A-Level Chemistry animations

Write the equilibrium before the expression

Fluorine is very electronegative, so it pulls the C–F bonding pair towards itself, and the pull is passed along the σ bonds (a negative inductive effect). When H⁺ is lost, this spreads out the negative charge of the ion, so the ion is more stable, more of the acid dissociates and the pKa is lower: CH₃COOH 4.76, CH₂ClCOOH 2.87, CH₂FCOOH 2.59, CHF₂COOH 1.24, CF₃COOH 0.23 (data-book values, 298 K).

Hydration of ethene

Ethene reacts with steam over a phosphoric acid catalyst, typically around 300 °C and 60–70 atm. The C=C π bond donates an electron pair to H⁺, giving the ethyl carbocation CH₃CH₂⁺, and a lone pair on a water oxygen then bonds to the positive carbon, forming the oxonium ion CH₃CH₂OH₂⁺. The O–H bond pair returns to O as H⁺ is lost, giving ethanol and regenerating the catalyst; overall the reaction is reversible.

Protonate OH before it leaves

In acid-catalysed elimination, a lone pair on the oxygen of propan-2-ol bonds to H⁺ from concentrated phosphoric or sulfuric acid, giving –OH₂⁺. The C–O bond pair then moves to O, so water leaves and the secondary carbocation CH₃CH⁺CH₃ forms. The pair from a C–H bond on the adjacent carbon forms the C=C as H⁺ is lost, giving propene and regenerating the acid catalyst.

RP5: distillation of a reaction product

The washed organic layer looks cloudy because it still holds tiny droplets of water. Anhydrous drying agent is added a little at a time and swirled: grains that take up water clump together, and once some grains move freely the liquid is clear and dry. The liquid is decanted off the solid and redistilled, collecting only the fraction that comes over within the specified range (shown here as 81–85 °C, either side of cyclohexene's boiling point of about 83 °C).

RP5: distillation of a reaction product

Simple distillation of the reaction product. Vapour reaching the side arm is measured by the thermometer, condenses in the counter-current condenser and collects as distillate in an unsealed receiver. Collect only the fraction boiling in the range your procedure specifies (cyclohexene boils at about 83 °C).

Distil an aldehyde; reflux for an acid

To make the carboxylic acid, the alcohol is heated under reflux with excess acidified potassium dichromate(VI). Vapour condenses in the vertical condenser and runs back, so the aldehyde formed first stays in the flask and is oxidised further. Cooling water enters at the bottom and leaves at the top, and the top stays open.

Distil an aldehyde; reflux for an acid

To stop at the aldehyde, the alcohol is warmed gently with a limited amount of acidified potassium dichromate(VI) and the aldehyde is distilled off as it forms, so it leaves the flask before it can be oxidised further. The thermometer bulb sits at the side arm, cooling water enters the condenser at the bottom, and the receiver is open to the air.

NaBH₄ supplies the hydride nucleophile

Aqueous NaBH₄ supplies the hydride ion, H⁻, which uses its lone pair to attack the δ+ carbonyl carbon while the C=O π pair moves onto oxygen; the planar carbonyl carbon becomes tetrahedral in an alkoxide intermediate with O⁻. An O⁻ lone pair then takes an H from water as that O–H bond pair returns to the water oxygen, giving the alcohol and OH⁻. Butan-2-one gives butan-2-ol, a secondary alcohol, and propanal gives propan-1-ol, a primary alcohol.

CN⁻ attacks through its carbon

The cyanide ion from KCN attacks ethanal through its carbon: one curly arrow goes from the lone pair on the C of :C≡N⁻ to the δ+ carbonyl carbon and another from the C=O π bond to O, giving CH₃CH(O⁻)CN, which H⁺ from dilute acid protonates to 2-hydroxypropanenitrile. Because the carbonyl group is planar, a red CN⁻ attacking from above and a blue CN⁻ attacking from below are equally likely (50% each). The two products, with their CN groups kept red and blue, are non-superimposable mirror images, so equal amounts of the two optical isomers form: a racemate.

