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

Options

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

Explanation

To solve the question step by step, we will analyze the reactions involving aniline (C₆H₅NH₂) and identify the products X and Y.

### Step 1: Identify the starting compound
The starting compound is aniline, which has the formula C₆H₅NH₂.

### Step 2: Reaction with NaNO₂ and HCl
When aniline (C₆H₅NH₂) is treated with sodium nitrite (NaNO₂) and hydrochloric acid (HCl) at a low temperature (0-5°C), it undergoes diazotization. The reaction can be represented as follows:

\[ C₆H₅NH₂ + NaNO₂ + HCl \rightarrow C₆H₅N₂Cl + NaCl + 2H₂O \]

The product formed in this reaction is benzene diazonium chloride (C₆H₅N₂Cl), which we will denote as X.

### Step 3: Identify Product X
From the above reaction, we can conclude that:
- **Product X = Benzene diazonium chloride (C₆H₅N₂Cl)**

### Step 4: Reaction of X with HNO₂ and CH₂O
Next, the benzene diazonium chloride (X) is reacted with nitrous acid (HNO₂) and formaldehyde (CH₂O). The reaction can be summarized as follows:

\[ C₆H₅N₂Cl + HNO₂ + CH₂O \rightarrow C₆H₅NO₂ + N₂ + HCl \]

In this reaction, the diazonium group (N₂) is lost, and a nitro group (NO₂) is introduced onto the benzene ring, resulting in the formation of nitrobenzene.

### Step 5: Identify Product Y
From the above reaction, we can conclude that:
- **Product Y = Nitrobenzene (C₆H₅NO₂)**

### Final Answer
Thus, the products X and Y are:
- **X = Benzene diazonium chloride (C₆H₅N₂Cl)**
- **Y = Nitrobenzene (C₆H₅NO₂)**

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

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