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 for each option

Option A:

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

Option B:

In $\text{SO}_2$, O has oxidation number $-2$. Change: $0 \to -2$ (decrease, O is oxidizing agent).

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).

Option D:

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

Option E:

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

Answer:

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