Compounds of \(\mathrm{Sc}^{3+}\) are not colored, but those of \(\mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) are. Why?

First, we need to find the electronic configurations of Sc, Ti, and V ions in their respective +3 oxidation states. For Scandium (Sc, atomic number 21): Neutral Sc atom has the electronic configuration: \[ \text{Sc} \: (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{1}) \] Upon losing three electrons to form \(\mathrm{Sc}^{3+}\), it becomes: \[ \text{Sc}^{3+} \: (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}) \] For Titanium (Ti, atomic number 22): Neutral Ti atom has the electronic configuration: \[ \text{Ti} \: (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{2}) \] Upon losing three electrons to form \(\mathrm{Ti}^{3+}\), it becomes: \[ \text{Ti}^{3+} \: (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{1}) \] For Vanadium (V, atomic number 23): Neutral V atom has the electronic configuration: \[ \text{V} \: (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{3}) \] Upon losing three electrons to form \(\mathrm{V}^{3+}\), it becomes: \[ \text{V}^{3+} \: (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{2}) \]

Transition metal ion compounds are often colored due to the presence of partially-filled d-orbitals. When light hits the compound, the energy from the absorbed light promotes an electron from a lower-energy d-orbital to a higher energy d-orbital. Since these orbitals have different energies, only specific wavelengths of light are absorbed, and the remaining wavelengths are reflected or transmitted, giving us the observed color.

Now that we have the electronic configurations for \(\mathrm{Sc}^{3+}\), \(\mathrm{Ti}^{3+}\), and \(\mathrm{V}^{3+}\), we can compare them and explain the difference in color. ① \(\mathrm{Sc}^{3+}\): The electronic configuration is \( (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}) \), which shows that all its d-orbitals are empty. Thus, there are no available electronic transitions within the d-orbitals that can absorb light, and the compound appears colorless. ② \(\mathrm{Ti}^{3+}\): The electronic configuration is \( (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{1}) \), which means it has one unpaired electron in the 3d subshell. When this ion forms a complex, electronic transitions between different energy d-orbitals are possible. These transitions absorb specific wavelengths of light, giving the compound its characteristic color. ③ \(\mathrm{V}^{3+}\): The electronic configuration is \( (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{2}) \), which shows that it also has partially-filled 3d orbitals. Similar to \(\mathrm{Ti}^{3+}\), electronic transitions within the d-orbitals can occur upon the formation of a complex, resulting in the absorption of light and a colored appearance. In conclusion, Scandium compounds are colorless because \(\mathrm{Sc}^{3+}\) has no unpaired electrons in its d-orbitals, which means no electronic transitions can occur. In contrast, \(\mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) ions have unpaired electrons in their d-orbitals, and electronic transitions within these orbitals can lead to the absorption of specific wavelengths of light and the resulting color in their compounds.