Question

oxygen is acting as an oxidizing agent in all of the following reactions excepta.\\( 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:

• Oxidizing agent: A substance that gets reduced (decrease in oxidation state).
• Oxidation state of O in \(O_2\): 0 (elemental form).

Reaction A: \(2\ C(s) + O_2(g)

ightarrow 2\ CO(g)\)

• O in \(O_2\) (0) → O in \(CO\) (-2). Oxidation state decreases (reduction) → \(O_2\) is oxidizing agent.

Reaction B: \(S(s) + O_2(g)

ightarrow SO_2(g)\)

• O in \(O_2\) (0) → O in \(SO_2\) (-2). Oxidation state decreases → \(O_2\) is oxidizing agent.

Reaction C: \(2\ F_2(g) + O_2(g)

ightarrow 2\ OF_2(g)\)

• O in \(O_2\) (0) → O in \(OF_2\) (+2). Oxidation state increases (oxidation) → \(O_2\) is oxidized, so it acts as a reducing agent (not oxidizing agent).

Reaction D: \(2\ Na(s) + O_2(g)

ightarrow Na_2O_2(s)\)

• O in \(O_2\) (0) → O in \(Na_2O_2\) (-1). Oxidation state decreases → \(O_2\) is oxidizing agent.

Reaction E: \(2\ Mg(s) + O_2(g)

ightarrow 2\ MgO(s)\)

• O in \(O_2\) (0) → O in \(MgO\) (-2). Oxidation state decreases → \(O_2\) is oxidizing agent.

Answer:

C. \(2\ F_2(g) + O_2(g)
ightarrow 2\ OF_2(g)\)