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(A) To determine the structure,we calculate the hybridization and number of lone pairs for each species:$1$. $BF_4^-$: Boron has $3$ valence electrons

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$BF_4^-$

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(A) To determine the structure,we calculate the hybridization and number of lone pairs for each species:$1$. $BF_4^-$: Boron has $3$ valence electrons. It forms $4$ bonds with $F$ atoms and gains $1$ electron from the negative charge. Total electron pairs = $(3+4+1)/2 = 4$. Hybridization is $sp^3$,which corresponds to a regular tetrahedral geometry.$2$. $SF_4$: Sulfur has $6$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $1$ lone pair. Total electron pairs = $5$ ($sp^3d$ hybridization). Due to the lone pair,it has a see-saw shape.$3$. $XeF_4$: Xenon has $8$ valence electrons. It forms $4$ bonds with $F$ atoms,leaving $2$ lone pairs. Total electron pairs = $6$ ($sp^3d^2$ hybridization). Due to the two lone pairs,it has a square planar geometry.$4$. $[Ni(CN)_4]^{2-}$: $Ni^{2+}$ has a $d^8$ configuration. $CN^-$ is a strong field ligand,causing pairing of electrons. Hybridization is $dsp^2$,resulting in a square planar geometry.Thus,only $BF_4^-$ has a regular tetrahedral structure.

Which one of the following has the regular tetrahedral structure?
(Atomic no. : $B = 5, S = 16, Ni = 28, Xe = 54$)

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Which one of the following has the regular tetrahedral structure?(Atomic no. : $B = 5, S = 16, Ni = 28, Xe = 54$)