Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Key Idea Bond order $=\frac{{\mathrm{N}}_{\mathrm{b}}-{\mathrm{N}}_{\mathrm{a}}}{2}$ where, ${\mathrm{N}}_{\mathrm{b}}=$ number of electrons in bonding $MO$ ${\mathrm{N}}_{\mathrm{a}}=$ number of electrons in anti bonding $MO$ $$\begin{array}{rl}& {\mathrm{N}}_2(7+7=14)=\sigma 1{\mathrm{s}}^2,\dot{\ast }1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\stackrel{\dot{\ast }}{\sigma }2{\mathrm{s}}^2,\pi 2{\mathrm{p}}_{\mathrm{x}}^2\\ & \approx \pi 2{\mathrm{p}}_{\mathrm{y}}^2,\sigma 2{\mathrm{p}}_{\mathrm{z}}^2\end{array}$$ $$\begin{array}{rl}& \mathrm{BO}=\frac{10-4}{2}=3\\ & {\mathrm{N}}_2(7+7+1=15)\\ & =\sigma 1{\mathrm{s}}^2,\dot{\ast }1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\\ & \stackrel{\ast }{\sigma }2{\mathrm{s}}^2,\sigma 2{\mathrm{p}}_z^2,\pi 2{\mathrm{p}}_{\mathrm{x}}^2\approx \pi 2{\mathrm{p}}_{\mathrm{y}}^2,\pi 2{\mathrm{p}}_{\mathrm{x}}^1\end{array}$$ $$\begin{array}{rl}& \mathrm{BO}=\frac{10-5}{2}=2.5\\ & {\mathrm{N}}_2^{2-}(7+7+2=16)\end{array}$$ $\begin{array}{rl}& =\sigma 1{\mathrm{s}}^2,\star 1{\mathrm{s}}^2,\sigma 2{\mathrm{s}}^2,\star 2{\mathrm{s}}^2,\sigma 2{\mathrm{p}}_{\mathrm{z}}^2,\pi 2{\mathrm{p}}_{\mathrm{z}}^2\\ & \approx \pi 2{\mathrm{p}}_{\mathrm{y}}^2,\pi 2{\mathrm{p}}_{\mathrm{x}}^1\approx \pi 2{\mathrm{p}}_{\mathrm{y}}^1\\ & \end{array}$ $$\mathrm{BO}=\frac{10-6}{2}=2$$ Hence, the increasing order of B.O is, $${\mathrm{N}}_2^{2-}<{\mathrm{N}}_2^{-}<{\mathrm{N}}_2$$