Q: What does H2 Chemistry Notes: Topic 2 - Chemical Bonding cover? A: Consolidate VSEPR shapes, bond polarity, intermolecular forces, and lattice energy trends for Core Idea 2 (Chemical Bonding) in the 2026 H2 Chemistry syllabus.
Chemical bonding connects atomic structure to macroscopic behaviour: lattice energies govern melting points, molecular geometry decides intermolecular forces, and bond polarity predicts reaction pathways. This note organises the skills required for Paper 2 and Paper 3 questions, supported by exam-style worked examples. For the full set of H2 Chemistry revision resources, refer to https://eclatinstitute.sg/blog/h2-chemistry-notes. For the full topic map and paper weightings, see our H2 Chemistry Syllabus 2026-27 overview.
Status: SEAB's current H2 Chemistry (9476) syllabus PDF is labelled for 2026, and the current Chemistry Data Booklet is labelled 8873/9476/9813 for use from 2026 in non-practical papers. Core Idea 2 expectations remain assessed across Papers 1-3.
The core idea is simple: Chemical bonding links particle structure to real properties such as shape, polarity, melting point, and solubility.
Use it as a working check: For explanation questions, name the bonding or force, compare its strength, then connect that strength to the property being asked.
Then go one layer deeper: Example: magnesium oxide has a much higher melting point than sodium chloride because 2+ and 2- ions give stronger ionic attractions than 1+ and 1- ions.
Quick revision box
What this topic tests: Shapes, polarity, intermolecular forces, and structure-property links.
Top mistakes to avoid: Confusing molecular shape with electron geometry; listing IMF without comparing strength; vague melting-point explanations.
20-minute sprint plan: 5 min VSEPR recap; 10 min polarity/IMF compare questions; 5 min structure-property summary.
Do not quote the ideal angle if lone pairs compress it.
Molecular polarity
Bond dipoles plus shape symmetry
Whether dipoles cancel or give a net dipole
Do not decide polarity from electronegativity alone.
Boiling point of simple molecules
IMF present in each substance
Relative IMF strength, molecular size, and surface area
Do not ignore London dispersion in polar molecules.
Melting point of ionic solids
Charge product and ionic radius
Lattice energy and strength of ionic attractions
Do not compare only ionic charge when ion size also changes.
Conductivity or malleability
Particle structure in the solid or liquid
Mobile ions, delocalised electrons, or fixed lattice positions
Do not say "has electrons" without explaining whether they are mobile.
Use this route map before writing the explanation. It keeps the answer tied to the property asked, instead of listing every bonding fact you remember.
1 Types of Chemical Bonds
Bonding model
Key features
Typical exam focus
Ionic
Electrostatic attraction between cations and anions in a lattice.
Explaining trends in melting point, solubility, conductivity.
Covalent
Shared pair(s) of electrons between atoms.
Bond polarity, molecular shape, sigma vs pi bonds.
Metallic
Lattice of positive ions in a sea of delocalised electrons.
Conductivity, malleability, strength across the periodic table.
1.1 Lattice Energy Trends
Lattice energy (lattice enthalpy of formation) measures enthalpy change when one mole of ionic solid forms from gaseous ions. A simplified Born-Landé equation highlights core dependencies:
Lattice energy increases when the ionic charge product is larger and the ion centres are closer together. Larger charge products and shorter separations make lattice formation more exothermic (or lattice dissociation more endothermic), raising melting points and hardness.
Lattice-energy comparison checkpoint
For ionic solids, compare charge before size. If the charge product is the same, compare ionic radius and distance between ion centres.
Comparison clue
First decision
Answer move
Common trap
MgO versus NaCl
Charge product differs: MgX2+ and OX2− give stronger attractions than NaX+ and ClX−.
MgO has much larger lattice energy and a higher melting point because oppositely charged ions attract more strongly.
Comparing only formula mass or ion count.
NaF versus NaCl
Charge product is the same.
FX− is smaller than ClX−
MgO versus CaO
Charge product is the same.
MgX2+ is smaller than CaX2+
Data table with melting points
Identify charge and size pattern first.
Link the higher melting point to stronger electrostatic attractions in the lattice.
Writing "stronger bonds" without naming ionic attractions between oppositely charged ions.
