Question

Which of the following is paramagnetic in nature ?
A.${N_2}$
B.${F_2}$
C.$K{O_2}$
D.None of these

Answer 1

This question is from molecular orbital theory where electrons are filled in bonding and anti bonding orbitals in increasing energy order . After filling electrons in orbitals if a compound contains any unpaired electron then it is paramagnetic and if there is no unpaired electron then the compound is diamagnetic .

Complete step by step answer:
During chemical bonding two types of forces come into nature one is attractive and second is repulsive . The force of attraction is between electrons and nuclei and force of repulsion is between nuclei of atoms . Due to these two forces two types of orbitals are formed during chemical bonding one is bonding orbital and second is anti bonding orbital . In these orbitals electrons are filled in increasing order of energy level in these bonding and anti bonding molecular orbitals .
The increasing energy order of orbitals is as follows :
$\sigma 1s < \sigma 1{s^ * } < \sigma 2s < \sigma 2{s^ * } < \sigma 2{p_x} < \pi 2{p_y} = \pi 2{p_z} < \pi 2{p_y}^ * = \pi 2{p_z}^ * < \sigma 2{p^ * }$ (for total numbers of electron greater than $14$ )
$\sigma 1s < \sigma 1{s^ * } < \sigma 2s < \sigma 2{s^ * } < \sigma 2{p_x} < \pi 2{p_y} = \pi 2{p_z} < \sigma 2{p^ * }_x < \pi 2{p_y}^ * = \pi 2{p_z}^ *$ (for total numbers of electron less than or equal to $14$ )
$*$: It is used for denoting anti bonding molecular orbital .
While filling these orbitals we should follow Hund’s rule , Afbau’s principle and Pauli’s exclusion principle .
Now after filling electrons in these molecular orbital in increasing energy order if there is any unpaired electron then the compound is called paramagnetic .
In ${N_2}$ there are total $14$ electrons if we fill these electrons for a total number of electrons $14$ we found that there are no unpaired electrons. So this is a diamagnetic compound .
In${F_2}$ total numbers of electrons is $18$ . After filling these electrons in increasing energy order again there is no unpaired electron .
In $K{O_2}$ $({K^ + }{O_2}^ - )$ , ${O_2}^ -$ has $17$ electrons , after filling these electrons in increasing order we find a unpaired electron in $\pi 2{P_y}^ *$ orbital .
So option (C) is the correct answer .

Note:
If the number of electrons in a bonding orbital is $a$ and that of anti bonding is $b$ then bond order is calculated by $\dfrac{{a - b}}{2}$ . If it is integer then the compound is diamagnetic else it is paramagnetic .