Effect of metal identity on CO bond length in carbonyl complexes

In organometallic carbonyl complexes, a back-donation effect occurs: the σ molecular orbital of CO yields electron density to an orbital of the appropriate metal atom, and in turn, a d orbital of appropriate symmetry yields electron density to the antibonding π* orbitals of CO.

On the basis of this model, the properties of carbonyls (e.g., the vibrational energy of the C–O bond or its bond length) can be explained. For example, in the following table, I understand that all of the bond lengths are longer than in free CO, because of this backdonation which decreases the C–O bond order. However, how can the differences between the three metal complexes be explained?

Species C–O bond length (pm)
$$\ce{CO}$$ 112.8
$$\ce{[Ti(CO)6]^2+}$$ 116.0
$$\ce{[Cr(CO)6]}$$ 114.0
$$\ce{[Ni(CO)4]}$$ 113.0
• There we go, that's a bit cleaner, and more to the point. Jul 7 at 13:09
• Jul 8 at 0:06

1 Answer

Let's look at the molecular orbital diagram calculated with JSmol.

As you know, the principal effect is back-donation from the metal to the π-antibonding orbitals of the carbon monoxide. But not all 3d transition metals are created equal. Earlier ones like titanium, with lower electronegativity and lower ionization energy, are better at this back-donation than later ones.