Formic acid (HCOOH) Lewis structure, molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polar or nonpolar
The simple chemical formula HCOOH represents formic acid, also known as methanoic acid. It is the simplest of carboxylic acids. The word ‘formic’ comes from the Latin formica which literally means ants. This name comes from the fact that formic acid was first isolated from ants.
Considering its multifold importance in chemistry, you will learn how to draw the Lewis dot structure of HCOOH, what is its molecular geometry or shape, electron geometry, bond angle, hybridization, formal charges, polarity, etc., all through this article.
So, continue reading!
Name of Molecule
Formic acid
Chemical formula
HCOOH
Molecular geometry of HCOOH
Trigonal planar
Electron geometry of HCOOH
Trigonal planar
Hybridization
sp2
Nature
Polar molecule
Bond angle
∠ (H-C-O) = 106° and 111°, ∠ (H-C=O) = 124° and ∠ (O-C=O)= 125°
The Lewis structure of formic acid (HCOOH) consists of three different elemental atoms, i.e., a carbon (C) atom, two hydrogens (H) atoms, and two atoms of oxygen (O).
The C-atom is present at the center. It is bonded to a hydrogen (H) atom and a hydroxyl (OH) functional group via single covalent bonds and to an oxygen (O) atom through a double covalent bond, respectively.
In this way, there are a total of 3 electron density regions or electron domains around the central C-atom in the HCOOH Lewis structure. All three electron domains are comprised of bond pairs; consequently, the central C-atom does not possess any lone pairs of electrons.
Steps for drawing the Lewis dot structure of HCOOH
1. Count the total valence electrons in HCOOH
The very first step while drawing the Lewis structure of HCOOH is to calculate the total valence electrons present in its concerned elemental atoms.
As three different elemental atoms are present in HCOOH, 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.
∴ The HCOOH molecule consists of 1 C-atom, 2 O-atoms, and 2 H-atoms. Therefore, the total valence electrons available for drawing the Lewis dot structure of HCOOH = 1(4) + 2(6) + 2(1) = 18 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.
Oxygen (E.N = 3.44) is more electronegative than both carbon and hydrogen. Hydrogen (E.N = 2.20) is less electronegative than carbon (E.N = 2.55). Still, it cannot be chosen as the central atom because a hydrogen (H) atom can accommodate only 2 electrons which denotes it can form a bond with a single adjacent atom only. This means H is always placed as an outer atom in a Lewis structure.
Consequently, the C-atom is placed at the center of the HCOOH Lewis structure while the O-atoms and the H-atoms are spread around it, as shown below.
3. Connect outer atoms with the central atom
In this step, the outer atoms are joined to the central C-atom using single straight lines.
But you must remember that an H-atom present next to the O-atom is only joined to the oxygen atom and not to the C-atom at the center. This is because an H-atom can form a single bond with one adjacent atom only in any Lewis structure. An oxygen atom is highly electronegative, so it readily forms a covalent chemical bond with its adjacent H-atom, not allowing the latter a chance to form a direct C-H bond.
All the other outer atoms, i.e., 1 H-atom and 2 O-atoms, are directly connected to the central C-atom, as shown below.
4. Complete the duplet and/or octet of the outer atoms
As we already identified, the hydrogen and oxygen atoms are the outer atoms in the Lewis dot structure of HCOOH.
Each hydrogen (H) atom requires a total of 2 valence electrons in order to achieve a stable duplet electronic configuration.
A C-H bond and an O-H single bond already represent 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 to that, an O-atom needs a total of 8 valence electrons to achieve a stable octet electronic configuration.
So first, see the O-atom marked as oxygen no.1. It is single-bonded to only one carbon atom.
The C-O bond represents 2 valence electrons, which denotes it still needs 6 more electrons to complete its octet. So these 6 electrons are placed as 3 lone pairs on the respective O-atom.
Now, if we see the O-atom marked as oxygen no. 2, it is single-bonded to an O-atom on one side and an H-atom on the other end. The C-H and O-H single bonds represent a total of 4 valence electrons around this O-atom which in turn hints at a deficiency of 4 electrons. So these 4 electrons are placed as 2 lone pairs around the respective O-atom.
