# Iodate ion (HCO3-) Ion Lewis Structure

Bicarbonate ion contains one carbon atom, three oxygen atoms and one hydrogen atom. Lewis structure of carbonate ion (HCO3-) contains one C=O bond, two C-O bonds and one O-H bond. There is -1 charge on one oxygen atom in HCO3-lewis structure.

## HCO3- lewis structure

There are three oxygen atoms around center carbon atom. One oxygen atom has made a double bond with carbon atom and hydrogen atom is joint with another oxygen atom. Other oxygen atom has a -1 charge.

## Steps of drawing lewis structure of HCO3-

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 HCO3- ion is an ion and there are five 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 carbon, hydrogen 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 HCO3-

There are three elements in bicarbonate ion; carbon, hydrogen and oxygen. Hydrogen is a group IA element in the periodic table and only has one electron in its last shell (valence shell).Carbon is a group IVA element in the periodic table and has four electrons in its last shell (valence shell). Also, Oxygen is a group VIA element in the periodic table and contains six electrons in its last shell. Now, we know how many electrons are there in valence shells of hydrogen, carbon and oxygen atoms.

• valence electrons given by hydrogen atom = 1 * 1 = 1
• valence electrons given by carbon atom = 4 * 1 = 4
• valence electrons given by oxygen atoms = 6 * 3 = 18

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

• Total valence electrons = 1 + 4 + 18 +1 = 24

### 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, HCO3- ion, Total pairs of electrons are twelve in their valence shells.

### Center atom of HCO3- ion

Hydrogen atom cannot be a center atom because hydrogen can only keep two electrons in last shell.

Because, carbon can show higher valance than oxygen, carbon has the higher priority to be the center atom in HCO3- ion. Also, carbon is more electropositive than oxygen, it again proves, carbon should be the center atom.

### Lone pairs on atoms

After determining the center atom and sketch of HCO3- ion, we can start to mark lone pairs on atoms. Remember that, there are total of twelve electron pairs.

• There are already four bonds in the drawn skeletal. So, now only eight 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. Oxygen atoms which are only connected to carbon atom will take three lone pairs. Then, six electron pairs are marked on those two oxygen atoms.
• Now, only electron pair is remaining and it is marked on remaining oxygen atom as following figure.

### Mark charges on atoms

Two oxygen atoms have -1 charge and carbon atom has +1 charge.

### Check the stability and minimize charges on atoms by converting lone pairs to bonds

Because three atoms have 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 can convert a lone pair of one oxygen atom to make double bond with carbon atom.

You can see, charges has been reduced in new structure than previous structure and we cannot convert more lone pairs to reduce charges.

Questions