Which of the following ions show higher spin only magnetic moment value?

Options

A. `Ti^(3+)`
B. `Mn^(2+)`
C. `Fe^(2+)`
D. `Co^(3+)`

Answer

To determine which of the given ions shows the highest spin-only magnetic moment value, we will analyze each ion's electronic configuration, count the number of unpaired electrons, and then use the formula for calculating the magnetic moment.

### Step-by-Step Solution:

1. **Identify the Ions**: We need to analyze the following ions:
- Ti³⁺
- Mn²⁺
- Fe²⁺
- Co³⁺

2. **Determine the Electronic Configuration**:
- **Titanium (Ti)**: Atomic number = 22
- Ground state configuration: [Ar] 3d² 4s²
- For Ti³⁺: Remove 3 electrons (2 from 4s and 1 from 3d)
- Configuration: 3d¹
- **Manganese (Mn)**: Atomic number = 25
- Ground state configuration: [Ar] 3d⁵ 4s²
- For Mn²⁺: Remove 2 electrons (both from 4s)
- Configuration: 3d⁵
- **Iron (Fe)**: Atomic number = 26
- Ground state configuration: [Ar] 3d⁶ 4s²
- For Fe²⁺: Remove 2 electrons (both from 4s)
- Configuration: 3d⁶
- **Cobalt (Co)**: Atomic number = 27
- Ground state configuration: [Ar] 3d⁷ 4s²
- For Co³⁺: Remove 3 electrons (2 from 4s and 1 from 3d)
- Configuration: 3d⁶

3. **Count the Unpaired Electrons**:
- **Ti³⁺ (3d¹)**: 1 unpaired electron
- **Mn²⁺ (3d⁵)**: 5 unpaired electrons
- **Fe²⁺ (3d⁶)**: 4 unpaired electrons
- **Co³⁺ (3d⁶)**: 4 unpaired electrons

4. **Calculate the Magnetic Moment**:
- The formula for spin-only magnetic moment (μ) is:
\[ μ = \sqrt{n(n + 2)} \]
- Where \( n \) is the number of unpaired electrons.
- **Ti³⁺**:
\[ μ = \sqrt{1(1 + 2)} = \sqrt{3} \]
- **Mn²⁺**:
\[ μ = \sqrt{5(5 + 2)} = \sqrt{35} \]
- **Fe²⁺**:
\[ μ = \sqrt{4(4 + 2)} = \sqrt{24} \]
- **Co³⁺**:
\[ μ = \sqrt{4(4 + 2)} = \sqrt{24} \]

5. **Compare the Magnetic Moments**:
- Ti³⁺: \( \sqrt{3} \)
- Mn²⁺: \( \sqrt{35} \)
- Fe²⁺: \( \sqrt{24} \)
- Co³⁺: \( \sqrt{24} \)

The highest magnetic moment value is from **Mn²⁺** with \( \sqrt{35} \).

### Conclusion:
The ion that shows the highest spin-only magnetic moment value is **Mn²⁺**.

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