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.
Which of the following is the correct order of reducing strength of hydrides of Group 16 elements?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(H_2O < H_2S < H_2Se < H_2Te\)
A hydride acts as a reducing agent by giving up hydrogen (or electrons). This becomes easier as the H–E bond becomes weaker. Down Group 16 the H–E bond dissociation enthalpy decreases, so the reducing character increases from H2S to H2Te, while water, with the strongest O–H bond, shows practically no reducing property:
\( \ce{H2O} \) < \( \ce{H2S} \) < \( \ce{H2Se} \) < \( \ce{H2Te} \)
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. … Owing to the decrease in bond (H–E) dissociation enthalpy down the group, the thermal stability of hydrides also decreases from H2O to H2Po. All the hydrides except water possess reducing property and this character increases from H2S to H2Te.”