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)$,where $V$ is the numb

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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)$,where $V$ is the number of valence electrons of the central atom ($8$ for $Xe$) and $M$ is the number of monovalent atoms bonded to it (Oxygen is divalent,so it does not contribute to $M$).$(1) XeO_3$: $V=8$,$M=0$. $	ext{Lone pairs} = \frac{1}{2} (8 - 0) = 4$ electrons,which is $1$ lone pair.$(2) XeF_6$: $V=8$,$M=6$. $	ext{Lone pairs} = \frac{1}{2} (8 - 6) = 2$ electrons,which is $1$ lone pair.$(3) XeF_2$: $V=8$,$M=2$. $	ext{Lone pairs} = \frac{1}{2} (8 - 2) = 6$ electrons,which is $3$ lone pairs.$(4) XeF_4$: $V=8$,$M=4$. $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 4$ electrons,which is $2$ lone pairs.$(5) XeOF_4$: $V=8$,$M=4$ (Oxygen is divalent). $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 4$ electrons,which is $1$ lone pair.Comparing the results,$XeO_3$,$XeF_6$,and $XeOF_4$ all have $1$ lone pair. Among the given options,the pair $(XeO_3, XeF_6)$ matches.

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

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