`[PdFClBrI]^(2-)` Number of Geometrical Isomers = n. For `[Fe(CN)_6]^(n-6)`, Determine the spin only magnetic moment and CFSE (Ignore the pairing energy)

Options

A. `1.73 B.M., -2delta_0`
B. `2.84 B.M., -1.6delta_0`
C. `0, -1.6delta_0`
D. `5.92 B.M., -2.4 delta_0`

Explanation

To solve the problem step by step, we will break it down into two parts: determining the number of geometrical isomers for the complex \([PdFClBrI]^{2-}\) and then calculating the spin-only magnetic moment and CFSE for the complex \([Fe(CN)_6]^{n-6}\).

### Step 1: Determine the number of geometrical isomers for \([PdFClBrI]^{2-}\)

1. **Identify the coordination number and geometry**:
- Palladium (Pd) typically forms square planar complexes when it is in the +2 oxidation state.
- The complex has four different ligands: F, Cl, Br, and I.

2. **Apply the formula for geometrical isomers**:
- For a square planar complex with four different ligands, the number of geometrical isomers can be determined using the formula:
\[ \text{Number of geometrical isomers} = \frac{n!}{(n-2)! \cdot 2!} \]
- Here, \(n\) is the number of different ligands (4 in this case).
- Thus, the number of geometrical isomers is:
\[ \frac{4!}{(4-2)! \cdot 2!} = \frac{24}{2 \cdot 2} = 6 \]
- However, we need to consider the symmetry of the ligands. After analyzing the possible arrangements, we find that there are actually 3 unique geometrical isomers.

3. **Conclusion**:
- The number of geometrical isomers \(n\) for \([PdFClBrI]^{2-}\) is 3.

### Step 2: Calculate the spin-only magnetic moment and CFSE for \([Fe(CN)_6]^{n-6}\)

1. **Determine the oxidation state of iron**:
- Given that \(n = 3\), we can substitute this into the complex:
\[ [Fe(CN)_6]^{3-6} = [Fe(CN)_6]^{-3} \]
- The charge of the complex is -3. Since CN is a strong field ligand with a charge of -1, the total contribution from the ligands is -6. Therefore, the oxidation state of iron is:
\[ x + (-6) = -3 \implies x = +3 \]
- Thus, iron is in the +3 oxidation state.

2. **Determine the electronic configuration of \(Fe^{3+}\)**:
- The electronic configuration of neutral iron (Fe) is \([Ar] 3d^6 4s^2\).
- For \(Fe^{3+}\), we remove three electrons (two from 4s and one from 3d):
\[ Fe^{3+} = [Ar] 3d^5 \]

3. **Determine the number of unpaired electrons**:
- In an octahedral field created by the six CN ligands, the \(3d\) orbitals split into \(t_{2g}\) and \(e_g\) levels.
- Since CN is a strong field ligand, it causes pairing of electrons in the \(t_{2g}\) orbitals.
- The \(3d^5\) configuration will fill the \(t_{2g}\) orbitals first:
- All five electrons will occupy the \(t_{2g}\) orbitals, resulting in 5 unpaired electrons.

4. **Calculate the spin-only magnetic moment**:
- The formula for the spin-only magnetic moment (\(\mu\)) is:
\[ \mu = \sqrt{n(n+2)} \text{ Bohr magneton} \]
- Where \(n\) is the number of unpaired electrons (5 in this case):
\[ \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ Bohr magneton} \]

5. **Calculate the CFSE (Crystal Field Stabilization Energy)**:
- The CFSE can be calculated using the formula:
\[ \text{CFSE} = (n_{t_{2g}} \times -0.4 \Delta_o) + (n_{e_g} \times 0.6 \Delta_o) \]
- For \(Fe^{3+}\) in \([Fe(CN)_6]^{-3}\):
- \(n_{t_{2g}} = 5\) and \(n_{e_g} = 0\):
\[ \text{CFSE} = (5 \times -0.4 \Delta_o) + (0 \times 0.6 \Delta_o) = -2.0 \Delta_o \]

### Final Results:
- The spin-only magnetic moment is approximately \(5.92 \text{ Bohr magneton}\).
- The CFSE is \(-2.0 \Delta_o\).