Which of the following compound have `sp^(3)d^(2)` hybridisation

Options

A. `BrF_(2^(-))`
B. `lCl_(4^(-))`
C. `lCl_(2^(-))`
D. `lF_(7)`

Answer

To determine which of the following compounds has `sp^3d^2` hybridization, we will follow these steps:

### Step 1: Identify the central atom and its valence electrons
- For each compound, identify the central atom and determine its valence electrons based on its position in the periodic table.

### Step 2: Determine the number of sigma bonds
- Count the number of sigma bonds formed by the central atom with surrounding atoms.

### Step 3: Count the lone pairs
- Calculate the number of lone pairs on the central atom by subtracting the number of sigma bonds from the total number of valence electrons (including any negative charge).

### Step 4: Calculate the steric number
- The steric number is the sum of the number of sigma bonds and lone pairs.
- Use the formula: Steric Number = Number of Sigma Bonds + Number of Lone Pairs.

### Step 5: Determine the hybridization
- Based on the steric number, determine the hybridization:
- Steric number 2: `sp`
- Steric number 3: `sp^2`
- Steric number 4: `sp^3`
- Steric number 5: `sp^3d`
- Steric number 6: `sp^3d^2`

### Step 6: Analyze the compounds
1. **For BrF2⁻:**
- Central atom: Bromine (Br) has 7 valence electrons.
- Bonds: 2 sigma bonds with fluorine.
- Remaining electrons: 7 - 2 = 5; with 1 negative charge, total = 6.
- Lone pairs: 6 / 2 = 3 lone pairs.
- Steric number = 2 (sigma bonds) + 3 (lone pairs) = 5.
- Hybridization: `sp^3d`.

2. **For ICl4⁻:**
- Central atom: Iodine (I) has 7 valence electrons.
- Bonds: 4 sigma bonds with chlorine.
- Remaining electrons: 7 - 4 = 3; with 1 negative charge, total = 4.
- Lone pairs: 4 / 2 = 2 lone pairs.
- Steric number = 4 (sigma bonds) + 2 (lone pairs) = 6.
- Hybridization: `sp^3d^2`.

3. **For ICl2⁻:**
- Central atom: Iodine (I) has 7 valence electrons.
- Bonds: 2 sigma bonds with chlorine.
- Remaining electrons: 7 - 2 = 5; with 1 negative charge, total = 6.
- Lone pairs: 6 / 2 = 3 lone pairs.
- Steric number = 2 (sigma bonds) + 3 (lone pairs) = 5.
- Hybridization: `sp^3d`.

4. **For IF7:**
- Central atom: Iodine (I) has 7 valence electrons.
- Bonds: 7 sigma bonds with fluorine.
- Remaining electrons: 7 - 7 = 0.
- Lone pairs: 0.
- Steric number = 7 (sigma bonds) + 0 (lone pairs) = 7.
- Hybridization: `sp^3d^3`.

### Conclusion:
The compound that has `sp^3d^2` hybridization is **ICL4⁻**.