1. Determine the metal oxidation state and $d$-electron count

For all complexes, each $Cl^-$ ligand has charge $-1$.

• A: $[CoCl_4]^{2-}$ $x+4(-1)=-2 \Rightarrow x=+2$ So, $Co^{2+}$ has configuration: $Co: [Ar]3d^74s^2 \Rightarrow Co^{2+}:[Ar]3d^7$

• B: $[FeCl_4]^{2-}$ $x+4(-1)=-2 \Rightarrow x=+2$ So, $Fe^{2+}$ is: $Fe:[Ar]3d^64s^2 \Rightarrow Fe^{2+}:[Ar]3d^6$

• C: $[NiCl_4]^{2-}$ $x+4(-1)=-2 \Rightarrow x=+2$ So, $Ni^{2+}$ is: $Ni:[Ar]3d^84s^2 \Rightarrow Ni^{2+}:[Ar]3d^8$

• D: $[PtCl_4]^{2-}$ $x+4(-1)=-2 \Rightarrow x=+2$ So, $Pt^{2+}$ is a $5d^8$ ion.

2. Use ligand strength and metal series to predict geometry

All complexes contain $Cl^-$, which is a weak field ligand.

For 4-coordinate complexes:

• $d^6$, $d^7$ with weak ligands generally prefer tetrahedral geometry.
• $3d^8$ ions with weak ligands such as $Ni^{2+}$ usually form tetrahedral complexes, e.g. $[NiCl_4]^{2-}$.
• $5d^8$ ions such as $Pt^{2+}$ strongly prefer square planar geometry because of larger crystal field splitting in $5d$ series.

3. Evaluate each option

• A: $[CoCl_4]^{2-}$ $Co^{2+}$ is $3d^7$ with weak field $Cl^-$ ligands $Rightarrow$ tetrahedral, not square planar.

• B: $[FeCl_4]^{2-}$ $Fe^{2+}$ is $3d^6$ with weak field $Cl^-$ ligands $Rightarrow$ tetrahedral, not square planar.

• C: $[NiCl_4]^{2-}$ $Ni^{2+}$ is $3d^8$, but with weak field $Cl^-$ and being a $3d$ metal, this complex is tetrahedral.

• D: $[PtCl_4]^{2-}$ $Pt^{2+}$ is $5d^8$ and characteristically forms square planar complexes.

4. Final answer

The complex with square planar geometry is: $\boxed{[PtCl_4]^{2-}}$ So, the correct option is D.