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(B) 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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$XeF_2, ICl$

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(B) 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,$N$ is the number of monovalent atoms,and $C$ is the charge.$A) PCl_5$: $P$ has $5$ valence electrons. Lone pairs = $\frac{1}{2}(5 - 5) = 0$. $BrF_5$: $Br$ has $7$ valence electrons. Lone pairs = $\frac{1}{2}(7 - 5) = 1$.$B) XeF_2$: $Xe$ has $8$ valence electrons. Lone pairs = $\frac{1}{2}(8 - 2) = 3$. $ICl$: $I$ has $7$ valence electrons. Lone pairs = $\frac{1}{2}(7 - 1) = 3$.$C) XeF_4$: $Xe$ has $8$ valence electrons. Lone pairs = $\frac{1}{2}(8 - 4) = 2$. $ClO_4^{-}$: $Cl$ has $7$ valence electrons. Lone pairs = $\frac{1}{2}(7 - 0 + 1) = 4$ (Oxygen is divalent,so $N=0$).$D) SCl_4$: $S$ has $6$ valence electrons. Lone pairs = $\frac{1}{2}(6 - 4) = 1$. $CH_4$: $C$ has $4$ valence electrons. Lone pairs = $\frac{1}{2}(4 - 4) = 0$.Thus,both $XeF_2$ and $ICl$ have $3$ lone pairs on their central atoms.

In which of the following pairs,the central atoms have the same number of lone pairs of electrons?

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