The correct order of energies of the molecular orbitals of the $\mathrm{N}{}_2$ molecule is:

Answer: (A) $\sigma 1s<{\sigma }^{\ast }1s<\sigma 2s<{\sigma }^{\ast }2s<(\pi 2p_x=\pi 2p_y)<\sigma 2p_z<({\pi }^{\ast }2p_x={\pi }^{\ast }2p_y)<{\sigma }^{\ast }2p_z$. \textbf{Answer:} (A) $\sigma 1s<{\sigma }^{\ast }1s<\sigma 2s<{\sigma }^{\ast }2s<(\pi 2p_x=\pi 2p_y)<\sigma 2p_z<({\pi }^{\ast }2p_x={\pi }^{\ast }2p_y)<{\sigma }^{\ast }2p_z$ \textbf{Solution:} For molecules with $14$ or fewer electrons ($\mathrm{N}{}_2$), the $\pi 2p$ orbitals lie below $\sigma 2p_z$, giving the order $\dots {\sigma }^{\ast }2s<(\pi 2p_x=\pi 2p_y)<\sigma 2p_z<({\pi }^{\ast }2p_x={\pi }^{\ast }2p_y)<{\sigma }^{\ast }2p_z$.