Bond dissociation enthalpy of (E-H) (E=element) are given below. Whic…

The correct answer is Option (4) → \(SbH_3\)

All the elements of Group 15 form hydrides of the type EH₃. A hydride acts as a reducing agent by giving up hydrogen, which needs the E–H bond to break. Down the group, the atom E becomes larger, its orbitals overlap less with the 1s orbital of hydrogen, and the E–H bond becomes weaker. The given bond dissociation enthalpies show this:

    \( \ce{NH3} \) (389) > \( \ce{PH3} \) (322) > \( \ce{AsH3} \) (297) > \( \ce{SbH3} \) (255) kJ mol⁻¹

The weaker the E–H bond, the less stable the hydride and the more easily it gives up hydrogen. So the reducing character increases from \( \ce{NH3} \) to \( \ce{SbH3} \) (and further to \( \ce{BiH3} \), the strongest of all). Among the given hydrides, SbH₃, with the lowest bond dissociation enthalpy, is the strongest reducing agent.

Why the other options are incorrect

• Option (1) \(NH_3\) — Incorrect. It has the strongest E–H bond (389 kJ mol⁻¹); ammonia is only a mild reducing agent.

• Option (2) \(PH_3\) — Incorrect. Its P–H bond (322 kJ mol⁻¹) is still stronger than the As–H and Sb–H bonds.

• Option (3) \(AsH_3\) — Incorrect. A stronger reducing agent than \( \ce{PH3} \), but its As–H bond (297 kJ mol⁻¹) is stronger than the Sb–H bond.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Reactivity towards Hydrogen