Educerie
Level

7 higher-level sections hidden.

Educerie · IB Diploma · Chemistry

Structure 2 Models of bonding and structure · S2.2 The covalent model

Level
SL and HL. Sections 12 to 18 are HL only. If you are SL, skip them; nothing in your papers tests them.
Themes (key concepts)
structure, and models. The covalent model explains a molecule's structure from one idea, the shared pair, and then asks how far that model stretches: to shapes, polarity and the forces between molecules, and at HL to bonds no single drawing can show.
The question this unit answers
what determines the covalent nature and properties of a substance?
Where it is examined
everywhere. Paper 1A multiple choice on shapes, angles, polarity and intermolecular forces (1 mark each); Paper 1B data on boiling points, solubility or chromatography; Paper 2 parts where you draw a Lewis formula (1–2 marks), predict a shape and angle (2 marks) or explain a difference in boiling point (3–4 marks). HL adds resonance, formal charge, σ and π bonds and hybridization, usually as 2- to 4-mark parts.

What you must be able to do

You must be able toLevelWhat it looks like in the exam
Describe a covalent bond, use the octet rule, and deduce Lewis formulas with up to four pairs per atom, including incomplete octetsSL, HL"Draw the Lewis formula of…" (1–2 marks)
Relate the number of shared pairs to bond length and strength; identify coordination bondsSL, HL"Explain why C≡C is shorter than C=C" (2 marks)
Predict electron domain geometry, molecular geometry and bond angles for up to four domainsSL, HL"Predict the shape and bond angle of…" (2 marks)
Deduce bond polarity from electronegativity, and molecular polarity from polarity plus shapeSL, HL"Explain why CO₂ is non-polar but H₂O is polar" (2–3 marks)
Describe and explain the structures and properties of diamond, graphite, graphene, fullerenes, silicon and SiO₂SL, HL"Explain why graphite conducts but diamond does not" (2–3 marks)
Deduce the intermolecular forces present, rank them, and explain volatility, conductivity and solubilitySL, HL"Explain the difference in boiling points" (3–4 marks)
Explain, calculate and interpret RF valuesSL, HLPaper 1B: measure, calculate RF, interpret (2–3 marks)
Include transition element complexes in coordination bondingHL onlyIdentify the ligand and the coordination bond
Deduce resonance structures; discuss the structure of benzene from evidenceHL only"Discuss the evidence for the structure of benzene" (3–4 marks)
Draw Lewis formulas and deduce shapes for five and six domainsHL only"Predict the shape of SF₄ and its bond angles" (2–3 marks)
Use formal charge to choose a preferred Lewis formulaHL onlyPaper 2 (2–4 marks), often sulfate
Deduce σ and π bonds; relate hybridization to Lewis formula and geometryHL only"State the hybridization of each carbon atom in…" (2–3 marks)

Before you start

You need electron configurations from S1.3, because the number of valence electrons decides everything here, and the ionic model from S2.1, because covalent substances are easiest to explain by contrast with it. Electronegativity is treated properly in Structure 3.1; here you only read its values from the data booklet.


1The idea in one paragraph

When two non-metal atoms meet, neither can take an electron from the other, so they share. A covalent bond is a pair of electrons held between two nuclei and attracted to both. Count the pairs round each atom and you have a Lewis formula. Let those pairs push each other as far apart as they can and you have the molecule's shape. Combine the shape with the unequal pull of different atoms on the pairs and you know whether the molecule is polar. That decides which intermolecular forces act between molecules, and those weak forces, not the strong bonds inside, decide how easily a molecular substance melts, boils and dissolves. A few substances, such as diamond, never form separate molecules at all: they are covalent networks, and they behave nothing like molecular substances.

2The covalent bond and the octet rule

A covalent bond is the electrostatic attraction between a shared pair of electrons and the positively charged nuclei of the two bonded atoms.

Figure 1 shows why that holds two atoms together. The pair sits between the nuclei and each nucleus is attracted to it, so the atoms cannot separate without pulling a nucleus away from a pair it is attracted to.

Figure 1 · What holds a covalent bond together Figure 1 · What holds a covalent bond together (a) The model: one pair, two nuclei + + the shared pair is attracted to both nuclei at once nucleus nucleus (b) Cl₂ as a dot-and-cross diagram Cl Cl one shared pair · each Cl now has eight outer electrons The bond is the attraction of both positive nuclei for the same negative pair between them.
Figure 1 · What holds a covalent bond together

Each chlorine atom in Figure 1(b) has seven outer electrons; by sharing one each, both reach eight. The octet rule describes that tendency: atoms tend to gain, lose or share electrons until they have eight in their valence shell, a noble gas configuration. Hydrogen's shell holds two.

Covalent bonding, unlike ionic bonding, can join atoms of the same element, because it needs no giver and taker. It is typical between non-metals. Noble gases rarely form covalent bonds because their valence shells are already full.

3Lewis formulas

A Lewis formula shows every valence electron: bonding pairs, usually as lines, and non-bonding pairs (lone pairs), as dots, crosses or lines. Five steps work every time.

  1. Count the valence electrons. For a negative ion add one per charge; for a positive ion take one away.
  2. Draw the skeleton. The atom forming most bonds goes in the middle (never hydrogen). Join each outer atom to it with a single bond.
  3. Complete the outer atoms' octets with lone pairs.
  4. Put leftover electrons on the central atom as lone pairs.
  5. If the central atom is still short of eight, turn a lone pair on an outer atom into another shared pair.

