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 will follow these 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 written as:
\[ B_2H_6 + 2NH_3 \rightarrow \text{Products} \]

3. **Determine the Products**:
- The reaction of diborane with ammonia leads to the formation of a complex. The products can be represented as:
\[ B_2H_6 + 2NH_3 \rightarrow 2BH_2NH_3 + BH_4^- \]
- Here, \(BH_2NH_3\) is a product formed along with the anion \(BH_4^-\).

4. **Rewrite the Reaction**:
- The initial product \(B_2H_6 \cdot 2NH_3\) can be expressed in terms of its components:
\[ B_2H_6 \cdot 2NH_3 \rightarrow 2BH_2NH_3 + BH_4^- \]

5. **Select the Correct Option**:
- From the options provided, the correct representation of the reaction is:
\[ BH_2NH_3 \cdot 2 + BH_4^- \]
- This corresponds to **Option 1**.

### Final Answer:
The reaction of ammonia with diborane can be represented as:
\[ B_2H_6 + 2NH_3 \rightarrow 2BH_2NH_3 + BH_4^- \]
Thus, the correct option is:
**Option 1: \(BH_2NH_3 \cdot 2 + BH_4^-\)**.