Hybridisation of iodine in `IF_(7)` molecule is

Options

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

Explanation

To determine the hybridization of iodine in the `IF_(7)` molecule, we will follow these steps:

### Step 1: Identify the central atom
The central atom in the `IF_(7)` molecule is iodine (I).

### Step 2: Count the number of bonds
In `IF_(7)`, iodine is bonded to seven fluorine (F) atoms. This means there are 7 sigma bonds formed between iodine and fluorine.

### Step 3: Determine the number of lone pairs
Iodine has 7 valence electrons (as it is in Group 17 of the periodic table). When it forms 7 bonds with fluorine, all of its valence electrons are used for bonding. Therefore, there are no lone pairs of electrons on the iodine atom in `IF_(7)`.

### Step 4: Calculate the steric number
The steric number is calculated using the formula:
\[ \text{Steric Number} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} + \text{Negative Charge} \]

For `IF_(7)`:
- Number of Sigma Bonds = 7 (from the 7 `I-F` bonds)
- Number of Lone Pairs = 0 (none left after bonding)
- Negative Charge = 0 (there are no negative charges)

So, the steric number is:
\[ \text{Steric Number} = 7 + 0 + 0 = 7 \]

### Step 5: Determine the hybridization
Based on the steric number:
- A steric number of 2 corresponds to sp hybridization.
- A steric number of 3 corresponds to sp² hybridization.
- A steric number of 4 corresponds to sp³ hybridization.
- A steric number of 5 corresponds to sp³d hybridization.
- A steric number of 6 corresponds to sp³d² hybridization.
- A steric number of 7 corresponds to sp³d³ hybridization.

Since the steric number for `IF_(7)` is 7, the hybridization of iodine in this molecule is **sp³d³**.

### Conclusion
The hybridization of iodine in the `IF_(7)` molecule is **sp³d³**.

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