Pick out the correct set of statements for Hl.
1. It reduces iodic acid to `I_(2)`
2. It reduces `H_(2)SO_(4)` to `SO_(2)`
3. It decolourises acidified `KMnO_(4)`
4. It liberates `I_(2)` with`CuSO_(4)` solution

Options

A. 1, 2, 3
B. 1, 2, 4
C. 1, 3, 4
D. 1, 2, 3, 4

Answer

To solve the question, we need to evaluate each statement regarding Hydroiodic acid (HI) and determine if they are correct.

**Step 1: Evaluate Statement 1**
- Statement: "It reduces iodic acid to I2."
- Reaction: HI can reduce iodic acid (HIO3) to iodine (I2). The balanced reaction is:
\[ 3 HI + HIO3 \rightarrow 3 H2O + I2 \]
- Conclusion: This statement is **correct**.

**Step 2: Evaluate Statement 2**
- Statement: "It reduces H2SO4 to SO2."
- Reaction: HI can reduce sulfuric acid (H2SO4) to sulfur dioxide (SO2). The balanced reaction is:
\[ 8 HI + H2SO4 \rightarrow 4 I2 + 2 SO2 + 4 H2O \]
- Conclusion: This statement is **correct**.

**Step 3: Evaluate Statement 3**
- Statement: "It decolourises acidified KMnO4."
- Reaction: HI can reduce potassium permanganate (KMnO4) in an acidic medium to manganese(II) ions (Mn2+), which is colorless. The balanced reaction is:
\[ 5 H2SO4 + 2 KMnO4 + 6 HI \rightarrow 2 MnSO4 + K2SO4 + 3 I2 + 8 H2O \]
- Conclusion: This statement is **correct**.

**Step 4: Evaluate Statement 4**
- Statement: "It liberates I2 with CuSO4 solution."
- Reaction: When HI reacts with copper(II) sulfate (CuSO4), it can produce iodine (I2). The reaction can be represented as:
\[ 2 HI + CuSO4 \rightarrow CuI2 + I2 \]
- Conclusion: This statement is **correct**.

**Final Conclusion:**
All four statements are correct regarding Hydroiodic acid (HI). Therefore, the correct answer is option number 4 (1, 2, 3, 4).

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