The oxidation number of sulphur in Fe2(SO4)3 is

Analyze the question: We need the oxidation number of sulphur (S) in iron(III) sulphate, Fe₂(SO₄)₃.

Concept: In a neutral compound, the oxidation numbers of all atoms add up to zero. In a polyatomic ion, they add up to the charge on the ion. Oxygen is usually -2.

Solution:

Method 1: Use the whole compound

• Fe₂(SO₄)₃ has 2 Fe atoms, 3 S atoms and 12 O atoms.
• Fe is +3 (iron(III)), so 2 × (+3) = +6
• O: 12 × (-2) = -24
• Let the oxidation number of S be x, so 3x is the total for sulphur.

+6 + 3x + (-24) = 0 3x = 18 x = +6

Method 2: Use the sulphate ion

• The sulphate ion is SO₄²⁻.
• x + 4(-2) = -2
• x - 8 = -2
• x = +6

Both methods agree.

Answer: The oxidation number of sulphur is +6, which is option D.

Tip: Sulphur in sulphates (SO₄²⁻) is always +6. In sulphites (SO₃²⁻) it is +4. In H₂S it is -2. Learn these three and you can answer many similar questions quickly. For any sulphate, just solve x + 4(-2) = -2.