# Why does the dipole moment order of group 15 hydrides increase down the group?

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?

• 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. Aug 16, 2019 at 16:56
• @porphyrin could you convert this into a proper answer for this question
– Stan
Aug 16, 2019 at 17:59

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