The Lanthanoid ion which forms coloured compounds typically does so because of the presence of unpaired electrons in its f-orbitals. These unpaired electrons can absorb visible light, resulting in coloured compounds. Let's analyze the given lanthanoid ions: 1. Yb2+( Ytterbium ) Ytterbium has an atomic number of 70 . In the Yb2+ ion, the electronic configuration would be [Xe]4f14. Since there are no unpaired electrons in the 4 f orbital (all 14 electrons are paired), Yb2+ would form colourless compounds. 2. La3+ (Lanthanum) Lanthanum has an atomic number of 57 . In the La3+ ion, the electronic configuration would be [Xe]4f0. Since there are no electrons in the 4 f orbital, La3+ would form colourless compounds. 3. Lu3+ (Lutetium) Lutetium has an atomic number of 71 . In the Lu3+ ion, the electronic configuration would be [Xe]4f14. As with Yb2+, all 14 electrons are paired, so Lu3+ would also form colourless compounds. 4. Pr3+ (Praseodymium) Praseodymium has an atomic number of 59 . In the Pr3+ ion, the electronic configuration would be [Xe]4f2. There are 2 unpaired electrons in the 4f orbital, which can interact with visible light and result in the formation of coloured compounds. Therefore, the correct answer is: Option D Pr3+