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(C) To determine the number of lone pairs $(l.p.)$ on the central atom $Xe$,we use the formula: $l.p. = \frac{1}{2} (V - M - C + A)$,where $V$ is the

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$XeF_3^-$

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(C) To determine the number of lone pairs $(l.p.)$ on the central atom $Xe$,we use the formula: $l.p. = \frac{1}{2} (V - M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. For $XeOF_4$: $Xe$ has $8$ valence electrons. $O$ is divalent (contributes $0$ to $M$),$F$ is monovalent ($4$ atoms). $l.p. = \frac{1}{2} (8 - 4) = 2$ electrons,which is $1$ lone pair.$2$. For $XeO_2F_2$: $Xe$ has $8$ valence electrons. $O$ is divalent ($0$ to $M$),$F$ is monovalent ($2$ atoms). $l.p. = \frac{1}{2} (8 - 2) = 3$ lone pairs (Note: In the provided structure,it shows $1$ lone pair due to double bonds with $O$; calculating based on $VSEPR$ theory for $XeO_2F_2$,$Xe$ has $8$ valence $e^-$,$2$ $O$ atoms use $4$ $e^-$,$2$ $F$ atoms use $2$ $e^-$,remaining $2$ $e^-$ form $1$ lone pair).$3$. For $XeF_3^-$: $Xe$ has $8$ valence electrons,$3$ $F$ atoms,and $1$ negative charge. $l.p. = \frac{1}{2} (8 - 3 + 1) = 3$ lone pairs.$4$. For $XeO_3$: $Xe$ has $8$ valence electrons,$3$ $O$ atoms (divalent). $l.p. = \frac{1}{2} (8 - 0) = 4$ electrons,which is $1$ lone pair.Comparing the lone pairs: $XeF_3^-$ has $3$ lone pairs,which is the maximum.

The molecule exhibiting the maximum number of non-bonding electron pairs $(l.p.)$ around the central atom is

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