Which molecule/ion out of the following does not contain unpaired electrons?

The correct option is A $O_2^{2-}$

Energy order for homo-diatomic molecules.

If the number of electrons are less than or equal to $14$

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

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

If the number of electrons are greater than $14$

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

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

Molecular orbital configuration for molecules given in options.

$\left(A\right)$ Number of electrons in $N_2^{+}=13(<14)$

$\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\pi 2px{)}^2\right(\pi 2py{)}^2(\sigma 2pz{)}^1$

Number of unpaired electrons $=1$

$\left(B\right)$ Number of electrons in $O_2=16(>14)\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2$

${\pi }^{\ast }2p_x^1{\pi }^{\ast }2p_y^1$

Number of unpaired electrons $=2$

$\left(C\right)$ Number of electrons in $O_2^{2-}=18(>14)$

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

${\pi }^{\ast }2p_x^2{\pi }^{\ast }2p_y^2$

Number of unpaired electrons $=0$

$\left(D\right)$ Number of electrons in $B_2=10(<14)$

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

Number of unpaired electrons $=2$

So, only $O_2^{2-}$does not have any unpaired electron.

Hence, the correct choice is $\left(C\right).$