Predict the number of unpaired electrons in the square planar `[Pt(CN)_(4)]^(2-)` ion.

Explanation

Pt in `[Pt(CN)_4]^(2-)` is in +2 state. Configration of Pt is `5d^9 6s^(1)`. Hence, the congiration of `Pt^(2+)" is " d^8` :

In case of square planar complexes, hybridization involved is `dsp^2`. Therefore , to make available one empty orbital for hybridization, two unpaired electrons in 3d orbitals get paired up.
Hence, there is no unpared electron in `[Pt(CN)_4]^(2-)`.