Which one of the following has a square planar geometry?

The geometry of a four-coordinate complex depends on three chief factors:
  • electronic configuration of the central metal ion
  • position of the metal in the transition series (3d, 4d, 5d)
  • strength of the surrounding ligands (spectro-chemical series)

For a $$d^8$$ configuration there are two possible geometries:
  • tetrahedral ($$sp^3$$ hybridisation)
  • square planar ($$dsp^2$$ hybridisation)

Among $$d^8$$ ions, elements of the 4d and 5d series (Pd(II), Pt(II), etc.) favour square-planar geometry because their crystal-field splitting energy (Δ0) is large; the extra stabilisation gained in a square-planar field outweighs the pairing energy. For 3d metals such as Ni(II) the same is true only when strong-field ligands (e.g. $$CN^-$$, CO) are present; with weak-field ligands like $$Cl^-$$, a tetrahedral arrangement is preferred.

Now examine each option.

Option A: $$[CoCl_4]^{2-}$$ contains Co(II) (3$$d^7$$). $$Cl^-$$ is a weak-field ligand, so the ion adopts $$sp^3$$ tetrahedral geometry.

Option B: $$[FeCl_4]^{2-}$$ contains Fe(II) (3$$d^6$$) with weak-field $$Cl^-$$; the complex is again tetrahedral.

Option C: $$[NiCl_4]^{2-}$$ contains Ni(II) (3$$d^8$$) but, as discussed, Ni(II) with weak-field $$Cl^-$$ forms a tetrahedral complex ($$sp^3$$).

Option D: $$[PtCl_4]^{2-}$$ contains Pt(II) (5$$d^8$$). Because Pt is a 5d metal, the splitting energy is large enough that dsp2 hybridisation is favoured even with the weak-field ligand $$Cl^-$$. Hence the complex is square planar.

Therefore, the only complex among the four that exhibits square-planar geometry is

Option D which is: $$[PtCl_4]^{2-}$$