2.2 The covalent model

Syllabus
First assessment 2025
Topic
2.2
Level
SL

Covalent Bonding

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.

Drawing Covalent Structures

Assessment in practice

Representative question

Question 1

[Maximum number: 1]

Draw the Lewis formula of the HCN molecule.

Bond Types and Strength

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.

Comparing Bond Strength

Assessment in practice

Representative question

Question 1

[Maximum number: 1]

Compare, giving a reason, the length of the bond between the carbon atoms in ethyne with that in ethane, C2H6\mathrm{C}_{2} \mathrm{H}_{6}.

Coordination Bonds

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.

Identifying Coordination Bonds

Assessment in practice

Representative question

Question 1

[Maximum number: 1]

State the precise type of bond formation between the cyanide ion and the iron ion.
bond

VSEPR Geometry

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.

Applying VSEPR

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Deduce the electron domain geometry and molecular geometry of SO2\mathrm{SO}_{2}.

Electron domain geometry:

Molecular domain geometry:

Bond Polarity

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.

Deducing Bond Dipoles

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Describe the nature of the bond between oxygen and phosphorus. Use sections 9 and 17 of the data booklet.

Molecular Polarity

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.

Predicting Molecular Polarity

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Explain the polarity of the SO2\mathrm{SO}_{2} molecule.

Covalent Network Structures

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.

Comparing Network Materials

Assessment in practice

Representative question

Question 1

[Maximum number: 4]

Identify three allotropes of carbon and describe their structures.

Intermolecular Forces

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.

Identifying IMFs

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Outline how a hydrogen bond is formed.

IMF Strength and Properties

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.

Explaining IMF-Dependent Properties

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Explain, in terms of the intermolecular forces present, the trend in the boiling points of the first four alkenes.

AlkeneBoiling point / K
ethene169
propene225
but-1-ene267
pent-1-ene303

Chromatography and Rf

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/distancetravelledbysolventfrontRf = distance travelled by component / distance travelled by solvent front

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.

Interpreting Chromatograms

Assessment in practice

Representative question

Question 1

[Maximum number: 2]

Explain how the separation of inks is achieved using paper chromatography.

The Covalent Model Summary

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.

Objective notes

10 learning objectives
2.2.1Covalent bonding• Electrostatic attraction between shared electron pair and nuclei• Octet rule: tendency to achieve 8 valence electrons• Lewis formulas (up to 4 electron pairs per atom)View2.2.2Bond types and strength• Single, double, triple bonds (1, 2, 3 shared pairs)• More bonds → shorter length, stronger bondView2.2.3Coordination bonds• Both electrons from same atom• Identify coordination bonds; include transition element complexes at HLView2.2.4VSEPR model• Predict molecular shapes from electron domain repulsion• Electron domain and molecular geometry (up to 4 domains)• Include bond angles and lone-pair effects for common shapesView2.2.5Bond polarity• Results from electronegativity differences• Deduce polar bonds from electronegativity values and show bond dipolesView2.2.6Molecular polarity• Depends on bond polarity + molecular geometry• Net dipole moment• Identify when bond dipoles cancel or produce a polar molecule/ionView2.2.7Covalent network structures• Carbon allotropes: diamond, graphite, fullerenes, graphene• Silicon and silicon dioxideView2.2.8Intermolecular forces (IMF)• London dispersion forces• Dipole-induced dipole• Dipole-dipole• Hydrogen bonding• Deduce IMF types from molecular size and polarityView2.2.9IMF strength and properties• Relative strength: London < dipole-dipole < hydrogen bonding• Effects on volatility, conductivity, solubility• Explain properties of covalent substances using IMF strength and molar massView2.2.10Chromatography• Separates components based on IMF attractions• Calculate and interpret RF values• Link mobile/stationary phase attraction to separation; operational details are not assessedView