Among the following the coloured compound is

Options

A. `CuCl`
B. `K_(3)[Cu(CN)_(4)]`
C. `CuF_(2)`
D. `[Cu(CH_(3)(CN)_(4)]BF_(4)`

Explanation

To determine which of the given compounds is colored, we need to analyze the oxidation states of copper in each compound and their electronic configurations. The presence of unpaired d-electrons in transition metal complexes is what typically leads to color due to d-d transitions.

### Step-by-Step Solution:

1. **Identify the Compounds:**
- CuCl
- K3Cu(CN)4
- CuF2
- Cu(CH3CN)4BF4

2. **Analyze CuCl:**
- In CuCl, copper is in the +1 oxidation state (Cu^+).
- The electronic configuration of Cu^+ (which has 29 electrons) is [Ar] 3d^10.
- Since Cu^+ has a full d-subshell (d^10), there are no unpaired electrons, and thus, it cannot undergo d-d transitions.
- **Conclusion:** CuCl is colorless.

3. **Analyze K3Cu(CN)4:**
- In K3Cu(CN)4, copper is also in the +1 oxidation state (Cu^+).
- Similar to CuCl, Cu^+ has a configuration of [Ar] 3d^10.
- Again, there are no unpaired electrons, leading to no d-d transitions.
- **Conclusion:** K3Cu(CN)4 is colorless.

4. **Analyze CuF2:**
- In CuF2, copper is in the +2 oxidation state (Cu^2+).
- The electronic configuration of Cu^2+ (which has 27 electrons) is [Ar] 3d^9.
- Cu^2+ has one unpaired electron in the d-orbital, allowing for d-d transitions.
- **Conclusion:** CuF2 is colored. In crystalline form, it appears blue.

5. **Analyze Cu(CH3CN)4BF4:**
- In Cu(CH3CN)4BF4, copper is again in the +1 oxidation state (Cu^+).
- As previously noted, Cu^+ has a configuration of [Ar] 3d^10.
- There are no unpaired electrons, resulting in no d-d transitions.
- **Conclusion:** Cu(CH3CN)4BF4 is colorless.

6. **Final Conclusion:**
- Among the given compounds, the only colored compound is **CuF2**.