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.


  1. J. Demuynck, and A. Veillard, Theor. chim. acta, 1973, 28 (3), 241-265.
  2. ChemSpider.com
  3. Raymond K. Sheline, J. Am. Chem. Soc., 1951, 73 (4), 1615–1618
  4. Dario Braga, Fabrizia Grepioni, and A. Guy Orpen, Organometallics, 1993, 12 (4), 1481–1483.
  5. Henry B. Chin, and Robert Bau, J. Am. Chem. Soc., 1976, 98 (9), 2434–2439.

Carbonyl is nowhere close to a bulky ligand. Apart from halides and hydrides it might as well be the smallest possible ligand.


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