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(D) To determine the number of square planar species,we analyze the geometry of each:$1$. $XeF_4$: $sp^3d^2$ hybridization with $2$ lone pairs,resulti

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$4$

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(D) To determine the number of square planar species,we analyze the geometry of each:$1$. $XeF_4$: $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar shape.$2$. $SF_4$: $sp^3d$ hybridization with $1$ lone pair,resulting in a see-saw shape.$3$. $SiF_4$: $sp^3$ hybridization,resulting in a tetrahedral shape.$4$. $BF_4^-$: $sp^3$ hybridization,resulting in a tetrahedral shape.$5$. $BrF_4^-$: $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar shape.$6$. $[Cu(NH_3)_4]^{2+}$: $dsp^2$ hybridization,resulting in a square planar shape.$7$. $[FeCl_4]^{2-}$: $sp^3$ hybridization,resulting in a tetrahedral shape.$8$. $[CoCl_4]^{2-}$: $sp^3$ hybridization,resulting in a tetrahedral shape.$9$. $[PtCl_4]^{2-}$: $dsp^2$ hybridization,resulting in a square planar shape.The species with a square planar shape are $XeF_4, BrF_4^-, [Cu(NH_3)_4]^{2+},$ and $[PtCl_4]^{2-}$.Therefore,the total number of square planar species is $4$.

Column-$I$ Complex	Column-$II$ Structure / Geometry / Property
$A$. $[Ni(CN)_4]^{2-}$	$I$. Tetrahedral / Paramagnetic
$B$. $[Ni(CO)_4]$	$II$. Tetrahedral / Diamagnetic
$C$. $[NiCl_4]^{2-}$	$III$. Square planar / Diamagnetic

$A$ list of species having the formula $XZ_4$ is given below.
$XeF_4, SF_4, SiF_4, BF_4^-, BrF_4^-, [Cu(NH_3)_4]^{2+}, [FeCl_4]^{2-}, [CoCl_4]^{2-}$ and $[PtCl_4]^{2-}$.
Defining shape on the basis of the location of $X$ and $Z$ atoms,the total number of species having a square planar shape is

Similar Questions

$Ni(CO)_4$ is

Match the following:

Column-$I$ Complex Column-$II$ Structure / Geometry / Property

$A$. $[Ni(CN)_4]^{2-}$ $I$. Tetrahedral / Paramagnetic

$B$. $[Ni(CO)_4]$ $II$. Tetrahedral / Diamagnetic

$C$. $[NiCl_4]^{2-}$ $III$. Square planar / Diamagnetic

The correct match is:

Which statement is incorrect regarding the complex $[Cr(NH_3)_6]Cl_3$?

An octahedral complex of $Ni(II)$ must be:

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$A$ list of species having the formula $XZ_4$ is given below.$XeF_4, SF_4, SiF_4, BF_4^-, BrF_4^-, [Cu(NH_3)_4]^{2+}, [FeCl_4]^{2-}, [CoCl_4]^{2-}$ and $[PtCl_4]^{2-}$.Defining shape on the basis of the location of $X$ and $Z$ atoms,the total number of species having a square planar shape is