One unpaired electron causes magnetic moment of 1.1 BM. The magnetic moment of `._(26)Fe^(2+)` is

Options

A. `4.4`
B. `5.5`
C. `2.2`
D. 0

Explanation

To find the magnetic moment of \( \text{Fe}^{2+} \), we can follow these steps:

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

### Step 2: Determine the Electron Configuration of \( \text{Fe}^{2+} \)
When iron loses 2 electrons to form \( \text{Fe}^{2+} \), the electrons are removed from the outermost shell, which is the 4s orbital. Thus, the configuration for \( \text{Fe}^{2+} \) becomes:
\[ \text{Fe}^{2+}: 1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^6 \]

### Step 3: Identify the Number of Unpaired Electrons
In the \( 3d \) subshell, we have 6 electrons. According to Hund's rule, the electrons will fill the orbitals singly first before pairing. The distribution of the 6 electrons in the \( 3d \) orbitals will be:
- 3d orbital filling: \( \uparrow \, \uparrow \, \uparrow \, \uparrow \, \downarrow \, \downarrow \)

This shows that there are 4 unpaired electrons in the \( 3d \) subshell.

### Step 4: Calculate the Magnetic Moment
The magnetic moment (\( \mu \)) can be calculated using the formula:
\[ \mu = \sqrt{n(n+2)} \]
where \( n \) is the number of unpaired electrons.

In this case, \( n = 4 \):
\[ \mu = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} \]
Calculating \( \sqrt{24} \):
\[ \mu \approx 4.9 \, \text{BM} \]

### Conclusion
The magnetic moment of \( \text{Fe}^{2+} \) is approximately \( 4.9 \, \text{BM} \).