Which of the following after reacting with KI do not remove iodine

Options

A. `CuSO_(4)`
B. `K_(2)Cr_(2)O_(7)`
C. `HNO_(3)`
D. `HCl`

Answer

To determine which of the given compounds does not remove iodine when reacting with potassium iodide (KI), we need to analyze the oxidation states of the elements in each compound. The key point is that a compound can only remove iodine if it is in its maximum oxidation state.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds we need to analyze are CuSO4, K2Cr2O7, HNO3, and HCl.

2. **Analyze CuSO4**:
- The oxidation state of Cu in CuSO4 can be calculated as follows:
- Let the oxidation state of Cu be \( x \).
- The oxidation state of sulfur (S) is +6, and oxygen (O) is -2.
- The equation becomes: \( x + 6 + (4 \times -2) = 0 \)
- Simplifying: \( x + 6 - 8 = 0 \) → \( x - 2 = 0 \) → \( x = +2 \).
- Maximum oxidation state of Cu is +2, so CuSO4 can remove iodine.

3. **Analyze K2Cr2O7**:
- The oxidation state of Cr in K2Cr2O7:
- Let the oxidation state of Cr be \( x \).
- The equation becomes: \( 2 + 2x + (7 \times -2) = 0 \)
- Simplifying: \( 2 + 2x - 14 = 0 \) → \( 2x - 12 = 0 \) → \( 2x = 12 \) → \( x = +6 \).
- Maximum oxidation state of Cr is +6, so K2Cr2O7 can remove iodine.

4. **Analyze HNO3**:
- The oxidation state of N in HNO3:
- Let the oxidation state of N be \( x \).
- The equation becomes: \( 1 + x + (3 \times -2) = 0 \)
- Simplifying: \( 1 + x - 6 = 0 \) → \( x - 5 = 0 \) → \( x = +5 \).
- Maximum oxidation state of N is +5, so HNO3 can also remove iodine.

5. **Analyze HCl**:
- The oxidation state of Cl in HCl:
- H is +1 and Cl is -1.
- The oxidation state of Cl is -1, which is its minimum oxidation state.
- Since Cl is in its minimum oxidation state, HCl cannot remove iodine.

6. **Conclusion**:
- The only compound that does not remove iodine when reacting with KI is **HCl**.

### Final Answer:
HCl does not remove iodine when reacting with KI.