Question

Oxygen molecule is:

• A. diamagnetic
• B. paramagnetic
• C. ferromagnetic
• D. ferromagnetic

Answer 1

Answer: (B)

According to MOT (molecular orbital theory) oxygen (at. no. $8$) has the following orbital configuration
$ O _{2}\left(16 e^{-}\right)=\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} $
$\left(\pi 2 p_{x}^{2} \pi 2 p_{y}^{2}\right) \pi^{*} 2 p_{x}^{1} \pi^{*} 2 p_{y}^{1}$
As oxygen has got two unpaired electrons, it is paramagnetic.
The sequence of energies of homonuclear diatomic molecule (except $O _{2}$ and $F _{2}$ ) is as follows
$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s$
$< \begin{bmatrix}\pi&2p_{x}\\ \pi&2p_{y}\end{bmatrix} < \sigma2p_{z} <\begin{bmatrix}\pi^{*}&2p_{x}\\ \pi^{*}&2p_{y}\end{bmatrix} < \sigma^{*} 2p_{z}$
For oxygen and fluorine
$\sigma 1 s<\sigma^{*} 1 s<\sigma 2 s<\sigma^{*} 2 s<\sigma 2 p_{z} $
$< \begin{bmatrix} \pi & 2 p_{x} \\\pi & 2 p_{y}\end{bmatrix} < \begin{bmatrix} \pi & 2 p_{x} \\ \pi & 2 p_{y} \end{bmatrix} < \sigma^{*} 2 p_{z}$