Question

The magnetic moment of $K{O_2}$ at room temperature is B.M.
A.1.41
B.1.73
C.2.23
D.2.64

Answer 1

Magnetic moment $= \mu = \sqrt n (n + 2)$ B.M. where, n = number of electrons present in the molecule. Here the value of n will be 1 because of the one unpaired electron in 2p molecular orbital.

Complete step by step answer:
Magnetic moments are often used in conjunction with electronic spectra to gain information about the oxidation state and stereochemistry of the central metal ion in coordination complexes.
In Potassium superoxide $K{O_2}$ , ${O^{2 - }}$ is superoxide ion. The superoxide $({O^{2 - }})$ is paramagnetic.
Electronic configuration of ${O^{2 - }}$ = $1{s^2}2{s^2}2{p^6}$
$K{O_2}$ for attaining 2 oxygen atoms, valency of k will be 2
Because of one unpaired electron $\left( {n = 1} \right)$ in 2p molecular orbital.
Magnetic moment $= \mu = \sqrt n (n + 2)$ B.M.
Where, n = number of unpaired electrons present in the molecule
$\to \mu = \surd 1\left( {1 + 2} \right) = \sqrt 3$ = 1.73 B.M.
The magnetic moment of $K{O_2}$ at room temperature is 1.73 B.M.

Therefore, the correct answer is option (B).

Note: $K{O_2}$ is a potent oxidizer, a yellow paramagnetic solid, if we leave the solid out in the air, it would soon decompose to a white solid due to Carbon dioxide or water vapour present in the air. It is a superoxide in which only one electron is released from the dioxygen atom and a superoxide ion is represented as ${O^{2 - }}$ . So, in $K{O_2}$ the oxygen atoms bear $\dfrac{{ - 1}}{2}$ oxidation state and they also behave as a free radical species, having an unpaired electron. $K{O_2}$ behaves as a paramagnetic molecule because of the presence of unpaired electrons or odd number of electrons in the anion