Carbon dioxide: 4 + 6 + 6 = 16 electrons, 8 pairs. O–C–O uses 2 pairs; completing both oxygens uses 6, so carbon is left with only 4 electrons. Move one lone pair from each oxygen into the bonds: O=C=O, and every atom has 8. Figure 2 shows this and seven more.

Figure 2 · Lewis formulas: every valence electron shown Figure 2 · Lewis formulas: every valence electron shown O H H H₂O · water 2 bonding, 2 lone pairs N H H H NH₃ · ammonia 3 bonding, 1 lone pair C O O CO₂ · carbon dioxide two double bonds C H N HCN · hydrogen cyanide a triple bond C C H H H H C₂H₄ · ethene organic: a C=C double bond C O H H H H CH₃OH · methanol organic: two lone pairs on O B F F F BF₃ · boron trifluoride only six electrons around B O H [ ] − OH⁻ · hydroxide ion 8 electrons: one gained A line is a shared pair; amber dots are lone pairs. Ions go in square brackets with the charge outside.
Figure 2 · Lewis formulas: every valence electron shown

Always check the total. Hydroxide has 6 + 1 + 1 = 8 valence electrons: one bond and three lone pairs, in square brackets with the charge outside.

Not every molecule obeys the octet rule. Boron in BF₃ has three valence electrons, forms three bonds and stops at six; beryllium in BeCl₂ stops at four. These incomplete octets are real, stable molecules and you must be able to draw them. Molecules with an odd number of electrons, such as NO, cannot give every atom eight, and atoms from period 3 down can hold more than eight (section 15, HL). The octet rule is a useful model for period 2 and a poor law.

4Single, double and triple bonds

A single bond is one shared pair, a double bond two, a triple bond three. More pairs between the same two nuclei means more negative charge between them, so the nuclei are attracted more strongly and pulled closer. Bond strength is measured by bond enthalpy, the energy to break one mole of the bond in the gas state (data booklet). Figure 3 shows the pattern.

Figure 3 · More shared pairs: a shorter, stronger bond Figure 3 · More shared pairs: a shorter, stronger bond (a) Bond length / pm 154 C–C 1 pair 134 C=C 2 pairs 120 C≡C 3 pairs (b) Bond enthalpy / kJ mol⁻¹ 346 C–C 1 pair 614 C=C 2 pairs 839 C≡C 3 pairs More electron density between the nuclei pulls them closer and holds them more tightly.
Figure 3 · More shared pairs: a shorter, stronger bond

More shared pairs: a shorter bond, and a stronger bond.

A double bond is not twice as strong as a single one: a clue that its two pairs are different, which HL section 17 explains.

5Coordination bonds

In a coordination bond both electrons of the shared pair come from the same atom. It needs a lone pair on one atom and room on the other. Figure 4 shows the standard cases.

Figure 4 · Coordination bonds: both electrons from one atom Figure 4 · Coordination bonds: both electrons from one atom (a) Ammonia accepts a proton N H H H + H⁺ N H H H H [ ] + the arrow points away from the atom that gave both electrons (b) Carbon monoxide C O two ordinary shared pairs and one pair given entirely by oxygen Once formed, a coordination bond is identical to any other covalent bond: all four N–H bonds in NH₄⁺ are the same.
Figure 4 · Coordination bonds: both electrons from one atom

H⁺ has no electrons, and ammonia's nitrogen has a lone pair, which becomes the shared pair of a new N–H bond: the ammonium ion, NH₄⁺. Water accepting H⁺ to make H₃O⁺ is the same. In carbon monoxide, oxygen supplies both electrons of the third bond. The bond is drawn as an arrow pointing away from the atom that supplied the pair. Once formed it is an ordinary covalent bond: all four N–H bonds in NH₄⁺ are identical.

6VSEPR: predicting the shape

The valence shell electron pair repulsion (VSEPR) model predicts shape from one principle: the groups of electrons round a central atom repel each other and get as far apart as possible.

Each group is an electron domain. A lone pair is one domain, a single bond is one, and a double or triple bond is also one, because all its electrons lie between the same two atoms. So carbon in CO₂ has two domains.

Draw the Lewis formula; count the domains round the central atom, which gives the electron domain geometry; then look only at where the atoms are, which gives the molecular geometry, the shape you name.

DomainsElectron domain geometryLone pairsMolecular geometryAngleExample
2linear0linear180°CO₂, HCN
3trigonal planar0trigonal planar120°BF₃, CH₂O
3trigonal planar1bentless than 120°SO₂, O₃
4tetrahedral0tetrahedral109.5°CH₄, NH₄⁺
4tetrahedral1trigonal pyramidalabout 107°NH₃, H₃O⁺
4tetrahedral2bentabout 104.5°H₂O
Figure 5 · Electron domains decide the shape Figure 5 · Electron domains decide the shape C O O 180° Linear CO₂ · 180° 2 domains, 0 lone pairs B F F F 120° Trigonal planar BF₃ · 120° 3 domains, 0 lone pairs S O O < 120° Bent (from 3 domains) SO₂ · less than 120° 3 domains, 1 lone pair C H H H H 109.5° Tetrahedral CH₄ · 109.5° 4 domains, 0 lone pairs N H H H 107° Trigonal pyramidal NH₃ · 107° 4 domains, 1 lone pair O H H 104.5° Bent (from 4 domains) H₂O · 104.5° 4 domains, 2 lone pairs C wedge: towards you C hashed: away from you lone pair: repels more than a bond Top row: 2 or 3 domains. Bottom row: 4 domains, and each lone pair closes the angle by about 2.5°.
Figure 5 · Electron domains decide the shape

Lone pairs close the angle. A lone pair is held by one nucleus only, so it sits closer to that atom and spreads wider than a bonding pair, and repels more: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair. In ammonia one lone pair squeezes the N–H bonds from 109.5° to about 107°; in water two squeeze to about 104.5°. Both still have four domains, a tetrahedral electron domain geometry; only the molecular geometry differs.

