S2.2 The covalent model | IB DP Chemistry

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🎯What you need to be able to do

Deduce Lewis formulas of molecules and ions for up to four electron pairs on each atom, including species with fewer than an octet.
Explain the relationship between number of bonds, bond length and bond strength.
Identify coordination bonds in compounds.
Predict electron domain geometry and molecular geometry for species with up to four electron domains, including the effect of non-bonding pairs and multiple bonds on bond angles.
Deduce bond polarity from electronegativity values, and molecular polarity from bond polarity plus geometry.
Describe the structures and explain the properties of silicon, silicon dioxide, diamond, graphite, fullerenes and graphene.
Deduce the types of intermolecular force present, rank their strengths, and use them to explain volatility, electrical conductivity and solubility.
Explain, calculate and interpret retardation factor \( R_{\mathrm{f}} \) values in chromatography.
AHL Deduce resonance structures, including benzene, and discuss the evidence for benzene’s structure.
AHL Draw Lewis formulas and deduce geometries for species with five and six electron domains.
AHL Apply formal charge to choose between possible Lewis formulas.
AHL Deduce the presence of sigma and pi bonds, and analyse sp, sp² and sp³ hybridization.

📚The chemistry

The covalent bond and Lewis formulas

A covalent bond is the electrostatic attraction between a shared pair of electrons and the positively charged nuclei of the two bonded atoms. Note that the definition is electrostatic, exactly as for the ionic bond — what differs is that here the electrons sit between two nuclei rather than being transferred.

The octet rule is the tendency of atoms to attain a valence shell of eight electrons. It is a rule of thumb with well-known exceptions, and the syllabus asks you to be able to state its limitations:

Fewer than an octet: \( \mathrm{BF_3} \) (six electrons around boron), \( \mathrm{BeCl_2} \) (four). These are electron-deficient and behave as Lewis acids (R3.4).
More than an octet AHL: \( \mathrm{PCl_5} \), \( \mathrm{SF_6} \) — only possible for period 3 and beyond, where empty d orbitals are available.
Odd numbers of electrons: \( \mathrm{NO} \), \( \mathrm{NO_2} \) — radicals, which cannot possibly have complete octets (R3.3).

A Lewis formula shows all the valence electrons, bonding and non-bonding. Electron pairs may be shown as dots, crosses or dashes, but lone pairs must be shown — a structure missing them is not a Lewis formula and will lose the mark. Method: count the total valence electrons (adding one per negative charge, subtracting one per positive), place the least electronegative atom in the centre, join with single bonds, then distribute the remainder as lone pairs to complete octets, forming double or triple bonds if you run short.

Bond order, length and strength

Bond order is the number of shared pairs: one for a single bond, two for a double, three for a triple. As bond order increases, more electron density lies between the nuclei, so the attraction is greater:

bond order ↑ → bond length ↓ → bond strength (bond enthalpy) ↑

Hence C–C (154 pm) > C=C (134 pm) > C≡C (120 pm) in length, and the reverse in strength. The relationship is not proportional — a double bond is not twice as strong as a single one — because the second bond is a \( \pi \) bond, which is weaker (see the AHL section).

Coordination bonds

A coordination bond (dative covalent bond) is a covalent bond in which both electrons of the shared pair come from the same atom. Once formed it is indistinguishable from any other covalent bond of that type; what differs is only its origin. It is conventionally drawn as an arrow pointing from the donor to the acceptor.

The standard examples: \( \mathrm{NH_4^{+}} \), where the nitrogen lone pair is donated to \( \mathrm{H^{+}} \); \( \mathrm{H_3O^{+}} \); and \( \mathrm{NH_3 \rightarrow BF_3} \), where nitrogen donates to electron-deficient boron. AHL Transition element complexes such as \( \mathrm{[Cu(H_2O)_6]^{2+}} \) are held entirely by coordination bonds from ligands, which is where this connects to S3.1.

VSEPR: shapes of molecules

Valence shell electron pair repulsion theory says that electron domains around a central atom arrange themselves as far apart as possible, because they repel. An electron domain is any region of electron density: a single bond, a double bond, a triple bond or a lone pair each count as one domain.

Two names are needed. The electron domain geometry counts all domains; the molecular geometry describes only where the atoms are. They differ whenever lone pairs are present.

