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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?
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
