Question

Shape of ${(Si{H_3})_3}N$ with respect to $N$ is:
A. Pyramidal
B. Tetrahedral
C. Trigonal planar
D. T-shaped

Answer 1

Here, nitrogen is a central atom with lone pair of electrons and silicon has vacant orbital due to which back bonding takes place and the hybridization of nitrogen also gets reduced by one. The hybridization of nitrogen is supposed to be $s{p^3}$ but in such a condition it will become $s{p^2}$ hybridized.

Complete step by step solution
As ${(Si{H_3})_3}N$ is a chemical compound named as trisilylamine. In this compound back bonding occurs because of two conditions that are firstly central atom (nitrogen) has lone pair of electrons and secondly silicon is placed adjacent to it with vacant $3d$ orbital which make them both capable of back bonding.
Generally, nitrogen is attached to three silicon atom so hybridization of nitrogen should be $s{p^3}$ but due to occurrence of back bonding that is $p\pi - d\pi$ coordinate bond its hybridization reduces to one and hence it become $s{p^2}$ hybridized.
So, the structure of ${(Si{H_3})_3}N$ is trigonal planar.
Hence, option C is right.

Note
Back bonding is also known as back donation, in back bonding electrons get moved from an atomic orbital on one atom to antibonding orbital of another atom on a pi-acceptor ligand. In the organometallic compounds of transition metals having multi-atomic ligands such as carbon monoxide or ethylene back bonding is very common. Electrons from the metal are used to bond to the ligand, in the process the metal releases its excess negative charge. Compounds where $\pi$ back bonding occurs include Zeise's salt. A transition metal in which synergic process takes place with donation of electrons from the filled $\pi - orbital$ or lone electron pair orbital of the ligand into an empty orbital of the metal, together with release that is back donation of electrons from an $nd - orbital$ of the metal into the empty ${\pi ^*} - antibonding$ orbital of the ligand.