Bicarbonate (HCO3-) lewis dot structure, molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, polarity
HCO3– is the chemical formula representing a polyatomic anion, i.e., the bicarbonate ion. It is an important chemical species that plays various roles in biological and chemical systems.
The bicarbonate ion serves as a key component in the acid-base buffering system, maintaining the human body’s pH. It is also crucial for CO2 transport through the bloodstream.
In this article, we will learn how to draw the Lewis dot structure of HCO3–, 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!
Name of the molecular ion
Bicarbonate
Chemical formula
HCO3–
Molecular geometry of HCO3–
Trigonal planar (w.r.t the C-atom)
Bent, angular or V-shaped (w.r.t the OH bonded O-atom)
The bicarbonate (HCO3–) ion comprises a carbon (C) atom at the center. It is double-covalently bonded to an oxygen (O) atom at one side and to another O-atom and an OH functional group vis single covalent bonds at the other two sides.
There is no lone pair of electrons on the central C-atom. However, the terminal O-atoms carry 2 or 3 lone pairs of electrons, respectively, in the HCO3– Lewis dot structure.
Come along and draw the Lewis dot structure of HCO3– with us by following the simple steps given below.
Steps for drawing the Lewis dot structure of HCO3–
1.Count the total valence electrons present in HCO3–
The three distinct elements present in HCO3– are carbon, hydrogen and oxygen.
Carbon (C) is present in Group IV A (or 14) of the Periodic Table of Elements. Thus, it has a total of 4 valence electrons in each atom.
In contrast, oxygen (O) is located in Group VI A (or 16), containing 6 valence electrons in each atom. However, 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 HCO3– ion comprises 1 C-atom, 1 H-atom and 3 O-atoms.
An important point to remember is that the HCO3– ion carries a negative (-1) charge, which means 1 extra valence electron is added in this Lewis structure.
∴ Therefore, the total valence electrons available for drawing the Lewis dot structure of HCO3– = 1(4) + 1(1) + 3(6) = 23 + 1 =24 valence electrons.
2. Find the least electronegative atom and place it at the center
By convention, the least electronegative atom out of all those available is chosen as the central atom while drawing the Lewis structure of a molecule or molecular ion.
The least electronegative atom can easily form covalent bonds with other atoms by sharing its electrons.
Hydrogen (E.N = 2.20) is less electronegative than both carbon (E.N = 2.55) and oxygen (E.N = 3.44).
However, the H-atom is an exception as it cannot be chosen as the central atom in any Lewis structure. It can accommodate a total of 2 valence electrons, forming a single covalent bond with 1 adjacent atom only.
Therefore, we select the second option, i.e., a C-atom as the central atom, while the 3 O-atoms and 1 H-atom occupy terminal positions in the HCO3– Lewis dot structure, as shown below.
3. Connect the outer atoms with the central atom
In this step, the outer atoms, i.e., 3 O-atoms, are joined to the central C-atom using single straight lines.
However, you must note that as we already identified, an H-atom can only form 1 bond; therefore, it is only joined to its adjacent O-atom and not to the central C-atom in HCO3– Lewis structure.
A straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons.
In the above structure, there are a total of 4 single bonds, i.e., 4(2) = 8 valence electrons are already consumed out of the 24 initially available.
Now let’s see in the next steps where to place the remaining 16 valence electrons in the HCO3– Lewis dot structure.
4. Complete the octet and/or duplet of the outer atoms
An O-atom needs a total of 8 valence electrons in order to achieve a stable octet electronic configuration.
A C-O bond represents 2 valence electrons already present around the oxygen atoms marked Oa and Ob in the Lewis dot structure obtained so far.
Therefore, the remaining 6 valence electrons are placed as 3 lone pairs around both these oxygen atoms to complete their octets.
In contrast, Ocis surrounded by 2 single bonds (C-O and O-H).
It is thus short of only 4 valence electrons to complete its octet. Therefore, these 4 valence electrons are placed as 2 lone pairs on Oc, as shown below.
The H-atom, on the other hand, already has a complete duplet, so we need not worry about this H-atom in HCO3– Lewis structure.
5. Complete the octet of the central atom and convert lone pairs into covalent bonds if necessary
Total valence electrons used till step 4 = 4 single bonds + 2(electrons placed around Oa) + electrons placed around Oc, shown as dots = 4(2) + 2(6) + 4 = 24 valence electrons.
