Identify a molecule having the highest number | vedclass

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(C) 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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$ICl_3$

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(C) 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).$1$. For $NH_3$: $N$ is $5$,$V = 5$,$N = 3$. $	ext{Lone pairs} = \frac{1}{2} (5 - 3) = 1$.$2$. For $SF_4$: $S$ is $6$,$V = 6$,$N = 4$. $	ext{Lone pairs} = \frac{1}{2} (6 - 4) = 1$.$3$. For $ICl_3$: $I$ is $7$,$V = 7$,$N = 3$. $	ext{Lone pairs} = \frac{1}{2} (7 - 3) = 2$.$4$. For $PCl_3$: $P$ is $5$,$V = 5$,$N = 3$. $	ext{Lone pairs} = \frac{1}{2} (5 - 3) = 1$.Thus,$ICl_3$ has the highest number of lone pairs $(2)$.

Identify a molecule having the highest number of lone pairs of electrons in the valence shell of the central atom.

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