Trivalent lanthanoid ions which do not show any colour, are

$La^{3+}$ and $Lu^{3+}$

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

To determine which trivalent lanthanoid ions do not show any color, we need to consider their electronic configurations and the presence of unfilled \(4f\) orbitals, which are responsible for the absorption of visible light and hence color.

Lanthanoid Series Overview

The lanthanides (or lanthanoids) consist of \(15\) elements, from lanthanum \((La)\) to lutetium \((Lu)\), with the common oxidation state being \(+3\) (trivalent). The electronic configuration for the trivalent lanthanides can be generalized as \([Xe] 4f^{n} 6s^2\), where \(n\) varies from 0 to 14 as we move across the series.

Analysis of Options

1.\( Nd^{3+}\) and \( Eu^{3+}\)

\( Nd^{3+}\) (Neodymium): Has a configuration of \([Xe] 4f^3\), which does have partially filled 4f orbitals. It shows color (purple).

\( Eu^{3+}\) (Europium): Configuration is \([Xe] 4f^7\), which is also partially filled. It shows color (red).

2. \( Er^{3+}\) and \( Pm^{3+}\)

\( Er^{3+}\) (Erbium): Configuration is \([Xe] 4f^{11}\), which is partially filled and shows color (pink).

\( Pm^{3+}\) (Promethium): Configuration is \([Xe] 4f^4\) and also shows color (green).

3. \( La^{3+}\) and \( Lu^{3+}\)

\( La^{3+}\) (Lanthanum): Configuration is \([Xe] 6s^2\), and it has no unpaired 4f electrons; it is colorless.

\( Lu^{3+}\) (Lutetium): Configuration is \([Xe] 4f^{14} 6s^2\), also has a filled 4f shell and is colorless.

4. \( Pr^{3+}\) and \( Dy^{3+}\)

\( Pr^{3+}\) (Praseodymium): Configuration is \([Xe] 4f^3\), which is partially filled and shows color (green).

\( Dy^{3+}\) (Dysprosium): Configuration is \([Xe] 4f^{9}\), which is also partially filled and shows color (yellow).

Conclusion

The only lanthanoid ions that do not show any color are:

\( La^{3+}\) (Lanthanum), \( Lu^{3+}\)(Lutetium)

Thus, the correct answer is:

\( La^{3+}\) and \( Lu^{3+}\)