Question

Arrange N₂⁺,N₂^- and N₂in the Increasing order of their bond length given that bond order is inversely proportional to bond length.

Answer 1

Orbital configuration of N₂
$\sigma 1s^2,\sigma *1s^2,\sigma 2s^2,\sigma *2s^2,\pi 2{p_y}^2\approx \pi 2{p_z}^2,\sigma 2{p_x}^2$

Orbital configuration of N₂⁺
$\sigma 1s^2,\sigma *1s^2,\sigma 2s^2,\sigma *2s^2,\pi 2{p_y}^2\approx \pi 2{p_z}^2,\sigma 2{p_x}^1$

Orbital configuration of N₂^-
$\sigma 1s^2,\sigma *1s^2,\sigma 2s^2,\sigma *2s^2,\pi 2{p_y}^2\approx \pi 2{p_z}^2,\sigma 2p_{x}2,\pi *2p{y}^1$

Bond order = $\frac{(no.ofelectronsinbondingorbitals-no.ofelectronsinantibondingorbitals)}{2}$

Bond order of N₂ = $\frac{(10-4)}{2}=3$

Bond order of N₂⁺ = $\frac{(9-4)}{2}=2.5$

Bond order of N₂^- = $\frac{(10-5)}{2}=2.5$

Increasing bond length-
N₂ < N₂⁺ $\approx$ N₂^-