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(A) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} 	imes (V - N)$,where $V$ is the nu

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$XeF_2$

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(A) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} 	imes (V - N)$,where $V$ is the number of valence electrons and $N$ is the number of bonding electrons.$1$. For $XeF_2$: $Xe$ has $8$ valence electrons. It forms $2$ bonds. Lone pairs = $\frac{1}{2} 	imes (8 - 2) = 3$.$2$. For $SF_4$: $S$ has $6$ valence electrons. It forms $4$ bonds. Lone pairs = $\frac{1}{2} 	imes (6 - 4) = 1$.$3$. For $ClF_3$: $Cl$ has $7$ valence electrons. It forms $3$ bonds. Lone pairs = $\frac{1}{2} 	imes (7 - 3) = 2$.$4$. For $H_2O$: $O$ has $6$ valence electrons. It forms $2$ bonds. Lone pairs = $\frac{1}{2} 	imes (6 - 2) = 2$.Comparing these,$XeF_2$ has the maximum number of lone pairs $(3)$.

Which of the following molecules has the maximum number of lone pairs?

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