Multiple bonds do the same, a little. A double bond puts four electrons in one domain and repels more than a single bond. In methanal, H₂C=O, the C=O domain pushes the C–H bonds together, so H–C–H is slightly less than 120°, about 116°.

How useful is VSEPR? Very, for a model this simple: it gets almost every shape right and every angle change in the right direction. It does not give exact angles (it cannot say why H₂S is about 92° while H₂O is 104.5°), and it says nothing about why the pairs form. It is a model of repulsion, not a theory of bonding.

7Polar bonds and polar molecules

Electronegativity is the ability of an atom to attract a shared pair towards itself. Data booklet values include F 4.0, O 3.4, Cl 3.2, N 3.0, C 2.6 and H 2.2. When bonded atoms differ, the pair is pulled towards the more electronegative one, which carries a small negative charge, δ−, leaving the other δ+. The bond is a polar covalent bond with a bond dipole, and the larger the difference, Δχ, the more polar it is.

BondΔχδ− endVerdict
Cl–Cl0neithernon-polar
C–H2.6 − 2.2 = 0.4Cslightly polar
H–Cl3.2 − 2.2 = 1.0Clpolar
O–H3.4 − 2.2 = 1.2Overy polar

A bond dipole is shown with δ+ and δ−, or with a dipole arrow pointing from δ+ to δ− with a cross stroke at the δ+ end.

A molecule can have polar bonds and still be non-polar. Bond dipoles have direction, and a molecule has a net dipole moment only if they do not cancel. Figure 6 shows the four cases to learn.

Figure 6 · Polar bonds, and whether they cancel Figure 6 · Polar bonds, and whether they cancel (a) CO₂: the dipoles cancel C O O δ− δ+ δ− equal and opposite: non-polar (b) H₂O: the dipoles add up O H H δ− δ+ δ+ net dipole bent, so they cannot cancel: polar (c) CCl₄: four dipoles cancel C Cl Cl Cl Cl symmetrical tetrahedron: non-polar (d) CHCl₃: one bond is different C H Cl Cl Cl net dipole the C–H bond breaks the symmetry: polar A polar molecule needs polar bonds and a shape in which their dipoles do not cancel. The dipole arrow points from δ+ to δ−; the cross stroke sits at the δ+ end.
Figure 6 · Polar bonds, and whether they cancel

Linear CO₂: two equal dipoles in opposite directions cancel, so it is non-polar. Bent H₂O: both dipoles point partly the same way and add, so it is polar. Tetrahedral CCl₄ cancels; CHCl₃, with one bond different, does not. BF₃ is non-polar and NH₃ polar for the same reasons. Ions work the same way: the three N–O dipoles in nitrate cancel by symmetry. Ask two questions in order: are there polar bonds, and if so, does the shape cancel them?

8Covalent network structures

Carbon and silicon form four covalent bonds and can keep doing so in every direction. A covalent network structure has no separate molecules: the whole crystal is held by covalent bonds. Carbon's allotropes, different structural forms of one element, have different bonding patterns and so different properties. Figure 7 shows them with silicon and silicon dioxide.

Figure 7 · Covalent network structures of carbon and silicon Figure 7 · Covalent network structures of carbon and silicon (a) Diamond (and silicon) each C bonded to four others, tetrahedral, 109.5°, in 3D (b) Graphite weak London forces between the layers layers of hexagons; each C bonded to three, one electron delocalized (c) Graphene a single layer of graphite, one atom thick (d) Fullerene, C₆₀ 60 C atoms in a closed cage: 12 pentagons and 20 hexagons. A molecule, not a giant network (e) Silicon dioxide, SiO₂ (drawn flat) Si Si Si Si Si Si Si Si Si each Si bonded to four O, each O to two Si (in 3D the four O round each Si are tetrahedral) Diamond, graphite, graphene, silicon and SiO₂ are giant networks. C₆₀ is a molecule.
Figure 7 · Covalent network structures of carbon and silicon
SubstanceStructureProperties, and why
Diamondeach C bonded tetrahedrally to four othersVery hard, very high melting point: many strong covalent bonds must break. No conduction: every valence electron is in a localized bond.
Graphitelayers of hexagons; each C bonded to three, its fourth electron delocalized over the layer; weak London forces between layersConducts along the layers: delocalized electrons move. Soft and slippery: layers slide. Very high melting point: strong bonds within layers.
Grapheneone layer of graphiteVery strong for its mass, excellent electrical and thermal conductor, very high melting point.
Fullerene, C₆₀a closed cage of 60 C (12 pentagons, 20 hexagons): a moleculePoor conductor: electrons cannot easily pass between molecules. Molecules held by London forces, so softer and more volatile than diamond, and soluble in some non-polar solvents.
Siliconthe diamond structureHard, high melting point (about 1410 °C); a semiconductor.
SiO₂ (quartz)each Si bonded tetrahedrally to four O, each O to two SiVery hard, melts above 1600 °C, insoluble, no conduction: no ions or delocalized electrons.

