Question

oxygen is acting as an oxidizing agent in all of the following reactions excepta.\\(\ce{2 c(s) + o2(g) -> 2 co(g)}\\)b.\\(\ce{s(s) + o2(g) -> so2(g)}\\)c.\\(\ce{2 f2(g) + o2(g) -> 2 of2(g)}\\)d.\\(\ce{2 na(s) + o2(g) -> na2o2(s)}\\)e.\\(\ce{2 mg(s) + o2(g) -> 2 mgo(s)}\\)

Explanation:

To determine when oxygen is not an oxidizing agent, we analyze the oxidation state of oxygen in each reaction:

• Option A: In \( \ce{O2} \), O has an oxidation state (OS) of 0. In \( \ce{CO} \), O has an OS of -2. Oxygen is reduced (OS decreases), so it acts as an oxidizing agent.
• Option B: In \( \ce{O2} \), O has OS 0. In \( \ce{SO2} \), O has OS -2. Oxygen is reduced, so it acts as an oxidizing agent.
• Option C: In \( \ce{O2} \), O has OS 0. In \( \ce{OF2} \), O has OS +2 (since F has OS -1, and \( 2(-1) + \text{OS of O} = 0 \) gives O as +2). Oxygen is oxidized (OS increases), so it acts as a reducing agent, not an oxidizing agent.
• Option D: In \( \ce{O2} \), O has OS 0. In \( \ce{Na2O2} \), O has OS -1 (since Na has OS +1, and \( 2(+1) + 2(\text{OS of O}) = 0 \) gives O as -1). Oxygen is reduced, so it acts as an oxidizing agent.
• Option E: In \( \ce{O2} \), O has OS 0. In \( \ce{MgO} \), O has OS -2. Oxygen is reduced, so it acts as an oxidizing agent.

Only in reaction C does oxygen act as a reducing agent (not an oxidizing agent).

Answer:

C. \( \ce{2 F2(g) + O2(g) → 2 OF2(g)} \)