# Will the bond order of CO be greater in its existence as carbon monoxide or in metal carbonyl?

$$\text{Bond order} = \frac{N(b) - N(a)}{2}$$

Where $N(b)$ is the number of electrons in bonding orbitals and $N(a)$ is the number of electrons in antibonding orbitals.

As in metal carbonyls, there is synergic effect during formation of the π back bond from metal to CO, so CO accepts electrons in its π* anti bonding orbital. As N(a) increases, Bond order decreases. So, CO in its free state has greater bond order in comparison and hence is more stable.
Would the above concept be correct?

In figure 1 you can see the molecular orbital scheme of a $\ce{[ML6]}$ complex including σ and π interactions. The class of ligands is not important; however, care should be taken when correctly assigning orbitals.

Figure 1: Octahedral $\ce{[ML6]}$ complex including σ and π ligand-metal interactions.

The first thing we should be doing is assigning the original, pre-bonding orbitals. The metal orbitals on the left are obvious. The ligand group orbitals of lower energy, $\mathrm{a_{1g} + e_g + t_{1u}}$ are the six ligand orbitals that interact with the metal in a σ fashion; in carbon monoxide, these represent the lone pair on carbon (the HOMO). The twelve p type orbitals, of which nine are omitted and the remaining transform as $\mathrm{t_{2g}}$ are π symmetric with respect to the $\ce{L\bond{->}M}$ bond and represent the $\pi^*_{\ce{C=O}}$ bonds of carbon monoxide; the LUMOs.

In a metal carbonyl complex, the left-hand side is usually higher in energy due to the metal’s lower oxidation state. Thus, the lower an orbital in the centre of the scheme is, the more ligand-centred it is.

While filling in the final complex’ electrons, we start from the bottom; first the twelve ligand electrons that represent the carbon monoxide lone pairs. Once we are done with this, the next orbitals happen to be $\mathrm{t_{2g}}$ — the bonding ones which have more ligand contribution. These can not only be seen as bonding with respect to the $\ce{M\bond{<-}L}$ bond but also antibonding with respect to the $\ce{C=O}$ bond. The maximum we can fill in here is 6 electrons. These six electrons reduce the $\ce{C#O}$ bond order as can be seen in the formula:

$$\text{B.O.} = \frac{(12 + 2\times12) - 6}{2\times6} = \frac{30}{12} = 2.5$$

The additional factor 6 in the denominator is because I am analysing six $\ce{C#O}$ bonds at the same time.

The result ($2.5$) clearly shows that the bond order is reduced with respect to noncoordinating $\ce{CO}$.

The discussion I did with an octahedral complex can be done with any coordination geometry; octahedra are just a very typical case.