`underset(("Lewis base"))(B_(2)H_(6)+2Lrarr2BH_(3)L)`
`L` would be `:`

Options

A. `NH_(3)`
B. `CH_(3)NH_(2)`
C. `(CH_(3))_(2)NH`
D. `(CH_(3))_(3)N`

Explanation

To solve the question regarding the reaction of diborane (B₂H₆) with a Lewis base (L) to form 2BH₃L, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- The reactant is diborane (B₂H₆).
- The product formed is 2BH₃L, where L represents a Lewis base.

2. **Understand the Nature of Diborane**:
- Diborane (B₂H₆) can undergo two types of cleavage: symmetrical cleavage and unsymmetrical cleavage.
- Symmetrical cleavage leads to the formation of 2BH₃L.
- Unsymmetrical cleavage leads to the formation of 1BH₂ and 1BH₄ units.

3. **Determine the Type of Cleavage**:
- Since the reaction produces 2BH₃L, it indicates that symmetrical cleavage is taking place.
- Symmetrical cleavage occurs in the presence of a bulky Lewis base.

4. **Identify Characteristics of Lewis Bases**:
- A Lewis base is a species that can donate an electron pair.
- In this context, we need to identify a Lewis base that is large and bulky to facilitate symmetrical cleavage.

5. **Evaluate the Given Options for L**:
- We need to analyze the options provided to find the most bulky Lewis base.
- The options include:
- NH₃ (Ammonia) - small
- CH₃NH₂ (Methylamine) - small
- CH₃₂NH (Dimethylamine) - small
- CH₃₂N (Dimethylamine) - bulky

6. **Select the Correct Lewis Base**:
- Among the options, CH₃₂N (Dimethylamine) is the most bulky.
- Therefore, L in this reaction is CH₃₂N.

7. **Conclusion**:
- The Lewis base (L) that leads to the formation of 2BH₃L from diborane (B₂H₆) is CH₃₂N.

### Final Answer:
L is CH₃₂N (Dimethylamine).