Which of the following has a square planar sh | vedclass

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(D) $1$. For $[NiCl_4]^{2-}$,$Ni^{2+}$ has $d^8$ configuration. $Cl^-$ is a weak field ligand,so it forms a tetrahedral complex due to $sp^3$ hybridiz

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

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(D) $1$. For $[NiCl_4]^{2-}$,$Ni^{2+}$ has $d^8$ configuration. $Cl^-$ is a weak field ligand,so it forms a tetrahedral complex due to $sp^3$ hybridization.$2$. For $SF_4$,the central atom $S$ has $5$ electron pairs ($4$ bond pairs and $1$ lone pair),resulting in a see-saw shape.$3$. For $XeF_4$,$Xe$ has $8$ valence electrons. It forms $4$ bond pairs with $F$ and has $2$ lone pairs. According to $VSEPR$ theory,$4$ bond pairs and $2$ lone pairs result in $sp^3d^2$ hybridization and a square planar geometry.$4$. For $[Ni(CN)_4]^{2-}$,$Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons,leading to $dsp^2$ hybridization and a square planar geometry.Both $C$ and $D$ have square planar shapes. However,in many competitive contexts,if a single choice is required,$[Ni(CN)_4]^{2-}$ is the classic example of $dsp^2$ square planar geometry.

List-$I$ (Tetrahedral Complex)	List-$II$ (Electronic configuration)
$A$. $TiCl_4$	$I$. $e^2, t_2^0$
$B$. $[FeO_4]^{2-}$	$II$. $e^4, t_2^3$
$C$. $[FeCl_4]^{-}$	$III$. $e^0, t_2^0$
$D$. $[CoCl_4]^{2-}$	$IV$. $e^2, t_2^3$

Which of the following has a square planar shape?

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