Draw and explain the Lewis structure for IF5. Does this molecule have a central atom that violates the octet rule? Explain.

To discuss the Lewis structure for IF₅ (iodine pentafluoride), we first need to consider the valence electrons of the constituent atoms. Iodine (I) belongs to group 17 and has 7 valence electrons, while each fluorine (F) atom, also from group 17, has 7 valence electrons as well.

In total, for IF₅, we have:

• Iodine: 7 valence electrons
• 5 Fluorines: 5 x 7 = 35 valence electrons

This gives us a total of 42 valence electrons (7 + 35) to work with.

To draw the Lewis structure:

1. Place iodine in the center as it is less electronegative than fluorine.
2. Surround the iodine with five fluorine atoms.
3. Draw single bonds from the iodine to each fluorine atom. Each bond uses 2 electrons, so 5 bonds will use 10 electrons (5 x 2 = 10).
4. Subtracting these from the total, we have 32 electrons remaining (42 – 10 = 32).
5. Now, distribute the remaining electrons to satisfy the octet rule for the fluorine atoms. Each fluorine needs 6 more electrons to complete its octet, using a total of 30 electrons (5 x 6 = 30).
6. This accounts for all 42 electrons: 10 in bonds, 30 on fluorines. Each fluorine will have 3 lone pairs.

The Lewis structure of IF₅ thus looks like this:

[F]   [F]
   |    |
[F]---[I]---[F]
   |    |
  [F]  

Now, regarding the central atom, iodine: it has 7 valence electrons, and in IF₅, it is surrounded by 5 bonding pairs (equivalent to 10 electrons). This results in a total of 17 electrons around iodine, which exceeds the octet rule (8 electrons). Therefore, iodine does violate the octet rule in this molecule.

This phenomenon is explained because iodine can expand its octet due to the availability of d orbitals in its higher energy levels. Hence, it can accommodate more than 8 electrons in its valence shell. This property is characteristic of several elements in periods 3 and beyond.