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

oxygen is acting as an oxidizing agent in all of the following reactions excepta.\\( 2 c(s) + o_2(g) \\rightarrow 2 co(g)\\)b.\\( s(s) + o_2(g) \\rightarrow so_2(g)\\)c.\\( 2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g)\\)d.\\( 2 na(s) + o_2(g) \\rightarrow na_2o_2(s)\\)e.\\( 2 mg(s) + o_2(g) \\rightarrow 2 mgo(s)\\)

oxygen is acting as an oxidizing agent in all of the following reactions excepta.\\( 2 c(s) + o_2(g) \\rightarrow 2 co(g)\\)b.\\( s(s) + o_2(g) \\rightarrow so_2(g)\\)c.\\( 2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g)\\)d.\\( 2 na(s) + o_2(g) \\rightarrow na_2o_2(s)\\)e.\\( 2 mg(s) + o_2(g) \\rightarrow 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 (oxidation state decreases), so it acts as 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\times(-1)+x = 0) gives (x = +2)). Here, oxygen's oxidation state increases (from 0 to +2), so it is oxidized. Thus, (O_2) is acting as a reducing agent, not an oxidizing agent.
  • Option D: In (O_2) (0), in (Na_2O_2), O has an oxidation state of -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))