Question

Give the electron geometry, molecular geometry, and hybridization for cbr−3. give the electron geometry, molecular geometry, and hybridization for . eg = trigonal pyramidal; mg = trigonal pyramidal; sp3 eg = octahedral; mg = octahedral, sp3 eg = trigonal planar; mg = trigonal planar; sp2 eg = tetrahedral; mg = linear; sp3 eg = tetrahedral; mg = trigonal pyramidal; sp3

Answer 1

Answer: Electronic geometry is tetrahedral, molecular geometry is trigonal pyramidal and hybridization is $sp_3$

Explanation: Formula used for calculating the no of electrons:

$:{\text{Number of electrons}} =\frac{1}{2}[V+N-C+A]$

where, V = number of valence electrons present in central atom i.e. carbon = 4

N = number of monovalent atoms bonded to central atom=3

C = charge of cation = 0

A = charge of anion = -1

$CBr_3^{-}:$

${\text{Number of electrons}} =\frac{1}{2}[4+3-0+1]=4$

The number of electrons is 4 that means the hybridization will be $sp^3$ and the electronic geometry of the molecule will be tetrahedral.

But as there are three atoms around the central carbon, the fourth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be trigonal pyramidal.