Chapter 10, Problem 10.101SP

Interpretation Introduction

Interpretation: Using a molecular orbital diagram, the properties such as the bond order and magnetism of ${\text{N}}_{\text{2}}^{\text{3-}}$ and ${\text{O}}_{\text{2}}^{\text{-}}$ have to be compared.

Concept Introduction:

Molecular orbitals:

Linear combinations of atomic orbitals are called as molecular orbitals. These molecular orbitals are generally divided into two types:

1. (i) Bonding orbitals
2. (ii) Antibonding orbitals

Molecular orbital theory:

During the formation of bond, the atomic orbitals of individual atom will combine together to form equal number of molecular orbital. These orbitals are arranged in the order of increasing energy level, while in case of valence bond theory, the individual atomic orbital will directly involve in bond formation.

Bond order:

Bond order of the molecule indicates the number of bonds present between the pair of atoms.

• Bond order is the measurement of the bond strength of a molecule.
• An increase in the bond order shows the increase in the bond strength and therefore, a stable compound is formed.
• Bond order can be calculated by using molecular orbital theory and molecular energy diagram.
• The formula for calculating bond order is:

$\text{Bond}\text{order}\text{=}\frac{\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\\ \text{bonding}\\ \text{molecular}\text{orbital}\end{array}\right)-\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\\ \text{in}\text{antibonding}\\ \text{molecular}\text{orbital}\end{array}\right)}{\text{2}}$

Magnetic properties:

Paramagnetic and diamagnetic:

The compounds in which all the electrons are paired in their orbitals are considered as diamagnetic compounds these compounds are repelled by the magnetic field. The paramagnetic compound contains one or more unpaired electrons in their orbital, and these compounds are attracted by the magnetic field.

Explanation of Solution

Comparison of $N_2^{3-}$ and $O_2^{-}$ based on the molecular orbital diagram:

• Isoelectric nature: From the molecular orbital diagrams of ${\text{N}}_{\text{2}}^{\text{3-}}$ and ${\text{O}}_{\text{2}}^{\text{-}}$, it is clear ${\text{N}}_{\text{2}}^{\text{3-}}$ ion is being isoelectric with ${\text{O}}_{\text{2}}^{\text{-}}$ ion which means both the ions have exactly the same electronic arrangement in their molecular orbitals.
• Magnetic properties: From the molecular orbital diagrams of ${\text{N}}_{\text{2}}^{\text{3-}}$ and ${\text{O}}_{\text{2}}^{\text{-}}$, it can be clearly observed that both ${\text{N}}_{\text{2}}^{\text{3-}}$ and ${\text{O}}_{\text{2}}^{\text{-}}$ have an unpaired electron in the anti-bonding pi-2p-orbitals. So, these two ions are paramagnetic in nature.
• Bond order of $N_2^{3-}$ ion:

The molecular orbitals of ${\text{N}}_{\text{2}}^{\text{3-}}$:

The molecular orbital configuration of ${\text{N}}_{\text{2}}^{\text{3-}}$ molecule is:

$\begin{array}{l}{\left(\text{σ1s}\right)}^{\text{2}}{\left(\text{σ*1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left(\text{σ*2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}\left({\text{π2p}}_{\text{x}}^{\text{2}}\text{=}{\text{π2p}}_{\text{y}}^{\text{2}}\right)\left({\text{π*2p}}_{\text{x}}^2\right)\left({\text{π*2p}}_{\text{y}}^{\text{1}}\right)\\ \text{*}-\text{Antibonding}\text{molecular}\text{orbital}\end{array}$

The bond order of ${\text{N}}_{\text{2}}^{\text{3-}}$ can be calculated as follows:

$\text{Bond}\text{order}\text{=}\frac{\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\\ \text{bonding}\\ \text{molecular}\text{orbital}\end{array}\right)-\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\\ \text{in}\text{antibonding}\\ \text{molecular}\text{orbital}\end{array}\right)}{\text{2}}$

$\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\left(\text{10}\right)-7}{\text{2}}\\ =\frac{3}{2}\\ =1.5\end{array}$

• Bond order of $O_2^{-}$ ion:

The molecular orbitals of ${\text{O}}_{\text{2}}^{\text{-}}$:

The molecular orbital configuration of ${\text{O}}_{\text{2}}^{\text{-}}$ molecule is:

$\begin{array}{l}{\left(\text{σ1s}\right)}^{\text{2}}{\left(\text{σ*1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left(\text{σ*2s}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}\left({\text{π2p}}_{\text{x}}^{\text{2}}\text{=}{\text{π2p}}_{\text{y}}^{\text{2}}\right)\left({\text{π*2p}}_{\text{x}}^2\right)\left({\text{π*2p}}_{\text{y}}^{\text{1}}\right)\\ \text{*}-\text{Antibonding}\text{molecular}\text{orbital}\end{array}$

The bond order of ${\text{O}}_{\text{2}}^{\text{-}}$ can be calculated as follows:

$\text{Bond}\text{order}\text{=}\frac{\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\text{in}\\ \text{bonding}\\ \text{molecular}\text{orbital}\end{array}\right)-\left(\begin{array}{l}\text{Number}\text{of}\text{electrons}\\ \text{in}\text{antibonding}\\ \text{molecular}\text{orbital}\end{array}\right)}{\text{2}}$

$\begin{array}{c}\text{Bond}\text{order}\text{=}\frac{\left(\text{10}\right)-7}{\text{2}}\\ =\frac{3}{2}\\ =1.5\end{array}$

Therefore, the bond orders of ${\text{N}}_{\text{2}}^{\text{3-}}$ and ${\text{O}}_{\text{2}}^{\text{-}}$ is being the same since both the ions are isoelectronic to each other.

The bond orders of ${\text{N}}_{\text{2}}^{\text{3-}}$ and ${\text{O}}_{\text{2}}^{\text{-}}$ is 1.5. So both ${\text{N}}_{\text{2}}^{\text{3-}}$ and ${\text{O}}_{\text{2}}^{\text{-}}$ ions can form only half of a single bond.