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](http://en.wikipedia.org/wiki/Isoelectronicity) 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.




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*References*


1. [J. Demuynck, and A. Veillard, *Theor. chim. acta*, 
**1973**, *28* (3), 241-265.](http://link.springer.com/article/10.1007%2FBF00533488)
2. [ChemSpider.com](http://www.chemspider.com/Chemical-Structure.19995981.html)
3. [Raymond K. Sheline, *J. Am. Chem. Soc.*, **1951**, *73* (4), 1615–1618](http://pubs.acs.org/doi/abs/10.1021/ja01148a060)
4. [Dario Braga, Fabrizia Grepioni, and A. Guy Orpen, *Organometallics*, **1993**, *12* (4), 1481–1483.](http://pubs.acs.org/doi/abs/10.1021/om00028a082)
5. [Henry B. Chin, and Robert Bau, *J. Am. Chem. Soc.*, **1976**, *98* (9), 2434–2439.](http://pubs.acs.org/doi/abs/10.1021/ja00425a009)




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*Note*  

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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