Question

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

Explanation:

Step1: Define oxidizing agent role

An oxidizing agent gains electrons, so its oxidation number decreases.

Step2: Find O oxidation number in reactants

In $\text{O}_2(g)$, oxidation number of O is $0$.

Step3: Calculate O oxidation number in products (Option A)

In $\text{CO}(g)$: let O = $x$, C = $+2$. $+2 + x = 0 \implies x = -2$. O is reduced.

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

In $\text{SO}_2(g)$: let O = $x$, S = $+4$. $+4 + 2x = 0 \implies x = -2$. O is reduced.

Step5: Calculate O oxidation number in products (Option C)

In $\text{OF}_2(g)$: let O = $x$, F = $-1$. $x + 2(-1) = 0 \implies x = +2$. O is oxidized.

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

In $\text{Na}_2\text{O}_2(s)$: let O = $x$, Na = $+1$. $2(+1) + 2x = 0 \implies x = -1$. O is reduced.

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

In $\text{MgO}(s)$: let O = $x$, Mg = $+2$. $+2 + x = 0 \implies x = -2$. O is reduced.

Answer:

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