The hybridization in interhalogen compound `AX_7` is 

Options

A. `sp^3 d^3`
B. `sp^3`
C. `sp^3 d`
D. `sp^3 d^2`

Answer

To determine the hybridization of the interhalogen compound \( AX_7 \), we can follow these steps:

### Step 1: Identify the central atom and terminal atoms
In the compound \( AX_7 \):
- \( A \) is the central atom (a halogen).
- \( X \) represents the terminal atoms (also halogens).

### Step 2: Count the number of terminal atoms
The formula \( AX_7 \) indicates that there are 7 terminal atoms (X).

### Step 3: Determine the hybridization based on the number of atoms
For a compound with 7 terminal atoms, we can use the following hybridization scheme:
- The central atom \( A \) will undergo hybridization to accommodate the 7 surrounding atoms.
- The hybridization can be determined by the formula \( n + m \), where \( n \) is the number of sigma bonds and \( m \) is the number of lone pairs on the central atom.

Since \( A \) is surrounded by 7 \( X \) atoms and assuming it has no lone pairs, we have:
- \( n = 7 \) (7 sigma bonds)
- \( m = 0 \) (no lone pairs)

Thus, the total number of hybrid orbitals required is \( n + m = 7 + 0 = 7 \).

### Step 4: Identify the type of hybridization
The hybridization that corresponds to 7 hybrid orbitals is \( sp^3d^3 \). This hybridization indicates that the geometry of the compound is pentagonal bipyramidal.

### Step 5: Conclusion
The hybridization of the interhalogen compound \( AX_7 \) is \( sp^3d^3 \).

### Final Answer:
The hybridization in the interhalogen compound \( AX_7 \) is \( sp^3d^3 \).

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