`[Fe(H_(2)O)_(6)]^(2+) and [Fe(CN)_(6)]^(4-)` differ in

Options

A. geometry, magnetic moment
B. geometry, hybridization
C. magnetic moment, colour
D. hybridisation, number of d - electrons

Answer

To solve the question regarding the differences between the complexes \([Fe(H_2O)_6]^{2+}\) and \([Fe(CN)_6]^{4-}\), we will analyze their properties step by step.

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

- For \([Fe(CN)_6]^{4-}\):
- The overall charge is -4.
- Each cyanide (\(CN^-\)) ligand has a charge of -1.
- Let the oxidation state of iron be \(x\).
- The equation is: \(x + 6(-1) = -4\) → \(x - 6 = -4\) → \(x = +2\).

Thus, the oxidation state of iron in both complexes is +2.

### Step 2: Determine the electronic configuration of iron in both complexes.
- Iron has an atomic number of 26, so its ground state electronic configuration is:
- \( [Ar] 3d^6 4s^2 \).
- For \([Fe(H_2O)_6]^{2+}\):
- Iron loses 2 electrons (from 4s) to form \(Fe^{2+}\):
- Configuration: \( [Ar] 3d^6 \).

- For \([Fe(CN)_6]^{4-}\):
- The same \(Fe^{2+}\) configuration applies:
- Configuration: \( [Ar] 3d^6 \).

### Step 3: Analyze the ligands and their field strength.
- Water (\(H_2O\)) is a weak field ligand, while cyanide (\(CN^-\)) is a strong field ligand.
- This difference affects the pairing of electrons in the d-orbitals.

### Step 4: Determine the hybridization and electron pairing in both complexes.
- For \([Fe(H_2O)_6]^{2+}\):
- As a weak field ligand, water does not cause pairing of the 3d electrons.
- The 3d orbitals will remain unpaired: \(3d^6\) → 4 unpaired electrons.
- Hybridization: \(sp^3d^2\).

- For \([Fe(CN)_6]^{4-}\):
- As a strong field ligand, cyanide causes pairing of the 3d electrons.
- The 3d orbitals will be fully paired: \(3d^6\) → 0 unpaired electrons.
- Hybridization: \(d^2sp^3\).

### Step 5: Calculate the magnetic moment for both complexes.
- The magnetic moment formula is given by:
\[ \mu = \sqrt{n(n + 2)} \]
where \(n\) is the number of unpaired electrons.

- For \([Fe(H_2O)_6]^{2+}\):
- \(n = 4\) (4 unpaired electrons).
- \(\mu = \sqrt{4(4 + 2)} = \sqrt{24} \approx 4.90 \, \text{BM}\).

- For \([Fe(CN)_6]^{4-}\):
- \(n = 0\) (0 unpaired electrons).
- \(\mu = \sqrt{0(0 + 2)} = 0 \, \text{BM}\).

### Step 6: Determine the color of both complexes.
- \([Fe(H_2O)_6]^{2+}\) has unpaired electrons, which allows it to absorb light and show a pale green color.
- \([Fe(CN)_6]^{4-}\) has no unpaired electrons, making it colorless.

### Conclusion:
The two complexes differ in:
1. **Magnetic moment**: \([Fe(H_2O)_6]^{2+}\) has a magnetic moment of approximately 4.90 BM, while \([Fe(CN)_6]^{4-}\) has a magnetic moment of 0 BM.
2. **Color**: \([Fe(H_2O)_6]^{2+}\) is pale green, while \([Fe(CN)_6]^{4-}\) is colorless.

### Final Answer:
The correct option is that they differ in **magnetic moment and color**.