The number of species from the following carr | vedclass

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

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$4$

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(A) To determine the number of lone pairs on the central atom $Xe$,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - M - C + A)$,where $V$ is the number of valence electrons of $Xe$ $(8)$,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. $XeF_5^{+}$: $\frac{1}{2} (8 - 5 - 1 + 0) = 1$ lone pair.$2$. $XeO_3$: $\frac{1}{2} (8 - 0 - 0 + 0) = 4$ electron pairs,$3$ are bonding,so $1$ lone pair.$3$. $XeO_2F_2$: $\frac{1}{2} (8 - 2 - 0 + 0) = 3$ bonding pairs,$1$ lone pair.$4$. $XeF_5^{-}$: $\frac{1}{2} (8 - 5 - 0 + 1) = 2$ lone pairs.$5$. $XeO_3F_2$: $\frac{1}{2} (8 - 2 - 0 + 0) = 5$ bonding pairs,$0$ lone pairs.$6$. $XeOF_4$: $\frac{1}{2} (8 - 4 - 0 + 0) = 5$ bonding pairs,$1$ lone pair.$7$. $XeF_4$: $\frac{1}{2} (8 - 4 - 0 + 0) = 4$ bonding pairs,$2$ lone pairs.The species with a single lone pair are $XeF_5^{+}$,$XeO_3$,$XeO_2F_2$,and $XeOF_4$. Thus,the total count is $4$.

The number of species from the following carrying a single lone pair on the central atom Xenon is $.....$: $XeF_5^{+}, XeO_3, XeO_2F_2, XeF_5^{-}, XeO_3F_2, XeOF_4, XeF_4$

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