The correct order of dipole moments is :

C H_{4} < N F_{3} < N H_{3} < H_{2} O

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N F_{3}, C H_{4} < N H_{3} < H_{2} O

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N H_{3} < N F_{3} < C H_{4} < H_{2} O

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H_{2} O < N H_{3} < N F_{3} < C H_{4}

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Solution

The correct option is B $C H_{4} < N F_{3} < N H_{3} < H_{2} O$
$C H_{4}$ : is symmetric with $0$ dipoled moment $N F_{3}$ ; and $N H_{3}$ : have similar geometry, $N H_{3}$ has greater bond angle, greater dipole moment $H_{2} O$ : $H - O$ bond is more polar, greater bond angle, greater dipole moment $C H_{4}$ ; $N F_{3} < N H_{3} < H_{2} O$

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