The ethanol lewis dot structure represents ethanol as the neutral alcohol CH3–CH2–O–H (molecular formula C2H6O) with all single bonds and two lone pairs on oxygen; every atom satisfies the duet rule (H) or the octet rule (C, O) with zero formal charge.
The molecular formula C2H6O corresponds to more than one constitutional isomer. The ethanol Lewis structure uses the connectivity CH3–CH2–OH (not the ether connectivity CH3–O–CH3).
Molecular formula and connectivity
Ethanol is the alcohol isomer of C2H6O with connectivity CH3–CH2–OH. Both carbon atoms form four single bonds (tetravalent carbon), and oxygen forms two single bonds (divalent oxygen) with two lone pairs.
Valence-electron accounting
The total number of valence electrons for a neutral ethanol molecule follows directly from group valences:
| Element | Count | Valence electrons per atom | Contribution |
|---|---|---|---|
| Carbon (C) | 2 | 4 | \(2 \cdot 4 = 8\) |
| Hydrogen (H) | 6 | 1 | \(6 \cdot 1 = 6\) |
| Oxygen (O) | 1 | 6 | \(1 \cdot 6 = 6\) |
| Total | 9 atoms | — | \(8 + 6 + 6 = 20\) |
The ethanol Lewis dot structure contains 8 single (sigma) bonds: 1 C–C, 5 C–H, 1 C–O, and 1 O–H. Bonding electrons account for \[ 8 \cdot 2 = 16 \text{ electrons}. \] The remaining electrons are \[ 20 - 16 = 4 \text{ electrons}, \] corresponding to two lone pairs on oxygen.
Octet and duet completion
- Hydrogen atoms reach a duet through one single bond (2 electrons around each H).
- Each carbon reaches an octet through four single bonds (8 electrons around each carbon center).
- Oxygen reaches an octet through two single bonds plus two lone pairs (4 bonding electrons + 4 nonbonding electrons).
Formal charges
Formal charge checks confirm the neutral distribution. The standard expression is \[ FC = V - \left(N + \frac{B}{2}\right), \] where \(V\) is valence electrons for the free atom, \(N\) is nonbonding electrons assigned to the atom, and \(B\) is bonding electrons shared in bonds.
| Atom type | \(V\) | \(N\) | \(B\) | Formal charge |
|---|---|---|---|---|
| Carbon (each) | \(4\) | \(0\) | \(8\) | \(4 - (0 + 8/2) = 0\) |
| Oxygen | \(6\) | \(4\) | \(4\) | \(6 - (4 + 4/2) = 0\) |
| Hydrogen (each) | \(1\) | \(0\) | \(2\) | \(1 - (0 + 2/2) = 0\) |
Local electron-domain geometry and shape
Around each carbon center, four electron domains (four sigma bonds) produce a tetrahedral electron-domain geometry with bond angles near \(109.5^\circ\) in an idealized model. Around oxygen, four electron domains (two bonds + two lone pairs) also produce a tetrahedral electron-domain arrangement, while the molecular shape at oxygen (considering atoms only) is bent; lone-pair repulsion compresses the H–O–C angle below \(109.5^\circ\).
Common pitfalls
- Oxygen lone pairs omitted; oxygen must carry two lone pairs in the neutral ethanol Lewis structure.
- Incorrect isomer drawn; CH3–O–CH3 corresponds to dimethyl ether, not ethanol.
- Multiple bonds introduced without justification; ethanol satisfies octets with single bonds and does not require double bonds or resonance.