What is the magnetic moment of `K_3[FeFe]`

Options

A. 5.91 BM
B. 4.89 BM
C. 3.87 BM
D. 6.92 BM

Explanation

To find the magnetic moment of the coordination compound \( K_3[FeF_6] \), we need to follow these steps:

### Step 1: Identify the Oxidation State of Iron
In the complex \( K_3[FeF_6] \), potassium (K) has a +1 charge, and fluoride (F) has a -1 charge. Since there are 6 fluoride ions, the total negative charge contributed by fluoride is \( 6 \times (-1) = -6 \).

The overall charge of the complex ion \( [FeF_6]^{3-} \) must balance the charge from potassium:
\[ x + (-6) = -3 \]
where \( x \) is the oxidation state of iron (Fe). Solving for \( x \):
\[ x - 6 = -3 \implies x = +3 \]
Thus, the oxidation state of iron in \( K_3[FeF_6] \) is +3.

### Step 2: Determine the Electron Configuration of Iron
Iron (Fe) has an atomic number of 26. The electron configuration of neutral iron is:
\[ [Ar] 3d^6 4s^2 \]
In the +3 oxidation state, iron loses two 4s electrons and one 3d electron, resulting in:
\[ Fe^{3+}: [Ar] 3d^5 \]

### Step 3: Identify the Geometry and Hybridization of the Complex
The complex \( [FeF_6]^{3-} \) is octahedral because it has six fluoride ligands. In an octahedral field, the 3d orbitals split into two sets: \( t_{2g} \) (lower energy) and \( e_g \) (higher energy).

### Step 4: Determine the Electron Configuration in the Octahedral Field
For \( Fe^{3+} \) with a \( 3d^5 \) configuration, all five electrons will occupy the \( t_{2g} \) orbitals due to Hund's rule, which states that electrons will fill degenerate orbitals singly before pairing up. Therefore, the electron configuration in the octahedral field will be:
\[ t_{2g}^5 \quad e_g^0 \]
This means all five \( t_{2g} \) orbitals are filled with one electron each, and there are no electrons in the \( e_g \) orbitals.

### Step 5: 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, since all five electrons are unpaired in the \( t_{2g} \) orbitals:
\[ n = 5 \]
Thus, substituting into the formula:
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92 \, \text{BM} \]

### Final Answer
The magnetic moment of \( K_3[FeF_6] \) is approximately \( 5.92 \, \text{BM} \).

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