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?

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    $\begingroup$ The oxidation state of the metal is a formal count not a real charge. $\endgroup$
    – Zhe
    Commented Sep 28, 2017 at 13:08
  • $\begingroup$ @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. $\endgroup$ Commented Sep 28, 2017 at 13:16
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    $\begingroup$ 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. $\endgroup$
    – Zhe
    Commented Sep 28, 2017 at 14:44
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    $\begingroup$ 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. $\endgroup$ Commented Sep 28, 2017 at 15:28

1 Answer 1


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).

This is true for most complexes. The metal tends to have rather high-lying orbitals while the ligands tend to donate rather low-lying ones to the overall complex. As long as this order is retained, there is no chemical redox behaviour going on.

For some systems, however, the orbitals are either so similar or even crossed-over that suddenly a metal orbital will become lower-lying than a ligand orbital. In this case, an electron that used to reside in a ligand orbital will relax into a metal orbital — a redox process. You can for example think of the reaction of iodide with copper(II) in this manner: copper(II) will immediately get reduced to copper(I) by iodide ions. While this is a unidirectional process, some systems are also known to bind ligands in a redox fashion reversively, meaning that the electrons will be exchanged both ways. The binding mode of oxygen to haemoglobin is understood in this manner.

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.

  • $\begingroup$ 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." $\endgroup$
    – MaxW
    Commented Sep 28, 2017 at 15:10
  • $\begingroup$ @MaxW As I tried to address when referring to the redox binding mode of oxygen to haemoglobin, lability does not mean that no redox process is involved. $\endgroup$
    – Jan
    Commented Sep 29, 2017 at 6:59
  • $\begingroup$ 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. $\endgroup$
    – Jan
    Commented Sep 29, 2017 at 7:09

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