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 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 for each reaction

Reaction A: $\text{CO}$

O oxidation number: $-2$ (decreases from $0$)

Reaction B: $\text{SO}_2$

O oxidation number: $-2$ (decreases from $0$)

Reaction C: $\text{OF}_2$

F oxidation number is $-1$. Let O oxidation number = $x$.
$x + 2(-1) = 0 \implies x = +2$ (increases from $0$)

Reaction D: $\text{Na}_2\text{O}_2$

Na oxidation number is $+1$. Let O oxidation number = $x$.
$2(+1) + 2x = 0 \implies x = -1$ (decreases from $0$)

Reaction E: $\text{MgO}$

O oxidation number: $-2$ (decreases from $0$)

Step4: Identify non-oxidizing agent case

Only in reaction C, O is oxidized, so $\text{O}_2$ is not an oxidizing agent.

Answer:

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