The ion that has ` sp ^ 3 d ^ 2 ` hybridization for the central atom, is :

Options

A. `[IC l_(4)]^(-)`
B. `IC l_(2)]^(-)`
C. `[Br F_(2)]^(-)`
D. `[IF_(6)]^(-)`

Explanation

To determine which ion has `sp^3d^2` hybridization for the central atom, we need to calculate the steric number (S) for each given ion using the formula:

\[ S = \frac{1}{2} \left( \text{Valence electrons of central atom} + \text{Monovalent atoms attached} - \text{Cationic charge} + \text{Anionic charge} \right) \]

Let's analyze each option step by step:

### Step 1: Analyze ICl₄⁻
- **Valence electrons of iodine (I)**: 7
- **Monovalent atoms attached (Cl)**: 4
- **Cationic charge**: 0
- **Anionic charge**: 1

Using the formula:
\[ S = \frac{1}{2} \left( 7 + 4 - 0 + 1 \right) = \frac{1}{2} \left( 12 \right) = 6 \]

For a steric number of 6, the hybridization is `sp^3d^2`.

### Step 2: Analyze ICl₂⁻
- **Valence electrons of iodine (I)**: 7
- **Monovalent atoms attached (Cl)**: 2
- **Cationic charge**: 0
- **Anionic charge**: 1

Using the formula:
\[ S = \frac{1}{2} \left( 7 + 2 - 0 + 1 \right) = \frac{1}{2} \left( 10 \right) = 5 \]

For a steric number of 5, the hybridization is `sp^3d`.

### Step 3: Analyze BrF₂⁻
- **Valence electrons of bromine (Br)**: 7
- **Monovalent atoms attached (F)**: 2
- **Cationic charge**: 0
- **Anionic charge**: 1

Using the formula:
\[ S = \frac{1}{2} \left( 7 + 2 - 0 + 1 \right) = \frac{1}{2} \left( 10 \right) = 5 \]

For a steric number of 5, the hybridization is `sp^3d`.

### Step 4: Analyze IF₆⁻
- **Valence electrons of iodine (I)**: 7
- **Monovalent atoms attached (F)**: 6
- **Cationic charge**: 0
- **Anionic charge**: 1

Using the formula:
\[ S = \frac{1}{2} \left( 7 + 6 - 0 + 1 \right) = \frac{1}{2} \left( 14 \right) = 7 \]

For a steric number of 7, the hybridization is `sp^3d^3`.

### Conclusion
From the calculations:
- **ICl₄⁻** has `sp^3d^2` hybridization.
- **ICl₂⁻** has `sp^3d` hybridization.
- **BrF₂⁻** has `sp^3d` hybridization.
- **IF₆⁻** has `sp^3d^3` hybridization.

Thus, the ion that has `sp^3d^2` hybridization for the central atom is **ICl₄⁻**.