The bond lengths of C 2 ,   N 2 and B 2 molecules are X 1 , X 2 and X 3 pm respectively. The correct…

Bond length is inversely related to bond order. The bond orders of given molecules can be calculated using molecular orbital theory.

$\mathrm{bond} \mathrm{order} =\frac{\mathrm{Number} \mathrm{of} \mathrm{BMO} {\mathrm{e}}^{-}-\mathrm{Number} \mathrm{of} \mathrm{ABMO} {\mathrm{e}}^{-}}{2}$

BMO = Bonding molecular orbital

ABMO = Anti bonding molecular orbital

The molecular orbital electronic configuration of ${\mathrm{B}}_2$ is $\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^1=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^1$

$\mathrm{BO} \mathrm{of} {\mathrm{B}}_2=\frac{6-4}{2}=1$

The molecular orbital electronic configuration of ${\mathrm{N}}_2$ is $\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2\mathrm{\sigma }2{{\mathrm{p}}_{\mathrm{z}}}^2$

$\mathrm{BO} \mathrm{of} {\mathrm{N}}_2=\frac{10-4}{2}=3$

The molecular orbital electronic configuration of ${\mathrm{C}}_2$ is $\mathrm{\sigma }1{\mathrm{s}}^2{\mathrm{\sigma }}^{*}1{\mathrm{s}}^2\mathrm{\sigma }2{\mathrm{s}}^2{\mathrm{\sigma }}^{*}2{\mathrm{s}}^2\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{x}}}^2=\mathrm{\pi }2{{\mathrm{p}}_{\mathrm{y}}}^2$

$\mathrm{BO} \mathrm{of} {\mathrm{C}}_2=\frac{8-4}{2}=2$

Hence, the order of their bond length is

${\mathrm{X}}_3\left({\mathrm{B}}_2\right)>{\mathrm{X}}_1\left({\mathrm{C}}_2\right)>{\mathrm{X}}_2\left({\mathrm{N}}_2\right)$