Hybridisation of iodine hepta fluoride molecule is

Options

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

Explanation

To determine the hybridization of the iodine heptafluoride (IF7) molecule, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the central atom and its valence electrons**:
- The central atom in IF7 is iodine (I). Iodine is in group 17 of the periodic table and has 7 valence electrons.

2. **Count the number of surrounding atoms**:
- In IF7, there are 7 fluorine (F) atoms bonded to the iodine atom.

3. **Determine the total number of electron pairs**:
- Each bond between iodine and fluorine involves a pair of electrons. Since there are 7 fluorine atoms, there will be 7 bonding pairs of electrons.

4. **Calculate the total number of orbitals needed for hybridization**:
- Iodine needs to accommodate 7 bonding pairs of electrons. Thus, it requires 7 hybrid orbitals.

5. **Identify the types of orbitals involved in hybridization**:
- To form 7 hybrid orbitals, iodine will use:
- 1 s orbital
- 3 p orbitals
- 3 d orbitals
- This gives a total of 1 + 3 + 3 = 7 orbitals.

6. **Determine the hybridization type**:
- The combination of these orbitals (1 s, 3 p, and 3 d) leads to the formation of sp³d³ hybrid orbitals.

7. **Conclude the hybridization**:
- Therefore, the hybridization of the iodine heptafluoride molecule (IF7) is **sp³d³**.

### Final Answer:
The hybridization of iodine heptafluoride (IF7) is **sp³d³**.

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