Find the bond order in ${\text{O}}_2$ molecule. 

O atom=1s² 2s² 2p⁴

$\therefore$ Total number of electrons in ${\mathrm{O}}_2$ molecule$=2\times 8=16$     

These 16 electrons occupy different molecular orbitals

Electron configuration of ${\mathrm{O}}_2$ molecule

$\left(\sigma 1s^2\right)\left(\sigma *1s^2\right)\left(\sigma 2s^2\right)\left(\sigma *2s^2\right)\left(\sigma 2p_z^2\right)\left(\pi 2p_x^2=\pi 2p_y^2\right)\left(\pi *2p_x^1=\pi *2p_y^1\right)$

Bond order is calculated as: $\mathrm{Bond} \mathrm{order}=\frac{\mathrm{Electrons} \mathrm{in} \mathrm{bonding} \mathrm{M}.\mathrm{O}-\mathrm{Electrons} \mathrm{in} \mathrm{antibonding} \mathrm{M}.\mathrm{O}.}{2}$

                      $=\frac{8-4}{2}=2$