I heard that $\ce{Fe}(\mathrm{III})$ is more common than $\ce{Fe}(\mathrm{II})$ but I've not heard a very clear explanation. Could someone please explain this incorporating electron configurations in their answer?

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    – hBy2Py
    Sep 16, 2015 at 0:48
  • $\begingroup$ Not in igneous rocks. Most iron there, in silicate minerals throughout the deep crust and mantle, is Fe(II). Need to weather the rocks and expose the iron to an oxidizing atmosphere to get substantial Fe(III). $\endgroup$ Aug 26 at 9:07

2 Answers 2


I think the best answer is not to try to describe this in terms of electron orbitals. The main factor is that Fe(II) is fairly easily oxidized by atmospheric oxygen to give Fe(III). Because we spend most of our time living in an oxygen atmosphere, Fe(III) is more common in our ordinary experience than Fe(II).

In environments with less oxygen, Fe(II) is very common. The mantle is full of Fe(II) minerals like olivine and ferropericlase. I don't know which species is actually the majority on earth, but I would bet on Fe(II).

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    $\begingroup$ How can we explain this on the basis of reduction potentials? For $\ce{Fe^3+ + e^-} \rightarrow Fe^{2+}$ is 0.77 and for $\ce {Fe^2+ + 2e^-} \rightarrow Fe$ is -0.44? $\endgroup$ Aug 24 at 5:43
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    $\begingroup$ @Proscionexium chemistry.stackexchange.com/a/44678/136839 is a good answer. $\endgroup$
    – anon
    Aug 24 at 11:15

The electron configuration of $\ce{Fe(II)}$ is $1s^22s^22p^6 3s^23p^6 3d^6$.

The electron configuration of $\ce{Fe(III)}$ is $1s^22s^22p^6 3s^23p^6 3d^5$.

The half filled orbital $3d$ is more stable than the same orbital filled with 6 electrons. So, $\ce{Fe(III)} $ ion is more common than $\ce{Fe(II)}$.

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    $\begingroup$ I’m not sure if this is more than half of the truth, but I’ll leave it to the true experts to tell me, whether I’m right or wrong. (Sure, this is taught in schools, but it seems to simple to me, and the half-filled shell concept has been deemed invalid in other answers that I don’t have on-hand atm.) $\endgroup$
    – Jan
    Sep 16, 2015 at 17:14
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    $\begingroup$ It's not even wrong, it's irrelevant, Real iron compounds don't have iron in a spherical environment, hence the orbital degeneracy pattern is different. For instance low spin iron II in an Oh environment has a completely filled set of orbitals. Just like Ne, so why isn't that the most "stable", whatever that means $\endgroup$
    – Ian Bush
    Aug 24 at 5:03

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