Question

Using the Lewis Dot Structure, what does IF6+ look like?

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Answer 1

Answer:

This ion is in an octahedral geometry.

See the diagram attached for the Lewis Dot Structure of the iodide hexafluoride cation $\rm {IF_6}^{+}$. (Created with Google Drawings.)

Note that in this diagram,

• A pair of double dots on an atom represent a lone pair of electrons.
• A single dash represents a single chemical bond.
• The square bracket and the superscript indicates that this structure is charged.

Explanation:

The iodine in $\rm {IF_6}^{+}$ forms an expanded octet. There are twelve valence electrons in total around this atom.

• Each of the six fluorine atom needs 8 - 7 = 1 electron to achieve an octet of eight electrons.
• The iodine atom needs 12 - 7 = 5 electrons to achieve an expanded octet of twelve electrons.
• The ion carries a positive charge of +1. Atoms in this ion lacks one extra electron.

Overall, there needs to be $6 \times 1 + 5 + 1= 12$ more electrons for seven atoms to achieve an octet. They will form half that number of chemical bonds. That's $12 / 2 = 6$ bonds.

Now consider: what will be the geometry of this ion? There are six chemical bonds but no lone pair around the central iodine atom. The six $\mathrm{I-F}$ bonds repel each other equally. They will stay as far apart from each other as possible. As a result, the shape of the ion will be octahedral. Each of the fluorine atoms occupies a vertex.