Give electron configuration, magnetic property bond order and energy diagram for oxygen $\left( {{{\rm{O}}_2}} \right)$ molecule.

$\mathrm{O}_{2}(\mathrm{Z}=8) 1 s^{2} 2 s^{2} 2 p^{4}$, So total electron in $\mathrm{O}_{2}=16 \& 12$ valence electrons are participate in bond. Electron configuration in $\mathrm{MO}$ for $\mathrm{O}_{2}$ :

$\mathrm{KK}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}=\left(\pi 2 p_{y}\right)^{2}\left(\pi^{*} 2 p_{x}\right)^{1}\left(\pi^{*} 2 p_{y}\right)^{1}$

Bond order $=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)$

$=\frac{1}{2}(10-6)=2 \quad\left(\right.$ Double bond $\left.\mathrm{O}_{2}\right)$

In it unpair $\bar{e}$ in $\pi_{2 p_{x}}^{*}$ and $\pi_{2 p_{y}^{\prime}}^{*}$ So paramagnetic Energy diagram for $\mathrm{O}_{2}$ molecule :