The correct statement among the following is:

🧠 Compare the geometry and basicity of $\mathrm{(SiH_3)_3N}$ vs $\mathrm{(CH_3)_3N}$ by considering pπ–dπ back-bonding.

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

• N: sp³ hybridized, pyramidal; lone pair in sp³ orbital → fully available for protonation
• Pyramidal, good base (pKₐ of conjugate acid ≈ 9.8)

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

• Si has empty d-orbitals; N lone pair (p-orbital) donates into Si 3d-orbitals → pπ–dπ back-bonding
• This delocalises N's lone pair into Si–N bonds → N becomes sp² hybridized (planar molecule)
• The lone pair is no longer concentrated on N → much less basic
• Structure: planar (like BF₃, where back-bonding also leads to sp²)

Summary:

$\mathrm{(SiH_3)_3N}$ is planar and less basic than $\mathrm{(CH_3)_3N}$

⚡ C has no empty d-orbitals → no back-bonding → lone pair stays on N → pyramidal and basic.

$\boxed{\text{Answer: } \mathrm{(SiH_3)_3N} \text{ is planar and less basic than } \mathrm{(CH_3)_3N}}$