SiO₂ looks like CO₂ on paper, but CO₂ is a gas of small molecules and SiO₂ a mineral. Its formula is only the 1 : 2 ratio in a network, like an ionic empirical formula.

9Intermolecular forces

The bonds inside a molecule are strong. The forces between molecules, intermolecular forces, are much weaker, and they are what you overcome when a molecular substance melts or boils: steam is still H₂O. Keep the words apart. A bond holds atoms together within a molecule or network; the forces here act between separate molecules.

London (dispersion) forces act between all molecules. Moving electrons make a molecule's cloud momentarily uneven, an instantaneous dipole, which distorts a neighbour's cloud into an induced dipole, and the two attract. More electrons make a more easily distorted cloud, so London forces grow with size. Shape matters too: pentane (36 °C) boils above its compact isomer 2,2-dimethylpropane (about 10 °C), because long molecules touch over more surface.

Dipole–induced dipole forces act when a polar molecule meets a non-polar one and distorts its cloud: mainly in mixtures.

Dipole–dipole forces act between polar molecules: the δ+ end of one attracts the δ− end of the next.

Hydrogen bonding occurs when hydrogen, covalently bonded to N, O or F, is attracted to a lone pair on an N, O or F atom of a neighbouring molecule (Figure 9). The bond to H is so polar that the hydrogen is left almost a bare proton, and the lone pair can approach it closely. IUPAC revised its definition of the hydrogen bond in 2011, as better instruments revealed a wider family of these interactions: definitions change when science can see more.

Van der Waals forces is the umbrella term for London, dipole–induced dipole and dipole–dipole forces. Figure 8 is the decision you make in an exam.

Figure 8 · Deducing the intermolecular forces in a covalent substance Figure 8 · Deducing the intermolecular forces in a covalent substance Every molecule London (dispersion) forces Is the molecule polar? polar bonds that do not cancel no London only e.g. CH₄, CO₂, I₂ yes + dipole–dipole forces e.g. HCl, CH₃Cl, propanone Is H bonded directly to N, O or F? and is there a lone pair on an N, O or F nearby? yes + hydrogen bonding e.g. H₂O, NH₃, HF, ethanol Mixtures only dipole–induced dipole a polar molecule next to a non-polar one The forces add up: a molecule with hydrogen bonding also has London and dipole–dipole forces. "Van der Waals forces" is the umbrella term for London, dipole–induced dipole and dipole–dipole.
Figure 8 · Deducing the intermolecular forces in a covalent substance
Figure 9 · A hydrogen bond between two water molecules Figure 9 · A hydrogen bond between two water molecules O H H O H H hydrogen bond δ− δ+ δ− O–H···O in a straight line (about 180°) the δ+ H of one molecule is attracted to a lone pair on the O of the next A hydrogen bond needs H covalently bonded to N, O or F, and a lone pair on an N, O or F nearby.
Figure 9 · A hydrogen bond between two water molecules

The condition catches students both ways. Methane has H, but on carbon: no hydrogen bonding. Methoxymethane, CH₃OCH₃, has O with lone pairs but no H on it, so its molecules cannot hydrogen bond with each other.

10Comparing forces, and the properties of covalent substances

For molecules of comparable molar mass:

London forces < dipole–dipole forces < hydrogen bonding

The condition matters. Iodine, I₂, with 106 electrons and London forces only, is a solid at room temperature, while water, with 10 electrons, is a liquid. Compare types of force only between molecules of similar size, and size only between molecules with the same type of force. Figure 10 does each.

Figure 10 · What boiling points reveal about intermolecular forces Figure 10 · What boiling points reveal about intermolecular forces (a) Three substances of similar molar mass 0 °C −42 °C C₃H₈ M = 44 London −24 °C CH₃OCH₃ M = 46 + dipole–dipole 78 °C C₂H₅OH M = 46 + hydrogen bonding (b) Straight-chain alkanes Boiling point / °C Number of carbon atoms −162 1 −89 2 −42 3 −1 4 36 5 69 6 −150 −100 −50 0 50 (a) Same size, stronger forces, higher boiling point. (b) Same forces, more electrons, stronger London forces.
Figure 10 · What boiling points reveal about intermolecular forces

In (a) the molar masses are 44 to 46 g mol⁻¹, so London forces are similar. Methoxymethane adds dipole–dipole forces and boils 18 °C above propane; ethanol adds hydrogen bonding and boils over 100 °C above methoxymethane. In (b) the alkanes are all non-polar, and the steady rise is London forces growing with the number of electrons.

Volatility. Simple molecular substances have low melting and boiling points, because only weak intermolecular forces are overcome. Network substances have very high ones, because covalent bonds must break.

Electrical conductivity. Almost all covalent substances are non-conductors in every state: no ions and no mobile delocalized electrons. Graphite and graphene are the exceptions. A few polar molecules react with water to form ions: HCl gas does not conduct, hydrochloric acid does.

Solubility. "Like dissolves like": a substance dissolves when it can form attractions with the solvent comparable to those it breaks. Small molecules that hydrogen bond, such as ethanol, glucose and ammonia, dissolve in water. Non-polar substances, such as iodine, hexane and oils, dissolve in non-polar solvents but barely in water, since they cannot replace the hydrogen bonds they would break between water molecules. Along the alcohols, solubility in water falls as the non-polar chain grows. Network structures dissolve in nothing.

