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 in which reaction oxygen is not acting as an oxidizing agent, we analyze the oxidation states of oxygen in each reaction:

Step 1: Recall the definition of an oxidizing agent

An oxidizing agent is a substance that gets reduced (its oxidation state decreases) during a reaction.

Step 2: Analyze Reaction A ($\boldsymbol{2C(s) + O_2(g)

ightarrow 2CO(g)}$)

• In $O_2$, the oxidation state of O is 0.
• In $CO$, the oxidation state of O is -2 (since C has an oxidation state of +2).
• The oxidation state of O decreases from 0 to -2, so $O_2$ is reduced (oxidizing agent).

Step 3: Analyze Reaction B ($\boldsymbol{S(s) + O_2(g)

ightarrow SO_2(g)}$)

• In $O_2$, O has an oxidation state of 0.
• In $SO_2$, O has an oxidation state of -2 (S has an oxidation state of +4).
• The oxidation state of O decreases from 0 to -2, so $O_2$ is reduced (oxidizing agent).

Step 4: Analyze Reaction C ($\boldsymbol{2F_2(g) + O_2(g)

ightarrow 2OF_2(g)}$)

• In $O_2$, O has an oxidation state of 0.
• In $OF_2$, F has an oxidation state of -1 (since F is more electronegative than O), so O has an oxidation state of +2.
• The oxidation state of O increases from 0 to +2, so $O_2$ is oxidized (not an oxidizing agent, but a reducing agent here).

Step 5: Analyze Reaction D ($\boldsymbol{2Na(s) + O_2(g)

ightarrow Na_2O_2(s)}$)

• In $O_2$, O has an oxidation state of 0.
• In $Na_2O_2$, O has an oxidation state of -1 (Na has an oxidation state of +1).
• The oxidation state of O decreases from 0 to -1, so $O_2$ is reduced (oxidizing agent).

Step 6: Analyze Reaction E ($\boldsymbol{2Mg(s) + O_2(g)

ightarrow 2MgO(s)}$)

• In $O_2$, O has an oxidation state of 0.
• In $MgO$, O has an oxidation state of -2 (Mg has an oxidation state of +2).
• The oxidation state of O decreases from 0 to -2, so $O_2$ is reduced (oxidizing agent).

Answer:

C. $\boldsymbol{2F_2(g) + O_2(g)
ightarrow 2OF_2(g)}$