Which of the following is not a square planar | vedclass

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(D) To determine the geometry,we look at the hybridization of the central metal ion:$1$. $[Cu(CN)_4]^{-2}$: $Cu^{+2}$ is $d^9$. It undergoes $dsp^2$ h

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$[Ni(Cl)_4]^{-2}$

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(D) To determine the geometry,we look at the hybridization of the central metal ion:$1$. $[Cu(CN)_4]^{-2}$: $Cu^{+2}$ is $d^9$. It undergoes $dsp^2$ hybridization,resulting in a square planar geometry.$2$. $[PtCl_4]^{-2}$: $Pt^{+2}$ is a $5d$ series element,which always forms square planar complexes due to high crystal field splitting energy.$3$. $[Cu(H_2O)_4]^{+2}$: This complex has a distorted octahedral geometry (Jahn-Teller distortion) in the solid state,often described as elongated octahedral,not square planar.$4$. $[Ni(Cl)_4]^{-2}$: $Ni^{+2}$ is $d^8$. $Cl^-$ is a weak field ligand,so it undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.Both $[Cu(H_2O)_4]^{+2}$ and $[Ni(Cl)_4]^{-2}$ are not square planar. However,in standard textbook contexts for this specific question,$[Ni(Cl)_4]^{-2}$ is the classic example of a tetrahedral complex. Given the options,$[Ni(Cl)_4]^{-2}$ is the most definitive tetrahedral complex.

Which of the following is not a square planar complex?

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