Problem
Determine the c2h4 lewis structure for ethene (ethylene), showing the correct bonding arrangement and verifying that the total valence-electron count, octet rule, and formal charges are satisfied.
Step-by-step solution
1) Count total valence electrons
Carbon (C) is in Group 14 and contributes 4 valence electrons per atom. Hydrogen (H) contributes 1 valence electron per atom.
| Element | Number of atoms | Valence electrons per atom | Total contributed |
|---|---|---|---|
| C | 2 | 4 | \(2 \cdot 4 = 8\) |
| H | 4 | 1 | \(4 \cdot 1 = 4\) |
| Total | \(8 + 4 = 12\) | ||
The molecule has \(12\) valence electrons available to place in the Lewis dot structure.
2) Choose a reasonable skeleton (connect atoms)
Hydrogen forms only one bond, so hydrogen atoms must be terminal. The carbon atoms form the backbone: place a C–C connection, then attach two H atoms to each carbon to satisfy hydrogen’s duet requirement.
3) Add bonds and account for electrons
Start by connecting the framework with single bonds, then adjust bonding to satisfy the octet rule for carbon.
- Four C–H single bonds use \(4 \cdot 2 = 8\) electrons.
- A C=C double bond uses \(4\) electrons (two shared pairs).
Total electrons used in bonds:
\[ 8 + 4 = 12 \]
All \(12\) valence electrons are accounted for by bonding pairs. No lone pairs are required on carbon in ethene. Each carbon attains an octet because it is surrounded by four shared pairs (two pairs in the double bond and one pair in each C–H bond), totaling \(8\) electrons around each carbon.
4) Check formal charges
Formal charge is computed by:
\[ \text{FC} = V - \left(N + \frac{B}{2}\right) \]
where \(V\) is valence electrons for the free atom, \(N\) is nonbonding electrons, and \(B\) is bonding electrons around that atom.
- For each carbon in C2H4: \(V = 4\), \(N = 0\), and the carbon participates in four bonds (one double bond counts as two bonds), so \(B = 8\). \[ \text{FC}_\mathrm{C} = 4 - \left(0 + \frac{8}{2}\right) = 4 - 4 = 0 \]
- For each hydrogen: \(V = 1\), \(N = 0\), \(B = 2\). \[ \text{FC}_\mathrm{H} = 1 - \left(0 + \frac{2}{2}\right) = 1 - 1 = 0 \]
Zero formal charges on all atoms confirms a stable Lewis structure for ethene.
5) Final Lewis structure interpretation
The correct c2h4 lewis structure is H2C=CH2: a C=C double bond with each carbon single-bonded to two hydrogens. Each carbon has three electron domains (two C–H bonds and one C=C domain), consistent with sp2 hybridization and an approximately trigonal planar arrangement around each carbon.
Visualization
Common checks and common mistakes
- Electron count check: the bonding pattern must use \(12\) electrons total; adding lone pairs on carbon would exceed the available valence electrons.
- Octet check: each carbon must have \(8\) electrons around it; a single C–C bond would leave each carbon short of an octet unless additional bonding is added.
- Hydrogen check: each hydrogen must have exactly one bond (a duet), never two bonds.