Which compound has the maximum number of lone | vedclass

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

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$[I_3]^-$

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(D) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{1}{2} (V - M - C + A)$,where $V$ is the number of valence electrons of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$A) [ClO_3]^-$: Central atom $Cl$ $(V=7)$. $O$ is divalent $(M=0)$. Anionic charge $A=1$. $	ext{Lone pairs} = \frac{1}{2} (7 - 0 - 0 + 1) = 4$ electrons,which corresponds to $1$ lone pair.$B) XeF_4$: Central atom $Xe$ $(V=8)$. $F$ is monovalent $(M=4)$. $	ext{Lone pairs} = \frac{1}{2} (8 - 4) = 2$ lone pairs.$C) SF_4$: Central atom $S$ $(V=6)$. $F$ is monovalent $(M=4)$. $	ext{Lone pairs} = \frac{1}{2} (6 - 4) = 1$ lone pair.$D) [I_3]^-$: Central atom $I$ $(V=7)$. Two $I$ atoms are attached $(M=2)$. Anionic charge $A=1$. $	ext{Lone pairs} = \frac{1}{2} (7 - 2 + 1) = 3$ lone pairs.Comparing the values,$[I_3]^-$ has the maximum number of lone pairs $(3)$.

Which compound has the maximum number of lone pairs of electrons on the central atom?

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