Ans: (A)
Hint : \(C_{2}: \sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2^{2} s \cdot 2 s^{2} \quad \pi 2 p^{2}=\pi 2^{2}{p_{y}}\)
There is no unpaired electron.
\(\mathrm{C}_{2}\) is diamagnetic in nature
\(\mathrm{B}_{2}: \sigma 1 \mathrm{~s}^{2} \sigma^{*} 1 \mathrm{~s}^{2} \quad \sigma 2 \mathrm{~s}^{2} \sigma^{*} 2 \mathrm{~s}^{2} \quad \pi 2 \mathrm{p}_{\mathrm{x}}^{1}=\pi 2 \mathrm{p}_{\mathrm{y}}^{1}\)
There is two unpaired electrons. B2 is paramagnetic in nature
