Carbonic acid (H2CO3) Lewis dot structure, molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polar or nonpolar
No one is unfamiliar with carbonic acid, the famous ingredient of fizzy drinks and sparkling water. The chemical formula for carbonic acid is H2CO3. It is a diprotic weak organic acid. It is also used in the chemical industry for precipitating ammonium salts, and cleaning lenses, in addition to having a prominent role in the blood buffer system of the human body.
Considering the multifold importance of carbonic acid (H2CO3), we have compiled this article for you so that you can learn how to draw the Lewis dot structure of H2CO3. What is its molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, polarity, etc.?
So without any further delay, let’s start reading and satisfying the inborn chemist within us all!
|Name of Molecule||Carbonic acid|
|Molecular geometry of H2CO3||Trigonal planar|
|Electron geometry of H2CO3||Trigonal planar|
∠H-O-C (106.8°), ∠O=C-O (125.5°) , ∠HO-C-O (109°)
|Total Valence electron in H2CO3||24|
|Overall Formal charge in H2CO3||Zero|
How to draw lewis structure of H2CO3?
The Lewis structure of carbonic acid (H2CO3) consists of a carbon (C) atom at the centre. It is double-covalently bonded to an oxygen (O) atom on one side and single-covalently bonded to two hydroxyl (OH) functional groups on the other two sides.
In this way, there are a total of three electron density regions surrounding the central C-atom in the H2CO3 Lewis structure. All three electron density regions or electron domains are constituted of bond pairs; thus, there is no lone pair of electrons on the central C-atom in this Lewis structure.
Drawing the Lewis dot structure of carbonic acid (H2CO3) is super easy to learn, especially if you follow the simple steps given below.
Steps for drawing the Lewis dot structure of H2CO3
1. Count the total valence electrons in H2CO3
The very first step while drawing the Lewis structure of H2CO3 is to find the total valence electrons present in its concerned elemental atoms.
As three different elemental atoms are present in H2CO3, so you first need to look for the position of these elements in the Periodic Table.
Carbon (C) belongs to Group IV A (or 14), so it has a total of 4 valence electrons. Oxygen (O) is present in Group VI A (or 16), so it has 6 valence electrons, while hydrogen (H) lies at the top of the Periodic Table containing a single valence electron only.
- Total number of valence electrons in hydrogen = 1
- Total number of valence electrons in carbon = 4
- Total number of valence electrons in oxygen = 6
∴ The H2CO3 molecule consists of 1 C-atom, 3 O-atoms, and 2 H-atoms. Therefore, the total valence electrons available for drawing the Lewis dot structure of H2CO3 = 1(4) + 3(6) + 2(1) = 24 valence electrons.
2. Choose the central atom
In this second step, usually the least electronegative atom out of all the concerned atoms is chosen as the central atom.
This is because the least electronegative atom is the one that is most likely to share its electrons with the atoms spread around it.
In H2CO3, oxygen is the most electronegative element, so an oxygen atom cannot be chosen as the central atom in the Lewis structure. Hydrogen is less electronegative than carbon. However, an H-atom has the ability to form a single covalent bond with an adjacent atom only by accommodating a total of 2 valence electrons. Therefore, H can also not be selected as the central atom in any Lewis structure.
Consequently, the C-atom is placed as the central atom in the H2CO3 Lewis structure, while the three O and two H atoms occupy terminal positions, as shown below.
3. Connect outer atoms with the central atom
In this step, we need to connect the outer atoms with the central atom of the molecule using single straight lines. As already identified, the O and H atoms are the outer atoms in the H2CO3 Lewis structure.
An important point to note is that the H-atoms can only form a single bond with their adjacent atom. Hence the two H-atoms are joined to their adjacent O-atom respectively and are not directly connected to the central C-atom.
In contrast, all three O-atoms are directly joined to the central C-atom using single straight lines, as shown below.
Each straight line represents a bond pair containing 2 electrons. There are a total of 5 straight lines in the above structure, which represents a total of 5(2) = 10 valence electrons.
10 valence electrons consumed out of the 24 initially available still leave behind 14 valence electrons that we need to accommodate in the above Lewis structure.
Let’s see through the following steps how we can do so.
4. Complete the duplet and/or octet of the outer atoms
The hydrogen and oxygen atoms are the outer atoms in the Lewis dot structure of H2CO3.
Each hydrogen (H) atom requires a total of 2 valence electrons in order to achieve a stable duplet electronic configuration.
An O-H single bond represents 2 valence electrons around each H-atom. This means both the H-atoms already have a complete duplet in the Lewis structure drawn till yet. Thus, we do not need to make any changes with regard to the hydrogen atoms in this structure.
In contrast, an O-atom needs a total of 8 valence electrons to achieve a stable octet electronic configuration.
There are two different types of O-atoms present in the above Lewis structure.
First, look at oxygen no.1; it only contains a C-O single bond denoting 2 valence electrons. It is thus short of 6 more electrons in order to complete its octet. Thus, these 6 electrons are placed as 3 lone pairs around this oxygen atom.
