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 : σ1s2, σ1s"2, σ2s2, σ2s"2, π2py2, π2pz2 σ2px2 Bond order = \ (\frac {N_b - N_a}2\) \ (=\frac {10 - 4}2\) \ (= 3\) The molecule has no unpaired electrons hence it is diamagnetic. In the question we are supposed to calculate the bond order of carbon monoxide molecule (CO) according to molecular orbital theory. We are very much familiar with the term bond order from the lower classes, first let’s briefly discuss a few points regarding bond order of a molecule. The bond order of CO is 3. The BO in CO+ is 3.5 because the order of the orbitals is not the same as it would be for a homonuclear diatomic molecule such as O2where both atoms are the same. Each pair of shared electrons constitutes a bond, so the bond order in carbon monoxide (CO) is 3. The bond order is determined by counting the number of electron pairs that are shared between the two atoms in a covalent bond. Single bonds have a bond order of 1, double bonds have a bond order of 2, and triple bonds have a bond order of 3.
