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(A) To determine the number of lone pairs on the central $Xe$ atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} 	imes (V - M - C + A)$,where

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$2$

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(A) To determine the number of lone pairs on the central $Xe$ atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} 	imes (V - M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. $XeO_{3}$: $Xe$ has $8$ valence electrons. $O$ is divalent,so $M=0$. Lone pairs = $\frac{1}{2} 	imes (8 - 0) = 4$ electrons = $1$ lone pair.$2$. $XeO_{2}F_{2}$: $Xe$ has $8$ valence electrons. $F$ is monovalent,so $M=2$. Lone pairs = $\frac{1}{2} 	imes (8 - 2) = 3$ electrons = $1$ lone pair.$3$. $XeO_{4}$: $Xe$ has $8$ valence electrons. $O$ is divalent,so $M=0$. Lone pairs = $\frac{1}{2} 	imes (8 - 0) = 4$ electrons = $0$ lone pairs.$4$. $XeO_{3}F_{2}$: $Xe$ has $8$ valence electrons. $F$ is monovalent,so $M=2$. Lone pairs = $\frac{1}{2} 	imes (8 - 2) = 3$ electrons = $0$ lone pairs.$5$. $Ba_{2}XeF_{4}$ contains the $[XeF_{6}]^{2-}$ ion (as $Ba^{2+}$ is the cation). For $[XeF_{6}]^{2-}$,$Xe$ has $8$ valence electrons,$F$ is monovalent $(M=6)$,and anionic charge is $2$ $(A=2)$. Lone pairs = $\frac{1}{2} 	imes (8 - 6 + 2) = 2$ electrons = $1$ lone pair.Thus,$XeO_{4}$ and $XeO_{3}F_{2}$ have zero lone pairs on the central $Xe$ atom. The count is $2$.

$A$. Trigonal planar	$I$. $PCl_5$
$B$. Tetrahedral	$II$. $SF_6$
$C$. Trigonal bipyramidal	$III$. $BF_3$
$D$. Octahedral	$IV$. $CCl_4$

Which of the following compounds have the number of lone pairs on the central $Xe$ atom equal to zero?
$XeO_{3}, XeO_{2}F_{2}, XeO_{4}, XeO_{3}F_{2}, Ba_{2}XeF_{4}$

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Which of the following compounds have the number of lone pairs on the central $Xe$ atom equal to zero?$XeO_{3}, XeO_{2}F_{2}, XeO_{4}, XeO_{3}F_{2}, Ba_{2}XeF_{4}$