Q.Explain why the $\text{H-P-H}$ bond angle in phosphine ($\approx 93.6^\circ$) is considerably smaller than the $\text{H-N-H}$ bond angle in ammonia ($\approx 107.8^\circ$).

[!TLDR] Phosphorus is larger and less electronegative than nitrogen, so its bonding pairs repel each other less, letting PH3's bonding use orbitals with far more p-character (closer to a 90 degree pure-p angle) than NH3's more purely sp3 bonding. [!ANSWER] PH3's smaller bond angle reflects phosphorus's larger siz …

Both $\text{NH}_3$ and $\text{PH}_3$ are trigonal pyramidal, $sp^3$-type molecules with one lone pair, so at first glance both might be expected to show a similar, near-tetrahedral bond angle. However, nitrogen is considerably smaller and more electronegative than phosphorus. In $\text{NH}_3$, nitrogen holds its three N-H bonding electron pairs relatively close and tightly, so these bonding pairs strongly repel one another (and the lone pair repels them even more strongly), pushing the molecule's hybridisation close to a genuine, near-ideal $sp^3$ description and giving a bond angle (107.8 degrees) reasonably close to the tetrahedral $109.5^\circ$. In $\text{PH}_3$, phosphorus is larger and less electronegative, so its P-H bonding pairs are held less tightly and sit farther from the central atom, reducing the mutual bond pair-bond pair (and lone pair-bond pair) repulsion considerably. With weaker inter-pair repulsion to overcome, phosphorus's bonding orbitals need less s-character mixed in to spread the bonds apart, and the P-H bonds end up using orbitals with much grea …

Compare the two central atoms' size/electronegativity, and connect weaker bonding-pair repulsion in the larger, less electrone …

Assuming both molecules must have identical bond angles because both are 'sp3 with one lone pair'; not connecting the angle difference …