2 domains
linear, 180°
\( \mathrm{CO_2} \), \( \mathrm{BeCl_2} \)

3 domains, 0 lone
trigonal planar, 120°
\( \mathrm{BF_3} \), \( \mathrm{CH_2O} \)

3 domains, 1 lone
bent (V-shaped), about 117°
\( \mathrm{SO_2} \), \( \mathrm{O_3} \)

4 domains, 0 lone
tetrahedral, 109.5°
\( \mathrm{CH_4} \), \( \mathrm{NH_4^{+}} \)

4 domains, 1 lone
trigonal pyramidal, about 107°
\( \mathrm{NH_3} \)

4 domains, 2 lone
bent (V-shaped), about 104.5°
\( \mathrm{H_2O} \)

Lone pairs repel more strongly than bonding pairs, because a lone pair is held by only one nucleus and so spreads out closer to the central atom. Each lone pair therefore squeezes the bond angle down by roughly 2.5° from the ideal: 109.5° in methane, 107° in ammonia (one lone pair), 104.5° in water (two). Multiple bonds, having greater electron density, also repel slightly more strongly than single bonds.

AHL With five domains the arrangement is trigonal bipyramidal (90° and 120°) — \( \mathrm{PCl_5} \); with a lone pair it becomes see-saw, then T-shaped, then linear. With six domains it is octahedral (90°) — \( \mathrm{SF_6} \); with lone pairs, square pyramidal then square planar.

Bond polarity and molecular polarity

Electronegativity is the ability of an atom to attract the shared pair in a covalent bond; values are in the data booklet. A difference in electronegativity makes the shared pair sit closer to one atom, giving that atom a partial negative charge \( \delta- \) and the other \( \delta+ \). The bond is then polar, and has a dipole, shown either with partial charges or as a vector arrow pointing towards \( \delta- \).

Molecular polarity depends on bond polarity and molecular geometry. A molecule is polar only if the individual bond dipoles do not cancel. This is a vector sum, so symmetry decides it:

\( \mathrm{CO_2} \) — two strongly polar C=O bonds, but the molecule is linear and the dipoles point in exactly opposite directions, so they cancel: the molecule is non-polar.
\( \mathrm{H_2O} \) — two polar O–H bonds, but the molecule is bent, so the dipoles do not cancel: polar.
\( \mathrm{CCl_4} \) — four polar bonds arranged tetrahedrally, symmetric, non-polar. \( \mathrm{CHCl_3} \) — symmetry broken, polar.

“It has polar bonds, so it is a polar molecule.” Carbon dioxide and tetrachloromethane both disprove this in one line. You must state the shape and then say whether the dipoles cancel. Any answer about molecular polarity that does not mention geometry is incomplete and will not get full marks.

Covalent network structures

Carbon and silicon form giant covalent networks: lattices in which every atom is covalently bonded to its neighbours throughout the structure. Because covalent bonds must be broken to melt them, all have very high melting points.

Diamond — each carbon bonded to four others tetrahedrally in a rigid 3D network. Extremely hard, very high melting point, and a non-conductor, because all four valence electrons are localised in bonds and none is free to move.
Graphite — each carbon bonded to three others in flat hexagonal layers, with the fourth electron delocalised between the layers. Consequences: it conducts electricity parallel to the layers, and it is soft and slippery because the layers are held to one another only by weak London forces and can slide.
Graphene — a single layer of graphite. Conducts extremely well, is transparent, and has an enormous strength-to-weight ratio.
Fullerenes — for example \( \mathrm{C_{60}} \), closed cages of hexagons and pentagons. Unlike the others these are discrete molecules, so they have low melting points and dissolve in some organic solvents.
Silicon — the diamond structure, but Si–Si bonds are weaker than C–C because silicon atoms are larger and the shared pair is further from the nuclei; hence the lower melting point. It is a semiconductor.
Silicon dioxide — each Si bonded to four O and each O to two Si, giving a 3D network with the empirical formula \( \mathrm{SiO_2} \). Hard, high melting, non-conducting. Compare with \( \mathrm{CO_2} \), a gas: carbon is small enough to form strong \( \pi \) bonds and so makes discrete O=C=O molecules, whereas silicon is not.

Allotropes are different structural forms of the same element, and their differing properties come entirely from differing bonding and structure — a point the syllabus makes explicitly.

