The energy of $\sigma 2{{\rm{p}}_{\rm{z}}}$ molecular orbital is greater than $\pi 2{{\rm{p}}_{\rm{x}}}$ and $\pi 2{{\rm{p}}_{\rm{y}}}$ molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behaviour of the following species : ${{\rm{N}}_2},{\rm{N}}_2^ + ,{\rm{N}}_2^ - ,{\rm{N}}_2^{2 + },$

Electronic configuration of N-atom $(\mathrm{Z}=7)$ is $1 s^{2} 2 s^{2} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z^{\prime}}^{1}$. Total number of electrons present in $\mathrm{N}_{2}$ molecule is 14,7 from each

N-atom. From the view of various rules for filling of molecular orbitals, the electronic configuration of $\mathrm{N}_{2}$ molecule will be $\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_{x}^{2}, \approx \pi^{*} 2 p_{y}^{2}, \sigma 2 p_{z}^{2}$

Comparative study of the relative stability and the magnetic behaviour of the following species

$(i)$ $\mathrm{N}_{2}$ molecule :

$\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi p_{x}^{2}=\pi 2 p_{y^{\prime}}^{2} \sigma 2 p_{z}^{2}$

Here, $\mathrm{N}_{b}=10, \mathrm{~N}_{a}=4$.

Hence, Bond order $=\frac{1}{2}\left(\mathrm{~N}_{b}-\mathrm{N}_{a}\right)=\frac{1}{2}(10-4)=3$

Hence, presence of no unpaired electron indicates it to be diamagnetic.

$(ii)$ $N_{2}^{+}$ions :

$\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_{x^{\prime}}^{2}=\pi 2 p_{y^{\prime}}^{2}, \sigma 2 p_{z}^{1}$

Here, $\mathrm{N}_{b}=9, \mathrm{~N}_{a}=4$ so that $\mathrm{BO}=\frac{1}{2}(9-4)=\frac{5}{2}=2.5$

Further, as $\mathrm{N}_{2}^{+}$ion has one unpaired electron in the $\sigma\left(2 p_{2}\right)$ orbital, therefore, it is paramagnetic in nature.

$(iii)$ $\mathrm{N}_{2}^{-}$ions :

$\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \pi 2 p_{x}^{2}, \approx \pi 2 p_{y}^{2}, \sigma 2 p_{z}^{2}, \pi^{*} 2 p_{z}^{1}$

Here, $\mathrm{N}_{b}=10, \mathrm{~N}_{a}=5$ so that $\mathrm{BO}=\frac{1}{2}(10-5)=\frac{5}{2}=2.5$

Again, as it has one unpaired electron in the $\pi^{*}\left(2 p_{x}\right)$ orbital, therefore, it is paramagnetic.