The structure of `IF_7` is

Options

A. square pyramid
B. trigonal bipyramid
C. octahedral
D. pentagonal bipyramid

Answer

To determine the structure of IF7 (iodine heptafluoride), we will follow a systematic approach involving electron count, hybridization, and geometry.

### Step-by-Step Solution:

1. **Determine the Valence Electrons:**
- Iodine (I) has 7 valence electrons.
- Each Fluorine (F) has 7 valence electrons, and since there are 7 fluorine atoms, they contribute a total of \(7 \times 7 = 49\) valence electrons.
- Total valence electrons = \(7 + 49 = 56\).

2. **Calculate the Number of Electron Pairs:**
- Each pair of electrons can be counted as one "domain" for hybridization.
- To find the number of electron pairs, divide the total number of valence electrons by 2:
\[ \text{Number of electron pairs} = \frac{56}{2} = 28. \]

3. **Determine the Hybridization:**
- The central atom (I) will use its valence electrons for bonding with the fluorine atoms.
- Since iodine is using 7 of its electrons to bond with 7 fluorine atoms, and it has 14 electrons in total (7 from itself and 7 from fluorine), we can deduce that it will undergo sp³d³ hybridization to accommodate these bonds.

4. **Identify the Geometry:**
- The hybridization sp³d³ corresponds to a geometry known as "pentagonal bipyramidal." This geometry is characterized by 5 atoms in a plane (forming a pentagon) and 2 axial atoms above and below the plane.

5. **Conclusion:**
- Therefore, the structure of IF7 is pentagonal bipyramidal.

### Final Answer:
The structure of IF7 is **pentagonal bipyramidal**.

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