Concept: The shape of a molecule is determined by the hybridization of the central atom and the number of lone pairs and bond pairs around it, according to VSEPR theory. A pyramidal shape typically arises from sp³ hybridization with one lone pair on the central atom.

Why (B) is correct:

Let's analyze the hybridization and geometry of the central atom in each option:

• A) $\mathrm{P}{\left({\mathrm{CH}}_3\right)}_3$: The central atom is Phosphorus (P). P has 5 valence electrons. It forms 3 single bonds with three ${\mathrm{CH}}_3$ groups. This leaves 5 - 3 = 2 electrons, which form one lone pair. Thus, P has 3 bond pairs and 1 lone pair. The steric number is 4, leading to sp³ hybridization and a trigonal pyramidal geometry.
• B) ${\left({\mathrm{SiH}}_3\right)}_3\mathrm{N}$: The central atom is Nitrogen (N). N has 5 valence electrons. It forms 3 single bonds with three ${\mathrm{SiH}}_3$ groups. In ${\left({\mathrm{SiH}}_3\right)}_3\mathrm{N}$, the lone pair on nitrogen is delocalized into the empty d-orbitals of silicon atoms (p$\pi$-d$\pi$ bonding). This delocalization reduces the electron density around nitrogen, making it sp² hybridized. With 3 bond pairs and effectively no localized lone pair, the geometry becomes trigonal planar. Therefore, it does not have a pyramidal shape.
• C) ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}$: The central atom is Nitrogen (N). N has 5 valence electrons. It forms 3 single bonds with three ${\mathrm{CH}}_3$ groups. This leaves 5 - 3 = 2 electrons, which form one lone pair. Thus, N has 3 bond pairs and 1 lone pair. The steric number is 4, leading to sp³ hybridization and a trigonal pyramidal geometry.
• D) $\mathrm{P}{\left({\mathrm{SiH}}_3\right)}_3$: The central atom is Phosphorus (P). P has 5 valence electrons. It forms 3 single bonds with three ${\mathrm{SiH}}_3$ groups. This leaves 5 - 3 = 2 electrons, which form one lone pair. Thus, P has 3 bond pairs and 1 lone pair. The steric number is 4, leading to sp³ hybridization and a trigonal pyramidal geometry. Unlike nitrogen in ${\left({\mathrm{SiH}}_3\right)}_3\mathrm{N}$, the lone pair on phosphorus is less available for p$\pi$-d$\pi$ bonding with silicon due to the larger size of P and less effective overlap, so it retains its pyramidal shape.

Option Analysis:

• A) $\mathrm{P}{\left({\mathrm{CH}}_3\right)}_3$: Has a pyramidal shape (sp³, 1 lone pair).
• B) ${\left({\mathrm{SiH}}_3\right)}_3\mathrm{N}$: Has a trigonal planar shape due to p$\pi$-d$\pi$ bonding (sp², no localized lone pair).
• C) ${\left({\mathrm{CH}}_3\right)}_3\mathrm{N}$: Has a pyramidal shape (sp³, 1 lone pair).
• D) $\mathrm{P}{\left({\mathrm{SiH}}_3\right)}_3$: Has a pyramidal shape (sp³, 1 lone pair).

Correct Answer: (B)