Question

The d orbital involved in the hybridization in the $PC{l_5}$ molecule is
A. $3{d_{{x^2} - {y^2}}}$
B. $3{d_{{z^2}}}$
C. $3{d_{xy}}$
D. $4{d_{{z^2}}}$

Answer 1

We know that hybridization is the mixing of orbitals of the nearly same energy. So we will find the hybridization of phosphorus pentachloride. Phosphorus is an element with atomic number 15 and 5 valence electrons. Chlorine has atomic number 17 and 7 valence electrons.
The hybridization of a molecule is given by the number of sigma bonds plus the number of lone pairs present in that molecule. From the lewis dot structure and Valence Bond Theory, the hybridization can be found out.

Complete step by step answer:
First, we will find out the hybridization of phosphorus pentachloride molecules by the given formula.
Hybridization is equal to the number of sigma bonds plus the lone pairs present in the molecule.
So here the number of sigma bonds in phosphorus pentachloride molecules is five and the number of lone pairs is zero because the valency of phosphorus is 5 and it's shared by 5 chlorine atoms. Therefore, no lone pairs.
The number of lone pairs is found out by subtracting the total number of bond pairs minus the valency of the central atom.
So the required hybridization is $5 + 0 = 5$ . So it $s{p^3}d$ and has a trigonal bipyramidal structure.
Now the electronic configuration of phosphorus is $[Ne]3{p^1}_x3p_y^13p_z^13{d^0}$
Here phosphorus pentachloride requires five electrons so it excites one electron from $3s$ to one of the empty $3d$ orbital which gives a trigonal bipyramidal geometry.
Now coming to the shape there is a criterion for hybridization. The hybrid orbital is formed in such a way that it has minimum repulsion between them and thus minimal energy.
So we have to choose a d orbital which shows minimum repulsion so ${d_{{z^2}}}$ is the most suitable one for hybridization as it has no x and y components so it will face less repulsion and will be suitable for hybridization.

So, the correct answer is Option B.

Note: The d orbitals are generally very large and have high energy to mix with s and p orbitals. The energy of orbitals depends on the mean radial distance. As d orbital has more radial distance than others so they are not into participating in the hybridization process but they do due to some other factors.