Identify the disproportionation reaction.

Options

A. `CH_(4)+2O_(2)rarr CO_(2)+2H_(2)O`
B. `CH_(4)+4Cl_(2)rarr C Cl_(4)+4HCl`
C. `2F_(2)+2OH^(-)rarr 2F^(-)+OF_(2)+H_(2)O`
D. `2NO_(2)+2OH^(-)rarr NO_(2)^(-)+NO_(3)+H_(2)O`

Explanation

To identify a disproportionation reaction, we need to look for a reaction in which a single element undergoes both oxidation and reduction simultaneously. Let's analyze the given reactions step by step.

### Step-by-Step Solution

1. **Analyze the first reaction:**
\[ \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \]
- **Oxidation states:**
- C in CH₄: +4
- H in CH₄: +1
- O in O₂: 0
- C in CO₂: +4
- H in H₂O: +1
- O in H₂O: -2
- **Conclusion:** Carbon (C) does not change its oxidation state, and oxygen (O) is reduced from 0 to -2. Since there is no oxidation occurring for the same element, this is not a disproportionation reaction.

**Hint:** Check if the same element changes its oxidation state in both oxidation and reduction.

2. **Analyze the second reaction:**
\[ \text{CH}_4 + 4\text{Cl}_2 \rightarrow \text{CCl}_4 + \text{HCl} \]
- **Oxidation states:**
- C in CH₄: +4
- H in CH₄: +1
- Cl in Cl₂: 0
- C in CCl₄: +4
- H in HCl: +1
- Cl in HCl: -1
- **Conclusion:** Chlorine (Cl) is reduced from 0 to -1, but there is no oxidation occurring for Cl. Therefore, this is also not a disproportionation reaction.

**Hint:** Identify if the same element is both oxidized and reduced in the reaction.

3. **Analyze the third reaction:**
\[ \text{F}_2 + \text{OH}^- \rightarrow 2\text{F}^- + \text{O} \]
- **Oxidation states:**
- F in F₂: 0
- O in OH⁻: -2
- H in OH⁻: +1
- F in F⁻: -1
- O in O: 0
- **Conclusion:** Fluorine (F) is reduced from 0 to -1, and oxygen (O) is oxidized from -2 to 0. However, since F is not undergoing oxidation, this is not a disproportionation reaction.

**Hint:** Ensure that the same element is involved in both oxidation and reduction processes.

4. **Analyze the fourth reaction:**
\[ \text{NO}_2 + \text{OH}^- \rightarrow \text{NO}_2^- + \text{NO}_3^- + \text{H}_2\text{O} \]
- **Oxidation states:**
- N in NO₂: +4
- O in OH⁻: -2
- H in OH⁻: +1
- N in NO₂⁻: +3
- N in NO₃⁻: +5
- **Conclusion:** Nitrogen (N) is reduced from +4 to +3 and oxidized from +4 to +5. Since the same element (N) is undergoing both oxidation and reduction, this is a disproportionation reaction.

**Hint:** Look for a single element that changes its oxidation state in both directions.

### Final Conclusion
The disproportionation reaction among the given options is:
\[ \text{NO}_2 + \text{OH}^- \rightarrow \text{NO}_2^- + \text{NO}_3^- + \text{H}_2\text{O} \]