The pair of xenon compounds which have the sa | vedclass

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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} (V - M - C + A)$,where $V$ is

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$XeO_3, XeF_6$

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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} (V - M - C + A)$,where $V$ is the number of valence electrons of $Xe$ $(8)$,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. For $XeO_3$: $Xe$ is bonded to $3$ oxygen atoms (divalent). $M=0$. $	ext{Lone pairs} = \frac{1}{2} (8 - 0) = 4$ electrons = $1$ lone pair.$2$. For $XeF_6$: $Xe$ is bonded to $6$ fluorine atoms (monovalent). $M=6$. $	ext{Lone pairs} = \frac{1}{2} (8 - 6) = 2$ electrons = $1$ lone pair.$3$. For $XeF_2$: $M=2$. $	ext{Lone pairs} = \frac{1}{2} (8 - 2) = 6$ electrons = $3$ lone pairs.$4$. For $XeF_4$: $M=4$. $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 4$ electrons = $2$ lone pairs.$5$. For $XeOF_4$: $Xe$ is bonded to $1$ oxygen (divalent) and $4$ fluorines (monovalent). $M=4$. $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 4$ electrons = $2$ lone pairs.Comparing the options:$XeO_3$ has $1$ lone pair.$XeF_6$ has $1$ lone pair.Thus,the pair $(XeO_3, XeF_6)$ has the same number of lone pairs.

The pair of xenon compounds which have the same number of lone pairs of electrons on the central atom is

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