Tetravalency of carbon; hybridization recap (sp, sp², sp³)
1. Tetravalency: WHY carbon makes 4 bonds
WHAT is the puzzle? Ground-state carbon () has configuration: Only two unpaired electrons () → it should be divalent (like in :CH₂, a carbene). Yet methane is CH₄.
HOW is the puzzle solved? (Derivation from first principles)
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Promotion (excitation): One electron is promoted to the empty : Now there are four unpaired electrons → four bonds possible.
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Energy cost vs payoff (the WHY it's worth it): Promotion costs energy ( kJ/mol to move ). But forming two extra C–H bonds releases ( kJ/mol). Net release ≫ cost, so the molecule is far more stable. Nature pays a small bill to earn a big refund.
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The new problem: We now have one orbital + three orbitals — but these have different shapes and energies. Yet all four C–H bonds in CH₄ are identical (same length 109 pm, same energy). So the orbitals must mix. That mixing is hybridization.
2. Hybridization: WHAT it is and HOW it works
Rules (the WHY behind each):
- Conservation of orbitals: atomic orbitals in → hybrid orbitals out. (You can't create or destroy orbitals.)
- Only the central atom hybridizes, using valence orbitals of similar energy.
- More -character → orbital held closer & tighter to nucleus → shorter, stronger bond, larger bond angle, higher electronegativity of that carbon.
3. The three carbon hybridizations
sp³ (4 hybrids)
- One + three → four orbitals.
- 25% s-character, angle , tetrahedral.
- Example: CH₄, ethane C–C, diamond. No leftover p orbitals → only σ bonds → single bonds only.
sp² (3 hybrids)
- One + two → three orbitals (planar, ); one unhybridized p left perpendicular.
- 33% s-character.
- That leftover p forms a π bond → presence of one double bond.
- Example: ethene C₂H₄, benzene, carbonyl C=O.
sp (2 hybrids)
- One + one → two orbitals (linear, ); two unhybridized p left.
- 50% s-character.
- Two p orbitals → two π bonds → a triple bond (or two cumulated doubles, as in allene C=C=C).
- Example: ethyne C₂H₂, CO₂, –C≡N.
4. Worked examples
5. Common mistakes (Steel-man + fix)
6. Active recall flashcards
Why is ground-state carbon expected to be divalent?
What makes carbon tetravalent despite that?
Define hybridization.
Formula for steric number?
Why don't π bonds count in SN?
SN 2, 3, 4 give which hybridizations and angles?
Order of s-character in sp, sp², sp³?
Effect of higher s-character on bond length and angle?
Hybridization of carbon in ethyne and why?
Hybridization of C in CO₂?
How many unhybridized p orbitals in sp, sp², sp³?
Why is terminal alkyne H acidic?
Recall Feynman: explain to a 12-year-old
Imagine carbon has four hands to hold onto other atoms. But at first it looks like it only has two free hands and two stuck together. With a tiny push (promotion), it frees them so all four hands are open. Then carbon shuffles its hands so they're all the same length and equally spread out — that shuffling is hybridization. If carbon spreads its hands into 4 directions like a pyramid, it's . Into a flat triangle, (and it keeps one extra hand for a double-grip). Into a straight line, (two extra hands for a triple-grip). The straighter the spread, the stronger and shorter the grip.
Connections
- Sigma and Pi bonds — orbital overlap
- VSEPR theory and molecular geometry
- Inductive effect and electronegativity of hybrid carbons
- Acidity of terminal alkynes
- Resonance and delocalization (sp² systems)
- Bond length and bond strength trends
- Aromaticity and benzene (all-sp² ring)
Concept Map
Hinglish (regional understanding)
Intuition Hinglish mein samjho
Dekho, carbon organic chemistry ka hero isliye hai kyunki woh chaar bonds banata hai (tetravalency). Lekin ground state mein carbon ke paas sirf do unpaired electrons hote hain, toh logically usse divalent hona chahiye. Solution kya hai? Promotion — ek electron ko mein bhej do, ab chaar unpaired electrons, chaar bonds ready. Yeh promotion energy maangta hai, par do extra bonds banne se jo energy release hoti hai woh kahin zyada hai, toh deal profitable hai. Nature thoda kharcha karke bada faayda kamati hai.
Ab problem yeh ki aur orbitals alag shape ke hain, par CH₄ ke saare bonds identical hain. Isliye orbitals mix ho jaate hain — yahi hybridization hai. Yaad rakhne ka simplest tareeka: Steric Number = sigma bonds + central atom ke lone pairs. SN 4 → (tetrahedral, 109.5°), SN 3 → (planar, 120°), SN 2 → (linear, 180°). Important baat: pi bonds ko mat ginna SN mein, kyunki woh pure p orbital se bante hain, hybrid ki zaroorat nahi.
Ek golden rule mnemonic: single bond → sp³, double bond → sp², triple bond → sp. Aur "more pi, less hybrid" — jitne zyada pi bonds, utne kam hybrid orbitals. s-character bhi yaad rakho: sp (50%) > sp² (33%) > sp³ (25%). Zyada s-character matlab electron nucleus ke paas, isliye chhota bond, strong bond, bada angle, aur zyada electronegative carbon — isi wajah se terminal alkyne ka H acidic hota hai. Bas SN nikaalo, aur poori geometry, bond length aur reactivity tumhare haath mein.