Question

If we consider no mixing of ${}^{\prime }2s^{\prime }$ and ${}^{\prime }2p^{\prime }$ orbitals, then the bond order and magnetic nature of the diatomic molecule $C_2$ is:

• A. $3$ and diamagnetic
• B. $2.5$ and diamagnetic
• C. $2$ and diamagnetic
• D. $2$ and paramagnetic

Answer 1

The correct option is D $2$ and paramagnetic

In absence of '2s' and '2p' orbitals mixing, the MOT configuration of $C_2$ would be:

$\sigma 1s^2{\sigma }^{\ast }{s}^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1$

Hence, it contains two unpaired electrons and so it is paramagnetic.

Bond order =$\frac{8-4}{2}=2$