In the following molecules: (i) XeO_3,(ii) Xe | vedclass

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(D) 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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$(i), (ii)$ and $(iii)$

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(D) 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,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge. For $Xe$ (valence electrons $V = 8$):CompoundCalculationLone Pairs$XeO_3$$\frac{1}{2}(8 - 0) = 4$ electron pairs; $3$ bonding pairs ($O$ is divalent) $ightarrow 4 - 3 = 1$$1$$XeOF_4$$\frac{1}{2}(8 - 4) = 2$ electron pairs; $5$ bonding pairs ($O$ is divalent,$F$ is monovalent) $ightarrow 6 - 5 = 1$$1$$XeF_6$$\frac{1}{2}(8 - 6) = 1$ electron pair$1$All three molecules $(i), (ii),$ and $(iii)$ have $1$ lone pair on the $Xe$ atom.

In the following molecules: $(i) XeO_3$,$(ii) XeOF_4$,$(iii) XeF_6$,which ones have the same number of lone pairs on $Xe$?

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