The spin only magnetic moment of `Fe^(2+)` ion (in BM) is approximately

Options

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

Explanation

To find the spin-only magnetic moment of the `Fe^(2+)` ion in Bohr magnetons (BM), we can follow these steps:

### Step 1: Determine the electron configuration of iron (Fe)
Iron (Fe) has an atomic number of 26. The electron configuration of neutral iron is:
\[ \text{Fe: } [\text{Ar}] 3d^6 4s^2 \]

### Step 2: Adjust the electron configuration for `Fe^(2+)`
The `Fe^(2+)` ion means that two electrons are removed from the neutral iron atom. Electrons are removed first from the outermost shell (4s) before the d-orbitals. Thus, the configuration for `Fe^(2+)` becomes:
\[ \text{Fe}^{2+}: [\text{Ar}] 3d^6 \]

### Step 3: Determine the number of unpaired electrons
To find the number of unpaired electrons in the `3d^6` configuration, we need to fill the d-orbitals according to Hund's rule. The filling of the `3d` orbitals is as follows:

1. Fill each of the five d-orbitals with one electron each first (this accounts for 5 electrons).
2. Then, add the sixth electron, which will pair up with one of the already filled orbitals.

The distribution of electrons in the `3d` orbitals will look like this:
- ↑ (1st orbital)
- ↑ (2nd orbital)
- ↑ (3rd orbital)
- ↑ (4th orbital)
- ↑↓ (5th orbital)

Thus, there are 4 unpaired electrons in the `Fe^(2+)` ion.

### Step 4: Use the formula for spin-only magnetic moment
The formula for calculating the spin-only magnetic moment (\( \mu_s \)) is given by:
\[ \mu_s = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

### Step 5: Substitute the value of \( n \)
Here, \( n = 4 \) (the number of unpaired electrons).
\[ \mu_s = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \]

### Step 6: Calculate the magnetic moment
Calculating \( \sqrt{24} \):
\[ \sqrt{24} \approx 4.89 \, \text{BM} \]

### Step 7: Round off the result
When rounding \( 4.89 \) to the nearest whole number, we get:
\[ \approx 5 \, \text{BM} \]

### Conclusion
The spin-only magnetic moment of the `Fe^(2+)` ion is approximately **5 Bohr magnetons (BM)**.