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

Dear user,Formation of [Ni(NH3)6]2+ :The oxidation state of nickel in [Ni(NH3)6]2+ ion is +2.The electronicconfiguration of Ni is 3d8 4s2.But,since the oxidation state of Ni in thecomplex is +2,its electronic configuration will be 3d8 and the orbital diagram for the Ni2+ can be represented as :                                3d                           4s         4p Ni atom in the :    | |               ground state Ni2+ ion :               3d                           4s         4p                               | |              Now,we know that all the complexes containing configuration of metal d7or d8 form outer orbital complex that is sp3d2 hybridisation.Therefore,drawing orbital diagram,sp3d2 hybridized                   3d                   4s        4p         4dorbitals of Ni2+ ion :    | |             Formation of[Ni(NH3)6]2+ :             3d                    4s        4p              4d                                  | |             Therefore six electron pairs donated by six NH3- ions to formouter octahedral complex that is sp3d2.

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 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.


The five 4d orbitals on nickel are empty, so we can form a set of empty sp3d2 hybrid orbitals by mixing the 4dx2-y2, 4dz2, 4s, 4px, 4py and 4pz 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.
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