The magnetic moment of an ion is `sqrt(24)` B.M. Then that ion may be

Options

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

Answer

To determine which ion has a magnetic moment of \( \sqrt{24} \) Bohr Magnetons (B.M.), we can follow these steps:

### Step 1: Understand the formula for magnetic moment
The magnetic moment (\( \mu \)) of an ion can be calculated using the formula:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 2: Set up the equation
Given that the magnetic moment is \( \sqrt{24} \) B.M., we can equate:
\[ \sqrt{n(n + 2)} = \sqrt{24} \]
Squaring both sides, we get:
\[ n(n + 2) = 24 \]

### Step 3: Solve the quadratic equation
Rearranging the equation:
\[ n^2 + 2n - 24 = 0 \]
Now we can factor this quadratic equation:
\[ (n + 6)(n - 4) = 0 \]
This gives us two possible solutions:
\[ n + 6 = 0 \quad \Rightarrow \quad n = -6 \quad (\text{not valid}) \]
\[ n - 4 = 0 \quad \Rightarrow \quad n = 4 \]
Thus, the number of unpaired electrons \( n \) is 4.

### Step 4: Identify the ions and their unpaired electrons
Now we will check the given ions to see which one has 4 unpaired electrons:
1. **Mn\(^{2+}\)**:
- Electron configuration: \( [Ar] \, 3d^5 \, 4s^0 \)
- Unpaired electrons: 5

2. **Fe\(^{2+}\)**:
- Electron configuration: \( [Ar] \, 3d^6 \, 4s^0 \)
- Unpaired electrons: 4

3. **Fe\(^{3+}\)**:
- Electron configuration: \( [Ar] \, 3d^5 \, 4s^0 \)
- Unpaired electrons: 5

4. **Cu\(^{2+}\)**:
- Electron configuration: \( [Ar] \, 3d^9 \, 4s^0 \)
- Unpaired electrons: 1

### Step 5: Conclusion
From the analysis, we find that only **Fe\(^{2+}\)** has 4 unpaired electrons, which corresponds to the magnetic moment of \( \sqrt{24} \) B.M.

### Final Answer:
The ion with a magnetic moment of \( \sqrt{24} \) B.M. is **Fe\(^{2+}\)**.

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