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(C) Let us analyze each molecule:$1$. $PCl_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs. The hybridisation is $sp^3d$ and the shape

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(C) Let us analyze each molecule:$1$. $PCl_5$: The central atom $P$ has $5$ bond pairs and $0$ lone pairs. The hybridisation is $sp^3d$ and the shape is trigonal bipyramidal.$2$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. The hybridisation is $dsp^2$ and the shape is square planar.$3$. $SF_6$: The central atom $S$ has $6$ bond pairs and $0$ lone pairs. The hybridisation is $sp^3d^2$ and the shape is octahedral. This is correct.$4$. $IF_3$: The central atom $I$ has $3$ bond pairs and $2$ lone pairs. The hybridisation is $sp^3d$ and the shape is bent $T$-shaped.Therefore,the correct set is $(c)$.

Molecule	Hybridisation of central atom	Shape
$(a)$ $PCl_5$	$dsp^3$	square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$	$sp^3$	tetrahedral
$(c)$ $SF_6$	$sp^3d^2$	octahedral
$(d)$ $IF_3$	$dsp^3$	pyramidal

Identify the correct set from the following table:
Molecule Hybridisation of central atom Shape
$(a)$ $PCl_5$ $dsp^3$ square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$ $sp^3$ tetrahedral
$(c)$ $SF_6$ $sp^3d^2$ octahedral
$(d)$ $IF_3$ $dsp^3$ pyramidal

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Identify the correct set from the following table:MoleculeHybridisation of central atomShape$(a)$ $PCl_5$$dsp^3$square pyramidal$(b)$ $[Ni(CN)_4]^{2-}$$sp^3$tetrahedral$(c)$ $SF_6$$sp^3d^2$octahedral$(d)$ $IF_3$$dsp^3$pyramidal