Number of unpaired electrons in N²⁺

The correct option is 1.

To find the number of unpaired electrons in the diatomic nitrogen cation ion $N_2^{+}$, we can use Molecular Orbital (MO) theory.

First, let's find the total number of electrons in $N_2^{+}$:
A neutral nitrogen atom (N) has an atomic number of 7, meaning it has 7 electrons.
Therefore, a neutral $N_2$ molecule has:
$7+7=14$ electrons.
The $N_2^{+}$ ion has a $+1$ charge, which means it has lost 1 electron:
Total electrons in $N_2^{+}$ = $14-1=13$ electrons.

Next, we write the molecular orbital configuration for a system with 13 electrons (where total electrons ≤ 14):
The increasing order of energy for molecular orbitals of such homonuclear diatomic molecules is:
$\sigma 1s<{\sigma }^{*}1s<\sigma 2s<{\sigma }^{*}2s<(\pi 2p_x=\pi 2p_y)<\sigma 2p_z<({\pi }^{*}2p_x={\pi }^{*}2p_y)<{\sigma }^{*}2p_z$

Filling the 13 electrons into these orbitals according to Hund's rule and Pauli's exclusion principle:
$\sigma 1s^2,{\sigma }^{*}1s^2,\sigma 2s^2,{\sigma }^{*}2s^2,\pi 2p_x^2=\pi 2p_y^2,\sigma 2p_z^1$

Let's count the electrons filled so far:
2 (in $\sigma 1s$) + 2 (in ${\sigma }^{*}1s$) + 2 (in $\sigma 2s$) + 2 (in ${\sigma }^{*}2s$) + 4 (in $\pi 2p_x$ and $\pi 2p_y$) + 1 (in $\sigma 2p_z$) = 13 electrons.

From the configuration above, we can see that all the inner molecular orbitals are fully filled with paired electrons, while the highest occupied molecular orbital ($\sigma 2p_z$) contains only 1 electron.

Therefore, there is exactly 1 unpaired electron in $N_2^{+}$.