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(A) To determine the geometry,we analyze the hybridization and lone pairs of the central atom:$1$. $XeF_4$: $Xe$ has $8$ valence electrons. It forms $

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$(i)$ and $(iv)$

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(A) To determine the geometry,we analyze the hybridization and lone pairs of the central atom:$1$. $XeF_4$: $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ and has $2$ lone pairs. Steric number = $4 + 2 = 6$ ($sp^3d^2$ hybridization). Due to $2$ lone pairs,it adopts a square planar geometry.$2$. $SF_4$: $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ and has $1$ lone pair. Steric number = $4 + 1 = 5$ ($sp^3d$ hybridization). It adopts a see-saw geometry.$3$. $[NiCl_4]^{2-}$: $Ni^{2+}$ is a $d^8$ ion. With a weak field ligand $(Cl^-)$,it undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry.$4$. $[PtCl_4]^{2-}$: $Pt^{2+}$ is a $5d$ series metal ion. It forms square planar complexes due to high crystal field splitting,resulting in $dsp^2$ hybridization.Therefore,the species with square planar geometry are $(i)$ $XeF_4$ and $(iv)$ $[PtCl_4]^{2-}$.

Among the following,the species having square planar geometry for the central atom are:
$(i)$ $XeF_4$
$(ii)$ $SF_4$
$(iii)$ $[NiCl_4]^{2-}$
$(iv)$ $[PtCl_4]^{2-}$

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Among the following,the species having square planar geometry for the central atom are:$(i)$ $XeF_4$$(ii)$ $SF_4$$(iii)$ $[NiCl_4]^{2-}$$(iv)$ $[PtCl_4]^{2-}$