Laboratory fractionation: explain each piece of apparatus

Fractional distillation of a hexane–heptane mixture. Vapour condenses and re-evaporates many times on the glass beads, so it becomes richer in the more volatile hexane as it rises. The thermometer at the side-arm entrance holds near 69 °C while the first fraction distils, then rises towards 98 °C, and the receiver is changed for each temperature range; the condenser only turns vapour back into liquid.

Crude oil is separated into fractions

Crude oil heated to about 350 °C enters near the bottom of a column that is hottest at the bottom and coolest at the top. Vapour rises through bubble-cap trays and each hydrocarbon condenses on the first tray cooler than its boiling point, so fractions with lower boiling ranges (shorter chains) are drawn off higher up; gases leave the top and residue that does not vaporise leaves the bottom. Only intermolecular forces are overcome: the molecules are separated, not broken.

Why alkane size matters

Each extra CH₂ in an unbranched alkane adds 8 electrons and more contact with neighbouring chains, so at any instant more induced-dipole attractions (London forces) act between the molecules: boiling temperatures rise from 112 K for methane to 447 K for decane and volatility falls. 2,2-Dimethylpropane has the same 42 electrons as pentane, but its compact shape touches its neighbours over a smaller area, so it boils lower (283 K against 309 K).

What a catalytic converter changes

A catalytic converter surface: CO and NO adsorb onto active sites, the N–O bond weakens and breaks, each CO combines with an adsorbed O atom to form CO₂ and two N atoms join as N₂, and the products desorb, freeing the sites. A lead atom bonded strongly to one site poisons it. Schematic: a simplified picture of the surface steps.

Further substitution reduces selectivity

The same chain reaction under the same UV lamp, with chlorine in excess (CH₄ : Cl₂ = 1 : 3). Radicals are still only a small fraction of the molecules, but more Cl₂ absorbs more UV, so more Cl• are made over the run, and as CH₃Cl builds up a Cl• increasingly takes its H from an already-chlorinated molecule. Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄, so when the Cl₂ is used up the mixture is mainly polysubstituted.

Further substitution reduces selectivity

Methane and chlorine (CH₄ : Cl₂ = 5 : 1) are sealed together under UV light. UV splits Cl₂ into Cl• radicals; in the two propagation steps a Cl• takes an H atom to form HCl and a carbon radical, which then takes a Cl atom from Cl₂ and regenerates Cl•. Because nearly every Cl• meets CH₄ rather than a chlorinated molecule, most of the product is CH₃Cl: excess methane favours monosubstitution but does not guarantee a pure product.

Propagation regenerates the chain carrier

UV light supplies the energy for homolytic fission of Cl–Cl: one fish-hook arrow goes to each atom, giving two chlorine radicals, Cl•, each with an unpaired electron and no charge. In propagation, Cl• removes a hydrogen atom from methane to give HCl and a methyl radical, CH₃•, which then reacts with Cl₂ to form chloromethane and regenerate Cl•, so the chain continues; adding the two steps cancels CH₃• and Cl• to give CH₄ + Cl₂ → CH₃Cl + HCl. In termination, two radicals combine (giving Cl₂, CH₃Cl or ethane), which removes chain carriers and explains why ethane can appear as an impurity.

Apply the pattern to another carbon position

Propane has two hydrogen environments. Cl• removes an H from C2, giving HCl and the radical CH₃C•HCH₃ with the dot on the carbon that lost H; that radical then takes Cl from Cl₂, so Cl is attached to C2 (2-chloropropane) and Cl• is regenerated. Choose C1 to compare terminal substitution via CH₃CH₂CH₂• to 1-chloropropane; product amounts depend on both the number of equivalent H atoms and how fast each is removed.

Build the HBr mechanism

The π bond in propene's C=C is a region of high electron density, and it donates an electron pair to the Hδ+ end of polar HBr while the H–Br bonding pair moves to Br. Adding H to the terminal carbon gives the secondary carbocation CH₃CH⁺CH₃, and a lone pair on Br⁻ then forms the C–Br bond. This major pathway gives 2-bromopropane; the primary carbocation route gives 1-bromopropane as the minor product.

