Generally transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured?

Options

A. `KMnO_(4)`
B. `Ce(SO_(4))_(2)`
C. `TiCl_(4)`
D. `Cu_(2)Cl_(2)`

Explanation

To determine which of the given compounds are colored, we need to analyze their electronic configurations, particularly focusing on the presence of unpaired electrons and the possibility of d-d transitions or charge transfer transitions.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds we are analyzing are KMnO4, cerium sulfate (Ce(SO4)2), titanium (Ti), and copper(I) (Cu+).

2. **Determine the Oxidation States**:
- **KMnO4**: Manganese (Mn) is in the +7 oxidation state.
- **Cerium Sulfate (Ce(SO4)2)**: Cerium (Ce) is in the +4 oxidation state.
- **Titanium (Ti)**: In this context, we consider titanium in the +4 oxidation state.
- **Copper(I) (Cu+)**: Copper is in the +1 oxidation state.

3. **Write the Electronic Configurations**:
- **KMnO4 (Mn^7+)**: Mn has the electron configuration [Ar] 4s² 3d⁵. In the +7 state, Mn loses all its 4s and 3d electrons, resulting in an electronic configuration of S0 D0 (no unpaired electrons).
- **Cerium Sulfate (Ce^4+)**: Ce has the electron configuration [Xe] 6s² 4f¹⁵. In the +4 state, it loses 4 electrons (2 from 6s and 2 from 4f), resulting in S0 D0 (no unpaired electrons).
- **Titanium (Ti^4+)**: Ti has the electron configuration [Ar] 4s² 3d². In the +4 state, it loses 4 electrons (2 from 4s and 2 from 3d), resulting in S0 D0 (no unpaired electrons).
- **Copper(I) (Cu^+)**: Cu has the electron configuration [Ar] 4s² 3d¹⁰. In the +1 state, it loses 1 electron (from 4s), resulting in an electronic configuration of S0 D10 (all paired electrons).

4. **Analyze the Possibility of Color**:
- **d-d Transitions**: For a compound to be colored due to d-d transitions, it must have unpaired electrons in its d-orbitals. Since all the compounds analyzed have either empty d-orbitals or fully filled d-orbitals, d-d transitions are not possible.
- **Charge Transfer Transitions**: Charge transfer can occur when there is a transfer of an electron between the metal and the ligand. KMnO4 and cerium sulfate are known to exhibit charge transfer transitions, which can lead to color.

5. **Conclusion**: Based on the analysis:
- **Colored Compounds**: KMnO4 and cerium sulfate are colored due to charge transfer transitions.
- **Non-Colored Compounds**: Titanium and copper(I) do not exhibit color due to the absence of unpaired electrons and d-d transitions.

### Final Answer:
The colored compounds among the given options are **KMnO4 and cerium sulfate**.