Which of the following statements are correct ?
(A) AsH3 > PH3 > NH3 (decreasing order of basic strength)
(B) H2O > NH3 > PH3 (decreasing order of boiling point)
(C) NO > NO2 > N2O3 (decreasing oxidation state of nitrogen)
(D) H4P2O6 > H4P2O5 > H3PO4 (decreasing oxidation state of phosphorus)
(E) H2O > H2S > H2Se (decreasing thermal stability)
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (B) and (E) only
Down a group of the p-block, the central atom becomes larger, so the E–H bond becomes longer and weaker. This one fact decides the basic strength and thermal stability of the hydrides; boiling points also depend on hydrogen bonding, and oxidation states are found by charge balance.
(A) \( \ce{AsH3} \) > \( \ce{PH3} \) > \( \ce{NH3} \) (decreasing order of basic strength) — Incorrect. The lone pair on the small N atom is the most available, so basicity decreases down the group: \( \ce{NH3} \) > \( \ce{PH3} \) > \( \ce{AsH3} \).
(B) \( \ce{H2O} \) > \( \ce{NH3} \) > \( \ce{PH3} \) (decreasing order of boiling point) — Correct. \( \ce{H2O} \) (373 K) forms the strongest and most extensive hydrogen bonds, \( \ce{NH3} \) (239.7 K) weaker ones, and \( \ce{PH3} \) (185.5 K) none.
(C) NO > \( \ce{NO2} \) > \( \ce{N2O3} \) (decreasing oxidation state of nitrogen) — Incorrect. NO: x − 2 = 0, x = +2; \( \ce{NO2} \): x − 4 = 0, x = +4; \( \ce{N2O3} \): 2x − 6 = 0, x = +3. The decreasing order is \( \ce{NO2} \) (+4) > \( \ce{N2O3} \) (+3) > NO (+2).
(D) \( \ce{H4P2O6} \) > \( \ce{H4P2O5} \) > \( \ce{H3PO4} \) (decreasing oxidation state of phosphorus) — Incorrect. \( \ce{H4P2O6} \): 4 + 2x − 12 = 0, x = +4; \( \ce{H4P2O5} \): 4 + 2x − 10 = 0, x = +3; \( \ce{H3PO4} \): 3 + x − 8 = 0, x = +5. \( \ce{H3PO4} \) has the highest, so the order is \( \ce{H3PO4} \) > \( \ce{H4P2O6} \) > \( \ce{H4P2O5} \).
(E) \( \ce{H2O} \) > \( \ce{H2S} \) > \( \ce{H2Se} \) (decreasing thermal stability) — Correct. The H–E bond dissociation enthalpy decreases from O to Se, so the thermal stability of the hydrides decreases from \( \ce{H2O} \) to \( \ce{H2Se} \).
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 towards Hydrogen – “… The stability of hydrides decreases from NH3 to BiH3 which can be observed from their bond dissociation enthalpy. Consequently, the reducing character of the hydrides increases. Ammonia is only a mild reducing agent while BiH3 is the strongest reducing agent amongst all the hydrides. Basicity also decreases in the order NH3 > PH3 > AsH3 > SbH3 ≥ BiH3.”
Reactivity with Hydrogen – “… 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. All the hydrides except water possess reducing property and this character increases from H2S to H2Te.”