Question

Trisilylamine $[\ddot N{(Si{H_3})_3}]$ has a
A. Planar geometry
B. Tetrahedral geometry
C. Pyramidal geometry
D. None of these

Answer 1

For solving this question, we need to refer to the valence bond theory which explains that the structure and magnetic properties of coordination compounds. According to valence bond theory, the metal ion or atom under the influence of ligands uses $\left( {n - 1} \right)d$, $ns$, $np$, and $d$ orbitals for hybridization. Thus, yields a set of equivalent orbitals that define geometry in terms of octahedral, tetrahedral, square planar etc. These hybridized orbitals can also overlap with the ligand orbitals to donate electron pairs for bonding.

Complete step by step answer:
A we know that the molecular formula of ${(Si{H_3})_3}N$ and it has $p\pi - p\pi$ bonding present. From the molecular formula we get to know that the concept of back bonding is followed here.

From the structure of the given compound, we know that Silicon has vacant $d$-orbitals. The lone pair on $N$ provides the vacant $d$-orbitals of $Si$ with the required lone pair of electrons. Hence, $N - Si{H_3}$ bond achieves a partial double bond character that is, its hybridization changes and becomes $s{p^2}$ which refers to the trigonal planar shape of the molecule. Now, with reference to the concept of back bonding as a sort of resonance between lone pair of electrons and the other atom with a vacant $d$ or $p$-orbital present in their central atom sulfur. Hence, gives a planar shape (option A)

$\therefore$ The option A is the correct answer .

Note:
As we know, the structure of trimethylamine is pyramidal in shape due to the presence of $p\pi - p\pi$ bonding like trisilylamine. The trimethylamine molecular formula is ${(C{H_3})_3}N$ where $N$ undergoes $s{p^3}$ hybridization and the shape is pyramidal. Therefore, it is more basic than the former compound.