Total valence electrons – electrons used till step 4 = 24– 24 = 0 valence electrons.
As all the valence electrons initially available for drawing the HCO3– Lewis structure are already consumed so there is no lone pair on the central C-atom.
However, this C-atom still has an incomplete octet.
To solve this problem, 1 lone pair from an adjacent O-atom (except OH-bonded oxygen) is converted into an additional covalent bond between the C-atom and the respective O-atom.
Now the central C-atom also has a complete octet with 2 single bonds + 1 double bond = 8 valence electrons surrounding it.
Hence, the final step is to check the stability of the HCO3– Lewis structure by applying the formal charge concept.
6. Check the stability of Lewis’s structure using the formal charge concept
The less the formal charge on the atoms of a molecule or molecular ion, 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 present on the HCO3– bonded 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
Non-bonding electrons = no lone pairs = 0 electrons
Formal charge = 1-0-2/2 = 1-0-1= 1-1 = 0
As per the above calculation, there are zero or no formal charges on either of the oxygen, carbon or hydrogen atoms in the HCO3– Lewis structure except the C-O bonded terminal oxygen, which carries a formal charge of -1. Hence, the overall charge present on the bicarbonate ion is -1.
This ensures that we have drawn the HCO3– Lewis structure correctly.
However, you must note that the following two resonance forms are possible for drawing HCO3– Lewis structure as the position of the pi-bonded electrons and thus, the -1 formal charge keeps revolving from one position to another on the molecular ion.
The actual bicarbonate ion structure is a hybrid of the above resonance forms, known as the resonance hybrid.
Now let us move ahead and discuss its molecular and electron geometry.
What are the electron and molecular geometry of HCO3-?
The molecular geometry or shape of the bicarbonate (HCO3–) ion is trigonal planar, identical to its ideal electronic geometry w.r.t the C-atom.
However, its shape is bent w.r.t the O-H bonded oxygen atom, as the presence of 2 lone pairs of electrons on the oxygen atom leads to strong lone pair-lone pair and lone pair-bond pair electronic repulsions, thus distorting the overall molecular shape.
Molecular geometry of HCO3–
The molecular geometry or shape of the bicarbonate (HCO3–) ion w.r.t the C-atom is trigonal planar.
To a carbon atom at the center, three oxygen atoms are directly attached like three vertices of an equilateral triangle. There is no lone pair of electrons on the central C-atom; hence no lone pair-lone pair or lone pair-bond pair electronic repulsions exist in the molecular ion w.r.t this carbon atom.
However, the presence of 2 lone pairs of electrons on the OH-bonded oxygen atom makes the bicarbonate ion occupy a bent, angular or V-shape w.r.t this O-atom, as shown below.
Electron geometry of HCO3–
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule or molecular ion containing a total of 3 electron density regions around the central atom is trigonal planar.
In HCO3–, the C-atom at the center is surrounded by 3 bond pairs, and it has no lone pair of electrons, making a total of 3 electron density regions. Hence, the ideal electron pair geometry of the HCO3– ion is trigonal planar.
An easy trick to finding a molecule’s electron and molecular geometry is using the AXN method.
AXN is a simple formula representing the number of bonded atoms and lone pairs present on the central atom.
It is used to predict the shape and geometry of a molecule or molecular ion using the VSEPR concept.
AXN notation for HCO3–
A in the AXN formula represents the central atom. In HCO3–, a carbon (C) atom is present at the center, so A = C.
X denotes the atoms bonded to the central atom. In HCO3–, 3 O-atoms are directly bonded to the central C-atom. So X = 3 for HCO3–.
N stands for the lone pairs present on the central atom. As per the Lewis structure of HCO3–, the central C-atom has no lone pair of electrons. Thus, N = 0 for HCO3–.
As a result, the AXN generic formula for HCO3– is AX3N0or simply AX3.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that the molecular geometry or shape of a molecule or molecular ion with an AX3 generic formula is identical to its electron geometry, i.e., trigonal planar, as we already noted down for the bicarbonate (HCO3–) ion.
Hybridization of HCO3–
The central C-atom is sp2 hybridized in HCO3–.
