To draw the Lewis structure for iodate ion (IO4–), we first need to count the total number of valence electrons available. Iodine (I) is in Group 17 and has 7 valence electrons. Oxygen (O) is in Group 16 and has 6 valence electrons, and since we have four oxygen atoms, that gives us 4 × 6 = 24 valence electrons from oxygen. Additionally, because the ion carries a -1 charge, we add 1 more electron to our total count.
Calculating the total, we have:
- 7 (from I) + 24 (from 4 O) + 1 (for the charge) = 32 valence electrons.
Next, we place iodine at the center as it is less electronegative than oxygen. We then connect iodine to each of the four oxygen atoms with a single bond. This uses 4 pairs (8 electrons) of our total, leaving us with 24 electrons still to place.
To satisfy the octet rule for each oxygen, we start placing lone pairs. Each oxygen atom will initially get 3 lone pairs (6 electrons each), which totals to 24 electrons (4 oxygens × 6 electrons). This gives each oxygen a complete octet:
- 8 electrons around each oxygen (2 from the bond with I and 6 from lone pairs).
Now, we check the iodine atom. Iodine is in period 5 and can expand its octet. After forming a single bond with each oxygen, it still has access to more electrons. We can delocalize some of the electrons to create double bonds between iodine and some of the oxygen atoms to distribute charges more evenly and satisfy formal charge considerations.
The final structure shows two double bonds between iodine and two of the oxygen atoms, with the other two oxygens being single-bonded with three lone pairs each. The overall ion has an overall -1 charge.
In summary, the Lewis structure of IO4– looks like this:
- Iodine at the center.
- Two oxygen atoms with double bonds to iodine.
- Two oxygen atoms with single bonds to iodine, each with three lone pairs.
This arrangement satisfies the necessary octet rule and the charge of the ion.