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 of identifying which pair of complexes has the same hybridization and the same Effective Atomic Number (EAN), we will follow these steps:

### Step 1: Understand the Concept of EAN
The Effective Atomic Number (EAN) is calculated using the formula:
\[ \text{EAN} = Z + 2 \times CN - OS \]
where:
- \( Z \) = Atomic number of the central metal atom
- \( CN \) = Coordination number (number of ligands attached)
- \( OS \) = Oxidation state of the central metal

### Step 2: Analyze the First Pair: Ni(CO)₄ and Ni(CN)₄²⁻
1. **For Ni(CO)₄**:
- Atomic number of Ni = 28
- Coordination number (CN) = 4 (since there are 4 CO ligands)
- Oxidation state (OS) = 0 (neutral complex)
- Calculate EAN:
\[ \text{EAN} = 28 + 2 \times 4 - 0 = 28 + 8 = 36 \]

2. **For Ni(CN)₄²⁻**:
- Atomic number of Ni = 28
- Coordination number (CN) = 4 (since there are 4 CN ligands)
- Oxidation state (OS) = +2 (since CN is a -1 ligand and the overall charge is -2)
- Calculate EAN:
\[ \text{EAN} = 28 + 2 \times 4 - 2 = 28 + 8 - 2 = 34 \]

### Step 3: Determine Hybridization
- **Ni(CO)₄**: The hybridization is \( sp^3 \) (due to 4 ligands).
- **Ni(CN)₄²⁻**: The hybridization is \( dsp^2 \) (due to pairing of electrons in the presence of a strong field ligand).

### Step 4: Conclusion for First Pair
- EANs are different (36 vs 34).
- Hybridizations are different (sp³ vs dsp²).

### Step 5: Analyze the Second Pair: FeF₆³⁻ and Fe(H₂O)₆³⁺
1. **For FeF₆³⁻**:
- Atomic number of Fe = 26
- Coordination number (CN) = 6 (since there are 6 F ligands)
- Oxidation state (OS) = +3
- Calculate EAN:
\[ \text{EAN} = 26 + 2 \times 6 - 3 = 26 + 12 - 3 = 35 \]

2. **For Fe(H₂O)₆³⁺**:
- Atomic number of Fe = 26
- Coordination number (CN) = 6 (since there are 6 H₂O ligands)
- Oxidation state (OS) = +3
- Calculate EAN:
\[ \text{EAN} = 26 + 2 \times 6 - 3 = 26 + 12 - 3 = 35 \]

### Step 6: Determine Hybridization
- **FeF₆³⁻**: The hybridization is \( sp^3d^2 \) (octahedral geometry).
- **Fe(H₂O)₆³⁺**: The hybridization is also \( sp^3d^2 \) (octahedral geometry).

### Step 7: Conclusion for Second Pair
- EANs are the same (35 vs 35).
- Hybridizations are the same (both are \( sp^3d^2 \)).

### Final Answer
The pair of complexes that have the same hybridization and the same values of EAN is **FeF₆³⁻ and Fe(H₂O)₆³⁺**.