Question

If we consider no mixing of '2s' and '2p' orbitals, then the bond order and magnetic nature of the diatomic molecule C${}_2$ would be:

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

Answer 1

Answer: (D)

2 and paramagnetic

In absence of '2s' and '2p' orbitals mixing, the MOT configuration of C${}_2$ would be:

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

Hence, it contains two unpaired electrons and so, it is paramagnetic.
Bond order $=\frac{8-4}{2}=2$