Hybridisation Made Simple: sp, sp², sp³ and Beyond for NEET & JEE
Carbon has the electron configuration 1s² 2s² 2p², with only two unpaired electrons, yet methane has four identical C–H bonds at 109.5°. How can one s orbital and three p orbitals produce four bonds that are exactly equal? Hybridisation is the model that answers this, and it is also a quick tool for predicting shape. It matters for both NEET and JEE.
Quick Summary
Hybridisation is the mixing of atomic orbitals of similar energy on the same atom to form the same number of equivalent hybrid orbitals.
Number of hybrid orbitals = number of orbitals mixed. One s + one p gives 2 sp; one s + two p gives 3 sp²; one s + three p gives 4 sp³.
Steric number (bond pairs + lone pairs on the central atom) 2, 3, 4, 5, 6 corresponds to sp, sp², sp³, sp³d, sp³d².
s-character is 50% in sp, 33.3% in sp², 25% in sp³. More s-character means a shorter bond and higher electronegativity of that carbon.
Only σ bonds and lone pairs count toward hybridisation. π bonds do not.
Why do we need the idea at all?
Start with what you know: an s orbital is spherical, and the three p orbitals point along the x, y and z axes at 90° to each other. If carbon used these directly in methane, three bonds would be p-type at 90° and one would be s-type, all different. Experiments show four identical bonds at 109.5°. So the model says that before bonding, the 2s and three 2p orbitals on carbon blend into four new, identical orbitals, called sp³ hybrids. Each has a large lobe on one side of the nucleus, which overlaps better with another atom's orbital than a plain s or p would, so the bond is stronger. Four equal orbitals repel each other and settle at the tetrahedral angle.
Notice the logic. Hybridisation is not a physical event you can watch. It is a bookkeeping model that explains geometry we already observe. This is why the practical method is to find the geometry first and read off the hybridisation from it.
The fast method: steric number
For the central atom, count steric number = (number of σ bonds) + (number of lone pairs). Here a σ bond is the single sigma connection to each neighbouring atom, and a double or triple bond still counts as just one σ bond plus π bonds.
Steric number 2: sp, linear, 180° (BeCl₂, CO₂, HC≡CH)
Steric number 3: sp², trigonal planar, 120° (BF₃, C₂H₄, SO₂ with a bent shape)
Steric number 4: sp³, tetrahedral, 109.5° (CH₄, NH₃, H₂O)
Steric number 5: sp³d, trigonal bipyramidal (PCl₅)
Steric number 6: sp³d², octahedral (SF₆)
Worked example 1: hybridisation of the central atom in NH₃ and H₂O
Nitrogen has 5 valence electrons. In NH₃ it forms 3 σ bonds with hydrogen, using 3 electrons. The remaining 2 electrons form 1 lone pair. Steric number = 3 + 1 = 4, so nitrogen is sp³. The four sp³ orbitals point to the corners of a tetrahedron, but only three hold bonds, so the molecule is pyramidal.
Oxygen has 6 valence electrons. In H₂O it forms 2 σ bonds, leaving 4 electrons, which is 2 lone pairs. Steric number = 2 + 2 = 4, again sp³. The shape is bent. The bond angle is about 104.5° in water and about 107° in ammonia, below 109.5°, because lone pairs repel more strongly than bond pairs and squeeze the bond angle.
Worked example 2: counting in a bigger molecule
Take propene, CH₃–CH=CH₂, and find the hybridisation of each carbon.
CH₃ carbon: 4 σ bonds (3 to H, 1 to the next C), no π bond. Steric number 4, so sp³.
Middle carbon (CH=): 3 σ bonds (1 to H, 1 to CH₃, 1 to the other C) and one π bond. Steric number 3, so sp².
Terminal CH₂: 3 σ bonds (2 to H, 1 to C) and one π bond. Steric number 3, so sp².
