Like in title: Why $\ce{Fe(CO)5}$ exists, but $\ce{Fe(NH3)5}$ doesn't?
Edit:
As Jan Dvorak pointed out, it may exist. If so: why is it so unstable?
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1$\begingroup$ How do you know it doesn't exist? We can make predictions it would be extremely unstable, we can predict it would spontaneously decompose in STP, we can grep the literature to verify noone has ever synthesized it, but that's all. Of course, we can still ask why is it so unstable. $\endgroup$– John DvorakJan 8, 2014 at 9:20
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$\begingroup$ @JanDvorak If you elaborate on that a bit, I think it will make for a good answer. $\endgroup$– jonscaFeb 27, 2014 at 20:35
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1 Answer
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While we cannot be sure that $\ce{Fe(NH3)5}$ does not exist, the bonding between ligand and metal offers a good explanation for why it is expected to be less stable than $\ce{Fe(CO)5}$.
In metal carbonyls like $\ce{Fe(CO)5}$, the bonding between CO and the metal consists of three single bonds, 1 $\sigma$ and 2 $\pi$ bonds. The $\sigma$ bond is formed by overlap of the carbon $sp$ hybrid orbital (which contains the lone pair) with an empty $d$ orbital of the metal. Electron density is donated from the orbital of CO into the metal orbital, so CO acts as a $\sigma$ donor. The two other $\pi$ bonds are formed by overlap of a filled metal $d$ orbital with a pair of $\pi^*$ molecular orbitals of CO. Electron density is donated from the metal orbital into the molecular orbitals of CO, so CO acts as a $\pi$ acceptor.
For the $\pi$ bonding to occur, the metal is required to have sufficient $d$ electrons, and thus preferably a lower oxidation state. This explains the relatively low oxidation state of iron (0) in $\ce{Fe(CO)5}$. The combination of one $\sigma$ and 2 $\pi$ bonds gives the M-CO bond a partial triple-bond character, while the C-O triple bond is weakened by the transfer of electron density into the $\pi^*$ orbitals (reference).
Because of its ability to serve as a $\sigma$ donor and $\pi$ acceptor simultaneously, CO is a stronger ligand than $\ce{NH3}$ which can only act as a $\sigma$ donor with its lone pair in the nitrogen $sp^3$ hybrid orbital. $\ce{NH3}$ lacks a $\pi$ system and therefore cannot act as a $\pi$ acceptor. Because $\pi$ backbonding is crucial for stabilization of a metal center in low oxidation state, ammonia is unlikely to form stable ammin complexes which are analogous to the corresponding metal carbonyls.
answered May 11, 2014 at 17:25