The shared pairs of electrons presentbetween the bonded atoms are called bond pairs. All valenceelectrons may not participate in bonding. The electron pairs that do notparticipate in bonding are called lone pairs of electrons.

For example, in C₂H₆ (ethane), there areseven bond pairs but no lone pair present.

In H₂O, there are two bondpairs and two lone pairs on the central atom (oxygen).

Question - 32 : -
Distinguishbetween a sigma and a pi bond.

Answer - 32 : -

Thefollowing are the differences between sigma and pi-bonds:

Sigma (σ) Bond	Pi (π) Bond
(a) It is formed by the end to end overlap of orbitals.	It is formed by the lateral overlap of orbitals.
(b) The orbitals involved in the overlapping are s–s, s–p, or p–p.	These bonds are formed by the overlap of p–p orbitals only.
(c) It is a strong bond.	It is weak bond.
(d) The electron cloud is symmetrical about the line joining the two nuclei.	The electron cloud is not symmetrical.
(e) It consists of one electron cloud, which is symmetrical about the internuclear axis.	There are two electron clouds lying above and below the plane of the atomic nuclei.
(f) Free rotation about σ bonds is possible.	Rotation is restricted in case of pi-bonds.

Question - 33 : -
Explain the formation of H₂ molecule on thebasis of valence bond theory.

Answer - 33 : -

Let us assume that two hydrogen atoms (A andB) with nuclei (N_A and N_B) and electrons (e_Aand e_B) are taken to undergo areaction to form a hydrogen molecule.

When A and B are at alarge distance, there is no interaction between them. As they begin to approacheach other, the attractive and repulsive forces start operating.

Attractive force arisesbetween:

(a) Nucleus of one atom and its own electroni.e., N_A – e_A and N_B – e_B.

(b) Nucleus of one atom and electron ofanother atom i.e., N_A – e_B and N_B – e_A.

Repulsive force arisesbetween:

(a) Electrons of two atoms i.e., e_A – e_B.

(b) Nuclei of two atoms i.e., N_A – N_B.

The force of attractionbrings the two atoms together, whereas the force of repulsion tends to pushthem apart.

Themagnitude of the attractive forces is more than that of the repulsive forces.Hence, the two atoms approach each other. As a result, the potential energydecreases. Finally, a state is reached when the attractive forces balance therepulsive forces and the system acquires minimum energy. This leads to theformation of a dihydrogen molecule.

Question - 34 : -
Writethe important conditions required for the linear combination of atomic orbitalsto form molecular orbitals.

Answer - 34 : -

The given conditionsshould be satisfied by atomic orbitals to form molecular orbitals:

(a) The combining atomic orbitals must havethe same or nearly the same energy. This means that in a homonuclear molecule,the 1s-atomic orbital of an atom can combine with the 1s-atomicorbital of another atom, and not with the 2s-orbital.

(b) The combining atomicorbitals must have proper orientations to ensure that the overlap is maximum.

(c)The extent of overlapping should be large.

Question - 35 : -
Use molecular orbital theory to explain whythe Be₂ molecule does not exist.

Answer - 35 : - The electronic configuration of Beryllium is

The molecular orbital electronicconfiguration for Be₂ molecule can be written as:

Hence, the bond order for Be₂is

Where,

Nb = Number ofelectrons in bonding orbitals

Na = Number ofelectrons in anti-bonding orbitals

Bond order of Be₂ = 0

Anegative or zero bond order means that the molecule is unstable. Hence, Be₂ molecule does not exist.

Question - 36 : -
Compare the relativestability of the following species and indicate their magnetic properties;
O₂,, (superoxide), (peroxide)

Answer - 36 : -

There are 16 electrons ina molecule of dioxygen, 8 from each oxygen atom. The electronic configurationof oxygen molecule can be written as:

Since the 1s orbital of eachoxygen atom is not involved in boding, the number of bonding electrons = 8= N_band the number of anti-bonding orbitals = 4= N_a.

