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:

Step1: Define oxidizing agent role

An oxidizing agent is reduced (gains electrons, oxidation number decreases).

Step2: Find O oxidation number in reactants

In $\text{O}_2$, oxidation number of O is $0$.

Step3: Calculate O oxidation number in products (Option A)

In $\text{CO}$, O has oxidation number $-2$. O is reduced.

Step4: Calculate O oxidation number in products (Option B)

In $\text{SO}_2$, O has oxidation number $-2$. O is reduced.

Step5: Calculate O oxidation number in products (Option C)

In $\text{OF}_2$, F has oxidation number $-1$. Let O's oxidation number be $x$:
$$x + 2(-1) = 0 \implies x = +2$$
O is oxidized (loses electrons), so it is not an oxidizing agent here.

Step6: Verify remaining options (D, E)

• In $\text{Na}_2\text{O}_2$, O has oxidation number $-1$ (reduced from $0$)
• In $\text{MgO}$, O has oxidation number $-2$ (reduced from $0$)

Answer:

C. $\ce{2 F_{2}(g) + O_{2}(g) -> 2 OF_{2}(g)}$