Explain the geometry of `[Co(NH_3)_6]^(3+)` on the basis of hybridisation. (Z of Co = 27)
Octahedral complex, `[Co(NH_3)_6]^(3+)`. In this complex ion, the oxidation state of cobalt is +3.
It has the electronic configuration as 3d6. This complex involves the d2sp3 hybridisation.
The six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. It proves that complex has octahedral geometry. An absence of unpaired electron makes this complex diamagnetic in nature.