Hybridisation of iodine in `IF_(5)` is

Options

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

Explanation

To determine the hybridization of iodine in IF₅, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Group and Valence Electrons:**
- Iodine (I) is in Group 17 of the periodic table, which means it has 7 valence electrons.

2. **Determine the Number of Bonds:**
- In IF₅, iodine forms 5 bonds with 5 fluorine atoms. Each bond is formed by sharing one electron from iodine with one electron from each fluorine.

3. **Calculate Remaining Electrons:**
- Since iodine has 7 valence electrons and it uses 5 of them to form bonds, it will have 2 electrons left. These remaining electrons will form a lone pair.

4. **Count Sigma Bonds and Lone Pairs:**
- In IF₅, there are 5 sigma bonds (one for each I-F bond) and 1 lone pair of electrons.

5. **Determine Hybridization:**
- The hybridization can be determined by adding the number of sigma bonds and lone pairs:
- Total = 5 (sigma bonds) + 1 (lone pair) = 6.
- The hybridization corresponding to 6 regions of electron density is sp³d².

6. **Identify the Molecular Geometry:**
- The presence of one lone pair and five bonded pairs leads to a square pyramidal shape for IF₅.

### Final Answer:
- The hybridization of iodine in IF₅ is **sp³d²**.