Question

The hybridization in $PC{{l}_{5}}$ molecule is:
A.$s{{p}^{3}}$
B.$s{{p}^{3}}{{d}^{2}}$
C.$s{{p}^{3}}d$
D.$s{{p}^{2}}$

Answer 1

Hint: In order to explain the characteristic geometrical shapes of polyatomic molecules like $C{{H}_{3}},N{{H}_{3}}$ and ${{H}_{2}}O$ etc. a concept called hybridisation was introduced by Linus Pauling According to him the atomic orbitals combine to form new set of equivalent orbitals known as hybrid orbitals or hybridised orbitals and the process is referred to as hybridisation.

Complete step by step answer:
VSEPR theory is defined as the electron pairs surrounding the central atom must be arranged in space as far apart as possible to minimize the electrostatic repulsion experienced between them.
A central atom can be defined as any atom that is bonded to two or more than two other atoms. The first and the most important rule of the VSEPR theory is that the bond angles about a central atom are those that minimize the total repulsion experienced between the Electron pairs in the atom’s valence shell.
Rules:
A bonding pair of electrons occupies less space than a lone pair of electrons because lone pair of electron is only influenced by one nucleus of the central atom, they are expected to occupy more space with a greater electron density because the bond pair of electrons have the influence of two nuclei.The decreasing order of repulsion is mentioned below
Lone pair-Lone pair repulsion $>$ Lone pair-Bond pair repulsion $>$ Bond pair- bond pair repulsion
Repulsion forces decrease sharply as interior angle increases.They are more strong at 90 degree, less strong at 120 degree and very weak at 180 degree.
Influence of a bonding electron pair decreases as electronegativity increases of an atom forming a molecule.
For the purpose of VSEPR bond theory, Multiple bonds behave equivalent to a single electron pair.
The two electron pair of a double bond occupies more space than one electron pair of a single bond.
The lone pair electrons repels bond pair electrons giving rise to some distortions in the molecular shape.
Hybridization of an atom is identified by calculating the sum of sigma bonds and lone pairs.
If a sum of sigma bonds and lone pairs is 2 $sp$ hybridization
If a sum of sigma bonds and lone pairs is 3 ,$s{{p}^{2}}$ hybridization
If a sum of sigma bonds and lone pairs is 4 , $s{{p}^{3}}$ hybridization
If a sum of sigma bonds and lone pairs is 5, $s{{p}^{3}}d$ hybridization
If a sum of sigma bonds and lone pairs is 6, $s{{p}^{3}}{{d}^{2}}$ hybridization.
Now $PC{{l}_{5}}$ contains 5 bond pairs and 0 lone pairs. So it is $s{{p}^{3}}d$ hybridized.

So C is the correct option.

Note: The shape of $PC{{l}_{5}}$ is Trigonal bipyramidal as it contains 5 bond pairs and 0 lone pairs.

Bond pair	Lone pair	Shape
7	0	Pentagonal bipyramidal
6	1	Distorted octahedral.