oxygen is acting as an oxidizing agent in all of the following reactions excepta.\n2 c(s) + o₂(g) → 2…

oxygen is acting as an oxidizing agent in all of the following reactions excepta.\n2 c(s) + o₂(g) → 2 co(g)\nb.\ns(s) + o₂(g) → so₂(g)\nc.\n2 f₂(g) + o₂(g) → 2 of₂(g)\nd.\n2 na(s) + o₂(g) → na₂o₂(s)\ne.\n2 mg(s) + o₂(g) → 2 mgo(s)

oxygen is acting as an oxidizing agent in all of the following reactions excepta.\n2 c(s) + o₂(g) → 2 co(g)\nb.\ns(s) + o₂(g) → so₂(g)\nc.\n2 f₂(g) + o₂(g) → 2 of₂(g)\nd.\n2 na(s) + o₂(g) → na₂o₂(s)\ne.\n2 mg(s) + o₂(g) → 2 mgo(s)

Answer

Brief Explanations:

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

  • Option A: In (O_2) (oxidation state = 0), in (CO), O has oxidation state -2. Oxygen is reduced (gains electrons), so it is an oxidizing agent.
  • Option B: In (O_2) (0), in (SO_2), O is -2. Oxygen is reduced, so it is an oxidizing agent.
  • Option C: In (O_2) (0), in (OF_2), O has an oxidation state of +2 (since F is -1, and (2(-1) + x = 0 \implies x = +2)). Oxygen is oxidized (loses electrons), so it is a reducing agent here, not an oxidizing agent.
  • Option D: In (O_2) (0), in (Na_2O_2), O has oxidation state -1. Oxygen is reduced, so it is an oxidizing agent.
  • Option E: In (O_2) (0), in (MgO), O is -2. Oxygen is reduced, so it is an oxidizing agent.

Answer:

C. (2 F_2(g) + O_2(g) \rightarrow 2 OF_2(g))