Draw the Lewis structure of bicarbonate HCO3 showing all possible resonance structures if there are any

The Lewis structure of bicarbonate (HCO₃^–) can be drawn by following a few systematic steps. First, we will determine the total number of valence electrons available for the bicarbonate ion. Carbon (C) has 4 valence electrons, each oxygen (O) has 6 valence electrons, and hydrogen (H) has 1 valence electron. The bicarbonate ion has a negative charge, which means we add one extra electron to our total count.

Total valence electrons = 4 (C) + 6 (O) + 6 (O) + 1 (H) + 1 (negative charge) = 18 valence electrons.

To draw the structure, we start with carbon in the center. Since carbon can form four bonds, we can connect it to one oxygen with a double bond, and the other oxygen with a single bond, where the single bond oxygen will carry the negative charge. The hydrogen atom is attached to the oxygen that has a single bond.

Here is the primary Lewis structure of bicarbonate:

   H
    |
   O      O
    ||    -
   C  <---->

In actual representation, this looks like this:

Next, we need to consider resonance structures. In bicarbonate, the double bond can be located between the carbon and either of the two oxygen atoms. This allows us to draw two additional resonance structures:

Resonance Structure 1:
   H          O
    |         ||
   O  <----> C
         - 

Resonance Structure 2:
   H
    |
   C   <----> O
    ||        
   O

Each of these resonance structures represents a valid arrangement of electrons, and the actual structure of the bicarbonate ion is a hybrid of these three forms. In the hybrid, the bond lengths to all oxygen atoms are equivalent, which is a key concept in resonance theory.

In summary, the Lewis structure of bicarbonate showcases a central carbon atom with one double bond to an oxygen and one single bond to another oxygen, which also connects to a hydrogen atom, while the structures allow for resonance that balances the charge and distribution of electrons in the molecule.