Arrange the bonds in order of increasing ioni | vedclass

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(C) The ionic character of a bond depends on the difference in electronegativity between the bonded atoms. Larger electronegativity difference leads t

Vedclass · Accepted Answer

$N_2 < SO_2 < ClF_3 < K_2O < LiF$

Vedclass · Answer

(C) The ionic character of a bond depends on the difference in electronegativity between the bonded atoms. Larger electronegativity difference leads to higher ionic character.$1$. $N_2$ (non-polar covalent,$\Delta EN = 0$)$2$. $SO_2$ (polar covalent,$\Delta EN \approx 0.9$)$3$. $ClF_3$ (polar covalent,$\Delta EN \approx 1.0$)$4$. $K_2O$ (ionic,$\Delta EN \approx 2.6$)$5$. $LiF$ (ionic,$\Delta EN \approx 3.0$)Therefore,the increasing order of ionic character is: $N_2 < SO_2 < ClF_3 < K_2O < LiF$.

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

Arrange the bonds in order of increasing ionic character in the molecules: $LiF$,$K_2O$,$N_2$,$SO_2$,and $ClF_3$.

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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)
$A. \ CH_3COCH_3$ $I. \ 2$
$B. \ CH_3CO^+$ $II. \ 0$
$C. \ CH_3CH_2^+$ $III. \ 1$
$IV. \ 3$

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