Explanation:
The electronic configuration of the given species is as follows:
\(\mathrm{O}_{2}: \sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2},\left\{\begin{array}{l}{\pi 2 p_{y}^{2},} \\ {\pi 2 p_{z}^{2},}\end{array}\left\{\begin{array}{l}{\pi^{*} 2 p_{y}^{1}} \\ {\pi^{*} 2 p_{z}^{1}}\end{array}\right.\right.\)
\(\mathrm{B}_{2}: \sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2},\left\{\begin{array}{l}{\pi 2 p_{y}^{1}} \\ {\pi 2 p_{z}^{1}}\end{array}\right.\)
\(\mathrm{NO}: \sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2},\left\{\begin{array}{l}{\pi 2 p_{y}^{2}, } \\ {\pi 2 p_{z}^{2},}\end{array}\left\{\begin{array}{l}{\pi^{*} 2 p_{y}^{1}} \\ {\pi^{*} 2 p_{z}^{0}}\end{array}\right.\right.\)
\(CO\): No unpaired electron is present in the molecule, therefore, is not paramagnetic.
