Difficulty: Intermediate | Prerequisites: Lewis structures, VSEPR geometry, basic periodic trends (electronegativity, atomic size)
This material sits at the boundary between general chemistry and organic/inorganic chemistry. You are expected to connect molecular shape (from VSEPR and Lewis structures) to two new questions: how do molecules stick to each other (intermolecular forces, boiling points), and how do atoms share electrons in bonds (hybridisation, sigma vs pi bonding)? If you are comfortable drawing Lewis structures and assigning VSEPR geometries, you have the foundation. If not, review those first. Everything here builds on the idea that molecular geometry determines physical and chemical properties.
Boiling points depend on the type and strength of intermolecular forces, with hydrogen bonding in water being the standout anomaly. Hybridisation (sp, sp2, sp3) describes how atomic orbitals mix before bonding, and the leftover unhybridised p orbitals form pi bonds. For conjugated systems like nonatetraene, count the atoms involved in double bonds to find how many p orbitals participate in pi bonding. Planarity follows from sp2 hybridisation across the molecule.
Intermolecular forces (IMFs)
Forces of attraction between molecules (as opposed to intramolecular bonds within a molecule). The three main types are London dispersion forces, dipole-dipole interactions, and hydrogen bonding. Think of these as the glue that holds molecules near each other in a liquid or solid.
London dispersion forces (van der Waals forces)
Weak, temporary attractions caused by instantaneous dipoles in electron clouds. Present in all molecules. Their strength increases with molecular size (more electrons, more polarisable). In simple terms, bigger molecules have stickier electron clouds.
Dipole-dipole interactions
Attractions between the positive end of one polar molecule and the negative end of another. Stronger than London forces alone for molecules of similar size. Think of it as magnets lining up, positive to negative.
Hydrogen bonding
An especially strong type of dipole-dipole interaction that occurs when H is bonded to N, O, or F. The small size and high electronegativity of these atoms creates an unusually strong partial positive charge on H. This is why water has an anomalously high boiling point compared to H2S, H2Se, and H2Te.
Hybridisation
The mixing of atomic orbitals (s, p, and sometimes d) on a single atom to form new, equivalent hybrid orbitals used for sigma bonding. sp3 = four hybrid orbitals (tetrahedral), sp2 = three (trigonal planar), sp = two (linear). In simple terms, the atom reshapes its orbitals to point towards the atoms it bonds with.
Sigma (σ) bond
A bond formed by head-on overlap of orbitals along the bond axis. Every single bond is a sigma bond. The first bond in any double or triple bond is also sigma.
Pi (π) bond
A bond formed by side-on overlap of unhybridised p orbitals above and below (or in front of and behind) the bond axis. The second bond in a double bond and the second and third bonds in a triple bond are pi bonds.
Unhybridised p orbital
A p orbital that was not mixed into the hybrid set and remains available for pi bonding. An sp2 carbon has one unhybridised p orbital; an sp carbon has two.
Conjugated system
A series of alternating single and double bonds, where the unhybridised p orbitals on adjacent atoms can overlap continuously. 1,3,5,7-nonatetraene is an example. Conjugation affects colour, stability, and reactivity.
Planar molecule
A molecule in which all atoms lie in the same plane. Planarity is required for effective pi overlap in conjugated systems. All sp2-hybridised carbons in a chain are planar; an sp3 carbon breaks planarity because of its tetrahedral geometry.
Boiling point depends on how strongly molecules attract each other. Stronger IMFs mean more energy is needed to pull molecules apart into the gas phase, so the boiling point is higher.
For the Group 16 hydrides (H2O, H2S, H2Se, H2Te):
H2S, H2Se, and H2Te have similar dipole moments. Among these three, the dominant force is London dispersion, which increases with molecular size. So the order is H2S < H2Se < H2Te.
H2O is the exception. Despite being the smallest molecule, it has the highest boiling point because of strong hydrogen bonding (O-H bonds are highly polar, and oxygen has lone pairs to accept hydrogen bonds).
Correct boiling point order (lowest to highest): H2S, H2Se, H2Te, H2O
Water's boiling point (100 °C) is far higher than you would predict from the trend of the heavier hydrides. H2S boils at -60 °C, H2Se at -41 °C, H2Te at -2 °C. If water followed the London-dispersion trend, it would boil well below 0 °C. Hydrogen bonding accounts for the difference.
To decide which IMFs are present, ask:
Is the molecule polar? If yes, dipole-dipole forces are present.
Does the molecule have H bonded to N, O, or F? If yes, hydrogen bonding is present (on top of dipole-dipole).
All molecules have London dispersion forces, regardless of polarity.
The hybridisation of an atom is determined by its steric number (the number of electron groups around it, counting bonds and lone pairs):
Steric number 4 → sp3 (tetrahedral, 109.5°)
Steric number 3 → sp2 (trigonal planar, 120°)
Steric number 2 → sp (linear, 180°)
For carbon specifically:
A carbon with four single bonds is sp3.
A carbon in a double bond is sp2.
A carbon in a triple bond is sp.
Nonatetraene has nine carbons in a chain with four double bonds: C=C-C=C-C=C-C=C-C.
