Question

Compound formed by $s{{p}^{3}}d$ hybridization will have which of the following structures?
(A) Trigonal bipyramidal
(B) T-shaped
(C) Linear
(D) Either of these depending on the number of lone pairs of electrons on the central atom.

Answer 1

The number of lone pairs and bond pairs present in the type of hybridization of the molecule will decide the arrangement of the atoms in the space and the structure of the molecule i.e. the steric number will decide the hybridization.

Complete step-by-step answer: In the question it is asked that for the given hybridization, in which structure will the atoms arranged in the space. First let us discuss some information on the $s{{p}^{3}}d$ hybridization. In $s{{p}^{3}}d$ hybridization, there is one s orbital, three p orbitals and one d orbital. All the orbitals are almost in similar energy and the five orbitals are intermixed to form five new hybrid orbitals which have the same energy and are identical. These orbitals are involved in bonding. The percentage of the s character in $s{{p}^{3}}d$ hybridization is about 20%, and p character is about 60% and d character is about 20%.
These five orbitals of the central atom bonds with five other atoms and from the five orbitals three orbitals are aligned in a trigonal plane. And the other two orbitals are arranged above and below the trigonal plane at right angles. The molecule which has $s{{p}^{3}}d$ hybridization will have trigonal bipyramidal geometry.

Therefore the correct answer for the above given question is option (A).

Additional Information:
One of the examples for a molecule which has $s{{p}^{3}}d$ hybridization is $PC{{l}_{5}}$ molecule.
In $PC{{l}_{5}}$, P is the central atom and the five Cl atoms combine with P through the five orbitals by sharing its electrons and forming five P-Cl sigma bonds.
Among the five bonds, three will be equatorial bonds, which are arranged in a trigonal plane and make an angle of ${{120}^{\circ }}$ with each other. And the two axial bonds will be arranged in the right angle fashion above and below the equatorial plane.

Note: Generally students get confused while calculating the steric number, as sometimes they count the pie bonds also for calculating the hybridization. But only the sigma bonds should be calculated. For every steric number there is always a geometry associated with the hybridization.