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 check the oxidation state of O. An oxidizing agent is reduced (oxidation state decreases).

• In options A, B, D, E: O in \(O_2\) has an oxidation state of 0. In the products (CO, \(SO_2\), \(Na_2O_2\), MgO), O has a negative oxidation state (e.g., -2 in CO, -1 in \(Na_2O_2\)), so O is reduced (oxidizing agent).
• In option C: \(O_2\) reacts with \(F_2\) to form \(OF_2\). F is more electronegative than O, so O has a +2 oxidation state in \(OF_2\) (while O was 0 in \(O_2\)). Here, O is oxidized (oxidation state increases), so \(O_2\) is a reducing agent, not an oxidizing agent.

Answer:

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