Question

A $ {\text{s}}{{\text{p}}^{\text{3}}}{\text{d}} $ hybridisation results in:
(A) A square planar molecule
(B) An octahedral molecule
(C) A trigonal bipyramidal molecule
(D) A tetrahedral molecule

Answer 1

Hint: To answer this question, you must recall the VSEPR theory which is the Valence Shell Electron Pair Repulsion theory. According to this theory, the hybrid bonding orbitals are arranged around the atom in such a way so that the bonds are as far from each other as possible. The bond pair bond pair repulsion is lower than the bond pair lone pair repulsion which is lesser than lone pair- lone pair.

Complete step by step solution:
The concept of mixing two or more atomic orbitals of similar energy levels to a new type of degenerate orbitals all having similar energies and shapes is known as hybridization. This intermixing of the atomic orbitals is based on the principles of quantum mechanics. Only the atomic orbitals having similar energy levels can be hybridized and both fully- filled and half- filled orbitals can also be hybridized, provided that they have similar energies.
 $ {\text{s}}{{\text{p}}^{\text{3}}}{\text{d}} $ hybridization is formed by the mixing of 1 s orbital, three p orbitals and one d- orbital to form 5 hybrid orbitals having similar energies and shape. The molecule assumes a trigonal bi pyramidal geometry. Three of the five orbitals are arranged in the same plane and are known as the equatorial orbitals. Two other bonds are present perpendicular to the equatorial plane and are known as the axial bonds.
The correct answer is C.

Note:
One might often make a mistake and consider pi bonds in the hybrid orbitals. Only sigma bonds participate in hybridization. While calculating hybridization, we should keep that in mind.