Which of the following ion has lowest magnetic moment?

Options

A. `Fe^(2+)`
B. `Mn^(2+)`
C. `Cr^(3+)`
D. `V^(3+)`

Answer

To determine which of the given ions has the lowest magnetic moment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Magnetic Moment**:
The magnetic moment of an ion is determined by the number of unpaired electrons in its electronic configuration. The formula for calculating the magnetic moment (μ) in Bohr magnetons is:
\[ μ = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

2. **Identify the Electronic Configurations**:
We need to find the electronic configurations of the ions provided in the question. The ions we will analyze are:
- Fe²⁺
- Mn²⁺
- Cr³⁺
- V³⁺

3. **Calculate for Each Ion**:
- **Fe²⁺**:
- Neutral Fe: [Ar] 3d⁶ 4s²
- Fe²⁺: [Ar] 3d⁶ (removing 2 electrons from 4s)
- Distribution in 3d: ↑↓ ↑↓ ↑↓ ↑ ↑ (4 unpaired electrons)
- \( n = 4 \)
- Magnetic moment:
\[ μ = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.89 \text{ Bohr magnetons} \]

- **Mn²⁺**:
- Neutral Mn: [Ar] 3d⁷ 4s²
- Mn²⁺: [Ar] 3d⁵ (removing 2 electrons)
- Distribution in 3d: ↑ ↑ ↑ ↑ ↑ (5 unpaired electrons)
- \( n = 5 \)
- Magnetic moment:
\[ μ = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.91 \text{ Bohr magnetons} \]

- **Cr³⁺**:
- Neutral Cr: [Ar] 3d⁵ 4s¹
- Cr³⁺: [Ar] 3d³ (removing 3 electrons)
- Distribution in 3d: ↑ ↑ ↑ (3 unpaired electrons)
- \( n = 3 \)
- Magnetic moment:
\[ μ = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \text{ Bohr magnetons} \]

- **V³⁺**:
- Neutral V: [Ar] 3d³ 4s²
- V³⁺: [Ar] 3d² (removing 3 electrons)
- Distribution in 3d: ↑ ↑ (2 unpaired electrons)
- \( n = 2 \)
- Magnetic moment:
\[ μ = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83 \text{ Bohr magnetons} \]

4. **Compare the Magnetic Moments**:
- Fe²⁺: 4.89 Bohr magnetons
- Mn²⁺: 5.91 Bohr magnetons
- Cr³⁺: 3.87 Bohr magnetons
- V³⁺: 2.83 Bohr magnetons

5. **Conclusion**:
The ion with the lowest magnetic moment is **V³⁺** with a magnetic moment of approximately 2.83 Bohr magnetons.

### Final Answer:
The ion with the lowest magnetic moment is **V³⁺**.