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 question of identifying the incorrect statements regarding the reactions of HI, we will analyze each statement step-by-step.

### Step-by-Step Solution:

1. **Analyze the First Statement:**
- The statement claims that HI reduces H2SO4 to SO2.
- Reaction: \( H_2SO_4 + 2 HI \rightarrow SO_2 + I_2 + 2 H_2O \)
- In this reaction, the oxidation state of sulfur changes from +6 in H2SO4 to +4 in SO2, indicating a reduction.
- **Conclusion:** This statement is **correct**.

2. **Analyze the Second Statement:**
- The statement claims that HI reacts with iodic acid (HIO3) to release I2 gas.
- Reaction: \( HIO_3 + 5 HI \rightarrow 3 I_2 + 3 H_2O \)
- The oxidation state of iodine changes from +5 in HIO3 to 0 in I2, indicating reduction.
- **Conclusion:** This statement is **correct**.

3. **Analyze the Third Statement:**
- The statement claims that HI does not decolorize acidic KMnO4 solution.
- Reaction: \( 10 HI + 3 H_2SO_4 + KMnO_4 \rightarrow K_2SO_4 + 2 MnSO_4 + 5 I_2 + 8 H_2O \)
- KMnO4 is purple in color and gets reduced to colorless MnSO4, indicating that the color is decolorized.
- **Conclusion:** This statement is **incorrect**.

4. **Analyze the Fourth Statement:**
- The statement claims that when CuSO4 reacts with HI, I2 gas is produced.
- Reaction: \( CuSO_4 + 4 HI \rightarrow CuI_2 + 2 H_2SO_4 \)
- On heating CuI2, it decomposes to produce Cu2I2 and I2.
- **Conclusion:** This statement is **correct**.

### Final Conclusion:
The incorrect statement among the options provided is the **third statement** which claims that HI does not decolorize acidic KMnO4 solution.

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