Answer the question on the basis of passage given below:
Nitrogen differs from rest of the members of group 15 due to its small size, high electronegativity, high ionization enthalpy and non-availability of d orbitals. Nitrogen has unique ability to form \(p\pi -p\pi \) multiple bonds with itself and with other elements having small size and high electronegativity. Heavier elements of this group do not form \(p\pi -p\pi \) bonds as their atomic orbitals are so large and diffuse that they cannot have effective overlapping.
Arrange the following in decreasing order of their basic strength :
A. \(PH_3\)
B. \(AsH_3\)
C. \(NH_3\)
D. \(SbH_3\)
E. \(BiH_3\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → C > A > B > D > E
The hydrides of group 15 (EH3) act as bases through the lone pair on the central atom. Down the group the atom becomes larger, so the lone pair is spread over a larger volume and is less available for donation. The basic strength therefore decreases from \( \ce{NH3} \) to \( \ce{BiH3} \).
(C) \( \ce{NH3} \) — first (most basic). N is the smallest atom, so its lone pair is the most concentrated and readily accepts a proton:
\( \displaystyle \ce{NH3 + H2O <=> NH4^+ + OH^-} \)
(A) \( \ce{PH3} \) — second. P is larger than N; \( \ce{PH3} \) is a weak base and forms \( \ce{PH4^+} \) only with strong acids such as HI.
(B) \( \ce{AsH3} \) — third. As is larger than P, so its lone pair is less available.
(D) \( \ce{SbH3} \) — fourth. Sb is larger still, so \( \ce{SbH3} \) is only about as basic as, or a little more basic than, \( \ce{BiH3} \).
(E) \( \ce{BiH3} \) — fifth (least basic). Bi is the largest atom of the group, so its lone pair is the most diffuse and the least available.
Note 1: 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):
Chemical Properties – “… (i) Reactivity towards hydrogen: All the elements of Group 15 form hydrides of the type EH3 where E = N, P, As, Sb or Bi. … 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.”
Note 2: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 02-07 S3 13-06 S3 12-06 S3 09-06 S3 | – |