Question

Which of the following species is not paramagnetic?
A. ${\rm{CO}}$
B. ${{\rm{O}}_{\rm{2}}}$
C. ${{\rm{B}}_{\rm{2}}}$
D. ${\rm{NO}}$

Answer 1

Hint:To know whether the species are paramagnetic or not, molecular orbital theory is used to identify the electronic configuration. As we know the electron may remain unpaired during the formation of the bond between two atoms, such species are called paramagnetic species.

Complete solution
We know that the magnetic character is categorized into three parts. One of them is diamagnetic, second is paramagnetic and third is ferromagnetic. Generally, the magnetic property is affected by the existence of magnetic fields on either one side of the magnetic bar.
As we all know, the compound which is said to be paramagnetic commonly has one or more unpaired electrons. In paramagnetism form, the substance is weakly attached by the given magnetic bar.
So, in the option A that is carbon monoxide. The electronic configuration is \[{\rm{1}}{{\rm{\sigma }}^{\rm{2}}}{\rm{1\sigma }}{{\rm{*}}^{\rm{2}}}{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}{\rm{2\sigma }}{{\rm{*}}^{\rm{2}}}{\rm{1}}{{\rm{\pi }}^{\rm{4}}}{\rm{3}}{{\rm{\sigma }}^{\rm{2}}}\] and it does not have any unpaired electrons. So it usually does not show the paramagnetic character.
In option B that is oxygen. The electronic configuration is \[{\left( {{{\rm{\sigma }}_{{\rm{2s}}}}} \right)^{\rm{2}}}{\left( {{{\rm{\sigma }}^{\rm{*}}}_{{\rm{2s}}}} \right)^{\rm{2}}}{\left( {{{\rm{\sigma }}_{{\rm{2p}}}}} \right)^{\rm{2}}}{\left( {{{\rm{\pi }}_{{\rm{2p}}}}} \right)^{\rm{4}}}{\left( {{{\rm{\pi }}^{\rm{*}}}_{{\rm{2p}}}} \right)^{\rm{2}}}\] and it has two unpaired electrons which are present in pi antibonding molecular orbitals. So, it usually shows the paramagnetic character.
Similarly, in option C that is bromine it also has two unpaired electrons which are present in the pi antibonding molecular orbitals. So, it also shows the paramagnetic character.
Similarly, in option D that is nitrogen monoxide. The electronic configuration is \[({{\rm{\sigma }}_{{\rm{2s}}}}{{\rm{)}}^{\rm{2}}}{{\rm{(}}{{\rm{\sigma }}^{\rm{*}}}_{{\rm{2s}}}{\rm{)}}^{\rm{2}}}{{\rm{(}}{{\rm{\pi }}_{{\rm{2px,y}}}}{\rm{)}}^{\rm{4}}}{{\rm{(}}{{\rm{\sigma }}_{{\rm{2pz}}}}{\rm{)}}^{\rm{2}}}{{\rm{(}}{{\rm{\pi }}^{\rm{*}}}_{{\rm{2pxy}}}{\rm{)}}^{\rm{1}}}\] and it has one unpaired electron in the pi antibonding molecular orbital.
Thus, only CO does not fulfil the condition of paramagnetic character that is the compound must have unpaired electrons. So it is not a paramagnetic species.

Therefore, the correct option for this given question is A that is carbon monoxide (CO).

Note:
The paramagnetic molecules always have the colour because they have unpaired electrons and they require less amount of energy for electron transition.