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

### Step 1: Identify the oxidation states of iron in both complexes.
- For \([Fe(H_2O)_6]^{2+}\):
- The overall charge is \(+2\).
- Since \(H_2O\) is a neutral ligand, the oxidation state of \(Fe\) is \(+2\).

- For \([Fe(CN)_6]^{4-}\):
- The overall charge is \(-4\).
- Each \(CN^-\) ligand has a charge of \(-1\), thus \(6 \times (-1) = -6\).
- Therefore, \(Fe\) must be \(+2\) to balance the charge: \(+2 + (-6) = -4\).

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

### Step 3: Determine the number of unpaired electrons in \(Fe^{2+}\).
- In the \(3d^6\) configuration, we need to consider the ligand field strength.
- \(H_2O\) is a weak field ligand, leading to a high-spin configuration:
- The \(3d\) orbitals will be filled as follows:
- \(t_{2g}^6\) (lower energy) and \(e_g^0\) (higher energy).
- This results in 4 unpaired electrons (as the first three fill singly before pairing).

### Step 4: Calculate the spin-only magnetic moment for \([Fe(H_2O)_6]^{2+}\).
- The formula for the spin-only magnetic moment (\(\mu\)) is:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.
- For \([Fe(H_2O)_6]^{2+}\), \(n = 4\):
\[ \mu = \sqrt{4(4 + 2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.9 \, \mu_B \]

### Step 5: Determine the number of unpaired electrons in \([Fe(CN)_6]^{4-}\).
- \(CN^-\) is a strong field ligand, leading to a low-spin configuration:
- The \(3d^6\) electrons will fill as follows:
- \(t_{2g}^6\) (lower energy) and \(e_g^0\) (higher energy).
- This results in 0 unpaired electrons.

### Step 6: Calculate the spin-only magnetic moment for \([Fe(CN)_6]^{4-}\).
- For \([Fe(CN)_6]^{4-}\), \(n = 0\):
\[ \mu = \sqrt{0(0 + 2)} = \sqrt{0} = 0 \, \mu_B \]

### Final Results:
- The spin-only magnetic moment for \([Fe(H_2O)_6]^{2+}\) is approximately \(4.9 \, \mu_B\).
- The spin-only magnetic moment for \([Fe(CN)_6]^{4-}\) is \(0 \, \mu_B\).