Question

Assuming (2s-2p) mixing is not operative, the paramagnetic species among the following is:
(A) $B{e_2}$
(B) ${B_2}$
(C) ${C_2}$
(D) ${N_2}$

Answer 1

Let us check the total number of electrons filled up in the bonding and antibonding orbitals. The species that are having unpaired electrons will thus be paramagnetic.

Complete step by step answer:
- Paramagnetism is a form of magnetism whereby some materials are weakly attracted by an externally applied magnetic field, and form internal, induced magnetic fields in the direction of the applied magnetic field.

-Thus, after the electronic configuration we can find out which species has unpaired electrons.
-Discussing option, A, we have $B{e_2}$ let us discuss its electronic configuration. In $B{e_2}$, we have 4 valence electrons.
$\sigma 2{S^2},{\sigma ^ * }2{S^2}$
Thus, there are no unpaired electrons that are available. Hence, $B{e_2}$ is not paramagnetic.
-Discussing option B, we have ${B_2}$ let us discuss its electronic configuration. In ${B_2}$, we have 6 valence electrons.
$\sigma 2{S^2},{\sigma ^ * }2{S^2},\sigma 2{P^2}$
Thus, there are no unpaired electrons available. Hence, ${B_2}$ is not paramagnetic.
-Discussing option C, we have ${C_2}$ let us discuss its electronic configuration. In ${C_2}$, we have 8 valence electrons.
$\sigma 2{S^2},{\sigma ^ * }2{S^2},\sigma 2{P^2},\pi 2{P^2}$
Thus, there are two unpaired electrons present. Hence, ${C_2}$ is paramagnetic.
-Discussing option D, we have ${N_2}$let us discuss its electronic configuration. In ${N_2}$, we have 10 valence electrons.
$\sigma 2{S^2},{\sigma ^ * }2{S^2},\sigma 2{P^2},\pi 2{P^4}$
Thus, there are no unpaired electrons again. Thus, ${N_2}$is not paramagnetic.

Clearly, the answer is option (C).

Note: Do not confuse paramagnetic with diamagnetic. Paramagnetism occurs when there are unpaired electrons in the compound. Diamagnetism means that the compound has no unpaired electrons, i.e. all electrons are completely paired.