Question

Paramagnetism of oxygen is explained on the basis of its electronic configuration of:

• A. $\left(2 \pi_{\pi}^{*}\right)\left(2 \pi_{y}\right)^{1}$
• B. $\left(\pi^{*} 2 p_{y}\right)^{1}\left(\pi^{*} 2 p_{z}^{1}\right)$
• C. $\left(2 \sigma_{s}^{*}\right)^{1}\left(2 \pi_{y}\right)^{1}$
• D. $\left(2 \sigma_{s}^{*}\right)^{1}\left(2 \pi_{y}\right)^{1}$

Answer 1

Answer: (B)

Write configuration of oxygen according to molecular orbital theory
$O_{2}=16=\sigma 1 s^{2}, \sigma 1^{*} s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2} $
$\pi 2 p_{y}{ }^{2}, \pi 2 p_{z}{ }^{2}, \pi^{*} 2 p_{y}{ }^{1}, \pi^{*} 2 p_{z}{ }^{1}$
$\therefore $ It has two unpaired electrons.
$\therefore O _{2}$ is paramagnetic.