Chemistry - CHEMICAL BONDING Question with Solution | TestHub

This can be done by the molecular orbital theory.

First, write the molecular orbital electronic configuration of ${\mathrm{N}}_2$molecule.

The total number of electrons present in the dinitrogen molecules is$=14$.

All the $14$ electrons distributed in the bonding and antibonding molecular orbitals given below.

${\mathrm{N}}_2\to {\mathrm{KK}}_0\sigma 2{\mathrm{s}}^2\mathrm{\sigma }*2{\mathrm{s}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2$

${\mathrm{N}}_2^{+}\to \mathrm{KK\sigma }2{\mathrm{s}}^2\mathrm{\sigma }*2{\mathrm{s}}^2\mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^{2}2{\mathrm{p}}_{\mathrm{z}}^1$

As per the molecular orbital electronic configuration, we can understand that the last electron is removing from the $\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}$ bonding molecular orbital.