oxygen is acting as an oxidizing agent in all of the following reactions except\na.\n$2 c(s) + o_2(g)…

oxygen is acting as an oxidizing agent in all of the following reactions except\na.\n$2 c(s) + o_2(g) \\rightarrow 2 co(g)$\n\nb.\n$s(s) + o_2(g) \\rightarrow so_2(g)$\n\nc.\n$2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g)$\n\nd.\n$2 na(s) + o_2(g) \\rightarrow na_2o_2(s)$\n\ne.\n$2 mg(s) + o_2(g) \\rightarrow 2 mgo(s)$

oxygen is acting as an oxidizing agent in all of the following reactions except\na.\n$2 c(s) + o_2(g) \\rightarrow 2 co(g)$\n\nb.\n$s(s) + o_2(g) \\rightarrow so_2(g)$\n\nc.\n$2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g)$\n\nd.\n$2 na(s) + o_2(g) \\rightarrow na_2o_2(s)$\n\ne.\n$2 mg(s) + o_2(g) \\rightarrow 2 mgo(s)$

Answer

Brief Explanations:

To determine when oxygen is not an oxidizing agent, we check the oxidation state change of oxygen. An oxidizing agent is reduced (oxidation state decreases).

  • In option A: C goes from 0 to +2 (oxidized), O in (O_2) (0) goes to -2 in CO (reduced) → O is oxidizing agent.
  • In option B: S goes from 0 to +4 (oxidized), O in (O_2) (0) goes to -2 in (SO_2) (reduced) → O is oxidizing agent.
  • In option C: F in (F_2) (0) goes to -1 in (OF_2) (reduced), O in (O_2) (0) goes to +2 in (OF_2) (oxidized) → O is a reducing agent (not oxidizing agent).
  • In option D: Na goes from 0 to +1 (oxidized), O in (O_2) (0) goes to -1 in (Na_2O_2) (reduced) → O is oxidizing agent.
  • In option E: Mg goes from 0 to +2 (oxidized), O in (O_2) (0) goes to -2 in MgO (reduced) → O is oxidizing agent.

Answer:

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