Arrange the following in the decreasing order of acidity:
(A) \(H_2O\)
(B) \(H_2S\)
(C) \(H_2Te\)
(D) \(H_2Se\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → C > D > B > A
Down group 16 the central atom becomes larger, the H–E bond becomes longer and weaker, and the hydride releases \( \ce{H^{+}} \) more easily. Bond dissociation enthalpies (kJ mol−1): H–O 463, H–S 347, H–Se 276, H–Te 238.
C. \( \ce{H2Te} \) — first (most acidic): weakest H–Te bond (238 kJ mol−1).
D. \( \ce{H2Se} \) — second: H–Se bond (276 kJ mol−1) weaker than H–S.
B. \( \ce{H2S} \) — third: H–S bond (347 kJ mol−1) weaker than H–O.
A. \( \ce{H2O} \) — last (least acidic): strongest H–O bond (463 kJ mol−1).
Shortcut: Only Option (2) starts with \( \ce{H2Te} \) (C) and ends with \( \ce{H2O} \) (A).
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 with Hydrogen
