What is the Lewis structure for the acetate ion (CH3COO-)?

The Lewis structure for the acetate ion (CH₃COO^–) can be drawn by following these steps:

1. Count the total number of valence electrons:
   • Carbon (C) has 4 valence electrons.
   • Hydrogen (H) has 1 valence electron.
   • Oxygen (O) has 6 valence electrons.

   For CH₃COO^–:

   • 2 Carbons: 2 × 4 = 8 electrons
   • 3 Hydrogens: 3 × 1 = 3 electrons
   • 2 Oxygens: 2 × 6 = 12 electrons
   • 1 extra electron for the negative charge: 1 electron

   Total valence electrons = 8 + 3 + 12 + 1 = 24 electrons

2. Draw the skeletal structure:

   The skeletal structure of CH₃COO^– is:

             H   O
             |   ||
           H-C-C-O
             |
             H
           

3. Distribute the electrons:

   Place the remaining electrons to satisfy the octet rule for each atom:

   • Each carbon atom should have 8 electrons.
   • Each oxygen atom should have 8 electrons.
   • Hydrogen atoms should have 2 electrons.

   The final Lewis structure is:

             H   O
             |   ||
           H-C-C-O
             |
             H
           

   With the double bond between the second carbon and one oxygen, and the single bond between the second carbon and the other oxygen. The negative charge is on the oxygen with the single bond.

4. Assign formal charges:

   Formal charges are calculated as:

   Formal charge = (Valence electrons) – (Non-bonding electrons) – (Bonding electrons / 2)

   For the oxygen with the double bond:

   • Valence electrons = 6
   • Non-bonding electrons = 4
   • Bonding electrons = 4
   • Formal charge = 6 – 4 – (4 / 2) = 0

   For the oxygen with the single bond:

   • Valence electrons = 6
   • Non-bonding electrons = 6
   • Bonding electrons = 2
   • Formal charge = 6 – 6 – (2 / 2) = -1

   For the carbons and hydrogens, the formal charges are 0.

This completes the Lewis structure for the acetate ion (CH₃COO^–).