In the formation of `XA_5` type interhalogen compound, X undergoes 

Options

A. `sp^3`, hybridisation
B. `sp^3 d`, hybridisation
C. `dsp^2`, hybridisation
D. `sp^3 d^2`, hybridisation

Answer

To solve the question regarding the formation of an `XA5` type interhalogen compound, we need to analyze the hybridization of the central atom `X`. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the Elements**:
- In the compound `XA5`, `X` is a halogen and `A` is also a halogen. Halogens include elements like fluorine (F), chlorine (Cl), bromine (Br), iodine (I), and astatine (At).

2. **Determine the Valence Electrons of `X`**:
- Halogens have 7 valence electrons. Therefore, `X` will have 7 electrons in its outer shell.

3. **Count the Number of Monovalent Atoms**:
- In the compound `XA5`, there are 5 halogen atoms `A`. Since each halogen atom is monovalent (can form one bond), we have a total of 5 bonds.

4. **Calculate the Hybridization**:
- The number of hybrid orbitals required can be determined using the formula:
\[ \text{Number of hybrid orbitals} = \frac{\text{Number of monovalent atoms} + \text{Number of lone pairs}}{2} \]
- Here, the number of monovalent atoms is 5 (from `A`), and `X` has 7 valence electrons. If `X` forms 5 bonds, it will have 2 lone pairs remaining (since 7 - 5 = 2).
- Therefore, the total number of orbitals involved in hybridization is:
\[ 5 \text{ (from A)} + 2 \text{ (lone pairs)} = 7 \]

5. **Determine the Hybridization Type**:
- The hybridization that corresponds to 7 orbitals is `sp^3d^2`. This means that `X` undergoes hybridization to form 7 hybrid orbitals.

6. **Conclusion**:
- Thus, in the formation of the `XA5` type interhalogen compound, `X` undergoes `sp^3d^2` hybridization.

### Final Answer:
In the formation of `XA5` type interhalogen compound, `X` undergoes **sp³d² hybridization**.