The least basic hydroxide amongst the following would be:

The correct answer is Option (4) → \(Yb(OH)_3\) \((Z = 70)\)

Across the lanthanoid series the atomic and ionic radii decrease steadily with increasing atomic number: the lanthanoid contraction. As the Ln³⁺ ion becomes smaller, it holds the \( \ce{OH^{-}} \) ions more strongly and the covalent character of the Ln–OH bond increases. The hydroxide then releases \( \ce{OH^{-}} \) ions less easily, so the basic character of Ln(OH)₃ decreases from La(OH)₃ to Lu(OH)₃.

Placing the four by atomic number: Nd (60) < Sm (62) < Er (68) < Yb (70), so basicity falls in the order \( \ce{Nd(OH)3} \) > \( \ce{Sm(OH)3} \) > \( \ce{Er(OH)3} \) > \( \ce{Yb(OH)3} \). Yb has the smallest Ln³⁺ ion and so the least basic hydroxide.

Why the other options are incorrect

• Option (1) \(Sm(OH)_3\) \((Z = 62)\) — Incorrect. \( \ce{Sm^{3+}} \) is larger than \( \ce{Er^{3+}} \) and \( \ce{Yb^{3+}} \), so \( \ce{Sm(OH)3} \) is more basic than both.

• Option (2) \(Nd(OH)_3\) \((Z = 60)\) — Incorrect. Nd has the lowest atomic number here, the largest ion and so the most basic hydroxide of the four.

• Option (3) \(Er(OH)_3\) \((Z = 68)\) — Incorrect. \( \ce{Er^{3+}} \) is still slightly larger than \( \ce{Yb^{3+}} \), so \( \ce{Er(OH)3} \) is a little more basic than \( \ce{Yb(OH)3} \).

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. The decrease in atomic radii (derived from the structures of metals) is not quite regular as it is regular in M³⁺ ions (Fig. 4.6).”