Chapter 9, Problem 56QP

Interpretation Introduction

Interpretation:

The reason for bond order of ${\text{N}}_{\text{2}}$ is greater than ${\text{N}}_{\text{2}}$ while that of ${\text{O}}_{\text{2}}$ is less than ${\text{O}}_{\text{2}}$ is to be explained.

Concept Introduction:

The combination of two atomic orbitals results in the formation of a bonding and an antibonding molecular orbital. Orbitals that lie on internuclear axis combine to form sigma σ molecular orbital, and orbitals parallel to each other combine to form $\pi$ molecular orbitals.

The molecular orbital formed by combination of 1s orbital forms bonding molecular orbital designated as ${\sigma }_{1s}$, and antibonding molecular orbital as ${\sigma }^{\ast }$. The 2s orbital forms corresponding molecular orbitals.

Bond order is determined by subtraction of the number of antibonding electrons from the number of bonding electrons and divide the obtained number by two.