Question

oxygen is acting as an oxidizing agent in all of the following reactions except
a.
$2 c(s) + o_2(g) \
ightarrow 2 co(g)$

b.
$s(s) + o_2(g) \
ightarrow so_2(g)$

c.
$2 f_2(g) + o_2(g) \
ightarrow 2 of_2(g)$

d.
$2 na(s) + o_2(g) \
ightarrow na_2o_2(s)$

e.
$2 mg(s) + o_2(g) \
ightarrow 2 mgo(s)$

Explanation:

To determine when oxygen is not an oxidizing agent, we analyze the oxidation state of oxygen in each reaction:

• 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 is reduced, so \(\text{O}_2\) is an oxidizing agent.

• 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 an oxidizing agent.

• 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, so \(\text{O}_2\) is a reducing agent (not an oxidizing agent here).

• 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, so \(\text{O}_2\) is an oxidizing agent.

• 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, so \(\text{O}_2\) is an oxidizing agent.

Only in reaction C, oxygen is oxidized (loses electrons), so it acts as a reducing agent, not an oxidizing agent.

Answer:

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