JEE Main · 2026 · 08 Apr 2026 · Shift 2 · p-Block Elements

Down group 15 the E\u2013H bond grows longer and weaker, so the hydrides become progressively less stable and give up hydrogen more readily; their reducing power therefore rises from $\mathrm{N}{\mathrm{H}}_3$ to $\mathrm{Bi}{\mathrm{H}}_3$ while their thermal stability falls in the same order. As the central atom becomes larger and less electronegative, its lone pair occupies a more diffuse orbital and is donated less readily, so the electron-pair donor tendency decreases from $\mathrm{N}{\mathrm{H}}_3$ to $\mathrm{Bi}{\mathrm{H}}_3$. The H\u2013E\u2013H bond angle also contracts along $\mathrm{N}{\mathrm{H}}_3(107^{\circ })>\mathrm{P}{\mathrm{H}}_3(\sim 94^{\circ })>\mathrm{As}{\mathrm{H}}_3(\sim 92^{\circ })>\mathrm{Sb}{\mathrm{H}}_3(\sim 91^{\circ })$ as the bonding pairs sit on a larger atom with weaker mutual repulsion. All four statements are correct.