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 a substance that gets reduced (its oxidation number decreases) during a reaction. We calculate oxidation numbers for O in each reactant and product.

Step2: Analyze Option A

Reactant O: $O_2$ has oxidation number 0. Product O: In CO, O is -2. Oxidation number decreases, so $O_2$ is oxidizing agent.

Step3: Analyze Option B

Reactant O: $O_2$ has oxidation number 0. Product O: In $SO_2$, O is -2. Oxidation number decreases, so $O_2$ is oxidizing agent.

Step4: Analyze Option C

Reactant O: $O_2$ has oxidation number 0. Product O: In $OF_2$, F is -1, so O is +2. Oxidation number increases, so $O_2$ is a reducing agent, not oxidizing.

Step5: Analyze Option D

Reactant O: $O_2$ has oxidation number 0. Product O: In $Na_2O_2$, O is -1. Oxidation number decreases, so $O_2$ is oxidizing agent.

Step6: Analyze Option E

Reactant O: $O_2$ has oxidation number 0. Product O: In MgO, O is -2. Oxidation number decreases, so $O_2$ is oxidizing agent.

Answer:

C. $2\ F_2(g) + O_2(g)
ightarrow 2\ OF_2(g)$