To draw the Lewis structure and show all ionic electron pairs and bond orders for acetic acid (CH₃COOH) and chloroacetic acid (CH₂ClCOOH), follow these steps:
Acetic Acid (CH₃COOH)
- 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 CH₃COOH: (4 × 1) + (3 × 1) + (4 × 1) + (6 × 2) = 24 valence electrons.
- Draw the skeletal structure:
Connect the atoms with single bonds: H₃C-COOH.
- Distribute the remaining electrons:
Place lone pairs on the oxygen atoms to satisfy the octet rule.
- Check for formal charges:
Ensure that the formal charges are minimized.
- Identify ionic electron pairs and bond orders:
The C=O bond is a double bond, and the C-O bond is a single bond. The C-H bonds are single bonds.
Chloroacetic Acid (CH₂ClCOOH)
- Count the total number of valence electrons:
- Carbon (C) has 4 valence electrons.
- Hydrogen (H) has 1 valence electron.
- Chlorine (Cl) has 7 valence electrons.
- Oxygen (O) has 6 valence electrons.
For CH₂ClCOOH: (4 × 1) + (2 × 1) + (7 × 1) + (4 × 1) + (6 × 2) = 30 valence electrons.
- Draw the skeletal structure:
Connect the atoms with single bonds: ClCH₂-COOH.
- Distribute the remaining electrons:
Place lone pairs on the oxygen and chlorine atoms to satisfy the octet rule.
- Check for formal charges:
Ensure that the formal charges are minimized.
- Identify ionic electron pairs and bond orders:
The C=O bond is a double bond, and the C-O bond is a single bond. The C-H and C-Cl bonds are single bonds.