# Lewis Acid / Base reaction for SO2 and H2O

I am having trouble with this question:

Identify the Lewis acid and Lewis base in each of the following reactions:

$\ce{SO2(g) + H2O(l) = H2SO3(aq)}$

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. So I tried to look at the lone pairs on the molecules. Since they both have lone pairs, I figured since $\ce{SO2}$ has more lone pairs, it would act as the Lewis base. Also in the final molecule, the hydrogen atoms would be bonded to two separate oxygen atoms, so I thought of them as migrating to the lone pairs on the $\ce{SO2}$ oxygen atoms.

However, the textbook says that $\ce{SO2}$ is the Lewis acid and $\ce{H2O}$ is the Lewis base.

Why would this be the case?

• Clearly it is $\ce{H2O}$ that donates an electron pair to from the bond? So it is the Lewis base. – Gert Dec 9 '17 at 21:25
• Can you explain why it must be $\ce{H2O}$ that donates the electron pair(s) and not $\ce{SO2}$? – user6615434 Dec 9 '17 at 22:07
• I'll try and answer that tomorrow. – Gert Dec 9 '17 at 22:48

The oxygen in $\ce{H2O}$ already has its orbitals ($2s^2$,$2p^4$) filled and since oxygen is in the second row of the periodic table it can't form hypervalent molecules (That means that the oxygen would have more than 8 valence electrons).
Because of this it is clear, that the $\ce{H2O}$ can't act as a Lewis acid as it can't have any more electrons. Therefore it has to be the Lewis-base which leaves $\ce{SO2}$ as the Lewis-acid.
Molecules that contain polar multiple bonds can function as Lewis acid because the central atom is electron deficient with a vacant orbital that can accept an electron pair. When $\ce{SO_2}$ dissolves in water , it forms the weak diprotic acid $\ce{H_2SO_3}$ (sulfurous acid) from Lewis acid/Lewis base reaction: $$\ce{O\bond{=}S-O +H_2O<=>H_2SO_3}$$
The $O$ atom of an $\ce{H_2O}$ molecule donates a lone pair to the $S$ of $\ce{SO_2}$,forming a new $\ce{S\bond{-}O}$sigma bond and breaking $\ce{S\bond{-}O}$ π bond