Question

how to find hybridisation of pbCl4?​

Answer 1

Explanation:

H=21[V+M−C+A]

where,

H= Number of orbitals involved in hybridization.

V=Valence electrons of a central atom.

M- Number of monovalent atoms linked to a central atom.

C= Charge of the cation.

A= Charge of an anion.

Consider the hybridization and shape of the options:-

A) PbCl4

H=21[4+4]=4.

⇒sp3 hybridized state.

Tetrahedral in shape.