The incorrect statement among the following is :

The lanthanoid (lanthanide) series consists of the 14 elements from lanthanum (La) to lutetium (Lu). In their most stable oxidation state they form trivalent ions, denoted $Ln^\{3+\}$.

These $Ln^\{3+\}$ ions have electrons occupying the 4f subshell. The 4f orbitals are partially shielded by the filled 5s and 5p orbitals, which makes the 4f‑4f electronic transitions only weakly allowed. Nevertheless, these f‑f transitions absorb visible light and give rise to characteristic colors for most lanthanoid ions.

Examples of colored trivalent lanthanoid ions include:

• $Ce^\{3+\}$ – pale yellow
• $Pr^\{3+\}$ – violet
• $Nd^\{3+\}$ – pale violet
• $Sm^\{3+\}$ – orange‑brown
• $Eu^\{3+\}$ – red
• $Gd^\{3+\}$ – faint yellow
• $Tb^\{3+\}$ – green
• $Dy^\{3+\}$ – yellowish‑green
• $Ho^\{3+\}$ – pink
• $Er^\{3+\}$ – pink‑red
• $Tm^\{3+\}$ – violet‑blue
• $Yb^\{3+\}$ – pale yellow‑green
• $Lu^\{3+\}$ – essentially colorless

Only the ions of lanthanum ($La^\{3+\}$) and lutetium ($Lu^\{3+\}$) have a completely filled or empty 4f subshell, so they do not show f‑f absorption and appear colorless. All other trivalent lanthanoid ions display some color in the solid state because of the 4f‑4f transitions.

Therefore, the statement “Most of the trivalent Lanthanoid ions are colorless in the solid state” is incorrect. The correct fact is that *most* trivalent lanthanoid ions are **colored**, with only a few (La³⁺, Lu³⁺, and sometimes Yb³⁺) being essentially colorless.