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?

Similar Questions

Generally,$FeCl_3 \cdot 6H_2O$ is represented as:

The geometry and magnetic behaviour of the complex $[Ni(CO)_4]$ are

The geometries of $[Ni(CO)_4]$,$[PtCl_4]^{2-}$ and $[Co(NH_3)_6]^{3+}$ respectively are

Which of the following statements are correct?
$i$. $[Mn(CN)_6]^{3-}, [MnCl_6]^{3-}$ both are paramagnetic
$ii$. Both $[NiCl_4]^{2-}, [Ni(CO)_4]$ are tetrahedral and diamagnetic
$iii$. $[Co(NH_3)_6]^{3+}$ is an inner orbital complex,but $[Ni(NH_3)_6]^{2+}$ is an outer orbital complex

Which of the following is true for the $[NiCl_4]^{2-}$ complex?

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