Question

Trisilylamine $[N{{(Si{{H}_{3}})}_{3}}]$ has a,
(A) Planar geometry
(B) Tetrahedral geometry
(C) Pyramidal geometry
(D) None of these

Answer 1

Hint: Trisilylamine has silicon atoms with vacant d-orbitals. The hybridisation of the central metal atom explains the structure and planarity of the molecule. Also resonance of lone pairs of nitrogen atoms plays a crucial role in this molecule.

Complete step by step answer:
Back Bonding is a type of resonance that is exhibited by several chemical compounds. It is known to provide increased stability of the compound. It takes place when the ligand has lone pairs of electrons present on them except bonding and metal atoms have high oxidation states. The electrons of ligands are shared with the metal atom after the formation of the sigma bond. Pi backbonding is a type of back bonding in which the electrons move from the atomic orbital of a given atom into the pi antibonding orbital on a ligand which is a pi-acceptor. Pi back bonding is very common in organometallic chemistry.
Trisilylamine has a planar geometry because it has backbonding. Silicon has vacant d-orbitals. The lone pair on the Nitrogen atom provides the vacant d-orbital of silicon with electrons. Hence the nitrogen silicon bond acquires partial double bond character, that the hybridization becomes $s{{p}^{2}}$. Molecules having $s{{p}^{2}}$ and $sp$ hybridization acquire planar geometry. And $s{{p}^{2}}$ hybridization shows trigonal planar geometry, therefore the shape of trisilylamine is trigonal planar.
Hence, the correct answer is A option.

note: In Trisilylamine, there is formation of $p\pi -d\pi $ bonding, due to the vacant 3d orbital of silicon. And trimethylamine ($N{{(C{{H}_{3}})}_{3}}$) has $p\pi -p\pi $bonding, between carbon and nitrogen atom and thus nitrogen have $s{{p}^{3}}$hybridization with pyramidal shape.