Which of the following molecules consists of | vedclass

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(B) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{V - B}{2}$,where $V$ is the number of val

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$PCl_3$

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(B) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{V - B}{2}$,where $V$ is the number of valence electrons of the central atom and $B$ is the number of bonding electrons (shared with surrounding atoms).$A) SCl_2$: Central atom $S$ (Group $16$) has $6$ valence electrons. It forms $2$ bonds. Lone pairs = $(6 - 2) / 2 = 2$.$B) PCl_3$: Central atom $P$ (Group $15$) has $5$ valence electrons. It forms $3$ bonds. Lone pairs = $(5 - 3) / 2 = 1$.$C) ClF_3$: Central atom $Cl$ (Group $17$) has $7$ valence electrons. It forms $3$ bonds. Lone pairs = $(7 - 3) / 2 = 2$.$D) XeF_4$: Central atom $Xe$ (Group $18$) has $8$ valence electrons. It forms $4$ bonds. Lone pairs = $(8 - 4) / 2 = 2$.Comparing the values,$PCl_3$ has the minimum number of lone pairs $(1)$.

List-$I$ (Compound)	List-$II$ (Structure)
$A$. $ClF_3$	$1$. Square planar
$B$. $PCl_5$	$2$. Tetrahedral
$C$. $IF_5$	$3$. Trigonal bipyramidal
$D$. $CCl_4$	$4$. Square pyramidal
$E$. $XeF_4$	$5$. $T$-shaped

Which of the following molecules consists of the minimum number of lone pairs of electrons in the valence shell of the central atom?

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