oxygen is acting as an oxidizing agent in all of the following reactions except\na.\n2 c(s) + o₂(g) → 2…

oxygen is acting as an oxidizing agent in all of the following reactions except\na.\n2 c(s) + o₂(g) → 2 co(g)\nb.\ns(s) + o₂(g) → so₂(g)\nc.\n2 f₂(g) + o₂(g) → 2 of₂(g)\nd.\n2 na(s) + o₂(g) → na₂o₂(s)\ne.\n2 mg(s) + o₂(g) → 2 mgo(s)

oxygen is acting as an oxidizing agent in all of the following reactions except\na.\n2 c(s) + o₂(g) → 2 co(g)\nb.\ns(s) + o₂(g) → so₂(g)\nc.\n2 f₂(g) + o₂(g) → 2 of₂(g)\nd.\n2 na(s) + o₂(g) → na₂o₂(s)\ne.\n2 mg(s) + o₂(g) → 2 mgo(s)

Answer

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) \rightarrow 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) \rightarrow \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) \rightarrow 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) \rightarrow \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) \rightarrow 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}) \rightarrow 2 \text{OF}_2(\text{g}))