Question

draw lewis structure of BrF5 and XeF2 using VSPER theory , predict their shapes

Answer 1

Answer: $BrF_5$ is square pyramidal and $XeF_2$ is linear in shape.

Formula used

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

where, V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

$BrF_5$

${\text{Number of electrons}} =\frac{1}{2}[7+5-0+0]=6$

The number of electrons is 6 that means the hybridization will be $sp^3d^2$ and the electronic geometry of the molecule will be octahedral.

But as there are five atoms around the central atom bromine, the sixth 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 square pyramidal.

2. $XeF_2$

${\text{Number of electrons}} =\frac{1}{2}[8+2-0+0]=5$

The number of electrons is 5 that means the hybridization will be $sp^3d$ and the electronic geometry of the molecule will be trigonal bipyramidal.

But as there are only two atoms around the central xenon, the rest three positions 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 linear.