Q.The HNH angle value is higher than HPH, HAsH and HSbH angles. Why? [Hint: Can be explained on the basis of sp3 hybridisation in NH3 and only s–p bonding between hydrogen and other elements of the group].

Step 1: Bonding in $NH_3$.

Nitrogen, being small with comparable 2s and 2p orbital energies, undergoes $sp^3$ hybridisation. Three of the four $sp^3$ hybrid orbitals form N–H sigma bonds, and the fourth holds the lone pair. An ideal tetrahedral arrangement would give 109.5°, but the lone pair (occupying more space, higher repulsion) compresses the bond angle slightly, giving the observed $\approx 107^\circ$.

Step 2: Bonding in $PH_3$, $AsH_3$, $SbH_3$.

As we go from P to Sb, the central atom becomes larger, and the energy gap between its valence s and p orbitals increases, making effective $s$–$p$ hybridisation unfavourable. These heavier hydrides bond using almost pure p-orbitals (which are mutually perpendicular, at 90°) rather than hybrid orbitals.

Step 3: Consequence for bond angle.

Since bonding is via nearly pure p-orbitals (not hybridised), the H–E–H angles in $PH_3$ (93.5°), $AsH_3$ (91.8°) and $SbH_3$ (91.3°) are all close to 90°, and noticeably smaller than $NH_3$'s 107°.

Step 4: Trend across the group. …

State that NH3 uses sp3 hybridisation (near-tetrahedral angle, reduced by lone pair repulsion) while the heavier hydrides bond via nearly pure, unhybridised p-orbitals (angle near 90 …

• Do not attribute the smaller angles in PH₃/AsH₃/SbH₃ to "more lone pair repulsion" — the real reason (per the given hint) is absence of hybridisation, i.e. bonding via pure p-orbitals, not extra repulsion.
• Do not forget that NH₃'s angle (107°) is itself reduced from the ideal tetrahedral 109.5° due to the lone pair — the value is not exactly 109.5°. …