Intermolecular forces

These are the forces between molecules, and they are far weaker than the covalent bonds within them. Given comparable molar mass, the ranking is:

London (dispersion) < dipole–induced dipole < dipole–dipole < hydrogen bonding

London (dispersion) forces act between all molecules, including non-polar ones. Random motion of electrons creates an instantaneous dipole, which induces a dipole in a neighbour. Strength increases with the number of electrons (so with molar mass) and with the surface area of contact — which is why a straight-chain hydrocarbon boils higher than its branched isomer.
Dipole–induced dipole — a polar molecule induces a dipole in a non-polar neighbour. This is why oxygen dissolves in water at all.
Dipole–dipole — permanent dipoles on polar molecules attract one another.
Hydrogen bonding — the strongest, occurring when hydrogen is covalently bonded to N, O or F and interacts with a lone pair on an N, O or F of a neighbouring molecule. Both conditions are required.

“Van der Waals forces” is used in this syllabus as an inclusive term covering London, dipole–induced dipole and dipole–dipole forces — but not hydrogen bonding. Use the specific name wherever you can.

These forces explain the physical properties of covalent substances. Volatility: the stronger the intermolecular forces, the more energy needed to separate the molecules, the higher the boiling point. Conductivity: simple covalent substances do not conduct, because they contain no ions and no delocalised electrons (graphite and graphene being the exceptions). Solubility: polar substances dissolve in polar solvents and non-polar in non-polar, because only then can the solute–solvent interactions replace the ones that must be broken.

“Boiling water breaks the O–H bonds.” It does not, and the distinction is worth marks on every paper. Boiling separates molecules from one another, so it breaks intermolecular forces — here, hydrogen bonds. The covalent O–H bonds inside each molecule survive intact; steam is still \( \mathrm{H_2O} \). Compare the numbers: a hydrogen bond is worth tens of kJ mol⁻¹, an O–H covalent bond about 463.

Chromatography

Chromatography separates the components of a mixture according to their relative attraction to a mobile phase and a stationary phase — and those attractions are the intermolecular forces just described, which is why the topic lives here rather than in S1.1.

\[ R_{\mathrm{f}} = \frac{\text{distance travelled by the component}}{\text{distance travelled by the solvent front}} \]

\( R_{\mathrm{f}} \) has no units and is always between 0 and 1. A component with a high \( R_{\mathrm{f}} \) has a stronger attraction to the mobile phase (and so is carried further); a low \( R_{\mathrm{f}} \) means a stronger attraction to the stationary phase. Under identical conditions \( R_{\mathrm{f}} \) is characteristic of a substance, so it can be used to identify components by comparison with standards. Knowledge of locating agents is not required, and the operational details of GC and HPLC are not assessed.

AHL Resonance and benzene

Resonance occurs when more than one valid Lewis formula can be drawn for a species, differing only in the placing of a multiple bond. The true structure is none of them: it is a single structure in which the electrons are delocalised, and the resonance structures are drawn with a double-headed arrow between them to indicate that.

Standard examples: ozone, the carbonate ion, the nitrate ion. The observable consequence is that all the bonds are identical and intermediate in length between a single and a double bond — in \( \mathrm{CO_3^{2-}} \) all three C–O bonds are the same, not two single and one double.

Benzene, \( \mathrm{C_6H_6} \), is the important case, and you should be able to give the evidence from both directions:

Physical evidence — all six C–C bonds are the same length, 140 pm, between a single bond (154) and a double bond (134). X-ray diffraction shows a regular planar hexagon with 120° angles.
Thermochemical evidence — the enthalpy of hydrogenation of benzene is considerably less exothermic than three times that of cyclohexene, so benzene is more stable than the hypothetical alternating structure by the resonance (delocalization) energy.
Chemical evidence — benzene does not decolourise bromine water in the dark, whereas alkenes do. It undergoes substitution rather than addition, because addition would destroy the delocalised system (R3.4).

AHL Formal charge

When several Lewis formulas are possible, formal charge chooses between them. For each atom:

\[ \text{formal charge} = (\text{valence electrons}) - \tfrac{1}{2}(\text{bonding electrons}) - (\text{non-bonding electrons}) \]

The preferred structure is the one with formal charges closest to zero, and with any negative formal charge on the most electronegative atom. This is how the expanded-octet structure of sulfate is preferred over the all-single-bond one.

