Which of the following complexes are square planar?
(A) \( [Ni(CN)_4]^{2-} \)
(B) \( [NiCl_4]^{2-} \)
(C) \( [PtCl_4]^{2-} \)
(D) \( [PdCl_4]^{2-} \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
A four-coordinate d8 metal ion can form a square planar complex (dsp2 hybridisation) or a tetrahedral one (sp3). The geometry depends on the crystal field splitting: when the splitting is large, the eight d electrons pair up in four d orbitals, one 3d (or 4d, 5d) orbital is left empty and dsp2 hybridisation gives a square planar complex. A strong field ligand, or a heavier 4d or 5d metal, produces such a large splitting.
(A) \( \ce{[Ni(CN)4]^2-} \) — Correct. Ni (Z = 28) is \( \mathrm{[Ar]\,3d^{8}\,4s^{2}} \); losing the two 4s electrons gives \( \ce{Ni^2+} \): \( \mathrm{3d^{8}} \). \( \ce{CN-} \) is a strong field ligand, so the electrons pair, one 3d orbital is free, and the complex is dsp2, square planar and diamagnetic.
(B) \( \ce{[NiCl4]^2-} \) — Incorrect. \( \ce{Ni^2+} \) is again \( \mathrm{3d^{8}} \), but \( \ce{Cl-} \) is a weak field ligand; the two unpaired electrons stay unpaired, no 3d orbital is free, and sp3 hybridisation gives a tetrahedral, paramagnetic complex.
(C) \( \ce{[PtCl4]^2-} \) — Correct. 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 a 5d metal the splitting is large even with \( \ce{Cl-} \), so the complex is square planar.
(D) \( \ce{[PdCl4]^2-} \) — Correct. Pd (Z = 46) is \( \mathrm{[Kr]\,4d^{10}} \); losing two 4d electrons gives \( \ce{Pd^2+} \): \( \mathrm{4d^{8}} \). As a 4d metal ion it also forms a square planar complex with \( \ce{Cl-} \).