Explain me the hybridisation of complex [Ni(NH₃)₆]²⁺
need the explaination with regard to Valence bond theory as soon as possible!!!

 We start, as always, by writing the configuration of the transition-metal ion.

Ni²⁺: [Ar] 3d⁸
This configuration creates a problem, because there are eight electrons in the 3d orbitals. Even if we invest the energy necessary to pair the 3d electrons, we can't find two empty 3d orbitals to use to form a set of d²sp³ hybrids.

SO

The five 4d orbitals on nickel are empty, so we can form a set of empty sp³d² hybrid orbitals by mixing the 4d_x²_-y², 4d_z², 4s, 4p_x, 4p_y and 4p_z orbitals. These hybrid orbitals then accept pairs of nonbonding electrons from six ammonia molecules to form a complex ion. 



The valence-bond theory therefore formally distinguishes between "inner-shell" complexes, which use 3d, 4s and 4p orbitals to form a set of d²sp³ hybrids, and "outer-shell" complexes, which use 4s, 4p and 4d orbitals to form sp³d² hybrid orbitals.