Formal charge assumes the bonding electrons are shared equally; oxidation state (S3.1) assumes they are transferred completely to the more electronegative atom. Both are bookkeeping devices, and the difference between their assumptions is a favourite short-answer question.

AHL Sigma bonds, pi bonds and hybridization

A sigma (\( \sigma \)) bond forms by head-on overlap of orbitals, with electron density concentrated along the bond axis. A pi (\( \pi \)) bond forms by sideways overlap of p orbitals, with electron density concentrated above and below the axis.

Counting is mechanical: a single bond is one \( \sigma \); a double bond is one \( \sigma \) and one \( \pi \); a triple bond is one \( \sigma \) and two \( \pi \). So ethyne, \( \mathrm{HC \equiv CH} \), has three \( \sigma \) and two \( \pi \). The \( \pi \) bond is weaker because sideways overlap is less effective, and it is what prevents rotation about a double bond — hence cis–trans isomerism in S3.2.

Hybridization is the mixing of atomic orbitals to form new hybrid orbitals for bonding. Only three types are required, and they map one-to-one onto the number of electron domains:

sp³
4 domains · tetrahedral · 109.5°
all single bonds — \( \mathrm{CH_4} \), \( \mathrm{NH_3} \), \( \mathrm{H_2O} \)

sp²
3 domains · trigonal planar · 120°
one double bond — ethene, benzene, \( \mathrm{BF_3} \)

sp
2 domains · linear · 180°
a triple bond or two double bonds — ethyne, \( \mathrm{CO_2} \)

The number in the label is the number of p orbitals used, so sp³ uses one s and three p to give four hybrids. Any p orbital not hybridised remains available to form a \( \pi \) bond — which is why an sp² carbon has exactly one \( \pi \) bond and an sp carbon has two.

✏️Worked example

Consider \( \mathrm{NH_3} \), \( \mathrm{BF_3} \) and \( \mathrm{SO_2} \).
(a) Draw the Lewis formula of each and deduce the electron domain geometry, molecular geometry and bond angle.
(b) State, with reasons, whether each molecule is polar.
(c) Explain why ammonia boils at −33 °C while phosphine, \( \mathrm{PH_3} \), of higher molar mass, boils at −88 °C.
(d) AHL State the hybridization of the central atom in each of the three molecules, and the number of \( \sigma \) and \( \pi \) bonds in \( \mathrm{SO_2} \).

(a) Count domains around the central atom in each case.

\( \mathrm{NH_3} \) — nitrogen has five valence electrons, three used in N–H bonds, leaving one lone pair. So three bonding domains plus one lone pair = 4 domains: electron domain geometry tetrahedral, molecular geometry trigonal pyramidal, bond angle about 107° (reduced from 109.5° by the greater repulsion of the lone pair).
\( \mathrm{BF_3} \) — boron has only three valence electrons, all used in bonds, so there is no lone pair and boron has only six electrons. Three domains: electron domain geometry and molecular geometry are both trigonal planar, bond angle 120°.
\( \mathrm{SO_2} \) — sulfur has six valence electrons; with two S=O bonds there is one lone pair left on sulfur. A double bond counts as one domain, so there are 3 domains: electron domain geometry trigonal planar, molecular geometry bent (V-shaped), bond angle about 117°.

(b) All three contain polar bonds, since in each case the two elements differ in electronegativity. The question is whether the dipoles cancel.

\( \mathrm{NH_3} \) — polar. The trigonal pyramidal shape is unsymmetrical, and the lone pair contributes to the dipole, so there is a net dipole pointing from the hydrogens towards the nitrogen.
\( \mathrm{BF_3} \) — non-polar. The three highly polar B–F bond dipoles are arranged symmetrically at 120° in a plane, and their vector sum is zero.
\( \mathrm{SO_2} \) — polar. The molecule is bent, so the two S=O dipoles do not cancel. (Compare \( \mathrm{CO_2} \), which is linear and therefore non-polar — the same bonds, a different shape, the opposite answer.)

(c) Boiling separates molecules, so it is the intermolecular forces that matter. Nitrogen is highly electronegative, so \( \mathrm{NH_3} \) molecules form hydrogen bonds to one another — hydrogen covalently bonded to N, interacting with a lone pair on the N of a neighbour. Phosphorus is much less electronegative, so \( \mathrm{PH_3} \) cannot hydrogen bond and is held only by weaker dipole–dipole and London forces. Hydrogen bonding is considerably stronger, so more energy is required to separate ammonia molecules, and it boils at the higher temperature despite its lower molar mass. The molar-mass trend, which would predict the opposite, is overwhelmed.

