The structure of \(\mathrm{N}\left(\mathrm{SiH}_{3}\right)_{3}\) involves a planar arrangement of \(\mathrm{N}\) and \(\mathrm{Si}\) atoms, whereas that of the related compound \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) has a pyramidal arrangement of N and \(\mathrm{C}\) atoms. Propose bonding schemes for these molecules that are consistent with this observation.

Identify the central atom in the molecule. In both compounds \(\mathrm{N}\left(\mathrm{SiH}_{3}\right)_{3}\) and \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\), Nitrogen (N) acts as the central atom.

Count the valence electrons for each central atom. The Nitrogen atom has 5 valence electrons. Each Hydrogen atom contributes 1 valence electron, and Carbon contributes 4, hence \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) has total of 8 valence electrons. Silicon in the case of \(\mathrm{N}\left(\mathrm{SiH}_{3}\right)_{3}\), contributes 4 valence electrons, which gives a total of 8 valence electrons.

The hybridization of an atom in a molecule can be calculated by the formula: \( \text{hybridization} = 1/2 * (\text{Number of valence electrons} + \text{Number of monovalent atoms attached to the atom} - \text{Charge on cation} + \text{Charge on anion})\). Applying this formula, both \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\) and \(\mathrm{N}\left(\mathrm{SiH}_{3}\right)_{3}\) show sp3 hybridization. However, due to the larger size and less electronegativity of Silicon as compared to Carbon, the bond pair-bond pair repulsion is much lesser, leading to a planar configuration in the case of \(\mathrm{N}\left(\mathrm{SiH}_{3}\right)_{3}\). Whereas in \(\mathrm{N}\left(\mathrm{CH}_{3}\right)_{3}\), due to higher electronegativity of Carbon, the bond pair-bond pair repulsion is greater, leading to a pyramidal shape.