In the series of reaction
`C_(6)H_(5)NH_(2)overset(NaNO_(2)//HCl)underset(0-5^(@)C)rarr X overset(HNO_(2))underset(H_(2)O)rarr Y+N_(2)+HCl`,
X and Y are respectively

Options

A. `C_(6)H_(5)-N=N-C_(6)H_(5), C_(6)H_(5)N_(2)^(+)Cl^(-)`
B. `C_(6)H_(5)N_(2)^(+)Cl^(-),C_(6)H_(5)-N=N-C_(6)H_(5)`
C. `C_(6)H_(5)N_(2)^(+)Cl^(-),C_(6)H_(5)NO_(2)`
D. `C_(6)H_(5)NO_(2),C_(6)H_(6)`

Explanation

To solve the given question, we will break down the reactions step by step.

### Step 1: Identify the starting compound
The starting compound is **C6H5NH2**, which is known as **aniline**.

### Step 2: Reaction with NaNO2 and HCl
When aniline (C6H5NH2) reacts with sodium nitrite (NaNO2) in the presence of hydrochloric acid (HCl) at a low temperature (0-5°C), a diazotization reaction occurs. This reaction forms a diazonium salt.

**Reaction:**
\[ \text{C6H5NH2} + \text{NaNO2} + \text{HCl} \rightarrow \text{C6H5N2Cl} + \text{NaCl} + \text{H2O} \]

The product formed here is **benzene diazonium chloride (C6H5N2Cl)**, which we will denote as **X**.

### Step 3: Reaction of X with HNO2 in water
Next, the benzene diazonium chloride (X) reacts with nitrous acid (HNO2) in the presence of water. In this reaction, the diazonium group (N2) is replaced by a hydroxyl group (-OH), leading to the formation of a phenol derivative.

**Reaction:**
\[ \text{C6H5N2Cl} + \text{HNO2} + \text{H2O} \rightarrow \text{C6H5OH} + \text{N2} + \text{HCl} \]

The product formed here is **nitrobenzene (C6H5NO2)**, which we will denote as **Y**.

### Summary of Products
- **X**: Benzene diazonium chloride (C6H5N2Cl)
- **Y**: Nitrobenzene (C6H5NO2)

### Final Answer
- **X**: Benzene diazonium chloride
- **Y**: Nitrobenzene