Which ion is colourless:

Options

A. `Cr^(+4)`
B. `Sc^(+3)`
C. `Ti^(+3)`
D. `V^(+3)`

Answer

To determine which ion is colorless among the given options, we need to analyze the electronic configurations of each ion and identify whether they have unpaired electrons or not. Ions with unpaired electrons are typically colored (paramagnetic), while those without unpaired electrons are colorless (diamagnetic).

### Step-by-Step Solution:

1. **Identify the Electronic Configuration of Each Element:**
- Chromium (Cr): Atomic number = 24
- Electronic configuration: [Ar] 3d^5 4s^2
- Scandium (Sc): Atomic number = 21
- Electronic configuration: [Ar] 3d^1 4s^2
- Titanium (Ti): Atomic number = 22
- Electronic configuration: [Ar] 3d^2 4s^2
- Vanadium (V): Atomic number = 23
- Electronic configuration: [Ar] 3d^3 4s^2

2. **Determine the Electronic Configuration of Each Ion:**
- Chromium (Cr^4+):
- Loses 4 electrons: [Ar] 3d^2 (2 unpaired electrons)
- Scandium (Sc^3+):
- Loses 3 electrons: [Ar] 3d^0 (0 unpaired electrons)
- Titanium (Ti^3+):
- Loses 3 electrons: [Ar] 3d^1 (1 unpaired electron)
- Vanadium (V^3+):
- Loses 3 electrons: [Ar] 3d^2 (2 unpaired electrons)

3. **Count Unpaired Electrons:**
- Cr^4+: 2 unpaired electrons (paramagnetic)
- Sc^3+: 0 unpaired electrons (diamagnetic)
- Ti^3+: 1 unpaired electron (paramagnetic)
- V^3+: 2 unpaired electrons (paramagnetic)

4. **Identify the Colorless Ion:**
- The only ion with no unpaired electrons is Sc^3+ (Scandium 3 positive), which is therefore colorless.

### Conclusion:
The correct answer to the question "Which ion is colorless?" is **Scandium 3 positive (Sc^3+)**.

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