Number of unpaired electrons in the square planar [Pt(CN)_4]^2- are:

The correct answer is Option (2) → Zero unpaired electron

A square planar complex uses dsp² hybrid orbitals. One inner d orbital must be empty for this, so the d electrons of the metal pair up in the remaining d orbitals.

First, the oxidation state of platinum in \( \ce{[Pt(CN)4]^2-} \), with cyanide as −1:

    \( \displaystyle x + 4(-1) = -2 \Rightarrow x = +2 \)

Pt (Z = 78) is \( \mathrm{[Xe]\,4f^{14}\,5d^{9}\,6s^{1}} \). Losing the 6s electron and one 5d electron gives \( \ce{Pt^2+} \): \( \mathrm{5d^{8}} \). For dsp² hybridisation the eight 5d electrons occupy four 5d orbitals as four pairs, leaving one 5d orbital free to combine with the 6s and two 6p orbitals. Cyanide is a strong field ligand and the 5d orbitals split strongly, so pairing is favoured.

So \( \ce{[Pt(CN)4]^2-} \) has zero unpaired electrons and is diamagnetic.