Question

For the complexes showing the square pyramidal structure, the d orbitals involved in the hybridisation is:
A. $$d_{x^2-y^2}$$
B. $$d_{x^2}$$
C. $$d_{xy}$$
D. $$d_{xz}$$

Answer 1

Hint:So to approach this question we should know the coordination number involved in this geometry. The coordination number is the total number of the atoms or the ions which have been bonded to the central atom in the compound. So the compound which has its geometry as square pyramidal structure and its coordination number is equal to 5. The number of ligands attached to the central metal atom is equal to 5.

Complete step-by-step answer: In the square pyramidal geometry is a molecular geometry in which the atoms are present and have been bonded with other atoms at the corners of the square present on the same plane about the central atom. Here the central atom tends to bond with 5 other atoms. So this geometry that is the square pyramidal will have the coordination number equal to 5. The bond angles involved in this structure are usually around the $${90}^o$$. So this means the orbitals which are involved must be oriented at the right angles and so thus lying at the same plane. The hybridisation involved for this structure is $$ds{p}^3$$. The $$p_x$$and the $$p_y$$orbitals which do lie on the x and y axis and the $$d_{x^2-y^2}$$orbitals also lies on the x and y axis. So these orbitals tend to lie in the same plane so the d orbital which is involved here is $$d_{x^2-y^2}$$.

So the correct answer is option A.

Note: If the $$d_{z^2}$$of the d orbital would have been involved in the hybridisation then the geometry would be trigonal bipyramidal. It has its atom bonded with the central atom in the axial and the equatorial positions. The equatorial bonded atoms have the bond angle equal to $${120}^o$$and the axial position has the angle equal to $${90}^o$$.