Question

Which of the following order of energies of molucular orbitals of $N_2$ is correct?

• A. $\left(\pi 2p_y\right)<\left(\sigma 2p_z\right)>\left({\pi }^{\ast }2p_x\right)\approx \left({\pi }^{\ast }2p_y\right)$
• B. $\left(\pi 2p_y\right)>\left(\sigma 2p_z\right)>\left({\pi }^{\ast }2p_x\right)\approx \left({\pi }^{\ast }2p_y\right)$
• C. $\left(\pi 2p_y\right)<\left(\sigma 2p_z\right)>\left({\pi }^{\ast }2p_x\right)\approx \left({\pi }^{\ast }2p_y\right)$
• D. $\left(\pi 2p_y\right)<\left(\sigma 2p_z\right)<\left({\pi }^{\ast }2p_x\right)\approx \left({\pi }^{\ast }2p_y\right)$

Answer 1

The correct option is D $\left(\pi 2p_y\right)<\left(\sigma 2p_z\right)<\left({\pi }^{\ast }2p_x\right)\approx \left({\pi }^{\ast }2p_y\right)$

Solution:

If the number of electrons in the molecule $\le$ 14, then increasing energy level is :

$\sigma 1s<\sigma {}^{\ast }1s<\sigma 2s<\sigma {}^{\ast }2s<\pi 2p_x=\pi 2p_y<\sigma 2p_z<\pi {}^{\ast }2p_x={\pi }^{\ast }2p_y<\sigma {}^{\ast }2p_z$

As $N_2$ having $14$ electrons, the order of energies is according to option $\left(A\right)$.