Generally transition elements from coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in solid state?

Options

A. `Ag_(2)SO_(4)`
B. `CuF_(2)`
C. `ZnF_(2)`
D. `Cu_(2)Cl_(2)`

Explanation

To determine which of the given compounds will be colored in the solid state, we need to analyze the electronic configurations of the transition metal ions present in each compound. Transition elements typically exhibit color due to the presence of unpaired electrons in their d-orbitals.

### Step-by-Step Solution:

1. **Identify the Compounds**: The compounds we need to analyze are Ag2SO4, CuSO4, ZnF2, and CuF2.

2. **Analyze Ag2SO4**:
- The silver ion (Ag⁺) has an electronic configuration of [Kr] 4d10 5s0.
- Since all electrons in the d-orbitals are paired (10 electrons in the 4d subshell), Ag⁺ has no unpaired electrons.
- **Conclusion**: Ag2SO4 is colorless in the solid state.

3. **Analyze CuSO4**:
- The copper ion (Cu²⁺) has an electronic configuration of [Ar] 3d9 4s0.
- In this case, there are 9 electrons in the d-orbitals, which means there is one unpaired electron.
- **Conclusion**: CuSO4 is colored in the solid state due to the presence of unpaired electrons.

4. **Analyze ZnF2**:
- The zinc ion (Zn²⁺) has an electronic configuration of [Ar] 3d10 4s0.
- All 10 electrons in the d-orbitals are paired, meaning there are no unpaired electrons.
- **Conclusion**: ZnF2 is colorless in the solid state.

5. **Analyze CuF2**:
- The copper ion (Cu⁺) has an electronic configuration of [Ar] 3d10 4s0.
- Similar to Zn²⁺, all 10 electrons in the d-orbitals are paired, indicating no unpaired electrons.
- **Conclusion**: CuF2 is colorless in the solid state.

### Final Answer:
The only compound that will be colored in the solid state is **CuSO4** due to the presence of unpaired electrons.