The correct order of dipole moment is

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(A) The dipole moment $(\mu)$ depends on the polarity of bonds and the geometry of the molecule.$CH_4$ is a non-polar tetrahedral molecule with $\mu =

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$CH_4 < NF_3 < NH_3 < H_2O$

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(A) The dipole moment $(\mu)$ depends on the polarity of bonds and the geometry of the molecule.$CH_4$ is a non-polar tetrahedral molecule with $\mu = 0 \ D$.In $NH_3$,the bond dipoles of $N-H$ bonds and the lone pair dipole are in the same direction,resulting in a high dipole moment $(\mu \approx 1.46 \ D)$.In $NF_3$,the bond dipoles of $N-F$ bonds and the lone pair dipole are in opposite directions,which partially cancel each other,resulting in a lower dipole moment $(\mu \approx 0.24 \ D)$.$H_2O$ has a bent geometry with two $O-H$ bonds and two lone pairs,resulting in a high dipole moment $(\mu \approx 1.85 \ D)$.Thus,the correct order is $CH_4 (0 \ D) < NF_3 (0.24 \ D) < NH_3 (1.46 \ D) < H_2O (1.85 \ D)$.

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