The energy of $\sigma 2p_z$ molecular orbital is greater than $\pi 2p_x$ and $\pi 2p_y$ 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 : $N_2, N_2^+, N_2^-, N_2^{2+}$

(N/A) The electronic configuration of $N$ atom $(Z=7)$ is $1s^2 2s^2 2p_x^1 2p_y^1 2p_z^1$. Total number of electrons in $N_2$ is $14$. The increasing order of energy levels for $N_2$ is: $\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$.
$(i)$ $N_2$ ($14$ $e^-$): Configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\sigma 2p_z)^2$. Bond Order $(BO)$ = $\frac{1}{2}(10-4) = 3$. Diamagnetic (no unpaired electrons).
$(ii)$ $N_2^+$ ($13$ $e^-$): Configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\sigma 2p_z)^1$. $BO = \frac{1}{2}(9-4) = 2.5$. Paramagnetic (one unpaired electron).
$(iii)$ $N_2^-$ ($15$ $e^-$): Configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\sigma 2p_z)^2, (\pi^* 2p_x)^1$. $BO = \frac{1}{2}(10-5) = 2.5$. Paramagnetic (one unpaired electron).
$(iv)$ $N_2^{2+}$ ($12$ $e^-$): Configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\pi 2p_x)^2, (\pi 2p_y)^2$. $BO = \frac{1}{2}(8-4) = 2$. Diamagnetic (no unpaired electrons).
Stability order based on $BO$: $N_2 > N_2^+ \approx N_2^- > N_2^{2+}$.