Question

Why is ${O_2}$ paramagnetic?

Answer 1

Hint: To attend this question, we have to recollect the molecular orbital theory. We can use the molecular orbital diagram of oxygen molecules to explain the paramagnetic behaviour of oxygen atoms. We also have to know that for the paramagnetic nature of a molecule, it should contain a minimum of one unpaired electron.

Complete answer:
The valence bond theory could not explain the paramagnetic nature of oxygen molecules. This is where the molecular orbital comes into picture. Based on valence bond theory, the electronic configuration of an atom of oxygen is $1{s^2}2{s^2}2{p^4}$. Valence bond theory explains that the bonds are formed by atomic orbital and because of this, electrons are paired at the course of overlapping and so, oxygen is a diamagnetic species. But experimentally, oxygen molecules are paramagnetic in nature and this is explained by molecular orbital theory.
Based on molecular orbital theory, orbitals are formed by overlapping atomic orbitals of oxygen atoms. We can write the electronic configuration as $\left( {\sigma 1{s^2}{\sigma ^*}1{s^2}} \right)\left( {\sigma 2{s^2}{\sigma ^*}2{s^2}} \right)\left( {\sigma 2{p_z}^2} \right)\left( {\pi 2{p_x}^2 = \pi 2{p_y}^2} \right)\left( {{\pi ^*}2{p_x}^1 = {\pi ^*}2{p_y}^1} \right)$.
From the above written electronic configuration, we can see that there are two electrons which go into two separate, pi degenerate orbitals. Due to the presence of two unpaired electrons, we can say that the oxygen molecule is paramagnetic in nature.
The reason why oxygen is paramagnetic is because of the presence of two unpaired electrons.

Note:
We also have to remember that we can calculate the bond order of oxygen molecules using the electronic configuration. We can define bond order as half the difference between electrons that are found in bonding orbitals and the electrons that are found in antibonding molecular orbitals.