Account for the following: [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is in outer orbital complex
In [Co(NH3)6]3+[Co(NH3)6]3+ the oxidation state of cobalt is +3+3, so its d-orbital will have six electrons making it a d 66 system. As NH3NH3 is a strong field ligand, the pairing will start in 3 d3 d electrons. The hybridization involves the two sub-shells of 3 d3 d, 4 s4 s and 4p4p empty orbitals. Therefore the hybridization is d2sp3d2sp3. Hence, [Co(NH3)6]3+[Co(NH3)6]3+ is an inner orbital octahedral complex. In[Ni(NH3)6]2+[Ni(NH3)6]2+ the oxidation state of nickel is +2+2 and its outer shell electronic configuration is 3 d84 s03 d84 s0, due to the presence of two unpaired electrons in 3 d3 d sub-shell. The hybridization involves, empty 4 s,4p4 s,4p and 4 d4 d orbitals, thats why it is an outer orbital octahedral complex.
