-n1 =8;

no. of bonds= n3/2=4

no. of non-bonding electron= n4

=(n1-n3)=0

no. of lone pairs= 0

Structure and bonding in CH4

C

H

H

H

H

It cannot explain

Limitations of Lewis theory

Odd electron species NO,NO2

Electron-deficient species BF3 ,BeCl2

Electron-rich species PCl5 ,SF6

1. Order of repulsionlp-lp > lp-bp > bp-bp

VSEPR theory

3. Decreasing order of repulsion, Triple bond > double bond > single bond.

2.lp-lp repulsion Electro-negativity of central atom

Electro-negativity of other atoms

Linear

Planar

Tetrahedral

Trigonal bipyramidal

Octahedral

PCl5

central atom is P.therefore, V= 5+5= 10 V/2=5

Shape will be trigonal bipyramidal.

Application of VSEPR theory

Central atom is S V= 6 + 6 = 12

V/2=6

No. of atoms attached to central atom is six.

Hence, shape is octahedral.

Shape of SF6

Question

The shape of CH3+ is likely to be

(a) Pyramidal (b) tetrahedral

(c) linear (d) planar

Illustrative Problem

According to VSEPR,

N = 4 +3 –1=7

N/2=3

the shape should be planar.

Solution:

Hence, the answer is (d).

Limitations

Cannot determine the shape

Multiple bonded species CO2,SO4

-2

Question

The shape of NH3 is very similar to

(a) CH4

(b) CH3

–

(c) BH3

(d) CH3+

Illustrative Problem

Hence, the answer is (b).

4

N 4 4for CH , 4

2 2

shape is tetrahedral

3CHN 4 3 1

for , 42 2

shape is pyramidal

3

N 5 3for NH , 4

2 2

shape is pyramidal

3

N 3 3for BH , 3

2 2

shape is planar

3CHN 4 3 1

for , 32 2

shape is planar

Solution

Class Test

Class Exercise - 1

Pi bond formation involves ______ overlap.

(a) s-p head-on (b) p-p head-on

(c) s-s head-on (d) p-p sideways

Solution:

Pi bond formation involves only sideways overlap of pand d-orbitals.

Hence the answer is (d)

Class Exercise - 2

Solution:

According to Lewis theory,n1 = 5 + 1 + 6 × 3 = 24n2 = 2 × 0 + 8 × 4 = 32n3 = n2 – n1; number of bonds = 3n

42

O — N — O

O

24 88

2

Number of lone pairs =

Formal charge on ‘N’ atom = 5 – 4 – 0 = +1

What is the formal charge on ‘N’ atom of ?3NO

Class Exercise - 3

Molecular structures of SF4 and XeF4

are (a) the same, with 2 and 1 lone pairs respectively (b) the same, with 1 lone pair each (c) different, with 0 and 2 lone pairs respectively (d) different with 1 and 2 lone pairs respectively

Solution:

For 4N 6 4

SF 52 2

Lone pair is one and the structure is trigonal bipyramidal.

Solution

Lone pairs are two and the structure is octahedral.

Hence, answer is (d)

For 4N 8 4

XeF , 62 2

Class Exercise - 4

Predict the geometry of H3O+ based on VSEPR theory.

Solution:

For H3O+, central atom is ‘O’

N 6 3 14

2 2

Since 3 atoms are attached to the central atom,geometry will be of pyramidal according toVSEPR to minimize lp-bp repulsion.

Class Exercise - 5

Solution:

Electronic configuration of Ca metal is 2, 8, 8, 2.While the configuration for Ca+2 is 2, 8, 8 which is a stable noble gas configuration.

Among Ca metal and Ca+2 the morereactive will be (Atomic No. of Ca is20)

(a) Calcium metal(b) Calcium ion(c) both are equally reactive (d) Cannot be predicted

Hence the answer is (a)

Class Exercise - 6

Pi-bonds in N2 and CN– are due to

(a) p-p overlap for both species(b) p-p and p-d overlap(c) d-d overlap for both species(d) p-d and p-p overlap

Solution:

Since Pi bonds are formed due to overlap of either p ord orbitals only. Both N and C-atoms do not have anyelectrons in d-orbitals. Hence, Pi bonds in both casesare obtained because of p-p overlap only.

Hence, the answer is (a).

Class Exercise - 7

The geometry of XeF2 accordingto VSEPR is

• angular• linear• pyramidal• None of these

Solution:

Number of atoms attached to the central atom is two.According to VSEPR theory geometry should be linear.

For XeF2N 8 2

52 2

Hence, the answer is (b)

Class Exercise - 8

Which of the following is a tri-atomicmolecule?

(a) Ammonia(b) Sulphur dioxide(c) Sulphur tri-oxide(d) Phosphine

Solution:

NH3 tetratomic moleculeSO2 tri-atomic moleculeSO3 tetratomic moleculePH3 tetratomic molecule

Hence the answer is (b)

Thank you