Which of the following pairs of compounds con | vedclass

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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} (V - N)$,where $V$ is the number of

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$BrF_5$ and $XeF_6$

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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} (V - N)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (monovalent atoms attached).$A$: $BrF_5$ ($Br$ has $7$ valence $e^-$,$5$ bonded $F$ atoms): $	ext{Lone pairs} = \frac{1}{2} (7 - 5) = 1$. $XeF_6$ ($Xe$ has $8$ valence $e^-$,$6$ bonded $F$ atoms): $	ext{Lone pairs} = \frac{1}{2} (8 - 6) = 1$.$B$: $ICl$ ($I$ has $7$ valence $e^-$,$1$ bonded $Cl$ atom): $	ext{Lone pairs} = \frac{1}{2} (7 - 1) = 3$. $H_2S$ ($S$ has $6$ valence $e^-$,$2$ bonded $H$ atoms): $	ext{Lone pairs} = \frac{1}{2} (6 - 2) = 2$.$C$: $ClF_3$ ($Cl$ has $7$ valence $e^-$,$3$ bonded $F$ atoms): $	ext{Lone pairs} = \frac{1}{2} (7 - 3) = 2$. $XeF_2$ ($Xe$ has $8$ valence $e^-$,$2$ bonded $F$ atoms): $	ext{Lone pairs} = \frac{1}{2} (8 - 2) = 3$.$D$: $IF_7$ ($I$ has $7$ valence $e^-$,$7$ bonded $F$ atoms): $	ext{Lone pairs} = \frac{1}{2} (7 - 7) = 0$. $XeF_4$ ($Xe$ has $8$ valence $e^-$,$4$ bonded $F$ atoms): $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 2$.Thus,$BrF_5$ and $XeF_6$ both have $1$ lone pair on their central atoms.

Which of the following pairs of compounds consists of an equal number of lone pairs of electrons in the valence shell of the central atom?

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