The order given in my book is

$$\ce{NH3} > \ce{SbH3} > \ce{AsH3} > \ce{PH3}$$

Phosphorous is more electronegative than arsenic and antimony. Then shouldn't the dipole moment be more in case of $\ce{PH3}?$

And why does the anomaly of $\ce{NH3}$ occur?

  • 1
    $\begingroup$ Geometry/symmetry also has a role in determining the dipole, for example, if the molecules were planar as is $\ce{BF3}$ the dipole would be zero. $\endgroup$
    – porphyrin
    Commented Aug 16, 2019 at 16:56
  • 2
    $\begingroup$ @porphyrin could you convert this into a proper answer for this question $\endgroup$
    – Stan
    Commented Aug 16, 2019 at 17:59

1 Answer 1


The query you have is a common mistake, and this is aksed as a very common question in examinations, well, I wouldn't blame you as the reason is simple but not easy to think of.

What you said about E.N. of phosphorous being more than arsenic and antimony is absolutely correct. But what you missed, is that lower down the group the direction of dipole itself reverses.

E.N. of $\ce{P}$ is $2.19$, while that of $\ce{As}$ is $2.18$ and of $\ce{Sb}$ is $2.05$

Whereas H as an E.N. of $2.20$. So hence now you can see clearly that your order is justified, as in actuality, H becomes the negative end of the dipole.

Another factor that further widens this gap is that the dipole vector in chemistry notations is from + to-, which means all vectors add up in case of $\ce{As}$ and Sb

  • $\begingroup$ Why should I not use doubt? @The_Vinz ? $\endgroup$ Commented Jan 17, 2020 at 16:14

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