Carbon’s Tetravalence and Molecular Shapes
🔑 Key Concepts for NEET
- Hybridization types (sp³, sp², sp) determine molecular shapes
- π bonds create reactive sites and restrict rotation in double bonds
- Higher s-character in hybrid orbitals = stronger bonds + higher electronegativity
- Counting σ and π bonds in organic molecules
- Predicting carbon hybridization in functional groups
💎 Why Carbon is Special
Carbon always forms 4 covalent bonds (tetravalence)! This shapes all organic molecules. Remember:
- Methane (CH4) → Carbon uses sp³ orbitals → Tetrahedral shape
- Ethene (C2H4) → Carbon uses sp² orbitals → Planar shape
- Ethyne (C2H2) → Carbon uses sp orbitals → Linear shape
⚡ Hybridization Controls Bond Strength
Mixing s and p orbitals (hybridization) changes bond properties:
Hybrid Orbital | s-character | Bond Length | Bond Strength |
---|---|---|---|
sp (e.g., C≡C) | 50% | Shortest 🔥 | Strongest 💪 |
sp² (e.g., C=C) | 33% | Medium | Medium |
sp³ (e.g., C-C) | 25% | Longest | Weakest |
More s-character = orbitals closer to nucleus → stronger/shorter bonds!
🧲 Hybridization Affects Electronegativity
Higher s-character makes carbon more electronegative:
sp (50% s) > sp² (33% s) > sp³ (25% s)
This influences how carbon interacts with other atoms!
🔄 π Bonds: The Reactive Sites
Double/triple bonds contain π bonds with special features:
- Formed by sideways overlap of p-orbitals
- Require atoms to be in same plane 📐
- Prevent rotation around double bonds (locked shape!)
- Electron cloud above/below bond plane → super reactive! ⚡
π bonds are attack hotspots in molecules!
✏️ Practice Problems
Problem 1: Count σ and π bonds
(a) HC≡C–CH=CH–CH3
Solution: σC-C: 4, σC-H: 6, πC≡C: 2, πC=C: 1 → Total π bonds = 3
(b) CH2=C=CH–CH3
Solution: σC-C: 3, σC-H: 6, π bonds: 2
Problem 2: Find carbon hybridization
(a) CH3Cl → sp³ (single bonds)
(b) (CH3)2C=O → sp³ (CH3), sp² (C=O)
(c) CH3–C≡N → sp³ (CH3), sp (–C≡N)
(d) HCONH2 → sp² (carbonyl carbon)
(e) CH3–CH=CH–C≡N → sp³, sp², sp², sp
Problem 3: Hybridization and shape
(a) H2C=O → sp² hybridized → Trigonal planar
(b) CH3F → sp³ hybridized → Tetrahedral
(c) HC≡N → sp hybridized → Linear
💡 Remember This!
Carbon’s 4-bond superpower comes from s+p orbital mixing. This controls shape, bond strength, and reactivity – the foundation of organic chemistry! 🧪