Arrange the following hydrides in increasing order of thermal stability.
A. H2O
B. H2Se
C. H2Po
D. H2Te
E. H2S
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → C < D < B < E < A
Down group 16 (O, S, Se, Te, Po) the size of the atom increases, so the E–H bond becomes longer and weaker. The H–E bond dissociation enthalpy decreases down the group, and the thermal stability of the hydrides decreases with it. The weakest bond gives the least stable hydride, which comes first in the increasing order.
A. H2O — last (most stable). Oxygen is the smallest atom, so the O–H bond is the strongest.
B. H2Se — third. Se is larger than S, so the Se–H bond is weaker than S–H but stronger than Te–H.
C. H2Po — first (least stable). Po is the largest atom of the group and forms the weakest bond with hydrogen.
D. H2Te — second. The Te–H bond is weaker than Se–H but stronger than Po–H.
E. H2S — fourth. The S–H bond is weaker than O–H but stronger than Se–H.
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.