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(A) The central atom in all these compounds is Xenon $(Xe)$, which has $8$ valence electrons.$1$. In $XeO_3$: $Xe$ forms $3$ double bonds with $3$ oxy

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$1, 1, 1$

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(A) The central atom in all these compounds is Xenon $(Xe)$, which has $8$ valence electrons.$1$. In $XeO_3$: $Xe$ forms $3$ double bonds with $3$ oxygen atoms. It uses $6$ electrons for bonding. Remaining electrons = $8 - 6 = 2$. Thus, $1$ lone pair.$2$. In $XeOF_4$: $Xe$ forms $4$ single bonds with $4$ fluorine atoms and $1$ double bond with $1$ oxygen atom. It uses $4 + 2 = 6$ electrons for bonding. Remaining electrons = $8 - 6 = 2$. Thus, $1$ lone pair.$3$. In $XeF_6$: $Xe$ forms $6$ single bonds with $6$ fluorine atoms. It uses $6$ electrons for bonding. Remaining electrons = $8 - 6 = 2$. Thus, $1$ lone pair.Therefore, the number of lone pairs is $1, 1, 1$.

The number of lone pairs of electrons on the central atom of $XeO_3$, $XeOF_4$ and $XeF_6$ respectively is

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