oxygen is acting as an oxidizing agent in all of the following reactions except\n\na. $2 \\text{ c(s)} +…
oxygen is acting as an oxidizing agent in all of the following reactions except\n\na. $2 \\text{ c(s)} + \\text{o}_2\\text{(g)} \\rightarrow 2 \\text{ co(g)}$\n\nb. $\\text{s(s)} + \\text{o}_2\\text{(g)} \\rightarrow \\text{so}_2\\text{(g)}$\n\nc. $2 \\text{ f}_2\\text{(g)} + \\text{o}_2\\text{(g)} \\rightarrow 2 \\text{ of}_2\\text{(g)}$\n\nd. $2 \\text{ na(s)} + \\text{o}_2\\text{(g)} \\rightarrow \\text{na}_2\\text{o}_2\\text{(s)}$\n\ne. $2 \\text{ mg(s)} + \\text{o}_2\\text{(g)} \\rightarrow 2 \\text{ mgo(s)}$
Answer
Explanation:
Step1: Define oxidizing agent
An oxidizing agent gains electrons and undergoes reduction, meaning its oxidation state decreases.
Step2: Analyze oxidation states in A
In $2\text{C} + \text{O}_2 \rightarrow 2\text{CO}$, Oxygen goes from $0$ to $-2$. It is reduced (oxidizing agent).
Step3: Analyze oxidation states in B
In $\text{S} + \text{O}_2 \rightarrow \text{SO}_2$, Oxygen goes from $0$ to $-2$. It is reduced (oxidizing agent).
Step4: Analyze oxidation states in C
In $2\text{F}_2 + \text{O}_2 \rightarrow 2\text{OF}_2$, Fluorine is more electronegative than Oxygen. Oxygen goes from $0$ to $+2$. It is oxidized (reducing agent).
Step5: Analyze oxidation states in D
In $2\text{Na} + \text{O}_2 \rightarrow \text{Na}_2\text{O}_2$, Oxygen goes from $0$ to $-1$. It is reduced (oxidizing agent).
Step6: Analyze oxidation states in E
In $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$, Oxygen goes from $0$ to $-2$. It is reduced (oxidizing agent).
Answer:
C. $2\text{F}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{OF}_2(\text{g})$