Question

If we consider no mixing of $‘2s^{\prime }$ and $‘2p^{\prime }$ orbital, 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 $‘2s^{\prime }$ and $‘2p^{\prime }$ orbitals, mixing the MOT configuration of $C_2$ would be;
$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^1=\pi 2p_y^1$. Hence, it contains two unpaired electrons and so it is paramagnetic.
Bond order $=\frac{8–4}{2}=2$