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?




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 $\ce{SO2}$ dissolves in water , it forms the weak diprotic acid $\ce{H2SO3}$ (sulfurous acid) from Lewis acid/Lewis base reaction:


$$\ce{O\bond{=}S\bond{=}O + H2O <=> H2SO3}$$


[![enter image description here](https://i.stack.imgur.com/uziWp.png)](https://i.stack.imgur.com/uziWp.png)


The $\ce{O}$ atom of $\ce{H2O}$ molecule donates a lone pair to the $\ce{S}$ of $\ce{SO2}$,forming a new $\ce{S\bond{-}O}$ $\sigma$-bond and breaking $\ce{S\bond{=}O}$ $\pi$-bond. 


 2

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](https://en.wikipedia.org/wiki/Octet_rule#Hypervalent_molecules) (That means that the oxygen would have more than 8 valence electrons).  

Sulfur on the other hand is in the third row of the periodic table and can therefore form such molecules. 


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.


 1

Oxygen is more electronegative than sulfur, so it pulls the shared valence electrons closer to itself. This causes sulfur to have a partial positive charge and hence act as the electrophile. Furthermore, sulfur is able to have more than 8 valence electrons as it is able to expand its octet by using the d orbitals.


 -1