`CuSO_(4)` is colourless while `CuSO_(4) 5H_(2)O` is coloured. Why?
`[Ti(H_(2)O)_(6)]^(3+)` is coloured while `[Sc(H_(2)O)_(6)]^(3+)` is colourless, why?

Explanation

To understand why `CuSO4` is colorless while `CuSO4·5H2O` is colored, and why `[Ti(H2O)6]^(3+)` is colored while `[Sc(H2O)6]^(3+)` is colorless, we need to analyze the electronic configurations and the role of ligands in these compounds.

### Step-by-Step Solution:

1. **Copper Sulfate (`CuSO4`) vs. Hydrated Copper Sulfate (`CuSO4·5H2O`)**:
- **Electronic Configuration of Copper**: Copper (Cu) has an atomic number of 29. Its electronic configuration is `[Ar] 3d^{10} 4s^1`. In its +2 oxidation state (as in `CuSO4`), the configuration becomes `3d^9`.
- **Colorless Nature of `CuSO4`**: In `CuSO4`, there are no water molecules acting as ligands. The absence of ligands means there is no splitting of the d-orbitals, and thus, no d-d transitions can occur. Therefore, `CuSO4` is colorless.
- **Colored Nature of `CuSO4·5H2O`**: In `CuSO4·5H2O`, the five water molecules act as ligands. The presence of these ligands splits the d-orbitals (due to ligand field theory), allowing for d-d transitions. The `Cu^2+` ion in this hydrated form has one unpaired electron, which results in the blue color of `CuSO4·5H2O`.
...