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Hybridisation questions appear in every chemistry paper, often as a one-mark item you should finish in under a minute. Most students know the definitions but count slowly or miscount lone pairs. This post explains what hybridisation is, gives one counting rule that works for almost every molecule you will meet, works ten examples, and flags the cases where the rule needs care.
What does the syllabus ask for?
The JEE (Main) 2026 syllabus includes "the concept of hybridization involving s, p and d orbitals" in chemical bonding, and "shapes of simple molecules - hybridization (s and p)" in organic chemistry. The NEET (UG) 2026 syllabus uses the same wording, and the JEE (Advanced) 2026 brochure lists "Hybridisation involving s, p and d orbitals only". Check the current documents for your exam.
What is hybridisation, and what is it not?
The NCERT chapter Chemical Bonding and Molecular Structure credits Linus Pauling with introducing the concept, which he set out in his 1931 paper The Nature of the Chemical Bond. Atomic orbitals of similar energy on one atom mix to form an equal number of equivalent hybrid orbitals, which point in directions that keep electron pairs apart. NCERT adds two points students often miss: promotion of an electron is not an essential condition, and filled orbitals can take part too.
Hybridisation describes a shape we already know from experiment. It is a model, not a step the atom "decides" to take. Research shows the topic is hard even for future teachers: Nakiboğlu (2003) tested 167 prospective chemistry teachers in Turkey, found misconceptions about atomic orbitals and hybridisation, and concluded that the atomic orbital concept is a key prerequisite. If s and p orbitals feel vague, fix that first.
How do you find the hybridisation quickly?
Count the electron domains on the atom: the number of σ bonds plus the number of lone pairs. A double or triple bond counts once, because it contains only one σ bond.
2 domains: sp (linear)
3 domains: sp² (trigonal planar)
4 domains: sp³ (tetrahedral)
5 domains: sp³d (trigonal bipyramidal)
6 domains: sp³d² (octahedral)
For a central atom bonded to monovalent atoms (H or halogens), a fast formula gives the count: H = ½(V + M − C + A), where V is the central atom's valence electrons, M the number of monovalent atoms, C the positive charge and A the negative charge. Oxygen and sulphur atoms attached to the centre are not counted in M.
Worked examples
XeF₄: ½(8 + 4) = 6, so sp³d². Four bonds and two lone pairs give a square planar shape.
ClF₃: ½(7 + 3) = 5, so sp³d. Three bonds and two lone pairs give a T shape.
NH₄⁺: ½(5 + 4 − 1) = 4, so sp³, tetrahedral.
CO₃²⁻: ½(4 + 0 + 2) = 3, so sp², trigonal planar.
I₃⁻: the central iodine has two terminal iodines, so ½(7 + 2 + 1) = 5, sp³d. With three lone pairs in the equatorial plane, the ion is linear.
BrF₅: ½(7 + 5) = 6, sp³d², square pyramidal, which matches NCERT's table.
PCl₅ and SF₆: 5 and 6, so sp³d and sp³d², as NCERT shows.
For carbon in organic molecules, count σ bonds and lone pairs on each carbon. In NCERT's organic chemistry chapter, Problem 8.2 gives CH₃CN as sp³ for the methyl carbon and sp for the nitrile carbon, and (CH₃)₂CO as sp³ and sp². The same chapter shows that a carbocation is sp² and trigonal planar, while a carbanion is generally sp³ with a distorted tetrahedral shape.
Why does hybridisation change bond length?
The more s character, the closer the electrons sit to the nucleus. NCERT notes that an sp³ orbital has 25 percent s character, while an sp orbital has 50 percent, which makes an sp carbon more electronegative than an sp² or sp³ carbon. It also gives the C–C bond in ethane as 154 pm and the C=C bond in ethene as 134 pm. Questions on acidity of ethyne, or on bond lengths, rest on this idea.
When does the counting rule need care?
Complexes. NCERT's table lists [Ni(CN)₄]²⁻ as dsp², square planar, and gives both sp³d² and d²sp³ for octahedral species. These come from valence bond theory in the coordination compounds chapter, not from the counting rule. Our guide to naming coordination compounds covers that chapter.
Lone pairs in conjugation. When a lone pair sits next to a double bond or inside an aromatic ring, the counting rule can mislead. In JEE Advanced questions, look at the whole structure before answering.
Shape versus hybridisation. NH₃ and H₂O are both sp³, but NCERT gives their bond angles as 107° and 104.5° because lone pairs repel more. The hybridisation is the same; the shapes differ. Our post on VSEPR and molecular shapes goes through the shapes in detail.
Which mistakes cost marks?
Counting a double bond as two domains.
Forgetting lone pairs on the central atom, especially in interhalogens and xenon fluorides.
Ignoring the charge on an ion.
Reporting the electron-domain geometry when the question asks for the shape.
A practice routine
Make a list of 20 species: five each with 3, 4, 5 and 6 domains, mixing neutral molecules and ions.
For each, write the domain count, the hybridisation and the shape, timing yourself. Aim for under 30 seconds each.
Take five organic molecules and label every carbon's hybridisation, as in NCERT Problem 8.2.
Once a week, add three new species you got wrong in a test.
If you are unsure why the orbitals mix in the first place, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can revise atomic orbitals before returning to hybridisation. For the next step in bonding theory, read our post on molecular orbital theory.
Right now, find the hybridisation of XeF₂, SF₄ and ICl₄⁻ with the formula, then check your shapes.
