Which of the following set has all the ions coloured ?

Options

A. `Cu^(2+),Fe^(2+), Co^(2+)`
B. `Cu^(2+), Sc^(3+), Mg^(2+)`
C. `Al^(3+), Zn^(2+), Co^(2+)`
D. `Fe^(3+), Sc^(3+), Ni^(2+)`

Answer

To determine which set of ions has all colored ions, we need to analyze the electronic configurations of the given ions and check for the presence of unpaired electrons. Colored ions typically have unpaired electrons in their d-orbitals.

### Step-by-Step Solution:

1. **Identify the Ions:**
Let's consider the ions provided in the options. We will analyze their electron configurations and count the number of unpaired electrons.

2. **Copper (Cu²⁺):**
- Atomic number of Copper (Cu) = 29
- Electron configuration of Cu = [Ar] 3d¹⁰ 4s¹
- For Cu²⁺, we remove two electrons:
- Cu²⁺ = [Ar] 3d⁹
- In 3d⁹, the filling is:
- 1, 2, 3, 4, 5, 6, 7, 8, 9 (1 unpaired electron)
- **Unpaired Electrons: 1**

3. **Iron (Fe²⁺):**
- Atomic number of Iron (Fe) = 26
- Electron configuration of Fe = [Ar] 3d⁶ 4s²
- For Fe²⁺, we remove two electrons:
- Fe²⁺ = [Ar] 3d⁶
- In 3d⁶, the filling is:
- 1, 2, 3, 4, 5, 6 (4 unpaired electrons)
- **Unpaired Electrons: 4**

4. **Cobalt (Co²⁺):**
- Atomic number of Cobalt (Co) = 27
- Electron configuration of Co = [Ar] 3d⁷ 4s²
- For Co²⁺, we remove two electrons:
- Co²⁺ = [Ar] 3d⁷
- In 3d⁷, the filling is:
- 1, 2, 3, 4, 5, 6, 7 (3 unpaired electrons)
- **Unpaired Electrons: 3**

5. **Scandium (Sc³⁺):**
- Atomic number of Scandium (Sc) = 21
- Electron configuration of Sc = [Ar] 3d¹ 4s²
- For Sc³⁺, we remove three electrons:
- Sc³⁺ = [Ar] 3d⁰
- **Unpaired Electrons: 0** (No color)

6. **Aluminum (Al³⁺):**
- Atomic number of Aluminum (Al) = 13
- Electron configuration of Al = [Ne] 3s² 3p¹
- For Al³⁺, we remove three electrons:
- Al³⁺ = [Ne] (fully filled)
- **Unpaired Electrons: 0** (No color)

7. **Zinc (Zn²⁺):**
- Atomic number of Zinc (Zn) = 30
- Electron configuration of Zn = [Ar] 3d¹⁰ 4s²
- For Zn²⁺, we remove two electrons:
- Zn²⁺ = [Ar] 3d¹⁰
- **Unpaired Electrons: 0** (No color)

### Conclusion:
From the analysis, we find that:
- Cu²⁺ has 1 unpaired electron (colored)
- Fe²⁺ has 4 unpaired electrons (colored)
- Co²⁺ has 3 unpaired electrons (colored)
- Sc³⁺ has 0 unpaired electrons (no color)
- Al³⁺ has 0 unpaired electrons (no color)
- Zn²⁺ has 0 unpaired electrons (no color)

Thus, the only set where all ions are colored is **Copper (Cu²⁺), Iron (Fe²⁺), and Cobalt (Co²⁺)**.

### Final Answer:
**Option 1: Copper (Cu²⁺), Iron (Fe²⁺), Cobalt (Co²⁺)** is the correct answer.