The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of `sp^(3)` -hybridisation in `NH_(3)` and only s-p bonding between H and other elements of the group].

Explanation

In the hydrides of group-15 elements, central atom is `sp^(3)` - hybridised. 3 out of 4 `sp^(3)` -orbitals form `3 E-H sigma` -bonds while the fourth one contains the lone pair of electrons.
Since the lone pair-bond pair repulsions are stronger than bond pair-bond pair repulsions, the bond angle of `NH_(3)` decreases from `109^(@)28'` to `107.8^(@)`. As we move from N to Sb, the electronegativity decreases. Consequently the bond pairs are aligned away from the central atom as the repulsion between the adjacent bond pairs decreases from `NH_(3)` to `SbH_(3)`. Hence, HNH bond angle is maximum `(107.8^(@))` followed by HPH `(93.6^(@))`, HAsH `(91.8^(@))` and HSbH `(91.3^(@))`.
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