Question
Compare the hybridization of carbon in ethane (C₂H₆), ethene (C₂H₄), and ethyne (C₂H₂). Explain the bond angles, bond lengths, and geometry in each case.
(NCERT Class 11, Chapter 4 — Chemical Bonding and Molecular Structure)
Solution — Step by Step
In ethane, each carbon forms 4 sigma bonds (3 C-H + 1 C-C). Carbon uses all four orbitals: one 2s and three 2p orbitals mix to form 4 sp³ hybrid orbitals.
- Geometry: Tetrahedral around each carbon
- Bond angle: 109.5°
- C-C bond length: 1.54 A (single bond — longest of the three)
- C-C bond: 1 sigma bond
In ethene, each carbon forms 3 sigma bonds (2 C-H + 1 C-C) and participates in 1 pi bond. One 2s and two 2p orbitals mix to form 3 sp² hybrid orbitals. The remaining unhybridized p orbital forms the pi bond by lateral overlap.
- Geometry: Trigonal planar around each carbon
- Bond angle: 120°
- C=C bond length: 1.34 A (shorter than ethane due to double bond)
- C=C bond: 1 sigma + 1 pi bond
In ethyne, each carbon forms 2 sigma bonds (1 C-H + 1 C-C) and participates in 2 pi bonds. One 2s and one 2p orbital mix to form 2 sp hybrid orbitals. The two remaining unhybridized p orbitals form two pi bonds (perpendicular to each other).
- Geometry: Linear
- Bond angle: 180°
- C≡C bond length: 1.20 A (shortest — triple bond)
- C≡C bond: 1 sigma + 2 pi bonds
Why This Works
The number of sigma bonds a carbon forms determines its hybridization: 4 sigma bonds → sp³, 3 sigma bonds → sp², 2 sigma bonds → sp. Pi bonds are always formed by unhybridized p orbitals through lateral (sideways) overlap.
As we go from sp³ → sp² → sp, the s-character increases (25% → 33% → 50%). More s-character means the electrons are held closer to the nucleus, making the bonds shorter and stronger. That’s why: C-C (1.54 A) > C=C (1.34 A) > C≡C (1.20 A).
The bond angle also increases with s-character: 109.5° → 120° → 180°. More s-character pushes the bonding pairs farther apart.
Alternative Method — Summary Table
| Property | Ethane (C₂H₆) | Ethene (C₂H₄) | Ethyne (C₂H₂) |
|---|---|---|---|
| Hybridization | sp³ | sp² | sp |
| s-character | 25% | 33.3% | 50% |
| Bond angle | 109.5° | 120° | 180° |
| Geometry | Tetrahedral | Trigonal planar | Linear |
| C-C bond length | 1.54 A | 1.34 A | 1.20 A |
| Sigma bonds per C | 4 | 3 | 2 |
| Pi bonds per C | 0 | 1 | 2 |
Quick rule for NEET: count the number of atoms directly bonded to a carbon + lone pairs on that carbon. That gives the number of hybrid orbitals needed: 4 → sp³, 3 → sp², 2 → sp. This works for any molecule, not just hydrocarbons.
Common Mistake
Students often say “ethene has sp² hybridization because it has a double bond.” This is backwards reasoning and can fail in complex cases. The correct logic: count sigma bonds + lone pairs on the atom → that determines hybridization. The double bond is a consequence, not a cause. Also, don’t confuse: the pi bond is NOT formed by hybrid orbitals — it’s formed by the unhybridized p orbital.