Which of the following gives colour with the water :

Options

A. `Cu^(+)`
B. `Cr^(3+)`
C. `Na^(+)`
D. None

Answer

To solve the question "Which of the following gives color with water?" we will analyze each of the given options: Copper (I) ion, Chromium (III) ion, Sodium ion, and None.

### Step-by-Step Solution:

1. **Analyze Copper (I) Ion (Cu⁺)**:
- Copper (Cu) is a d-block element with the electronic configuration of [Ar] 4s¹ 3d¹⁰.
- For Cu⁺, the configuration becomes [Ar] 3d¹⁰ (since one electron is removed from the 4s orbital).
- In this configuration, there are no unpaired electrons and no empty d-orbitals available for d-d transitions.
- **Conclusion**: Cu⁺ does not exhibit color in solution as there are no d-d transitions possible.

2. **Analyze Chromium (III) Ion (Cr³⁺)**:
- Chromium (Cr) is also a d-block element with the electronic configuration of [Ar] 4s¹ 3d⁵.
- For Cr³⁺, the configuration becomes [Ar] 3d³ (after losing three electrons).
- In this case, there are unpaired electrons in the 3d orbitals, allowing for d-d transitions.
- **Conclusion**: Cr³⁺ can exhibit color in solution due to possible d-d transitions.

3. **Analyze Sodium Ion (Na⁺)**:
- Sodium (Na) has the electronic configuration of [Ne] 3s¹.
- For Na⁺, the configuration becomes [Ne] (after losing one electron from the 3s orbital).
- The resulting configuration has no d-orbitals, and the transitions that could occur (like from 2p to 3s) require a very high amount of energy.
- **Conclusion**: Na⁺ does not exhibit color in solution as the transitions do not fall within the visible spectrum.

4. **Final Conclusion**:
- Among the options provided, only Chromium (III) ion (Cr³⁺) gives color when dissolved in water due to the presence of unpaired electrons and possible d-d transitions.

### Answer:
The correct answer is **Chromium (III) ion (Cr³⁺)**.