Pick out the incorrect statements for HI.

Options

A. It reduces `H_2SO_4 "to " SO_2`
B. It reduces iodic acid to `I_2`
C. It does not decolourise acidified `KMnO_4`
D. It liberates `I_2` with `CuSO_4` solution

Answer

To solve the problem of identifying the incorrect statement regarding hydrogen iodide (HI), we will analyze each statement provided in the question.

### Step-by-Step Solution:

1. **Analyze Statement 1: "HI reduces H2SO4 to SO2."**
- **Reaction:** H2SO4 + HI → SO2 + I2 + 2H2O
- **Oxidation States:**
- In H2SO4, sulfur (S) has an oxidation state of +6.
- In SO2, sulfur (S) has an oxidation state of +4.
- **Conclusion:** Since the oxidation state of sulfur decreases from +6 to +4, it indicates that sulfur is being reduced. Thus, this statement is **correct**.

2. **Analyze Statement 2: "HI reduces iodic acid to iodine."**
- **Reaction:** HIO3 + HI → I2 + 3H2O
- **Oxidation States:**
- In HIO3, iodine (I) has an oxidation state of +5.
- In I2, iodine (I) has an oxidation state of 0.
- **Conclusion:** Since the oxidation state of iodine decreases from +5 to 0, it indicates that iodine is being reduced. Thus, this statement is **correct**.

3. **Analyze Statement 3: "HI does not decolorize acidified KMnO4."**
- **Reaction:** KMnO4 + HI → K2SO4 + MnSO4 + H2O + I2
- **Color Change:** KMnO4 is violet, and MnSO4 is colorless.
- **Conclusion:** Since KMnO4 changes from violet to colorless, it indicates that HI does decolorize KMnO4. Therefore, this statement is **incorrect**.

4. **Analyze Statement 4: "HI liberates iodine with copper sulfate."**
- **Reaction:** CuSO4 + 4HI → 2CuI + 2H2SO4
- **Conclusion:** The reaction produces iodine (I2) gas. Thus, this statement is **correct**.

### Final Conclusion:
The only incorrect statement is **Statement 3: "HI does not decolorize acidified KMnO4."**