What is the shape of IOF4+?
square planar
square pyramidal
trigonal bipyramidal
see-saw
octahedral
The Correct Answer and Explanation is :
The shape of IOF₄⁺ (Iodine Oxytetrafluoride cation) is square pyramidal.
Explanation (300 Words)
To determine the molecular shape of IOF₄⁺, we follow the VSEPR (Valence Shell Electron Pair Repulsion) Theory:
- Valence Electrons Calculation:
- Iodine (I): Group 17 → 7 valence electrons.
- Oxygen (O): Group 16 → 6 valence electrons.
- Fluorine (F): Group 17 → 7 valence electrons each (4 Fluorine atoms → 4 × 7 = 28).
- Total electrons before charge adjustment:
[
7 (I) + 6 (O) + 28 (F) = 41
] - Charge on the ion: +1, meaning one electron is lost, so the total count is 40 valence electrons.
- Lewis Structure and Steric Number:
- Iodine (I) is the central atom, bonded to one oxygen (O) and four fluorine (F) atoms.
- Five regions of electron density (four single bonds with F, one double bond with O).
- One lone pair remains on iodine.
- Molecular Geometry:
- With five electron groups, the electron geometry is trigonal bipyramidal.
- The presence of one lone pair makes the actual molecular shape square pyramidal, where four fluorine atoms form a square base and the oxygen occupies an axial position.
- Confirmation:
- This shape is consistent with other known iodine pentahalides and follows expected bond angles (approx. 90° in the square base and <90° between axial and base atoms).
Thus, the correct answer is square pyramidal.
Now, I will generate an image of the molecular structure.
Here is the 3D molecular model of IOF₄⁺ in its square pyramidal shape. The structure shows the iodine atom at the center, bonded to four fluorine atoms forming a square base, with an oxygen atom positioned at the top. Let me know if you need further modifications or explanations!