2.2 The covalent model
- Syllabus
- First assessment 2025
- Topic
- 2.2
- Level
- SL
A covalent bond is the electrostatic attraction between a shared electron pair and the nuclei of the bonded atoms. Lewis formulas show valence electrons and shared pairs.

Use the octet tendency to place shared and lone pairs, while remembering that the assessed Lewis scope allows up to four electron pairs around an atom.
Build a Lewis formula by counting all valence electrons, choosing a skeleton, completing outer octets and placing any remainder on the central atom. Then recount electrons and formal charge; a shared pair contributes to the attraction between both nuclei rather than belonging exclusively to either atom.
Charged-species check: NH₄⁺ has 5 + 4(1) − 1 = 8 valence electrons, so draw four N–H shared pairs, no lone pair on N, enclose the ion in brackets and write the overall + charge. A final audit must match the available electron total, complete the required outer-shell arrangements and reproduce the species charge; lone pairs, shared pairs and formal charges answer different parts of that audit.
1 mark
Draw the Lewis formula of the HCN molecule.
| Bond | Shared pairs | Relative length and strength |
|---|---|---|
| Single | 1 | Longest and weakest of the three |
| Double | 2 | Shorter and stronger |
| Triple | 3 | Shortest and strongest |
More shared electron pairs increase electron density between nuclei, so the bond becomes shorter and stronger.
For the same pair of atoms, increasing bond order generally shortens and strengthens the bond because more electron density lies between the nuclei. Use that comparison locally: bond enthalpy also depends on the atoms and molecular environment, so any triple bond is not automatically stronger than every unrelated single bond.
1 mark
Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6.
In a coordination bond, both electrons in the shared pair come from the same atom. The bond can be represented by an arrow from the electron-pair donor to the acceptor.

Identify the donor atom or ligand and the acceptor, including coordination bonds in transition-element complexes at HL.
In NH₃→BF₃ the nitrogen lone pair supplies both bonding electrons, so the arrow starts at N and ends at B. The arrow records how the pair originated; after formation it is a shared covalent pair, and in a complex the same donor logic identifies each ligand–metal bond.
1 mark
State the precise type of bond formation between the cyanide ion and the iron ion.
bond
VSEPR predicts molecular shape by arranging electron domains around a central atom to minimize repulsion. Lone pairs repel more strongly than bonding pairs and therefore alter common bond angles.



| Step | Decision |
|---|---|
| 1 | Count bonding and lone-pair electron domains |
| 2 | Assign electron-domain geometry |
| 3 | Ignore lone pairs when naming molecular geometry |
| 4 | Adjust expected bond angles for lone-pair repulsion |
NH₃ has four electron domains around N, so its electron-domain geometry is tetrahedral but its molecular shape is trigonal pyramidal; the lone pair compresses the H–N–H angle below 109.5°. Count a double bond as one domain and distinguish electron geometry from the shape named using atoms only.
2 marks
Deduce the electron domain geometry and molecular geometry of SO2.
Electron domain geometry:
Molecular domain geometry:
A bond is polar when a difference in electronegativity gives unequal sharing of the bonding electrons. The more electronegative atom carries partial negative character.

Compare electronegativities, assign partial charges, and draw the bond-dipole arrow toward the more electronegative atom.
For H–Cl, chlorine is more electronegative, so label Hδ⁺–Clδ⁻ and point the dipole arrow toward Cl. Electronegativity difference predicts unequal sharing within that bond; it does not by itself decide the polarity of the whole molecule.
2 marks
Describe the nature of the bond between oxygen and phosphorus. Use sections 9 and 17 of the data booklet.
Molecular polarity depends on both the polarity of individual bonds and the three-dimensional geometry of the molecule or ion. Bond dipoles can cancel or produce a net dipole moment.

