Which of the following species have a square | vedclass

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(A) $1$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The hybridization is $sp^

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$(i)$ and $(iv)$

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(A) $1$. $XeF_4$: The central atom $Xe$ has $8$ valence electrons. It forms $4$ bonds with $F$ atoms and has $2$ lone pairs. The hybridization is $sp^3d^2$,resulting in a square planar geometry.$2$. $SF_4$: The central atom $S$ has $6$ valence electrons. It forms $4$ bonds with $F$ atoms and has $1$ lone pair. The hybridization is $sp^3d$,resulting in a see-saw geometry.$3$. $[NiCl_4]^{2-}$: $Ni$ is in the $+2$ oxidation state ($d^8$ configuration). $Cl^-$ is a weak field ligand,so it forms a tetrahedral complex ($sp^3$ hybridization).$4$. $[PtCl_4]^{2-}$: $Pt$ is in the $+2$ oxidation state ($5d^8$ configuration). $Cl^-$ is a ligand,but $Pt^{2+}$ is a $5d$ series metal,which always forms square planar complexes ($dsp^2$ hybridization) due to high crystal field splitting.Therefore,$(i)$ and $(iv)$ have square planar geometry.

Which of the following species have a square planar geometry for the central atom?
$(i) XeF_4 \ (ii) SF_4 \ (iii) [NiCl_4]^{2-} \ (iv) [PtCl_4]^{2-}$

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Which of the following species have a square planar geometry for the central atom?$(i) XeF_4 \ (ii) SF_4 \ (iii) [NiCl_4]^{2-} \ (iv) [PtCl_4]^{2-}$