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I have encountered this equation: $$\ce{SO4^2- -> S2O8^2-}$$ In this reaction 2 of the oxygen ions change from $-2$ to $-1$ so they are getting oxidized.

It is because two oxygen ions are connected together.

A question rises: why can't we oxidize oxygen ions in $\ce{NO3-}$ or any other?

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    $\begingroup$ $\ce{NO3-}$ can become $\ce{NO2}$, which according to this post, has a canonical form containing a peroxide bridge causing the oxygen to have oxidation states of $-1$. $\endgroup$
    – DHMO
    Commented Sep 21, 2016 at 15:27

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Oxygen can be oxidized. However, it is uncommon due to its high electronegativity. In fact, it is highest second only to fluorine. This makes it extremely difficult to remove electrons from oxygen, thus making it hard to oxidize.

One definitive example of oxygen being oxidized is oxygen difluoride. This exception is even listed on many lists of oxidation state rules (see: oxidation states).

And of course, if you go beyond stable molecules, it is possible to ionize oxygen with an electrical charge.

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  • $\begingroup$ What I basically meant is oxygen in ions, In NO3 nitrogen cant oxidizes (reached its highest oxidation state) but Oxygen is not. So from your answer, We can force oxygen to be oxidized, right? $\endgroup$
    – Biker
    Commented Sep 21, 2016 at 17:05
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    $\begingroup$ Of course we can. Just electrolyze nearly any salt with oxygen-containing anion, and that would be it. $\endgroup$
    – Ivan Neretin
    Commented Sep 21, 2016 at 18:32
  • $\begingroup$ @Biker I don't know if I understand your follow up. This might be interesting to you, though: But see: en.wikipedia.org/wiki/Fluorine_nitrate $\endgroup$
    – user7652
    Commented Sep 21, 2016 at 21:39
  • $\begingroup$ It is just because our exams gives us a multiple choice: Which of the following cant be oxidized? and the answer is NO3- because N is +5 They seem to exclude O-2 for a reason but it can be oxidized under certain cases like Fluorine or electrolysis. We can exclude electrolysis because it requires energy and maybe they are referencing that it must happen spontaneous so we are left with fluorine. That is why I asked if we can oxidize oxygen ion or no. $\endgroup$
    – Biker
    Commented Sep 22, 2016 at 2:28
  • $\begingroup$ @Biker Based on that, it sounds like they're asking about the molecule as a whole, not the atoms in it. Although it is a bit of a suspicious question if that's the exact wording, because it does sound confusing. It might just be a case of poor wording; you could ask the teacher for an explanation of what was intended. $\endgroup$
    – user7652
    Commented Sep 22, 2016 at 14:17
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In the equation $\ce{2SO4^2- -> S2O8^2- + 2e-}$ sulfate is being oxidized.
This is different from oxygen being oxidized.

The statement in the OP "In this reaction 2 of the oxygen ions change from $-2$ to $-1$ " is incorrect. Instead, each sulfate ion, taken as a whole, loses 1 electron. There is no reason to ascribe the loss to oxygen, rather than sulfur.

The analogy for nitrate would be $\ce{2NO3- -> N2O6 + 2e-}$ , which is discussed on page 86 here

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  • $\begingroup$ Umm because Sulfur can't be oxidized? It has reached its highest oxidation state and if the ion loses electrons then it must be the oxygen not the sulfur that was my reasoning. $\endgroup$
    – Biker
    Commented Sep 22, 2016 at 14:45
  • $\begingroup$ @Biker in some artificial book-keeping system we can pretend that in sulfate S is +6 and each O is -2, but what is it in the real world? "In $\ce{SO4 ^{2-}}$ there is 0.5 of negative charge on each oxygen and no charge on sulfur." scitation.aip.org/content/aip/journal/jcp/45/6/10.1063/… $\endgroup$
    – DavePhD
    Commented Sep 22, 2016 at 15:06
  • $\begingroup$ @Biker and "the peroxide O-O bond in $\ce{S2O8 ^{-2}}$ is nonpolar with a negative Mulliken charges of 0.65 on both peroxide oxygen atoms, as determined by our AM1/COSMO computations." pubs.acs.org/doi/abs/10.1021/jp2041845 So O going from -2 to -1 is just imagination, just as the IUPAC definition of oxidation state is "the charge an atom might be imagined to have" goldbook.iupac.org/O04365.html $\endgroup$
    – DavePhD
    Commented Sep 22, 2016 at 15:44
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Oxygen is also oxidized by platinum hexafluoride, where the platinum (in a high oxidation state) rather than the fluorine is the oxidizing agent, see https://en.wikipedia.org/wiki/Platinum_hexafluoride. The reaction is represented as

$\ce{O2 +PtF6 -> O2+PtF6-}.$

This reaction heralded the possibility of similarly oxidizing noble gases, which busted the "inert gas" myth (at least for the heavier noble gases).

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