(d) Hybridization follows the number of electron domains directly: \( \mathrm{NH_3} \) has four domains, so nitrogen is sp³; \( \mathrm{BF_3} \) has three, so boron is sp²; \( \mathrm{SO_2} \) has three, so sulfur is sp². In \( \mathrm{SO_2} \), each S=O double bond is one \( \sigma \) plus one \( \pi \), giving 2 \( \sigma \) and 2 \( \pi \) bonds.

Check it. Two independent checks. First, hybridization and electron domain geometry must always agree — four domains means tetrahedral means sp³, three means trigonal planar means sp², two means linear means sp. If your answers to (a) and (d) disagree for any molecule, one of them is wrong. Second, for polarity, ask whether the central atom has a lone pair: a central atom with a lone pair and identical outer atoms is almost always polar, because the lone pair breaks the symmetry. \( \mathrm{BF_3} \) has none and is non-polar; \( \mathrm{NH_3} \) and \( \mathrm{SO_2} \) each have one and are polar.

Counting a double bond as two electron domains. It is one. Treat \( \mathrm{SO_2} \) as four domains and you predict a bent molecule with an angle near 104° and sp³ sulfur — the shape name happens to come out right, so the error hides, but the angle and the hybridization are both wrong and two marks go. The rule is that a domain is a region of electron density, and a double bond, however many electrons it holds, occupies one region.

📝Practise

Work through these on paper, then reveal the answer.

1. Draw Lewis formulas for \( \mathrm{CO_2} \), \( \mathrm{H_2O} \), \( \mathrm{NH_4^{+}} \) and \( \mathrm{BeCl_2} \), and state the molecular geometry and bond angle of each.

\( \mathrm{CO_2} \): O=C=O, with two lone pairs on each oxygen and none on carbon. Two domains → linear, 180°. \( \mathrm{H_2O} \): two O–H single bonds and two lone pairs on oxygen. Four domains, two lone → bent, about 104.5°. \( \mathrm{NH_4^{+}} \): four N–H bonds, no lone pair on nitrogen (one bond is a coordination bond, but that changes nothing about the shape); the whole species carries a positive charge, shown in square brackets. Four domains, none lone → tetrahedral, 109.5°. \( \mathrm{BeCl_2} \): two Be–Cl bonds, three lone pairs on each chlorine, and only four electrons around beryllium — an exception to the octet rule. Two domains → linear, 180°.

2. Explain why \( \mathrm{CCl_4} \) is non-polar while \( \mathrm{CHCl_3} \) is polar, even though the C–Cl bond is polar in both.

In both molecules chlorine is considerably more electronegative than carbon, so each C–Cl bond is polar with \( \delta- \) on chlorine. In \( \mathrm{CCl_4} \) the four identical polar bonds point towards the corners of a regular tetrahedron. The arrangement is completely symmetrical, so the four bond dipoles cancel exactly as vectors and the molecule has no net dipole — it is non-polar. In \( \mathrm{CHCl_3} \) one chlorine is replaced by hydrogen. The C–H bond is only very slightly polar (carbon and hydrogen have similar electronegativities) and points in the opposite direction to the resultant of the three C–Cl dipoles. The symmetry is broken, the dipoles no longer cancel, and there is a net dipole pointing away from the hydrogen: the molecule is polar. The lesson is that polarity requires an examination of shape, not just of bonds.

3. Diamond and graphite are both allotropes of carbon. Explain why graphite conducts electricity and is soft, while diamond does neither.

In diamond, every carbon atom forms four covalent bonds to four other carbons in a rigid three-dimensional tetrahedral network. All four valence electrons on every atom are localised in those bonds, so there are no mobile charge carriers and diamond does not conduct. There are also no planes of weakness — deforming the structure requires strong covalent bonds to be broken in every direction — so it is extremely hard. In graphite, each carbon forms only three covalent bonds, in flat hexagonal layers. The fourth valence electron is delocalised and free to move parallel to the layers, so graphite conducts electricity in that direction. Between the layers there are only weak London (dispersion) forces, so the layers can slide over one another readily, making graphite soft and a good lubricant. Both substances have very high melting points, because melting either requires covalent bonds within the layers or the network to be broken.

