Chemistry - CHEMICAL BONDING Question with Solution | TestHub

$\text{Bond}\text{order}=\frac{1}{2}\left({\text{N}}_{\text{b}}-{\text{N}}_{\text{a}}\right)$
${\text{N}}_{\text{b}}\to$Total number of electrons in bonding molecular orbits.
${\text{N}}_{\text{a}}\text{=}$Total number of electrons in anti-bonding molecular orbits.
$\text{NO}\to {\text{σ}}_{\text{1s}}^{\text{2}}{\text{σ}}_{\text{1s}}^{\text{*2}}{\text{σ}}_{\text{1s}}^{\text{2}}{\text{σ}}_{\text{2s}}^{\text{*2}}{\text{σ}}_{{\text{2p}}_{\text{z}}}^{\text{2}}{\text{π}}_{{\text{2p}}_{\text{x}}}^{\text{2}}\text{=}{\text{π}}_{{\text{2p}}_{\text{y}}}^{\text{2}}$
${\text{π}}_{{\text{2p}}_{\text{x}}}^{\text{*1}}\text{=}{\text{π}}_{{\text{2p}}_{\text{y}}}^{\text{*0}}\Rightarrow \text{Bond}\text{order}\text{of}\text{NO}=\frac{1}{2}\left(10-5\right)=2.50$