# Lewis Structure of H2CO3 (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of H2CO3?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of H2CO3 (along with images).

So, if you are ready to go with these 6 simple steps, then let’s dive right into it!

Lewis structure of H2CO3 (Carbonic acid) contains one double bond between the Carbon atom (C) & one Oxygen atom (O) and the rest other atoms are single bonded with each other. The Carbon atom (C) is at the center and it is surrounded by one Oxygen atom (O) and two O-H bonds. All the three Oxygen atoms have 2 lone pairs.

Let’s draw and understand this lewis dot structure step by step.

(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of H2CO3).

## 6 Steps to Draw the Lewis Structure of H2CO3

### Step #1: Calculate the total number of valence electrons

Here, the given molecule is H2CO3 (carbonic acid). In order to draw the lewis structure of H2CO3, first of all you have to find the total number of valence electrons present in the H2CO3 molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).

So, let’s calculate this first.

Calculation of valence electrons in H2CO3

• For Hydrogen:

Hydrogen is a group 1 element on the periodic table.

Hence, the valence electron present in hydrogen is 1 (see below image).

• For Carbon:

Carbon is a group 14 element on the periodic table.

Hence, the valence electrons present in carbon is 4 (see below image).

• For Oxygen:

Oxygen is a group 16 element on the periodic table.

Hence, the valence electron present in oxygen is 6 (see below image).

Hence in a H2CO3 molecule,

Valence electrons given by each Hydrogen (H) atom = 1
Valence electrons given by Carbon (C) atom = 4
Valence electrons given by each Oxygen (O) atom = 6
So, total number of Valence electrons in H2CO3 molecule = 1(2) + 4 + 6(3) = 24

### Step #2: Select the center atom

While selecting the atom, always put the least electronegative atom at the center.

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table).

Here in the H2CO3 molecule, if we compare the carbon atom (C), oxygen atom (O) and hydrogen atom (H), then hydrogen is less electronegative than carbon and oxygen. But as per the rule, we have to keep hydrogen outside.

So, carbon (which is less electronegative than oxygen) should be placed in the center and the remaining oxygen atom as well as two OH groups will surround it.

### Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of H2CO3 molecule, put the two electrons (i.e electron pair) between the carbon-oxygen atoms and oxygen-hydrogen atoms to represent a chemical bond between them.

These pairs of electrons present between the Carbon-Oxygen atoms as well as between the Oxygen & Hydrogen atoms form a chemical bond, which bonds these atoms with each other in a H2CO3 molecule.

### Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom

Don’t worry, I’ll explain!

In the Lewis structure of H2CO3, the outer atoms are hydrogen atom as well as oxygen atom.

Hydrogen already has a duplet (see below image).

So now, you have to complete the octet on oxygen atom (because oxygen requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that the oxygen atom forms an octet.

Also, all the 24 valence electrons of H2CO3 molecule (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.

Hence there is no change in the above sketch.

Let’s move to the next step.

### Step #5: Check whether the central atom has octet or not. If it does not have an octet, then move the electron pair from the outer atom to form a double bond or triple bond

In this step, we have to check whether the central atom (i.e carbon) has an octet or not.

In simple words, we have to check whether the central Carbon (C) atom has 8 electrons or not.

As you can see from the above image, the central atom (i.e carbon) has only 6 electrons. So it does not fulfill the octet rule.

Now, in order to fulfill the octet of a carbon atom, we have to move the electron pair from the outer atom (i.e oxygen atom) to form a double bond.

Now you can see from the above image that the central atom (i.e carbon), is having 8 electrons. So it fulfills the octet rule.

### Step #6: Final step – Check the stability of lewis structure by calculating the formal charge on each atom

Now, you have come to the final step and here you have to check the formal charge on oxygen atom (O), carbon atom (C) as well as each hydrogen atom (H).

For that, you need to remember the formula of formal charge;

Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2

• For Hydrogen:
Valence electron = 1 (as it is in group 1)
Nonbonding electrons = 0
Bonding electrons = 2
• For Carbon:
Valence electrons = 6 (as it is in group 16)
Nonbonding electrons = 0
Bonding electrons = 8
• For double bonded Oxygen:
Valence electrons = 6 (as it is in group 16)
Nonbonding electrons = 4
Bonding electrons = 4
• For single bonded Oxygen:
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 4
Bonding electrons = 4

So you can see above that the formal charges on Carbon, Oxygen as well as Hydrogen are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of H2CO3 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of H2CO3 represents the single bond ( | ). So the above lewis dot structure of H2CO3 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of SBr2
Lewis Structure of HOCl
Lewis Structure of C6H6 (Benzene)
Lewis Structure of NBr3
Lewis Structure of SeF4