# Why do nonmetal oxides react with water to form acidic compounds? [closed]

I am confused about the molecular basis by which nonmetal oxides react with water to form acidic compounds. For example, I understand that sulfur trioxide reacts with water to form sulfuric acid as shown in the following reaction:

SO3 (g) + H2O (l) --> H2SO4 (aq)

I recognize that the polarized nature of the O-H bond in sulfuric acid makes it thermodynamically favorable for H2O to abstract a proton and subsequently form an acidic solution with the formation of H3O+. However, I fail to recognize how reactions like this with a nonmetal oxide proceeds to form an acidic oxide. Namely, I fail to understand the role of the covalent bond in the nonmetal oxide for the mechanism of this reaction. I recognize that metal oxides dissolve in water to form the highly polarizing, high charge density ion O2-, which polarizes the O-H bond in H2O to form hydroxide. Can someone give a similar explanation for nonmetal oxides and their dominant acidic nature in water?

## closed as too broad by MaxW, Mithoron, Mathew Mahindaratne, airhuff, aventurinNov 11 '18 at 13:09

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Well, in terms of this reaction, sulfur in $$\ce{SO3}$$ acts as a Lewis acid for an incoming $$\ce{OH-}$$ nucleophile. The reaction mechanism seems to be any old nucleophilic addition reaction:
• $$\ce{H2O <=> H+ + OH-}$$ : (Water can disassociate into its ions)
• $$\ce{SO3 + OH- -> SO2(OH)O-}$$ : (I couldn't decide on whether this is the right way to write this intermediate, but I'll be happy to add a diagram for clarity)
• $$\ce{SO2(OH)O- + H+ -> SO2(OH)2}$$ : (This is the final step of the mechanism to yield $$\ce{H2SO4}$$)