Question
Write molecular orbital configuration of O2. Predict its magnetic behaviour and calculate its bond order.

Answer

O2(16),
Molecular orbital configuration:
$(\sigma\text{ls})^2(\sigma*\text{ls})^2(\sigma2\text{s})^2(\sigma*2\text{s})^2(\sigma2\text{P}_\text{z})^2(\pi2\text{p}_\text{x}^2=\pi2\text{p}_\text{y}^2)$
$(\pi*2\text{p}_\text{x}^1=\pi*2\text{p}^1_\text{y})$
It is paramagnetic in nature due to presence of two unpaired electrons.
$\text{B.O.}=\frac12(\text{N}_\text{b}-\text{N}_\text{a})=\frac12(10-6)=\frac42=2$

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