The hybridization states of the central atom in the complexes `[Fe(CN)_(6)]^(3-),[Fe(CN)_(6)]^(4-)` and `[Co(NO_(2))_(6)]^(3-)` are:

Options

A. `d^(2)sp^(3),sp^(3)d^(2) and dsp^(2)` respectively
B. `d^(2)sp^(3),sp^(3)d^(2) and sp^(3) d^(2)` respectively
C. `d^(2)sp^(3),sp^(3)d^(2) and d^(2)sp^(3)` respectively
D. all `d^(2)sp^(3)`

Answer

To determine the hybridization states of the central atom in the complexes \([Fe(CN)_{6}]^{3-}\), \([Fe(CN)_{6}]^{4-}\), and \([Co(NO_{2})_{6}]^{3-}\), we will follow these steps:

### Step 1: Determine the oxidation state of the central metal ion.

1. **For \([Fe(CN)_{6}]^{3-}\)**:
- Let the oxidation state of Fe be \(x\).
- The charge of CN is \(-1\), and there are 6 CN ligands.
- The equation becomes: \(x + 6(-1) = -3\).
- Solving gives: \(x - 6 = -3 \Rightarrow x = +3\).

2. **For \([Fe(CN)_{6}]^{4-}\)**:
- Let the oxidation state of Fe be \(x\).
- The equation becomes: \(x + 6(-1) = -4\).
- Solving gives: \(x - 6 = -4 \Rightarrow x = +2\).

3. **For \([Co(NO_{2})_{6}]^{3-}\)**:
- Let the oxidation state of Co be \(x\).
- The charge of NO2 is \(-1\), and there are 6 NO2 ligands.
- The equation becomes: \(x + 6(-1) = -3\).
- Solving gives: \(x - 6 = -3 \Rightarrow x = +3\).

### Step 2: Write the electron configuration for the central metal ion in each oxidation state.

1. **For \(Fe^{3+}\)**:
- Atomic number of Fe = 26.
- Ground state configuration: \([Ar] 4s^2 3d^6\).
- For \(Fe^{3+}\): Remove 2 electrons from 4s and 1 from 3d → \([Ar] 3d^5\).

2. **For \(Fe^{2+}\)**:
- For \(Fe^{2+}\): Remove 2 electrons from 4s → \([Ar] 3d^6\).

3. **For \(Co^{3+}\)**:
- Atomic number of Co = 27.
- Ground state configuration: \([Ar] 4s^2 3d^7\).
- For \(Co^{3+}\): Remove 2 electrons from 4s and 1 from 3d → \([Ar] 3d^6\).

### Step 3: Determine the hybridization based on the ligand field.

1. **For \([Fe(CN)_{6}]^{3-}\)**:
- CN is a strong field ligand, which causes pairing of electrons.
- The \(3d^5\) configuration will pair up: \(3d\) becomes fully paired.
- Hybridization: \(d^2sp^3\) (6 orbitals needed for 6 ligands).

2. **For \([Fe(CN)_{6}]^{4-}\)**:
- CN is still a strong field ligand.
- The \(3d^6\) configuration will also pair up.
- Hybridization: \(d^2sp^3\) (6 orbitals needed for 6 ligands).

3. **For \([Co(NO_{2})_{6}]^{3-}\)**:
- NO2 is also a strong field ligand.
- The \(3d^6\) configuration will pair up.
- Hybridization: \(d^2sp^3\) (6 orbitals needed for 6 ligands).

### Final Answer:
- The hybridization states for the complexes are:
- \([Fe(CN)_{6}]^{3-}\): \(d^2sp^3\)
- \([Fe(CN)_{6}]^{4-}\): \(d^2sp^3\)
- \([Co(NO_{2})_{6}]^{3-}\): \(d^2sp^3\)