In the interhalogen compound `AB_(n)`, what is the maximum value of n ?

Explanation

To determine the maximum value of \( n \) in the interhalogen compound \( AB_n \), we need to analyze the bonding capabilities of the halogens involved, particularly focusing on the central atom \( A \).

### Step-by-Step Solution:

1. **Identify the Central Atom**: In an interhalogen compound of the type \( AB_n \), \( A \) is typically a heavier halogen, while \( B \) is a lighter halogen. For this example, we will consider iodine (I) as the central atom \( A \) and fluorine (F) as the surrounding atom \( B \).

2. **Determine the Valence Electrons of the Central Atom**: The atomic number of iodine is 53. Its electron configuration can be expressed as:
\[ \text{I: } [Kr] 4d^{10} 5s^2 5p^5 \]
This indicates that iodine has 7 valence electrons (2 in the 5s orbital and 5 in the 5p orbital).

3. **Calculate the Maximum Number of Bonds**: The maximum number of bonds that the central atom can form is directly related to the number of valence electrons available for bonding. Since iodine has 7 valence electrons, it can form a maximum of 7 bonds with surrounding atoms.

4. **Example of an Interhalogen Compound**: A well-known example of an interhalogen compound is iodine heptafluoride, \( IF_7 \). In this compound:
- Iodine (I) acts as the central atom (A).
- Fluorine (F) acts as the surrounding atom (B).
- The formula \( IF_7 \) indicates that \( n = 7 \).

5. **Hybridization**: To accommodate the formation of 7 bonds, iodine undergoes hybridization. The hybridization state can be described as \( sp^3d^3 \), which involves the mixing of one s orbital, three p orbitals, and three d orbitals to create 7 equivalent hybrid orbitals.

6. **Conclusion**: Therefore, the maximum value of \( n \) in the interhalogen compound \( AB_n \) is 7, as demonstrated by the compound \( IF_7 \).

### Final Answer:
The maximum value of \( n \) in the interhalogen compound \( AB_n \) is 7.

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