To draw the Lewis structure for bicarbonate (HCO3–), we start by counting the total number of valence electrons. Carbon (C) has 4 valence electrons, oxygen (O) has 6, and hydrogen (H) has 1. We have 3 oxygen atoms and one hydrogen atom, so:
Total Valence Electrons: 4 (C) + 1 (H) + 3 × 6 (O) = 4 + 1 + 18 = 23 electrons
Since HCO3 has a -1 charge, we add one additional electron, giving us a total of 24 valence electrons.
Now, let’s draw the structure. Carbon is the central atom, bonded to three oxygen atoms. One of the oxygen atoms will have a double bond to the carbon to satisfy the octet rule for carbon, while the remaining two oxygen atoms will be single bonded and have a -1 charge overall:
This represents the Lewis structure of HCO3–.
Next, we determine the formal charge of each atom:
- Carbon (C):
Formal Charge = Valence Electrons – (Number of Bonds + Lone Pair Electrons) = 4 – (4 + 0) = 0 - Oxygen (O with double bond):
Formal Charge = 6 – (2 + 4) = 0 - Oxygen (O with -1 charge):
Formal Charge = 6 – (1 + 6) = -1 - Oxygen (O with single bond):
Formal Charge = 6 – (1 + 6) = -1 - Hydrogen (H):
Formal Charge = 1 – (1 + 0) = 0
The formal charges on the oxygen atoms indicate that one of the oxygens holds a negative charge, resulting in the overall -1 charge of the bicarbonate ion. Therefore, the final formal charges are:
C: 0, O (double bond): 0, O (single bond): -1, O (single bond): -1, H: 0.