Product (X and Y) of the following reaction (1 and 2) are
(1) `underset("(Cold and dilute)")(2NaOH)+Cl_(2)rarr NaCl_(2)+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 solve the problem, we need to analyze the two reactions given and identify the products (X and Y) formed in each case.

### Step 1: Analyze the first reaction
The first reaction is:
\[ 2 \text{NaOH (cold and dilute)} + \text{Cl}_2 \rightarrow \text{NaCl} + X + \text{H}_2\text{O} \]

In this reaction, sodium hydroxide (NaOH) is cold and dilute. When NaOH reacts with chlorine gas (Cl₂) under these conditions, it forms sodium chloride (NaCl) and sodium hypochlorite (NaOCl) as products.

Thus, we can write:
\[ X = \text{NaOCl} \]

### Step 2: Analyze the second reaction
The second reaction is:
\[ 6 \text{NaOH (hot and concentrated)} + 3 \text{Cl}_2 \rightarrow \text{NaCl} + Y + 3 \text{H}_2\text{O} \]

In this case, sodium hydroxide is hot and concentrated. Under these conditions, the reaction of NaOH with Cl₂ leads to the formation of sodium chloride (NaCl) and sodium chlorate (NaClO₃).

Thus, we can write:
\[ Y = \text{NaClO}_3 \]

### Conclusion
From the analysis of both reactions, we find:
- Product \( X \) is sodium hypochlorite (\( \text{NaOCl} \)).
- Product \( Y \) is sodium chlorate (\( \text{NaClO}_3 \)).

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