Reaction of ammonia with diborane gives initially `B_(2) H_(6) 2NH_(3)` which can also be written as

Options

A. `[BH_(2) (NH_(3))_(2)]^(+) [BH_(4)]^(-)`
B. `[BH_4]^(+) [BH_(2) (NH_(3))_(2)]^(+)`
C. `[BH_(3) NH_(3)]^(+) [BH_(4)]^(-)`
D. `[B_(2) N_(2) H_(6)]^(+) [H_(3)]^(-)`

Answer

To solve the question regarding the reaction of ammonia with diborane, we can break it down into clear steps:

### Step-by-Step Solution:

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

2. **Write the Initial Reaction**: When diborane reacts with ammonia, the initial reaction can be represented as:
\[ \text{B}_2\text{H}_6 + 2\text{NH}_3 \rightarrow \text{B}_2\text{H}_6 \cdot 2\text{NH}_3 \]
This indicates that diborane forms a complex with two molecules of ammonia.

3. **Understand the Structure of Diborane**: Diborane (B2H6) has a unique structure that includes "banana bonds" between the boron atoms. This structure is crucial for understanding how it interacts with ammonia.

4. **Reaction Mechanism**: In the reaction, the B-H bonds in diborane break, allowing the ammonia molecules to coordinate with the boron atoms. This leads to the formation of:
\[ \text{BH}_2\text{NH}_3 \text{ (twice)} + \text{BH}_4^- \]
Here, each boron atom is now bonded to an NH3 molecule, and one boron atom forms a tetrahydroborate ion (BH4^-).

5. **Final Representation**: The final products can be represented as:
\[ \text{B}_2\text{H}_6 \cdot 2\text{NH}_3 \rightarrow 2\text{BH}_2\text{NH}_3 + \text{BH}_4^- \]
This shows the breakdown of the initial complex into more stable products.

### Final Answer:
The reaction of ammonia with diborane can be represented as:
\[ \text{B}_2\text{H}_6 \cdot 2\text{NH}_3 \rightarrow 2\text{BH}_2\text{NH}_3 + \text{BH}_4^- \]