Question

Discuss the hybridization in case of $PC{l_5}$. Why are the axial bonds longer as compared to equatorial bonds?

Answer 1

Phosphorus is an element of nitrogen group and ground state electronic configuration of Phosphorus atom is $[Ne]3{s^2}3{p^3}$. Repulsive forces may make some bonds longer due to bond pair-bond pair repulsion.

Complete answer:
We know that ground state electronic configuration of Phosphorus atom is $[Ne]3{s^2}3{p^3}$ and excited state electronic configuration is $[Ne]3{s^1}3{p^3}3{d^1}$. It is shown below.


So, each chlorine atom will donate one electron each to each hybrid orbitals and for new five sigma bonds. So, we can see that there are five orbitals involved in the hybridization and the orbitals are one s, three p and one d-orbital. So, we can say that the molecule has $s{p^3}d$ hybridization. This molecule favours the Pentagonal bipyramidal shape. Let’s see the shape of the molecule first.

As you can see that three chlorine atoms are in one plane and the other two are forming a perpendicular line to the plane.
Here, the chlorine atoms that are present in the plane are called equatorial atoms and the atoms that are in perpendicular line, are called axial atoms.
Note, these axial atoms feel more repulsive forces with three adjacent chlorine atoms, hence they try to be as far as possible from the equatorial chlorine atoms.
That is the reason why the axial bonds are longer compared to equatorial bonds.

Additional Information:
- In $s{p^3}d$ hybridization, compound can have either trigonal bipyramidal or square pyramidal geometry.
- If a compound is having trigonal bipyramidal geometry, then it uses ${d_{{z^2}}}$ orbital as one of the d-orbitals to form bonds. If a compound has square pyramidal geometry, then ${d_{{x^2} - {y^2}}}$ is used as one of the d-orbitals to make bonds.

Note: Note that $ds{p^3}$ is different from $s{p^3}d$ hybridization. So, do not get confused between them. One needs to remember that $PC{l_5}$ does not have square bipyramidal geometry as it is also a possible geometry for compounds that have $s{p^3}d$ hybridization.