From `B_(2)H_(6)`, all the following can be prepared except

Options

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

Answer

To solve the question "From \( B_2H_6 \), all the following can be prepared except," we will analyze the possible reactions of diborane (\( B_2H_6 \)) with various reagents.

### Step-by-Step Solution:

1. **Identify the Reactions of \( B_2H_6 \)**:
- Diborane can react with water, ammonia, trimethylamine, and sodium hydride. We will evaluate each reaction to see if the products can be formed.

2. **Reaction with Water**:
- When \( B_2H_6 \) reacts with water, it undergoes hydrolysis to form boric acid (\( H_3BO_3 \)) and hydrogen gas.
- **Equation**:
\[ B_2H_6 + 6H_2O \rightarrow 2H_3BO_3 + 6H_2 \]
- **Conclusion**: This reaction can produce \( H_3BO_3 \).

3. **Reaction with Ammonia**:
- When \( B_2H_6 \) reacts with ammonia (\( NH_3 \)), it forms a complex involving boron and ammonia.
- **Equation**:
\[ B_2H_6 + 2NH_3 \rightarrow H_2B(NH_3)_2^+ + BH_4^- \]
- **Conclusion**: This reaction can produce a boron-ammonia complex.

4. **Reaction with Trimethylamine**:
- When \( B_2H_6 \) reacts with trimethylamine (\( (CH_3)_3N \)), the bulky nature of trimethylamine makes it difficult for the reaction to occur.
- **Conclusion**: No stable product can be formed due to steric hindrance. This is the option that cannot be prepared.

5. **Reaction with Sodium Hydride**:
- When \( B_2H_6 \) reacts with sodium hydride (\( NaH \)), it forms sodium borohydride (\( NaBH_4 \)).
- **Equation**:
\[ B_2H_6 + 2NaH \rightarrow 2NaBH_4 \]
- **Conclusion**: This reaction can produce sodium borohydride.

### Final Answer:
From the analysis, the only option that cannot be prepared from \( B_2H_6 \) is the one involving trimethylamine. Therefore, the answer is:

**Option 3 cannot be prepared.**