IB Chemistry HLTopic 2 — Models of Bonding & StructurePaper 1 & 2Higher level~10 min read
Hybridisation
Carbon’s electron configuration says it should form two bonds. It forms four, and all four are identical. Hybridisation is the model that fixes that contradiction, and once you have it, predicting the shape and the bonding of any organic molecule becomes almost automatic.
📘 What you need to know
Hybridisation is the mixing of atomic orbitals in the same shell to form new, equivalent hybrid orbitals used for bonding.
Carbon’s ground state is 1s2 2s2 2p2, which shows only two unpaired electrons.
One 2s electron is promoted to the empty 2p orbital, giving four unpaired electrons.
sp3: one s + three p → four hybrid orbitals, tetrahedral, 109.5°, all single bonds.
sp2: one s + two p → three hybrid orbitals, trigonal planar, 120°, one p orbital left over for a π bond.
sp: one s + one p → two hybrid orbitals, linear, 180°, two p orbitals left over for two π bonds.
Hybrid orbitals only ever form σ bonds. Leftover unhybridised p orbitals form the π bonds.
The problem hybridisation solves
Write out carbon’s ground state configuration and draw the orbital diagram. The 1s and 2s orbitals are full, and the two electrons in the 2p subshell sit in separate orbitals. That is two unpaired electrons, so carbon should form two bonds — and methane should be CH2.
The fix is in two steps. First one 2s electron is promoted into the empty 2p orbital, which costs a little energy but gives four unpaired electrons. Then the 2s and the three 2p orbitals mix to give four identical orbitals, all with the same energy and shape. Forming four bonds instead of two releases far more energy than the promotion cost, so it is well worth it.
Promoting, then mixing
The four sp3 orbitals are drawn at the same height because they are genuinely identical in energy. That is the whole point — it is why the four C–H bonds in methane are indistinguishable.
The three hybridisation states
Carbon does not always use all three p orbitals. How many it mixes in depends on how many σ bonds it needs to make, and whatever is left over goes into π bonds.
sp³, sp² and sp side by side
Read the middle row as the answer to “what shape?” and the row below it as the answer to “what bonds?”. Both come from the same single fact: how many p orbitals were mixed in.
🤔 Why mix orbitals at all?
An unmixed 2s orbital is spherical and the three 2p orbitals point along three axes at 90° to each other. Bonds built straight from those would not be equivalent, and they would sit at 90° rather than the 109.5° we measure. Mixing produces four identical orbitals pointing at the corners of a tetrahedron — as far apart as possible, which is exactly what VSEPR says should happen.
Working out the hybridisation
You do not need to think about orbitals at all in an exam. Just count the electron domains, exactly as you did for VSEPR.
Electron domains
Hybridisation
Shape of the hybrids
Leftover p orbitals
Example
4
sp3
tetrahedral
0
CH4, NH3, H2O
3
sp2
trigonal planar
1
C2H4, benzene, CH2O
2
sp
linear
2
C2H2, CO2, HCN
🧠 The examiner’s shortcut for carbon
Carbon always forms four bonds. Look at the highest bond order on that carbon: only single bonds → sp3; one double bond → sp2; a triple bond (or two doubles, as in CO2) → sp. It takes two seconds and it is never wrong for carbon.
Lone pairs count too. Nitrogen in ammonia has three bonds and one lone pair — four domains, so sp3. Oxygen in water has two bonds and two lone pairs — also four domains, also sp3. The hybrid orbitals hold the lone pairs just as happily as they hold bonding pairs.
A useful sanity check: an sp2 carbon must be part of a double bond, and every sp2 or sp carbon lies in a flat region of the molecule. If you have decided a carbon is sp2 but cannot find its π bond, recount.
Beyond carbon
The same idea applies to other elements. In period 3, the 3s and 3p orbitals hybridise in the same way, and when the octet is expanded the vacant 3d orbitals can be brought in as well. The IB focuses on carbon, but you should recognise that hybridisation is a general model rather than something peculiar to organic chemistry.
Worked examples
WE 1
State the hybridisation of each carbon in propene, CH3CH=CH2
Step 1: take each carbon separately
Carbon 1 (the CH3): four single bonds → 4 domains
Carbon 2 and carbon 3: each has one double bond → 3 domainsStep 2: convert domains to hybridisation4 domains → sp³3 domains → sp²CH₃ carbon is sp³; both C=C carbons are sp²the two sp² carbons and the four atoms attached to them all lie in one plane
WE 2
Explain the bonding in ethyne, C2H2, in terms of hybridisation [3]
Mark 1: state the hybridisation
Each carbon has 2 electron domains, so each is sp hybridised, giving two hybrid orbitals at 180°.
Mark 2: account for the σ bonds
One sp orbital forms a σ bond to hydrogen and one forms a σ bond to the other carbon: 3 σ bonds in total.
Mark 3: account for the π bonds
Each carbon keeps two unhybridised p orbitals, which overlap sideways to give two perpendicular π bonds.
Linear, with 3 σ and 2 π bondsC≡C is 1 σ + 2 π, so the molecule is 3 σ and 2 π overall
WE 3
Deduce the hybridisation of the nitrogen atom in the ammonium ion, NH4+
Step 1: count the domains
Four N–H bonds and no lone pair left, so 4 electron domains.
Step 2: convert to hybridisation4 domains → sp³Step 3: check it against the shape
sp3 predicts tetrahedral at 109.5°, which is exactly what NH4+ is.
Nitrogen is sp³ hybridisedammonia itself is also sp³ — the lone pair still occupies a hybrid orbital
💡 Exam tips
Count electron domains, not bonds. Lone pairs occupy hybrid orbitals too.
Write the superscript properly: sp3, sp2, sp. “sp3” without the superscript can cost you.
Link hybridisation to shape and bond angle in the same sentence when the question allows it.
Remember that hybrid orbitals only form σ bonds; the π bonds come from unhybridised p orbitals.
For a molecule with several carbons, state each one separately — they are often different.
Mention promotion if asked to explain why carbon forms four bonds, and say the energy is repaid.
⚠ Common mix-ups
Counting a double bond as two domains. It is one domain, so an alkene carbon is sp2, not sp.
Forgetting lone pairs. Water is sp3 even though oxygen only makes two bonds.
Saying π bonds come from hybrid orbitals. They come from the leftover unhybridised p orbitals.
Thinking promotion is the whole explanation. Promotion gives four unpaired electrons; hybridisation is what makes the four bonds identical.
Giving one hybridisation for a whole molecule when different carbons differ.
Confusing sp2 with sp3 in benzene. Every ring carbon is sp2, which is why the ring is flat.
That completes The Covalent Model. Up next: The Metallic Model — the third and final type of bonding, where the electrons are shared not between two atoms but across the entire structure.
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