Question

oxygen is acting as an oxidizing agent in all of the following reactions excepta.\\( 2 c(s) + o_2(g) \
ightarrow 2 co(g)\\)b.\\( s(s) + o_2(g) \
ightarrow so_2(g)\\)c.\\( 2 f_2(g) + o_2(g) \
ightarrow 2 of_2(g)\\)d.\\( 2 na(s) + o_2(g) \
ightarrow na_2o_2(s)\\)e.\\( 2 mg(s) + o_2(g) \
ightarrow 2 mgo(s)\\)

Explanation:

Step1: Define oxidizing agent

An oxidizing agent gains electrons (oxidation number decreases).

Step2: Find O oxidation number in reactants

In $\text{O}_2$, oxidation number of O = $0$.

Step3: Calculate O oxidation number in products for each reaction

Reaction A: $\text{CO}$

Let O oxidation number = $x$. $\text{C}$ is $+2$, so $+2 + x = 0 \implies x = -2$.

Reaction B: $\text{SO}_2$

Let O oxidation number = $x$. $\text{S}$ is $+4$, so $+4 + 2x = 0 \implies x = -2$.

Reaction C: $\text{OF}_2$

Let O oxidation number = $x$. $\text{F}$ is $-1$, so $x + 2(-1) = 0 \implies x = +2$.

Reaction D: $\text{Na}_2\text{O}_2$

Let O oxidation number = $x$. $\text{Na}$ is $+1$, so $2(+1) + 2x = 0 \implies x = -1$.

Reaction E: $\text{MgO}$

Let O oxidation number = $x$. $\text{Mg}$ is $+2$, so $+2 + x = 0 \implies x = -2$.

Step4: Identify non-oxidizing agent case

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

Answer:

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