Product (X and Y) of the following reaction (1 and 2) are
(1) `underset("(Cold and dilute)")(2NaOH)+Cl_(2)rarr NaCl +X+H_(2)O`
(2) `underset("(Hot and conc.)")(6NaOH)+3Cl_(2)rarr NaCl+Y+3H_(2)O`

Options

A. `X=NaClO_(3) and Y=NaOCl`
B. `X=NaOCl and Y=NaClO_(3)`
C. `X=NaHClO_(3) and Y=NaOCl`
D. `X=NaClO_(3) and Y=NaHClO_(3)`

Answer

To determine the products (X and Y) of the given reactions, we need to analyze each reaction step by step.

### Reaction 1:
**Cold and Dilute NaOH with Cl₂**
The reaction is:
\[ 2 \text{NaOH} + \text{Cl}_2 \rightarrow \text{NaCl} + X + \text{H}_2\text{O} \]

1. **Identify the type of reaction**: Chlorine reacts with cold and dilute sodium hydroxide. This is a disproportionation reaction where chlorine is both oxidized and reduced.

2. **Determine the products**: In this reaction, chlorine (Cl₂) is reduced to chloride ion (Cl⁻) and oxidized to hypochlorite ion (ClO⁻). Therefore, the products are sodium chloride (NaCl) and sodium hypochlorite (NaOCl).

3. **Assign the product X**: From the above, we can conclude that:
\[ X = \text{NaOCl} \]

### Reaction 2:
**Hot and Concentrated NaOH with Cl₂**
The reaction is:
\[ 6 \text{NaOH} + 3 \text{Cl}_2 \rightarrow \text{NaCl} + Y + 3 \text{H}_2\text{O} \]

1. **Identify the type of reaction**: Chlorine reacts with hot and concentrated sodium hydroxide. This reaction also involves disproportionation but leads to different products due to the conditions.

2. **Determine the products**: In this case, chlorine is reduced to chloride ion (Cl⁻) and oxidized to chlorate ion (ClO₃⁻). Thus, the products are sodium chloride (NaCl) and sodium chlorate (NaClO₃).

3. **Assign the product Y**: Therefore, we can conclude that:
\[ Y = \text{NaClO}_3 \]

### Summary of Products:
- Product X from Reaction 1 is Sodium Hypochlorite (NaOCl).
- Product Y from Reaction 2 is Sodium Chlorate (NaClO₃).

### Final Answer:
- \( X = \text{NaOCl} \)
- \( Y = \text{NaClO}_3 \)

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