HCO3- Lewis structure

HCO₃^– (bicarbonate) has one hydrogen atom, one carbon atom, and three oxygen atoms.

In the HCO₃^– Lewis structure, there is one double bond and two single bonds around the carbon atom, with three oxygen atoms attached to it. The oxygen atom with a double bond has two lone pairs, the right oxygen atom (with which the hydrogen atom is attached) also has two lone pairs, and the left oxygen atom with a single bond has three lone pairs.

Also, there is a negative (-1) charge on the left oxygen atom.

Alternative method: Lewis structure of HCO₃^–

Rough sketch

• First, determine the total number of valence electrons

In the periodic table, hydrogen lies in group 1, carbon lies in group 14, and oxygen lies in group 16.

Hence, hydrogen has one valence electron, carbon has four valence electrons, and oxygen has six valence electrons.

Since HCO₃^– has one hydrogen atom, one carbon atom, and three oxygen atoms, so…

Valence electrons of one hydrogen atom = 1 × 1 = 1
Valence electrons of one carbon atom = 4 × 1 = 4
Valence electrons of three oxygen atoms = 6 × 3 = 18

Now the HCO₃^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 1 + 4 + 18 + 1 = 24

Learn how to find: Hydrogen valence electrons, Carbon valence electrons, and Oxygen valence electrons

• Second, find the total electron pairs

We have a total of 24 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 24 ÷ 2 = 12

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Now we have to choose the central atom from carbon and oxygen. Place the least electronegative atom at the center.

Since carbon is less electronegative than oxygen, assume that the central atom is carbon.

Therefore, place carbon in the center and hydrogen and oxygen on either side.

• And finally, draw the rough sketch

Lone pair

Here, we have a total of 12 electron pairs. And four bonds are already marked. So we have to only mark the remaining eight electron pairs as lone pairs on the sketch.

Also remember that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And both (carbon and oxygen) are the period 2 elements, so they can not keep more than 8 electrons in their last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are hydrogen and oxygens. But no need to mark on hydrogen, because hydrogen already has two electrons.

So for top oxygen and left oxygen, there are three lone pairs, for right oxygen, there are two lone pairs, and for carbon, there is zero lone pair because all eight electron pairs are over.

Mark the lone pairs on the sketch as follows:

Formal charge

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For carbon atom, formal charge = 4 – 0 – ½ (6) = +1

For top oxygen and left oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

For right oxygen atom, formal charge = 6 – 4 – ½ (4) = 0

Here, both carbon and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both carbon and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

Convert a lone pair of the top oxygen atom to make a new C — O bond with the carbon atom as follows:

Final structure

The final structure of HCO₃^– features a central carbon atom connected to three oxygen atoms, with one of those oxygen atoms further bonded to a hydrogen atom. Within this layout, the central carbon forms a double bond with one oxygen and single bonds with the other two, ensuring it satisfies the octet rule. The oxygen atom involved in the double bond maintains two lone pairs, the oxygen bonded to hydrogen also retains two lone pairs, and the third oxygen atom carries three lone pairs to fulfill its octet. This configuration is the most stable because it minimizes formal charges while keeping the negative charge on the most electronegative atom. Consequently, this specific electronic distribution serves as the definitive and most accurate Lewis representation of the bicarbonate ion.

To complete the representation, draw square brackets around the entire Lewis structure and place a “-” or “-1” sign as a superscript outside the upper right bracket. This notation signifies that the negative charge is a property of the whole ion.

Next: HNO₂ Lewis structure

External video

• HCO3- Lewis Structure: How to Draw the Lewis Structure for HCO3- – YouTube • Wayne Breslyn

External links

• https://www.chemistryscl.com/general/HCO3–lewis-structure/index.php
• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/HCO3–lewis-structure.html
• https://lambdageeks.com/hco3-lewis-structure/