Of the following species,the one having a squ | vedclass

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(C) To determine the geometry,we calculate the hybridization of the central atom:$1$. $NH_4^+$: $N$ has $5$ valence electrons. Steric number = $0.5

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$XeF_4$

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(C) To determine the geometry,we calculate the hybridization of the central atom:$1$. $NH_4^+$: $N$ has $5$ valence electrons. Steric number = $0.5 	imes (5 + 4 - 1) = 4$ ($sp^3$ hybridization,tetrahedral).$2$. $BF_4^-$: $B$ has $3$ valence electrons. Steric number = $0.5 	imes (3 + 4 + 1) = 4$ ($sp^3$ hybridization,tetrahedral).$3$. $XeF_4$: $Xe$ has $8$ valence electrons. Steric number = $0.5 	imes (8 + 4) = 6$ ($sp^3d^2$ hybridization). With $4$ bond pairs and $2$ lone pairs,the geometry is square planar.$4$. $SCl_4$: $S$ has $6$ valence electrons. Steric number = $0.5 	imes (6 + 4) = 5$ ($sp^3d$ hybridization). With $4$ bond pairs and $1$ lone pair,the geometry is see-saw.Thus,$XeF_4$ has a square planar structure.

Of the following species,the one having a square planar structure is:

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