Dec 10, 2017 · $\ce{SO2(g) + H2O(l) = H2SO3(aq)}$ What I tried: A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor. The other problems of this type involved cations or anions, so I was able to identify the cation as the Lewis acid and the anion as the Lewis base. However, this question does not involve ions.

Aug 16, 2015 · Therefore we expect $\ce{SO2}$ to have the largest bond angle of the four molecules, and this is indeed the case. $\ce{H2O}$ and $\ce{NH3}$ are hydrides of the same period so we can use the first rule to determine that $\ce{H2O}$ has a smaller bond angle. Now we just have to decide whether $\ce{H2O}$ or $\ce{H2S}$ has a smaller bond angle.

Aug 26, 2020 · Majority of texts (at high school level) say that when $\ce{SO2}$ is dissolved in water sulphurous acid $\ce{H2SO3}$ is formed. However my text book states that it forms clathrate $\ce{SO2.6H2O}$ when dissolved in water, and can significantly exist as $\ce{H+ + HSO3-}$. I found similar finding on Wikipedia:

Jun 13, 2015 · Upon adding $\ce{SO2}$ gas to water, it shows its acidic oxide character by hydrolyzing to sulfurous acid: $$\ce{SO2(g) + H2O(l) <=> H2SO3(aq)}$$ $\ce{H2SO3}$ is a weak diprotic acid in water, with the following acid dissociation equilibria:

Jan 7, 2016 · Orthocresol already sufficiently pointed out, that half-reactions are a book-keeping tool for electrons. When talking about electrolysis there might be coincidental overlap between half-reactions and the overall mechanism, but as the wording is supposed to indicate, this is not necessarily true.

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Mar 1, 2018 · Since $\ce{SO2}$ is a strong reducing agent, it reduces $\ce{Mn}$ as denoted by the equation $$\ce{5SO2 + 2KMnO4 + 2H2O -> 2H2SO4 + 2MnSO4 + K2SO4}$$ Both the reactions are viable. Because of the formation of $ \ce{MnSO4}$ and the disappearance of $ \ce{KMnO4}$, decolorisation occurs.

May 27, 2019 · $\ce{so2 + h2o = h+ + hso3-}$ $\ce{HOI + HSO3- -> H+ + I- + HSO4-}$ However, the reaction could also proceed to some extent via the following reactions following a disproportionation of hypoiodous acid:

Sep 15, 2022 · This is indeed very close to 120∘ so you can say that the VSEPR method gives a good approximation of the SO2 bond angle. The essence of my question revolves around the fact that lone and/or bond pairs separated by an angle of 120 degrees do not create repulsions , so considering that, why is the $\ce{O-S-O}$ bond angle compressed down to 119?

And finally, we have another decomposition namely that of sulfurous acid to sulfur dioxide $\ce{SO2}$ and water. This can be seen as the back-reaction of a Lewis acid-base reaction wherein water is the liberated Lewis base and sulfur dioxide the liberated Lewis acid.