Among the following pair of complexes in which case the central atoms are having some hybridisation and have same values of E.A.N. also.

Options

A. `[Ni(CO)_(4)] and [Ni(CN)_(4)]^(2-)`
B. `[Fe(F)_(6)]^(3-) and [Fe(H_(2)O)_(6)]^(3+)`
C. `[Fe(CN)_(6)]^(3-) and [Fe(CN)_(6)]^(4-)`
D. `[Cu(CN)_(4)]^(2-) and [Cu(CN)_(4)]^(3-)`

Answer

To solve the problem, we need to determine which pair of complexes has the same hybridization and the same effective atomic number (E.A.N.).

### Step-by-Step Solution:

1. **Understanding E.A.N.**:
The formula for the effective atomic number (E.A.N.) is given by:
\[ \text{E.A.N.} = \text{Atomic Number} + 2 \times \text{Coordination Number} - \text{Oxidation State} \]

2. **Analyzing the First Option: Ni(CO)₄**:
- **Oxidation State**: For Ni(CO)₄, CO is a neutral ligand, so the oxidation state of Ni is 0.
- **Electron Configuration**: Ni has an atomic number of 28, so its configuration is [Ar] 4s² 3d⁸.
- **Hybridization**: With a coordination number of 4 and no pairing, the hybridization is sp³.
- **E.A.N. Calculation**:
\[ \text{E.A.N.} = 28 + 2 \times 4 - 0 = 28 + 8 = 36 \]

3. **Analyzing the Second Option: Ni(CN)₄²⁻**:
- **Oxidation State**: For Ni(CN)₄²⁻, CN is a -1 ligand. Let the oxidation state of Ni be x:
\[ x + 4(-1) = -2 \implies x = +2 \]
- **Electron Configuration**: Ni²⁺ has a configuration of [Ar] 3d⁸.
- **Hybridization**: With a coordination number of 4 and pairing due to CN being a strong field ligand, the hybridization is dsp².
- **E.A.N. Calculation**:
\[ \text{E.A.N.} = 28 + 2 \times 4 - 2 = 28 + 8 - 2 = 34 \]

4. **Analyzing the Third Option: FeF₆³⁻**:
- **Oxidation State**: For FeF₆³⁻, F is a -1 ligand. Let the oxidation state of Fe be x:
\[ x + 6(-1) = -3 \implies x = +3 \]
- **Electron Configuration**: Fe³⁺ has a configuration of [Ar] 3d⁵.
- **Hybridization**: With a coordination number of 6, the hybridization is sp³d².
- **E.A.N. Calculation**:
\[ \text{E.A.N.} = 26 + 2 \times 6 - 3 = 26 + 12 - 3 = 35 \]

5. **Analyzing the Fourth Option: Fe(H₂O)₆³⁺**:
- **Oxidation State**: For Fe(H₂O)₆³⁺, H₂O is a neutral ligand, so the oxidation state of Fe is +3.
- **Electron Configuration**: Fe³⁺ has a configuration of [Ar] 3d⁵.
- **Hybridization**: With a coordination number of 6, the hybridization is also sp³d².
- **E.A.N. Calculation**:
\[ \text{E.A.N.} = 26 + 2 \times 6 - 3 = 26 + 12 - 3 = 35 \]

6. **Conclusion**:
- The complexes FeF₆³⁻ and Fe(H₂O)₆³⁺ both have the same hybridization (sp³d²) and the same E.A.N. (35).
- Therefore, the correct answer is **Option B**: FeF₆³⁻ and Fe(H₂O)₆³⁺.