Solveeit Logo
Login

Question

Question: \({K_2}O\,\) is paramagnetic while \(K{O_2}\) and \({K_2}{O_2}\) are diamagnetic. Answer whether ...

K2O {K_2}O\, is paramagnetic while KO2K{O_2} and K2O2{K_2}{O_2} are diamagnetic.
Answer whether the above statement is true or false. If true enter 11 , else 00.

Explanation

Solution

The compounds having unpaired electrons in their valence orbitals show paramagnetic behavior and the compounds with paired electrons in their valence orbitals are diamagnetic in nature. One can write the configuration for each structure to determine the number of unpaired electrons and their nature.

Complete step by step answer: 1) We can observe that in all the compounds K2O, KO2{K_2}O,\,K{O_2} and K2O2{K_2}{O_2} the oxidation state of potassium i.e. (K)\left( K \right) is +1 + 1 but oxygen has varying oxidation states. We can write the molecular orbital (MO) configuration of oxygen in all the three given molecules on the basis of molecular orbital theory taking into account that oxygen has eight electrons.
2) First let's analyze the structure K2O {K_2}O\, where oxygen is present O2−{O^{2 - }}. The number of electrons present on an oxygen atom is 1010 where eight from oxygen and two from the negative charge on oxygen. Since it is an atom and its atomic configuration can be written as:
1s2,2s2,2px2,2py2,2pz21{s^2},2{s^2},2{p_x}^2,2{p_y}^2,2{p_z}^2
After analyzing the electronic configuration of the oxygen the electrons present in each orbital are paired hence it is diamagnetic.
3) In the structure KO2K{O_2} Oxygen is present in O2−1O_2^{ - 1}. The number of electrons present is 1717 where the sixteen electrons are from two oxygen atoms and one from the negative charge. Its MO configuration can be written as:
(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(π∗2x)2(π∗2py)1{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^ * }1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^ * }2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {{\pi ^ * }{2_x}} \right)^2}{\left( {{\pi ^ * }2{p_y}} \right)^1}
After analyzing the MO configuration of the oxygen there is one unpaired electron present in the last orbital hence it shows paramagnetic nature.
4) In the structure K2O2{K_2}{O_2} Oxygen is present in O22−O_2^{2 - }. The number of electrons present is 1818 where the sixteen from two oxygen atoms and two from the negative charge. Its MO configuration can be written as:
(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)2(π∗2py)2{\left( {\sigma 1s} \right)^2}{\left( {\sigma * 1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {\sigma * 2s} \right)^2}{\left( {\sigma 2{p_z}} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi * 2{p_x}} \right)^2}{\left( {\pi * 2{p_y}} \right)^2}
After analyzing the MO configuration of the oxygen there is no unpaired electron present hence it shows diamagnetic nature.
5) As it is clear from the above-written configurations oxygen atoms only KO2K{O_2} have unpaired electrons. Thus, it is paramagnetic in nature. We can say that the statement K2O {K_2}O\, is paramagnetic while KO2K{O_2} and K2O2{K_2}{O_2} are diamagnetic is false.

Note: Molecular orbital theory (written as MOT sometimes) is used to predict the electronic structures of molecules and their bond orders as well as magnetic properties. The paramagnetic substance shows a form of magnetism in which it is weakly attracted by an external magnetic field and then forms internal induced magnetic fields in that same direction. In simple words, the paramagnetic substances show the induced magnetism when they are in other substance's magnetic fields. In the case of diamagnetic materials which get repelled by a magnetic field due to unpaired electrons.