11Chromatography

Chromatography separates a mixture using a stationary phase that stays put and a mobile phase that moves. Each component is attracted to both through intermolecular forces, and the balance decides how far it travels. In paper chromatography the stationary phase is water held in the cellulose fibres; in thin-layer chromatography (TLC) it is a thin layer of silica or alumina. Both are polar. The mobile phase is the solvent.

The mixture is spotted on a pencil baseline (ink would itself separate), the plate stands in solvent below the baseline, and the solvent rises until it is stopped and its solvent front marked. A component attracted more to the mobile phase than the stationary phase travels further; with a polar stationary phase, less polar components usually travel further.

RF = distance travelled by the component ÷ distance travelled by the solvent front

Both distances are measured from the baseline, the first to the centre of the spot. RF has no units and lies between 0 and 1 (Figure 11).

Figure 11 · Measuring RF on a thin-layer chromatogram Figure 11 · Measuring RF on a thin-layer chromatogram solvent front baseline (pencil) M A B a b RF = a ÷ b a: baseline to the centre of the spot b: baseline to the solvent front M, the mixture, has two spots. Its upper spot matches A: same RF, same conditions. Its lower spot matches neither. A spot that travels further is attracted more to the mobile phase than to the stationary phase.
Figure 11 · Measuring RF on a thin-layer chromatogram
spot centre 3.6 cm, solvent front 8.0 cm
RF = 3.6 ÷ 8.0 = 0.45no units; two significant figures, as the data

A component is identified by matching its RF with a known substance run under the same conditions: same stationary phase, solvent and temperature. Two substances can share an RF in one solvent, so a match is evidence, not proof; a second solvent strengthens it.

12HLCoordination bonds in transition element complexes

SL students can skip to section 19.

Transition element ions have empty orbitals that accept lone pairs. Molecules or ions with a lone pair, ligands, form coordination bonds to the ion, making a complex. In [Cu(H₂O)₆]²⁺ six water molecules each donate a lone pair from oxygen to Cu²⁺: six coordination bonds, arranged octahedrally. Ammonia, chloride and cyanide are other common ligands. The ligand is a Lewis base and the metal ion a Lewis acid, which is how Reactivity 3.4 describes the same bond.

13HLResonance and delocalization

The carbonate ion, CO₃²⁻, has 4 + 3 × 6 + 2 = 24 valence electrons. Carbon needs one C=O double bond, but it could go to any of the three oxygens (Figure 12).

Figure 12 · The carbonate ion: three resonance structures, one real ion (HL) Figure 12 · The carbonate ion: three resonance structures, one real ion (HL) C O O O [ ] 2− C O O O [ ] 2− C O O O [ ] 2− ↔ ↔ the double bond could be drawn in any of three places; none of the three is the real ion The resonance hybrid: charge and π electrons spread over all three C–O bonds C O −⅔ O −⅔ O −⅔ all three C–O bonds equal: bond order 1⅓, length between a C–O and a C=O bond Draw every resonance structure, joined by double-headed arrows. The real ion is the average of them.
Figure 12 · The carbonate ion: three resonance structures, one real ion (HL)

These are resonance structures: Lewis formulas differing only in where the double bond and charges sit. Experiment says none is right: all three C–O bonds are the same length, between single and double. The real ion is the resonance hybrid, with the π electrons spread over all three bonds. Electrons spread over more than two atoms are delocalized, and delocalization lowers energy.

Resonance structures are joined by ↔, which never means an equilibrium: the ion does not flip between forms. The hybrid's bond order is bonding pairs over bonds: carbonate has 4 pairs over 3 bonds, 1⅓. Ozone (O=O–O ↔ O–O=O, two bonds of order 1.5), nitrate, nitrite and carboxylate ions such as ethanoate behave the same way.

14HLBenzene

Benzene, C₆H₆, is a ring of six carbons each bonded to one H. Kekulé's structure alternates single and double bonds, and Figure 13(a) shows it can be drawn two ways: the signature of resonance.

Figure 13 · Benzene: the structure, and the energy evidence for it (HL) Figure 13 · Benzene: the structure, and the energy evidence for it (HL) (a) Two Kekulé structures and the real molecule ↔ = the circle: six π electrons delocalized round the whole ring all six C–C bonds 139 pm a C–C single bond 154 pm a C=C double bond 134 pm planar, every angle 120°, every bond the same (b) Enthalpy of hydrogenation Kekulé structure + 3H₂ (if it had three C=C bonds) benzene + 3H₂ cyclohexane, C₆H₁₂ 3 × (−120) = −360 predicted −208 measured about 150 kJ mol⁻¹ more stable enthalpy Equal bond lengths and the missing 150 kJ mol⁻¹ both say the same thing: the π electrons are delocalized.
Figure 13 · Benzene: the structure, and the energy evidence for it (HL)

"Discuss the structure of benzene" asks you to weigh the evidence against three separate double bonds. Use at least two strands.

Physical: bond lengths and shape. X-ray diffraction shows all six C–C bonds are 139 pm, between C–C (154 pm) and C=C (134 pm). Kekulé predicts three short and three long. The molecule is planar with every angle 120°.

Thermochemical: hydrogenation. Hydrogenating cyclohexene's one C=C releases about 120 kJ mol⁻¹, so three C=C bonds should release about 360 kJ mol⁻¹. Benzene releases only about 208 kJ mol⁻¹: it is about 150 kJ mol⁻¹ more stable than Kekulé predicts, the stabilization from delocalization (Figure 13(b)).

