`C_(6)H_(5)Cl underset(Cu_(2)O, 200^(@)C)overset(NH_(3)) to X, X underset(0-5^(@)C)overset(HNO_(3)) to Z, X+Z to A`, the `sigma` and `pi` bonds in 'A' are

Options

A. `25 sigma, 6pi`
B. `25 sigma, 7 pi`
C. `27 sigma, 7pi`
D. `27 sigma, 6pi`

Explanation

To solve the problem step by step, we will analyze the reactions and determine the structure of compound A, then count the sigma and pi bonds.

### Step 1: Reaction of Chlorobenzene with Ammonia
Chlorobenzene (C₆H₅Cl) reacts with ammonia (NH₃) in the presence of copper(I) oxide (Cu₂O) at 200°C. This reaction leads to the formation of aniline (C₆H₅NH₂).

**Product X: Aniline (C₆H₅NH₂)**

### Step 2: Reaction of Aniline with Nitrous Acid
Next, aniline (X) is treated with nitrous acid (HNO₂) at low temperatures (0-5°C). This reaction results in the formation of benzene diazonium chloride (C₆H₅N₂Cl).

**Product Z: Benzene Diazonium Chloride (C₆H₅N₂Cl)**

### Step 3: Reaction of Aniline and Benzene Diazonium Chloride
Now, we need to consider the reaction between aniline (X) and benzene diazonium chloride (Z). Aniline acts as a nucleophile, while the diazonium compound acts as an electrophile. The reaction typically occurs at the para position due to steric hindrance.

The product formed (A) will be para-substituted aniline derivative:

**Product A: Para-Aminoazobenzene (C₆H₄(NH₂)N₂C₆H₅)**

### Step 4: Counting Sigma and Pi Bonds in Compound A
Now, we will count the sigma (σ) and pi (π) bonds in the final product A.

1. **Sigma Bonds (σ)**:
- Each C-C bond in the benzene rings contributes 1 sigma bond.
- Each C-H bond contributes 1 sigma bond.
- Each C-N bond contributes 1 sigma bond.
- Totaling these, we find:
- 6 C-C bonds in the two benzene rings = 6 σ bonds
- 4 C-H bonds in para-substituted aniline = 4 σ bonds
- 1 C-N bond = 1 σ bond
- Total σ bonds = 6 + 4 + 1 = 11 σ bonds.

2. **Pi Bonds (π)**:
- Each double bond in the benzene rings contributes 1 pi bond.
- Totaling these, we find:
- 3 double bonds in the two benzene rings = 3 π bonds.
- Total π bonds = 3.

### Final Count:
- **Total Sigma Bonds (σ)**: 11
- **Total Pi Bonds (π)**: 3

### Conclusion:
The final answer is:
- **Sigma bonds in A = 11**
- **Pi bonds in A = 3**