This is the IF5 Lewis structure.
For IF5, we have a total of 42 valence electrons.
Iodine is the least electronegative.
We'll put that at the center, and then we'll
put Fluorines around the outside.
Next, we'll draw a single bond between the
Iodine and each of the Fluorines to form chemical
bonds.
So we've used a total of 2, 4, 6, 8, 10 valence
electrons.
Let's complete the octets for the Fluorines.
So we have 10, 12, 14, and 40.
So we've used 40 valence electrons.
We had 42 to start out with.
All the Fluorines have octets.
Iodine's in period 5 on the periodic table,
so it can have more than eight valence electrons.
So let's just put that last pair on the Iodine
here.
Now Iodine has 2, 4, 6, 8, 10, 12, but that's
fine.
If you check the formal charges for each of
the atoms in this molecule, you'll find that
they're zero.
So this is the Lewis structure for IF5.
This is Dr. B., and thanks for watching.
