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

OR

Amongst [Fe(C₂O₄)₃]^3- and [Fe(NH₃)₆]³⁺ which is more stable and why?

Pt is in +2 oxidation state. So, electronic configuration of Pt²⁺ will be d⁸. (Since, it is a part of Ni group)

Orbitals of Pt²⁺

As CN is a strong field ligand, it will pair up the 2 unpaired electrons and thus no unpaired electrons will be left.

Therefore, no. of unpaired electrons in [Pt(CN)₄]^2- will be 0.

OR

In both the complexes Fe is in +3 oxidation state. But in one complex bidentate ligand is present i.e. C₂O₄^2-and in another neutral unidentate ligand is present i.e. NH_3.

Due to the property of chelation shown by bidentate ligand (C₂O₄^2-) in first complex, it becomes more stable.