Question

Which of the following species represent the example of $ds{{p}^{2}}$ hybridization?
A. ${{[Fe{{(CN)}_{6}}]}^{3-}}$
B. ${{[Ni{{(CN)}_{4}}]}^{2-}}$
C. ${{[Ag{{(CN)}_{2}}]}^{-}}$
D. ${{[Co{{(CN)}_{6}}]}^{3-}}$

Answer 1

When two atomic orbitals combine to form hybrid orbital in a molecule then redistribution of the energy of orbitals of individual atoms produces orbitals of equivalent energy and the new orbital formed are known as hybrid orbitals and the phenomenon is known as hybridization.

Complete answer:
According to valence bond theory the metal atom or ion in presence of ligands can use its outer orbitals for hybridization which yield a set of equivalent orbitals of definite geometry like octahedral, tetrahedral, square planar and so on. These hybrid orbitals are allowed to overlap with ligand orbitals which can easily donate electron pairs for bonding.
Out of all these orbitals, ${{[Ni{{(CN)}_{4}}]}^{2-}}$ shows $ds{{p}^{2}}$ hybridization this can be calculated by noting the pairing of electrons in the given shell and we know that cyanide ion is said to be strong field ligand which favors pairing so pairing of electrons takes place in 3d orbital due to which it contains $ds{{p}^{2}}$ hybridization which is known as inner orbital hybridization and due to this hybridization it attains square planar geometry.
Hence from the above discussion we conclude that option B is the correct answer to other forms of outer orbital hybridization.

Note:
The shape and the hybridization of any compound can be concluded with the help of a theory given by the German Physicists called valence bond theory. This theory explains the electronic structure of the molecule formed by the overlapping of atomic orbitals.