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This seemed like a trivial question to me... until I began to think about it.

The "usual" criteria for greater acidity are larger sizes of the atom attached to the acidic proton within the same group(e.g. hydrogen selenide over hydrogen sulfide over water) and inductive (and/or resonance, the latter of which usually dominates) stabilisation(e.g. phenol over ethanol); hydrogen peroxide has no reason to be more acidic than water by the first criterion and the second criterion would destabilise the anion instead, with the hyperconjugative electron donation from the oxygen-hydrogen bonding orbital far outweighing the negative-hyperconjugative electron withdrawal into the oxygen-hydrogen antibonding orbital.

So why is hydrogen peroxide acidic in aqueous solution, when it has no reason to be more acidic than water and every reason to be less acidic than water?

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  • $\begingroup$ related chemistry.stackexchange.com/questions/154346/… $\endgroup$
    – Mithoron
    Jul 27 '21 at 14:16
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    $\begingroup$ Does this answer your question? Is there a general consensus on the causes of the alpha-effect? $\endgroup$
    – Mithoron
    Jul 27 '21 at 14:16
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    $\begingroup$ It's too technical for me to understand properly... $\endgroup$ Jul 27 '21 at 14:36
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    $\begingroup$ Well, out of the top of my head, I don't see why we couldn't just say hydroxyl inductively withdraws better then hydrogen and call it a day. The more you'd want to get into details the more muddled it would become, I think. $\endgroup$
    – Mithoron
    Jul 27 '21 at 16:41
  • $\begingroup$ Yeah that might be right... but doesn't resonance usually dominate over inductivity when it comes to cases like this? $\endgroup$ Jul 28 '21 at 2:47
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The simple size comparison is not accurate. In hydra-acids you can often draw a correlation between the size and strength of $\ce{H-A}$ bond. However $\ce{H2O2}$ consists of only $\ce{O-H}$ bond with respect to hydrogen and hence has a comparable bond strength with respect to $\ce{H2O}$ regarding acid-base reactions.

$\ce{OOH^{-}}$ is not destabilized by resonance effects as $\ce{O^{-}}$ lacks vacant orbitals to accept $\ce{-OH}$'s lone pair. As a result the effect is mostly determined by inductive effect (($-I$) of $\ce{-OH}$ is larger than $\ce{-H}$.)

Negative hyperconjugative effect ($-H$) effect can't occur because it requires polarity of substituent here $\ce{H}$ to be more electronegative than $\ce{O}$ in $\ce{-OH}$.

One more interesting point (again in favor of making $\ce{H2O2}$ more acidic) is the gauche effect due to which lone pair of one oxygen donates electron density to $\sigma *$ of $\ce{O-H}$ bond thereby further weakening the $\ce{O-H}$ by reducing bond order.

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    $\begingroup$ He doesn't talk about "normal" hyperconjugation, but stereoelectronic effect, with lone pair and antibonding orbital. BTW why people say "no vacant orbitals" when there are antibonding orbitals, like, everywhere? $\endgroup$
    – Mithoron
    Jul 27 '21 at 17:53
  • $\begingroup$ I upvoted your answer, but the gauche effect itself is due to hyperconjugation. $\endgroup$ Jul 28 '21 at 1:21
  • $\begingroup$ @JustSomeNerdyTeen I don't disagree though it is really just hyperconjugation but with ABMO. If you feel like this answered your question, consider ticking the check mark. $\endgroup$ Jul 28 '21 at 2:27
  • $\begingroup$ Can I know the reason of downvote? Is there anything wrong to change here? $\endgroup$ Jul 28 '21 at 2:27
  • $\begingroup$ @Mithoron Good point. I edited the answer to reflect that. $\endgroup$ Jul 28 '21 at 2:29

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