Now you may look at oxygen no.2 and no.3. Both these O-atoms are identical, i.e., both contain a C-O and an O-H bond. 2 single bonds mean 4 valence electrons; thus, each O-atom has a deficiency of 4 more electrons for it to complete its octet. Thus, 2 lone pairs of electrons are placed around these two O-atoms, as shown below.
5. Complete the octet of the central atom by converting a lone pair into a covalent bond if necessary
- Total valence electrons used till step 4 = 5 single bonds + electrons placed around oxygen no.1 + 2 (electrons placed around oxygen no.2) = 5(2) +6+2(4) = 24 valence electrons.
- Total valence electrons – electrons used till step 4 = 24– 24 = 0 valence electrons.
As all the 24 valence electrons initially available to draw the H2CO3 Lewis structure are already consumed, thus there is no lone pair on the central C-atom.
However, a problem here is that the central C-atom in the above structure has a total of 3 single bonds around it which denotes 6 valence electrons only. It is thus short of 2 more electrons in order to gain a complete octet.
This problem can be solved by converting a lone pair on an outer O-atom into an extra covalent bond between the central C-atom and the respective O-atom.
Out of the three O-atoms available, it is fair to choose the O-atom containing an extra lone pair compared to the other two O-atoms. So a lone pair from the O-atom containing three lone pairs of electrons are converted into a bond pair, as shown below.
In this way, the central C-atom now has a complete octet with 2 single bonds + 1 double bond.
So, our final step is to check the stability of the H2CO3 Lewis structure obtained above. We can do that by using the formal charge concept.
6. Check the stability of the H2CO3 Lewis structure using the formal charge concept
The fewer formal charges present on the atoms of a molecule, the better the stability of its Lewis structure.
The formal charges can be calculated using the formula given below.
- Formal charge = [ valence electrons – nonbonding electrons- ½ (bonding electrons)]
Now let us use this formula and the Lewis structure obtained in step 5 to determine the formal charges on H2CO3 atoms.
For carbon atom
- Valence electrons of carbon = 4
- Bonding electrons = 2 single bonds + 1 double bond = 2(2) + 4 = 8 electrons
- Non-bonding electrons = no lone pair = 0 electrons
- Formal charge = 4-0-8/2 = 4-0-4 = 4-4 = 0
For C-O-H bonded oxygen atoms
- Valence electrons of oxygen = 6
- Bonding electrons = 2 single bonds = 2(2) = 4 electrons
- Non-bonding electrons = 2 lone pairs = 2 (2) = 4 electrons
- Formal charge = 6-4-4/2 = 6-4-2 = 6-6 = 0
For C=O bonded oxygen atom
- Valence electrons of oxygen = 6
- Bonding electrons = 1 double bond = 4 electrons
- Non-bonding electrons = 2 lone pairs = 2(2) = 4 electrons
- Formal charge = 6-4-4/2 = 6-4-2 = 6-6 = 0
The above calculation shows that there are no or zero formal charges on all the bonded atoms in the H2CO3 Lewis structure. Thus, the overall charge present on the molecule is also zero.
In conclusion, we have obtained the correct and most stable Lewis representation of the carbonic acid molecule.
The following three resonance forms are possible for drawing H2CO3 molecules. Each resonance form is a way of representing the Lewis structure of a molecule.
However, as you can see, specific formal charges are present in resonance forms 2 and 3; thus, they are certainly less stable than the most stable resonance form of carbonic acid, i.e., structure 1, identical to what we obtained above.
Also check –
What are the electron and molecular geometry of H2CO3?
The carbonic acid (H2CO3) molecule possesses an identical electron and molecular geometry or shape, i.e., trigonal planar. To a C-atom at the center, one O-atom and two OH groups are attached, which makes a total of 3 electron density regions or electron domains around it.
There is no lone pair of electrons on the central C-atom, so no distortion is witnessed in the shape and geometry of H2CO3.
Molecular geometry of H2CO3
The molecular geometry or shape of carbonic acid (H2CO3) is trigonal planar.
The absence of any lone pair of electrons on the central C-atom means no lone pair-lone pair and lone pair-bond pair electronic repulsions exist in the molecule.
A bond pair-bond pair repulsive effect is present. The OH functional group present around the C-atom is considered a single electron domain. A double-bonded O-atom and two single-bonded OH groups occupy the three vertices of an equilateral triangle in a perfectly symmetrical arrangement, forming a trigonal planar molecular geometry and shape, as shown in the figure below.
Electron geometry of H2CO3
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule containing a total of 3 electron density regions around the central atom is trigonal planar.
In H2CO3, one O-atom and two OH groups make a total of 3 electron density regions around the central C-atom. Thus, its electron geometry is also trigonal planar.
A shortcut to finding the electron and the molecular geometry of a molecule is by using the AXN method.
AXN is a simple formula to represent the number of atoms bonded to the central atom in a molecule and the number of lone pairs present on it.
It is used to predict the shape and geometry of a molecule based on the VSEPR concept.
AXN notation for H2CO3 molecule
- A in the AXN formula represents the central atom. In H2CO3, a carbon (C) atom is present at the center, so A = C.
