Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Number of${\mathrm{e}}^{-}$in$\mathrm{NO}=7+8=15$

$\mathrm{NO}=\mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma } }1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma } }2{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{P}}_{\mathrm{z}}^2<\mathrm{\pi }2{\mathrm{P}}_{\mathrm{y}}^2=\mathrm{\pi }2{\mathrm{Px}}^2<\stackrel{*}{\mathrm{\pi }}2{\mathrm{P}}_{\mathrm{y}}^1=\stackrel{*}{\mathrm{\pi } }2{\mathrm{P}}_{\mathrm{x}}^0$

Bond order$=\frac{\text{(number of electron in bonding molecular orbital − number of electron} \mathrm{in} \mathrm{anti} \mathrm{bonding} \mathrm{molecular} \mathrm{orbital})}{2}$

$\mathrm{BO}=\frac{6-1}{2}=2.5$.

Number of${\mathrm{e}}^{-}$in$\mathrm{CN}=6+7=13$

$\mathrm{NO}=\mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma } }1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma } }2{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{P}}_{\mathrm{z}}^2<\mathrm{\pi }2{\mathrm{P}}_{\mathrm{y}}^2=\mathrm{\pi }2{\mathrm{Px}}^2<\stackrel{*}{\mathrm{\pi }}2{\mathrm{P}}_{\mathrm{y}}^1=\stackrel{*}{\mathrm{\pi } }2{\mathrm{P}}_{\mathrm{x}}^0$.

Bond order$=\frac{5}{2}=2.5$

Number of${\mathrm{e}}^{-}$in${\mathrm{CN}}^{+}=6+7-1=12$

$\mathrm{NO}=\mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma } }1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma } }2{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{P}}_{\mathrm{z}}^2<\mathrm{\pi }2{\mathrm{P}}_{\mathrm{y}}^2=\mathrm{\pi }2{\mathrm{Px}}^2<\stackrel{*}{\mathrm{\pi }}2{\mathrm{P}}_{\mathrm{y}}^1=\stackrel{*}{\mathrm{\pi } }2{\mathrm{P}}_{\mathrm{x}}^0$

Bond order$=\frac{4-0}{2}=2$

${\mathrm{CN}}^{-}=$number of${\mathrm{e}}^{-}$in${\mathrm{CN}}^{-}=6+7+1=14$

${\mathrm{CN}}^{-}=\mathrm{\sigma }1{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma } }1{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{s}}^2<\stackrel{*}{\mathrm{\sigma } }2{\mathrm{s}}^2<\mathrm{\sigma }2{\mathrm{P}}_{\mathrm{z}}^2<\mathrm{\pi }2{\mathrm{P}}_{\mathrm{y}}^2=\mathrm{\pi }2{\mathrm{Px}}^2<\stackrel{*}{\mathrm{\pi }}2{\mathrm{P}}_{\mathrm{y}}^1=\stackrel{*}{\mathrm{\pi } }2{\mathrm{P}}_{\mathrm{x}}^0$

$\mathrm{Bond} \mathrm{order}=\frac{6-0}{2}=3$.

${\mathrm{CN}}^{-}$as highest bond order.