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.

List-$I$ (Complex)	List-$II$ (Hybridization)
$A. Ni(CO)_4$	$I. sp^3d^2$
$B. [Ni(CN)_4]^{2-}$	$II. d^2sp^3$
$C. [Co(NH_3)_6]^{3+}$	$III. dsp^2$
$D. [CoF_6]^{3-}$	$IV. sp^3$

Which of the following is not a square planar complex?

Similar Questions

Define the following term: Coordination polyhedron.

How many electrons are present in the $3d$ orbital of the tetrahedral $K_2[NiCl_4]$ complex?

In $[NiCl_4]^{2-}$,the number of unpaired electrons is

Match the complexes in List-$I$ with their hybridization in List-$II$.
List-$I$ (Complex) List-$II$ (Hybridization)
$A. Ni(CO)_4$ $I. sp^3d^2$
$B. [Ni(CN)_4]^{2-}$ $II. d^2sp^3$
$C. [Co(NH_3)_6]^{3+}$ $III. dsp^2$
$D. [CoF_6]^{3-}$ $IV. sp^3$

The shape of pentacarbonyl iron,$Fe(CO)_5$,is:

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