Question

The correct statement among the following is:
(A) ${(Si{H_3})_3}N$ is pyramidal and more basic than ${(C{H_3})_3}N$
(B) ${(Si{H_3})_3}N$ is planar and more basic than ${(C{H_3})_3}N$
(C) ${(Si{H_3})_3}N$ is pyramidal and less basic than ${(C{H_3})_3}N$
(D) ${(Si{H_3})_3}N$ is planar and less basic than ${(C{H_3})_3}N$

Answer 1

Hint:. One of the compounds has $d\pi - p\pi $ back bonding in their structure. Find the hybridization of the compound in order to predict the shape of the molecule. The $s{p^3}$ hybrid molecule is pyramidal and $s{p^2}$ hybrid molecule is planar in shape.

Complete step by step answer:
We will examine both the shape as well as the basicity of the given two compounds in order to find the correct answer.
- ${(C{H_3})_3}N$ is trimethylamine and it has a pyramidal shape. The central atom is nitrogen. The electronic configuration of nitrogen is $[He]2{s^2}2{p^3}$.
- So, one electron each is provided by methyl groups and it forms a bond with three p-orbitals. The pair of electrons in s-orbital acts as a lone pair of electrons. Thus, the hybridization in ${(C{H_3})_3}N$ is $s{p^3}$.
- In ${(Si{H_3})_3}N$, the hybridization of nitrogen atoms is $s{p^2}$. Actually it has a lone pair in a 2p orbital. This gets transferred to the empty d-orbital of Silicon atom. So, because of its $s{p^2}$ hybridization, it has planar shape.
- Due to no back bonding, trimethylamine (${(C{H_3})_3}N$) is more basic than ${(Si{H_3})_3}N$.
Thus, we can conclude that ${(C{H_3})_3}N$ is pyramidal and ${(Si{H_3})_3}N$ is planar. ${(Si{H_3})_3}N$ is less basic than ${(C{H_3})_3}N$ .
So, the correct answer is “Option D”.

Note: Do not assume that same as ${(C{H_3})_3}N$, ${(Si{H_3})_3}N$ also has $s{p^3}$ hybridization. Actually due to $d\pi - p\pi $ backbonding, ${(Si{H_3})_3}N$ has $s{p^3}$ hybridization. This happens because Si has vacant d-orbitals where C-atom does not have any d-orbital in its valence shell.