Among the following molecules:(i) XeO_3(ii) X | vedclass

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

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$(i)$,$(ii)$ and $(iii)$

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(D) To determine the number of lone pairs on the central $Xe$ atom in each molecule,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.$1.$ For $XeO_3$:$Xe$ has $8$ valence electrons. Oxygen is divalent,so $M = 0$.$	ext{Lone pairs} = \frac{1}{2} 	imes (8 - 0) = 4$ electrons,which corresponds to $1$ lone pair.$2.$ For $XeOF_4$:$Xe$ has $8$ valence electrons. $F$ is monovalent $(M = 4)$,$O$ is divalent $(M = 0)$.$	ext{Lone pairs} = \frac{1}{2} 	imes (8 - 4) = 2$ electrons,which corresponds to $1$ lone pair.$3.$ For $XeF_6$:$Xe$ has $8$ valence electrons. $F$ is monovalent $(M = 6)$.$	ext{Lone pairs} = \frac{1}{2} 	imes (8 - 6) = 2$ electrons,which corresponds to $1$ lone pair.Since all three molecules ($XeO_3$,$XeOF_4$,and $XeF_6$) have $1$ lone pair on the $Xe$ atom,the correct option is $(D)$.

Among the following molecules:
$(i)$ $XeO_3$
$(ii)$ $XeOF_4$
$(iii)$ $XeF_6$
Those having the same number of lone pairs on $Xe$ are:

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Among the following molecules:$(i)$ $XeO_3$$(ii)$ $XeOF_4$$(iii)$ $XeF_6$Those having the same number of lone pairs on $Xe$ are: