Explain on the basis of V.B, theory that `[Ni(CN)_4]^(2-)` ion with square planar structure is diamagnetic and the `[NiCl_4]^2-` ion with tetrahedral geometry is paramagnetic.

Explanation

In .[Ni(CN)]]^3-, Ni. is in .+2. oxidaion state and has .d^8. configuration.
Figure
For square planar shape, the hybridization is .dsp^2 .. Hence, the unpaired electrons in .3 ~d. orbital pair up to make one d-orbital.empty for dsp .^2. hybridisation:
These .dsp^2. hybrid orbitals. accommódate four pàir. of electrons from .4 CN^-. groups and the resulting squarẻ planar complex is diamagnetic since no unpaired electron is present. For tetrahedral shape of [NiCl .._4]^2-., The hybridisation is .5 p^3 .. Hence the unpaired elections in .3 ~d. orbital. do not pair. up. The available orbitals for sp. hybridisation are one .4 s. and three .4 p .. The complex is paramagnetic since it contains two unpaired clectrons.
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