Answer:-

Salts of Sc^3+, Ti^4+, and V^5+ ions are colorless because of their electronic configurations and the absence of unpaired electrons in their d orbitals.

1. Scandium (Sc^3+): Scandium has the electron configuration [Ar] 3d^1 4s^2 when it forms the Sc^3+ ion. This ion loses three electrons from the 3d orbital, leaving it with an empty 3d orbital. Since there are no unpaired electrons in the d orbitals, Sc^3+ ions do not absorb visible light, resulting in a colorless appearance.

2. Titanium (Ti^4+): Titanium forms the Ti^4+ ion by losing all four of its valence electrons. The electron configuration of Ti^4+ is [Ar] 3d^0 4s^0. In this configuration, all the d orbitals are completely filled, and there are no unpaired electrons. As a result, Ti^4+ ions are also colorless.

3. Vanadium (V^5+): Vanadium forms the V^5+ ion by losing all five of its valence electrons. The electron configuration of V^5+ is [Ar] 3d^0 4s^0. Similar to Ti^4+, all the d orbitals in V^5+ are fully occupied, and there are no unpaired electrons, making V^5+ ions colorless.

In general, the color of a transition metal ion or compound is often related to the presence of unpaired electrons in the d orbitals, which can absorb specific wavelengths of visible light and give the compound its characteristic color. When there are no unpaired electrons in the d orbitals, as is the case with Sc^3+, Ti^4+, and V^5+, the compound appears colorless because it does not absorb visible light in the visible spectrum.