Match the following species in List-I with th | vedclass

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(A) To determine the number of lone pairs,we analyze the structure of each species:$A. \ CH_3COCH_3$ (Acetone): The oxygen atom has two lone pairs. So

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$A-I, B-III, C-II$

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(A) To determine the number of lone pairs,we analyze the structure of each species:$A. \ CH_3COCH_3$ (Acetone): The oxygen atom has two lone pairs. So,$A \rightarrow I$.$B. \ CH_3CO^+$ (Acylium ion): The oxygen atom is bonded to carbon by a triple bond $(CH_3-C \equiv O^+)$. The oxygen atom has one lone pair. So,$B \rightarrow III$.$C. \ CH_3CH_2^+$ (Ethyl carbocation): All valence electrons of carbon are involved in bonding (three $C-H$ bonds and one $C-C$ bond). There are no lone pairs on the carbocationic carbon. So,$C \rightarrow II$.Thus,the correct match is $A-I, B-III, C-II$.

List-$I$ (Species)	List-$II$ (Number of lone pairs)
$A. \ CH_3COCH_3$	$I. \ 2$
$B. \ CH_3CO^+$	$II. \ 0$
$C. \ CH_3CH_2^+$	$III. \ 1$
	$IV. \ 3$

Match the following species in List-$I$ with the number of lone pairs in List-$II$.
List-$I$ (Species) List-$II$ (Number of lone pairs)
$A. \ CH_3COCH_3$ $I. \ 2$
$B. \ CH_3CO^+$ $II. \ 0$
$C. \ CH_3CH_2^+$ $III. \ 1$
$IV. \ 3$

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Match the following species in List-$I$ with the number of lone pairs in List-$II$.List-$I$ (Species)List-$II$ (Number of lone pairs)$A. \ CH_3COCH_3$$I. \ 2$$B. \ CH_3CO^+$$II. \ 0$$C. \ CH_3CH_2^+$$III. \ 1$$IV. \ 3$