Question

oxygen is acting as an oxidizing agent in all of the following reactions except
a. $2\\ c(s) + o_2(g) \
ightarrow 2\\ co(g)$
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 $\ce{O2(g)}$, oxidation number of O is $0$.

Step3: Calculate O oxidation number in products for each reaction

Reaction A: $\ce{2 C(s) + O2(g) -> 2 CO(g)}$

In $\ce{CO}$, let O oxidation number be $x$. $\ce{C}$ is $+2$, so $+2 + x = 0 \implies x = -2$. O is reduced.

Reaction B: $\ce{S(s) + O2(g) -> SO2(g)}$

In $\ce{SO2}$, let O oxidation number be $x$. $\ce{S}$ is $+4$, so $+4 + 2x = 0 \implies x = -2$. O is reduced.

Reaction C: $\ce{2 F2(g) + O2(g) -> 2 OF2(g)}$

In $\ce{OF2}$, let O oxidation number be $x$. $\ce{F}$ is $-1$, so $x + 2(-1) = 0 \implies x = +2$. O is oxidized.

Reaction D: $\ce{2 Na(s) + O2(g) -> Na2O2(s)}$

In $\ce{Na2O2}$, let O oxidation number be $x$. $\ce{Na}$ is $+1$, so $2(+1) + 2x = 0 \implies x = -1$. O is reduced.

Reaction E: $\ce{2 Mg(s) + O2(g) -> 2 MgO(s)}$

In $\ce{MgO}$, let O oxidation number be $x$. $\ce{Mg}$ is $+2$, so $+2 + x = 0 \implies x = -2$. O is reduced.

Step4: Identify non-oxidizing agent case

Only in reaction C, O is oxidized, so $\ce{O2}$ is not an oxidizing agent.

Answer:

C. $\ce{2 F2(g) + O2(g) -> 2 OF2(g)}$