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 in which reaction oxygen is not acting as an oxidizing agent, we analyze the oxidation states of oxygen in each reaction:

• Oxidizing agent: A substance that gets reduced (oxidation state decreases).
• Reduction: Gain of electrons (oxidation state decreases).
• Oxidation: Loss of electrons (oxidation state increases).

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

ightarrow 2\ CO(g)} \)

• Oxidation state of \( O \) in \( O_2 \): \( 0 \).
• Oxidation state of \( O \) in \( CO \): \( -2 \).
• \( O \) gains electrons (oxidation state decreases from \( 0 \) to \( -2 \)) → \( O_2 \) is reduced (oxidizing agent).

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

ightarrow SO_2(g)} \)

• Oxidation state of \( O \) in \( O_2 \): \( 0 \).
• Oxidation state of \( O \) in \( SO_2 \): \( -2 \).
• \( O \) gains electrons (oxidation state decreases from \( 0 \) to \( -2 \)) → \( O_2 \) is reduced (oxidizing agent).

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

ightarrow 2\ OF_2(g)} \)

• Oxidation state of \( O \) in \( O_2 \): \( 0 \).
• Oxidation state of \( O \) in \( OF_2 \): Fluorine is more electronegative than oxygen, so \( F \) has \( -1 \). Let oxidation state of \( O \) be \( x \). Then \( x + 2(-1) = 0 \) → \( x = +2 \).
• \( O \) loses electrons (oxidation state increases from \( 0 \) to \( +2 \)) → \( O_2 \) is oxidized (not an oxidizing agent).

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

ightarrow Na_2O_2(s)} \)

• Oxidation state of \( O \) in \( O_2 \): \( 0 \).
• Oxidation state of \( O \) in \( Na_2O_2 \): \( Na \) is \( +1 \). Let oxidation state of \( O \) be \( x \). Then \( 2(+1) + 2x = 0 \) → \( x = -1 \).
• \( O \) gains electrons (oxidation state decreases from \( 0 \) to \( -1 \)) → \( O_2 \) is reduced (oxidizing agent).

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

ightarrow 2\ MgO(s)} \)

• Oxidation state of \( O \) in \( O_2 \): \( 0 \).
• Oxidation state of \( O \) in \( MgO \): \( -2 \).
• \( O \) gains electrons (oxidation state decreases from \( 0 \) to \( -2 \)) → \( O_2 \) is reduced (oxidizing agent).

In reaction C, oxygen’s oxidation state increases (it is oxidized), so \( O_2 \) is not acting as an oxidizing agent here.

Answer:

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