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

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$A-5, B-1, C-2, D-3$

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(C) To determine the number of lone pairs on the central atom,we use the formula: $	ext{Lone pairs} = \frac{V - M}{2}$,where $V$ is the number of valence electrons of the central atom and $M$ is the number of monovalent atoms bonded to it.For $A$. $NH_3$: Central atom $N$ has $5$ valence electrons and $3$ $H$ atoms attached. $	ext{Lone pairs} = \frac{5-3}{2} = 1$. Thus,$A-5$.For $B$. $H_2O$: Central atom $O$ has $6$ valence electrons and $2$ $H$ atoms attached. $	ext{Lone pairs} = \frac{6-2}{2} = 2$. Thus,$B-1$.For $C$. $XeF_2$: Central atom $Xe$ has $8$ valence electrons and $2$ $F$ atoms attached. $	ext{Lone pairs} = \frac{8-2}{2} = 3$. Thus,$C-2$.For $D$. $CH_4$: Central atom $C$ has $4$ valence electrons and $4$ $H$ atoms attached. $	ext{Lone pairs} = \frac{4-4}{2} = 0$. Thus,$D-3$.Therefore,the correct match is $A-5, B-1, C-2, D-3$.

Column-$I$ (Molecules)	Column-$II$ (Number of lone pairs on central atom)
$A$. $NH_3$	$1$. Two
$B$. $H_2O$	$2$. Three
$C$. $XeF_2$	$3$. Zero
$D$. $CH_4$	$4$. Four
	$5$. One

Match the following molecules in Column-$I$ with the number of lone pairs on their central atom in Column-$II$.
Column-$I$ (Molecules) Column-$II$ (Number of lone pairs on central atom)
$A$. $NH_3$ $1$. Two
$B$. $H_2O$ $2$. Three
$C$. $XeF_2$ $3$. Zero
$D$. $CH_4$ $4$. Four
$5$. One

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Match the following molecules in Column-$I$ with the number of lone pairs on their central atom in Column-$II$.Column-$I$ (Molecules)Column-$II$ (Number of lone pairs on central atom)$A$. $NH_3$$1$. Two$B$. $H_2O$$2$. Three$C$. $XeF_2$$3$. Zero$D$. $CH_4$$4$. Four$5$. One