# Why does Fe(CO)₄ adopt a tetrahedral, as opposed to square planar, geometry?

Why does $\ce{Fe(CO)4}$ adopt a tetrahedral, as opposed to square planar, geometry? Here's my analysis:

Factors for square planar:

• coordination: 4

• $\ce{Fe(0)}$: d8

• $\ce{CO}$: strong field ligand (gives large d-splitting)

Factors for tetrahedral:

• oxidation state: 0 (gives weak d-splitting)

• $\ce{CO}$: relatively bulky (?) ligand

Other thoughts:

The general guidelines for determining whether or not a row 1 transition metal complex is square planar or tetrahedral (given it has 4-coordination and d8 electrons), are the strength and bulkiness of the ligand. Generally, strong field ligands like $\ce{CO}$ and $\ce{CN^{-}}$ gives square planar complexes. In fact, $\ce{[Ni(CN)4]^{2-}}$ is square planar. Thus, the only way I could explain why $\ce{Fe(CO)4}$ would be tetrahedral would be because of its low oxidation state. Is this correct?

The hypothetical $\ce{Fe(CO)4}$ complex is isoelectronic to the observed $\ce{[Ni(CN)4]^{2-}}$ complex and therefore would also be square planar.[1,2] Both are 16 electron complexes, i.e. one of the valence orbitals of the central atom remains unoccupied. Your assessment for $\ce{Fe^{\pm0}}$ as $\mathrm{d^8}$ is absolutely correct, also that carbonyl is a good ligand causing a strong field and hence a strong splitting. In a tetrahedral complex there would be triply degenerated HOMO, which would only be occupied by four electrons. This arrangement cannot be more stable than a planar complex. However, the iron tetracarbonyl complex as a singular entity has not been observed (to my current knowledge). The observed structures for the compound are various dimers and trimers.[3]
There is also a neutral iron carbonyl complex, but it binds five instead of only four ligands. Therefore $\ce{Fe(CO)5}$ is a trigonal bipyramidal 18 electron complex.[4]
The observed $\ce{[Fe(CO)4]^{2-}}$ complex[5] is isoelectronic to $\ce{Ni(CO)4}$ which is a tetrahedral 18 electron complex.[1,4] Without going into more detail, the iron in this complex is of course $\ce{Fe^{-II}}$ and therefore $\mathrm{d^{10}}$. A square planar arrangement would no longer be more stable and a complex with a higher symmetry and more degeneracy is preferable.