Actually no, because the deprotonated form (superoxide) is a radical as well, so nothing is gained in terms of stability by deprotonation on that
front.
I think the reason for the increased acidity of HO2 compared to H2O2 has to do with the fact that the bond order of the O-O bond in H2O2 is 1, whereas
in superoxide it is 1.5. This means that p orbitals are being used up to create a pi bond between the oxygens, and as such the hybridisation of the
oxygens will be that of less p character and more s character. As s orbitals are on average closer to the nucleus than p orbitals, increased s
character means that the electrons will be held onto more tightly, and less available for bonding. As such, any proton bound to the oxygen will have a
weaker bond and be more easily removed. This is the same phenomenon that makes alkynes stronger acids than alkenes, which are in turn stronger acids
than alkanes. |