How many orbitals are involved in the hybridisation of idione in `IF_(7)` ?

Explanation

To determine how many orbitals are involved in the hybridization of iodine in the compound IF₇, we can follow these steps:

### Step 1: Identify the central atom and its valence electrons
The central atom in IF₇ is iodine (I). The atomic number of iodine is 53, and its electron configuration can be written as:
- **Krypton (Kr)**: 4d¹⁰ 5s² 5p⁵

From this configuration, we can see that iodine has a total of **7 valence electrons** (2 from 5s and 5 from 5p).

### Step 2: Determine the type of hybridization
In IF₇, iodine is surrounded by 7 fluorine atoms. To accommodate these 7 bonds, iodine will undergo hybridization. The hybridization can be determined by the number of orbitals that need to be mixed to form new hybrid orbitals.

### Step 3: Calculate the number of orbitals involved
To form 7 bonds with fluorine atoms, iodine needs to use:
- 1 s orbital
- 3 p orbitals
- 3 d orbitals

This gives us a total of:
- **1 (s) + 3 (p) + 3 (d) = 7 orbitals**

### Step 4: Identify the hybridization type
The hybridization that involves 1 s orbital, 3 p orbitals, and 3 d orbitals is referred to as **sp³d³ hybridization**.

### Conclusion
Thus, the number of orbitals involved in the hybridization of iodine in IF₇ is **7**.

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