Solution:
Paramagnetic species contains unpaired electrons in their molecular orbital electronic configuration.
Molecular orbital configuration of the given species is as
$$
\begin{aligned}
&C O(6+8=14)=\sigma 1 s_{, \sigma}^{2} 1 s^{2} \\
&\sigma 2 s^{2}, \sigma_{\sigma} 2 s^{2}, \pi 2 P_{X}^{2} \approx \pi 2 P_{Y}^{2}, \sigma 2 P_{Z}^{2}
\end{aligned}
$$
(All the electrons are paired so, it is diamagnetic).
$$
\begin{gathered}
O_{2}^{-}(8+8+1=17) \\
=\sigma 1 s^{2} * 1 s^{2}, \sigma 2 s^{2}{ }_{* \sigma}^{*} 2 s^{2}, \sigma 2 P_{Z}^{2}, \pi 2 P_{X}^{2} \approx \pi 2 P_{Y}^{2} \\
\underset{\pi}{*} 2 P_{X}^{2} \approx_{\pi}^{*} 2 P_{Y}^{1}
\end{gathered}
$$
(It contains one unpaired electron so, it is paramagnetic.)
\(C N-(6+7+1=14)=\) same as \(\mathrm{CO}\)
\(\mathrm{NO}^{+}(7+8-1)=\) same as \(\mathrm{CO}\)
Thus, among the given species only \(\mathrm{O}_{2}^{-}\)is paramagnetic.
