How Does sp2 Hybridisation Lead to the Double Bond in Ethene?
The Hybridisation Of C2H4 is a fundamental concept in JEE Main Chemistry, as it explains both the bonding and geometric arrangement in ethene (C2H4). Understanding hybridisation helps predict molecular shapes, bond angles, and the stability of many organic molecules. Ethene’s symmetrical, two-dimensional structure is a classic example of sp2 hybridisation, with direct links to sigma and pi bond formation. Rapid revision of this concept is essential for mastering hydrocarbons and organic chemistry theory.
What is Hybridisation and Why Does it Matter for Ethene?
Hybridisation is the process by which atomic orbitals in an atom mix to form new, equivalent hybrid orbitals suitable for the pairing of electrons to form chemical bonds. In organic molecules like C2H4, it clarifies both connectivity (bond types) and geometry (shape), both of which frequently appear in JEE Main questions. Identifying the correct hybridisation gives a direct shortcut to predicting bond angles and the presence of sigma (σ) or pi (π) bonds.
Types of Hybridisation: sp, sp2, and sp3
Hybrid orbitals are classified based on the number and type of atomic orbitals that combine. Recognise these key types for JEE:
- sp hybridisation: 1 s + 1 p, linear shape, 180° bond angle (e.g., C2H2).
- sp2 hybridisation: 1 s + 2 p, trigonal planar, 120° bond angle (used in C2H4).
- sp3 hybridisation: 1 s + 3 p, tetrahedral, 109.5° bond angle (e.g., C2H6).
Stepwise Explanation: Hybridisation Of C2H4
To find the Hybridisation Of C2H4, analyse the carbon atoms in ethene. Each carbon has electron configuration 1s2 2s2 2p2. In C2H4, each carbon forms three sigma (σ) bonds—two with hydrogen atoms and one with the other carbon. To do this, one 2s and two 2p orbitals mix to create three sp2 hybrid orbitals. The unhybridised 2p orbital on each carbon remains orthogonal to the plane, ready to form a π bond.
- Step 1: Excite the carbon atom (promote an electron from 2s to 2p).
- Step 2: Mix one 2s and two 2p orbitals to form three planar sp2 hybrids.
- Step 3: Arrange these hybrids at 120°, ensuring maximum separation.
- Step 4: The remaining 2p orbital (unhybridised) will laterally overlap to form a π bond.
This model explains why the Hybridisation Of C2H4 is sp2 and gives the molecule its planar, triangular symmetry.
Bonding and Structure in Ethene: sigma and pi Bonds
Each carbon atom in C2H4 forms three σ bonds using the three sp2 hybrid orbitals: two bonds to hydrogen and one to the other carbon. The unhybridised p orbitals on both carbons overlap sideways, creating the π bond of the C=C double bond. Thus, a double bond in ethene consists of one strong σ bond (head-on overlap) and one weaker π bond (sidewise overlap).
- 4 C–H σ bonds (sp2–1s overlap)
- 1 C–C σ bond (sp2–sp2 overlap)
- 1 C–C π bond (p–p lateral overlap)
The presence of the π bond restricts rotation around the double bond, explaining why ethene is rigid and planar—this is vital in many organic mechanisms and reaction questions found in JEE Main.
Shape and Bond Angle: Trigonal Planar Geometry in Hybridisation Of C2H4
Because the three sp2 hybrid orbitals in each carbon are arranged at 120°, the shape of C2H4 is trigonal planar. All atoms (four hydrogens and two carbons) lie in the same plane. The bond angles are all about 120°. This matches the predictions from Valence Shell Electron Pair Repulsion (VSEPR) theory, which also frequently appears on JEE papers.
Comparison Table: Ethane vs Ethene vs Ethyne Hybridisation
| Molecule | Formula | Hybridisation | Geometry | Bond Angle |
|---|---|---|---|---|
| Ethane | C2H6 | sp3 | Tetrahedral | 109.5° |
| Ethene | C2H4 | sp2 | Trigonal planar | 120° |
| Ethyne | C2H2 | sp | Linear | 180° |
Notice that as the number of σ bonds decreases and the number of π bonds increases (from ethane to ethyne), the hybridisation changes from sp3 to sp2 to sp, resulting in different geometries and bond angles. This progression is a common multiple-choice trap in JEE exams. Explore more with this difference between sp, sp2, and sp3 hybridisation resource.
Hybridisation Of C2H4: Formula and Quick Application Tips
Use this fast formula to determine hybridisation in C2H4 and similar molecules:
- H = Number of σ-bonded atoms + Single lone pairs
- For each carbon in C2H4: H = 3 (2 hydrogens + 1 carbon), so sp2 hybridisation
- If H = 2 → sp; H = 3 → sp2; H = 4 → sp3
Shortcut: Count the number of groups (atoms/lone pairs) attached to the central atom. This rule is exam-friendly for JEE Main and matches NCERT reasoning. For more examples, visit hybridization of carbon or hybridization in molecules topics on Vedantu.
