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 O in each reaction:

• Option A: In \(O_2\) (oxidation state = 0), in \(CO\), O has oxidation state -2. O is reduced (gains electrons), so \(O_2\) is an oxidizing agent.
• Option B: In \(O_2\) (0), in \(SO_2\), O is -2. O is reduced, so \(O_2\) is an oxidizing agent.
• Option C: In \(O_2\) (0), in \(OF_2\), O has oxidation state +2 (since F is -1, and \(2\times(+2)+2\times(-1)=0\) for \(OF_2\)). O is oxidized (loses electrons), so \(O_2\) is a reducing agent here, not an oxidizing agent.
• Option D: In \(O_2\) (0), in \(Na_2O_2\), O has oxidation state -1 (since Na is +1, and \(2\times(+1)+2\times(-1)=0\) for \(Na_2O_2\)). O is reduced (gains electrons from 0 to -1), so \(O_2\) is an oxidizing agent.
• Option E: In \(O_2\) (0), in \(MgO\), O is -2. O is reduced, so \(O_2\) is an oxidizing agent.

Only in reaction C, oxygen is oxidized (acts as a reducing agent), so it is not acting as an oxidizing agent here.

Answer:

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