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$

$A. NO_3^-$	$1. sp^3$
$B. NH_2^-$	$2. sp$
$C. SCN^-$	$3. sp^3d$
$D. ICl_2^-$	$4. sp^2$

$Column\, I$	$Column\, II$
$(A)\, XeF_6$	$(i)\, \text{Distorted octahedral}$
$(B)\, XeO_3$	$(ii)\, \text{Square planar}$
$(C)\, XeOF_4$	$(iii)\, \text{Pyramidal}$
$(D)\, XeF_4$	$(iv)\, \text{Square pyramidal}$

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

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Match the following ions with the corresponding hybridisation of their central atoms.
$A. NO_3^-$ $1. sp^3$
$B. NH_2^-$ $2. sp$
$C. SCN^-$ $3. sp^3d$
$D. ICl_2^-$ $4. sp^2$

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