Question

Explain hybridisation of central atom in: $I{{F}_{5}}$

Answer 1

The hybridisation of any molecule can be predominantly determined by the presence of lone pairs of electrons; the central metal atom have and the number of sigma bonds it forms with other elements, responsible for forming a molecule.

Complete answer:
Let us see what we exactly mean by the term hybridisation and then discuss its types;
Hybridisation-
It is basically an extension to the Valence Bond Theory. It is the concept of mixing of atomic orbitals into new hybrid orbitals having different energies, shapes, etc. These newly formed orbitals influence the molecular geometry and bonding properties.
There are five types of hybridisation depending upon the bonding pattern the molecule follows-
They are $sp$ , $s{{p}^{2}}$ , $s{{p}^{3}}$, $s{{p}^{3}}d$ , $s{{p}^{3}}{{d}^{2}}$ .
One basic technique to know the hybridisation of any given molecule is the summation of sigma bonds it has and lone pairs present on the central metal atom-
The summation of the number of sigma bonds the central metal forms and the presence of lone pair of electrons on it gives the hybridisation of the molecule as follows-
If summation = 2 then, hybridisation = $sp$.
If summation = 3 then, hybridisation = $s{{p}^{2}}$.
If summation = 4 then, hybridisation = $s{{p}^{3}}$.
If summation = 5 then, hybridisation = $s{{p}^{3}}d$.
If summation = 6 then, hybridisation = $s{{p}^{3}}{{d}^{2}}$.
Now, looking towards the illustration we can say that,
Molecule - $I{{F}_{5}}$
The central atom of I has 7 valence electrons. Out of those, 5 are engaged into sigma bonding with F. Thus, there is one lone pair of electrons present on I.
Thus, the summation will be 6 (5 sigma bonds and 1 lone pair of electrons) which eventually gives the $s{{p}^{3}}{{d}^{2}}$ hybridisation of $I{{F}_{5}}$ molecule.

Note:
$I{{F}_{5}}$molecule has $s{{p}^{3}}{{d}^{2}}$ hybridisation thus, it will be square pyramidal in shape and its geometry will be octahedral.
Do note that the summation we find is known as the steric number of a molecule. Here, the summation is 6 i.e. the steric number of $I{{F}_{5}}$ is 6.