Least basic compound among the following would be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(Lu(OH)_3\)
Across the lanthanoid series the size of the M3+ ion decreases steadily from \( \ce{La^{3+}} \) to \( \ce{Lu^{3+}} \) (lanthanoid contraction). As the cation becomes smaller, its pull on the \( \ce{OH^{-}} \) group increases, the M–OH bond becomes more covalent and the hydroxide gives \( \ce{OH^{-}} \) ions less readily. So the basic character of Ln(OH)3 decreases from La(OH)3 to Lu(OH)3:
\( \ce{La(OH)3} \) > \( \ce{Ce(OH)3} \) > \( \ce{Gd(OH)3} \) > \( \ce{Lu(OH)3} \)
Lu is the last lanthanoid with the smallest Ln3+ ion, so Lu(OH)3 is the least basic.
Why the other options are incorrect
Option (1) La(OH)3 — Incorrect. \( \ce{La^{3+}} \) is the largest ion of the series, so \( \ce{La(OH)3} \) is the most basic.
Option (3) Ce(OH)3 — Incorrect. Ce comes second in the series; \( \ce{Ce^{3+}} \) is still large and its hydroxide is strongly basic.
Option (4) Gd(OH)3 — Incorrect. Gd lies in the middle of the series, so its basicity is between those of \( \ce{La(OH)3} \) and \( \ce{Lu(OH)3} \).
NCERT (2026-27 edition):
Atomic and Ionic Sizes – “The overall decrease in atomic and ionic radii from lanthanum to lutetium (the lanthanoid contraction) is a unique feature in the chemistry of the lanthanoids. It has far reaching consequences in the chemistry of the third transition series of the elements. … There is fairly regular decrease in the sizes with increasing atomic number.”