What makes an alkene reactive?

A C=C double bond is one σ bond plus one π bond. Use the View buttons to see the planar σ framework (about 120° at each carbon), the parallel p orbitals, and their sideways overlap into one π bond with electron density above and below the plane. Restricted rotation shows that twisting one CH₂ group removes the overlap. Drag to rotate the molecule.

From one double bond to a long chain

Nine monomers join one at a time: the π bond of each C=C breaks, C=C becomes C–C, and new C–C bonds join neighbouring units, so the two former double-bond carbons stay adjacent and side groups end up on alternate backbone carbons. A real chain has thousands of units, so the shorthand follows: one repeat unit in brackets, with bonds through both brackets and n at the bottom right, named from its monomer. Choose ethene, propene, chloroethene, tetrafluoroethene or phenylethene above the figure.

From one double bond to a long chain

Beyond the specification, this shows how PVC chains grow by free-radical addition polymerisation: heat splits a peroxide initiator, ROOR, homolytically into RO• radicals, and the chain-end radical then adds chloroethene molecules one at a time, regenerating a radical at the chain end after each addition. Termination joins two radicals; each fish-hook arrow shows the movement of one electron.

A non-polar reagent can become polarised

Br₂ is non-polar, but as it approaches the electron-rich C=C of but-2-ene the π electrons repel its bonding pair, inducing Brδ+–Brδ−. Arrows from the C=C bond to Brδ+ and from the Br–Br bond to Brδ− give a carbocation and Br⁻, and a lone pair on Br⁻ then bonds to the positive carbon, giving 2,3-dibromobutane with one Br on each former C=C carbon. This is the school-level model; the intermediate is better described as a bridged bromonium ion.

Concentrated sulfuric acid adds across C=C

Concentrated sulfuric acid is drawn as H–O–S(=O)₂–OH so the polar Hδ+–Oδ− bond is visible. Curly arrows from ethene's C=C to the δ+ H and from that H–O bond to O give a carbocation and the hydrogensulfate ion, HSO₄⁻; a lone pair on the negatively charged O of HSO₄⁻ then bonds to the positive carbon, giving ethyl hydrogensulfate, CH₃CH₂OSO₃H, in which carbon is bonded to O, not S. Warming this with water gives ethanol and regenerates the sulfuric acid.

A base uses its lone pair to accept H⁺

A base uses the nitrogen lone pair to accept H⁺. In the established A-level model the ethyl group pushes electron density towards N, so ethylamine's lone pair is more available than ammonia's, while in phenylamine the lone pair overlaps the ring's π system and is partly delocalised, so it is less available. All three still accept H⁺ to form an ammonium ion with four N bonds and a + charge, and base strength follows ethylamine > ammonia > phenylamine in this aqueous comparison.

An acyl group makes an amide, not a more substituted amine

The nitrogen lone pair of ethylamine attacks the δ+ carbonyl carbon of ethanoyl chloride as the C=O π pair moves to O, giving a tetrahedral intermediate with O⁻ and a positive N. An O⁻ lone pair reforms C=O as chloride leaves, and a second ethylamine molecule removes one N–H proton, giving N-ethylethanamide and ethylammonium chloride. This nucleophilic addition–elimination makes an amide, not a more substituted amine.

Primary, secondary, tertiary, then quaternary

Bromoethane is heated with excess ammonia (NH₃ : CH₃CH₂Br = 6 : 1) in ethanol in a sealed vessel. When a nitrogen lone pair meets the δ+ carbon from the side opposite Br, SN2 substitution forms the C–N bond and releases Br⁻, and another NH₃ removes H⁺ from the nitrogen, forming NH₄⁺. Bromoethane nearly always meets NH₃ before an amine, so ethylamine is the main product, although further substitution is not impossible.

