Let us consider the Caro's Acid: $\mathrm{H_2SO_5 \equiv SO_3H-OOH}$. It is widely known that the hydrogen tied with the $\mathrm{-OO-}$ group is pretty much unlikely to dissociate: $$ \text {Terminal dissociation: }\mathrm{SO_3H - OOH \rightleftharpoons H^+ + [SO_3-OOH]^-} \\ \text{An unlikely step: }\mathrm{[SO_3-OOH]^- \rightleftharpoons H^+ +[SO_3-OO]^{2-}} $$ The first time I encountered that fact I was presented a pretty inconclusive piece of argumentation: it was stated that $ \mathrm{-OO-} $ group is way more electronegative than $ \mathrm{-O-} $, which somehow brings about an obstacle for the dissociation.

Unfortunately, it was a long time ago and I cannot remember the exact details. I stumbled upon this problem not so long ago. Having a better background in chemistry, I attempted to apply the Sanders' electronegativity balance theory, but the result was pretty inane in terms of our chemical reality: $\chi_{eq}(\mathrm{H_2SO_4}) \approx 2.89894; \chi_{eq}(\mathrm{H_2SO_5}) \approx 2.96803 $. In other words: $ \chi_{eq}(\mathrm{H_2SO_5}) > \chi_{eq}(\mathrm{H_2SO_4}) $.

What am I doing wrong? Is it possible to tweak my attempt to use Sanders' theory? Is there any other approach to this problem?

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    $\begingroup$ It's just not delocalised there, that's it about it. $\endgroup$
    – Mithoron
    Commented Apr 11, 2020 at 18:51
  • $\begingroup$ @Mithoron well, that’s a good point, but the localization analysis is, in fact, just another way to look at electronegativity computational models. I mean, the resulting electronegativity does depend on the electron-nucleus relations (i.e. ionization and electron affinity energy). So, to be honest, this does not really answer my question. I may have worded it a bit clumsily: I’m interested in the computational aspect of the problem (and the application of the Sanders’ theory, to be more precise). $\endgroup$ Commented Apr 11, 2020 at 21:06
  • $\begingroup$ So the question: why is it a peroxide instead of e.g. an oxide? $\endgroup$
    – Greg
    Commented May 17, 2020 at 2:17
  • $\begingroup$ Hmmm... What is the "Sanders' electronegativity balance theory" you are referring to? I believe most of us in the community are not well-versed with this theory. Perhaps, you could provide a reference for us to look at so that we can learn about this theory? $\endgroup$ Commented May 17, 2020 at 12:32
  • $\begingroup$ Based on this reference (dash.harvard.edu/bitstream/handle/1/11913975/…), the first $\ce {pK_a}$ of the acid is 0.4, which is higher than that of sulfuric acid's corresponding first $\ce {pK_a}$ value of around -3. It is not surprising due to the fact that there is a greater extent of delocalisation of the negative charge in the sulfate conjugate base compared to in the peroxymonosulfate conjugate base. Note that the 2 $\ce {O}$'s in the peroxy group are not able to participate in resonance. $\endgroup$ Commented May 17, 2020 at 12:51

1 Answer 1


One possibility is the formation of an intra-"molecular" hydrogen bond between the peroxy-bonded hydrogen and a third oxygen atom. The structure shown in Wikipedia for the acid suggests this interaction. The involved hydrogen is curled up into the five-membered ring formed by such a hydrogen bond, while the other hydrogen atom is freely exposed and ready to launch. Potassium peroxymonosulfate corroborates this; it shows only the non-peroxo hydrogen dissociated.

Hydrogen peroxide does not form this sort of intramolecular hydrogen bond, and as we expect from induction considerations the proton dissociates from this molecule more readily than from water.


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