Out of the following ions `Ti^(3+), V^(3+), Cu^(+), Sc^(3+), Mn^(3+) and Co^(2+)` the colourless ions will be

Options

A. `Cu^(+), Sc^(3+)`
B. `Ti^(3+), V^(3+)`
C. `Cu^(+), Co^(2+)`
D. `Sc^(3+), Fe^(3+)`

Answer

To determine which of the given ions are colorless, we need to analyze their electronic configurations and the presence of d-electrons, as the color of transition metal ions is primarily due to d-d transitions. Here’s a step-by-step solution:

### Step 1: Identify the electronic configurations of the ions
We will write the electronic configurations for each of the ions provided in the question:

1. **Ti^(3+)**: The atomic number of Titanium (Ti) is 22. The electronic configuration of Ti is [Ar] 3d² 4s². For Ti^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in:
- **Ti^(3+)**: [Ar] 3d¹

2. **V^(3+)**: The atomic number of Vanadium (V) is 23. The electronic configuration of V is [Ar] 3d³ 4s². For V^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in:
- **V^(3+)**: [Ar] 3d²

3. **Cu^(+)**: The atomic number of Copper (Cu) is 29. The electronic configuration of Cu is [Ar] 3d¹⁰ 4s¹. For Cu^(+), we remove 1 electron (from 4s), resulting in:
- **Cu^(+): [Ar] 3d¹⁰**

4. **Sc^(3+)**: The atomic number of Scandium (Sc) is 21. The electronic configuration of Sc is [Ar] 3d¹ 4s². For Sc^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in:
- **Sc^(3+)**: [Ar] 3d⁰

5. **Mn^(3+)**: The atomic number of Manganese (Mn) is 25. The electronic configuration of Mn is [Ar] 3d⁵ 4s². For Mn^(3+), we remove 3 electrons (2 from 4s and 1 from 3d), resulting in:
- **Mn^(3+)**: [Ar] 3d⁴

6. **Co^(2+)**: The atomic number of Cobalt (Co) is 27. The electronic configuration of Co is [Ar] 3d⁷ 4s². For Co^(2+), we remove 2 electrons (2 from 4s), resulting in:
- **Co^(2+)**: [Ar] 3d⁷

### Step 2: Determine the presence of d-electrons
Now, we will analyze the presence of d-electrons in each ion:

- **Ti^(3+)**: 1 d-electron (colored)
- **V^(3+)**: 2 d-electrons (colored)
- **Cu^(+)**: 10 d-electrons (colorless, fully filled)
- **Sc^(3+)**: 0 d-electrons (colorless, empty)
- **Mn^(3+)**: 4 d-electrons (colored)
- **Co^(2+)**: 7 d-electrons (colored)

### Step 3: Identify colorless ions
From the analysis, the colorless ions are those with either a fully filled d-orbital or an empty d-orbital.

- **Colorless ions**:
- **Cu^(+)**: 3d¹⁰ (fully filled)
- **Sc^(3+)**: 3d⁰ (empty)

### Conclusion
The colorless ions from the given list are **Cu^(+)** and **Sc^(3+)**.

### Final Answer
The colorless ions are: **Cu^(+) and Sc^(3+)**.

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