Worked check: MgO has 2+ and 2− ions, while NaCl has 1+ and 1− ions. The larger charge product in MgO gives stronger electrostatic attraction between ions, so more energy is needed to separate the lattice.
Misconception check: lattice energy is not decided by molecular mass. Ionic charge and ion-centre distance are the first comparison points.
1.2 Covalent Bond Strength and Length
Stronger bonds have higher bond dissociation enthalpies and shorter lengths. The syllabus expects students to relate strength to orbital overlap: sigma bonds involve head-on overlap (stronger), whereas pi bonds involve side-on overlap (weaker). Recall that double bonds comprise one sigma and one pi bond.
When numerical comparisons are needed, use the bond enthalpies tabulated in the SEAB Chemistry Data Booklet (exams from 2026) instead of memorised values.
2 Molecular Geometry and Polarity
2.1 VSEPR Workflow
Determine the central atom's steric number: count bonded pairs + lone pairs.
Assign electron-domain geometry (e.g. tetrahedral for steric number 4).
Deduce molecular shape by considering only bonding pairs.
Judge polarity: combine vector sum of bond dipoles; consider symmetry.
Lone-pair checkpoint
After counting steric number, separate electron-domain geometry from molecular shape before discussing polarity. Lone pairs still repel, but they are not drawn as atoms in the final molecular shape.
Central-atom count
Electron-domain geometry
Molecular shape to name
Explanation move
4 bonded pairs, 0 lone pairs
Tetrahedral
Tetrahedral
Symmetrical examples often have cancelling dipoles if all outer atoms are identical.
3 bonded pairs, 1 lone pair
Tetrahedral
Trigonal pyramidal
Lone-pair repulsion compresses the bond angle and can give a net dipole.
2 bonded pairs, 2 lone pairs
Tetrahedral
Bent
Two lone pairs compress the bond angle further and usually prevent dipole cancellation.
2 bonded pairs, 3 lone pairs
Trigonal bipyramidal
Linear
Lone pairs occupy positions that minimise repulsion, leaving the two bonds opposite each other.
Misconception check: a molecule can have tetrahedral electron-domain geometry without having tetrahedral molecular shape. State both when the question asks for shape and bond angle.
Steric number
Electron-domain geometry
Common shapes
Bond angles
3
Trigonal planar
Trigonal planar, bent
120∘, <120∘
4
Tetrahedral
Tetrahedral, trigonal pyramidal, bent
109.5∘,107∘,104.5∘
5
Trigonal bipyramidal
Seesaw, T-shaped, linear
120∘ and 90∘ combinations
6
Octahedral
Square pyramidal, square planar
90∘
Bond angles above are standard reference values; actual angles vary with lone-pair repulsion and the central atom/ligands (so use them as a starting point, not a memorised “exact” for every molecule).
CO2: linear molecule with dipoles that cancel.
NH3: trigonal pyramidal and polar due to lone pair.
2.2 Bond Polarity and Molecular Dipole
Even if bonds are polar, molecules can be non-polar when dipoles cancel (e.g. COX2). Use vector addition language: “the dipoles are equal and opposite along the linear axis.” For molecules such as NHX3, asymmetry yields a net dipole, linking to hydrogen bonding strength.
When writing IMF explanations, point to the trigonal pyramidal NHX3 sketch above: that shape gives a net dipole and the nitrogen lone pair enables hydrogen-bond interactions.
Polarity decision checkpoint
Decide molecular polarity in two passes: first check whether the individual bonds are polar, then check whether the bond dipoles cancel because of shape. Do not stop after comparing electronegativity.
Molecule
Bond-level check
Shape check
Polarity conclusion
Common trap
COX2
Each C=O bond is polar.
Linear molecule, so the two equal bond dipoles point in opposite directions.
Non-polar molecule overall.
Calling it polar because it contains polar bonds.
BFX3
Each B−F bond is polar.
Trigonal planar and symmetrical, so the three bond dipoles cancel.
Non-polar molecule overall.
Ignoring three-way symmetry and adding only two arrows mentally.
NHX3
Each N−H bond is polar.
Trigonal pyramidal shape is not symmetrical because of the lone pair.
Polar molecule with a net dipole.
Treating it like a flat trigonal planar molecule.
CHX3Cl
The C−Cl bond is strongly polar.
Tetrahedral shape has different atoms around carbon, so dipoles do not cancel.
Polar molecule overall.
