Question

Rationalize why the geometry about the carbons in R2C=CR2 is planar, but the geometry about the E elements R2E=ER2 becomes increasingly pyramidal for E = Si, Ge, Sn, Pb.

Does entropic changes favour the left or right side of the equation in the equilibrium of R2E=ER2 à 2 R2E: (Sorry, can

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Answer 1

The optimized geometry of heavy group-14 ethylenes (dimetallenes; H2E=EH2,( E = Si, Ge, Sn and Pb) is trans-bent in contrast to planar ethylene (E = C) due to  due to significant mixing between the π*(SOMO) and a σ*(Si-H) orbital with proper symmetry representation. The mono-anion of disilene [H2Si=SiH2]•− adopts trans- and cis-bent geometries due to significant mixing between the π*(SOMO) and a σ*(Si-H) orbital with proper symmetry representation. Applicability of the model to heavy ethylenes with bulky substituents is less satisfactory because the steric effects are the controlling factor determining the geometry.

while,  E = Si, Ge, Sn, Pb

Structure: trans-bent

The trans- and cis-bent geometries of neutral dimetallenes may be more stable than the planar geometry but with very small energy gains, while these geometries should be largely stabilized with significant bent angles and stabilization energies in dimetallene anions.