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(A) In $BF_3$,the dipole moment is zero due to its symmetrical trigonal planar structure,where the vector sum of all bond dipoles is zero.In $H_2S$ an

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$BF_3 < H_2S < H_2O$

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(A) In $BF_3$,the dipole moment is zero due to its symmetrical trigonal planar structure,where the vector sum of all bond dipoles is zero.In $H_2S$ and $H_2O$,the molecules have a bent structure,resulting in a net dipole moment.Since oxygen is more electronegative than sulfur,the polarity of the $O-H$ bond is greater than that of the $S-H$ bond.Therefore,the dipole moment of $H_2O$ is greater than that of $H_2S$.Thus,the correct order of increasing dipole moment is: $BF_3 < H_2S < H_2O$.

Arrange the following in order of increasing dipole moment : $H_2O$,$H_2S$,$BF_3$.

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