In this way, both the outer O-atoms now have a complete octet in addition to a complete duplet of the H-atoms.
5. Complete the octet of the central atom by converting a lone pair into a covalent bond
Total valence electrons used till step 4 = 4 single bonds + electrons placed around the oxygen shown as dots = 4(2) + 10 = 18 valence electrons.
Total valence electrons – electrons used till step 4 = 18 – 18 = 0 valence electrons.
As all the valence electrons initially present for drawing the HCOOH Lewis structure are already consumed so there is no lone pair on the central C-atom in this structure.
However, a problem here is that this central C-atom only has a total of 3 single bonds around it which means 6 valence electrons. So it is still short of 2 more electrons to complete its octet.
But don’t worry at all because we have an easy solution to this problem. The solution is to convert a lone pair present on the O-atom containing 3 lone pairs into an extra chemical bond between the central C-atom and that O-atom.
This way, the central C-atom also has a complete octet comprised of 2 single bonds + 1 double bond.
Now that our problem is solved let’s check the stability of the HCOOH Lewis structure obtained above by applying the formal charge concept.
6. Check the stability of Lewis’s 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.
The above calculation shows that zero formal charges are present on the bonded atoms in the HCOOH Lewis structure. Thus it is a stable structure, and the good news is that we have drawn it correctly.
What are the electron and molecular geometry of HCOOH?
Formic acid (HCOOH) possesses an identical electron and molecular geometry or shape, i.e., trigonal planar. To a C-atom at the center, one H-atom, one O-atom, and one OH group 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 HCOOH.
Molecular geometry of HCOOH
The molecular geometry or shape of formic acid (HCOOH) 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, which makes the bonded electron pairs occupy the three vertices of an equilateral triangle in a perfectly symmetrical molecular arrangement, as shown in the figure below.
Electron geometry of HCOOH
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 HCOOH, there are 2 single bonds and 1 double bond around the central carbon atom, which makes a total of 3 electron density regions. 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 HCOOH molecule
A in the AXN formula represents the central atom. In HCOOH, a carbon (C) atom is present at the center, so A = C.
X denotes the atoms bonded to the central atom. In HCOOH, 1 H-atom, 1 O-atom, and 1 OH group are directly bonded to the central C-atom. The OH group is considered 1 region of electron density. In short, X = 3 for HCOOH.
N stands for the lone pairs present on the central atom. As per the Lewis structure of HCOOH, there are no lone pairs of electrons on the central carbon; hence N = 0.
As a result, the AXN generic formula for HCOOH 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 formic acid (HCOOH).
Hybridization of HCOOH
The central C-atom is sp2 hybridized in HCOOH.
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.
HCOOH hybridization can also be determined from its steric number. The steric number of the central C-atom in HCOOH is 3, so it has sp2 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The HCOOH bond angle
The ideal bond angle in a symmetrical trigonal planar molecule is 120°. However, it is due to the different types of bonds present in HCOOH that four different bond angles exist in this molecule. In ascending order, the C-O-H bond angles are 106 ° and 111°, the H-C=O bond angle is 124°, and the O=C-O bond angle is 125° respectively.
Similarly, experimental studies revealed multiple bond lengths present in a formic acid molecule, including the O-H bond length (97 pm), the C-H bond length (110 pm), the C=O bond length (123 pm) and the longest C-O bond length (132 pm), as shown in the figure below.
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 HCOOH, four different types of covalent bonds are present, i.e., a C-H bond, a C=O bond, a C-O bond, and an O-H bond.
A small electronegativity difference of 0.35 units exists between the carbon (E.N = 2.55) and hydrogen (E.N = 2.20) atoms. So the C-H bond is only weakly polar. Contrarily, a higher 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.
Above all, the O-H bond present in HCOOH is extremely polar as an 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 HCOOH.
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 HCOOH is a polar molecule (net µ = 1.42 D).
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/
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