Total σ bonds: 6 C–H bonds plus 2 C–C bonds = 8 σ. Total π bonds: 1. So propene has 8 σ and 1 π bond. This count is a favourite exam question.
s-character and what it changes
An sp orbital is one part s to one part p, so s-character = 1/2 = 50%. For sp² it is 1/3 ≈ 33.3%, and for sp³ it is 1/4 = 25%. The s orbital sits closer to the nucleus, so more s-character pulls electrons in more tightly. Consequences you can use directly:
Bond length falls as s-character rises: C–C (sp³–sp³) is longer than C=C, which is longer than C≡C. Specifically, the C–H bond in ethyne is shorter than in ethene, which is shorter than in ethane.
Electronegativity of carbon rises in the order sp³ < sp² < sp.
Acidity of C–H rises with s-character, so ethyne is far more acidic than ethene or ethane.
Key Insight: Never count π bonds when finding hybridisation, and never forget lone pairs. A reliable shortcut is to ask: how many atoms are attached plus how many lone pairs? That single number is the steric number and gives the hybridisation directly. For ions use the same method after adjusting the electron count: in NH₄⁺ nitrogen has 4 σ bonds and no lone pair (sp³), while in NH₂⁻ it has 2 σ bonds and 2 lone pairs (also sp³).
Mistakes students actually make
Counting a double bond as two bonds. Only one of them is σ; the other is π and does not affect hybridisation.
Ignoring lone pairs, which is why H₂O gets wrongly called sp².
Assuming hybridisation always equals shape. NH₃ is sp³ but pyramidal, not tetrahedral. Hybridisation tells you the arrangement of electron pairs; the lone pairs decide the molecular shape.
Applying sp³d and sp³d² to second-period elements. Nitrogen and oxygen have no d orbitals in their valence shell, so they cannot expand their octet.
Forgetting that carbon in a carbocation is sp² (3 σ bonds, no lone pair), while a carbanion carbon is sp³ (3 σ bonds plus a lone pair).
How it is asked in NEET and JEE
Direct: "The hybridisation of the central atom in XeF₄ is..." (6 pairs, so sp³d²).
Counting: number of sp² carbons, or σ and π bonds in a given organic molecule.
Comparative: arrange by bond length, bond angle or s-character.
Match-the-column: species against shape and hybridisation, such as PCl₅ (sp³d, trigonal bipyramidal) and SF₆ (sp³d², octahedral).
Conceptual: why the bond angle in H₂O is less than in NH₃, which is a lone-pair repulsion argument, not a hybridisation one.
When you practise these on Sukriti NeetJee or from any problem set, write the steric number next to each central atom before choosing an answer. It prevents most errors.
Frequently Asked Questions
Is hybridisation a real physical process?
No. It is a mathematical model that combines atomic orbitals to explain observed bond angles and equal bond strengths. It is useful and examined, but the atom does not literally pass through a promotion step in time.
Can orbitals of different energies hybridise?
Only orbitals of comparable energy mix effectively, such as 2s with 2p, or 3s, 3p and 3d. That is why the 1s orbital is never involved in hybridisation of carbon.
What is the hybridisation of carbon in CO₂ and why?
It is sp. Carbon makes two σ bonds, one to each oxygen, and has no lone pair, so the steric number is 2. The two double bonds each contain one σ and one π bond, and the molecule is linear.
Why is the bond angle in NH₃ less than 109.5° if nitrogen is sp³?
The lone pair occupies more space than a bonding pair and repels the three bond pairs more strongly. This squeezes the H–N–H angle to about 107°. The hybridisation stays sp³.
How does hybridisation affect acidity of hydrocarbons?
The more s-character the carbon has, the more tightly it holds the electrons in the C–H bond, so the conjugate carbanion is more stable. That is why the order of acidity is ethyne > ethene > ethane.
If any step here felt unclear, ask in the comments and I will work through it with you, and share this with a fellow aspirant who still mixes up shape and hybridisation.
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