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

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

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

  • Option A: In (2C(s)+O_2(g)\rightarrow2CO(g)), C goes from 0 to +2 (oxidized), O in (O_2) (0) goes to -2 (reduced). So (O_2) is an oxidizing agent.
  • Option B: In (S(s)+O_2(g)\rightarrow SO_2(g)), S goes from 0 to +4 (oxidized), O in (O_2) (0) goes to -2 (reduced). So (O_2) is an oxidizing agent.
  • Option C: In (2F_2(g)+O_2(g)\rightarrow2OF_2(g)), F in (F_2) (0) goes to -1 (reduced), O in (O_2) (0) goes to +2 (oxidized). Here, (O_2) is oxidized (loses electrons), so it acts as a reducing agent, not an oxidizing agent.
  • Option D: In (2Na(s)+O_2(g)\rightarrow Na_2O_2(s)), Na goes from 0 to +1 (oxidized), O in (O_2) (0) goes to -1 (reduced). So (O_2) is an oxidizing agent.
  • Option E: In (2Mg(s)+O_2(g)\rightarrow2MgO(s)), Mg goes from 0 to +2 (oxidized), O in (O_2) (0) goes to -2 (reduced). So (O_2) is an oxidizing agent.

Answer:

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