Question

Which of the ${O_2}^ - $, ${O_2}^{ - 2}$, $C{N^ - }$ and ${N_2}$ is/are paramagnetic?
(A) ${N_2}$
(B) ${O_2}^{ - 2}$
(C) $C{N^ - }$
(D) ${O_2}^ - $

Answer 1

Hint: Here we will be using the concept of Molecular Orbital Theory (MOT). So, using this theory write the electronic configuration of all these compounds and see which of them has unpaired electrons.
(Paramagnetic are the ones with unpaired electrons and diamagnetic are the ones with all electrons paired)

Complete answer:
-First of all let us see what Molecular Orbital Theory (MOT) is.
When the Valence Bond Theory failed to explain the bonding pattern of certain complex
molecules or how certain molecules contain two or more equivalent bonds like the bonds of resonance stabilised molecules, the MOT came in its place. It helps to explain such complex bonding between atoms and to predict the distribution of electrons.
This in turn helps in predicting the molecular properties like bond order, magnetism, shape, etc.
According to this theory:
The atomic orbitals combine to form molecular orbitals. These molecular orbitals are of 2 types: bonding and antibonding molecular orbitals. Electrons prefer staying in the molecular orbitals since the molecular bonds have lower potential energy in atomic orbitals.
The order of increasing energy of molecular orbitals according to this theory is:
 $\sigma 1s,{\sigma ^*}1s,\sigma 2s,{\sigma ^*}2s,\sigma 2{p_z},\pi 2{p_x} = \pi 2{p_y},{\pi ^*}2{p_x} = {\pi ^*}2{p_y},{\sigma ^*}2{p_z}$ and so on.
Now let’s start writing the configurations of various given compounds using the above given energy sequence of molecular orbitals.
For ${N_2}$ : total number of electrons = 14
Configuration: ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}{(\sigma 2{p_z})^2}$
Here there are no unpaired electrons and so ${N_2}$ is not paramagnetic. It is diamagnetic,

 -For ${O_2}^{ - 2}$ : total number of electrons = 18 Configuration: ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\sigma 2{p_z})^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}{({\pi ^*}2{p_x})^2}{({\pi ^*}2{p_y})^2}$
Here also there are no unpaired electrons and so ${O_2}^{ - 2}$ is also diamagnetic.
  -For $C{N^ - }$ : total number of electrons = 14
Configuration: ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\sigma 2{p_z})^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}$
 Here also we can see that there are no unpaired electrons and so $C{N^ - }$ is diamagnetic.
 -For ${O_2}^ - $ : total number of electrons = 17
 Configuration: ${(\sigma 1s)^2}{({\sigma ^*}1s)^2}{(\sigma 2s)^2}{({\sigma ^*}2s)^2}{(\sigma 2{p_z})^2}{(\pi 2{p_x})^2}{(\pi 2{p_y})^2}{({\pi ^*}2{p_x})^2}{({\pi ^*}2{p_y})^1}$
Here we can see that there is 1 unpaired electron in the last antibonding molecular orbital. So, we can now say that ${O_2}^ - $ is paramagnetic in nature.
-So, the correct option is : (D)${O_2}^ - $ .

Note: The most delicate part of this question is writing the electronic configuration. So, while doing so always keep in mind to fill the electrons according to the increasing energy level of the molecular orbitals.