Sulphur (IV)oxide is a strong reducing agent in the presence of water due to the formation of

Analysis

The question asks why sulphur(IV) oxide ($SO_2$) acts as a strong reducing agent when water is present. We need to identify the specific chemical species formed during this reaction that facilitates the reduction process.

Concept Explanation

When sulphur(IV) oxide ($SO_2$) dissolves in water, it reacts to form trioxosulphuric(IV) acid ($H_2SO_3$):

In this reaction, the sulphur atom in $H_2SO_3$ has an oxidation state of +4. Because sulphur can be oxidized to a higher oxidation state (like +6 in sulphate ions), $H_2SO_3$ (or the resulting trioxosulphate(IV) ions, $SO_3^{2-}$) readily donates electrons to other substances, thereby acting as a reducing agent.

The reducing property is specifically attributed to the formation of the trioxosulphate(IV) species, which is easily oxidized to tetraoxosulphate(VI) ($SO_4^{2-}$).

Step-by-Step Solution

1. Reaction with water: $SO_2$ reacts with water to form $H_2SO_3$.
2. Dissociation: $H_2SO_3$ dissociates in water to produce hydrogen ions and trioxosulphate(IV) ions ($SO_3^{2-}$).
3. Redox behavior: The trioxosulphate(IV) ion ($SO_3^{2-}$) is unstable in the presence of oxidizing agents and is easily oxidized to the more stable tetraoxosulphate(VI) ion ($SO_4^{2-}$).
4. Conclusion: Because it is easily oxidized, it forces the other substance to be reduced. Therefore, the presence of the trioxosulphate(IV) species is the reason for its reducing power.

Identifying the Answer

Looking at the options provided:

• A is incorrect because hydroxide ions are associated with bases.
• B is incorrect as sulphur(VI) oxide is the product of oxidation, not the cause of the reducing power.
• C is incorrect as hydrogen sulphide is a different compound entirely.
• D correctly identifies the species formed that facilitates the reduction.

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Exam Tip: Whenever you see "Sulphur(IV) oxide" and "reducing agent" in a question, always look for options involving "trioxosulphate(IV)" or "sulphite." Remember that "IV" usually corresponds to the lower oxidation state, which is ready to be oxidized to the "VI" state.