Describe the hydration of \(\mathrm{SO}_{2}\) as a Lewis acid-base reaction. (Hint: Refer to the discussion of the hydration of \(\mathrm{CO}_{2}\) in Section \(\left.15.12 .\right)\)

Short Answer

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In the hydration of \(\mathrm{SO}_{2}\), \(\mathrm{SO}_{2}\) acts as a Lewis acid and water (\(H_{2}O\)) as a Lewis base. The oxygen atom of \(H_{2}O\) donates an electron pair to \(\mathrm{SO}_{2}\), forming sulphurous acid (\(H_{2}SO_{3}\)). This process is a representation of a Lewis acid-base reaction.

Step by step solution

01

Identifying Lewis Acids and Bases

In a Lewis acid-base reaction, a Lewis acid is a species that accepts an electron pair, while a Lewis base is a species that donates an electron pair. In the case of the hydration of \(\mathrm{SO}_{2}\), water (\(H_{2}O\)) acts as the Lewis base and \(\mathrm{SO}_{2}\) as the Lewis acid since the latter accepts the electron pair from the water molecule.
02

Writing Down the Reaction

Now that the Lewis acid and base have been identified, the hydration reaction can be written as follows: \[\mathrm{SO}_{2} + H_{2}O \rightarrow H_{2}SO_{3}\]where \(\mathrm{SO}_{2}\) (Lewis acid) accepts an electron pair from \(H_{2}O\) (Lewis base) to form \(H_{2}SO_{3}\) (sulphurous acid).
03

Detailing the Bond Formation

In detail, the oxygen atom of water molecule donates its electron pair to form a bond with the sulfur atom of \(\mathrm{SO}_{2}\). Thus, the sulfur of \(\mathrm{SO}_{2}\) shares its vacant orbital with water's electron pair, leading to the formation of sulphurous acid (\(H_{2}SO_{3}\)).

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