Question

oxygen is acting as an oxidizing agent in all of the following reactions except
a.
2 c(s) + o₂(g) → 2 co(g)
b.
s(s) + o₂(g) → so₂(g)
c.
2 f₂(g) + o₂(g) → 2 of₂(g)
d.
2 na(s) + o₂(g) → na₂o₂(s)
e.
2 mg(s) + o₂(g) → 2 mgo(s)

Explanation:

Step1: Define oxidizing agent

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

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

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

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

In $\text{OF}_2(g)$, F has oxidation number $-1$. Let O oxidation number = $x$.
$x + 2(-1) = 0 \implies x = +2$.
Change: $0
ightarrow +2$ (increase, O is reducing agent).

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

In $\text{Na}_2\text{O}_2(s)$, Na has oxidation number $+1$. Let O oxidation number = $x$.
$2(+1) + 2x = 0 \implies x = -1$.
Change: $0
ightarrow -1$ (decrease, O is oxidizing agent).

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

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

Answer:

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