Chemical: reactions and isomers. Alkenes decolourise bromine water by addition at room temperature. Benzene does not; with a catalyst it reacts by substitution, keeping its delocalized ring intact. Kekulé also predicts two different 1,2-disubstituted benzenes (substituents joined by a single or a double bond); only one exists.

So the six π electrons are delocalized in a ring above and below the plane of the carbons, drawn as a circle in the hexagon, and all six bonds have order 1.5.

15HLExpanded octets

Atoms in period 3 and below can hold more than eight valence electrons, an expanded octet: phosphorus in PCl₅ has 10, sulfur in SF₆ has 12. Period 2 atoms cannot; their second shell holds only eight.

The Lewis method is unchanged, except that leftover electrons go on the central atom beyond eight. SF₄: 6 + 4 × 7 = 34 electrons, 17 pairs: four S–F bonds, twelve lone pairs on the fluorines, and one lone pair on sulfur. XeF₄: 8 + 4 × 7 = 36 electrons, 18 pairs: four bonds, twelve pairs on fluorine, two lone pairs on xenon (Figure 14).

Figure 14 · Lewis formulas with more than eight electrons round the centre (HL) Figure 14 · Lewis formulas with more than eight electrons round the centre (HL) P Cl Cl Cl Cl Cl PCl₅ 5 bonding pairs: 10 electrons S F F F F SF₄ 4 bonding + 1 lone pair: 10 electrons Xe F F F F XeF₄ 4 bonding + 2 lone pairs: 12 electrons Only atoms from period 3 downwards can do this. Every terminal atom still has its full octet.
Figure 14 · Lewis formulas with more than eight electrons round the centre (HL)

Five domains form a trigonal bipyramid: three equatorial positions at 120° round the middle, two axial positions at 90° to them. Six form an octahedron, all at 90°. Lone pairs take the positions with most room: equatorial with five domains; with six, the second lone pair goes opposite the first.

Figure 15 · Shapes with five and six electron domains (HL) Figure 15 · Shapes with five and six electron domains (HL) P Cl Cl Cl Cl Cl Trigonal bipyramidal PCl₅ · 90° and 120° 5 domains, 0 lone pairs S F F F F Seesaw SF₄ · < 90° and < 120° 5 domains, 1 lone pair Cl F F F T-shaped ClF₃ · < 90° 5 domains, 2 lone pairs Xe F F Linear XeF₂ · 180° 5 domains, 3 lone pairs S F F F F F F Octahedral SF₆ · 90° 6 domains, 0 lone pairs Br F F F F F Square pyramidal BrF₅ · < 90° 6 domains, 1 lone pair Xe F F F F Square planar XeF₄ · 90° 6 domains, 2 lone pairs With five domains, lone pairs take the equatorial positions (120° apart, more room). With six, a second lone pair goes opposite the first. Only lone pairs on the central atom are drawn.
Figure 15 · Shapes with five and six electron domains (HL)
DomainsLone pairsShapeAnglesExample
50trigonal bipyramidal90° and 120°PCl₅
51seesawless than 90° and 120°SF₄
52T-shapedless than 90°ClF₃
53linear180°XeF₂
60octahedral90°SF₆
61square pyramidalless than 90°BrF₅
62square planar90°XeF₄

16HLFormal charge

When several Lewis formulas obey the counting rules, formal charge chooses between them. It is the charge an atom would carry if every bonding pair were shared exactly equally.

Formal charge = valence electrons − non-bonding electrons − ½ × bonding electrons

Prefer the formula with formal charges closest to zero, with any negative charge on the most electronegative atom. The formal charges always add up to the species' overall charge: a check on your arithmetic.

O=C=O: C = 4 − 0 − ½(8) = 0 each O = 6 − 4 − ½(4) = 0
O≡C–O: C = 4 − 0 − ½(8) = 0 O(triple) = 6 − 2 − ½(6) = +1 O(single) = 6 − 6 − ½(2) = −1

All zeros wins, so O=C=O is preferred. The guide's own example is sulfate: (a) four S–O single bonds, or (b) two S=O and two S–O, with sulfur expanding to 12 electrons.

(a) S = 6 − 0 − ½(8) = +2 each O = 6 − 6 − ½(2) = −1 total −2
(b) S = 6 − 0 − ½(12) = 0 O in S=O = 6 − 4 − ½(4) = 0 O in S–O = −1 total −2

Structure (b) puts sulfur at 0 and has only two non-zero charges, so formal charge prefers it. Formal charge and oxidation state (Reactivity 3.2) make opposite assumptions: formal charge treats every bond as perfectly shared, oxidation state as fully ionic, the pair going to the more electronegative atom. Neither is the atom's real charge: in (b) sulfur's formal charge is 0 but its oxidation state is +6.

17HLSigma and pi bonds

Orbitals overlap to form bonds in two ways (Figure 16).

Figure 16 · Sigma bonds overlap end-on; pi bonds overlap side-on (HL) Figure 16 · Sigma bonds overlap end-on; pi bonds overlap side-on (HL) (a) s + s → σ e.g. H–H (b) p + p end-on → σ e.g. the bond in Cl–Cl (c) p + p side-on → π density above and below the axis, none on it single bond 1 σ double bond 1 σ + 1 π triple bond 1 σ + 2 π Amber marks the overlap, where the shared pair sits. The dashed line is the bond axis between the nuclei.
Figure 16 · Sigma bonds overlap end-on; pi bonds overlap side-on (HL)

A sigma bond (σ) forms by head-on overlap along the bond axis, the line between the nuclei, with the electron density concentrated on that axis: s with s (H₂), s with p (HCl), p with p end-on (Cl₂), or hybrid orbitals. A pi bond (π) forms by sideways overlap of two parallel p orbitals, with the density above and below the axis and none on it. A π bond forms only alongside a σ bond.

