Metal carbonyl bonding

I have been going through this topic in my coordination chemistry chapter. I read that in metal carbonyls, -CO group acts as a ligand and donates its lone pair of electrons into a vacant orbital of the metal to form a sigma bond. Also, there's a pi bond formation by the donation of electrons from the filled d orbital of the metal into pi* MO of the -CO molecule, and this pi bond strengthens the metal to ligand bonding.

But filling of electrons in the antibonding MO means decrease of bond order and correspondingly decrease of bond strength and bond lengths. It sounds contradictory to what's written in my book.

Where am I making a mistake?

• That's right, it weakens the C-O bond, but strengthens the metal-C one. – Ivan Neretin Feb 2 '16 at 15:01

Carbon monoxide bonds to transition metals using "synergistic π* back-bonding." The bonding has three components, giving rise to a partial triple bond. A sigma bond arises from overlap of the nonbonding (or weakly anti-bonding) sp-hybridized electron pair on carbon with a blend of d-, s-, and p-orbitals on the metal. A pair of π bonds arises from overlap of filled d-orbitals on the metal with a pair of π-antibonding orbitals projecting from the carbon atom of the $\ce{CO}$. The latter kind of binding requires that the metal have d-electrons, and that the metal is in a relatively low oxidation state ($<+2$) which makes the back donation process favorable.
As electrons from the metal fill the π-antibonding orbital of $\ce{CO}$, they weaken the carbon-oxygen bond compared with free carbon monoxide, while the metal-carbon bond is strengthened. Because of the multiple bond character of the $\ce{M-CO}$ linkage, the distance between the metal and carbon atom is relatively short, often $< 1.8~\mathrm{Å}$, about $0.2\ \mathrm{Å}$ shorter than a metal-alkyl bond.
• Not all carbonyl complexes will be acurately represented by a $\ce{M#C-O}$ depiction as your description would imply. In fact, $\ce{M=C=O}$ is much more common. Otherwise a good answer. (Please note that the unit is Å, no Â ;)) – Jan Feb 2 '16 at 22:49