Question

oxygen is acting as an oxidizing agent in all of the following reactions except
a. $ce{2 c(s) + o_{2}(g) -> 2 co(g)}$
b. $ce{s(s) + o_{2}(g) -> so_{2}(g)}$
c. $ce{2 f_{2}(g) + o_{2}(g) -> 2 of_{2}(g)}$
d. $ce{2 na(s) + o_{2}(g) -> na_{2}o_{2}(s)}$
e. $ce{2 mg(s) + o_{2}(g) -> 2 mgo(s)}$

Explanation:

Step1: Define oxidizing agent

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 = $0$.

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

In $\text{CO}$, O oxidation number = $-2$.
Change: $0 \to -2$ (reduction; O is oxidizing agent).

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

In $\text{SO}_2$, O oxidation number = $-2$.
Change: $0 \to -2$ (reduction; O is oxidizing agent).

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

In $\text{OF}_2$, F oxidation number = $-1$. Let O oxidation number = $x$.
$x + 2(-1) = 0 \implies x = +2$.
Change: $0 \to +2$ (oxidation; O is not oxidizing agent).

Step6: Verify remaining options (D,E)

In $\text{Na}_2\text{O}_2$, O oxidation number = $-1$; in $\text{MgO}$, O = $-2$. Both are reductions from $0$.

Answer:

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