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

Correct answer: Trioxosulphate(IV) salt

Explanation: Sulphur(IV) oxide, SO₂, dissolves in water to form sulphurous acid (H₂SO₃), which ionizes to give trioxosulphate(IV) (sulphite) ions, HSO₃⁻ / SO₃²⁻. These ions are readily oxidized to sulphate (SO₄²⁻), so they act as good reducing agents. Hence the reducing behaviour of SO₂ in water is due to the formation of sulphite (trioxosulphate(IV)).

• Hydroxide ion — Incorrect. SO₂ forms an acidic solution, not OH⁻.
• Sulphur(VI) oxide — Incorrect. That’s SO₃, the oxidized product, not what imparts reducing power.
• Hydrogen sulphide — Incorrect. H₂S is a reducing agent but is not formed when SO₂ dissolves in water.
• Trioxosulphate(IV) salt — Correct. Formation of sulphite/bisulphite ions gives the reducing property.