Chapter 9, Problem 9.5P

Interpretation Introduction

Interpretation:

The reason why the bond energy of ${\text{N}}_{\text{2}}$ is greater than the ${\text{N}}_{\text{2}}{}^{\text{+}}$ ion has to be explained using MOT theory.

Concept Introduction:

Molecular orbital theory:

Molecular orbital theory suggests the combination of all atomic orbitals having comparable energy and proper symmetry.

Postulates of MOT is,

• Atomic orbitals of same energy and proper symmetry combine together to form molecular orbitals.
• The movement of electrons in a molecular orbital is influenced by all the nuclei of combining atoms.
• The number of molecular orbitals formed is equal to the number of combining atomic orbital when two atomic orbitals combine two molecular orbitals are formed.  One molecular orbital has high energy than the corresponding atomic orbitals and is called antibonding orbital and the other one with lower energy is called bonding orbital.
• In molecular orbitals, the electrons are filled according to the Pauli’s exculsion principle, Aufbau principle and the Hund’s rule.

Bond order $\text{=}\frac{\left({\text{N}}_{\text{b}}\right)\text{-}\left({\text{N}}_{\text{a}}\right)}{\text{2}}$

Where, ${\text{N}}_{\text{b}}$ is number of electrons in bonding orbital.

${\text{N}}_{\text{a}}$ is number of electrons in antibonding orbital.

Explanation of Solution

The bond energy of ${\text{N}}_{\text{2}}$ is greater than the ${\text{N}}_{\text{2}}{}^{\text{+}}$ ion this can be explained based on their bond order.  Bond order is directly proportional to the bond energy thus, higher the bond order higher will be the bond energy.

The atomic number of $\text{N}$ is seven.

The electronic configuration of $\text{N}$=${\text{1s}}^{\text{2}}{\text{2s}}^{\text{2}}{\text{2p}}^{\text{3}}$

The molecular configuration of ${\text{N}}_{\text{2}}$ is,

  ${\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^{\text{2}}{\left({\text{σ2p}}_{\text{z}}\right)}^{\text{2}}$

MO diagram of ${\text{N}}_{\text{2}}$ is,

  Figure 1:  MO of nitrogen

Bond order can be calculated as,

  Bond order $\text{=}\frac{\left({\text{N}}_{\text{b}}\right)\text{-}\left({\text{N}}_{\text{a}}\right)}{\text{2}}$

There are ten electrons in bonding orbital and four electrons antibonding orbitals.

  $\begin{array}{l}\text{Bond order}\text{=}\frac{\left(\text{10}\right)\text{-}\left(\text{4}\right)}{\text{2}}\\ \text{ =}\frac{\text{6}}{\text{2}}\\ \text{=3}\end{array}$

Bond order of ${\text{N}}_{\text{2}}$ is three.

${\text{N}}_{\text{2}}{}^{\text{+}}$ ion is formed when ${\text{N}}_{\text{2}}$ molecule loses its one electron from the outermost orbital.  Thus, there are nine electrons in the bonding orbitals and 4 electrons in the antibonding orbitals.

The molecular configuration of ${\text{N}}_{\text{2}}{}^{\text{+}}$ is,

  ${\left(\text{σ1s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{1s}\right)}^{\text{2}}{\left(\text{σ2s}\right)}^{\text{2}}{\left({\text{σ}}^{\text{*}}\text{2s}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{x}}\right)}^{\text{2}}{\left({\text{π2p}}_{\text{y}}\right)}^2{\left({\text{σ2p}}_{\text{z}}\right)}^1$

MO diagram for nitronium ion is,

Figure 2:  MO of nitronium ion

Bond order of nitronium ion is,

  $\begin{array}{l}\text{Bond order}\text{=}\frac{\left(\text{9}\right)\text{-}\left(\text{4}\right)}{\text{2}}\\ \text{ =}\frac{\text{5}}{\text{2}}\\ \text{=2}\text{.5}\end{array}$

Bond order of nitronium ion is 2.5.

Hence, ${\text{N}}_{\text{2}}$ has the highest bond order and its bond energy is greater than the nitronium ion.