Redox reaction reduces sulfate to sulfide

Steps:

• Concentrated H2SO4 reacts with NaI, where I⁻ acts as a strong reducing agent.
• I⁻ reduces SO₄²⁻ to H₂S (sulfur oxidation state -2) and forms HI.
• Excess H₂SO4 oxidizes HI to I₂, producing SO₂ (sulfur +4) as a byproduct.
• Products containing sulfur are H₂S and SO₂; lowest oxidation number is -2 in H₂S.

Why A is correct:

• Oxidation number rules assign -2 to S in H₂S, as H is +1 and the molecule is neutral.

Why the others are wrong:

• B: 0 occurs in elemental S, not formed here as the primary lowest state.
• C: +4 is in SO₂, higher than -2 in H₂S.
• D: +6 is in original H₂SO₄, not a reduced product.

Final answer: A