The ion with highest magnetic moment is

Options

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

Answer

To determine which ion has the highest magnetic moment, we need to calculate the magnetic moment for each ion based on the number of unpaired electrons. The formula for calculating the magnetic moment (μ) is:

\[ \mu = \sqrt{n(n + 2)} \]

where \( n \) is the number of unpaired electrons.

### Step-by-Step Solution:

1. **Identify the Ions**: We will consider the following ions: V³⁺, Cr³⁺, Mn³⁺, and Co³⁺.

2. **Determine the Electronic Configuration**:
- **V³⁺ (Vanadium)**:
- Atomic number of V = 23.
- Electronic configuration: [Ar] 4s² 3d³.
- For V³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
- Configuration: [Ar] 3d².
- Unpaired electrons: 2.

- **Cr³⁺ (Chromium)**:
- Atomic number of Cr = 24.
- Electronic configuration: [Ar] 4s² 3d⁵.
- For Cr³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
- Configuration: [Ar] 3d³.
- Unpaired electrons: 3.

- **Mn³⁺ (Manganese)**:
- Atomic number of Mn = 25.
- Electronic configuration: [Ar] 4s² 3d⁵.
- For Mn³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
- Configuration: [Ar] 3d⁴.
- Unpaired electrons: 4.

- **Co³⁺ (Cobalt)**:
- Atomic number of Co = 27.
- Electronic configuration: [Ar] 4s² 3d⁷.
- For Co³⁺, we remove 3 electrons (2 from 4s and 1 from 3d):
- Configuration: [Ar] 3d⁶.
- Unpaired electrons: 4.

3. **Calculate the Magnetic Moment for Each Ion**:
- **V³⁺**:
\[ \mu = \sqrt{2(2 + 2)} = \sqrt{8} \approx 2.83 \]

- **Cr³⁺**:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{15} \approx 3.87 \]

- **Mn³⁺**:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.90 \]

- **Co³⁺**:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.90 \]

4. **Compare the Magnetic Moments**:
- V³⁺: 2.83
- Cr³⁺: 3.87
- Mn³⁺: 4.90
- Co³⁺: 4.90

5. **Conclusion**: The ion with the highest magnetic moment is **Mn³⁺** (Manganese).

### Final Answer:
The ion with the highest magnetic moment is **Mn³⁺**.