Question

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

Short answer

Compounds of Sc^3+ are colorless because they have no electrons in their 3d orbitals, preventing electron transitions between different energy levels and absorption of visible light. However, Ti^3+ and V^3+ compounds are colored because they have electrons in their 3d orbitals, allowing electron transitions between energy levels and absorption of visible light.

Step-by-step solution

Scandium (Sc)

Scandium has an atomic number of 21, and its electron configuration is \([Ar]3d^14s^2\). When it forms a +3 ion, three electrons are lost from the 3d and 4s orbitals, giving Sc a \([Ar]3d^0\) configuration.

Titanium (Ti)

Titanium has an atomic number of 22, so its electron configuration is \([Ar]3d^24s^2\). When it forms a +3 ion, three electrons are lost from the 3d and 4s orbitals, giving Ti a \([Ar]3d^1\) configuration.

Vanadium (V)

Vanadium has an atomic number of 23, and its electron configuration is \([Ar]3d^34s^2\). When it forms a +3 ion, three electrons are lost from the 3d and 4s orbitals, giving V a \([Ar]3d^2\) configuration.

Visualizing difference in energy levels between ions

In Sc^3+ ions, there are no electrons in the 3d orbital, which means there is no possibility of electron transitions between the lower energy levels and the higher energy levels of its d orbitals. Therefore, no visible light is absorbed, and Sc^3+ compounds appear colorless. However, Ti^3+ and V^3+ ions have electrons in their 3d orbitals, so electron transitions can occur between different energy levels of the 3d orbitals. These transitions absorb visible light, which results in colored compounds. In summary, compounds of Sc^3+ are colorless because there are no electrons in their 3d orbitals, while Ti^3+ and V^3+ compounds are colored due to electron transitions in their 3d orbitals.