7 higher-level sections hidden.
Educerie · IB Diploma · Chemistry
Structure 2 Models of bonding and structure · S2.2 The covalent model
What you must be able to do
| You must be able to | Level | What 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 octets | SL, HL | "Draw the Lewis formula of…" (1–2 marks) |
| Relate the number of shared pairs to bond length and strength; identify coordination bonds | SL, 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 domains | SL, HL | "Predict the shape and bond angle of…" (2 marks) |
| Deduce bond polarity from electronegativity, and molecular polarity from polarity plus shape | SL, 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 SiO₂ | SL, HL | "Explain why graphite conducts but diamond does not" (2–3 marks) |
| Deduce the intermolecular forces present, rank them, and explain volatility, conductivity and solubility | SL, HL | "Explain the difference in boiling points" (3–4 marks) |
| Explain, calculate and interpret RF values | SL, HL | Paper 1B: measure, calculate RF, interpret (2–3 marks) |
| Include transition element complexes in coordination bonding | HL only | Identify the ligand and the coordination bond |
| Deduce resonance structures; discuss the structure of benzene from evidence | HL only | "Discuss the evidence for the structure of benzene" (3–4 marks) |
| Draw Lewis formulas and deduce shapes for five and six domains | HL only | "Predict the shape of SF₄ and its bond angles" (2–3 marks) |
| Use formal charge to choose a preferred Lewis formula | HL only | Paper 2 (2–4 marks), often sulfate |
| Deduce σ and π bonds; relate hybridization to Lewis formula and geometry | HL 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.
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.
- Count the valence electrons. For a negative ion add one per charge; for a positive ion take one away.
- Draw the skeleton. The atom forming most bonds goes in the middle (never hydrogen). Join each outer atom to it with a single bond.
- Complete the outer atoms' octets with lone pairs.
- Put leftover electrons on the central atom as lone pairs.
- 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.
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.
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.
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.
| Domains | Electron domain geometry | Lone pairs | Molecular geometry | Angle | Example |
|---|---|---|---|---|---|
| 2 | linear | 0 | linear | 180° | CO₂, HCN |
| 3 | trigonal planar | 0 | trigonal planar | 120° | BF₃, CH₂O |
| 3 | trigonal planar | 1 | bent | less than 120° | SO₂, O₃ |
| 4 | tetrahedral | 0 | tetrahedral | 109.5° | CH₄, NH₄⁺ |
| 4 | tetrahedral | 1 | trigonal pyramidal | about 107° | NH₃, H₃O⁺ |
| 4 | tetrahedral | 2 | bent | about 104.5° | H₂O |
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.
| Bond | Δχ | δ− end | Verdict |
|---|---|---|---|
| Cl–Cl | 0 | neither | non-polar |
| C–H | 2.6 − 2.2 = 0.4 | C | slightly polar |
| H–Cl | 3.2 − 2.2 = 1.0 | Cl | polar |
| O–H | 3.4 − 2.2 = 1.2 | O | very 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.
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.
| Substance | Structure | Properties, and why |
|---|---|---|
| Diamond | each C bonded tetrahedrally to four others | Very hard, very high melting point: many strong covalent bonds must break. No conduction: every valence electron is in a localized bond. |
| Graphite | layers of hexagons; each C bonded to three, its fourth electron delocalized over the layer; weak London forces between layers | Conducts along the layers: delocalized electrons move. Soft and slippery: layers slide. Very high melting point: strong bonds within layers. |
| Graphene | one layer of graphite | Very strong for its mass, excellent electrical and thermal conductor, very high melting point. |
| Fullerene, C₆₀ | a closed cage of 60 C (12 pentagons, 20 hexagons): a molecule | Poor 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. |
| Silicon | the diamond structure | Hard, high melting point (about 1410 °C); a semiconductor. |
| SiO₂ (quartz) | each Si bonded tetrahedrally to four O, each O to two Si | Very 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.
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.
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).
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).
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.
"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).
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.
| Domains | Lone pairs | Shape | Angles | Example |
|---|---|---|---|---|
| 5 | 0 | trigonal bipyramidal | 90° and 120° | PCl₅ |
| 5 | 1 | seesaw | less than 90° and 120° | SF₄ |
| 5 | 2 | T-shaped | less than 90° | ClF₃ |
| 5 | 3 | linear | 180° | XeF₂ |
| 6 | 0 | octahedral | 90° | SF₆ |
| 6 | 1 | square pyramidal | less than 90° | BrF₅ |
| 6 | 2 | square planar | 90° | 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.
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.
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).
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).
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.
| Domains | Geometry | Hybridization | Spare p orbitals | Examples |
|---|---|---|---|---|
| 4 | tetrahedral, 109.5° | sp³ | 0 | C in CH₄; N in NH₃; O in H₂O |
| 3 | trigonal planar, 120° | sp² | 1 | C in C₂H₄ and C=O; B in BF₃; C in benzene |
| 2 | linear, 180° | sp | 2 | C 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
- Lewis formulas: every valence electron, lone pairs included; ions in square brackets with the charge outside. Count the total before you finish.
- Shapes: wedges and hashed lines for 3D, lone pairs on the central atom, the bond angle written on the diagram.
- 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).
- Dipoles: δ+ and δ− on the right atoms, or dipole arrows pointing to δ−; show the net dipole of a polar molecule.
- 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.
- Chromatograms: pencil baseline, solvent front marked, distances from the baseline to the centre of each spot.
- 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
| Compound | Structure | Molar mass / g mol⁻¹ | Boiling point / °C |
|---|---|---|---|
| butane | CH₃CH₂CH₂CH₃ | 58 | −1 |
| propanone | CH₃COCH₃ | 58 | 56 |
| propan-1-ol | CH₃CH₂CH₂OH | 60 | 97 |
(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.