Which of the following pairs are not both coloured in aqueous solution?

Options

A. `Sc^(3+), CO^(3+)`
B. `Ni^(2+), Cu^(2+)`
C. `Ni^(2+), Ti^(3+)`
D. `Sc^(3+), Ti^(4+)`

Answer

To solve the question of which pairs are not both colored in aqueous solution, we need to analyze the electronic configurations of the given ions and determine the presence of unpaired electrons, as unpaired electrons are responsible for the color in transition metal complexes.

### Step-by-Step Solution:

1. **Identify the Ions**: The ions we need to analyze are Scandium (Sc³⁺), Cobalt (Co³⁺), Nickel (Ni²⁺), Copper (Cu²⁺), Titanium (Ti³⁺), and Titanium (Ti⁴⁺).

2. **Determine Electronic Configurations**:
- **Scandium (Sc)**: Atomic number 21 → Configuration: [Ar] 3d¹ 4s²
- Sc³⁺: Loses 3 electrons → [Ar] 3d⁰ 4s⁰ (Colorless, no unpaired electrons)
- **Cobalt (Co)**: Atomic number 27 → Configuration: [Ar] 3d⁷ 4s²
- Co³⁺: Loses 3 electrons → [Ar] 3d⁶ 4s⁰ (Colored, has unpaired electrons)
- **Nickel (Ni)**: Atomic number 28 → Configuration: [Ar] 3d⁸ 4s²
- Ni²⁺: Loses 2 electrons → [Ar] 3d⁸ 4s⁰ (Colored, has unpaired electrons)
- **Copper (Cu)**: Atomic number 29 → Configuration: [Ar] 3d¹⁰ 4s¹
- Cu²⁺: Loses 2 electrons → [Ar] 3d⁹ 4s⁰ (Colored, has unpaired electrons)
- **Titanium (Ti)**: Atomic number 22 → Configuration: [Ar] 3d² 4s²
- Ti³⁺: Loses 3 electrons → [Ar] 3d¹ 4s⁰ (Colored, has unpaired electrons)
- Ti⁴⁺: Loses 4 electrons → [Ar] 3d⁰ 4s⁰ (Colorless, no unpaired electrons)

3. **Analyze the Pairs**:
- **Pair 1**: Sc³⁺ (Colorless) and Co³⁺ (Colored) → Not both colored
- **Pair 2**: Ni²⁺ (Colored) and Cu²⁺ (Colored) → Both colored
- **Pair 3**: Ti³⁺ (Colored) and Ti⁴⁺ (Colorless) → Not both colored
- **Pair 4**: Co³⁺ (Colored) and Ti⁴⁺ (Colorless) → Not both colored

4. **Conclusion**: The pairs that are not both colored in aqueous solution are:
- Sc³⁺ and Co³⁺
- Ti³⁺ and Ti⁴⁺
- Co³⁺ and Ti⁴⁺

Thus, the answer to the question is **D option: Scandium 3+ and Titanium 4+**.