Chemistry · Chemical Bonding · NEET
Count how many orbitals mixed to make the hybrid. sp is made from 1 s + 1 p = 2 orbitals, so s-character = 1/2 = 50%. sp2 is 1 s + 2 p = 3 orbitals, so 1/3 = 33.33%. sp3 is 1 s + 3 p = 4 orbitals, so 1/4 = 25%. The p-character is just 100% minus the s-character. So sp = 50% s and 50% p, sp2 = 33% s and 67% p, sp3 = 25% s and 75% p.
Because s-character = (number of s orbitals) divided by (total orbitals mixed). In every case only ONE s orbital is used. In sp you mix 1 s with 1 p, so the single s is half of the mix = 50%. In sp3 you mix 1 s with 3 p orbitals, so the single s is only 1 out of 4 = 25%. More p orbitals in the mix means the s is a smaller share.
sp has the highest s-character (50%), sp2 is in the middle (33%), and sp3 has the lowest (25%). Order of s-character: sp > sp2 > sp3. This order is very important in NEET because many properties (electronegativity, bond strength, acidity) follow the same order.
More s-character means the electrons stay closer to the nucleus, because an s orbital is nearer the nucleus than a p orbital. So a carbon atom with more s-character pulls bonding electrons harder, which means higher electronegativity. Order of electronegativity of carbon: sp > sp2 > sp3. This is exactly why the H on an sp carbon (alkyne) is more acidic than on sp3 carbon (alkane).
An orbital with more s-character is smaller and closer to the nucleus, so it forms a shorter and stronger bond. That is why an sp carbon bond is shorter and stronger than an sp2 bond, and an sp2 bond is shorter and stronger than an sp3 bond. Short bond = strong bond here. NCERT states this directly in the bonding chapter.
Yes, for sp, sp2 and sp3 orbitals there are no d orbitals involved, so s-character + p-character = 100%. If you know s = 25% (sp3), then p = 75%. For d-mixed hybrids like sp3d you also include d-character, but those are not asked as simple percentages in NEET the same way.
Which of the following molecules represents the order of hybridisation sp2, sp2, sp, sp from left to right atoms?
The hybridisations of the atomic orbitals of nitrogen in NO2+, NO3- and NH4+ respectively are:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
sp = 50% s and 50% p, sp2 = 33.33% s and 66.67% p, sp3 = 25% s and 75% p. Only one s orbital is used each time, so the s-character equals 1 divided by the total number of orbitals mixed.
sp2 has more s-character (33.33%) than sp3 (25%). Because sp2 mixes only 2 p orbitals while sp3 mixes 3 p orbitals, the single s orbital is a larger share in sp2.
An s orbital sits closer to the nucleus than a p orbital. When a hybrid has more s-character, its electrons are held nearer the nucleus, so the atom pulls shared electrons more strongly, giving higher electronegativity. Order: sp > sp2 > sp3.
Yes. More s-character makes the orbital smaller and closer to the nucleus, so it forms a shorter and stronger bond. So an sp C-H bond is shorter and stronger than an sp2 or sp3 C-H bond.
NEET rarely asks the percentage directly. Instead it tests the consequences: which C-H is most acidic, which carbon is most electronegative, or which bond is shortest. All of these follow the s-character order sp > sp2 > sp3, so learning this ladder answers many questions.
