Explain `[Co(NH_(3))_(6)]^(3+)` is an inner orbital complex whereas `[Ni(NH_(3))_(6)]^(2+)` is an outer oubital complex.

Explanation

In `[Co(NH_(3))_(6)]^(3+)`, oxidation state of `Co` is `+3` with electronic configuration `3d^(6)4s^(0)`. In presence of `NH_(3)` (strong ligand), 3d - electrons pair up, leaving two d - orbitals empty. So `d^(2)sp^(3)` hybridisation occurs involving appropriate number of 3d, 4 s and 4p - orbitals. `Co^(3+)` then combines with six `NH_(3)` molecules to form the inner orbital complex `[Co(NH_(3))_(6)]^(3+)`

In `[Ni(NH_(3))_(6)]^(2+)`, oxidation state of Ni is `+2`, with electronic configuration `3d^(8)4s^(0)`. In presence of `NH_(3)`, 3d - electrons do not pair up as it is impossible to have two d - orbitals empty. Thus `sp^(3)d^(2)` hybridisation occurs involving appropriate number of `4s, 4p` and 4d- orbitals. Then `Ni^(2+)` combines with six `NH_(3)` molecules to form the outer orbital complex `[Ni(NH_(3))_(6)]^(2+)` having an octahedral structure.