Question

Which of the following is paramagnetic?
A) ${B_2}$
B) $CO$
C) ${O_2}^{2 - }$
D) $N{O^ + }$

Answer 1

This question is based on the molecular orbital theory. In order to identify the paramagnetic nature of the given compounds, we need to check the molecular orbital configuration. If all the orbitals are fully occupied, then the compound is diamagnetic and if there are unpaired electrons left then the compound is paramagnetic.

Complete step by step solution:
Given to us are four compounds whose magnetic properties have to be found out. In order to do that we calculate the total number of electrons present in each compound and write its molecular orbital configuration. For these compounds to be paramagnetic, they have to have one or more unpaired electrons.
Let us write the orbital configurations for each compound.
A) Given compound is ${B_2}$
Each Boron atom contains five electrons so the total number of electron in this compound would be $10$
Now, we write its orbital configuration as $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2{p_x}^1 = \pi 2{p_y}^1$
There are two unpaired electrons present in this compound and hence it is paramagnetic.
B) Similarly, the total number of electrons present in $CO$ are $6 + 8 = 14$
We write the configuration as $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2{p_x}^2 = \pi 2{p_y}^2,\sigma 2{p_z}^2$
There are no unpaired electrons and hence $CO$ is not paramagnetic.
C) Total number of electrons present in ${O_2}^{2 - }$ are $8 + 8 + 2 = 20$
We can write its configuration as $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2{p_x}^2 = \pi 2{p_y}^2,\sigma 2{p_z}^2,{\pi ^*}2{p_x}^2 = {\pi ^*}2{p_y}^2$ and hence it is not paramagnetic.
D) Total number of electrons present in $N{O^ + }$ are $7 + 8 - 1 = 14$
We write its configuration as $\sigma 1{s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\pi 2{p_x}^2 = \pi 2{p_y}^2,\sigma 2{p_z}^2$ and hence it is not paramagnetic.

Therefore the paramagnetic compound is ${B_2}$ i.e. option A.

Note: It is to be noted that when writing the molecular orbital configuration of any compound, the electrons in the p orbital start pairing only after both ${p_x}$ and ${p_y}$ have one orbital each which is why in the configuration of ${B_2}$ the electrons remain unpaired contributing to its paramagnetic nature.