`HI` can be prepared by all the following methods except

Options

A. `Pl_3+H_2O`
B. `Kl_+H_2SO_4`
C. `H_2+I_2 overset"Pt"to`
D. `I_2+H_2S`

Explanation

To determine which method does not produce HI (hydroiodic acid), we can analyze each option provided in the question.

### Step-by-Step Solution:

1. **Understanding the Options**:
We need to evaluate each method mentioned in the options to see if it produces HI.

2. **Option A: Reaction of PI3 with H2O**:
- The reaction of phosphorus triiodide (PI3) with water (H2O) produces phosphoric acid (H3PO4) and phosphorous acid (H3PO3), along with HI.
- The balanced equation is:
\[ PI3 + 3H2O \rightarrow H3PO4 + H3PO3 + 3HI \]
- Here, HI is produced.

3. **Option B: Reaction of KI with H2SO4**:
- The reaction of potassium iodide (KI) with sulfuric acid (H2SO4) leads to the oxidation of iodide ions (I-) to iodine (I2).
- The balanced equation is:
\[ 8KI + 5H2SO4 \rightarrow 4K2SO4 + 4I2 + 4H2O \]
- In this reaction, HI is not produced because I- is oxidized to I2.

4. **Option C: Reaction of H2 with I2**:
- The direct reaction of hydrogen (H2) with iodine (I2) produces HI.
- The balanced equation is:
\[ H2 + I2 \rightarrow 2HI \]
- Here, HI is produced.

5. **Option D: Reaction of I2 with H2S**:
- The reaction of iodine (I2) with hydrogen sulfide (H2S) produces HI and sulfur (S).
- The balanced equation is:
\[ I2 + H2S \rightarrow 2HI + S \]
- Here, HI is produced.

6. **Conclusion**:
- From the analysis, we find that HI is produced in options A, C, and D, but not in option B.
- Therefore, the answer is **Option B**.

### Final Answer:
**HI can be prepared by all the following methods except B (Reaction of KI with H2SO4).**

---