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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?

Kanghun Kim
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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.

napstablook
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    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? – Mithoron Jul 27 '21 at 17:53
  • I upvoted your answer, but the gauche effect itself is due to hyperconjugation. – Kanghun Kim Jul 28 '21 at 01:21
  • @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. – napstablook Jul 28 '21 at 02:27