Which of the following d - block ions show the maximum value of the magnetic moment (Bohr's moment)?

Options

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

Answer

To solve the question about which d-block ion shows the maximum value of the magnetic moment (Bohr's moment), we will follow these steps:

### Step 1: Understand the formula for magnetic moment
The magnetic moment (μ) is calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \, \text{Bohr magnetons} \]
where \( n \) is the number of unpaired electrons in the ion.

### Step 2: Determine the electronic configurations of the given ions
We will analyze the electronic configurations of each ion provided in the options:

1. **Mn²⁺**:
- Atomic number of Mn = 25
- Electronic configuration of Mn = [Ar] 3d⁵ 4s²
- For Mn²⁺, we remove two electrons (from 4s first):
- Configuration: 3d⁵
- Number of unpaired electrons (n) = 5

2. **Fe²⁺**:
- Atomic number of Fe = 26
- Electronic configuration of Fe = [Ar] 3d⁶ 4s²
- For Fe²⁺, we remove two electrons:
- Configuration: 3d⁶
- Number of unpaired electrons (n) = 4 (since two electrons will pair up)

3. **Ni²⁺**:
- Atomic number of Ni = 28
- Electronic configuration of Ni = [Ar] 3d⁸ 4s²
- For Ni²⁺, we remove two electrons:
- Configuration: 3d⁸
- Number of unpaired electrons (n) = 2

4. **Co²⁺**:
- Atomic number of Co = 27
- Electronic configuration of Co = [Ar] 3d⁷ 4s²
- For Co²⁺, we remove two electrons:
- Configuration: 3d⁷
- Number of unpaired electrons (n) = 3

### Step 3: Calculate the magnetic moments for each ion
Using the number of unpaired electrons calculated above, we can find the magnetic moments:

1. **Mn²⁺**:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \, \text{Bohr magnetons} \]

2. **Fe²⁺**:
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \, \text{Bohr magnetons} \]

3. **Ni²⁺**:
\[ \mu = \sqrt{2(2 + 2)} = \sqrt{2 \times 4} = \sqrt{8} \, \text{Bohr magnetons} \]

4. **Co²⁺**:
\[ \mu = \sqrt{3(3 + 2)} = \sqrt{3 \times 5} = \sqrt{15} \, \text{Bohr magnetons} \]

### Step 4: Compare the magnetic moments
- Mn²⁺: \(\sqrt{35}\)
- Fe²⁺: \(\sqrt{24}\)
- Ni²⁺: \(\sqrt{8}\)
- Co²⁺: \(\sqrt{15}\)

### Conclusion
The ion with the maximum number of unpaired electrons and thus the maximum magnetic moment is **Mn²⁺**.

### Final Answer
**Mn²⁺ shows the maximum value of the magnetic moment (Bohr's moment).**

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