The color of the aqueous solutions of transition metal cations is generally due to electronic transitions among the d-orbitals as a consequence of ligand field splitting. Cations can absorb light of certain energies resulting in electronic transitions, which is what imparts color to their solutions. Let's consider each of the given options: Option A:Ti3+ (Titanium in +3 oxidation state) has one d electron ( 3d1 configuration). This cation can undergo d−d transitions and is not colorless; it usually appears violet in aqueous solutions. Option B:V3+ (Vanadium in +3 oxidation state) has two d electrons ( 3d2 configuration). Like titanium, vanadium can also undergo d-d transitions and is typically green or blue in solutions, so it is not colorless either. Option C : Sc3+ (Scandium in +3 oxidation state) has an electronic configuration of 3d0 after it has lost its three valence electrons. Since it has no d electrons, it cannot have d-d transitions and thus generally forms colorless solutions in water. Scandium is the correct answer to the question. Option D:Cu2+ (Copper in +2 oxidation state) has one d electron ( 3d9 configuration). It absorbs light in the red region, which makes its aqueous solution appear blue, so it is definitely not colorless. Therefore, the cation that produces a colorless aqueous solution in its respective oxidation state is: Option C: Sc3+.