To draw the Lewis structure for IOF5, we first need to determine the total number of valence electrons in the molecule. Iodine (I) has 7 valence electrons, oxygen (O) has 6, and each fluorine (F) has 7. Thus, the total number of valence electrons is:
- 1 Iodine: 7 electrons
- 1 Oxygen: 6 electrons
- 5 Fluorine: 5 x 7 = 35 electrons
Total = 7 + 6 + 35 = 48 valence electrons.
Next, we place iodine in the center as the central atom and surround it with the five fluorine atoms and one oxygen atom. Each fluorine and oxygen will be connected to iodine with a single bond initially. Therefore, 6 bonds use up 12 electrons (2 electrons per bond), leaving us with 36 electrons to distribute.
Since each fluorine atom needs 3 lone pairs (6 electrons) to complete their octet, we place 3 lone pairs on each of the five fluorine atoms, consuming 30 electrons. This uses up all but 6 of our original 48 electrons. The remaining 6 electrons can be distributed as three lone pairs around the central iodine atom.
The final Lewis structure shows iodine at the center with a single bond to each fluorine and one bond to the oxygen, leading to the following arrangement:
F
|
F–I–O
|
F
F
F
Now, to determine the electron and molecular geometries, we consider the regions of electron density around the iodine atom. The iodine atom has 5 bond pairs (from the five fluorine atoms) and 3 lone pairs (from its own lone pairs). This results in a total of 8 regions of electron density.
Using VSEPR theory, we can deduce that the electron geometry around iodine is octahedral due to the 8 regions of electron density. However, the molecular geometry focuses on the atoms only, which, due to the presence of lone pairs, results in a square pyramidal shape.
In summary:
- Electron Geometry: Octahedral
- Molecular Geometry: Square Pyramidal