Which of the following arrangements of molecu | vedclass

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(A) $1$. $BF_3$ is a planar molecule with a trigonal planar geometry. The bond moments of the three $B-F$ bonds cancel each other out,resulting in a n

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$BF_3 < NF_3 < NH_3$

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(A) $1$. $BF_3$ is a planar molecule with a trigonal planar geometry. The bond moments of the three $B-F$ bonds cancel each other out,resulting in a net dipole moment of $0 \ D$.$2$. In $NH_3$ and $NF_3$,both have a pyramidal geometry with a lone pair on the nitrogen atom.$3$. In $NH_3$,the direction of the dipole moment of the $N-H$ bonds and the lone pair are in the same direction,which adds up to a higher value $(1.46 \ D)$.$4$. In $NF_3$,the $N-F$ bond moments are in the opposite direction to the lone pair moment,which partially cancels out,resulting in a lower value $(0.24 \ D)$.$5$. Therefore,the correct order of dipole moment is $BF_3 (0 \ D) < NF_3 (0.24 \ D) < NH_3 (1.46 \ D)$.

List-$I$ (Molecules)	List-$II$ (Dipole moment)
$A$. $HBr$	$I$. $1.04$
$B$. $H_2S$	$II$. $0$
$C$. $NH_3$	$III$. $0.79$
$D$. $CHCl_3$	$IV$. $0.95$
	$V$. $1.47$

Which of the following arrangements of molecules is correct with respect to their dipole moment?

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Match the following molecules in List-$I$ with their respective dipole moments in List-$II$.
List-$I$ (Molecules) List-$II$ (Dipole moment)
$A$. $HBr$ $I$. $1.04$
$B$. $H_2S$ $II$. $0$
$C$. $NH_3$ $III$. $0.79$
$D$. $CHCl_3$ $IV$. $0.95$
$V$. $1.47$

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