The correct order of dipole moment 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 A
$C H_{4} < N F_{3} < N H_{3} < H_{2} O$

$: C H_{4}$ is symmetrical $\Rightarrow$ zero dipole moment.

In NF3, moment by lone pair is opposite to bond pair moment.

[F is more electronegative than N]

In $N H_{3}$ , moment by lone pair is in the direction of bond pair moment.

In water, both the moments add up.

Thus, the order is

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

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