So you have seen the above image by now, right?
Let me explain the above image in short.
H2CO3 lewis structure has a Carbon atom (C) at the center which is surrounded by one Oxygen atom (O) and two O-H groups. There is 1 double bond between the Carbon atom (C) & Oxygen atom (O) and the rest other atoms have a single bond. There are 2 lone pairs on all three Oxygen atoms (O).
If you haven’t understood anything from the above image of H2CO3 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of H2CO3.
So let’s move to the steps of drawing the lewis structure of H2CO3 molecule.
Steps of drawing H2CO3 lewis structure
Step 1: Find the total valence electrons in H2CO3 molecule
In order to find the total valence electrons in a H2CO3 molecule, first of all you should know the valence electrons present in hydrogen atom, carbon atom as well as oxygen atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)
Here, I’ll tell you how you can easily find the valence electrons of hydrogen, carbon as well as oxygen using a periodic table.
Total valence electrons in H2CO3 molecule
→ Valence electrons given by hydrogen atom:
Hydrogen is group 1 element on the periodic table. [1] Hence the valence electron present in hydrogen is 1.
You can see that only 1 valence electron is present in the hydrogen atom as shown in the above image.
→ Valence electrons given by carbon atom:
Carbon is group 14 element on the periodic table. [2] Hence the valence electrons present in carbon is 4.
You can see the 4 valence electrons present in the carbon atom as shown in the above image.
→ Valence electrons given by oxygen atom:
Oxygen is group 16 element on the periodic table. [3] Hence the valence electrons present in oxygen is 6.
You can see the 6 valence electrons present in the oxygen atom as shown in the above image.
Hence,
Total valence electrons in H2CO3 molecule = valence electrons given by 2 hydrogen atoms + valence electrons given by 1 carbon atom + valence electrons given by 3 oxygen atoms = 1(2) + 4 + 6(3) = 24.
Step 2: Select the central atom
For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.
(Remember: If hydrogen is present in the given molecule, then always put hydrogen outside.)
Now here the given molecule is H2CO3 and it contains hydrogen atom (H), carbon atom (C) and oxygen atoms (O).
So as per the rule we have to keep hydrogen outside.
Now, you can see the electronegativity values of carbon atom (C) and oxygen atom (O) in the above periodic table.
If we compare the electronegativity values of carbon (C) and oxygen (O) then the carbon atom is less electronegative.
So here the carbon atom (C) is the center atom and the oxygen atoms (O) are the outside atom.
Step 3: Connect each atoms by putting an electron pair between them
Now in the H2CO3 molecule, you have to put the electron pairs between the oxygen (O) & hydrogen (H) atom and between the oxygen (O) & carbon (C) atoms.
This indicates that these atoms are chemically bonded with each other in a H2CO3 molecule.
Step 4: Make the outer atoms stable
Now in this step, you have to check the stability of the outer atoms.
Here in the sketch of H2CO3 molecule, you can see that the outer atoms are hydrogen and oxygen atoms.
These hydrogen and oxygen atoms are forming a duplet and octet respectively and hence they are stable.
Also, in step 1 we have calculated the total number of valence electrons present in the H2CO3 molecule.
The H2CO3 molecule has a total 24 valence electrons and all these valence electrons are used in the above sketch.
Hence there are no remaining electron pairs to be kept on the central carbon atom.
So now let’s proceed to the next step.
Step 5: Check the octet on the central atom. If it does not have octet, then shift the lone pair to form a double bond or triple bond.
In this step, you have to check whether the central carbon atom (C) is stable or not.
In order to check the stability of the central carbon (C) atom, we have to check whether it is forming an octet or not.
Unfortunately, the carbon atom is not forming an octet here. Carbon has only 6 electrons and it is unstable.
Now to make this carbon atom stable, you have to shift the electron pair from the outer oxygen atom so that the carbon atom can have 8 electrons (i.e octet).
After shifting this electron pair, the central carbon atom will get 2 more electrons and thus its total electrons will become 8.
You can see from the above picture that the carbon atom is forming an octet as it has 8 electrons.
Now let’s proceed to the final step to check whether the above lewis structure is stable or not.
Step 6: Check the stability of lewis structure
Now you have come to the final step in which you have to check the stability of lewis structure of H2CO3.
The stability of lewis structure can be checked by using a concept of formal charge.
In short, now you have to find the formal charge on hydrogen (H) atoms, carbon (C) atom as well as oxygen (O) atoms present in the H2CO3 molecule.
For calculating the formal charge, you have to use the following formula;
Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons
You can see the number of bonding electrons and nonbonding electrons for each atom of H2CO3 molecule in the image given below.
For Hydrogen (H) atom:
Valence electron = 1 (because hydrogen is in group 1)
Bonding electrons = 2
Nonbonding electrons = 0
For Carbon (C) atom:
Valence electrons = 4 (because carbon is in group 14)
Bonding electrons = 8
Nonbonding electrons = 0
For double bonded Oxygen (O) atom:
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 4
Nonbonding electrons = 4
For single bonded Oxygen (O) atom:
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 4
Nonbonding electrons = 4
Formal charge | = | Valence electrons | – | (Bonding electrons)/2 | – | Nonbonding electrons | ||
H | = | 1 | – | 2/2 | – | 0 | = | 0 |
C | = | 4 | – | 8/2 | – | 0 | = | 0 |
O (double bonded) | = | 6 | – | 4/2 | – | 4 | = | 0 |
O (single bonded) | = | 6 | – | 4/2 | – | 4 | = | 0 |
From the above calculations of formal charge, you can see that the hydrogen (H) atoms, carbon (C) atom as well as oxygen (O) atoms have a “zero” formal charge.
This indicates that the above lewis structure of H2CO3 is stable and there is no further change in the above structure of H2CO3.
In the above lewis dot structure of H2CO3, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of H2CO3.
I hope you have completely understood all the above steps.
For more practice and better understanding, you can try other lewis structures listed below.
Try (or at least See) these lewis structures for better understanding:
HOCl Lewis Structure | C6H6 (Benzene) Lewis Structure |
NBr3 Lewis Structure | SeF4 Lewis Structure |
H3PO4 Lewis Structure | H2Se Lewis Structure |
Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.
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