Question

The sp$^{3}$d$^{2}$ hybridisation of a central atom of a molecule would lead to:
A.Square planar geometry
B.Tetrahedral geometry
C.Trigonal bipyramidal geometry
D.Octahedral geometry

Answer 1

Hint: As we know, hybridization takes place in the atomic orbitals having equal energies. The sp$^{3}$d$^{2}$ hybridization is found in a number of molecules, like sulphur hexafluoride, from the name we can say there is involvement of 6 electron groups. So, the shape of molecules could be known.

Complete step by step answer:
-First, let us know about the sp$^{3}$d$^{2}$ hybridization. It is useful as the shape of a molecule can be predicted whether it has tetrahedral, square planar, trigonal bipyramidal, or the linear shape.
-Now, if we consider the sp$^{3}$d$^{2}$ hybridization, in this hybridisation, it has 1s, 3p, and 2d orbitals, which undergo intermixing. These 6 orbitals are inclined at an angle of 90to one another, and also directed towards the corners of an octahedron.
-If we look at the given options, first is square planar geometry, it is found in sp$^{3}$ hybridization, second is tetrahedral geometry it is too found in sp$^{3}$ hybridisation, and the third one is trigonal bipyramidal it is found in sp$^{3}$d hybridization.
-Thus, we can conclude that the sp$^{3}$d$^{2}$ hybridisation of a central atom molecule would lead to octahedral geometry. The correct option is D.

Note: Don’t get confused between the different types of hybridization. The d$^{2}$ sp$^{3}$ hybridisation of a central atom of a molecule has octahedral geometry in some cases. There is no need of confusion as this depends upon the ligands attached to the central atom.