Question

If Pauli exclusion principle is not application and one orbital has three $e^{-}$ of $N_2$ molecule is present in:

• A. $\sigma \left(2s\right)$ orbital
• B. ${\sigma }^{\ast }\left(2s\right)$ orbital
• C. ${\pi }^{\ast }2p_y$ orbital`
• D. $\pi 2p_y$ orbital

Answer 1

Answer: (D)

$\pi 2p_y$ orbital

If Pauli exclusion principle is not present and one orbital can accommodate $3$ electrons, then the electronic configuration of nitrogen molecule will be $\sigma l{s}^3<\stackrel{\ast }{\sigma }(ls{)}^3<\sigma 2s^3<\stackrel{\ast }{\sigma }(2s{)}^3<\pi 2p{y}^2$.
Thus, the last electron enters in $\pi 2p_y$ orbital.