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)$

Question

Accepted Answer

# Explanation:
To determine in which reaction oxygen is not acting as an oxidizing agent, we analyze the oxidation states of oxygen in each reaction. An oxidizing agent is reduced (its oxidation state decreases).

## Step 1: Recall Oxidation State Rules
- For elemental oxygen ($O_2$), the oxidation state is $0$.
- In compounds, oxygen typically has an oxidation state of $-2$, except in peroxides (where it is $-1$) or when combined with fluorine (where it can have positive oxidation states).

## Step 2: Analyze Reaction A
Reaction: $2C(s) + O_2(g)
ightarrow 2CO(g)$
- In $O_2$, oxidation state of $O = 0$.
- In $CO$, oxidation state of $O = -2$.
- Oxidation state of $O$ decreases (from $0$ to $-2$), so $O_2$ is reduced (oxidizing agent).

## Step 3: Analyze Reaction B
Reaction: $S(s) + O_2(g)
ightarrow SO_2(g)$
- In $O_2$, $O = 0$.
- In $SO_2$, $O = -2$.
- Oxidation state of $O$ decreases (from $0$ to $-2$), so $O_2$ is reduced (oxidizing agent).

## Step 4: Analyze Reaction C
Reaction: $2F_2(g) + O_2(g)
ightarrow 2OF_2(g)$
- In $O_2$, $O = 0$.
- In $OF_2$, let the oxidation state of $O$ be $x$. Fluorine has an oxidation state of $-1$ (since it is more electronegative than oxygen here). So, $x + 2(-1) = 0 \implies x = +2$.
- Oxidation state of $O$ increases (from $0$ to $+2$), so $O_2$ is oxidized (not an oxidizing agent here).

## Step 5: Analyze Reaction D
Reaction: $2Na(s) + O_2(g)
ightarrow Na_2O_2(s)$
- In $O_2$, $O = 0$.
- In $Na_2O_2$ (peroxide), oxidation state of $O = -1$.
- Oxidation state of $O$ decreases (from $0$ to $-1$), so $O_2$ is reduced (oxidizing agent).

## Step 6: Analyze Reaction E
Reaction: $2Mg(s) + O_2(g)
ightarrow 2MgO(s)$
- In $O_2$, $O = 0$.
- In $MgO$, $O = -2$.
- Oxidation state of $O$ decreases (from $0$ to $-2$), so $O_2$ is reduced (oxidizing agent).

# Answer:
C. $2 F_2(g) + O_2(g)
ightarrow 2 OF_2(g)$

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QUESTION IMAGE

Question

Explanation:

Step 1: Recall Oxidation State Rules

Step 2: Analyze Reaction A

Step 3: Analyze Reaction B

Step 4: Analyze Reaction C

Step 5: Analyze Reaction D

Step 6: Analyze Reaction E

Answer: