Organic Chemistry · Structure and Bonding
sp2 Hybrid Orbitals and the Structure of Ethylene
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In 30 seconds
Ethylene (ethene, CH2=CH2) introduces the double bond and the first break from tetrahedral carbon. Each carbon in ethylene has only three electron groups: two C–H σ bonds and one C–C σ bond. (The second line of the double bond is a π bond The sideways-overlap bond above and below the σ bond; it locks the alkene flat. Full entry →, which does not count as a separate electron group.) Mixing one 2s orbital with two 2p orbitals gives three sp² hybrid orbitals arranged at 120° in a single plane — trigonal planar Shape with three groups at 120° in one plane Full entry → — and leaves one 2p orbital unhybridized, sticking straight up and down out of that plane. Head-on overlaps build the σ framework (one C–C and four C–H bonds), while side-by-side overlap of the two leftover p orbitals forms a π bond with electron density above and below the molecular plane. A double bond is therefore one σ bond plus one π bond. The π bond locks the molecule flat, prevents rotation about the C=C axis, and shortens the bond to 1.33 Å.
Why this matters
Alkenes are among the most reactive functional groups in organic chemistry, and the π bond is the reason: its electrons sit exposed above and below the plane, where they act as the electron-rich site attacked in addition reactions (Chapters 7–8). The planarity of sp² carbon underlies conjugated systems, aromatic rings, and the flat shapes of DNA bases and many drug molecules. restricted rotation Inability to rotate freely about a double bond Full entry → about C=C produces cis/trans stereoisomers — molecules with identical formulas but different shapes, and therefore different properties and biological effects. Ethylene itself is a plant hormone that triggers fruit ripening and is the feedstock for polyethylene, the world's most common plastic.
The college version
Core Concepts
Three electron groups: the sp² recipe
Count electron groups on one carbon of ethylene: two C–H σ bonds plus one C–C σ bond = three. Three groups → sp²: mix one s with two p orbitals. The result is three equivalent sp² orbitals, each 33% s and 67% p, lying in one plane at 120° from one another.
The σ framework and the leftover p orbital
The sp² orbitals overlap head-on: sp²–sp² for the C–C bond and sp²–1s for each C–H bond. Because all three sp² orbitals lie in one plane, the entire σ framework is planar — all six atoms of ethylene sit in a single plane. The unhybridized 2p orbital on each carbon is perpendicular to that plane, one lobe above and one below.
The π bond: side-by-side overlap
The two parallel p orbitals overlap sideways, above and below the plane, forming a π bond. Unlike a σ bond, a π bond has no electron density along the internuclear axis; its electrons occupy two lobes, one over and one under the molecular plane. A double bond is the combination: one σ (the strong, on-axis bond) plus one π (the weaker, off-axis bond).
Consequences: planarity, restricted rotation, shorter bond
The π bond dictates the molecule's behavior. Rotating one CH2 end relative to the other would destroy the parallel alignment of the p orbitals and break the π bond, so rotation about C=C is strongly restricted — a sharp contrast with the freely rotating C–C single bond of ethane. The extra bonding also shortens the bond: C=C is 1.33 Å versus 1.54 Å for C–C, and the higher s character (33% vs. 25%) strengthens it further.
Angles and lengths
The ideal sp² angle is 120°. In ethylene the measured H–C–H angle is about 117° (slightly compressed) and the C–C–H angles about 121°; the C=C bond is 1.33 Å and C–H bonds about 1.08 Å. The deviations from the 120° ideal are small, and the trigonal-planar label remains accurate.
How It Works / Step-by-Step Process
- Draw the Lewis structure of CH2=CH2.
- Count electron groups per carbon: three (two C–H σ + one C–C σ); π bonds do not count.
- Assign sp²; mix one s + two p → three sp² orbitals at 120°; leave one p orbital unhybridized.
- Overlap sp² orbitals head-on to build the planar σ framework (C–C and C–H bonds).
- Overlap the leftover p orbitals side-by-side to form the π bond above and below the plane.
