oxygen is acting as an oxidizing agent in all of the following reactions except\n\na. $2\\text{c}(\\text{s})…

Explanation:

Step1: Define oxidizing agent

An oxidizing agent gains electrons and its oxidation state decreases.

Step2: Analyze reaction A

Carbon state changes from $0$ to $+2$; Oxygen from $0$ to $-2$. $$2\text{C}(s) + \text{O}_2(g) \rightarrow 2\text{CO}(g)$$

Step3: Analyze reaction B

Sulfur state changes from $0$ to $+4$; Oxygen from $0$ to $-2$. $$\text{S}(s) + \text{O}_2(g) \rightarrow \text{SO}_2(g)$$

Step4: Analyze reaction C

Fluorine is more electronegative than oxygen. Oxygen state changes from $0$ to $+2$. $$2\text{F}_2(g) + \text{O}_2(g) \rightarrow 2\text{OF}_2(g)$$

Step5: Analyze reaction D

Sodium state changes from $0$ to $+1$; Oxygen from $0$ to $-1$. $$2\text{Na}(s) + \text{O}_2(g) \rightarrow \text{Na}_2\text{O}_2(s)$$

Step6: Analyze reaction E

Magnesium state changes from $0$ to $+2$; Oxygen from $0$ to $-2$. $$2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s)$$

Step7: Identify the exception

In reaction C, oxygen is oxidized, making it the reducing agent.

Answer:

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