Bond order according to Lewis : The bond order is given by the number of bonds (shared  electron pair) between the two atoms in a molecule.

e.g. : Bond order in $\mathrm{H}_{2}, \mathrm{~F}_{2}, \mathrm{Cl}_{2}, \mathrm{HCl}$ is one. In $\mathrm{O}_{2}$ bond order two and in $\mathrm{N}_{2}$ three because shared electron pairs between two atoms is respectively 2 and $3 .$

According to MO theory bond order: $\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)$

Where, $\mathrm{N}_{\mathrm{b}}=$ Total no. of bonding $\mathrm{e}^{-}$of $\mathrm{BMO}$

$\mathrm{N}_{\mathrm{a}}=$ Total no. of anti bonding $\mathrm{e}^{-}$of $\mathrm{ABMO}$

Bond order : The difference of electron in MO and ABMO orbitals divided by 2 is called Bond order.

e.g. : Bond order in $\mathrm{H}_{2}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)$

$=\frac{1}{2}(2-0)=1$

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

$=\frac{1}{2}(10-8)=1$

Bond order $\mathrm{BO}$ in $\mathrm{O}_{2}=\frac{1}{2}(10-6)=2$

Bond order $\mathrm{BO}$ in $\mathrm{N}_{2}=\frac{1}{2}(10-4)=3$