When finely divided iron combines with carbon monoxide, we get the complex $\ce{Fe(CO)_5}$.

Why the number 5? It doesn't even relate to the orbital theory wherein we can see the number of empty and filled orbitals.

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    $\begingroup$ Although not accurate, the concept of EAN yielding a noble gas configuration can be applied. $\endgroup$ Jun 27 '21 at 12:16

In organometallic chemistry there is something known as the 18-electron rule, which essentially suggests that stable complexes have an 'electron count' of 18. A general discussion of electron counting is beyond the scope of the current answer, but in the specific case of $\ce{[Fe(CO)5]}$, we have $\ce{Fe(0)}$ which accounts for 8 electrons, and five $\ce{CO}$ ligands which provide 2 electrons each, for a total of 18.

What is the basis behind this rule? I don't like the argument based on 'noble gas configuration': that makes little sense because the orbital structure is not the same as a plain atom.

A better argument is to draw out an MO diagram for the complex. The metal 3d, 4s, and 4p orbitals ($5 + 1 + 3 = 9$ orbitals in total) are involved in bonding with the ligands, and can form bonding + antibonding MOs. Generally there will be 9 bonding MOs which are low in energy, and these can be filled with 18 electrons.

For example, here's the standard MO diagram for an octahedral complex (from Wikimedia Commons). I'm aware that $\ce{[Fe(CO)5]}$ isn't octahedral, but the same principles apply; or you can imagine for a while that we're talking about $\ce{[Cr(CO)6]}$ or $\ce{[V(CO)6]-}$, which are octahedral and obey the 18-electron rule.

The 'low-energy' MOs are the bottom 6, plus the $\mathrm{t_{2g}}$ orbitals which are unfilled in this diagram. Generally the 18-electron rule holds true for strong-field ligands, especially π-acceptors like $\ce{CO}$. The reason for this is because these ligands push the $\mathrm{t_{2g}}$ orbitals down in energy (which is not illustrated in the diagram here, but can be found in any standard inorganic chemistry textbook).

Octahedral MO diagram

  • $\begingroup$ "3d, 4s, and 4p orbitals (5+1+3=9 orbitals in total) are involved in bonding with the ligands", umm, I'd expect you to say this rule is dumb, because while that was an idea, it's thorough debunked. $\endgroup$
    – Mithoron
    Jun 27 '21 at 19:14
  • $\begingroup$ @Mithoron I'm not sure what you're trying to say; the 4s and 4p orbitals are the a1g and t1u orbitals in the diagram above. Of course, the resulting MOs are mostly ligand-based; but it doesn't affect the conclusion that there are 9 bonding MOs (with π-acceptor ligands). $\endgroup$
    – orthocresol
    Jun 27 '21 at 20:07
  • $\begingroup$ I mean this 18 electron rule smells very much of this dsp hybridisation nonsense. Wasn't even a 12 electron rule coined because of that? $\endgroup$
    – Mithoron
    Jun 27 '21 at 21:09
  • $\begingroup$ I don't know about any 12e rule, although the Wikipedia page on the 18e rule seems to mention something about it. I never mentioned hybridisation in my post, so I think it's unfair to criticise me on this basis; if you want a $D_\mathrm{3h}$ trigonal bipyramidal analysis, you can look in pages 471–473 of Albright et al. "Orbital Interactions in Chemistry" 2nd ed. Exactly the same text that I wrote applies: the metal 3d, 4s and 4p orbitals overlap with ligands, giving rise to 9 relatively low-lying MOs, just that there isn't as nice a 1-to-1 correspondence as in the octahedral case. $\endgroup$
    – orthocresol
    Jun 27 '21 at 21:59
  • $\begingroup$ To be clear, this is the sort of thing that we agree is outdated, etc. etc.; but I don't think I ever said anything remotely like that. $\endgroup$
    – orthocresol
    Jun 27 '21 at 22:15

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