The aqua ion which will be almost colorless is-

Options

A. `Cu^(2+)`
B. `Cr^(2+)`
C. `Ti^(4+)`
D. `Mn^(2+)`

Answer

To determine which aqua ion is almost colorless, we need to analyze the electronic configurations of the given ions and check for the presence of unpaired electrons. The presence of unpaired electrons in the d-orbitals can lead to d-d transitions, which result in color. If all electrons are paired, the ion will be colorless.

### Step-by-Step Solution:

1. **Understand the Concept of Aqua Ions**:
Aqua ions are metal ions that are surrounded by water molecules acting as ligands. The color of these ions depends on the electronic configuration and the presence of unpaired electrons.

**Hint**: Remember that unpaired electrons can lead to color due to d-d transitions.

2. **Identify the Ions**:
The ions we are considering are Cu²⁺, Cr²⁺, Ti⁴⁺, and Mn²⁺.

3. **Write the Electronic Configuration**:
- **Cu²⁺**:
- Electronic configuration: 3d⁹ 4s⁰
- Unpaired electrons: 1 (since there are 9 electrons in 3d, one will be unpaired)
- **Cr²⁺**:
- Electronic configuration: 3d⁴ 4s⁰
- Unpaired electrons: 4 (since there are 4 electrons in 3d, all will be unpaired)
- **Ti⁴⁺**:
- Electronic configuration: 3d⁰ 4s⁰
- Unpaired electrons: 0 (no electrons in 3d)
- **Mn²⁺**:
- Electronic configuration: 3d⁵ 4s⁰
- Unpaired electrons: 5 (since there are 5 electrons in 3d, all will be unpaired)

**Hint**: Focus on counting the number of unpaired electrons in each configuration.

4. **Determine Color Based on Unpaired Electrons**:
- **Cu²⁺**: Colorful (1 unpaired electron)
- **Cr²⁺**: Colorful (4 unpaired electrons)
- **Ti⁴⁺**: Colorless (0 unpaired electrons)
- **Mn²⁺**: Colorful (5 unpaired electrons)

**Hint**: Recall that ions with no unpaired electrons will not exhibit color.

5. **Conclusion**:
The aqua ion that is almost colorless is **Ti⁴⁺** since it has no unpaired electrons, and therefore, d-d transitions cannot occur.

**Final Answer**: Ti⁴⁺ is the aqua ion that will be almost colorless.