`BrF_(3)` and `IF_(5)` both exist in ionic form, in solid state. What are the hybridizations of cationic parts of both, respectively ?

Options

A. `sp^(3), sp^(3)`
B. `sp^(3)d, sp^(3)d`
C. `sp^(3), sp^(3)d`
D. `sp^(2), sp^(3)`

Explanation

To determine the hybridizations of the cationic parts of BrF3 and IF5, we will follow a systematic approach using the formula for calculating the number of hybrid orbitals.

### Step-by-Step Solution:

1. **Identify the Cationic Forms**:
- For BrF3, the cationic part is BrF2+.
- For IF5, the cationic part is IF4+.

2. **Determine the Valence Electrons**:
- Bromine (Br) has 7 valence electrons.
- Iodine (I) has 7 valence electrons.

3. **Count the Monovalent Atoms Attached**:
- In BrF2+, there are 2 fluorine (F) atoms attached.
- In IF4+, there are 4 fluorine (F) atoms attached.

4. **Identify the Charge on the Cation**:
- For BrF2+, the charge is +1.
- For IF4+, the charge is +1.

5. **Using the Hybridization Formula**:
The formula to calculate the number of hybrid orbitals (x) is:
\[ x = \frac{1}{2} \left( \text{Valence electrons on central atom} + \text{Monovalent atoms attached} - \text{Charge on cation} + \text{Charge on anion} \right) \]

6. **Calculate for BrF2+**:
- Valence electrons = 7 (for Br)
- Monovalent atoms attached = 2 (F)
- Charge on cation = +1
- Charge on anion = 0 (since we are only considering the cation)
\[ x = \frac{1}{2} \left( 7 + 2 - 1 + 0 \right) = \frac{1}{2} \left( 8 \right) = 4 \]
- Since x = 4, the hybridization is sp³.

7. **Calculate for IF4+**:
- Valence electrons = 7 (for I)
- Monovalent atoms attached = 4 (F)
- Charge on cation = +1
- Charge on anion = 0
\[ x = \frac{1}{2} \left( 7 + 4 - 1 + 0 \right) = \frac{1}{2} \left( 10 \right) = 5 \]
- Since x = 5, the hybridization is sp³d.

### Final Answer:
- The hybridization of the cationic part of BrF3 (BrF2+) is **sp³**.
- The hybridization of the cationic part of IF5 (IF4+) is **sp³d**.