Which is colourless in `H_2O` ?

Options

A. `Ti^(3+)`
B. `V^(3+)`
C. `Cr^(3+)`
D. `Sc^(3+)`

Explanation

To determine which ion is colorless in water among Titanium (Ti), Vanadium (V), Chromium (Cr), and Scandium (Sc), we need to analyze the electronic configurations of their respective ions in the +3 oxidation state.

### Step-by-Step Solution:

1. **Identify the Atomic Numbers and Electronic Configurations:**
- Titanium (Ti): Atomic number 22, electronic configuration: [Ar] 3d² 4s²
- Vanadium (V): Atomic number 23, electronic configuration: [Ar] 3d³ 4s²
- Chromium (Cr): Atomic number 24, electronic configuration: [Ar] 3d⁵ 4s¹
- Scandium (Sc): Atomic number 21, electronic configuration: [Ar] 3d¹ 4s²

2. **Determine the Electronic Configuration of Each Ion in +3 State:**
- For Ti³⁺: Remove 3 electrons (2 from 4s and 1 from 3d)
- Configuration: [Ar] 3d¹ 4s⁰
- For V³⁺: Remove 3 electrons (2 from 4s and 1 from 3d)
- Configuration: [Ar] 3d² 4s⁰
- For Cr³⁺: Remove 3 electrons (1 from 4s and 2 from 3d)
- Configuration: [Ar] 3d³ 4s⁰
- For Sc³⁺: Remove 3 electrons (2 from 4s and 1 from 3d)
- Configuration: [Ar] 3d⁰ 4s⁰

3. **Analyze the Number of Unpaired Electrons:**
- Ti³⁺: 1 unpaired electron (3d¹)
- V³⁺: 2 unpaired electrons (3d²)
- Cr³⁺: 3 unpaired electrons (3d³)
- Sc³⁺: 0 unpaired electrons (3d⁰)

4. **Determine Color in Aqueous Solution:**
- Ions with unpaired electrons typically exhibit color due to d-d transitions.
- Ti³⁺, V³⁺, and Cr³⁺ have unpaired electrons and will show color in solution.
- Sc³⁺ has no unpaired electrons and will not show color in solution.

5. **Conclusion:**
- The ion that is colorless in water is **Scandium (Sc³⁺)**.