- X denotes the atoms bonded to the central atom. In H2CO3, 1 O-atom and 2 OH groups are directly bonded to the central C-atom. The OH group is considered 1 region of electron density. In short, X = 3 for H2CO3.
- N stands for the lone pairs present on the central atom. As per the Lewis structure of H2CO3, there are no lone pairs of electrons on the central carbon; hence N = 0.
As a result, the AXN generic formula for H2CO3 is AX3.
Now, you may have a quick look at the VSEPR chart below.
The VSEPR chart confirms that the molecular geometry or shape of a molecule with an AX3 generic formula is identical to its ideal electron pair geometry, i.e., trigonal planar, as we already noted down for the carbonic acid (H2CO3) molecule.
Hybridization of H2CO3
The central C-atom is sp2 hybridized in H2CO3.
The electronic configuration of carbon is 1s2 2s2 2p2.
During chemical bonding, the 2s atomic orbital of carbon hybridizes with two of its 2p atomic orbitals to produce three sp2 hybrid orbitals.
Each sp2 hybrid orbital is equivalent and contains a single electron only. It possesses a 33.3 % s-character and a 67.7 % p-character.
The sp2 hybrid orbitals of carbon overlap with the orbitals of adjacent atoms to form the required sigma (σ) bonds.
An unhybridized p-orbital of the central carbon atom overlaps with the p-orbital of the double-bonded O-atom to form the C=O pi (π) bond.
Refer to the figure drawn below.
H2CO3 hybridization can also be determined from its steric number. The steric number of the central C-atom in H2CO3 is 3, so it has sp2 hybridization.
The H2CO3 bond angle
The ideal bond angle in a symmetrical trigonal planar shape is 120°. However, it is due to the different types of bonds present in H2CO3 that more than one different types of bond angle are present in it.
In ascending order, the H-O-C bond angle is 106.8°, and the O-C-O and O=C-O bond angles are 109°and 125.5°, respectively.
Similarly, the different bond lengths present in H2CO3 are 120 pm (C=O), 133 pm (C-O), and 96 pm (O-H).
Also check:- How to find bond angle?
Is H2CO3 polar or nonpolar?
Pauling’s electronegativity scale states that a covalent chemical bond is polar if the bonded atoms have an electronegativity difference between 0.5 to 1.6 units.
In H2CO3, three different types of covalent bonds are present, i.e., a C=O bond, a C-O bond, and an O-H bond.
An electronegativity difference of 0.89 units exists between a carbon and an oxygen (E.N = 3.44) atom. So both C-O and C=O bonds are strongly polar.
Conversely, the O-H bonds present in H2CO3 are extremely polar as an even higher electronegativity difference of 1.24 units exists between an oxygen and a hydrogen atom.
The highly electronegative oxygen atoms strongly attract the shared electron cloud from each bond of H2CO3.
A partial negative (δ–) charge develops on each O-atom while the C –atom and H-atoms acquire partial positive (δ+) charges.
The individual dipole moments of these polar bonds do not get canceled in the molecule overall. Thus, it is due to this non-uniform electron cloud distribution that H2CO3 is a polar molecule (net µ > 0).
Read in detail–
What is the Lewis structure for H2CO3?
There is no lone pair on the central C-atom or any of the H-atoms in the H2CO3 Lewis structure.
What is the molecular geometry or shape of H2CO3?
The carbonic acid (H2CO3) molecule has a trigonal planar molecular geometry or shape. To one C-atom, one O-atom and two OH-groups are directly attached, and there is no lone pair of electrons on the central C-atom; thus, no distortion is witnessed in its shape and/or geometry.
Is the molecular shape of H2CO3 different from its ideal electron pair geometry as per the VSEPR concept?
No. H2CO3 has an identical electron and molecular geometry or shape.
As per the VSEPR concept, the AXN generic formula for H2CO3 is AX3N0 or simply AX3.
The absence of any lone pair of electrons on the central C-atom leads to an undistorted shape and electronic geometry.
How is the shape of H2CO3 different from that of HCOOH?
Both H2CO3 and HCOOH have a similar trigonal planar shape as there are a total of three electron domains around the central C-atom, and it has no lone pair of electrons in each case, as shown below.
Why is H2CO3 polar even though it has an ideal trigonal planar shape?
The polarity of H2CO3 is due to the presence of three highly electronegative oxygen atoms. The O-atoms strongly attract the C-O, C=O and O-H bonded electrons.
The charge distribution stays non-uniform in the molecule. It is thus polar (net µ > 0).
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- The total number of valence electrons available for drawing carbonic acid (H2CO3) Lewis structure is 24.
- H2CO3 has an identical electron and molecular geometry or shape, i.e., trigonal planar.
- The H2CO3 molecule has sp2 hybridization.
- Multiple bond angles and bond lengths are present in the carbonic acid molecule.
- H2CO3 is a polar molecule due to three highly electronegative O-atoms and an unbalanced charge distribution overall (net µ > 0).
- Zero formal charges present on all the bonded atoms account for the extraordinary stability of the H2CO3 Lewis structure drawn in this article.
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