Answer:
Option A,B
Explanation:
C2 (6+6=12)= $\sigma 1s^{2},\sigma ^{*}1s^{2}, \sigma 2s^{2},\sigma ^{*}2s^{2},\pi2p_x^2=\pi 2p_y^2$
Since all the electrons are paired, it is a diamagnetic species.
N2 (7+7=14) $\sigma 1s^{2},\sigma ^{*}1s^{2}, \sigma 2s^{2},\sigma ^{*}2s^{2},\pi 2p_x^2=\pi 2p_y^2, \sigma 2P_{z}^{2}$
It is also a diamagnetic species because of the absence of unpaired electrons.
O2 (8+8=16)
or S2 = $\sigma 1s^{2},\sigma ^{*}1s^{2}, \sigma 2s^{2},\sigma ^{*}2s^{2},\sigma 2p_{z}^{2},\pi2p_{x}^{2}=\pi 2p_y^2 $
$\pi ^{*}2p_{x}^{1}=\pi ^{*}2p_{y}^{1}$
Due to the presence of two unpaired electrons, O2 and S2 both are paramagnetic molecules.