Hybridisation of each carbon (left to right): sp2, sp2, sp2, sp2, sp2, sp2, sp2, sp2, sp3
The first eight carbons are each part of a double bond, so they are all sp2. The ninth (terminal) carbon has only single bonds, making it sp3.
Each sp2 carbon has one unhybridised p orbital available for pi bonding. Each double bond uses two p orbitals (one from each carbon) for one pi bond.
Nonatetraene has four double bonds, involving eight sp2 carbons. That gives 8 unhybridised p orbitals involved in pi bonding.
The ninth carbon is sp3, so it contributes no unhybridised p orbitals to pi bonding.
Identify every atom involved in a pi bond (double or triple bond).
Each such atom contributes one unhybridised p orbital per pi bond it participates in.
For a simple conjugated polyene, the number of unhybridised p orbitals = 2 x (number of double bonds).
A molecule is completely planar when every atom in it lies in the same plane. This happens when all carbons (and other heavy atoms) are sp2 or sp hybridised. An sp3 carbon has tetrahedral geometry, which forces substituents out of the plane.
From the exam question on four molecules (I through IV):
Molecule I (a conjugated diene with substituents): contains sp3 carbons, so not fully planar.
Molecule II (cis-1,2-dichloroethene-type structure): all atoms around the double bond are sp2, so planar.
Molecule III (another dichloroethene isomer): likewise all sp2 carbons, planar.
Molecule IV (butatriene, H2C=C=C=CH2): cumulated double bonds, all carbons are sp2 or sp, all atoms in one plane.
The correct answer is II, III, and IV are completely planar.
Free rotation around single bonds means that a molecule with an sp3 centre can adopt conformations that are temporarily planar, but the question asks which are planar "all the time." Only molecules with no sp3 centres and no conformational flexibility that breaks planarity qualify.
Students often think water should have the lowest boiling point among Group 16 hydrides because it is the smallest molecule. It does not. Hydrogen bonding overrides the size trend.
Students confuse intramolecular bonds (covalent bonds holding atoms together within a molecule) with intermolecular forces (attractions between separate molecules). Boiling point is about breaking IMFs, not covalent bonds.
Students sometimes count the number of pi bonds and report that as the number of unhybridised p orbitals. Each pi bond involves two p orbitals (one per atom), so the count is always double the number of pi bonds in a simple polyene.
Assuming that any molecule with double bonds is automatically planar. A molecule can have double bonds and still be non-planar if it also contains sp3 centres that force atoms out of the plane.
⚠️ Ordering boiling points is a near-certain exam question. Know the anomaly of water and be able to explain it in one sentence.
⚠️ Hybridisation assignment for every carbon in a chain is commonly tested. Work left to right, check each carbon's bonding environment.
⚠️ Counting unhybridised p orbitals: the exam will ask you to add up all p orbitals on all atoms involved in pi bonding. Do not just count the double bonds.
⚠️ Planarity questions test whether you can identify sp3 centres that break the plane. If every carbon is sp2 or sp, the molecule is planar.
True or false: H2O has the lowest boiling point among H2O, H2S, H2Se, H2Te. (False, it has the highest.)
An sp2 carbon has ___ unhybridised p orbital(s). (One.)
True or false: London dispersion forces are only present in nonpolar molecules. (False, they are present in all molecules.)
Fill in the blank: The hybridisation of a carbon in a double bond is ___. (sp2.)
True or false: A molecule with all sp2 carbons is always planar. (True, assuming no other non-planar centres.)
Q: Order H2O, H2S, H2Se, and H2Te by boiling point from lowest to highest.
A: H2S < H2Se < H2Te < H2O. The first three follow the London dispersion trend (larger molecule = higher bp). Water is highest because of hydrogen bonding.
Q: How many unhybridised p orbitals are involved in pi bonding in 1,3,5,7-nonatetraene?
A: 8. There are four double bonds, each involving two p orbitals (one per carbon), and all eight sp2 carbons contribute one p orbital each.
Q: What is the hybridisation of each carbon in 1,3,5,7-nonatetraene, left to right?
A: sp2, sp2, sp2, sp2, sp2, sp2, sp2, sp2, sp3. The eight carbons in double bonds are sp2; the terminal CH3 carbon is sp3.
Q: Which molecules are completely planar: a conjugated diene with sp3 centres (I), cis-dichloroethene (II), trans-dichloroethene (III), butatriene (IV)?
A: II, III, and IV. Molecule I has sp3 carbons that prevent all atoms from lying in one plane.
Q: Explain why H2O has a much higher boiling point than H2S, even though H2S is a larger molecule.
A: H2O forms strong hydrogen bonds (O is highly electronegative and has lone pairs). H2S cannot form hydrogen bonds of comparable strength because S is much less electronegative than O, so its IMFs are limited to weaker dipole-dipole and London forces.
This material connects directly to molecular orbital theory (covered in a separate set of notes), which provides an alternative explanation for bonding and magnetic properties. The intermolecular forces concepts here also underpin later discussions of solubility, phase diagrams, and colligative properties. Hybridisation is the foundation for understanding organic reaction mechanisms, where knowing whether a carbon is sp2 or sp3 determines geometry and reactivity.
Related Terms / Search Tags
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