Oxidation reaction involves loss of electrons, and reduction reaction involves gain of electrons. The reaction in which a species disproportinates into two oxidation states ( lower and higher) is called disproportionation reaction.
Which of the following statements is wrong?

Options

A. Acidified `KMnO_(4)` solutions decolourises on the addition of sodium oxalate.
B. In the reaction between `Br_(2)` and `CsI, Br_(2)` is an oxidising agent and `CsI` is a reducing agent.
C. In the reaction `2K_(2)S_()O_(3)+I_(2)rarr2KI+K_(2)S_(4)O_(6)`, the change in the oxidation number of `S` is `0.5`.
D. `C` has the same oxidation number in both `CH_(4)` and `CO_(2)`

Explanation

To determine which of the statements is wrong regarding oxidation and reduction reactions, we will analyze each statement step by step.

### Step 1: Analyze the first statement
**Statement 1:** Acidified KMnO4 solution decolorizes on addition of sodium oxalate.

- **Reaction:**
\[ \text{KMnO}_4 + \text{C}_2\text{O}_4^{2-} \rightarrow \text{Mn}^{2+} + \text{CO}_2 \]
- **Oxidation states:**
- In KMnO4, Mn has an oxidation state of +7.
- In Mn²⁺, Mn has an oxidation state of +2.
- **Conclusion:** The purple color of KMnO4 disappears as Mn is reduced from +7 to +2. Thus, this statement is **correct**.

### Step 2: Analyze the second statement
**Statement 2:** In the reaction between bromine and cesium iodide, Br2 is an oxidizing agent and cesium iodide is a reducing agent.

- **Reaction:**
\[ \text{Br}_2 + \text{CsI} \rightarrow \text{CsBr} + \text{I}_2 \]
- **Oxidation states:**
- Br2 has an oxidation state of 0 and is reduced to Br⁻ (oxidation state -1).
- I⁻ is oxidized to I2 (oxidation state 0).
- **Conclusion:** Br2 is indeed the oxidizing agent and CsI (iodide) is the reducing agent. Thus, this statement is **correct**.

### Step 3: Analyze the third statement
**Statement 3:** In the reaction \(2 \text{K}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow 2 \text{KI} + \text{K}_2\text{S}_4\text{O}_6\), the change of oxidation number of sulfur is 0.5.

- **Oxidation states:**
- In \( \text{S}_2\text{O}_3^{2-} \), let the oxidation state of sulfur be x:
\[ 2x + 3(-2) = -2 \implies 2x - 6 = -2 \implies 2x = 4 \implies x = +2 \]
- In \( \text{S}_4\text{O}_6^{2-} \):
\[ 4x + 6(-2) = -2 \implies 4x - 12 = -2 \implies 4x = 10 \implies x = 2.5 \]
- **Change in oxidation state:**
\[ 2.5 - 2 = 0.5 \]
- **Conclusion:** The change in oxidation state of sulfur is indeed 0.5. Thus, this statement is **correct**.

### Step 4: Analyze the fourth statement
**Statement 4:** Carbon has the same oxidation number in both CH4 and CO2.

- **Oxidation states:**
- In CH4:
\[ x + 4(+1) = 0 \implies x = -4 \]
- In CO2:
\[ x + 2(-2) = 0 \implies x = +4 \]
- **Conclusion:** The oxidation state of carbon in CH4 is -4 and in CO2 is +4, which are not the same. Thus, this statement is **wrong**.

### Final Answer
The wrong statement is **Statement 4**: Carbon has the same oxidation number in both CH4 and CO2.

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