The sum of the number of lone pairs of electr | vedclass

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(B) To find the number of lone pairs on the central atom $(Xe)$ in each molecule:$1$. In $XeO_3$: The $Xe$ atom has $8$ valence electrons. It forms $3

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$3$

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(B) To find the number of lone pairs on the central atom $(Xe)$ in each molecule:$1$. In $XeO_3$: The $Xe$ atom has $8$ valence electrons. It forms $3$ double bonds with $3$ oxygen atoms (using $6$ electrons). Thus,$8 - 6 = 2$ electrons remain,which form $1$ lone pair.$2$. In $XeOF_4$: The $Xe$ atom has $8$ valence electrons. It forms $1$ double bond with oxygen ($2$ electrons) and $4$ single bonds with fluorine atoms ($4$ electrons). Total electrons used = $6$. Thus,$8 - 6 = 2$ electrons remain,which form $1$ lone pair.$3$. In $XeF_6$: The $Xe$ atom has $8$ valence electrons. It forms $6$ single bonds with fluorine atoms ($6$ electrons). Thus,$8 - 6 = 2$ electrons remain,which form $1$ lone pair.Sum of lone pairs = $1 + 1 + 1 = 3$.

The sum of the number of lone pairs of electrons present on the central atoms of $XeO_3$,$XeOF_4$,and $XeF_6$ is $.......$

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