Match the following molecules in Column I wit | vedclass

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(A) 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

Vedclass · Accepted Answer

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

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(A) 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.$A. \ NH_3$: Central atom $N$ has $5$ valence electrons. It is bonded to $3$ $H$ atoms. $	ext{Lone pairs} = \frac{5-3}{2} = 1$ (Option $5$).$B. \ H_2O$: Central atom $O$ has $6$ valence electrons. It is bonded to $2$ $H$ atoms. $	ext{Lone pairs} = \frac{6-2}{2} = 2$ (Option $1$).$C. \ XeF_2$: Central atom $Xe$ has $8$ valence electrons. It is bonded to $2$ $F$ atoms. $	ext{Lone pairs} = \frac{8-2}{2} = 3$ (Option $2$).$D. \ CH_4$: Central atom $C$ has $4$ valence electrons. It is bonded to $4$ $H$ atoms. $	ext{Lone pairs} = \frac{4-4}{2} = 0$ (Option $3$).Therefore,the correct matching is $A-5, B-1, C-2, D-3$.

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

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