Question

State True or False:
${O_2},\;O_2^{ - \;\;} and \;O_2^{2 - }$ are paramagnetic species.
A. True
B. False

Answer 1

To know whether the species are paramagnetic or not, a molecular orbital diagram is required to be drawn with 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 step by step answer:
The diagram contains bonding and non-bonding orbitals, and it is indicated by the sigma and sigma star or pi and pi star. The electrons get filled in each orbital as shown in the configuration and it follows Aufbau, Pauli and Hund’s rule for writing electronic configuration. The negative ions mean the electrons are accepted and the number of electrons increases.
If unpaired electrons are present in an ion/molecule, then it is paramagnetic. If all electrons are paired, then the ion/molecule is diamagnetic.

The MO electronic configurations of oxygen, oxide, and peroxide are given as follows:

\Rightarrow {O_2}:KK{\left( {\sigma 2s} \right)^2}{(\sigma * 2s)^2}{\left( {\sigma 2px} \right)^2}{\left( {\pi 2py} \right)^2}{\left( {\pi 2pz} \right)^2}{(\pi * 2py)^1}{(\pi * 2pz)^1}\;$$ 2 unpaired electrons. $$\Rightarrow O_2^ - :{\rm{ }}KK{\left( {\sigma 2s} \right)^2}{(\sigma * 2s)^2}{\left( {\sigma 2px} \right)^2}{\left( {\pi 2py} \right)^2}{\left( {\pi 2pz} \right)^2}{(\pi * 2py)^2}{(\pi * 2pz)^1}\;$$ 1 unpaired electron.

\Rightarrow{O_2^{2 - }:KK{{\left( {\sigma 2s} \right)}^2}{{(\sigma * 2s)}^2}{{\left( {\sigma 2px} \right)}^2}{{\left( {\pi 2py} \right)}^2}{{\left( {\pi 2pz} \right)}^2}{{(\pi * 2py)}^2}(\pi * 2pz){}^2;}$0 unpaired electron. The given statement is False.${O_2},;O_2^{ - ;;}$are paramagnetic while$O_2^{2-}$$ is diamagnetic.

Hence, the correct option is (B).

Note: The MOT diagram is very useful to know bond order, bond length and magnetic property (paramagnetic or diamagnetic character) of atoms combined to form molecules.