# Why don't ligands affect the oxidation state of the central metal ion?

If $\ce{Fe^3+}$ were complexed with 6 $\ce{H2O}$ ligands, the resulting complex ion would be $\ce{[Fe(H2O)6]^3+}$, and the central metal ion would still have the same oxidation state of +3 as it did before complexation. However, didn't the $\ce{Fe^3+}$ ion receive 12 electrons (2 from each water molecule) in forming the complex? Is this not a kind of reduction?

Also, is the positive charge of the complex ion still localized on the metal ion, or is it spread over the entirety of the complex?

• The oxidation state of the metal is a formal count not a real charge. – Zhe Sep 28 '17 at 13:08
• @Zhe I acknowledge that, and I had asked a separate subquestion regarding the real charge of the complex. But whenever something is reduced, their oxidation state goes down, no? So my main question is essentially why $\ce{Fe^3+}$ isn't reduced by being coordinated by ligands. – lightweaver Sep 28 '17 at 13:16
• Even if the end result is a reduction, it doesn't matter because the oxidation state is just computed from a formula that doesn't account for that. The difference is that water is approximated as a neutral ligand whereas something like a chloride is approximated as an anionic ligand. The former does not affect the oxidation state of the metal ion whereas the latter does. There's nothing to be confused about here because we just created a definition, and then, we applied it. Definitions aren't reality. – Zhe Sep 28 '17 at 14:44
• You don't become any richer when the banks lend you money. Same thing here, with metal ion instead of you and electrons instead of money. – Ivan Neretin Sep 28 '17 at 15:28

If you take a glimpse at the molecular orbital scheme, you will notice that the electrons donated by the water molecules still reside in orbitals that are largely ligand-centred (and have a relatively low energy). On the other hand, iron(III) also has five d electron itself; these reside in the $$\mathrm{t_{2g}}$$ and $$\mathrm{e_g}^*$$ orbitals, which are largely metal-centred (and have a relatively high energy).
Finally, sometimes there is a change in oxidation state that is merely formal. Tetracarbonylhydridoiron(0) $$\ce{[Fe(CO)4H]}$$ is such a case. It can be protonated to formally give tetracarbonyldihydridoiron(II) $$\ce{[Fe(CO)4H2]+}$$ or deprotonated to give tetracarbonylferrate(–II) $$\ce{[Fe(CO)4]-}$$. Both changes in oxidation state are merely formal and chemically all that is happening is a protonation/deprotonation.
• Another point here is that ligands are labile. The $\ce{H2O}$ molecules aren't "glued" to the complex but "swap" in and out. The $\ce{Fe^{3+}}$ can also have ligands other than $\ce{H2O}$ if there any are in the solution. This ligand swapping is why it is called a "complex." – MaxW Sep 28 '17 at 15:10
• Another thing: Even if the ligands are firmly glued to the central metal and no swapping occurs at any reasonable timescale (eg $\ce{[Fe(CN)6]^4-}$) it is called a complex. – Jan Sep 29 '17 at 7:09