Which of the following is coloured compound?

Options

A. `CuF_(2)`
B. `CuI`
C. `NaCl`
D. `MgCl_(2)`

Answer

To determine which of the given compounds is a colored compound, we need to analyze the electronic configurations of the metal ions in each compound and check for the presence of unpaired electrons. Colored compounds typically arise from the presence of unpaired d-electrons that can undergo electronic transitions when exposed to visible light.

### Step-by-Step Solution:

1. **Identify the Compounds**: We are given several compounds, including CuF2, CuI, NaCl, and MgCl2.

2. **Determine the Oxidation States**:
- For CuF2, copper (Cu) is in the +2 oxidation state.
- For CuI, copper (Cu) is in the +1 oxidation state.
- For NaCl, sodium (Na) is in the +1 oxidation state, and chlorine (Cl) is in the -1 oxidation state.
- For MgCl2, magnesium (Mg) is in the +2 oxidation state.

3. **Electronic Configurations**:
- **Cu^2+ (from CuF2)**: The electronic configuration of Cu is [Ar] 3d^10 4s^1. For Cu^2+, it loses two electrons, resulting in [Ar] 3d^9. This means there is one unpaired electron in the 3d orbital.
- **Cu^+ (from CuI)**: The electronic configuration for Cu^+ is [Ar] 3d^10. All electrons are paired, so there are no unpaired electrons.
- **Na+ (from NaCl)**: The electronic configuration for Na is [Ne] 3s^1. For Na^+, it loses one electron, resulting in [Ne]. All electrons are paired.
- **Mg^2+ (from MgCl2)**: The electronic configuration for Mg is [Ne] 3s^2. For Mg^2+, it loses two electrons, resulting in [Ne]. All electrons are paired.

4. **Check for Unpaired Electrons**:
- **Cu^2+ (CuF2)**: Has 1 unpaired electron (colored).
- **Cu^+ (CuI)**: Has 0 unpaired electrons (not colored).
- **Na+ (NaCl)**: Has 0 unpaired electrons (not colored).
- **Mg^2+ (MgCl2)**: Has 0 unpaired electrons (not colored).

5. **Conclusion**: The only compound with unpaired electrons is CuF2, which means it is the colored compound.

### Final Answer:
**CuF2 is the colored compound.**