\(\mathrm{H}_2-(\sigma l s)^2\)
\(\mathrm{H}_2\) does not have any unpaired electron hence diamagnetic in nature.
\(\mathrm{N}_2 \rightarrow K K^{\prime}(\sigma 2 \mathrm{~s})^2\left(\sigma^* 2 \mathrm{~s}\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\left(\sigma p_z\right)^2\)
\(\mathrm{N}_2\) does not have any unpaired electron hence diamagnetic in nature.
\(\begin{array}{r}
\mathrm{O}_2 \rightarrow K K^1(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2 \left(\pi^{\star} 2 p_x\right)^1\left(\pi^{\star} 2 p_y\right)^1
\end{array}\)
\(\mathrm{O}_2\) have two unpaired electrons hence paramagnetic in nature.
\(\mathrm{N}_2^{+} \rightarrow K K^1(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2\left(\pi 2 p_z\right)^1\)
\(\mathrm{N}_2\) have one unpaired electron, so it paramagnetic in nature.
\(\begin{aligned}
& \mathrm{O}_2^{+} \rightarrow K K^{\prime}(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2 \left(\pi_2^* p_x\right)^1\left(\pi^* 2 p_y\right)^0
\end{aligned}\)
\(\mathrm{O}_2^{+}\)has unpaired electron, so it is paramagnetic in nature.
\(\begin{array}{r}
\mathrm{O}_2^{-} \rightarrow K^{\prime}(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2 \left(\pi^* 2 p_x\right)^2\left(\pi^* 2 p_y\right)^1 \\
\mathrm{~F}_2 \rightarrow K^{\prime}(\sigma 2 s)^2\left(\sigma^* 2 s\right)^2\left(\sigma 2 p_z\right)^2\left(\pi 2 p_x\right)^2\left(\pi 2 p_y\right)^2 \left(\pi^* 2 p_x\right)^2\left(\pi^* 2 p_y\right)^2
\end{array}\)
\(\mathrm{F}_2\) does not have unpaired electron, hence diamagnetic in nature.
Hence, the molecules, \(\mathrm{N}_2^{+}, \mathrm{O}_2, \mathrm{O}_2^{+}, \mathrm{O}_2^{-}\)and \(\mathrm{H}_2, \mathrm{~N}_2, \mathrm{~F}_2\) are paramagnetic and diamagnetic respectively due to presence of unpaired and paired electrons. Hence, option (b) is the correct answer.