Question

The paramagnetic property of the oxygen molecule is due to the presence of unpaired electron present in

• A. $\left(\sigma 2 p _{x}\right)^{1}$ and $\left(\sigma^{*} 2 p_{x}\right)^{1}$
• B. $\left(\sigma 2 p_{x}\right)^{1}$ and $\left(\pi 2 p_{y}\right)^{1}$
• C. $\left(\pi^{*} 2 p_{y}\right)^{1}$ and $\left(\pi^{*} 2 p_{x}\right)^{1}$
• D. $\left(\pi^{*} 2 p_{y}\right)^{1}$ and $\left(\pi 2 p_{y}\right)^{1}$

Answer 1

Answer: (C)

The electronic configuration of $O _{2}$ molecule on the basic of MOT is :
$(\sigma 1 s)^{2}\left(\sigma^{*} 1 s\right)^{2}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\sigma 2 p_{z}\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\pi^{*} 2 p_{x}\right)^{1}\left(\pi^{*} 2 p_{y}\right)^{1}$
Thus, due to the presence of an unpaired electron each in $\left(\pi^{*} 2 p_{y}\right)^{1}$ and $\left(\pi^{*} 2 p_{x}\right)^{1}, O_{2}$ molecule is paramagnetic.