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 hydrogen and other elements of the group].

(A) The hydrides of Group $15$ elements are $NH_{3}, PH_{3}, AsH_{3}$ and $SbH_{3}$.
The bond angles are: $NH_{3} (107^{\circ}), PH_{3} (92^{\circ}), AsH_{3} (91^{\circ}), SbH_{3} (90^{\circ})$.
In $NH_{3}$,the central atom $N$ undergoes $sp^{3}$ hybridization,leading to a tetrahedral geometry with a bond angle of $107^{\circ}$ due to the lone pair-bond pair repulsion.
For the heavier elements $(P, As, Sb)$,the electronegativity decreases down the group. As a result,the bond pairs are further away from the central atom,and the bond angles approach $90^{\circ}$ because the bonding involves almost pure $p$-orbitals of the central atom with little to no hybridization.