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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)$,where $V$ is the number of

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$2, 1, 2$ and $0$

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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)$,where $V$ is the number of valence electrons of the central atom and $N$ is the number of bonding electrons (or atoms attached).$1$. For $SCl_2$: Sulfur $(S)$ has $6$ valence electrons. It forms $2$ bonds with $Cl$ atoms. Lone pairs = $\frac{1}{2} (6 - 2) = 2$.$2$. For $O_3$: The central Oxygen atom has $6$ valence electrons. It forms $2$ bonds with one oxygen atom and $1$ coordinate bond with another. The central atom has $1$ lone pair.$3$. For $ClF_3$: Chlorine $(Cl)$ has $7$ valence electrons. It forms $3$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (7 - 3) = 2$.$4$. For $SF_6$: Sulfur $(S)$ has $6$ valence electrons. It forms $6$ bonds with $F$ atoms. Lone pairs = $\frac{1}{2} (6 - 6) = 0$.Thus,the number of lone pairs are $2, 1, 2$ and $0$ respectively. The correct option is $B$.

Molecule	$P-X$ (axial) bond length / $P-X$ (equatorial) bond length
$PF_5$	$a / b$
$PF_4CH_3$	$c / d$
$PF_3(CH_3)_2$	$e / f$
$PCl_5$	$g / h$

The number of lone pairs of electrons on the central atom of $SCl_2$,$O_3$,$ClF_3$,and $SF_6$,respectively,are:

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Consider the following information $(X = F$ or $Cl)$:

Molecule $P-X$ (axial) bond length / $P-X$ (equatorial) bond length

$PF_5$ $a / b$

$PF_4CH_3$ $c / d$

$PF_3(CH_3)_2$ $e / f$

$PCl_5$ $g / h$

According to the given information,choose the incorrect order of bond lengths.

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