Which of the following is the correct order o | vedclass

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(D) $1$. The dipole moment of $BF_3$ is $0 \ D$ because it has a planar trigonal geometry where the bond dipoles cancel each other out.$2$. Both $NH_3

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$NH_3 > NF_3 > BF_3$

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(D) $1$. The dipole moment of $BF_3$ is $0 \ D$ because it has a planar trigonal geometry where the bond dipoles cancel each other out.$2$. Both $NH_3$ and $NF_3$ 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,leading to a high dipole moment $(1.46 \ D)$.$4$. In $NF_3$,the $N-F$ bond dipoles are in the opposite direction to the lone pair dipole,which reduces the net dipole moment $(0.24 \ D)$.$5$. Therefore,the correct order is $NH_3 > NF_3 > BF_3$.

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 is the correct order of dipole moment for the given molecules?

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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)
$A$. $HBr$ $I$. $1.04$
$B$. $H_2S$ $II$. $0$
$C$. $NH_3$ $III$. $0.79$
$D$. $CHCl_3$ $IV$. $0.95$
$V$. $1.47$

Number of compounds from the following with zero dipole moment is . . . . . .
$HF, H_2, H_2S, CO_2, NH_3, BF_3, CH_4, CHCl_3, SiF_4, H_2O, BeF_2$

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