Using VBT, explainthe geometry and magnetic property of `[Ni(CN)_(4)]^(-2)`. (Atomic Number of Ni=28).

Explanation

The atomic no. of Ni is 28 and its oxidation state in this complex is +2
Electronic configuration of `N i = [Ar]3d^8 4s^2`

In this complex electrons are forced to pair up in against to the Hund.s rule in the presence of strong ligand `CN^(-)`
One 3d, one 4s & two 4p orbitals undergoes `dsp^2` hybridization to give four equivalent vacant hybrid orbitals

These four hybrid orbitals are occupied by four lone pairs of electrons from four `CN^(-)` ligands to form `[Ni(CN)_4]^(2-)`
Formation of `[Ni(CN)_4]^(2-)`

Since the hybridization is `dsp^2`, the eometry is Square planar.
The complex is diamagnetic due to the absence of unpaired electrons  .