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 state of oxygen in each reaction:

Step 1: Recall the concept of oxidizing agent

An oxidizing agent is a substance that gets reduced (its oxidation state decreases) in a reaction. So we need to check the oxidation state of O in each reactant and product.

Step 2: Analyze Reaction A

In \( O_2(g) \), the oxidation state of O is 0. In \( CO(g) \), the oxidation state of O is -2. So O is reduced (oxidation state decreases from 0 to -2), so \( O_2 \) is an oxidizing agent.

Step 3: Analyze Reaction B

In \( O_2(g) \), oxidation state of O is 0. In \( SO_2(g) \), oxidation state of O is -2. O is reduced, so \( O_2 \) is an oxidizing agent.

Step 4: Analyze Reaction C

In \( O_2(g) \), oxidation state of O is 0. In \( OF_2(g) \), let the oxidation state of O be \( x \). Since F has an oxidation state of -1, we have \( x + 2(-1)=0 \Rightarrow x = +2 \). So the oxidation state of O increases from 0 to +2, meaning O is oxidized (not reduced), so \( O_2 \) is not acting as an oxidizing agent here.

Step 5: Analyze Reaction D

In \( O_2(g) \), oxidation state of O is 0. In \( Na_2O_2(s) \), let the oxidation state of O be \( x \). Na has an oxidation state of +1, so \( 2(+1) + 2x = 0 \Rightarrow 2x = -2 \Rightarrow x = -1 \). So O is reduced (oxidation state decreases from 0 to -1), so \( O_2 \) is an oxidizing agent.

Step 6: Analyze Reaction E

In \( O_2(g) \), oxidation state of O is 0. In \( MgO(s) \), oxidation state of O is -2. O is reduced, so \( O_2 \) is an oxidizing agent.

Answer:

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