Worked check: CClX4 is non-polar because four identical C−Cl bond dipoles cancel in a symmetrical tetrahedral shape. CHClX3 is polar because one C−H bond and three C−Cl bonds make the tetrahedron unsymmetrical, so the dipoles do not cancel.
Misconception check: "has polar bonds" and "is a polar molecule" are different claims. A polar molecule needs a non-zero resultant dipole after shape is considered.
3 Intermolecular Forces (IMF)
IMF
Requirements
Relative strength
Example
London dispersion
Present in all molecules; induced dipoles.
Weakest; increases with molecular mass and surface area.
IX2
Permanent dipole-dipole
Polar molecules with permanent dipoles.
Medium.
CHX3Cl
Hydrogen bonding
H bonded to N, O, or F; lone pair on adjacent molecule.
Strongest IMF; drives high boiling points.
HX2O
Candidates should support IMF arguments with both electron distribution (polarity) and ability to form specific interactions (e.g. availability of lone pairs).
IMF comparison checkpoint
When comparing boiling points of simple molecular substances, identify every IMF present, then decide which difference actually explains the trend.
Comparison clue
First check
Stronger explanation to write
Trap to avoid
HX2O has a higher boiling point than HX2S
Is H bonded to N, O, or F?
Water forms hydrogen bonds between molecules; hydrogen sulfide has weaker permanent dipole-dipole forces and London dispersion forces.
Saying both are polar, so their boiling points should be similar.
CHX3CHX2OH
IX2 has a higher boiling point than ClX2
CHX3Cl has a higher boiling point than CHX4
Worked check: if a question asks why NHX3 boils at a higher temperature than PHX3, start with hydrogen bonding in NHX3 because N-H bonds and the nitrogen lone pair allow strong intermolecular hydrogen bonds. Then contrast PHX3, which relies mainly on weaker dipole-dipole and London forces.
Misconception check: London dispersion forces are present in all simple molecular substances. The question is not whether they exist, but whether they are the dominant difference in the comparison.
Structure-property explanation checkpoint
For structure-property questions, identify the particle arrangement before naming the force. The same word "strong" is too vague unless you say what particles are attracted and what must be overcome.
Structure clue
Particles or layers to picture
Property explanation move
Common trap
Giant ionic lattice
Alternating cations and anions in fixed positions
High melting point comes from strong electrostatic attractions throughout the lattice; molten or aqueous samples conduct because ions can move.
Saying a solid ionic compound conducts because it "has ions" without checking mobility.
Giant covalent network
Atoms joined by many covalent bonds across a lattice
High melting point comes from many strong covalent bonds that must be broken.
Explaining it using intermolecular forces when there are no discrete molecules.
Simple molecular substance
Separate molecules with covalent bonds inside each molecule
Low boiling point usually comes from weaker intermolecular forces between molecules; covalent bonds inside molecules are not broken during boiling.
Saying boiling breaks covalent bonds inside each molecule.
Metallic lattice
Positive ions in a lattice with delocalised electrons
Electrical conductivity comes from mobile delocalised electrons; malleability comes from layers sliding while attraction to electrons remains.
Saying metals conduct because atoms move through the solid.
Worked check: diamond and silicon dioxide have high melting points because many covalent bonds across their giant covalent networks must be broken. Carbon dioxide has covalent bonds inside each molecule, but sublimation or boiling separates molecules by overcoming weaker intermolecular forces, so the structure-property explanation is different.
Misconception check: first decide whether the substance contains discrete molecules. If it does not, do not use simple molecular boiling-point language.
4 Worked Multi-Part Example
Question:
Consider PClX3 and PClX5.
Predict the shapes using VSEPR.
Explain why PClX5 exists whereas NClX5 does not.
Compare the melting points of PClX3 and PClX5
Solution outline:
PClX3: Steric number 4 (three bonding pairs + one lone pair) → trigonal pyramidal; bond angle slightly less than 109.5∘ due to lone-pair repulsion.
PClX5: Steric number 5 → trigonal bipyramidal.
Phosphorus forms a stable hypervalent PClX5 because its larger valence shell accommodates five bonding pairs with lower electron-pair repulsion (described by three-centre four-electron bonding); nitrogen is too small to stabilise five N−Cl bonds, so NClX5
PClX5 forms an ionic lattice in the solid state ([PClX4]X+[PClX6]X−
Each explanation links shape, orbital availability, and lattice structure - the exact reasoning markers want.
