Diborane reacts with ammonia under different conditions to give :

Options

A. `B_(2)H_(6).2NH_(3)`
B. `B_(12)H_(12)`
C. `B_(3)N_(3)H_(6)`
D. `(BN)_(x)`

Answer

To solve the question regarding the reaction of diborane (B2H6) with ammonia (NH3) under different conditions, we will analyze the reactions based on the amount of ammonia and the temperature conditions.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this scenario are diborane (B2H6) and ammonia (NH3).

2. **Condition 1: Ammonia in Less Amount at Low Temperature**:
- When diborane reacts with a small amount of ammonia at low temperatures, the product formed is a complex known as B2H6·2NH3.
- This is a stable adduct where diborane coordinates with ammonia.

3. **Condition 2: Ammonia in Excess at High Temperature**:
- If ammonia is present in excess and the temperature is high, the reaction leads to the formation of BN, which is referred to as inorganic graphite.
- This product is formed due to the polymerization of boron and nitrogen under these conditions.

4. **Condition 3: Ammonia and Diborane in a 1:2 Ratio at High Temperature**:
- When diborane and ammonia are mixed in a 1:2 ratio and subjected to high temperatures, the product formed is B3N3H6, commonly known as inorganic benzene.
- This compound has a cyclic structure similar to benzene but contains boron and nitrogen.

5. **Conclusion**:
- The reactions yield different products based on the conditions of temperature and the ratio of reactants. The products formed are:
- B2H6·2NH3 (low temperature, less ammonia)
- BN (high temperature, excess ammonia)
- B3N3H6 (high temperature, 1:2 ratio of diborane to ammonia)

### Final Answer:
The products formed when diborane reacts with ammonia under different conditions are:
- B2H6·2NH3
- BN (inorganic graphite)
- B3N3H6 (inorganic benzene)