Concept: The geometry around a central atom is determined by the number of bond pairs and lone pairs of electrons, as predicted by VSEPR theory. However, in some cases, back-bonding can alter the expected geometry.

Why (A) is correct:

In trisilylamine, $\mathrm{N}({\mathrm{SiH}}_3{)}_3$, the nitrogen atom has a lone pair and is bonded to three ${\mathrm{SiH}}_3$ groups. According to VSEPR theory, with three bond pairs and one lone pair, the expected geometry around nitrogen would be pyramidal (like ammonia). However, silicon has vacant d-orbitals, and the lone pair on nitrogen can be delocalized into these vacant d-orbitals of silicon through $\mathrm{p}\pi -\mathrm{d}\pi$ back-bonding. This back-bonding gives the nitrogen atom a partial positive charge and reduces the electron density of the lone pair, making it less stereochemically active. As a result, the molecule adopts a planar geometry around the nitrogen atom, similar to ${\mathrm{BF}}_3$ or ${\mathrm{CO}}_3^{2-}$.

Option Analysis:

• A) Planar geometry: This is correct due to $\mathrm{p}\pi -\mathrm{d}\pi$ back-bonding between nitrogen's lone pair and silicon's vacant d-orbitals.
• B) Tetrahedral geometry: This would be expected if nitrogen formed four single bonds with no lone pairs, or if the lone pair was stereochemically active and the molecule was distorted tetrahedral (like water).
• C) Pyramidal geometry: This is the expected geometry for a central atom with three bond pairs and one lone pair (e.g., ammonia, ${\mathrm{NH}}_3$), but it is altered in trisilylamine due to back-bonding.
• D) None of these: Incorrect, as planar geometry is the correct description.

Correct Answer: (A)