Primary, secondary, tertiary, then quaternary

The same reaction with bromoethane in excess (NH₃ : CH₃CH₂Br = 1 : 3). Ethylamine still has a lone pair on nitrogen, so, surrounded by bromoethane, it is attacked again, giving diethylamine, then triethylamine and some tetraethylammonium bromide, a quaternary salt with no N–H to lose. Each substitution also releases H⁺, which ends up on a nitrogen, so the reaction stops when no free lone pair remains, leaving a mixture of amines, mostly as their ammonium salts.

First form a C–N bond, then remove H⁺

Bromoethane is heated with excess ammonia in ethanol. The lone pair on the N of NH₃ attacks the δ+ carbon from the side opposite Br as the C–Br pair moves to Br, giving CH₃CH₂NH₃⁺ (N with four bonds and a + charge) and Br⁻. A second NH₃ then acts as a base: its lone pair takes H⁺ from that N as the N–H pair returns to N, giving ethylamine and NH₄⁺.

A peptide link is an amide link

Condensation joins the –COOH of one amino acid to the –NH₂ of the next: the OH from the carboxyl group and an H from the amino group leave as H₂O, and the carbonyl carbon bonds directly to nitrogen in a –C(=O)–NH– peptide link. Gly–Ala and Ala–Gly have the same formula but different sequences, and Gly–Ala–Ser has two peptide links, so two H₂O form. Hydrolysis adds water back across each link.

Covalent links build each strand

A DNA nucleotide contains a phosphate group, 2-deoxyribose (H rather than OH at the 2′ carbon) and a base bonded through a ring nitrogen to C1′ (N9 in adenine). Each phosphate links the 3′ oxygen of one sugar to the 5′ oxygen of the next, making a covalent sugar–phosphate backbone; the bases point inwards and pair by hydrogen bonds. The two strands run in opposite directions, so 5′-AGTC-3′ pairs with 3′-TCAG-5′. Drag the double helix to rotate it.

A platinum–nitrogen bond can disrupt DNA replication

Cisplatin is square planar Pt(II), with the two NH₃ ligands adjacent and the two Cl⁻ ligands adjacent, so the complex is neutral overall. In the simplified scheme a Cl⁻ ligand is replaced by water, then a lone pair on N7 of guanine forms a coordinate bond to Pt; the second Cl⁻ is replaced in the same way by N7 of an adjacent guanine. The cross-linked DNA is distorted and cannot be replicated, and DNA in healthy cells can be affected too.

Donor and acceptor positions determine complementarity

Adenine pairs with thymine through two hydrogen bonds and guanine with cytosine through three. Each hydrogen bond runs from an H already covalently bonded to N towards a lone pair on an N or O of the partner base, so complementarity depends on matching donor and acceptor positions. The sugar is attached through N9 of the purine and N1 of the pyrimidine, away from the pairing edge.

Blocking binding can reduce catalytic activity

The active site is chiral, with a charged group, a hydrogen-bond acceptor and a non-polar pocket fixed in 3D. One enantiomer can place its –COO⁻, O–H and –CH₃ groups at all three positions at once; its mirror image, however it is turned, can match at most two. An inhibitor with a similar shape can occupy the active site, so the substrate cannot bind there while it is bound.

Backbone hydrogen bonds organise helices and sheets

Primary structure is the sequence of residues joined by covalent peptide links. In an α-helix each backbone C=O oxygen hydrogen-bonds to the N–H hydrogen four residues further on; in a β-pleated sheet, C=O···H–N hydrogen bonds hold neighbouring backbone segments side by side. Tertiary structure is the overall fold, held by side-chain hydrogen bonds, ionic attractions, covalent S–S bridges and clustering of non-polar groups. Drag the helix or sheet to rotate it.

Practical: heating to constant mass

Hydrated copper(II) sulfate is heated in a weighed crucible with the lid ajar, gently at first: water vapour escapes and the blue crystals turn white from the hot base and walls inwards. After each heating the crucible is cooled in a desiccator and the same crucible, lid and contents are reweighed, and the cycle is repeated until successive readings agree (constant mass); the masses shown are illustrative. Here 2.50 g leaves 1.60 g, giving x = 0.0500 ÷ 0.01003 ≈ 5; constant mass alone does not identify the product or rule out decomposition.

