Colour of `La^(2+)` is due to

Options

A. d-d transition.
B. f-f transition.
C. charge transfer
D. None of these.

Answer

To determine the color of \( \text{La}^{2+} \), we need to analyze the electronic configuration and the factors that contribute to the color of transition metal ions.

### Step-by-Step Solution:

1. **Identify the oxidation state of Lanthanum**:
- Lanthanum (La) has an atomic number of 57. In its elemental form, it has the electronic configuration of \([Xe] 6s^2\).
- When it loses two electrons to form \( \text{La}^{2+} \), the configuration becomes \([Xe]\) (removing the two 6s electrons).

2. **Determine the presence of unpaired electrons**:
- The electronic configuration of \( \text{La}^{2+} \) is \([Xe]\), which means it has no unpaired electrons in the f-orbitals (since the f-orbitals are not occupied in this case).
- The absence of unpaired electrons suggests that there are no d-d transitions or f-f transitions that could lead to color.

3. **Consider the possible transitions**:
- The color of a compound is often due to electronic transitions. In transition metals, these transitions can be d-d transitions or charge transfer transitions.
- In lanthanides, the color is typically due to f-f transitions when there are unpaired electrons in the f-orbitals. However, \( \text{La}^{2+} \) does not have unpaired electrons.

4. **Conclusion on the color of \( \text{La}^{2+} \)**:
- Since \( \text{La}^{2+} \) has no unpaired electrons and does not participate in f-f transitions, it does not exhibit color in the visible spectrum.
- Therefore, the color of \( \text{La}^{2+} \) is due to the absence of electronic transitions that would result in color, which can be interpreted as "none of these" in the context of the given options.

### Final Answer:
The color of \( \text{La}^{2+} \) is due to **none of these** (option 4).