Question

Draw MO diagram of CO and calculate its bond order.

Answer 1

Bonding in some hetero nuclear di-atomic molecules:

Molecular orbital diagram of Carbon monoxide molecule (CO)

Electronic configuration of C atom: 1s² 2s² 2p²

Electronic configuration of O atom: 1s² 2s² 2p⁴

Electronic configuration of CO molecule :

σ_1s², σ_1s^*2, σ_2s², σ_2s^*2, π_2py², π_2pz² σ_2px²

Bond order = \(\frac{N_b - N_a}2\)

\(=\frac{10 - 4}2\)

\(= 3\)

The molecule has no unpaired electrons hence it is diamagnetic.

Answer 2

1. Electronic configuration of C atom: 1s² 2s²2p²

Electronic configuration of O atom: 1s² 2s² 2p⁴

2. Electronic configuration of CO molecule is:

σ1s² σ*1s² σ2s² σ*2s² π2p_y² π2p_z² π2p_x²

3. Bond order = 

= \(\frac{N_b-N_a}{2}\) = \(\frac{10-4}{2}\) = 3

4. Molecule has no unpaired electron, hence it is diamagnetic.