Solved Example: Applying Hybridisation Of C2H4 in JEE-Style Questions
- What is the hybridisation of each carbon atom in C2H4? Solution: Each carbon is bonded to three atoms (2 H + 1 C) and has no lone pairs, so the hybridisation is sp2.
- How many sigma and pi bonds are present in C2H4? Solution: There are 5 σ bonds (4 C–H + 1 C–C) and 1 π bond (in the C=C).
- What is the molecular geometry of C2H4? Solution: Both carbons have sp2 hybridisation, so the shape is trigonal planar with bond angles of 120°.
These types of bonded atom counts and geometry predictions are regular features in JEE Main chemical bonding and molecular structure questions.
Where to Revise Next: Related Hybridisation and Hydrocarbon Resources
Summary: Hybridisation Of C2H4 Key Points for JEE
- C2H4 (ethene) has sp2 hybridisation in both carbon atoms.
- Each carbon forms three sigma bonds (using sp2 orbitals) and has one unhybridised p orbital for pi bonding.
- The shape is trigonal planar, all atoms are in a single plane, and bond angle is 120°.
- C=C double bond = 1 σ + 1 π bond, explaining restricted rotation and rigidity.
- Quick trick: Number of σ-bonded atoms + lone pairs = hybridisation type (2 → sp; 3 → sp2; 4 → sp3).
- Understanding Hybridisation Of C2H4 provides a foundation for analysing more advanced hydrocarbons and reactions for JEE Main Chemistry.
This concise revision of Hybridisation Of C2H4 provides the key tools you need for problem-solving, quick recall, and visible mastery of molecular shapes, all tuned to JEE Main Chemistry questions and best practices. For deeper exploration and stepwise video explanations, Vedantu’s resources cover linked topics in detail to tighten your organic chemistry preparation.
FAQs on Hybridisation of C2H4: Concept, Structure & Diagrams
1. What is the hybridization of C2H4?
C2H4 (ethene) exhibits sp2 hybridisation in both its carbon atoms. In this molecule:
- Each carbon forms three sp2 hybrid orbitals for sigma bonds (two with hydrogen, one with another carbon).
- One unhybridised p orbital remains on each carbon atom to form a pi bond.
- This leads to a planar structure with approximately 120° bond angles.
2. Is ethene sp2 or sp3 hybridised?
Ethene (C2H4) is sp2 hybridised, not sp3. The reasons are:
- Each carbon in C2H4 undergoes sp2 hybridisation to make three sigma bonds.
- The leftover p orbital forms a pi bond, which is characteristic of alkenes like ethene.
3. What is the hybridization trick of C2H4?
The hybridisation trick for C2H4 is to count sigma bonds + lone pairs on the carbon atom:
- Each carbon in C2H4 forms three sigma bonds (two C–H, one C–C) and has zero lone pairs.
- Three regions of electron density indicate sp2 hybridisation.
4. What is the shape and bond angle of C2H4?
C2H4 (ethene) has a trigonal planar shape with bond angles close to 120°.
- Each carbon atom's sp2 hybrid orbitals are arranged in a planar triangle.
- The molecule is flat, and all H–C–H and H–C–C angles are about 120°.
5. What is the main difference between the hybridisation of C2H6 and C2H4?
C2H6 uses sp3 hybridisation, while C2H4 uses sp2 hybridisation. The differences are:
- C2H6 (ethane): All carbons are sp3 hybridised with four sigma bonds and tetrahedral geometry (bond angle ~109.5°).
- C2H4 (ethene): All carbons are sp2 hybridised, forming one double bond (1 sigma + 1 pi) and trigonal planar geometry (bond angle ~120°).
6. How many sigma and pi bonds are present in C2H4?
C2H4 contains five sigma bonds and one pi bond.
- There are four C–H sigma bonds and one C–C sigma bond.
- One pi bond exists between the two carbon atoms as part of the double bond.
7. Can C2H4 have sp3 hybridisation in any of its carbon atoms?
No, C2H4 cannot have sp3 hybridisation in its carbon atoms under normal conditions.
- The presence of a double bond restricts each carbon to three sigma bonds, making sp2 hybridisation necessary.
- sp3 hybridisation only occurs when a carbon forms four sigma bonds, as in alkanes like ethane.
8. What type of shape is C2H4?
C2H4 (ethene) has a trigonal planar molecular shape.
- Each carbon's bonded atoms are arranged in a flat, 120° angle.
- This geometry results from sp2 hybridisation.
9. Does hybridisation affect the strength or length of the double bond in C2H4?
Yes, hybridisation directly affects both the strength and length of the double bond in C2H4.
- sp2 hybrid orbitals have more s-character than sp3, leading to shorter, stronger sigma bonds.
- The formation of a pi bond (via unhybridised p orbitals) further strengthens the double bond, making it shorter and stronger than single bonds in alkanes.
10. Can the hybridisation of C2H4 be determined using a simple formula or rule?
Yes, the hybridisation of C2H4 can be determined by using the steric number (sigma bonds + lone pairs):
- For each carbon in C2H4: 3 sigma bonds + 0 lone pairs = 3.
- Steric number 3 corresponds to sp2 hybridisation.
