The hybridization of `IF_7` is

Options

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

Explanation

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

### Step 1: Determine 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.

### Step 2: Count the number of bonds
In IF₇, iodine is bonded to 7 fluorine (F) atoms. Each bond between iodine and fluorine is a single bond (sigma bond). Therefore, there are 7 sigma bonds.

### Step 3: Calculate the steric number
The steric number is calculated as the number of sigma bonds plus the number of lone pairs on the central atom. In this case, iodine has 7 sigma bonds and no lone pairs (since all valence electrons are used in bonding). Thus, the steric number is:
\[ \text{Steric Number} = \text{Number of Sigma Bonds} + \text{Number of Lone Pairs} = 7 + 0 = 7 \]

### Step 4: Determine the hybridization
The hybridization can be determined from the steric number:
- 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 of IF₇ is 7, the hybridization is sp³d³.

### Step 5: Identify the molecular geometry
With sp³d³ hybridization and a steric number of 7, the molecular geometry of IF₇ is a pentagonal bipyramid.

### Final Answer
The hybridization of IF₇ is **sp³d³**.

---