The bond angle difference between NH_3 (107.3 degrees) and PH_3 (93.6 degrees) cannot be explained by lone pair repulsion alone, because both molecules have the same number of lone pairs (one) and the same number of bonds (three). The key difference lies in the degree of hybridisation. Nitrogen, being small and highly electronegative, undergoes effective sp^3 hybridisation, giving bond angles close to the tetrahedral angle (109.5 degrees), slightly reduced to 107.3 degrees by lone pair repulsion. Phosphorus, being a larger and less electronegative atom in the third period, has 3s and 3p orbitals that are much more diffuse and similar in energy to each other in terms of effective overlap. The 3s orbital of phosphorus is stabilised by the nucleus and has poor tendency to hybridise with 3p orbitals. As a result, P-H bonds in PH_3 form using nearly pure 3p orbitals, which are oriented at approximately 90 degrees to each other (the natural p-orbital angle), giving a bond angle close to 93.6 degrees. While nitrogen's greater electronegativity also plays a role, it does not directly cause the 90-degree-like angles seen in PH_3. Option 'Larger atomic radius' is a contributory factor but does not directly explain the near-90-degree angles. Option 'NH_3 has a lone pair while PH_3 does not' is factually wrong; both have one lone pair. This is a classic NCERT JEE periodicity-and-bonding question. Plausibility: the ~90 degree angles in PH_3 (and similar H_2S, H_2Se) across period 3 consistently indicate use of near-pure p orbitals, not sp^3 hybridisation.