Which of the following is correct statement regarding `N(CH_(3))_(3)` and `N(SiH_(3))_(3)`?

Options

A. `N(SiH_(3))_(3)` is planar and more basic than `N(CH_(3))_(3)`
B. `N(SiH_(3))_(3)` is planar and less basic than `N(CH_(3))_(3)`
C. `N(CH_(3))_(3)` is planar and more basic than `N(SiH_(3))_(3)`
D. `N(CH_(3))_(3)` is pyramidal and less basic than `N(SiH_(3))_(3)`

Answer

To analyze the question regarding the compounds \( N(CH_3)_3 \) (trimethylamine) and \( N(SiH_3)_3 \) (trimethylsilane), we will follow these steps:

### Step 1: Identify the Structures
- **Trimethylamine \( N(CH_3)_3 \)**: This molecule has a nitrogen atom bonded to three methyl groups and has a lone pair of electrons on the nitrogen. The structure can be represented as:

\[ \text{N} \quad \text{CH}_3 \quad \text{CH}_3 \quad \text{CH}_3 \]

- **Trimethylsilane \( N(SiH_3)_3 \)**: This molecule has a nitrogen atom bonded to three silane groups. The structure can be represented as:

\[ \text{N} \quad \text{SiH}_3 \quad \text{SiH}_3 \quad \text{SiH}_3 \]

### Step 2: Analyze the Bonding
- In \( N(CH_3)_3 \), the nitrogen has a lone pair that is not involved in bonding with any other atom, making it a good base.
- In \( N(SiH_3)_3 \), the silicon atom has vacant d-orbitals, allowing for back-bonding from the lone pair on nitrogen to silicon. This results in a partial double bond character between nitrogen and silicon.

### Step 3: Determine Basicity
- The presence of back-bonding in \( N(SiH_3)_3 \) reduces the availability of the lone pair on nitrogen for protonation, making it less basic.
- In contrast, \( N(CH_3)_3 \) does not have back-bonding, thus the lone pair is more available for donation, making it more basic.

### Step 4: Conclusion
Based on the analysis, we can conclude:
- \( N(CH_3)_3 \) is more basic than \( N(SiH_3)_3 \) due to the absence of back-bonding in the former.

### Final Answer
The correct statement regarding \( N(CH_3)_3 \) and \( N(SiH_3)_3 \) is that **\( N(CH_3)_3 \) is more basic than \( N(SiH_3)_3 \)**.

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