Answer:
Option C
Explanation:
Plan This problem can be solved by using the concept involved in molecular orbital theory . Write the molecular orbital electronic configuration keeping in mind that there is no 2s-2p mixing then if highest occupied molecular orbital contain unpaired electron then molecule is paramagnetic otherwise diamagnetic
Assuming that no 2s-2p mixing takes place the molecular orbital electronic configuration can be written in the following sequence of energy levels of molecular orbitals
$\sigma 1s,\sigma*1s,\sigma2s,\sigma*2s, \sigma 2p_{z},\pi2p_{x}\equiv\pi2p_{y},\pi *2p_{x}\equiv\pi*2p_{y},\sigma*2p_{z}$
(a) $Be_{2}\rightarrow \sigma 1s^{2},\sigma*1s^{2},\sigma2s^{2},\sigma*2s^{2}$ (diamagnetic)
(b) $B_{2}\rightarrow \sigma 1s^{2},\sigma*1s^{2},\sigma2s^{2},\sigma*2s^{2},\sigma 2p_z^2 ,_{\pi 2p_y^0 } ^{ \pi 2p_x^0}$
(diamagnetic)
(c) $C_{2}\rightarrow \sigma 1s^{2},\sigma*1s^{2},\sigma2s^{2},\sigma*2s^{2}, \sigma 2p_z^2, _{\pi 2p_y^1 } ^{ \pi 2p_x^1},$
$_{\pi*2p_y^0}^{\pi*2p_x^0} ,\sigma*2p_z^0 (paramagnetic) $
(d) $N_{2}\rightarrow \sigma 1s^{2},\sigma*1s^{2},\sigma2s^{2},\sigma*2s^{2}, \sigma2p_z^2,_{\pi 2p_y^2}^{\pi2p_x^2},$
$_{\pi*2p_y^0}^{\pi*2p_x^0} ,\sigma*2p_z^0 (diamagnetic) $
Hence (c) is the correct choice
