The oxidation number of sulphur is +4 in

Analyze the question: We need the compound in which sulphur has an oxidation number of +4.

Concept: In a neutral compound, the oxidation numbers of all atoms add up to zero. Common values: H = +1, O = −2, Na = +1.

Solution:

• A. Na₂S₂O₃ (the option is typed "Na2S>2O3", which is a typo for sodium thiosulphate): 2(+1) + 2(S) + 3(−2) = 0 2 + 2S − 6 = 0, so 2S = +4 and S = +2

• B. H₂SO₃: 2(+1) + S + 3(−2) = 0 2 + S − 6 = 0, so S = +4 ✔

• C. H₂SO₄: 2(+1) + S + 4(−2) = 0 2 + S − 8 = 0, so S = +6

• D. SO₃: S + 3(−2) = 0, so S = +6

Answer: Only H₂SO₃ (sulphurous acid) has sulphur with an oxidation number of +4.

Tip: The "-ous" acid, sulphurous acid (H₂SO₃), has sulphur at +4, and the "-ic" acid, sulphuric acid (H₂SO₄), has sulphur at +6. Also remember that the "ite" ending (sulphite, SO₃²⁻) means +4 and the "ate" ending (sulphate, SO₄²⁻) means +6.