Question

According to molecular orbital theory, which among the following is diamagnetic species?

• A. $B_2$
• B. $O_2$
• C. $NO$
• D. $C_2$

Answer 1

The correct option is D $C_2$

$B_2:\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2(\pi 2p_x^1=\pi 2p_y^1)$
$O_2:\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_x^2=\pi 2p_y^2\left)\right({\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1)$
$NO:\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2\left(\sigma 2p_z{)}^2\right(\pi 2p_x^2=\pi 2p_y^2\left)\right({\pi }^{\ast }2p_x^1)$
$C_2:\left(\sigma 1s{)}^2\right({\sigma }^{\ast }1s{)}^2\left(\sigma 2s{)}^2\right({\sigma }^{\ast }2s{)}^2(\pi 2p_x^2=\pi 2p_y^2)$

$C_2$ is diamagnetic in nature as it does not contain unpaired electron(s) in its molecular orbital.