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 role

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