What will be the 'spin only' magnetic moment of the complex formed when Fe `(SCN)_(3)` reacts with solution containing excess F- ?

Options

A. 2.83 BM
B. 3.87 BM
C. 5.92 BM
D. 1.73 BM

Explanation

To solve the problem regarding the 'spin only' magnetic moment of the complex formed when Fe(SCN)₃ reacts with excess F⁻, we will follow these steps:

### Step 1: Identify the initial complex and its components
The initial complex is Fe(SCN)₃, where Fe is iron and SCN⁻ is thiocyanate.

### Step 2: Determine the oxidation state of iron in Fe(SCN)₃
In Fe(SCN)₃, SCN has a charge of -1. Therefore, the overall charge of the complex is -3 (since there are three SCN⁻ ligands). Let the oxidation state of iron be \( x \).

The equation can be set up as:
\[ x + 3(-1) = -3 \]
\[ x - 3 = -3 \]
\[ x = 0 \]

Thus, the oxidation state of iron in Fe(SCN)₃ is 0.

### Step 3: Determine the new complex formed with excess F⁻
When Fe(SCN)₃ reacts with excess F⁻, the thiocyanate ligands are replaced by fluoride ions. The new complex formed is [FeF₆]³⁻.

### Step 4: Determine the oxidation state of iron in [FeF₆]³⁻
The charge of each fluoride ion (F⁻) is -1. Since there are six fluoride ions, the overall charge contributed by fluoride is -6. Let the oxidation state of iron in this complex be \( y \).

The equation can be set up as:
\[ y + 6(-1) = -3 \]
\[ y - 6 = -3 \]
\[ y = +3 \]

Thus, the oxidation state of iron in [FeF₆]³⁻ is +3.

### Step 5: Determine the electron configuration of Fe³⁺
The atomic number of iron (Fe) is 26. The electron configuration of neutral iron is:
\[ \text{Fe: } [Ar] 3d^6 4s^2 \]

For Fe³⁺, we remove three electrons (two from 4s and one from 3d):
\[ \text{Fe}^{3+}: [Ar] 3d^5 \]

### Step 6: Determine the number of unpaired electrons
In the 3d subshell, for Fe³⁺, we have:
\[ 3d^5 \]
This means all five d electrons are unpaired.

### Step 7: Calculate the spin-only magnetic moment
The formula for the spin-only magnetic moment (\( \mu \)) is given by:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.

Here, \( n = 5 \):
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \]

Calculating this gives:
\[ \mu \approx 5.92 \, \text{BM} \]

### Conclusion
The 'spin only' magnetic moment of the complex formed when Fe(SCN)₃ reacts with excess F⁻ is approximately 5.92 Bohr Magnetons (BM).

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