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$2 c(s) + o_2(g) \\rightarrow 2 co(g)$\nb.\n$s(s) + o_2(g) \\rightarrow so_2(g)$\nc.\n$2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g)$\nd.\n$2 na(s) + o_2(g) \\rightarrow na_2o_2(s)$\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$2 c(s) + o_2(g) \\rightarrow 2 co(g)$\nb.\n$s(s) + o_2(g) \\rightarrow so_2(g)$\nc.\n$2 f_2(g) + o_2(g) \\rightarrow 2 of_2(g)$\nd.\n$2 na(s) + o_2(g) \\rightarrow na_2o_2(s)$\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 analyze the oxidation states:

  • In an oxidizing agent, the element is reduced (oxidation state decreases).
  • For reaction C: (2 \text{F}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2 \text{OF}_2(\text{g}))
    • Oxidation state of F in (\text{F}_2) is 0, in (\text{OF}_2) F is -1 (so F is reduced? No, wait: F is more electronegative than O. In (\text{OF}_2), O has an oxidation state of +2 (since F is -1, 2*(-1) + x = 0 → x = +2).
    • Oxidation state of O in (\text{O}_2) is 0, in (\text{OF}_2) O is +2. So O is oxidized (oxidation state increases from 0 to +2), meaning O is acting as a reducing agent here, not an oxidizing agent.
  • For other reactions (A, B, D, E):
    • In A: C (0) → CO (+2), O (0) → CO (-2) (O is reduced, oxidizing agent).
    • In B: S (0) → SO₂ (+4), O (0) → SO₂ (-2) (O is reduced, oxidizing agent).
    • In D: Na (0) → Na₂O₂ (+1), O (0) → Na₂O₂ (-1) (O is reduced, oxidizing agent).
    • In E: Mg (0) → MgO (+2), O (0) → MgO (-2) (O is reduced, oxidizing agent).

Answer:

C. (2 \text{F}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2 \text{OF}_2(\text{g}))