`overset(H_(2)+Ni)(leftarrow) PhNO_(2) overset(LAH)(rarr)(A)`.
The products `(A)` and `(B)` are:

Options

A. a.`PhNH_(2)`, `PhNH_(2)`
B. b.`Ph-N=N-Ph`, `Ph-N=N-Ph`
C. c.`Ph-N=N-Ph, PhNH_(2)`
D. d.`Ph-NH-NH-Ph, PhNH_(2)`

Explanation

To solve the question regarding the reduction of nitrobenzene using two different reducing agents, we can break down the process step by step.

### Step-by-Step Solution:

1. **Identify the Reactant:**
The starting compound is nitrobenzene (PhNO₂). It has a nitro group (-NO₂) attached to a benzene ring.

2. **First Reducing Agent - Lithium Aluminum Hydride (LAH):**
- Lithium aluminum hydride (LAH) is a strong reducing agent.
- When nitrobenzene is treated with LAH, it undergoes reduction.
- The nitro group (-NO₂) is reduced to an amine group (-NH₂), resulting in the formation of **azobenzene** (Ph-N=Ph).
- Therefore, the product (A) from this reaction is **azobenzene**.

3. **Second Reducing Agent - Catalytic Hydrogenation (H₂ with Ni):**
- The second method involves catalytic hydrogenation using hydrogen gas (H₂) in the presence of nickel (Ni) as a catalyst.
- In this reaction, the nitro group (-NO₂) is reduced to an amine group (-NH₂) as well.
- The product formed from this reaction is **aniline** (Ph-NH₂).
- Therefore, the product (B) from this reaction is **aniline**.

4. **Summarize the Products:**
- Product (A) from the reaction with LAH is **azobenzene**.
- Product (B) from the reaction with H₂ and Ni is **aniline**.

### Final Answer:
- Product (A) is **azobenzene**.
- Product (B) is **aniline**.