Bondorder

= 2

Similarly,the electronic configuration of can be written as:

N_b = 8

N_a = 3

Bondorder of

= 2.5

Electronicconfiguration of ion will be:

N_b = 8

N_a = 5

Bondorder of =

= 1.5

Electronicconfiguration of ion will be:

N_b = 8

N_a = 6

Bondorder of

= 1

Bonddissociation energy is directly proportional to bond order. Thus, the higherthe bond order, the greater will be the stability. On this basis, the order ofstability is

Question - 37 : -
Writethe significance of a plus and a minus sign shown in representing the orbitals.

Answer - 37 : -

Molecularorbitals are represented by wave functions. A plus sign in an orbital indicatesa positive wave function while a minus sign in an orbital represents a negativewave function.

Question - 38 : -
Describe the hybridisation in case of PCl₅. Why are the axial bondslonger as compared to equatorial bonds?

Answer - 38 : -

The ground state andexcited state outer electronic configurations of phosphorus (Z = 15) are:

Phosphorus atom is sp³d hybridized in theexcited state. These orbitals are filled by the electron pairs donated by fiveCl atoms as:

PCl₅

The five sp³d hybrid orbitals aredirected towards the five corners of the trigonal bipyramidals. Hence, thegeometry of PCl₅ can be represented as:

There are five P–Cl sigma bonds in PCl₅. Three P–Cl bonds lie inone plane and make an angle of 120° with each other. These bonds are calledequatorial bonds.

The remaining two P–Clbonds lie above and below the equatorial plane and make an angle of 90° withthe plane. These bonds are called axial bonds.

Asthe axial bond pairs suffer more repulsion from the equatorial bond pairs,axial bonds are slightly longer than equatorial bonds.

Question - 39 : -
Definehydrogen bond. Is it weaker or stronger than the van der Waals forces?

Answer - 39 : -

A hydrogen bond is definedas an attractive force acting between the hydrogen attached to anelectronegative atom of one molecule and an electronegative atom of a differentmolecule (may be of the same kind).

Due to a differencebetween electronegativities, the bond pair between hydrogen and theelectronegative atom gets drifted far away from the hydrogen atom. As a result,a hydrogen atom becomes electropositive with respect to the other atom andacquires a positive charge.

The magnitude of H-bondingis maximum in the solid state and minimum in the gaseous state.

There are two types ofH-bonds:

(i) Intermolecular H-bond e.g., HF, H₂O etc.

(ii) Intramolecular H-bonde.g., o-nitrophenol

Hydrogenbonds are stronger than Van der Walls forces since hydrogen bonds are regardedas an extreme form of dipole-dipole interaction.

Question - 40 : -
What is meant by the term bond order?Calculate the bond order of: N₂, O₂, and.

Answer - 40 : -

Bond order is defined asone half of the difference between the number of electrons present in thebonding and anti-bonding orbitals of a molecule.

If N_a is equal to thenumber of electrons in an anti-bonding orbital, then N_b is equal to thenumber of electrons in a bonding orbital.

Bondorder =

If N_b > N_a, then the molecule issaid be stable. However, if N_b ≤ N_a, then the molecule isconsidered to be unstable.

Bond order of N₂ can be calculatedfrom its electronic configuration as:

Number of bonding electrons, N_b = 10

Number of anti-bonding electrons, N_a = 4

Bondorder of nitrogen molecule

= 3

There are 16 electrons ina dioxygen molecule, 8 from each oxygen atom. The electronic configuration ofoxygen molecule can be written as:

Since the 1s orbital of eachoxygen atom is not involved in boding, the number of bonding electrons = 8= N_band the number of anti-bonding electrons = 4= N_a.

Bondorder

= 2

Hence, the bond order ofoxygen molecule is 2.

Similarly,the electronic configuration of can be written as:

N_b = 8

N_a = 3

Bondorder of

= 2.5

Thus,the bond order of

is 2.5.
Theelectronic configuration of

ion will be:

N_b = 8

N_a = 5

Bondorder of

= 1.5

Thus,the bond order of

ionis 1.5.

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