Q.11 N(CH3)3 and N(SiH3)3 are congeners, but around N-atom the former has pyramidal geometry whereas the latter is nearly planar. The bonding responsible for planarity of N(SiH3)3 is
(A) pπ–pπ
(B) pπ–dπ
(C) dπ–dπ
(D) δ
Introduction
Questions related to molecular geometry, hybridization,
and orbital interactions are very common in JEE, NEET, and other
competitive chemistry examinations.
In this article, we explain why N(CH3)3 shows
pyramidal geometry, whereas N(SiH3)3 is nearly planar,
and identify the exact type of bonding responsible for this difference.
Key Concept Used
Molecular shape depends on hybridization, lone pair repulsion, and the availability
of vacant orbitals for back bonding. Delocalization of lone pair electrons can
significantly alter geometry.
Structural Analysis
1️⃣ Geometry of N(CH3)3
Nitrogen is sp3 hybridized with one lone pair.
Carbon atoms do not have vacant orbitals for back bonding.
Strong lone pair–bond pair repulsion results in a
trigonal pyramidal geometry.
2️⃣ Geometry of N(SiH3)3
Silicon atoms possess vacant 3d orbitals.
The lone pair on nitrogen (2p orbital) can be donated into these vacant orbitals.
This delocalization reduces lone pair repulsion and makes the hybridization
approach sp2, resulting in a
nearly planar structure.
Bonding Responsible for Planarity
The planarity of N(SiH3)3 is due to
pπ–dπ back bonding.
Nitrogen donates electron density from its p orbital to the vacant d orbitals
of silicon, stabilizing the planar geometry.
Correct Answer
pπ–dπ bonding
Option (B)
Quick Exam Tip
Whenever nitrogen is bonded to silicon and planarity is mentioned,
always think of pπ–dπ back bonding.
Conclusion
The difference in geometry between N(CH3)3 and
N(SiH3)3 arises due to orbital interactions.
Vacant d orbitals of silicon allow back bonding with nitrogen,
reducing lone pair repulsion and leading to planarity.