So a single bond is 1 σ, a double bond 1 σ + 1 π, a triple bond 1 σ + 2 π. Sideways overlap is smaller, so π is weaker than σ: that is why C=C is less than twice C–C and why alkenes react by breaking the π bond. Counting: ethene has 5 σ and 1 π; HCN and CO₂ each have 2 σ and 2 π; benzene has 12 σ plus six delocalized π electrons.

18HLHybridization

Carbon, 1s² 2s² 2p², has only two unpaired electrons, yet forms four identical bonds at 109.5° in methane. Hybridization, the mixing of atomic orbitals on one atom into new hybrid orbitals, reconciles the two (Figure 17).

Figure 17 · Hybridization of carbon's orbitals (HL) Figure 17 · Hybridization of carbon's orbitals (HL) energy Ground state 2s 2p One 2s electron promoted 2s 2p mix Mix, three ways sp³ · 4 domains · tetrahedral, 109.5° sp² · 3 domains · trigonal planar, 120° p sp · 2 domains · linear, 180° p p left for one π bond left for two π bonds Hybrid orbitals form σ bonds and hold lone pairs. Unhybridized p orbitals, left over, form π bonds.
Figure 17 · Hybridization of carbon's orbitals (HL)

One 2s electron is promoted to the empty 2p orbital, then the orbitals mix. Mixing the 2s with all three 2p gives four sp³ orbitals pointing to a tetrahedron's corners. Mixing it with two gives three sp² orbitals in a plane at 120°, leaving one p orbital at right angles for a π bond. Mixing it with one gives two sp orbitals at 180°, leaving two p orbitals for two π bonds. Hybrid orbitals form σ bonds or hold lone pairs.

DomainsGeometryHybridizationSpare p orbitalsExamples
4tetrahedral, 109.5°sp³0C in CH₄; N in NH₃; O in H₂O
3trigonal planar, 120°sp²1C in C₂H₄ and C=O; B in BF₃; C in benzene
2linear, 180°sp2C in CO₂, HCN and C₂H₂

Count domains, not bonds. In ethanal, CH₃CHO, the CH₃ carbon is sp³ and the CHO carbon sp². The table runs both ways: from hybridization you can predict the geometry, and from geometry the hybridization. Hybridization describes the geometry rather than causing it; the model was built so the orbitals point where experiment says the bonds are.

19Where marks are lost

Leaving lone pairs off a Lewis formula. H–O–H alone scores zero; oxygen's two lone pairs are part of the answer.

Counting a double bond as two domains. CO₂ has two domains round carbon, so it is linear.

Naming the electron domain geometry as the shape. Water's domains are tetrahedral; its shape is bent.

Judging molecular polarity by the bonds alone. CO₂ and CCl₄ have polar bonds and are non-polar molecules.

Saying boiling breaks covalent bonds. Boiling a molecular substance overcomes intermolecular forces; only network substances need bonds broken.

Giving hydrogen bonding to any molecule containing H. The H must be bonded to N, O or F.

Comparing force types across very different sizes. The order London < dipole–dipole < hydrogen bonding holds at comparable molar mass only.

Measuring RF from the plate's edge or to the top of the spot. Both distances start at the baseline; the spot's ends at its centre.

HL: treating resonance structures as forms that interconvert. The hybrid is one structure; ↔ is not an equilibrium sign.

20Draw it right

  1. Lewis formulas: every valence electron, lone pairs included; ions in square brackets with the charge outside. Count the total before you finish.
  2. Shapes: wedges and hashed lines for 3D, lone pairs on the central atom, the bond angle written on the diagram.
  3. Name the shape in the IB's words: linear, trigonal planar, bent, tetrahedral, trigonal pyramidal (HL: trigonal bipyramidal, seesaw, T-shaped, octahedral, square pyramidal, square planar).
  4. Dipoles: δ+ and δ− on the right atoms, or dipole arrows pointing to δ−; show the net dipole of a polar molecule.
  5. Hydrogen bonds: a dashed line from H to a lone pair on N, O or F of the next molecule, O–H···O roughly straight, δ+/δ− labelled.
  6. Chromatograms: pencil baseline, solvent front marked, distances from the baseline to the centre of each spot.
  7. HL resonance: every structure in full, joined by ↔, each carrying the same overall charge.

21Try it

Marks in brackets. Answers and marker's notes are at the end.

Q1. Two multiple-choice items. 2 marks

(a) Which molecule is polar? A. CO₂ · B. BF₃ · C. CCl₄ · D. NH₃ 1 mark

(b) Which substance conducts electricity as a solid? A. diamond · B. graphite · C. silicon dioxide · D. iodine 1 mark

Q2. Methanal has the formula CH₂O. 4 marks

(a) Draw the Lewis formula of methanal. 1 mark

(b) Predict the shape of the molecule and the H–C–H bond angle, and explain your prediction. 3 marks

Q3. The table gives data for three compounds. 5 marks

CompoundStructureMolar mass / g mol⁻¹Boiling point / °C
butaneCH₃CH₂CH₂CH₃58−1
propanoneCH₃COCH₃5856
propan-1-olCH₃CH₂CH₂OH6097

(a) State why these three compounds are suitable for comparing types of intermolecular force. 1 mark

