${{\rm{N}}_2},{{\rm{O}}_2}{\rm{,O}}_2^ + $ અને ${\rm{O}}_2^ - $ ના બંધક્રમાંક ગણો.

In $\mathrm{N}_{2} \mathrm{Z}=7$, so total electron $=14$

Electron configuration of in $MO$ for $\mathrm{N}_{2}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\sigma 2 p_{z}\right)^{2}$ $\mathrm{BO}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(10-4)=3$ (Triple bond)

Calculate the bond order of $\mathrm{O}_{2}$

In $\mathrm{O}_{2}, \mathrm{Z}=8 \mathrm{So}$, Total electron $=16$

Electron configuration in $MO$ for $\mathrm{O}_{2}:(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}$

$\left(\pi^{*} 2 p_{x}\right)^{1}$ $\left(\pi^{*} 2 p_{y}\right)^{1}$

$\mathrm{BO}=\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right)=\frac{1}{2}(10-6)=($ Double bond $)$

Total electron in $\mathrm{O}_{2}=16$ and total electron in

$\mathrm{O}_{2}^{+}=15$

Electron configuration in $MO$ for $\mathrm{O}_{2}:\left(\sigma_{1 s}\right)^{2}\left(\sigma_{1 s}^{*}\right)^{2}\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p_{z}}\right)^{2}\left(\pi_{2 p_{x}}\right)^{2}\left(\pi_{2 p_{y}}\right)^{2}\left(\pi_{2 p_{x}}^{*}\right)^{1}$

$\left(\pi_{2 p_{y}}^{*}\right)^{0}$ BO $=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(10-5)=2.5$

Calculate the bond order of $\mathrm{O}_{2}^{-}$

Total electron in $\mathrm{O}_{2}^{-}=16+1=17$

Electron configuration in $MO$ for $\mathrm{O}_{2}^{-}:\left(\sigma_{1 s}\right)^{2}\left(\sigma_{1 s}^{*}\right)^{2}\left(\sigma_{2 s}\right)^{2}\left(\sigma_{2 s}^{*}\right)^{2}\left(\sigma_{2 p_{z}}\right)^{2}\left(\pi_{2 p_{x}}\right)^{2}\left(\pi_{2 p_{y}}\right)^{2}\left(\pi_{2 p_{x}}^{*}\right)^{2}$

$\left(\pi_{2 p_{y}}^{*}\right)^{1}$

$\begin{aligned} \mathrm{BO} =\frac{1}{2}\left(\mathrm{~N}_{\mathrm{b}}-\mathrm{N}_{\mathrm{a}}\right) \\ =\frac{1}{2}(10-7)=\frac{3}{2}=1.5 \end{aligned}$