Answer the question on basis of passage given below:
All the members of Group 15 elements form covalent hydrides with hydrogen as \(NH_3, PH_3, AsH_3, SbH_3, BiH_3\). Group 16 elements form covalent hydrides with hydrogen as \(H_2O, H_2S, H_2Se, H_2Te, H_2Po\). Group 17 elements form covalent hydrides with hydrogen as \(HF, HCl, HBr, HI\). Based on these facts answer the following question.
The correct decreasing order of bond dissociation enthalpies for the following compounds is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(H_2O > H_2S > H_2Se > H_2Te\)
The H–E bond in the Group 16 hydrides is formed by overlap of the 1s orbital of hydrogen with an orbital of E. Down the group the atom E becomes larger (O < S < Se < Te), its bonding orbital becomes more diffuse, the overlap with the small 1s orbital becomes poorer and the bond becomes longer and weaker. So the H–E bond dissociation enthalpy decreases down the group:
\( \ce{H2O} \) > \( \ce{H2S} \) > \( \ce{H2Se} \) > \( \ce{H2Te} \)
The same fall in bond enthalpy explains why the thermal stability of these hydrides decreases and their acidic character increases from \( \ce{H2O} \) to \( \ce{H2Te} \).
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 = 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. Owing to the decrease in bond (H–E) dissociation enthalpy down the group, the thermal stability of hydrides also decreases from H2O to H2Po.”