# Analysis of metal carbonyl bonding in Fe(CO)5 using molecular orbitals

I am really struggling with this question we have been set in class:

Metal-carbonyl bonding: Conceptionally, one can build up $$\ce{[Fe(CO)5]}$$ from the $$\ce{Fe(CO)4}$$ fragment plus free $$\ce{CO}$$ in order to analyse the bonding between the two moieties. Consider first the HOMO and LUMO of free $$\ce{CO}$$: Describe both MOs with respect to their $$\ce{C-O}$$ bonding character (bonding/anti-bonding). Note the type of bond that would be formed if $$\ce{CO}$$ bonded to another moiety through the HOMO and LUMO respectively.

Predict how the $$\ce{C-O}$$ bond itself would be affected (i) if electrons were donated into the LUMO (e.g., from a metal centre); (ii) if electrons were withdrawn from the HOMO (e.g., by a metal centre).

Next consider the $$\ce{Fe(CO)4}$$ complex fragment: Does its LUMO possess significant amplitude at the location where the additional $$\ce{CO}$$ will bind? Does the LUMO’S shape and symmetry (by visual inspection) match the symmetry of the HOMO of $$\ce{CO}$$? Conversely, does the HOMO of the complex fragment match the LUMO of $$\ce{CO}$$?

Based on these considerations, describe the bonding and orbital interactions between the $$\ce{Fe(CO)4}$$ fragment and $$\ce{CO}$$. Predict how binding to the metal centre would affect the structure and vibrational frequency of $$\ce{CO}$$.

Compare your prediction with the structural and vibrational data obtained for $$\ce{CO}$$. $$\ce{[Fe(CO)5]}$$ and $$\ce{[Fe(CO)4]^{2-}}$$and rationalise the observed trends.

I know that the HOMO of $$\ce{CO}$$ is 3σ which is primarily carbon in character and the LUMO is $$2π^*$$ which is also more carbon in character.

Symmetry is one of my worst topics so if someone could give me a hand on this and could even suggest some further reading material it would be awesome :)

• I love symmetry and I can probably help you. Just to check how much you understood though, can you show a molecular orbital diagram of how you derived the HOMO and LUMO of CO? If you don't know how to do that, I will have to make my answer a lot longer... Apr 5 '16 at 15:27