4. Rank butane, propan-1-ol and propanone in order of increasing boiling point, naming the dominant intermolecular force in each. Their molar masses are 58.14, 60.11 and 58.08 g mol⁻¹.

The molar masses are deliberately almost identical, so London forces are comparable and the difference is entirely in the additional intermolecular forces. Butane, \( \mathrm{C_4H_{10}} \), is a non-polar hydrocarbon with only London (dispersion) forces — the weakest set, so the lowest boiling point (−0.5 °C). Propanone, \( \mathrm{CH_3COCH_3} \), is polar because of the C=O bond, so it has dipole–dipole forces in addition to London forces (56 °C). Propan-1-ol, \( \mathrm{CH_3CH_2CH_2OH} \), has an O–H group, so its molecules hydrogen bond to one another — the strongest of the three (97 °C). Order: butane < propanone < propan-1-ol. Note that propanone has a lone pair on oxygen and can accept a hydrogen bond from water, which is why it is miscible with water even though it cannot hydrogen bond to itself.

5. In a paper chromatogram, the solvent front travelled 8.4 cm and two spots were seen at 6.3 cm and 2.1 cm from the origin. Calculate both \( R_{\mathrm{f}} \) values and explain what they indicate about the two components.

\( R_{\mathrm{f}} = \dfrac{\text{distance moved by component}}{\text{distance moved by solvent front}} \). Spot 1: \( \dfrac{6.3}{8.4} = \mathbf{0.75} \). Spot 2: \( \dfrac{2.1}{8.4} = \mathbf{0.25} \). \( R_{\mathrm{f}} \) has no units and must lie between 0 and 1. The component with the higher \( R_{\mathrm{f}} \) (0.75) has the stronger attraction to the mobile phase and a weaker attraction to the stationary phase, so it spends more of its time moving and is carried further. The component with \( R_{\mathrm{f}} = 0.25 \) is more strongly attracted to the stationary phase and is retained near the origin. The attractions involved are intermolecular forces, so in a polar stationary phase the more polar component will have the lower \( R_{\mathrm{f}} \). Because \( R_{\mathrm{f}} \) is characteristic of a substance only for a given solvent, stationary phase and temperature, identification requires a known standard run on the same plate.

6. AHL For benzene, \( \mathrm{C_6H_6} \): (a) describe the bonding, (b) give two pieces of evidence that it is not cyclohexa-1,3,5-triene, and (c) state the hybridization of each carbon and the total number of \( \sigma \) bonds.

(a) Each carbon is bonded to two other carbons and one hydrogen by \( \sigma \) bonds, giving a planar regular hexagon with 120° angles. Each carbon retains one unhybridised p orbital perpendicular to the ring, and these six p orbitals overlap sideways all the way round, producing a delocalised \( \pi \) system above and below the plane of the ring — not three localised double bonds. (b) Any two of: bond lengths — all six C–C bonds are identical at 140 pm, between a single bond (154 pm) and a double bond (134 pm), whereas the alternating structure would show two distinct lengths; enthalpy of hydrogenation — the measured value is about 150 kJ mol⁻¹ less exothermic than three times that of cyclohexene, showing benzene is more stable than the localised structure by that delocalization energy; chemical behaviour — benzene does not decolourise bromine water in the dark and undergoes substitution rather than addition, unlike an alkene; and isomer counting — only three isomers of dibromobenzene exist, whereas the alternating structure predicts four. (c) Each carbon has three electron domains and is therefore sp² hybridized. \( \sigma \) bonds: six C–C plus six C–H = 12 \( \sigma \) bonds, with the remaining six electrons in the delocalised \( \pi \) system.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

PhET — Molecule Shapes and Molecule Polarity. The first lets you add and remove lone pairs and watch the angles change; the second shows the net dipole arrow appearing and vanishing as you alter electronegativity and geometry. Between them they make VSEPR and polarity almost impossible to get wrong.
MolView or any web-based 3D molecule builder — the syllabus asks you to construct models, real or virtual, and rotating a tetrahedral molecule is worth more than reading about it.
RSC Learn Chemistry — paper and thin-layer chromatography practicals, including how to run a standard alongside an unknown.