Question

How many hybrid orbitals are present in the molecule $PC{l_5}$?

Answer 1

The concept of hybridization is based on VSEPR theory. The VSEPR theory says that during a bond formation mixing of orbitals takes place.

Complete step by step answer:
The given compound is phosphorus pentachloride. It is an inorganic compound composed of phosphorus and chlorine atoms. Phosphorus is the least electronegative atom between phosphorus and chlorine, so it is the central atom according to VSEPR theory.
Phosphorus is an element in the periodic table with atomic number $15$ and electronic configuration $\left[ {Ne} \right]3{s^2}3{p^3}$. The outermost shell of phosphorus is $3$ which has five electrons of which two electrons are present in $3s$-orbital and three electrons are present in $3p$-orbital. Phosphorus also has outer $3d$-orbitals which are used for bond formation.
In $PC{l_5}$, five $P - Cl$ bonds are present so the $P$ atom needs five orbitals to form the five $P - Cl$ bonds. In order to form five bonds the $P$ atom needs one $3s$ and three $3p$ orbitals and one $3d$ orbital. The hybridization of the $P$ atom is $s{p^3}d$. This is also evaluated using the VSEP number.
VSEPR number = $\dfrac{1}{2}$ (number of electrons from $P$ atom + number of bonded $Cl$ atoms)
$= \dfrac{1}{2}(5 + 5) = 5.$
The VSEP number equal to $5$ belongs to $s{p^3}d$ hybridization. Thus the geometry of the molecule is trigonal bipyramidal. Three chlorine atoms are positioned in the equatorial positions and two chlorine atoms are positioned in two axial positions of the TBP structure.
Thus all the hybridized orbitals of phosphorus atoms are used up in $PC{l_5}$ and no hybrid orbitals remain free.

Note:
The concept of hybridization is very helpful in the determination of molecular structures of several molecules. The VSEPR is the short form of Valence Shell Electron Pair Repulsion Theory. The interactions between the bond pairs and lone pairs of electrons are considered for obtaining the exact structure.