Predict the number of unpaired electrons in the square planar `[Pt(CN)_4]^(2-)` ion.

Explanation

`._(78)Pt` lies in Group 10 with the configuration `5d^(9)6s^(1)`. Hence, `Pt^(2+)` has the configuration `d^(8)`

For square planar shape, the hybridisation is `dsp^(2)`. Hence, the unpaired electrons in 5d pair up to make one d-orbital empty for `dsp^(2)` hybridization. Thus, there is no unpaired electron.