The correct statement among the following is:

🧠 In $\mathrm{(SiH_3)_3N}$, silicon has vacant 3d-orbitals that can accept the lone pair from nitrogen, creating pπ–dπ back-bonding.

🗺️ $\mathrm{(CH_3)_3N}$ (trimethylamine):

• N is sp³ hybridized → pyramidal
• Lone pair in sp³ orbital, fully localized on N → freely available for protonation
• Pyramidal, stronger base

$\mathrm{(SiH_3)_3N}$ (trisilylamine):

• Si has empty 3d-orbitals; N donates its lone pair into Si 3d via pπ–dπ overlap
• N becomes sp² hybridized (lone pair occupies p-orbital that overlaps with Si d-orbitals → donated away)
• Molecule becomes planar (all N–Si bonds and the N lone pair coplanar)
• Lone pair is delocalized → N is much less basic

⚡ Analogous to why N(BF₃)₃ is a weaker base than N(CH₃)₃ — back-donation to empty orbitals removes electron density from N.

$\boxed{\text{Answer: planar and less basic than } \mathrm{(CH_3)_3N}}$