Generally transition elements form coloured salts due to the presence of unpaired electrons. Which of the following compounds will be coloured in the 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 metals present in each compound and check for the presence of unpaired electrons. Compounds with unpaired electrons typically exhibit color due to d-d transitions.

### Step-by-Step Solution:

1. **Identify the Compounds and Their Transition Metals:**
- The compounds to analyze are:
- AgSO4
- CuF2
- ZnF2
- Cu2Cl2

2. **Analyze AgSO4 (Silver Sulfate):**
- Silver (Ag) in AgSO4 exists as Ag⁺.
- The electronic configuration of Ag is:
\[ \text{Ag: [Kr] 4d}^{10} 5s^1 \]
- When Ag loses one electron to become Ag⁺, it loses the 5s electron:
\[ \text{Ag}^+: [Kr] 4d^{10} \]
- The d-orbital is fully filled (10 electrons), which means all electrons are paired.
- **Conclusion:** AgSO4 is colorless (diamagnetic).

3. **Analyze CuF2 (Copper(II) Fluoride):**
- Copper (Cu) in CuF2 exists as Cu²⁺.
- The electronic configuration of Cu is:
\[ \text{Cu: [Ar] 3d}^{10} 4s^1 \]
- For Cu²⁺, it loses two electrons (one from 4s and one from 3d):
\[ \text{Cu}^{2+}: [Ar] 3d^{9} \]
- There is one unpaired electron in the 3d subshell.
- **Conclusion:** CuF2 is colored due to the presence of unpaired electrons.

4. **Analyze ZnF2 (Zinc Fluoride):**
- Zinc (Zn) in ZnF2 exists as Zn²⁺.
- The electronic configuration of Zn is:
\[ \text{Zn: [Ar] 3d}^{10} 4s^2 \]
- For Zn²⁺, it loses two electrons from the 4s subshell:
\[ \text{Zn}^{2+}: [Ar] 3d^{10} \]
- The d-orbital is fully filled (10 electrons), meaning all electrons are paired.
- **Conclusion:** ZnF2 is colorless (diamagnetic).

5. **Analyze Cu2Cl2 (Copper(I) Chloride):**
- Copper (Cu) in Cu2Cl2 exists as Cu⁺.
- The electronic configuration of Cu is:
\[ \text{Cu: [Ar] 3d}^{10} 4s^1 \]
- For Cu⁺, it loses one electron (from 4s):
\[ \text{Cu}^+: [Ar] 3d^{10} \]
- The d-orbital is fully filled (10 electrons), meaning all electrons are paired.
- **Conclusion:** Cu2Cl2 is colorless (diamagnetic).

### Final Conclusion:
Among the compounds analyzed, only **CuF2** is colored in the solid state due to the presence of unpaired electrons in the Cu²⁺ ion.