Which of the following is a diamagnetic ion?

Options

A. `Co^(2+)`
B. `Cu^(2+)`
C. `Mn^(2+)`
D. `Sc^(3+)`

Explanation

To determine which of the given ions is diamagnetic, we need to analyze the electronic configurations of each ion and check for the presence of unpaired electrons. A diamagnetic substance is one that has all its electrons paired.

### Step-by-Step Solution:

1. **Understand Diamagnetism**:
- Diamagnetic substances have no unpaired electrons. When placed in a magnetic field, they are weakly repelled.

2. **Identify the Ions**:
- We have four ions to consider:
- A: Cobalt (Co²⁺)
- B: Copper (Cu²⁺)
- C: Manganese (Mn²⁺)
- D: Scandium (Sc³⁺)

3. **Determine the Atomic Numbers and Electronic Configurations**:
- **Cobalt (Co)**: Atomic number = 27
- Electronic configuration: [Ar] 3d⁷ 4s²
- For Co²⁺: Remove 2 electrons (from 4s and 3d)
- Co²⁺ configuration: [Ar] 3d⁷
- **Copper (Cu)**: Atomic number = 29
- Electronic configuration: [Ar] 3d¹⁰ 4s¹
- For Cu²⁺: Remove 2 electrons (from 4s and 3d)
- Cu²⁺ configuration: [Ar] 3d⁹
- **Manganese (Mn)**: Atomic number = 25
- Electronic configuration: [Ar] 3d⁵ 4s²
- For Mn²⁺: Remove 2 electrons (from 4s)
- Mn²⁺ configuration: [Ar] 3d⁵
- **Scandium (Sc)**: Atomic number = 21
- Electronic configuration: [Ar] 3d¹ 4s²
- For Sc³⁺: Remove 3 electrons (2 from 4s and 1 from 3d)
- Sc³⁺ configuration: [Ar] 3d⁰

4. **Count Unpaired Electrons**:
- **Co²⁺ (3d⁷)**: 3 unpaired electrons
- **Cu²⁺ (3d⁹)**: 1 unpaired electron
- **Mn²⁺ (3d⁵)**: 5 unpaired electrons
- **Sc³⁺ (3d⁰)**: 0 unpaired electrons

5. **Identify the Diamagnetic Ion**:
- Since Sc³⁺ has 0 unpaired electrons, it is diamagnetic.

### Conclusion:
The diamagnetic ion among the options provided is **D: Scandium 3+**.

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