(b) Explain the order of the boiling points. 4 marks

Q4. A food colouring is run on a TLC plate. The solvent front is 6.4 cm above the baseline. The colouring gives two spots, with centres 1.6 cm and 4.8 cm above the baseline. 3 marks

(a) Calculate the RF value of each spot. 2 marks

(b) The stationary phase is polar. Suggest which component is the more polar. 1 mark

Q5 (HL). Dinitrogen monoxide, N₂O, has its atoms in the order N–N–O. Two Lewis formulas are N=N=O and N≡N–O, each atom with a full octet. 4 marks

(a) Calculate the formal charge on each atom in both structures. 3 marks

(b) Deduce which structure is preferred. 1 mark

Q6 (HL). Ethanenitrile is CH₃CN, with the atoms joined C–C≡N. 4 marks

(a) Deduce the number of σ bonds and π bonds in the molecule. 2 marks

(b) State the hybridization of each carbon atom and the bond angle around each. 2 marks

22In one breath

A covalent bond is the attraction between a shared pair and both nuclei; atoms share to reach an octet, though boron and beryllium stop short. More shared pairs make a shorter, stronger bond; a coordination bond takes both electrons from one atom. VSEPR spreads domains (a multiple bond counts as one) to 180°, 120° or 109.5°, and lone pairs repel most, closing the angle to 107° and 104.5°. Polar bonds make a polar molecule only if the shape stops their dipoles cancelling. Diamond, graphite, graphene, silicon and SiO₂ are covalent networks with high melting points; graphite and graphene conduct through delocalized electrons; C₆₀ is a molecule. Between molecules act London forces, dipole–induced dipole, dipole–dipole and hydrogen bonding (H on N, O or F to a lone pair on N, O or F), weakest to strongest at comparable molar mass, and they set volatility and solubility. RF is spot distance over solvent distance, both from the baseline. HL: ligands give lone pairs to metal ions; resonance means delocalized electrons and equal bonds, and benzene's equal bonds, low hydrogenation enthalpy and substitution chemistry prove its ring; period 3 atoms expand their octets into five- and six-domain shapes; formal charge picks the formula with charges nearest zero; single bonds are σ, double σ + π, triple σ + 2π; four, three and two domains mean sp³, sp² and sp.


Answers

Q1. (a) D. NH₃ is trigonal pyramidal, so its N–H dipoles do not cancel; the other three are symmetrical. (b) B. Graphite's delocalized electrons move along its layers; diamond and SiO₂ hold every electron in localized bonds, and iodine is molecular. 1 for each. A and C in (a) come from judging by bond polarity alone.

Q2. (a) C in the centre, single bonds to two H, a double bond to O, two lone pairs on O: 12 valence electrons (4 + 2 + 6). (b) Carbon has three electron domains and no lone pairs, so the molecule is trigonal planar. The C=O double bond holds more electrons and repels more than a single bond, so H–C–H is slightly less than 120° (about 116°). 1 for the Lewis formula with both lone pairs on O; 1 for three domains round C; 1 for trigonal planar; 1 for an angle below 120° with the double-bond reason. Exactly 120° with correct reasoning scores 3. Missing lone pairs scores 0 for (a).

Q3. (a) Their molar masses are almost equal, so their London forces are similar and differences in boiling point come from other forces. (b) Butane is non-polar with only London forces, the weakest, so it boils lowest. Propanone has a polar C=O bond and is a polar molecule, so it also has dipole–dipole forces and more energy is needed to separate its molecules. Propan-1-ol has H bonded to O, so its molecules hydrogen bond to each other, the strongest of the three, and it boils highest. 1 for (a); in (b) 1 for butane London only, 1 for propanone dipole–dipole with the reason, 1 for propan-1-ol hydrogen bonding with the reason, 1 for stronger forces needing more energy to separate the molecules. "Breaking bonds" loses the last mark.

Q4. (a) Lower spot: 1.6 ÷ 6.4 = 0.25. Upper spot: 4.8 ÷ 6.4 = 0.75. (b) The component with RF = 0.25: it is attracted more strongly to the polar stationary phase, so it travels less far. 1 for each RF with no units, 1 for (b) with the reason. An RF above 1 or with units scores 0.

Q5 (HL). (a) N=N=O: terminal N = 5 − 4 − ½(4) = −1; central N = 5 − 0 − ½(8) = +1; O = 6 − 4 − ½(4) = 0. N≡N–O: terminal N = 5 − 2 − ½(6) = 0; central N = +1; O = 6 − 6 − ½(2) = −1. Each set sums to 0, the charge on the molecule. (b) Both have two non-zero formal charges, so electronegativity decides: N≡N–O is preferred, because its negative formal charge is on oxygen, the most electronegative atom. 1 for each correct set in (a), 1 for all six values correct, 1 for (b) with the electronegativity reason. A choice with no reason scores 0 for (b).

Q6 (HL). (a) Three C–H, one C–C and the σ within C≡N: 5 σ bonds; the triple bond also has 2 π bonds. (b) The CH₃ carbon has four domains: sp³, 109.5°. The C≡N carbon has two: sp, 180°. 1 for 5 σ, 1 for 2 π, 1 for sp³ with 109.5°, 1 for sp with 180°. Counting a triple bond as three σ bonds loses both marks in (a).


Educerie · written from the published IB Diploma Programme Chemistry guide, first assessment 2025, section S2.2 The covalent model. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

Mocks: in the future, hold tight!