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(D) To determine the hybridization,we calculate the steric number $(SN)$ for the central atom: $SN = 	ext{Number of sigma bonds} + 	ext{Number of lo

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$PF_5$

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(D) To determine the hybridization,we calculate the steric number $(SN)$ for the central atom: $SN = 	ext{Number of sigma bonds} + 	ext{Number of lone pairs}$.$(a)$ In $BrF_5$,$Br$ has $7$ valence electrons. It forms $5$ bonds with $F$ atoms and has $1$ lone pair. $SN = 5 + 1 = 6$,which corresponds to $sp^3d^2$ hybridization.$(b)$ In $SF_6$,$S$ has $6$ valence electrons. It forms $6$ bonds with $F$ atoms and has $0$ lone pairs. $SN = 6 + 0 = 6$,which corresponds to $sp^3d^2$ hybridization.$(c)$ In $[CrF_6]^{3-}$,$Cr^{3+}$ has $d^3$ configuration. It forms $6$ coordinate bonds with $F^-$ ligands. The hybridization is $d^2sp^3$ (inner orbital complex),which is equivalent to $sp^3d^2$ in terms of geometry (octahedral).$(d)$ In $PF_5$,$P$ has $5$ valence electrons. It forms $5$ bonds with $F$ atoms and has $0$ lone pairs. $SN = 5 + 0 = 5$,which corresponds to $sp^3d$ hybridization.Therefore,$PF_5$ does not display $sp^3d^2$ hybridization.

$sp^3d^2$ hybridization is not displayed by

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