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In detail, what I really mean is which electron in which orbital is being taken when a oxidation happen? Let me give a example: chromate ion(II)

Two electrons are taken but from which orbital:
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For transition metals (specially for those on the second and third rows), like chromium, or iron, which electrons are taken away in oxidation is not straightforwardly deduced. The ones that will be lost are those which are higher in energy and lead to a more stable situation.

For iron, which is $\mathrm{[Ar]3d^6 4s^2}$, the higher energy electrons are $\mathrm{4s^2}$, and those that will be removed on oxidation, leading to a $\mathrm{d^6}$ configuration for iron (II).

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  • $\begingroup$ Can you answered to the example? The case of iron is too easy. Of course electrons in s orbitals will be removed. $\endgroup$ Nov 22, 2014 at 18:32
  • $\begingroup$ I took "not straightforwardly deduced" as having to look up... $\endgroup$ Nov 22, 2014 at 18:34
  • $\begingroup$ For the first transition row, the ion configuration can be predicted. Sorry, I forgot to answer to your example. Cr is $\mathrm{[Ar]3d^54s^1}$, the first eletron to be taken away will be the higher in energy one, ie, the only s electron. Then you have to eliminate another one, and you only have d electrons left, so the $\ce{Cr^2+}$ configuration will be $\mathrm{[Ar]3d^4}$. And yes, many transition metals of the 2nd and 3rd rows, will require you to look them up. $\endgroup$ Nov 22, 2014 at 18:39
  • $\begingroup$ One of the d orbital electrons will be taken out, in terms of orbital diagram, does it matter to take out the electron from the first or the last box? $\endgroup$ Nov 22, 2014 at 20:00
  • $\begingroup$ For the isolated ion, which is the case, it does not matter, all the five d orbitals have the same energy. $\endgroup$ Nov 22, 2014 at 20:35

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