Why is carbon tetravalent if ground state has only two unpaired electrons? Also my sigma-pi count for CH2=CH-C≡CH is off
Aarav Kulkarni11th Sample
18 Aug 2026 · 1 min read
ok so I'm making a checklist for the Organic Basic Principles chapter and I got stuck on page one, which is embarrassing.
Carbon's ground state is , so only two unpaired electrons. Then why is it tetravalent and not divalent? NCERT says one 2s electron gets promoted to the empty 2p orbital, but promotion costs energy, so where does that energy come from? Feels like carbon is paying for something it didn't need.
Second thing. I tried a count question: for I wrote the hybridisation as sp2, sp2, sp, sp, which I think is right. But then I counted sigma and pi bonds and got 8 sigma and 3 pi, and the answer key at the back says something different. I counted every C–H, every C–C once, and then added the extra bonds in the double and triple bond. Where am I double counting?
Also, is the bond angle exactly 109.5° in methane and exactly 120° in ethene? My teacher said "approximately" and now I'm doubting everything lol.
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[+][–]Kavya NaikDropperSample· 2mo ago
The energy thing confused me too in my first year. Promotion does cost energy, but carbon then makes four bonds instead of two, and every bond formed releases energy. The energy released by two extra bonds is way more than the promotion cost, so the molecule ends up much more stable overall. So carbon isn't losing anything, it's a net profit.
After promotion the config is and these four orbitals mix (sp3) to give four equivalent orbitals. That's why all four C–H bonds in methane are identical, even though one s and three p orbitals were used.
[+][–]Arjun Nair12thSample· 2mo ago
Your hybridisation is right. The count is where it slipped. Draw it out with all atoms numbered:
C1 = CH2, C2 = CH, C3 = C (no H!), C4 = CH.
C–H bonds: 2 + 1 + 0 + 1 = 4 (all sigma). C1=C2: 1 sigma + 1 pi. C2–C3: 1 sigma. C3≡C4: 1 sigma + 2 pi.
Sigma = 4 + 1 + 1 + 1 = 7. Pi = 1 + 2 = 3. Total bonds = 10, which you can check as 7 + 3.
[+][–]Aarav Kulkarni11thSample· 2mo agoAuthor
oh no. I gave C3 a hydrogen. I thought of it as "CH≡" for some reason. Thanks, 7 and 3 it is 😭
[+][–]Diya Menon11thSample· 2mo ago
Omg I just learned this too! I think hybridisation is when the s orbital of hydrogen mixes with the p orbitals of carbon, that's how you get four equal bonds? That's how my teacher's drawing looked anyway.
[+][–]Rahul DubeyDropperSample· 2mo ago
Diya, small correction. Hybridisation is mixing of orbitals of the same atom with similar energies, before bonding. Carbon's own 2s and 2p mix. Hydrogen's 1s overlaps with the resulting sp3 orbital afterwards, and that overlap is the bond. Mixing and overlapping are different steps.
A quick way to read hybridisation off a structure: count sigma bonds plus lone pairs on the atom. 4 means sp3, 3 means sp2, 2 means sp. In your molecule C1 and C2 have 3 sigma bonds each, C3 and C4 have 2 each. Done.
And yes "approximately" is the honest answer. Ideal values are 109.5° (sp3), 120° (sp2), 180° (sp), but real molecules deviate when atoms or lone pairs differ. In ethene the H–C–H angle is a bit under 120° because the double bond region repels more.
[+][–]Diya Menon11thSample· 2mo ago
ahh okay that makes sense, thank you rahul bhaiya. I was mixing the two steps.
[+][–]Pooja Iyer12thSample· 2mo ago
For NEET they like the angle-orbital link and the s-character link. s-character is 25% in sp3, 33% in sp2, 50% in sp. More s-character means the electrons are held closer to the nucleus, so the carbon is more electronegative, and C–H bond gets shorter. That is why sp C–H (in ethyne) is shorter than sp3 C–H (in ethane), and why terminal alkynes are acidic. Free marks in a PYQ-style question 😌