Common Confusions
| Do Not Confuse | With | The Difference |
|---|---|---|
| Double bond = two σ bonds | Double bond = one σ + one π | Only one bond can lie on the internuclear axis; the second is a π bond |
| π electrons on the bond axis | π electrons above and below the plane | σ is on-axis; π is off-axis |
| Free rotation about C=C | Restricted rotation | Rotating would break the π bond, which costs substantial energy |
| sp² (planar) | sp³ (tetrahedral) | 3 electron groups vs. 4 |
| Counting π bonds as electron groups | Counting only σ bonds and lone pairs | A double bond contributes just ONE group (its σ bond) |
Eli explains
The same idea, in plain words
Explain it like I’m 10
Picture a child holding three equal arms spread flat like a peace sign, plus a fourth arm pointing straight up out of the paper. Two such children join their three flat arms — that is the strong σ framework. Then their up-pointing arms reach over and grab each other, one hand above the table and one below: that second grab is the π bond. Because the up-arms are holding on, the children cannot spin around, and the whole toy stays flat.
Worked example
Example 1: Counting σ and π bonds in ethylene
In CH2=CH2 there are four C–H bonds, one C–C bond, and one π bond. Each carbon contributes three σ bonds (two C–H plus one C–C), consistent with sp². Total per molecule: 5 σ bonds + 1 π bond = 6 bonding interactions. The π bond exists because each carbon keeps one unhybridized p orbital aligned parallel to its partner's.
Example 2: Converting the C=C bond length (dimensional analysis)
The C=C bond in ethylene is 1.33 Å. Convert to picometers using 1 Å = 100 pm:
1.33 Å × 100 pm1 Å = 133 pm
Then to nanometers, using 1 nm = 1000 pm:
133 pm × 1 nm1000 pm = 0.133 nm
Compare with ethane's C–C bond (154 pm): the double bond is about 21 pm shorter.
Example 3: Percent s character of sp²
Compute the s character of an sp² orbital with the formula % s = (s orbitals mixed ÷ total orbitals mixed) × 100%:
% s character = 13 × 100% ≈ 33%
The jump from 25% (sp³) to 33% (sp²) means the sp² C–H bonds of ethylene are shorter and stronger than the sp³ C–H bonds of ethane — part of why the C=C bond is so much shorter than C–C.
Key takeaways
- 3 electron groups → sp², trigonal planar, 120°.
- sp² = one s + two p mixed; 33% s character; one p orbital left unhybridized.
- A double bond = one σ + one π; π bonds form only from parallel, unhybridized p orbitals.
- π electron density sits above and below the plane — exposed and reactive.
- Rotation about C=C is restricted; rotating would break the π bond.
- C=C ≈ 1.33 Å, shorter than C–C ≈ 1.54 Å.
- All six atoms of ethylene are coplanar; H–C–H ≈ 117° (near the 120° ideal).
Check yourself
6 review questions from the chapter. Try each one, then open the answer.
How many electron groups does each carbon in ethylene have, and what hybridization results?
Show answer
Three electron groups (two C–H σ + one C–C σ; the π bond is not counted), so each carbon is sp² with trigonal-planar geometry.
What two components make up a C=C double bond?
Show answer
One σ bond (head-on sp²–sp² overlap) plus one π bond (side-by-side p-orbital overlap).
Where is the π electron density located in ethylene?
Show answer
Above and below the molecular plane, in two lobes; there is no π density on the internuclear axis.
Why can't ethylene rotate about its C=C bond?
Show answer
Rotation would misalign the parallel p orbitals and break the π bond; the molecule is locked planar.
How does the C=C bond length compare with the C–C single-bond length?
Show answer
It is shorter: C=C ≈ 1.33 Å (133 pm) vs. C–C ≈ 1.54 Å (154 pm).
What are the approximate bond angles around an sp² carbon?
Show answer
About 120° (ideal sp²); measured H–C–H in ethylene is ≈ 117°.
Study tools & related lessonsKey vocabulary · Related
Key vocabulary
- sp² hybridization
- Mixing one s + two p orbitals → three planar orbitals at 120°
- π (pi) bond
- Bond formed by side-by-side overlap of parallel p orbitals
- unhybridized p orbital
- p orbital left out of the hybridization mixing
- trigonal planar
- Shape with three groups at 120° in one plane
- restricted rotation
- Inability to rotate freely about a double bond
- alkene
- Hydrocarbon containing a C=C double bond
- π bond
- The sideways-overlap bond above and below the σ bond; it locks the alkene flat.
- sp³ hybridization
- Mixing one s + three p orbitals to make four equivalent orbitals
Sources & references
This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.
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