The electronic configuration of carbon is 1s2 2s2 2p2.
During chemical bonding in HCO3–, one 2s electron of carbon gets unpaired and shifts to its empty 2p atomic orbital.
As a result, the half-filled 2s orbital of carbon hybridizes with its two 2p orbitals to produce three sp2 hybrid orbitals.
Each sp2 hybrid orbital possesses a 33.3 % s-character and a 66.6 % p-character.
Thus, carbon uses these sp2 hybrid orbitals to form C-O sigma bonds by overlapping with the hybrid orbitals of adjacent oxygen atoms, one on each side of the bicarbonate ion.
On the other hand, carbon uses its unhybridized p-orbital to form a C=O pi bond by p-p orbital overlap.
Refer to the figure drawn below.
Another shortcut to finding the hybridization present in a molecule or molecular ion is using its steric number against the table below.
The steric number of the C-atom in HCO3– is 3, so it has sp2 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
The bond angle of HCO3–
The O-C=O bonded atoms form an ideal bond angle of 120° in the symmetrical trigonal planar shape while the C-O-H bond angle is reduced to 104.5° due to the bent or V-shape of HCO3– w.r.t the oxygen atom.
As per Pauling’s electronegativity scale, a polar covalent bond is formed between two dissimilar atoms with an electronegativity difference between 0.4 and 1.6 units.
The two main types of bonds present in HCO3– are C-O (or C=O) and O-H bonds.
The C-O (or C=O) bond is polar as a specific electronegativity difference of 0.89 units is present between the covalently bonded carbon (E.N = 2.55) and oxygen (E.N = 3.44) atoms.
In contrast, an even higher electronegativity difference of 1.24 units exists between an oxygen and a hydrogen (E.N = 2.20) atom in an O-H bond.
As a result, the C-atom and H-atom gain partial positive (δ+) charges while the O-atoms obtain partial negative (δ–) charges in HCO3–, respectively.
The unequal C-O and O-H dipole moments do not get canceled in the asymmetric bent shape of HCO3– w.r.t the O-atom. Therefore, the bicarbonate (HCO3–) ion is overall polar (net µ> 0).
The Lewis dot structure of the bicarbonate (HCO3–) ion displays a total of 24 valence electrons, i.e., 24/2 = 12 electron pairs.
Out of the 12 electron pairs, there are 5 bond pairs and 7 lone pairs of electrons.
A C-atom at the center is double-covalently bonded to an O-atom on one side and single-covalently bonded to another O-atom and an OH functional group on the other two sides.
There is no lone pair of electrons on the central C-atom, while the terminal O-atoms carry 2 and 3 lone pairs of electrons, respectively, in HCO3– Lewis dot structure.
What is the molecular shape of HCO3–?
HCO3– ion has a trigonal planar molecular shape w.r.t the central carbon atom while its shape is bent, angular or V-shaped w.r.t the OH-bonded oxygen atom.
What is the electron geometry of HCO3–?
There are 3 bond pairs surrounding the central C-atom in HCO3– which has no lone pair of electrons.
Therefore, the AXN generic formula for HCO3– w.r.t the central C-atom is AX3, so its ideal electronic geometry is trigonal planar.
Why is the VSEPR shape of HCO3– w.r.t O-atom different from its shape w.r.t C-atom?
There is no lone pair of electrons on the central C-atom in HCO3– while the OH-bonded O-atom carries 2 lone pairs of electrons.
Strong lone pair-lone pair and lone pair-bond pair electronic repulsions exist in the molecular ion w.r.t O-atom, which distorts the molecular shape of HCO3– w.r.t this oxygen atom, and it occupies a bent, angular or V-shape.
How is the shape of HCO3– different from that of H2CO3?
Carbonic acid (H2CO3) and the bicarbonate (HCO3–) ion possess similar shapes, i.e., trigonal planar w.r.t the central C-atom and bent, angular or V-shaped w.r.t the O-atoms.
How is the shape of H3PO4 similar to or different from that of HCO3–?
The shape of phosphoric acid (H3PO4) is tetrahedral w.r.t the central P-atom while it is bent, angular or V-shaped w.r.t the O-atoms.
It is quite different from the molecular shape of HCO3– i.e., trigonal planar w.r.t the carbon atom at the center.
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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