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

Explanation

In all the hybrides of group `15` elements, the central atoms is `sp^(3)-` hybridized. Three of the four `sp^(3)-` orbitals form three `E-H` bonds while the fourth contains the lone pair of electrons as shown below :

Since the lone pair-bond pair repulsions are stronger then the bond pair-bond pair repulsion, 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, force of repulsion between the adjacent bond pairs 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^(@))`