MCQ
Arrange the following in order of increasing dipole moment : $H_2O, H_2S, BF_3.$
  • $BF_3 < H_2S < H_2O$
  • B
    $H_2S < BF_3 < H_2O$
  • C
    $H_2O < H_2S < BF_3$
  • D
    $BF_3 < H_2O < H_2S$

Answer

Correct option: A.
$BF_3 < H_2S < H_2O$
a
In $BF _3$, dipole moment is zero due to its symmetrical structure. Summations of all dipoles are zero.

In $H _2 S$ and $H _2 O$ due to unsymmetrical structure net $+ve$ dipole is there. $H _2 O$ has a higher dipole due to the higher electronegativity of oxygen than sulphur.

Hence, the correct order is: $BF _3\,<\, H _2 S \,<\, H _2 O$.

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