Which of the following is not formed in the below reaction
`O_(3) + Kl + D_(2)O rarr`

Options

A. KOH
B. `O_(2)`
C. `I_(2)`
D. KOD

Answer

To solve the question regarding the reaction of ozone (O₃) with potassium iodide (KI) in heavy water (D₂O), we need to analyze the products formed in the reaction.

### Step-by-Step Solution:

1. **Identify Reactants**: The reactants in the given reaction are ozone (O₃), potassium iodide (KI), and heavy water (D₂O).

2. **Understand the Reaction**: When ozone reacts with potassium iodide in heavy water, it typically leads to the formation of several products. Ozone is a strong oxidizing agent and can oxidize iodide ions (I⁻) to iodine (I₂).

3. **Determine Possible Products**:
- **Oxygen (O₂)**: Ozone decomposes to form molecular oxygen (O₂).
- **Iodine (I₂)**: The iodide ions are oxidized to form iodine (I₂).
- **Potassium Oxide (KOD)**: The potassium from potassium iodide can combine with deuterium oxide (D₂O) to form potassium deuteroxide (KOD).

4. **List the Products**: Based on the reaction, the expected products are:
- O₂ (oxygen)
- I₂ (iodine)
- KOD (potassium deuteroxide)

5. **Identify the Product Not Formed**: Among the products listed, we need to identify which one is not formed in this reaction. The options provided are:
- KOH (potassium hydroxide)
- O₂ (oxygen)
- I₂ (iodine)
- KOD (potassium deuteroxide)

6. **Conclusion**: KOH is not formed in this reaction. Instead, KOD is formed due to the reaction with heavy water (D₂O). Therefore, the answer to the question is KOH.

### Final Answer:
KOH is not formed in the reaction.

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