Practical: heating to constant mass

Magnesium oxidation: the crucible sits in a pipe-clay triangle on a tripod above the Bunsen burner, and the lid is raised briefly with tongs now and then to admit oxygen while limiting the loss of magnesium oxide smoke. The crucible is cooled and reweighed until the mass is constant; here 0.486 g of Mg gains 0.320 g of oxygen, a 1 : 1 amount ratio, so the formula is MgO (illustrative masses). Wear eye protection and do not stare at burning magnesium.

Gas-syringe practicals

Magnesium is kept in a small tube until the flask is sealed, then the flask is tilted so it falls into excess hydrochloric acid. The hydrogen made pushes the plunger out and the volume is read from the syringe scale. When it levels off, the final volume and the room temperature and pressure give the amount of gas from pV = nRT: here 48.0 cm³ at 101 kPa and 293 K is 1.99 × 10⁻³ mol of H₂.

Making a standard solution (Required Practical 1)

One acid sample followed through the whole method: weigh by difference, dissolve in deionised water, transfer through a funnel and rinse the beaker, rod and funnel so the clinging HCl reaches the 250 cm³ volumetric flask, make up to the mark and invert, then pipette a 25.0 cm³ aliquot and titrate it with NaOH. The 7 HCl found in the aliquot scale up by 250.0 / 25.0 = 10 to the 70 in the original sample. Use the numbered steps to jump to any stage.

Ionic equations

Nine neutralisations, from 1 : 1 to 2 : 3 and 3 : 2. Each H⁺ from the acid joins one OH⁻ from the base to make one water molecule, so the H⁺ donated, the OH⁻ used and the water formed always match, and the ions left over make the salt. Matching the H⁺ per acid molecule with the OH⁻ per base formula unit gives the coefficients in the balanced equation; H₃PO₄ is a weak acid and Al(OH)₃ an insoluble base, so those steps show the reacting ratios for complete neutralisation.

Titration method and why

Hydrochloric acid runs from the burette into a 25.0 cm³ aliquot of sodium hydroxide solution containing methyl orange, while the flask is swirled on a white tile. Each addition turns the indicator red where the acid lands and the colour fades as it mixes in. Near the end point the acid is added dropwise, and the end point is the first permanent orange; the bottom of the meniscus is read at eye level to 2 d.p.

Indicators and end points

A particle model of the end point. Hydrochloric acid drips from the burette one drop at a time into 12 NaOH formula units with phenolphthalein; each H⁺ reacts with an OH⁻ to form water while Na⁺ and Cl⁻ stay as spectator ions. The solution stays pink while any OH⁻ remains and turns colourless once all 12 have reacted: 12 HCl : 12 NaOH = 1 : 1. In this model the colour changes exactly at equivalence; in a real titration a suitable indicator changes within the steep pH change near equivalence, so the end point is very close to it.

A dissolved metal ion has water ligands

In each aqua ion six water molecules donate a lone pair from oxygen to the metal ion, giving six coordinate bonds in an octahedral arrangement. Al³⁺ is not a transition metal ion but forms the same kind of complex. The complex keeps the charge of the metal ion because water is neutral.

Compare reactions giving the same final product

Every route ends at cyclohexane. Hydrogenating one C=C releases about 120 kJ mol⁻¹, so a localised cyclohexa-1,3,5-triene would be expected to release 3 × 120 = 360 kJ mol⁻¹, but benzene releases only about 208 kJ mol⁻¹. Its hydrogenation is less exothermic, so benzene starts lower in enthalpy: about 360 − 208 = 152 kJ mol⁻¹ of stabilisation from delocalisation, compared with the hypothetical model.

Six carbons share a delocalised π system

Benzene is a flat, regular hexagon: each carbon forms three σ bonds at 120°, and all six C–C bonds are the same length (0.140 nm, between C=C and C–C). Use the View buttons to see the six parallel p orbitals, then watch them overlap sideways with their neighbours all round the ring and merge into two continuous rings of delocalised π electron density, one above and one below the plane, and see how the circle in the skeletal formula represents them. Drag to rotate the molecule.

