\(\mathrm{O}_{2}\)
Total electrons \(=8+8=16\)
Electronic configuration
$$
\begin{aligned}
&8^{0}=1 s^{2}, 2 s^{2} 2 p^{4} \\
&8^{0}=1 s^{2}, 2 s^{2} 2 p^{4}
\end{aligned}
$$
Molecular orbital configuration,
$$
\begin{aligned}
&\left(\sigma^{b} 1 s\right)^{2} \cdot\left(\sigma^{\prime} 1 s\right)^{2},\left(\sigma^{b} 2 s\right)^{2} \cdot\left(\sigma^{*} 2 s\right)^{2},\left(\sigma^{\prime} 2 \mathrm{px}\right)^{2}, \\
&\left(\pi^{3} 2 p y\right)^{2}\left(\pi^{b} 2 p z\right)^{2},\left(\pi^{2} 2 p y\right)^{\prime}\left(\pi^{\prime} 2 p z\right)^{l},\left(\sigma^{\prime} 2 p x\right)^{0}
\end{aligned}
$$
Since, there are two unpaired electrons in molecular orbitals of \(\mathrm{O}_{2}\) molecule, so it is paramagnetic in nature.
