Question

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

Short answer

In conclusion, \(\mathrm{Sc}^{3+}\) compounds are colorless due to the absence of unpaired electrons in the d-orbitals, preventing electronic transitions and the absorption of light. On the other hand, \(\mathrm{Ti}^{3+}\) and \(\mathrm{V}^{3+}\) ions have unpaired electrons in their d-orbitals, allowing electronic transitions to occur upon the formation of complexes. As a result, these compounds can absorb specific wavelengths of light, leading to their characteristic color.

Step-by-step solution

Determine the electronic configurations of Sc, Ti, and V ions

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: \[ \text{V}^{3+} \: (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{2}) \]

Understanding the cause of color in transition metal ion compounds

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.

Comparing electronic configurations and explaining color differences

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.