Which of the following is coloured compound?

Options

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

Answer

To determine which of the following compounds is colored, we need to analyze the electronic configurations of the metal ions in each compound and check for the presence of unpaired electrons. Unpaired electrons are responsible for the absorption of visible light, leading to color.

### Step-by-Step Solution:

1. **Analyze CuF2:**
- Copper (Cu) in CuF2 exists in the +2 oxidation state (Cu²⁺).
- The electronic configuration of neutral copper is:
\[ [Ar] 4s^1 3d^{10} \]
- Upon losing two electrons to become Cu²⁺, the configuration becomes:
\[ [Ar] 3d^9 \]
- In the 3d subshell, there are 9 electrons. When these electrons are arranged, we find that there is one unpaired electron in the 3d orbitals.
- Since there is an unpaired electron, CuF2 can absorb visible light and is therefore colored.

2. **Analyze CuI:**
- In CuI, copper exists in the +1 oxidation state (Cu⁺).
- The electronic configuration for Cu⁺ is:
\[ [Ar] 4s^0 3d^{10} \]
- Here, all 10 electrons in the 3d subshell are paired.
- Since there are no unpaired electrons, CuI does not absorb visible light and is colorless.

3. **Analyze NaCl:**
- Sodium (Na) in NaCl exists in the +1 oxidation state (Na⁺).
- The electronic configuration for Na⁺ is:
\[ [Ne] 3s^0 \]
- All electrons are paired, and there are no unpaired electrons.
- Therefore, NaCl is colorless.

4. **Analyze MgCl2:**
- Magnesium (Mg) in MgCl2 exists in the +2 oxidation state (Mg²⁺).
- The electronic configuration for Mg²⁺ is:
\[ [Ne] 3s^0 \]
- Again, all electrons are paired, and there are no unpaired electrons.
- Thus, MgCl2 is also colorless.

### Conclusion:
The only compound among the options that is colored is **CuF2** due to the presence of an unpaired electron in the Cu²⁺ ion.

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

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