Attack the positively charged acyl carbon

Anhydrous AlCl₃ accepts a lone pair from the Cl of propanoyl chloride as the C–Cl bond pair moves onto Cl, forming AlCl₄⁻ and the acylium ion CH₃CH₂–C⁺=O. A curly arrow from the delocalised π electrons of benzene to the acyl carbon (not the O) gives a non-aromatic intermediate with H and –C(=O)CH₂CH₃ on the attacked carbon, a horseshoe over the other five carbons and an overall + charge. The C–H bond pair returns into the ring to give C₆H₅COCH₂CH₃, and AlCl₄⁻ takes the H⁺ to form HCl, regenerating AlCl₃.

Loss of a proton restores delocalisation

Concentrated nitric and sulfuric acids form the nitronium ion, NO₂⁺, whose positive charge is on N. A curly arrow from the delocalised π electrons of benzene to N forms a C–N bond, giving a positively charged intermediate in which the attacked carbon carries both H and NO₂ and a partial (horseshoe) ring covers the other five carbons. A second arrow returns the C–H bond pair to the ring as H⁺ is lost, restoring the delocalised ring in nitrobenzene; HSO₄⁻ accepts the H⁺, regenerating the H₂SO₄ catalyst.

Developing the model of the atom

Alpha particles from a source in a lead shield are fired at thin gold foil, and each one that reaches the zinc sulfide screen makes a tiny flash. Most pass straight through, a few are deflected through large angles and a very small number bounce back. The magnified view shows why: the atom is mostly empty space, and the positive alpha particles are repelled by a tiny, dense, positive nucleus.

Shells, sub-shells and orbitals

Each dot cloud shows where an electron is most likely to be found. The 1s and 2s orbitals are spherical, with 2s larger than 1s. Each 2p orbital is a dumbbell with a nodal plane through the nucleus, and the three 2p orbitals lie along x, y and z at right angles. Each orbital holds a maximum of 2 electrons with opposite spins. Drag the diagram to rotate it.

Filling order and rules

Electrons fill the sub-shells in the order 1s 2s 2p 3s 3p 4s 3d 4p: lowest energy first, one electron in each equal-energy orbital with parallel spins before any pairing, and at most two electrons per orbital with opposite spins. 4s fills before 3d, but configurations are written with 3d before 4s, and 4s electrons are the first lost when transition-metal ions form. Chromium ([Ar] 3d⁵ 4s¹) and copper ([Ar] 3d¹⁰ 4s¹) are the two exceptions to learn.

Second ionisation energy across elements

For a second ionisation energy, start from the 1+ ion. Na⁺ is [Ne], so its next electron comes from 2p in an inner shell, close to the nucleus and less shielded: sodium is the peak. Mg⁺ is [Ne] 3s¹, so magnesium's value is far lower; across the rest of the period the values rise, with a dip at Si (Si⁺ loses a 3p electron) and only a small rise at Cl (Cl⁺ has a paired 3p electron). The graph is drawn to scale, but only its shape is needed.

Trend across Period 3

First ionisation energy rises across Period 3 because the nuclear charge increases while the electron is removed from the same shell with similar shielding. It dips at Al, whose electron is removed from the higher-energy 3p sub-shell rather than 3s, and at S, where repulsion between the two electrons paired in one 3p orbital makes one of them easier to remove. The Period selector shows the matching Be → B and N → O dips in Period 2.

Big jumps and electron shells

Each electron is removed from an increasingly positive ion, so successive ionisation energies rise steadily within a shell. The first electron taken from a new, inner shell is much closer to the nucleus and less shielded, giving a big jump: for aluminium it comes after three removals (580, 1820, 2750, then 11 600 kJ mol⁻¹ to 3 s.f.), so Al has three outer-shell electrons and is in Group 13. Use only the first big jump; later jumps mark deeper shells.

