What is the value of spin only magnetic moment of anionic and cationic part of complex `[Fe(H_(2)O)_(6)]_(2)`
`[Fe(CN)_(6)]`

Options

A. 4.9 B.M & zero
B. zero and 4.9 B.M
C. 2.9 B.M. and 0
D. 0 and 2.9 B.M.

Answer

To determine the spin-only magnetic moment of the cationic and anionic parts of the complex \([Fe(H_2O)_6]^{2+} [Fe(CN)_6]^{4-}\), we will follow these steps:

### Step 1: Identify the Oxidation States
1. **Cationic Part**: \([Fe(H_2O)_6]^{2+}\)
- Water (\(H_2O\)) is a neutral ligand, so the oxidation state of Fe in this complex is +2.

2. **Anionic Part**: \([Fe(CN)_6]^{4-}\)
- Each cyanide ion (\(CN^-\)) has a charge of -1. Therefore, for 6 cyanide ions, the total charge is -6.
- Let the oxidation state of Fe be \(x\). The equation is:
\[ x + 6(-1) = -4 \implies x - 6 = -4 \implies x = +2 \]
- Thus, Fe is also in the +2 oxidation state in this complex.

### Step 2: Determine the Electron Configuration
- The atomic number of Fe is 26. The electron configuration of neutral Fe is:
\[ [Ar] 3d^6 4s^2 \]
- For \(Fe^{2+}\), we remove two electrons (from the 4s orbital):
\[ Fe^{2+}: [Ar] 3d^6 \]

### Step 3: Count Unpaired Electrons in Cationic Part
- In \([Fe(H_2O)_6]^{2+}\):
- Water is a weak field ligand, meaning it does not cause pairing of electrons.
- The \(3d^6\) configuration will have 4 unpaired electrons (following Hund's rule).

### Step 4: Calculate Spin-Only Magnetic Moment for Cationic Part
- The formula for the spin-only magnetic moment (\(\mu\)) is:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.
- For the cationic part:
\[ n = 4 \implies \mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.9 \, B_m \]

### Step 5: Count Unpaired Electrons in Anionic Part
- In \([Fe(CN)_6]^{4-}\):
- Cyanide is a strong field ligand, which causes pairing of electrons.
- Therefore, all 6 electrons in the \(3d\) orbitals will pair up, resulting in 0 unpaired electrons.

### Step 6: Calculate Spin-Only Magnetic Moment for Anionic Part
- For the anionic part:
\[ n = 0 \implies \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \, B_m \]

### Conclusion
- The spin-only magnetic moment for the cationic part \([Fe(H_2O)_6]^{2+}\) is approximately \(4.9 \, B_m\).
- The spin-only magnetic moment for the anionic part \([Fe(CN)_6]^{4-}\) is \(0 \, B_m\).

### Final Answer
- The values are:
- Cationic part: \(4.9 \, B_m\)
- Anionic part: \(0 \, B_m\)