The spin only magnetic moment of `Fe^(3+)` ion (inBM) is approximately

Options

A. 4
B. 7
C. 5
D. 6

Explanation

To calculate the spin-only magnetic moment of the `Fe^(3+)` ion in Bohr magnetons (BM), we can follow these steps:

### Step 1: Determine the Electron Configuration of `Fe^(3+)`
Iron (Fe) has an atomic number of 26. The electron configuration of neutral iron is:
\[ \text{Fe: } [\text{Ar}] 3d^6 4s^2 \]
When iron loses three electrons to form `Fe^(3+)`, it loses the two 4s electrons and one 3d electron. Thus, the electron configuration for `Fe^(3+)` becomes:
\[ \text{Fe}^{3+}: [\text{Ar}] 3d^5 \]

### Step 2: Identify the Number of Unpaired Electrons
In the `3d^5` configuration, the five electrons will occupy the five d orbitals. According to Hund's rule, each orbital will get one electron before any pairing occurs. Therefore, all five electrons in `3d^5` are unpaired.

Thus, the number of unpaired electrons (n) is:
\[ n = 5 \]

### Step 3: Use the Formula for Spin-Only Magnetic Moment
The formula for the spin-only magnetic moment (μ) is given by:
\[ \mu = \sqrt{n(n + 2)} \]
Substituting the value of n:
\[ \mu = \sqrt{5(5 + 2)} \]
\[ \mu = \sqrt{5 \times 7} \]
\[ \mu = \sqrt{35} \]

### Step 4: Calculate the Value
Now, we calculate the square root of 35:
\[ \mu \approx 5.92 \, \text{BM} \]

### Step 5: Round to Appropriate Significant Figures
The value can be rounded to:
\[ \mu \approx 6 \, \text{BM} \]

### Conclusion
The spin-only magnetic moment of `Fe^(3+)` ion is approximately **6 Bohr magnetons (BM)**.

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