Question

oxygen is acting as an oxidizing agent in all of the following reactions except
a.
$2\\ \mathrm{c}(\mathrm{s}) + \mathrm{o}_2(\mathrm{g}) \
ightarrow 2\\ \mathrm{co}(\mathrm{g})$

b.
$\mathrm{s}(\mathrm{s}) + \mathrm{o}_2(\mathrm{g}) \
ightarrow \mathrm{so}_2(\mathrm{g})$

c.
$2\\ \mathrm{f}_2(\mathrm{g}) + \mathrm{o}_2(\mathrm{g}) \
ightarrow 2\\ \mathrm{of}_2(\mathrm{g})$

d.
$2\\ \mathrm{na}(\mathrm{s}) + \mathrm{o}_2(\mathrm{g}) \
ightarrow \mathrm{na}_2\mathrm{o}_2(\mathrm{s})$

e.
$2\\ \mathrm{mg}(\mathrm{s}) + \mathrm{o}_2(\mathrm{g}) \
ightarrow 2\\ \mathrm{mgo}(\mathrm{s})$

Explanation:

Step1: Recall oxidizing agent definition

An oxidizing agent is reduced (gains electrons, oxidation state decreases).

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 CO (oxidation state -2). O is reduced, so \(\text{O}_2\) is an 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, \(\text{O}_2\) is an 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 is oxidized (loses electrons, oxidation state increases), so \(\text{O}_2\) is a reducing agent, not an 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 is reduced, \(\text{O}_2\) is an 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 is reduced, \(\text{O}_2\) is an oxidizing agent.

Answer:

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