Which of the following ion has maximum value of magnetic moment `

Options

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

Answer

To determine which ion has the maximum value of magnetic moment, we will use the formula for magnetic moment:

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

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

### Step-by-Step Solution:

1. **Identify the Ions and Their Electron Configurations**:
- **Cu²⁺ (Copper(II))**: The electron configuration of Cu is [Ar] 3d¹⁰ 4s¹. For Cu²⁺, it loses 2 electrons: [Ar] 3d⁹. Thus, it has 1 unpaired electron.
- **Mn³⁺ (Manganese(III))**: The electron configuration of Mn is [Ar] 3d⁵ 4s². For Mn³⁺, it loses 3 electrons: [Ar] 3d⁴. Thus, it has 4 unpaired electrons.
- **Fe³⁺ (Iron(III))**: The electron configuration of Fe is [Ar] 3d⁶ 4s². For Fe³⁺, it loses 3 electrons: [Ar] 3d⁵. Thus, it has 5 unpaired electrons.
- **V³⁺ (Vanadium(III))**: The electron configuration of V is [Ar] 3d³ 4s². For V³⁺, it loses 3 electrons: [Ar] 3d³. Thus, it has 3 unpaired electrons.

2. **Calculate the Magnetic Moment for Each Ion**:
- **Cu²⁺**:
\[ n = 1 \quad \Rightarrow \quad \mu = \sqrt{1(1 + 2)} = \sqrt{3} \]
- **Mn³⁺**:
\[ n = 4 \quad \Rightarrow \quad \mu = \sqrt{4(4 + 2)} = \sqrt{24} = 2\sqrt{6} \]
- **Fe³⁺**:
\[ n = 5 \quad \Rightarrow \quad \mu = \sqrt{5(5 + 2)} = \sqrt{35} \]
- **V³⁺**:
\[ n = 3 \quad \Rightarrow \quad \mu = \sqrt{3(3 + 2)} = \sqrt{15} \]

3. **Compare the Magnetic Moments**:
- Cu²⁺: \(\sqrt{3}\)
- Mn³⁺: \(2\sqrt{6}\)
- Fe³⁺: \(\sqrt{35}\)
- V³⁺: \(\sqrt{15}\)

4. **Determine the Maximum**:
- To find which is maximum, we can compare the values:
- \(\sqrt{3} \approx 1.73\)
- \(2\sqrt{6} \approx 4.90\)
- \(\sqrt{35} \approx 5.92\)
- \(\sqrt{15} \approx 3.87\)

The highest value is for **Fe³⁺** with a magnetic moment of \(\sqrt{35}\).

### Conclusion:
The ion with the maximum value of magnetic moment is **Fe³⁺**.