Draw or infer the geometry, place each bond dipole, and check whether the vector sum is zero. Do not decide molecular polarity from a single bond alone.
CO₂ contains polar C=O bonds, but their equal opposite dipoles cancel in a linear molecule. In bent H₂O they do not cancel, so the molecule has a net dipole. Always establish the three-dimensional geometry before adding dipoles as vectors.
2 marks
Explain the polarity of the SO2 molecule.
| Material | Structural evidence | Property or use explained |
|---|---|---|
| Diamond | each C covalently bonded in a rigid 3D network | very hard; high melting point; no mobile charge carriers |
| Graphite | strong covalent sheets with delocalized electrons; weak attractions between sheets | conducts along sheets; layers slide, so it is soft/lubricating |
| Graphene | one atom-thick covalent sheet with delocalized electrons | strong, light and electrically conducting |
| Fullerenes | finite carbon cages or tubes rather than an infinite 3D network | molecular shape and intermolecular contacts give properties distinct from diamond/graphite |
| Silicon | extended covalent structure with limited charge mobility | semiconductor behaviour; detailed doping is outside this card |
| Silicon dioxide | 3D Si–O covalent network, not discrete SiO₂ molecules | hard and high-melting because many strong covalent bonds must be overcome |
Decide conductivity by available mobile charges, not by the word covalent alone.

Explain a network material property by connecting the structure and bonding arrangement to the relevant mobility, strength, or dimensional feature.
Diamond is hard because each carbon is held in a three-dimensional covalent network, while graphite conducts along layers through delocalized electrons and its layers can slide. Silicon dioxide is also an extended network: describe network atoms, not discrete SiO₂ molecules, when explaining its high melting point.
4 marks
Identify three allotropes of carbon and describe their structures.
| Force | Evidence used to identify it |
|---|---|
| London dispersion | Present; increases with molecular size and electron count |
| Dipole-induced dipole | A permanent dipole induces a dipole in a neighbour |
| Dipole-dipole | Permanent dipoles attract |
| Hydrogen bonding | Hydrogen bonded to a strongly electronegative atom creates the required interaction |


Start with molecular size and polarity, then check for the structural requirement for hydrogen bonding. More than one IMF can be present.
All molecules have London dispersion forces. Add permanent dipole–dipole attraction when a net molecular dipole exists, and add hydrogen bonding only when the required H–N, H–O or H–F environment and an acceptor lone pair are present. Name every relevant force before deciding which dominates.
2 marks
Outline how a hydrogen bond is formed.
For the relative comparison in this topic, London dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonding. Molecular size also affects dispersion strength.
Stronger intermolecular attractions generally reduce volatility. Explain conductivity and solubility by considering whether charged particles are available and whether solute–solvent attractions are favourable.
Compare like evidence: pentane has stronger dispersion forces and a higher boiling point than butane because its electron cloud is larger. The simple London < dipole–dipole < hydrogen-bond ordering is a guide for comparable molecules, not a rule that ignores molecular size and the number of interaction sites.
2 marks
Explain, in terms of the intermolecular forces present, the trend in the boiling points of the first four alkenes.
| Alkene | Boiling point / K |
|---|---|
| ethene | 169 |
| propene | 225 |
| but-1-ene | 267 |
| pent-1-ene | 303 |
Chromatography separates components because they have different attractions to the stationary and mobile phases. A component that is more strongly attracted to the mobile phase travels farther.

Rf=distancetravelledbycomponent/distancetravelledbysolventfront
Rf is a dimensionless ratio of distances measured from the same origin under the same conditions. Because a component cannot pass the solvent front, a valid result lies from 0 to 1; a value above 1 signals a distance or origin error. Operational details beyond the separation principle are not assessed here.
If a spot moves 3.2 cm while the solvent front moves 8.0 cm, Rf = 0.40. A larger Rf means greater relative affinity for the mobile phase under those conditions, but values from different solvents, stationary phases or temperatures are not directly interchangeable.
2 marks
Explain how the separation of inks is achieved using paper chromatography.
Retrieve the covalent pathway: shared pairs and bond order lead to geometry, polarity and molecular polarity; structure determines network properties, IMF behaviour and chromatography; HL representations extend to resonance, formal charge, sigma/pi bonds and hybridization.
Check the representation first, then count domains, apply geometry, identify polarity or forces, and connect the structure to the requested property or HL bonding description.