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 states:

• In an oxidizing agent, the element is reduced (oxidation state decreases).
• For reaction C: \(2 \text{F}_2(\text{g}) + \text{O}_2(\text{g})

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

• Oxidation state of F in \(\text{F}_2\) is 0, in \(\text{OF}_2\) F is -1 (so F is reduced? No, wait: F is more electronegative than O. In \(\text{OF}_2\), O has an oxidation state of +2 (since F is -1, 2*(-1) + x = 0 → x = +2).
• Oxidation state of O in \(\text{O}_2\) is 0, in \(\text{OF}_2\) O is +2. So O is oxidized (oxidation state increases from 0 to +2), meaning O is acting as a reducing agent here, not an oxidizing agent.
• For other reactions (A, B, D, E):
• In A: C (0) → CO (+2), O (0) → CO (-2) (O is reduced, oxidizing agent).
• In B: S (0) → SO₂ (+4), O (0) → SO₂ (-2) (O is reduced, oxidizing agent).
• In D: Na (0) → Na₂O₂ (+1), O (0) → Na₂O₂ (-1) (O is reduced, oxidizing agent).
• In E: Mg (0) → MgO (+2), O (0) → MgO (-2) (O is reduced, oxidizing agent).

Answer:

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