Sulphur has an oxidation number of +4 in

Analyze the question: We need the compound in which sulphur has an oxidation number of +4. We work it out by making the oxidation numbers in each compound add up to zero.

Key rules:

• Oxygen is usually -2
• Hydrogen is +1 (with non-metals)
• Group 1 metals (Na) are +1
• Group 2 metals (Mg) are +2
• The sum of oxidation numbers in a neutral compound is 0

Step-by-step solution:

A. H₂S: 2(+1) + S = 0, so S = -2

B. H₂SO₄: 2(+1) + S + 4(-2) = 0, so 2 + S - 8 = 0, giving S = +6

C. MgSO₄: (+2) + S + 4(-2) = 0, so 2 + S - 8 = 0, giving S = +6

D. Na₂SO₃: 2(+1) + S + 3(-2) = 0, so 2 + S - 6 = 0, giving S = +4 ✔

E. SO₃: S + 3(-2) = 0, so S = +6

Tip: Names ending in -ite (sulphite, SO₃²⁻) usually have the element in a lower oxidation state than those ending in -ate (sulphate, SO₄²⁻). So sulphite gives S = +4, while sulphate gives S = +6. Spotting "-ite" can save you time in the exam.