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 for each reaction

• Reaction A: In $\text{CO}$, O has oxidation number $-2$.
• Reaction B: In $\text{SO}_2$, O has oxidation number $-2$.
• Reaction C: In $\text{OF}_2$, F has oxidation number $-1$, so O has oxidation number $+2$.
• Reaction D: In $\text{Na}_2\text{O}_2$, Na has oxidation number $+1$, so O has oxidation number $-1$.
• Reaction E: In $\text{MgO}$, O has oxidation number $-2$.

Step4: Identify non-oxidizing agent case

Only in reaction C, O's oxidation number increases from $0$ to $+2$, so $\text{O}_2$ acts as a reducing agent, not an oxidizing agent.

Answer:

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