PCl3: trigonal pyramidal molecular species.
PCl5: framework used in shape and bonding comparisons.
5 Exam Tactics
Paper 1: Expect geometry MCQs where lone pairs must be counted correctly. Sketch quick Lewis structures to avoid errors.
Paper 2: Structured questions often request energy comparisons (e.g. lattice energies of MgO vs NaF). Reference ionic charge and ionic radius explicitly, stating relative charge density.
Paper 3: Long questions integrate bonding with properties. For example, deducing why silicon dioxide is a solid network while carbon dioxide remains gaseous at RTP - discuss covalent network vs discrete molecules.
Paper 4: Practical planning tasks may need justification of solvents based on polarity or explanation of observed conductivity based on ionic vs molecular structure.
6 Frequent Misconceptions
Confusing bond polarity with molecular polarity: Always consider shape.
Forgetting lone-pair repulsion: Bond angles decrease with each lone pair; quote specific angle adjustments.
Misstating metallic bonding explanations: Highlight delocalised electrons and cation lattice rather than vague “sea of electrons” statements.
Overcomplicating ‘expanded octet’ explanations: Keep to the syllabus framing (electron-pair repulsion + size/energetics) and write clearly rather than memorising contested orbital stories.
7 Quick Drill Set
Compare the melting points of MgClX2, NaCl, and AlX2OX3 using lattice energy arguments.
Sketch and label the shapes of SFX4, IFX5
Explain why graphite conducts electricity in the plane but diamond does not, referencing bonding and structure.
Worked solutions should mention the bonding model, structural arrangement, and the inter-particle forces in each case.
Common exam mistakes
Reporting molecular shape instead of electron-domain geometry: VSEPR requires you to state both - electron geometry first, then molecular shape after discounting lone pairs. Giving only the molecular shape misses marks when the question asks to "describe the shape".
Ignoring lone pairs when predicting bond angles: Bond angles in molecules such as NHX3 and HX2O are reduced below the tetrahedral ideal; failing to acknowledge lone-pair repulsion gives the wrong predicted angle.
Confusing bond polarity with molecular polarity: A molecule with polar bonds can still be non-polar if the bond dipoles cancel by symmetry (e.g. COX2, CClX4
Listing intermolecular forces without ranking strength: Questions that compare boiling points require you to identify the type of IMF and justify which is stronger; a bare list without comparative reasoning will not score explanation marks.
Overlooking London dispersion forces in polar molecules: All molecules experience London dispersion; omitting them when explaining properties of larger polar molecules gives an incomplete answer.
Mixing up sigma and pi bond counts: A C=C double bond contains one sigma and one pi bond; a triple bond contains one sigma and two pi bonds - reversing these is a common Paper 2 error.
Vague metallic bonding language: Phrases like "sea of electrons" alone are insufficient; state "a lattice of positive ions surrounded by delocalised electrons" and link this directly to the property asked about.
Frequently asked questions
Is VSEPR theory needed for every bonding question? VSEPR is the required model for predicting molecular geometry in H2 Chemistry. You are expected to apply the steric-number workflow for any molecule or ion with a central atom, and to link shape to polarity and intermolecular forces where relevant.
How do I decide which intermolecular force to discuss in a melting or boiling point comparison? First identify the types of IMF present in each substance. Then compare strength: hydrogen bonding > permanent dipole-dipole > London dispersion (though London forces can dominate for large non-polar molecules). Always justify which force is stronger rather than simply naming it.
What is the difference between lattice energy and bond energy? Lattice energy refers to the electrostatic energy of an ionic solid (enthalpy change when one mole of ionic compound forms from gaseous ions), while bond energy (bond dissociation enthalpy) refers to breaking a covalent bond in a gaseous molecule. They are assessed in different calculation types - Born-Haber cycles vs Hess' law bond-energy calculations.
Do I need to draw Lewis structures in the exam? Yes. Drawing a Lewis structure (showing all bonding pairs and lone pairs) is often the first step in VSEPR questions and is sometimes directly awarded marks. Always include lone pairs on the central atom and check the total valence electron count.
Struggling with Chemical Bonding? Our H2 Chemistry tuition programme covers this topic with structured practice, Paper 4 practical drills, and worked exam solutions.
