Question

What of the following molecule is paramagnetic?
A) Fluorine
B) Oxygen
C) Nitrogen
D) Hydrogen

Answer 1

We have all learned about molecular orbital theory. It considers the wave nature of atoms and it is the first theory that says bond order can be fractional. To determine if the given molecule is paramagnetic or not, write the electronic configuration of the given molecule using molecular orbital theory.

Complete step-by-step answer:
As we know that particle (ion, atom, or molecule) can be paramagnetic or diamagnetic. If all the electrons in the orbitals are paired then it is called diamagnetic and if there are unpaired electrons, it is called paramagnetic. To determine if the molecule is paramagnetic or not, it is important that we know the electronic configuration of that molecule.
We also know that the molecular orbital theory considers the wave nature of atoms and it is the first theory that says bond order can be fractional. In this, atomic orbitals combine to form new orbitals that are bonding and antibonding orbitals. The number of molecular orbitals formed is equal to the number of atomic orbitals combining. Now filling of these molecular orbitals occurs according to certain rules.
Aufbau Principle: According to the Aufbau principle, orbitals are filled in order of increasing energy. This means the orbital with the lowest energy will be filled first.
$\sigma 1s < \sigma *1s < \sigma 2s < \sigma *2s < \left( {\Pi 2{p_x} = \Pi 2{p_y}} \right) < \sigma 2{p_z} < \left( {{\Pi ^*}2{p_x} = {\Pi ^*}2{p_y}} \right) < {\sigma ^*}2{p_z}$
Here there are two types of orbitals that are bonding and antibonding orbital. $\sigma ,\Pi$ are bonding molecular orbitals whereas $\sigma *,\Pi *$ are anti-bonding molecular orbitals.
Now filling electrons according to the rules we will get:
Fluorine: $\sigma 1{s^2},\sigma *1{s^2},\sigma 2{s^2},\sigma *2{s^2},\sigma 2p_x^2\left( {\Pi 2p_y^2 = \Pi 2p_z^2} \right),\left( {{\Pi ^*}2p_y^2 = {\Pi ^*}2p_z^2} \right),{\sigma ^*}2p_x^0$
Oxygen: $\sigma 1{s^2},\sigma *1{s^2},\sigma 2{s^2},\sigma *2{s^2},\sigma 2p_x^2\left( {\Pi 2p_y^2 = \Pi 2p_z^2} \right),\left( {{\Pi ^*}2p_y^1 = {\Pi ^*}2p_z^1} \right),{\sigma ^*}2p_x^0$
Nitrogen: $\sigma 1{s^2},\sigma *1{s^2},\sigma 2{s^2},\sigma *2{s^2},\sigma 2p_x^2\left( {\Pi 2p_y^2 = \Pi 2p_z^2} \right)$
Hydrogen: $\sigma 1{s^2},\sigma *1{s^0}$

Now, from the electronic configuration we can conclude that there are $2$ unpaired electrons present in the oxygen, hence the oxygen molecule is paramagnetic in nature. And the fluorine, nitrogen and hydrogen all possess the paired electrons and hence they are diamagnetic in nature.
Therefore, the correct answer is option (B).

Note: It is important to note that there are two types of orbitals that are bonding and antibonding orbital. The energy of both orbitals is different. The energy of bonding molecular orbital is lower than antibonding molecular orbital. So we must fill electrons according to the Aufbau principle.