The four carbon-to-hydrogen bonds in an ethylene molecule (C2H4)

Question

BlackTom AI · Accepted Answer

Each carbon in ethylene is sp2-hybridized, so the geometry around each carbon is trigonal planar and all bonded atoms lie in the same plane.

1) Relevant knowledge points and concepts
- Hybridization: A carbon with a double bond (C=C) uses sp2 hybrid orbitals for sigma bonds and retains one unhybridized p orbital for the pi bond.
- Electron-domain geometry: Three regions of electron density (two C–H sigma bonds + one C–C sigma bond) around each carbon → trigonal planar (≈120°).
- Pi bond requirement: The unhybridized p orbitals on the two carbons must be parallel to overlap and form the pi bond; this enforces a common plane for the sp2 orbitals and the attached atoms.

2) Step-by-step reasoning
- Start with one carbon in C2H4: it is bonded to two H atoms and the other carbon (three sigma bonds) and participates in one pi bond. That corresponds to sp2 hybridization (three sp2 orbitals + one p orbital).
- The three sp2 orbitals arrange themselves 120° apart in a plane, forming sigma bonds to two hydrogens and the sigma component of the C–C bond.
- The second carbon is identical, so both carbons and their attached hydrogens occupy a single plane to allow the unhybridized p orbitals to be parallel and form the pi bond.
- Therefore all four C–H bonds (two on each carbon) lie in that same molecular plane.

3) Intermediate steps / counts
- Electron domains per carbon = 3 → sp2 hybridization.
- Geometry per carbon = trigonal planar → bond angles ≈120°.
- Pi bond formed by overlap of perpendicular p orbitals requires planarity for optimal overlap.

4) Why the correct answer is correct
Because sp2 hybridization and the need for parallel p orbitals to form the pi bond force each carbon and its two hydrogens into a single plane; thus all four C–H bonds are coplanar.

5) Why the other options are incorrect
- “Point to the corners of a tetrahedron”: describes sp3 geometry (tetrahedral), which applies to single-bonded carbons (e.g., CH4 or ethane carbons), not sp2 carbons in C2H4.
- “Are arranged in a linear fashion”: linear (180°) applies to sp hybridization (e.g., CO2-like), not to ethylene.
- “Are positioned along the three co-ordinate axes”: arbitrary and not chemically meaningful here; the real geometry is trigonal planar, not aligned with three orthogonal axes.

Conclusion: All reasoning confirms the correct answer is “lie in one plane,” matching the provided answer.

The four carbon-to-hydrogen bonds in an ethylene molecule (C2H4)Single choice

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