Which of the following have `sp^(3)` d hybridisation ? .

Options

A. `SF_(4)`
B. `BrCI_(3)`
C. `XeOF_(2)`
D. `H_(3)O^(o+)`

Explanation

To determine which of the given species have `sp^3d` hybridization, we will follow these steps:

### Step 1: Identify the species and calculate the total number of valence electrons (V) for each.

1. **For SF4**:
- Sulfur (S) has 6 valence electrons.
- Each Fluorine (F) has 7 valence electrons, and there are 4 Fluorine atoms.
- Total V = 6 + (4 × 7) = 6 + 28 = 34.

2. **For BrCl3**:
- Bromine (Br) has 7 valence electrons.
- Each Chlorine (Cl) has 7 valence electrons, and there are 3 Chlorine atoms.
- Total V = 7 + (3 × 7) = 7 + 21 = 28.

3. **For XeOF2**:
- Xenon (Xe) has 8 valence electrons.
- Oxygen (O) has 6 valence electrons.
- Each Fluorine (F) has 7 valence electrons, and there are 2 Fluorine atoms.
- Total V = 8 + 6 + (2 × 7) = 8 + 6 + 14 = 28.

4. **For H3O+**:
- Oxygen (O) has 6 valence electrons.
- Each Hydrogen (H) has 1 valence electron, and there are 3 Hydrogen atoms.
- Since it has a positive charge, we subtract 1 electron.
- Total V = 6 + (3 × 1) - 1 = 6 + 3 - 1 = 8.

### Step 2: Calculate the number of bond pairs and lone pairs.

To find the hybridization, we will divide the total number of valence electrons (V) by 8.

1. **For SF4**:
- V = 34.
- Number of bond pairs = V / 8 = 34 / 8 = 4 (quotient).
- Remainder = 34 % 8 = 2, which corresponds to 1 lone pair (2/2 = 1).
- Hybridization = `sp^3d`.

2. **For BrCl3**:
- V = 28.
- Number of bond pairs = V / 8 = 28 / 8 = 3 (quotient).
- Remainder = 28 % 8 = 4, which corresponds to 2 lone pairs (4/2 = 2).
- Hybridization = `sp^3`.

3. **For XeOF2**:
- V = 28.
- Number of bond pairs = V / 8 = 28 / 8 = 3 (quotient).
- Remainder = 28 % 8 = 4, which corresponds to 2 lone pairs (4/2 = 2).
- Hybridization = `sp^3d`.

4. **For H3O+**:
- V = 8.
- Number of bond pairs = V / 8 = 8 / 8 = 1 (quotient).
- Remainder = 8 % 8 = 0, which corresponds to 0 lone pairs.
- Hybridization = `sp^3`.

### Step 3: Conclusion

From the calculations:
- SF4 has `sp^3d` hybridization.
- BrCl3 has `sp^3` hybridization.
- XeOF2 has `sp^3d` hybridization.
- H3O+ has `sp^3` hybridization.

Thus, the species that have `sp^3d` hybridization are **SF4 and XeOF2**.