oxygen is acting as an oxidizing agent in all of the following reactions except\na. $ce{2 c(s) + o_{2}(g) ->…

oxygen is acting as an oxidizing agent in all of the following reactions except\na. $ce{2 c(s) + o_{2}(g) -> 2 co(g)}$\nb. $ce{s(s) + o_{2}(g) -> so_{2}(g)}$\nc. $ce{2 f_{2}(g) + o_{2}(g) -> 2 of_{2}(g)}$\nd. $ce{2 na(s) + o_{2}(g) -> na_{2}o_{2}(s)}$\ne. $ce{2 mg(s) + o_{2}(g) -> 2 mgo(s)}$

oxygen is acting as an oxidizing agent in all of the following reactions except\na. $ce{2 c(s) + o_{2}(g) -> 2 co(g)}$\nb. $ce{s(s) + o_{2}(g) -> so_{2}(g)}$\nc. $ce{2 f_{2}(g) + o_{2}(g) -> 2 of_{2}(g)}$\nd. $ce{2 na(s) + o_{2}(g) -> na_{2}o_{2}(s)}$\ne. $ce{2 mg(s) + o_{2}(g) -> 2 mgo(s)}$

Answer

Explanation:

Step1: Define oxidizing agent

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 $-2$. Change: $0 \to -2$ (decrease, O is oxidizing agent).

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

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

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

In $\text{Na}_2\text{O}_2$, Na has oxidation number $+1$. Let O oxidation number = $x$: $$2(+1) + 2x = 0 \implies x = -1$$ Change: $0 \to -1$ (decrease, O is oxidizing agent).

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

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

Answer:

C. $\ce{2 F_{2}(g) + O_{2}(g) -> 2 OF_{2}(g)}$