What a mass spectrometer is for

Positive ions are accelerated by an electric field towards a negatively charged plate, so every 1+ ion has the same kinetic energy. They then drift through the evacuated, field-free flight tube, where lighter ions move faster and reach the detector first. Use the selector to compare electron impact (chlorine: X(g) + e⁻ → X⁺(g) + 2e⁻, with some fragmentation) and electrospray (aspartame: M(g) + H⁺ → [M+H]⁺(g), so m/z = Mr + 1).

Dative covalent bonding

Dative covalent bond formation in 3D: a lone pair on N (or O) is donated to the empty orbital of H⁺, or of boron in BF₃, and the new bond is drawn as an arrow from donor to acceptor. NH₄⁺ ends up tetrahedral (109.5°) with all four N–H bonds identical, and the + charge belongs to the whole ion.

Ionic bonding and lattices

Giant ionic lattices of sodium chloride and magnesium oxide: each ion is surrounded by six ions of opposite charge, and the attraction acts in all directions throughout the lattice. Mg²⁺ and O²⁻ have larger charges and are smaller (all ions are drawn at about 40% of their ionic radii so the lattice shows), so the attractions in MgO are stronger and its melting point is much higher.

Metallic bonding

Metallic bonding: a giant lattice of positive ions in a sea of delocalised electrons, held together by the electrostatic attraction between the ions and those electrons. Magnesium gives two delocalised electrons per atom and forms smaller Mg²⁺ ions, so its attraction is stronger and it melts at a higher temperature than sodium. Drag to rotate.

London forces

Each extra CH₂ in an unbranched alkane adds 8 electrons and more contact with neighbouring chains, so at any instant more induced-dipole attractions (London forces) act between the molecules: boiling points rise from 112 K for methane to 447 K for decane and volatility falls. 2,2-Dimethylpropane has the same 42 electrons as pentane, but its compact shape touches its neighbours over a smaller area, so it boils lower (283 K against 309 K).

London forces

Drawn to one scale, the halogen molecules from F₂ (18 electrons) to I₂ (106) have larger outer electron clouds that are more easily distorted. In each pair an instantaneous dipole in one molecule induces a dipole in its neighbour, and both the shifts and the attraction grow down the group, so more energy is needed to separate the molecules: boiling points rise from 85 K to 457 K. At 298 K fluorine and chlorine are gases, bromine is a liquid and iodine is a solid.

London forces

At an instant the electron density in one I₂ molecule piles up at one end, an instantaneous dipole (δ− where it builds up, δ+ where it thins), although the molecule stays neutral. This repels electron density in the neighbouring molecule, inducing a dipole lined up δ+ to δ−, and the two attract: a London force. Pulling back to six molecules and then to solid iodine shows such correlated, short-lived dipoles forming and vanishing everywhere; they are strong because each I₂ has 106 electrons in an easily polarised cloud.

Polar bonds

Each bonding pair is drawn as dots of electron density. When the two atoms differ in electronegativity, the dots stream along the bond towards the more electronegative atom, which becomes δ− (H–Cl, and more so H–F); in H–H the pair stays evenly shared. Choose H₂O to see both O–H pairs drawn towards O in a bent molecule, giving an overall dipole, or CO₂, where two equal bond dipoles cancel.

Polar and non-polar molecules

Oxygen (3.44) is more electronegative than hydrogen (2.20), so the dots of both O–H bonding pairs stream towards O: O becomes δ− and each H δ+. Water is bent, so the two bond dipoles add up to an overall dipole; choose CO₂ to see two equal bond dipoles cancel in a linear molecule.

Hydrogen bonding

Each O–H bonding pair is drawn towards O, leaving the H strongly δ+ with little electron density around it. A hydrogen bond (dotted line) forms when this δ+ H lines up with a lone pair on the O of another molecule, O–H···O close to 180°; each water molecule can donate two and accept two, giving the open network of ice. Choose Why N, O or F? to see that a C–H bond is almost non-polar, so it cannot form a hydrogen bond.

