# Iodate ion (IO3-) Ion Lewis Structure

Iodate ion contains one iodine and three oxygen atoms. Lewis structure of iodate ion (IO3-) contains two I=O bonds and one I-O bond. There is -1 charge on oxygen atom in IO3-lewis structure.

## IO3- lewis structure Oxygen atoms have made bonds with center iodine atom. From those bonds, there are two double bonds and one single bond in the IO3- lewis structure. Also, there is one lone pair exist on iodine atom -1 charge exists on one oxygen atom in the lewis structure of IO3- ion.

## Steps of drawing lewis structure of IO3-

When we draw a lewis structure, there are several guidelines to follow. Number of steps can be changed according the complexity of the molecule or ion. Because IO3- ion is an ion and there are four atoms, we will have more steps than drawing a simple molecule. However those all steps are mentioned and explained in detail in this tutorial for your knowledge.

1. Find total number of electrons of the valance shells of iodine atom and oxygen atoms
2. Determine total electrons pairs existing as lone pairs and bonds
3. Deciding center atom selection
4. Mark lone pairs on atoms
5. Mark charges on atoms if there are charges.
6. Check the stability and minimize charges on atoms by converting lone pairs to bonds to obtain best lewis structure.

### Total number of electrons of the valance shells of IO3-

There are two elements in iodate ion; iodine and oxygen. Oxygen is a group VIA element in the periodic table and has six electrons in its last shell (valence shell). Iodine is a group VIIA element in the periodic table and contains seven electrons in its last shell. Now, we know how many electrons are there in valence shells of oxygen and iodine atoms.

• valence electrons given by oxygen atoms = 6 * 3 = 18
• valence electrons given by iodine atom = 7 * 1 = 7

• Due to -1 charge, one electron should be added to the total number of valence electrons.

• Total valence electrons = 18 + 7 + 1 = 26

### Total valence electrons pairs

Total valance electrons pairs = σ bonds + π bonds + lone pairs at valence shells

Total electron pairs are determined by dividing the number total valence electrons by two. For, IO3- ion, Total pairs of electrons are thirteen in their valence shells.

### Center atom of IO3- ion

Because, iodine can show higher valance than oxygen, iodine has the higher priority to be the center atom in IO3-. Also, iodine is more electropositive than oxygen, it again proves, iodine should be the center atom. ### Lone pairs on atoms

After determining the center atom and sketch of IO3- ion, we can start to mark lone pairs on atoms. Remember that, there are total of thirteen electron pairs.

• There are already three bonds in the drawn skeletal. So, now only ten electron pairs are remaining to mark as lone pairs.
• Usually, those remaining electron pairs should be started to mark on outside atoms. Therefore, we can start to mark those remaining electrons pairs on oxygen atoms. One oxygen atom will take three lone pairs. Then, nine electron pairs are marked on three oxygen atoms.
• Now, one electron pair is remaining and it is marked on iodine atom as following figure. ### Mark charges on atoms

Each oxygen atom has -1 charge and iodine atom has +2 charge. ### Check the stability and minimize charges on atoms by converting lone pairs to bonds

Because every atom has charges, above structure is not stable. Therefore, we should try to reduce charges by converting lone pairs to bonds to find the most stable structures.

As below, we convert lone pairs of oxygen atoms to bonds in two steps. You can see, charges has been reduced in this structure than previous structures.

Questions