The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

Explanation

In all hydrides of group 15 elements, the central atom is `sp^(3)`-hybridized. Three of the four `sp^(3)`-orbitals form three E-H, `sigma`-bonds while the fourth contains the lone pair of electrons as shown below :
Since the lone pair-bond pair repulsions are stronger than the bond pair-bond pair repulsions, therefore, in `NH_(3)` the bond angle decreases from `109^(@)`-28' to `107.8^(@)`. As we move from N to P to As to Sb, the electronegativity of the central atom goes on decreasing. As a result, bond pairs of electrons, lie away and away from the central atom. In other words, forces of repulsion between the adjacent bond pair goes on decreasing and consequently the bond angles keep on decreasing from `NH_(3)` to `SbH_(3)`. In other words, HNH bond angle is maximum `(107.8^(@))` followed by HPH `(93.6^(@))`, followed by HAsH `(91.8^(@))` and HSbH `(91.3^(@))`