Question

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

b.
s(s) + o₂(g) → so₂(g)

c.
2 f₂(g) + o₂(g) → 2 of₂(g)

d.
2 na(s) + o₂(g) → na₂o₂(s)

e.
2 mg(s) + o₂(g) → 2 mgo(s)

Explanation:

To determine when oxygen is not an oxidizing agent, we analyze the oxidation state changes:

• Option A: In \(2C(s)+O_2(g)

ightarrow2CO(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)

ightarrow 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)

ightarrow2OF_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)

ightarrow 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)

ightarrow2MgO(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. \(2 F_2(g) + O_2(g)
ightarrow 2 OF_2(g)\)