Deviation in bond angles (from standard $109°28'$ ) in hydrides of group 15 [duplicate]

Here are the approximate bond angles for first three members of group15.

$\ce{NH3}$: $107°$

$\ce{PH3}$: $94°$

$\ce{AsH3}$: $92°$

As we move down the group the bond angles approach towards $90°$. I don't understand why there is such a major deviation from the standard angle of $109°28'$.

In the case of $\ce{NH3}$, the deviation is justified because of the presence of lone pair(lp-bp repulsions) , but for others, this much deviation is quite unexpected.

What causes this large deviation in angle?

• In the same direction ($\ce{N -> P -> As}$), not only the atoms include more nucleons and electrons (increase of mass), but the atomic radii increase, too ($65, 100, \pu{115 pm}$) which contributes to the reason. (consulted source: en.wikipedia.org/wiki/Atomic_radii_of_the_elements_(data_page)) – Buttonwood Jun 20 '17 at 11:53
• Also incorrect is the assumption that VSEPR works in general. – Zhe Jun 20 '17 at 13:43