Draw the best Lewis structure for the acetate ion, CH3CO2⁻, including any important resonance structures.

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Draw the best Lewis structure for the acetate ion, CH3CO2⁻, including any important resonance structures. What is the formal charge on the terminal double bonded O atom? +2 – 2 0 + 1 – 1

The Correct Answer and Explanation is:

To draw the best Lewis structure for the acetate ion, CH₃CO₂⁻, we need to follow the steps for constructing a Lewis structure and ensure that we account for resonance and formal charges.

Step 1: Count the total number of valence electrons

  • Carbon (C) has 4 valence electrons.
  • Each hydrogen (H) has 1 valence electron, and there are 3 hydrogens in CH₃.
  • Oxygen (O) has 6 valence electrons, and there are 2 oxygens in the acetate ion.
  • The ion has a negative charge, so we add one extra electron.

The total number of valence electrons is:
4+3(1)+2(6)+1=14 valence electrons.4 + 3(1) + 2(6) + 1 = 14 \text{ valence electrons.}4+3(1)+2(6)+1=14 valence electrons.

Step 2: Arrange the atoms

  • The carbon (C) will be in the center, as it is less electronegative than oxygen.
  • The hydrogen atoms are bonded to the carbon atom, forming CH₃.
  • The two oxygen atoms will be attached to the carbon, one through a single bond and the other through a double bond. This accounts for the charge distribution.

Step 3: Draw the single bonds and fill in the valence electrons

  • Draw single bonds between the carbon and each oxygen, and between the carbon and the hydrogens.
  • The remaining electrons are placed as lone pairs on the oxygen atoms.

Step 4: Ensure all atoms have full octets (except hydrogen)

  • The oxygen atoms in the acetate ion have lone pairs to complete their octets.
  • The carbon atom shares electrons to complete its octet.

Step 5: Check resonance

  • The acetate ion has resonance structures. The double bond can switch between the two oxygen atoms, meaning that both oxygens share the double bond and the formal charge distribution alternates between them. This gives the ion its resonance nature.

Formal Charge Calculation:

For the terminal oxygen atoms (with a double bond), the formal charge is calculated as:Formal charge=Valence electrons−(Lone pair electrons+Bonding electrons2)\text{Formal charge} = \text{Valence electrons} – \left( \text{Lone pair electrons} + \frac{\text{Bonding electrons}}{2} \right)Formal charge=Valence electrons−(Lone pair electrons+2Bonding electrons​)

For each oxygen:

  • Valence electrons = 6
  • Lone pair electrons = 4 (since each oxygen has two lone pairs)
  • Bonding electrons = 4 (2 electrons from the double bond)

Formal charge=6−(4+42)=6−6=0\text{Formal charge} = 6 – (4 + \frac{4}{2}) = 6 – 6 = 0Formal charge=6−(4+24​)=6−6=0

Thus, the formal charge on the double-bonded oxygen is 0.

Best Lewis Structure:

The best Lewis structure is one where the negative charge is placed on the oxygen atom with a single bond (because oxygen is more electronegative and better able to handle the negative charge). The structure is as follows:

mathematicaCopyEdit       H
       |
H - C - O⁻
       |
       O

Resonance Structures:

There are two important resonance structures where the double bond alternates between the two oxygens:

  1. First structure: The first oxygen has a single bond with the carbon and a negative charge, while the second oxygen has a double bond with carbon and no charge.
  2. Second structure: The second oxygen has the negative charge and a single bond, while the first oxygen has a double bond.

The formal charge on the double-bonded oxygen is 0.

This structure minimizes the formal charges across the atoms and ensures the best resonance hybrid.

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