Solveeit Logo
Login

Question

Question: Why \(K{O_2}\) is paramagnetic?...

Why KO2K{O_2} is paramagnetic?

Explanation

Solution

We know that the substances with odd numbers of electrons are paramagnetic and the substances with an even number of electrons are diamagnetic. Before that we have to write molecular orbital configuration for KO2K{O_2} .

Complete step by step answer:
We know that O2−{O_2}^ - is superoxide. First, calculate the total number of electrons in the anion.
We must need to remember that the total number of electrons in O2−{O_2}^ - is 1717 .
We know that for the elements which have more than 14 electrons the electrons enter the molecular energy level as follows,
KK<(σ2s)<(σ∗2s)<(σ2px)<(π2py)=(π2pz)<(π∗2py)(π∗2pz)<(σ2px)KK < \left( {\sigma 2s} \right) < \left( {{\sigma ^*}2s} \right) < \left( {\sigma 2{p_x}} \right) < \left( {\pi 2{p_y}} \right) = \left( {\pi 2{p_z}} \right) < \left( {\pi *2{p_y}} \right)\left( {\pi *2{p_z}} \right) < \left( {\sigma 2{p_x}} \right)
Write the molecular orbital configuration of superoxide anion,
We can write the molecular orbital configuration of O2−{O_2}^ - as,
KK(σ2s)2(σ∗2s)2(σ2px)2(π2py)2(π2pz)2(π∗2py)2(π∗2pz)1KK{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\sigma 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {\pi *2{p_y}} \right)^2}{\left( {\pi *2{p_z}} \right)^1}
We know that the substances with an odd number of electrons are paramagnetic.
From molecular orbital configuration, it is clear that Potassium superoxide is a paramagnetic compound since it has one electron in its antibonding orbital.

Note:
By drawing the Lewis structure of a molecule we can determine the number of unpaired electrons in the molecule.
We know that the Lewis structure is the simple representation of the number of valence electrons in a molecule. To draw the Lewis structures of a molecule first calculate the total number of valence electrons in the atom.
To calculate the valence electrons add the group number of the element and charge on the atom. The valence electron in the oxygen atom is 6. Thus the number of valence electrons is 13(2×6+1).13\left( {2 \times 6 + 1} \right).
We need to remember that the superoxide anion has an unpaired electron. Thus it is a paramagnetic compound.