Elements of group- 15, 16, and 17 form compounds with hydrogen with general formula \(MH_3\), \(H_2M\), and \(HX\) (called hydrogen halides). The stability of these hydrides, along with \(M-H\) bond strength and basic nature decreases down the group. Boiling point of these hydrides mainly depends on the strength of their intermolecular forces.
Which of the following is most basic among the following hydrides of group 16?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → H2O
The acidic character of these hydrides rises from \( \ce{H2O} \) to \( \ce{H2Te} \) as the H–E bond weakens, so basic character runs the other way: \( \ce{H2O} \) is the least acidic, and its oxygen has lone pairs that readily accept a proton (\( \ce{H2O} \) + \( \ce{H^{+}} \) → \( \ce{H3O^{+}} \)). So H2O is the most basic of these hydrides.
Why the other options are incorrect
Option (1) H2Te — Incorrect. The weakest H–E bond makes it the most acidic, i.e. the least basic.
Option (2) H2Se — Incorrect. More acidic than \( \ce{H2S} \) and \( \ce{H2O} \), so less basic.
Option (3) H2S — Incorrect. A weak acid in water; more acidic and less basic than \( \ce{H2O} \).
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 – “All the elements of Group 16 form hydrides of the type H2E (E = O, S, Se, Te, Po). Some properties of hydrides are given in Table 7.7. Their acidic character increases from H2O to H2Te. The increase in acidic character can be explained in terms of decrease in bond (H–E) dissociation enthalpy down the group.”