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: Recall oxidizing agent definition

An oxidizing agent is reduced (gains electrons, oxidation state decreases). We analyze each reaction's oxidation states.

Step2: Analyze Option A

In \(2\text{C}(s) + \text{O}_2(g)
ightarrow 2\text{CO}(g)\):

• O in \(\text{O}_2\) (oxidation state = 0) becomes O in \(\text{CO}\) (oxidation state = -2). O gains electrons (reduced), so \(\text{O}_2\) is oxidizing agent.

Step3: Analyze Option B

In \(\text{S}(s) + \text{O}_2(g)
ightarrow \text{SO}_2(g)\):

• O in \(\text{O}_2\) (0) becomes O in \(\text{SO}_2\) (-2). O is reduced, so \(\text{O}_2\) is oxidizing agent.

Step4: Analyze Option C

In \(2\text{F}_2(g) + \text{O}_2(g)
ightarrow 2\text{OF}_2(g)\):

• O in \(\text{O}_2\) (0) becomes O in \(\text{OF}_2\) (+2). O loses electrons (oxidized), so \(\text{O}_2\) is reducing agent (not oxidizing agent here).

Step5: Analyze Option D

In \(2\text{Na}(s) + \text{O}_2(g)
ightarrow \text{Na}_2\text{O}_2(s)\):

• O in \(\text{O}_2\) (0) becomes O in \(\text{Na}_2\text{O}_2\) (-1). O gains electrons (reduced), so \(\text{O}_2\) is oxidizing agent.

Step6: Analyze Option E

In \(2\text{Mg}(s) + \text{O}_2(g)
ightarrow 2\text{MgO}(s)\):

• O in \(\text{O}_2\) (0) becomes O in \(\text{MgO}\) (-2). O gains electrons (reduced), so \(\text{O}_2\) is oxidizing agent.

Answer:

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