Question

2) Give the number of lone pairs around the central atom and the molecular geometry of IF5. A) 0 lone pairs, square pyramidal D) 1 lone pair, square pyramidal B) 0 lone pairs, trigonal bipyramidal E) 2 lone pairs, pentagonal C) 1 lone pair, octahedral

Answer 1

Answer: D) 1 lone pair, square pyramidal

Explanation:

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

where,

V = number of valence electrons present in central atom  = 7

N = number of monovalent atoms bonded to central atom  = 5

C = charge of cation  = 0

A = charge of anion  = 0

Now we have to determine the hybridization of the $IF_5$  molecule.

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

Bond pair electrons = 5

Lone pair electrons = 6-5 = 1

The number of electrons are 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 iodine atom, 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.