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:

• In an oxidizing agent, the element (oxygen here) gains electrons, so its oxidation state decreases.
• Option A: In \(O_2\) (oxidation state = 0), in \(CO\), O has oxidation state -2. Oxidation state of O decreases (0 → -2), so \(O_2\) is an oxidizing agent.
• Option B: In \(O_2\) (0), in \(SO_2\), O is -2. Oxidation state decreases, \(O_2\) is an oxidizing agent.
• Option C: In \(O_2\) (0), in \(OF_2\), F is -1 (more electronegative than O), so O has oxidation state +2. Oxidation state of O increases (0 → +2), meaning O loses electrons. Thus, \(O_2\) is a reducing agent here, not an oxidizing agent.
• Option D: In \(O_2\) (0), in \(Na_2O_2\), O has oxidation state -1. Oxidation state decreases, \(O_2\) is an oxidizing agent.
• Option E: In \(O_2\) (0), in \(MgO\), O is -2. Oxidation state decreases, \(O_2\) is an oxidizing agent.

Answer:

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