Trivalent lanthanoid ions which do not show any colour, are

The correct answer is Option (3) → \(La^{3+}\) and \(Lu^{3+}\)

The colour of trivalent lanthanoid ions comes from f–f transitions: an electron in a 4f orbital absorbs visible light and moves to another 4f orbital. An ion with no 4f electron (4f⁰) or a full 4f subshell (4f¹⁴) cannot do this, so it is colourless.

The Ln³⁺ ion is formed by losing the two 6s electrons and one more (5d, or else 4f):

• La (Z = 57) [Xe]5d¹6s² → \( \ce{La^{3+}} \): 4f⁰, colourless.

• Lu (Z = 71) [Xe]4f¹⁴5d¹6s² → \( \ce{Lu^{3+}} \): 4f¹⁴, colourless.

So the trivalent ions that show no colour are La³⁺ and Lu³⁺.

Why the other options are incorrect

• Option (1) \(Nd^{3+}\) and \(Eu^{3+}\) — Incorrect. Nd (Z = 60) [Xe]4f⁴6s² gives \( \ce{Nd^{3+}} \) 4f³, and Eu (Z = 63) [Xe]4f⁷6s² gives \( \ce{Eu^{3+}} \) 4f⁶; both have partly filled 4f orbitals and are coloured.

• Option (2) \(Er^{3+}\) and \(Pm^{3+}\) — Incorrect. Er (Z = 68) [Xe]4f¹²6s² gives \( \ce{Er^{3+}} \) 4f¹¹, and Pm (Z = 61) [Xe]4f⁵6s² gives \( \ce{Pm^{3+}} \) 4f⁴; both are coloured.

• Option (4) \(Pr^{3+}\) and \(Dy^{3+}\) — Incorrect. Pr (Z = 59) [Xe]4f³6s² gives \( \ce{Pr^{3+}} \) 4f² (green), and Dy (Z = 66) [Xe]4f¹⁰6s² gives \( \ce{Dy^{3+}} \) 4f⁹; both are coloured.

NCERT (2026-27 edition):
General Characteristics – “… Many trivalent lanthanoid ions are coloured both in the solid state and in aqueous solutions. Colour of these ions may be attributed to the presence of f electrons. Neither La³⁺ nor Lu³⁺ ion shows any colour but the rest do so.”