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

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$XeF_5^+$

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(C) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - N - C)$,where $V$ is the number of valence electrons of the central atom,$N$ is the number of monovalent atoms attached,and $C$ is the charge on the species.$1$. For $BrF_3$: Central atom $Br$ $(V=7)$,$N=3$,$C=0$. $	ext{Lone pairs} = \frac{1}{2} (7 - 3 - 0) = 2$.$2$. For $BrF_4^-$: Central atom $Br$ $(V=7)$,$N=4$,$C=-1$. $	ext{Lone pairs} = \frac{1}{2} (7 - 4 + 1) = 2$.$3$. For $XeF_5^+$: Central atom $Xe$ $(V=8)$,$N=5$,$C=+1$. $	ext{Lone pairs} = \frac{1}{2} (8 - 5 - 1) = 1$.$4$. For $I_3^-$: Central atom $I$ $(V=7)$,$N=2$,$C=-1$. $	ext{Lone pairs} = \frac{1}{2} (7 - 2 + 1) = 3$.Comparing the values,$XeF_5^+$ has the minimum number of lone pairs $(1)$.

Which of the following molecules/species has the minimum number of lone pairs on its central atom?

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