Question

oxygen is acting as an oxidizing agent in all of the following reactions except
a. \\(2c(s) + o_2(g) \
ightarrow 2co(g)\\)
b. \\(s(s) + o_2(g) \
ightarrow so_2(g)\\)
c. \\(2f_2(g) + o_2(g) \
ightarrow 2of_2(g)\\)
d. \\(2na(s) + o_2(g) \
ightarrow na_2o_2(s)\\)
e. \\(2mg(s) + o_2(g) \
ightarrow 2mgo(s)\\)

Explanation:

To determine in which reaction oxygen is not acting as an oxidizing agent, we analyze the oxidation state changes of oxygen:

Step 1: Recall the role of an oxidizing agent

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

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

ightarrow 2CO(g)}$)

• Oxidation state of O in $O_2$: 0
• Oxidation state of O in $CO$: -2
• Oxygen is reduced (oxidation state decreases from 0 to -2), so it acts as an oxidizing agent.

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

ightarrow SO_2(g)}$)

• Oxidation state of O in $O_2$: 0
• Oxidation state of O in $SO_2$: -2
• Oxygen is reduced (oxidation state decreases from 0 to -2), so it acts as an oxidizing agent.

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

ightarrow 2OF_2(g)}$)

• Oxidation state of O in $O_2$: 0
• Oxidation state of O in $OF_2$: +2 (since F has an oxidation state of -1, and for $OF_2$, let O be $x$: $x + 2(-1) = 0 \implies x = +2$)
• Oxygen is oxidized (oxidation state increases from 0 to +2), so it acts as a reducing agent (not an oxidizing agent).

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

ightarrow Na_2O_2(s)}$)

• Oxidation state of O in $O_2$: 0
• Oxidation state of O in $Na_2O_2$: -1 (since Na has an oxidation state of +1, and for $Na_2O_2$, let O be $x$: $2(+1) + 2x = 0 \implies x = -1$)
• Oxygen is reduced (oxidation state decreases from 0 to -1), so it acts as an oxidizing agent.

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

ightarrow 2MgO(s)}$)

• Oxidation state of O in $O_2$: 0
• Oxidation state of O in $MgO$: -2
• Oxygen is reduced (oxidation state decreases from 0 to -2), so it acts as an oxidizing agent.

Answer:

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