Chemistry · Hydrocarbons · NEET
A carbon-carbon triple bond has exactly 1 sigma bond and 2 pi bonds. The two sp hybrid orbitals overlap head-on (axial) to make the 1 sigma bond. The two leftover unhybridised p orbitals on each carbon overlap sideways to make 2 pi bonds. So the total is 1 sigma + 2 pi. This is a very common NEET fact to remember.
Each triple-bonded carbon in an alkyne is sp hybridised. Count the sigma bonds and lone pairs on that carbon: it has 2 sigma bonds (or 1 sigma + 1 lone pair) and no lone pair extra, giving 2 electron groups. Two groups means sp hybridisation. In ethyne (HC≡CH), each carbon makes 1 sigma to H and 1 sigma to the other carbon = 2 sigma = sp.
The two sp hybrid orbitals of a carbon point in exactly opposite directions to stay as far apart as possible. NCERT says the two sp hybrid orbitals are oriented in opposite direction. Opposite directions means the angle between them is 180 degrees. That is why H-C≡C is a straight line and the whole ethyne molecule is linear.
A triple-bond carbon is sp hybridised, NOT sp2. Easy rule: single bond area = sp3, double bond carbon = sp2, triple bond carbon = sp. Students often mix this up. Remember: the number after sp equals (number of pi bonds on that carbon). Triple bond carbon has 2 pi bonds but is still called sp (2 electron groups, no superscript for 2 sigma-domains).
Ethyne, HC≡CH, is linear. All four atoms (H, C, C, H) lie in one straight line because every carbon is sp hybridised with 180 degree bond angles. The C-C triple bond is short (about 120 pm) and strong because it has three shared electron pairs holding the carbons close together.
Add the number of sigma bonds and lone pairs on that carbon. 4 groups = sp3, 3 groups = sp2, 2 groups = sp. A triple-bond carbon has 2 sigma bonds (the two pi bonds do NOT count as separate groups), so 2 groups = sp. This trick works for the whole 2018 NEET question about the order sp2, sp2, sp, sp.
Which of the following molecules represents the order of hybridisation sp2, sp2, sp, sp from left to right atoms?
Among the compounds shown, the correct order of bond dissociation energy of the marked C-H bond is: (I) benzene, aryl sp2 C-H; (II) phenylacetylene C6H5-C≡C-H, terminal alkynyl sp C-H; (III) an sp3 C-H. Choose the correct order.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Ethyne (HC≡CH) has 3 sigma bonds (two C-H sigma bonds + one C-C sigma bond) and 2 pi bonds (both from the C≡C triple bond). So 3 sigma + 2 pi in the whole molecule.
The two unhybridised p orbitals on each triple-bond carbon overlap sideways (laterally) to form the 2 pi bonds. Each carbon keeps two p orbitals unhybridised because only one s and one p orbital mix to make the sp hybrids.
More shared electron pairs pull the two carbons closer. A triple bond (3 pairs) is shorter (about 120 pm) than a double bond (about 134 pm) and single bond (about 154 pm). Shorter bonds are also stronger, so the triple bond has high bond energy.
No. sp hybridisation happens on any atom with 2 electron groups. It occurs in alkynes (C≡C), in nitriles (C≡N), and in CO2. But for the Hydrocarbons chapter, the key example is the triple-bond carbon of an alkyne.
The sp carbon has 50% s-character, so it holds electrons close to the nucleus. This makes the terminal C-H easier to lose as H+ and the resulting carbanion more stable, so terminal alkynes are weakly acidic. This idea leads into the next topic on acidity of terminal alkynes.
