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

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

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(D) To determine the number of lone pairs on $Xe$,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (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.$(i) XeO_3$: $Xe$ has $8$ valence electrons. Oxygen is divalent,so $M=0$. $	ext{Lone pairs} = \frac{1}{2} (8 - 0) = 4$ electrons,which corresponds to $1$ lone pair.$(ii) XeOF_4$: $Xe$ has $8$ valence electrons. $F$ is monovalent $(M=4)$,$O$ is divalent. $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 2$ electrons,which corresponds to $1$ lone pair.$(iii) XeF_6$: $Xe$ has $8$ valence electrons. $F$ is monovalent $(M=6)$. $	ext{Lone pairs} = \frac{1}{2} (8 - 6) = 2$ electrons,which corresponds to $1$ lone pair.All three molecules $(i), (ii),$ and $(iii)$ have $1$ lone pair on the $Xe$ atom. Therefore,the correct option is $(D)$.

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

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