Question

The correct statement among the following is

• A. ${(SiH3)3N}$ is pyramidal and more basic than ${(CH3)3N}$
• B. ${(SiH3)3N}$ is planar and more basic than ${(CH3)3N}$
• C. ${(SiH3)3N}$ is pyramidal and less basic than ${(CH3)3N}$
• D. ${(SiH3)3N}$ is planar and less basic than ${(CH3)3N}$

Answer 1

The Correct Option is D

 To determine the correct statement, we need to analyze the structure and basicity of ${(SiH3)3N}$ and ${(CH3)3N}$. Let's go through the concepts step by step:

1. **Structure of ${(CH3)3N}:**
   • Trimethylamine, ${(CH3)3N}$, has a pyramidal shape due to the sp³ hybridization of the nitrogen atom. The lone pair on the nitrogen creates a tetrahedral electron-pair geometry, but the molecular shape is pyramidal.
2. **Structure of ${(SiH3)3N}:**
   • In contrast, trisilylamine, ${(SiH3)3N}$, tends to be planar. The larger size of silicon compared to carbon, and its lower ability to participate in effective hyperconjugation or back-bonding with nitrogen, leads to a planar structure where the lone pair is delocalized over the Si-N-Si bonds.
3. **Basicity Comparison:**
   • Basicity is often related to the availability of the lone pair of electrons that can be donated. In ${(CH3)3N}$, the lone pair is more localized and available for donation, making it a stronger base.
   • In ${(SiH3)3N}$, the lone pair on nitrogen is less available because of delocalization over the Si-N-Si bonds, making it less basic than trimethylamine.

Based on the structural and electronic considerations above, the correct statement is: \({(SiH3)3N}\) is planar and less basic than \({(CH3)3N}\).

**Conclusion:**

The option "\({(SiH3)3N}\) is planar and less basic than \({(CH3)3N}\)" is correct because of the delocalization of the lone pair in the planar structure of ${(SiH3)3N}$, reducing its basicity compared to the pyramidal and more localized lone pair in ${(CH3)3N}$.