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 gains electrons, so its 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 $-2$. $0 \to -2$ (decrease, O is oxidizing agent).

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

In $\text{SO}_2$, O has oxidation number $-2$. $0 \to -2$ (decrease, O is oxidizing agent).

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

In $\text{OF}_2$, F has oxidation number $-1$. Let O oxidation number = $x$. $x + 2(-1) = 0 \implies x = +2$. $0 \to +2$ (increase, O is reducing agent).

Step6: Calculate O oxidation number in products (Option D)

In $\text{Na}_2\text{O}_2$, O has oxidation number $-1$. $0 \to -1$ (decrease, O is oxidizing agent).

Step7: Calculate O oxidation number in products (Option E)

In $\text{MgO}$, O has oxidation number $-2$. $0 \to -2$ (decrease, O is oxidizing agent).

Answer:

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