Which of the following molecules has sp^3d^2 | vedclass

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(C) To determine the hybridisation and shape,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2}(V + M - C + A)$,where $V$ is t

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$ICl_4^-$

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(C) To determine the hybridisation and shape,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge. For $ICl_4^-$: $V = 7$ (for $I$),$M = 4$ (for $Cl$),$A = 1$. $SN = \frac{1}{2}(7 + 4 + 1) = 6$. $A$ steric number of $6$ corresponds to $sp^3d^2$ hybridisation. With $4$ bond pairs and $2$ lone pairs,the geometry is octahedral and the shape is square planar. Chemical speciesHybridisation$SF_4$$sp^3d$$BF_4^-$$sp^3$$ICl_4^-$$sp^3d^2$$XeO_2F_2$$sp^3d$

$LIST$-$I$ (Hybridization)	$LIST$-$II$ (Orientation in Space)
$A$. $sp^3$	$I$. Trigonal bipyramidal
$B$. $dsp^2$	$II$. Octahedral
$C$. $sp^3 d$	$III$. Tetrahedral
$D$. $sp^3 d^2$	$IV$. Square planar

Which of the following molecules has $sp^3d^2$ hybridisation and a square planar shape?

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Match List-$I$ with List-$II$ :

$LIST$-$I$ (Hybridization) $LIST$-$II$ (Orientation in Space)

$A$. $sp^3$ $I$. Trigonal bipyramidal

$B$. $dsp^2$ $II$. Octahedral

$C$. $sp^3 d$ $III$. Tetrahedral

$D$. $sp^3 d^2$ $IV$. Square planar

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