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: $\text{CO}$

Let oxidation number of O be $x$. For $\text{CO}$, $\text{C}$ has $+2$, so $+2 + x = 0 \implies x = -2$. O is reduced.

Reaction B: $\text{SO}_2$

Let oxidation number of O be $x$. For $\text{SO}_2$, $\text{S}$ has $+4$, so $+4 + 2x = 0 \implies x = -2$. O is reduced.

Reaction C: $\text{OF}_2$

Let oxidation number of O be $x$. For $\text{OF}_2$, $\text{F}$ has $-1$, so $x + 2(-1) = 0 \implies x = +2$. O is oxidized.

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

Let oxidation number of O be $x$. For $\text{Na}_2\text{O}_2$, $\text{Na}$ has $+1$, so $2(+1) + 2x = 0 \implies x = -1$. O is reduced.

Reaction E: $\text{MgO}$

Let oxidation number of O be $x$. For $\text{MgO}$, $\text{Mg}$ has $+2$, so $+2 + x = 0 \implies x = -2$. O is reduced.

Step4: Identify non-oxidizing agent case

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

Answer:

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