When `Cl_2` gas is passed through cold and dilute solution of KOH, following compound is formed

Options

A. `KClO_3`
B. `KClO`
C. `KClO_2`
D. `KClO_4`

Explanation

To determine the compound formed when chlorine gas (`Cl_2`) is passed through a cold and dilute solution of potassium hydroxide (KOH), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reactants**:
- We have chlorine gas (`Cl_2`) and a cold, dilute solution of potassium hydroxide (KOH).

2. **Identifying the Reaction Type**:
- The reaction between chlorine gas and KOH can be classified as a redox reaction where chlorine is reduced and the hydroxide ions are oxidized.

3. **Writing the Reaction Equation**:
- When chlorine gas is passed through cold and dilute KOH, the reaction can be represented as:
\[ Cl_2 + 2 KOH \rightarrow KCl + KClO + H_2O \]
- In this reaction, potassium chloride (KCl) and potassium hypochlorite (KClO) are formed along with water.

4. **Identifying the Products**:
- From the reaction, we see that potassium hypochlorite (KClO) is one of the products formed.

5. **Conclusion**:
- Therefore, the compound formed when chlorine gas is passed through cold and dilute KOH is potassium hypochlorite (KClO).

### Final Answer:
The compound formed is **KClO**.

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