Find total electrons in $N_2^+$

Each nitrogen atom has atomic number $7$, so neutral $N_2$ has $7+7=14$ electrons.

Since $N_2^+$ is a cation with one positive charge, it has lost one electron: $14-1=13$ electrons.

Write the molecular orbital order for $N_2$-type molecules

For diatomic molecules up to nitrogen, the MO energy order is: $\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^* 2p_x = \pi^* 2p_y < \sigma^* 2p_z$

Fill 13 electrons in this order

Filling electrons one by one:

• $\sigma 1s^2$
• $\sigma^{*}1s^2$
• $\sigma 2s^2$
• $\sigma^{*}2s^2$
• $\pi 2p_x^2, \pi 2p_y^2$
  together gives $4$ electrons in $\pi 2p$

Total used so far: $2+2+2+2+4=12$

One electron remains, and it goes into the next orbital, which is $\sigma 2p_z$.

So configuration near the valence level is: $\pi 2p_x^2\,\pi 2p_y^2\,\sigma 2p_z^1$

Count electrons in $\sigma 2p_z$

Hence, the number of electrons in the $\sigma 2p_z$ molecular orbital is: $1$

Match with options

• A: $0$
• B: $1$ ✅
• C: $2$
• D: $3$

Therefore, the correct option is B.