Give the molecular orbital energy diagram of N₂ and O₂. Calculate respective bond order write the magnetic nature of N₂ and O₂ molecules.

MODE of N₂ :
$\text{ E. }cof\mathrm{N}:1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}_{\mathrm{x}}^{1}2{\mathrm{p}}_{\mathrm{y}}^{1}2{\mathrm{p}}_{\mathrm{z}}^1$
The electronic configuration of N₂ is
${\sigma }_{1{\mathrm{s}}^2}<\mathrm{\sigma }{*}_{1{\mathrm{s}}^2}<{\mathrm{\sigma }}_{2{\mathrm{s}}^2}<\mathrm{\sigma }{*}_{2{\mathrm{s}}^2}<{\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{y}}^2}={\mathrm{\pi }}_{2{\mathrm{p}}_{\mathrm{z}}^2}<{{}^{\mathrm{\sigma }}2\mathrm{p}}_{\mathrm{x}}^2$
$\text{ Bond order }=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}$
$=\frac{10-4}{2}=\frac{6}{2}=3$
$(\mathrm{N}\equiv \mathrm{N})$
It is diamagnetic due to absence of unpaired

MOED of O₂ :
$\mathrm{O}:1{ \mathrm{s}}^{2}2{ \mathrm{s}}^{2}2{\mathrm{p}}_{\mathrm{x}}^{1}2{\mathrm{p}}_{\mathrm{y}}^{1}2{\mathrm{p}}_{\mathrm{z}}^1$
The electronic distribution in O₂ is
$$\begin{gathered} {\mathrm{\sigma }}_{1{\mathrm{s}}^2}^{ }<{\mathrm{\sigma }}_{1{ \mathrm{s}}^2}<{\mathrm{\sigma }}_{2{ \mathrm{s}}^2}<{\mathrm{\sigma }}_{2{ \mathrm{s}}^2}^{*}<{\mathrm{\sigma }}_{2{\mathrm{p}}_{\mathrm{x}}^2} \\ <\mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2=\mathrm{\pi }2{\mathrm{p}}_{\mathrm{z}}^2<{\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{y}}^1={\mathrm{\pi }}^{*}2{\mathrm{p}}_{\mathrm{z}}^1 \end{gathered}$$
Bond order =
$\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}$
$\begin{array}{l}(\mathrm{O}=\mathrm{O})\\ =\frac{10-6}{2}=\frac{4}{2}=2\end{array}$

O₂ is paramagnetic in nature due to the presence of unpaired electrons.