The expected spin only magnetic mometum for `[Fe(CN)_(6)]^(4-)` and `[FeF_(6)]^(3+)` are

Options

A. 1.73 and 1.73 B.M
B. 1.73 and 5.93 B.M
C. 0.0 and 1.73 B.M
D. 0.0 and 5.92 B.M

Explanation

To determine the expected spin-only magnetic moments for the complexes \([Fe(CN)_6]^{4-}\) and \([FeF_6]^{3+}\), we will follow these steps:

### Step 1: Determine the oxidation state of Iron in \([Fe(CN)_6]^{4-}\)

- The cyanide ion \((CN^-)\) has a charge of -1. Since there are 6 cyanide ions, the total negative charge contributed by cyanide is \(-6\).
- The overall charge of the complex is -4. Therefore, we can set up the equation:
\[ x + 6(-1) = -4 \]
where \(x\) is the oxidation state of iron.
- Solving for \(x\):
\[ x - 6 = -4 \implies x = +2 \]
- Thus, the oxidation state of iron in \([Fe(CN)_6]^{4-}\) is +2.

### Step 2: Determine the electronic configuration of Iron in \([Fe(CN)_6]^{4-}\)

- The electronic configuration of iron in the +2 oxidation state is:
\[ [Ar] 3d^6 \]

### Step 3: Analyze the ligand field strength of cyanide

- Cyanide \((CN^-)\) is a strong field ligand, which means it will cause the pairing of electrons in the d-orbitals.
- For \(3d^6\) in a strong field, the electrons will pair up:
- The configuration will be: \( \uparrow\downarrow \, \uparrow\downarrow \, \uparrow \)
- This results in 0 unpaired electrons.

### Step 4: Calculate the spin-only magnetic moment for \([Fe(CN)_6]^{4-}\)

- The formula for spin-only magnetic moment \(\mu\) is:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.
- Since \(n = 0\):
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \]

### Step 5: Determine the oxidation state of Iron in \([FeF_6]^{3+}\)

- The fluoride ion \((F^-)\) has a charge of -1. Since there are 6 fluoride ions, the total negative charge contributed by fluoride is \(-6\).
- The overall charge of the complex is +3. Therefore, we can set up the equation:
\[ x + 6(-1) = +3 \]
where \(x\) is the oxidation state of iron.
- Solving for \(x\):
\[ x - 6 = 3 \implies x = +6 \]
- Thus, the oxidation state of iron in \([FeF_6]^{3+}\) is +3.

### Step 6: Determine the electronic configuration of Iron in \([FeF_6]^{3+}\)

- The electronic configuration of iron in the +3 oxidation state is:
\[ [Ar] 3d^5 \]

### Step 7: Analyze the ligand field strength of fluoride

- Fluoride \((F^-)\) is a weak field ligand, which means it will not cause pairing of electrons.
- For \(3d^5\) in a weak field, the electrons will remain unpaired:
- The configuration will be: \( \uparrow \, \uparrow \, \uparrow \, \uparrow \, \uparrow \)
- This results in 5 unpaired electrons.

### Step 8: Calculate the spin-only magnetic moment for \([FeF_6]^{3+}\)

- Using the formula for spin-only magnetic moment:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n = 5\):
\[ \mu = \sqrt{5(5 + 2)} = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92 \]

### Final Results

- The expected spin-only magnetic moment for \([Fe(CN)_6]^{4-}\) is **0**.
- The expected spin-only magnetic moment for \([FeF_6]^{3+}\) is approximately **5.92**.

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