# Oxalic Acid (H2C2O4) Lewis Structure, Steps of Drawing

Oxalic acid (H2C2O4) is a dibasic weak acid. There are six atoms in H2C2O4 molecule and carbon, oxygen, hydrogen as elements. In this tutorial, we will study how to draw the lewis structure of H2C2O4 step by step.

## H2C2O4 lewis structure

In oxalic acid (H2C2O4) lewis structure, there are two C=O bonds, one C-C bond, two C-O bonds, two O-H bonds. There are no charges on atoms in H2C2O4 lewis structure.

There are two lone pairs on each oxygen atoms. But, there are no lone pairs on carbon atoms.

### Oxalic acid (H2C2O4)

Oxalic acid is an odorless white solid at room temperature. It is highly soluble in water and give a weak acidic solution.

## Steps of drawing lewis structure of H2C2O4

There are several steps to complete the lewis structure of H2C2O4. Each step of drawing the lewis structure is explained in detail in this tutorial.

1. Find total number of electrons of the valance shells of carbon, oxygen and hydrogen atoms
2. Determine total electrons pairs as lone pairs and bonds
3. Find center atom and draw basic skeletal structure
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 H2C2O4

There are three elements in H2C2O4 molecule; carbon, oxygen and hydrogen.

• Carbon atom has 4 electrons in its last shell because it is a IV group element.
• Oxygen is a group VIA element in the periodic table and contains 6 electrons in its last shell.
• Hydrogen, a group IA (alkali metal series group) has only 1 electron in its last shell.

Now we know how many electrons are included in valence shells of each atom and can determine how many valence electrons exist in H2C2O4 molecule.

• valence electrons given by carbon atoms = 4 * 2 = 8
• valence electrons given by oxygen atoms = 6 * 4 = 24
• valence electrons given by hydrogen atoms = 1 * 2 = 12

• Total valence electrons = 8 + 24 + 2 = 34

### 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, H2C2O4 molecule, total pairs of electrons are 17 (34/2) in their valence shells.

### Center atom selection and draw basic skeletal of H2C2O4

To be the center atom, ability of having greater valance and being most electropositive element in the molecule are important facts.

Because, H2C2O4 is a bit complex molecule, it takes some time to select the center atom if you are a beginner to lewis structure drawing.

We can directly eliminate hydrogen from the center atom selection because hydrogen cannot keep more than two electrons in its valence shell.

• Considering greater valence - Oxygen's's highest valence is 2 and carbon's highest valence is 4. From that case, we can suggest that carbon has the higher potential to be the center atom.
• Most electropositive element - Carbon's electronegativity is 2.55 and oxygen's electronegativity is 3.44 according to the Pauling's electronegativity scale. Therefore, carbon is more electropositive than oxygen. In that case, we can suggest that carbon has the potential to be the center atom.
• Because both above facts suggest that carbon atom should be the center atom, we can select carbon as the center atom.
• As well as, because there are two carbon atoms, we assume that there may be two center atoms. (When molecules gets larger and become too much complex, they have several or lot of center atoms)
• From above finding and assumptions, we can draw a skeletal structure for H2C2O4 molecule as below.

### Mark lone pairs on atoms

After deciding the center atom and drawing of skeletal structure of H2C2O4 molecule, we can start to mark lone pairs on atoms. Remember that, there are total of 11 electron pairs to mark between (and on) atoms as bonds and lone pairs.

• There are already 7 bonds in the above drawn skeletal structure. Now only 10 (17-7) electron pairs are remaining to mark on atoms.
• Usually, those remaining electron pairs should be started to mark on outside atoms.
• Because, there are already two electrons in the valence shell of hydrogen atom, more lone pairs cannot be marked on hydrogen atoms.
• Therefore, mark those electrons pairs on each oxygen atom;
• An oxygen atom which has joint a with hydrogen atom will take two more lone pairs and other similar oxygen atom also will accept two more lone pairs. Now. four lone pairs have been marked on those two oxygen atoms. Now, six more lone pairs are remaining to mark on atoms.
• There are two more oxygen atoms and each one will accept three lone pairs. So, total of six lone pairs are marked on those two oxygen atoms too.
• Then, we have finished the step of marking lone pairs on atoms.

### Mark charges on atoms if there are

If there are charges on atoms in above drawn structure, mark them now.

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

There are charges on four atoms (two carbon atoms and two oxygen atoms) and those four atoms are located very close to each other. Because of that, that structure is not very stable.

So, convert a lone pair on an oxygen atom to make a bond with carbon atom as following figure.

Now, you will get a much stable structure because charges of new structure is comparably lower than previous structure. But, when we observe more carefully, it is clear that we can reduce charges furthermore.

Latest structure is much more stable than previous two structures because there are no charges on any atom. So, we have obtained the lewis structure of H2C2O4 molecule.

Questions