In the reaction 2H2S(g) + SO2(g) → 2H2SO(I)+3S(s)SO2is acting as _agent

Analyze the question: The reaction is really 2H₂S(g) + SO₂(g) → 2H₂O(l) + 3S(s). The question has typos ("2H2SO(I)" should be 2H₂O(l)). We need to decide the role of SO₂.

Concept: An oxidizing agent gains electrons, so its oxidation number decreases. A reducing agent loses electrons, so its oxidation number increases.

Solution:

1. Oxidation number of S in H₂S: H is +1, so S is -2.
2. Oxidation number of S in SO₂: O is -2, so S is +4.
3. Oxidation number of S in elemental S: 0.

Changes:

• H₂S: S goes from -2 to 0. This is an increase, so H₂S is oxidized and acts as the reducing agent.
• SO₂: S goes from +4 to 0. This is a decrease, so SO₂ is reduced and acts as the oxidizing agent.

Electron check: 2 S atoms each lose 2 electrons (4 e⁻ lost in total), and 1 S atom gains 4 electrons. The electrons balance.

Answer: SO₂ is acting as an oxidizing agent, which is option C.

Tip: Remember "OIL RIG": Oxidation Is Loss, Reduction Is Gain (of electrons). The substance that is reduced is the oxidizing agent. Always work out the oxidation number of the key element before and after the reaction.