# H2CO3 Lewis structure

H2CO3 (carbonic acid) has two hydrogen atoms, one carbon atom, and three oxygen atoms.

In the H2CO3 Lewis structure, there is one double bond and two single bonds around the carbon atom, with three oxygen atoms attached to it. The oxygen atom with a double bond has two lone pairs, and the left oxygen and right oxygen atom (with which the hydrogen atom is attached) also has two lone pairs.

Contents

## Steps

To properly draw the H2CO3 Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary
#4 Minimize formal charges by converting lone pairs of the atoms
#5 Repeat step 4 if necessary, until all charges are minimized

Let’s break down each step in more detail.

### #1 Draw a rough sketch of the structure

• First, determine the total number of valence electrons

In the periodic table, hydrogen lies in group 1, carbon lies in group 14, and oxygen lies in group 16.

Hence, hydrogen has one valence electron, carbon has four valence electrons, and oxygen has six valence electrons.

Since H2CO3 has two hydrogen atoms, one carbon atom, and three oxygen atoms, so…

Valence electrons of two hydrogen atoms = 1 × 2 = 2
Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of three oxygen atoms = 6 × 3 = 18

And the total valence electrons = 2 + 4 + 18 = 24

• Second, find the total electron pairs

We have a total of 24 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 24 ÷ 2 = 12

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Now we have to choose the central atom from carbon and oxygen. Place the least electronegative atom at the center.

Since carbon is less electronegative than oxygen, assume that the central atom is carbon.

Therefore, place carbon in the center and hydrogen and oxygen on either side.

• And finally, draw the rough sketch

### #2 Next, indicate lone pairs on the atoms

Here, we have a total of 12 electron pairs. And five bonds are already marked. So we have to only mark the remaining seven electron pairs as lone pairs on the sketch.

Also remember that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And both (carbon and oxygen) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogens and oxygens. But no need to mark on hydrogen, because each hydrogen has already two electrons.

So for top oxygen, there are three lone pairs, for left oxygen and right oxygen, there are two lone pairs, and for carbon, there is zero lone pair because all seven electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 Indicate formal charges on the atoms, if necessary

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For each hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For carbon atom, formal charge = 4 – 0 – ½ (6) = +1

For top oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

For left oxygen and right oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

Here, both carbon and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both carbon and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Minimize formal charges by converting lone pairs of the atoms

Convert a lone pair of the top oxygen atom to make a new C — O bond with the carbon atom as follows:

In the above structure, you can see that the central atom (carbon) forms an octet. The outside atoms (oxygens) also form an octet, and both hydrogens form a duet. Hence, the octet rule and duet rule are satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of H2CO3.

Next: ICl5 Lewis structure