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

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$(i)$,$(iii)$,$(iv)$

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(A) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pair} = \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.$(i)$ $SO_2$: $S$ has $6$ valence electrons. It forms $2$ double bonds with $O$ atoms,so $N = 4$. $	ext{Lone pair} = \frac{1}{2} (6 - 4) = 1$.$(ii)$ $XeF_4$: $Xe$ has $8$ valence electrons. It forms $4$ single bonds with $F$ atoms,so $N = 4$. $	ext{Lone pair} = \frac{1}{2} (8 - 4) = 2$.$(iii)$ $PbCl_2$: $Pb$ (Group $14$) has $4$ valence electrons. It forms $2$ single bonds with $Cl$ atoms,so $N = 2$. $	ext{Lone pair} = \frac{1}{2} (4 - 2) = 1$.$(iv)$ $SF_4$: $S$ has $6$ valence electrons. It forms $4$ single bonds with $F$ atoms,so $N = 4$. $	ext{Lone pair} = \frac{1}{2} (6 - 4) = 1$.$(v)$ $ClF_3$: $Cl$ has $7$ valence electrons. It forms $3$ single bonds with $F$ atoms,so $N = 3$. $	ext{Lone pair} = \frac{1}{2} (7 - 3) = 2$.Thus,molecules $(i)$,$(iii)$,and $(iv)$ have only one lone pair on their central atoms.

Which set of the following molecules has only one lone pair of electrons on their respective central atoms?
$(i)$ $SO_2$
$(ii)$ $XeF_4$
$(iii)$ $PbCl_2$
$(iv)$ $SF_4$
$(v)$ $ClF_3$

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Which set of the following molecules has only one lone pair of electrons on their respective central atoms?$(i)$ $SO_2$$(ii)$ $XeF_4$$(iii)$ $PbCl_2$$(iv)$ $SF_4$$(v)$ $ClF_3$