Polar and non-polar molecules

Bond dipoles add as vectors in three dimensions. The equal dipoles of linear CO₂, trigonal planar BF₃, tetrahedral CCl₄ and square planar XeF₄ cancel, so these molecules are non-polar; in CH₃Cl, bent H₂O and pyramidal NH₃ they do not, leaving a net dipole. The C–H bonds are treated as almost non-polar.

Ice and anomalous hydrides

Boiling points of the Group 14–17 hydrides. Down each group boiling points rise as London forces increase, but NH₃, H₂O and HF boil far above the trend (the hollow markers continue the period 3 to 4 trend back to period 2) because they also form hydrogen bonds; CH₄ cannot, so Group 14 rises smoothly.

Ice and anomalous hydrides

Hydrogen bonding in ice: each water molecule donates two hydrogen bonds through its δ+ H atoms and accepts two at the lone pairs on its O, each O–H···O close to 180°, so every molecule has four neighbours arranged tetrahedrally in an open hexagonal network. On melting some hydrogen bonds break and the network partly collapses, so liquid water packs more closely and is denser than ice.

Five and six regions with lone pairs

Choose a shape to see its example molecule, its bonding pairs and lone pairs, and its bond angles. The faint outline shows the arrangement of all the electron regions (linear, trigonal planar, tetrahedral, trigonal bipyramidal or octahedral); lone pairs take some of its positions (equatorial with five regions, opposite each other with six), so the shape name describes only where the atoms are. Drag to rotate the molecule, or switch to Wedge and dash to see the same orientation drawn with in-plane lines, wedges and dashed wedges.

Drawing shapes

NH₃, H₂O, KrF₂ and XeF₄ in 3D with their lone pairs shown on the central atom. One lone pair makes NH₃ trigonal pyramidal (about 107°) and two make H₂O bent (about 104.5°); KrF₂'s three lone pairs are equatorial, leaving a linear molecule (180°), and XeF₄'s two lone pairs sit opposite each other, leaving a square planar molecule (90°). Drag to rotate, or switch to Wedge and dash to see how each is drawn with wedges and dashed wedges.

The six required crystals

At an instant the electron density in one I₂ molecule piles up at one end, an instantaneous dipole (δ− where it builds up, δ+ where it thins), although the molecule stays neutral. This repels electron density in the neighbouring molecule, inducing a dipole lined up δ+ to δ−, and the two attract: a London force. Pulling back to six molecules and then to solid iodine shows such correlated, short-lived dipoles forming and vanishing everywhere; they are strong because each I₂ has 106 electrons in an easily polarised cloud.

Conductivity

Diamond and graphite compared. In diamond every outer electron is in a localised C–C bond, so it does not conduct; in graphite each carbon forms three bonds and its fourth outer electron is delocalised across the layer, so graphite conducts along the layers. The layers slide over each other because only weak London forces hold them together.

The six required crystals

The required crystal structures in 3D: giant ionic sodium chloride, giant metallic magnesium, macromolecular diamond and graphite (and silicon dioxide), and simple molecular iodine and ice. Each panel names the particles and the force that is overcome on melting; the Bonding view shows those forces. Drag to rotate.

Weak acids still react with carbonate

Ethanoic acid only partially ionises in water. In the ethanoate ion the two C–O bonds become equal in length and the negative charge is spread over both oxygen atoms; the two localised contributors describe this one delocalised ion (overall −1), which does not switch between them. Choose Hydrogen bonding to see O–Hδ+···:Oδ− hydrogen bonds between two acid molecules and with water.

Weak acids still react with carbonate

Ethanoic acid only partially dissociates, and its conjugate base, the ethanoate ion, is stabilised by delocalisation. The p orbitals on the carboxylate carbon and both oxygens are parallel and overlap sideways into one π system above and below the O–C–O plane, so the negative charge is spread over both oxygens rather than confined to one, and the two C–O bonds are the same length (0.126 nm). Use the View buttons to switch between the separate p orbitals and the delocalised cloud, or compare the localised picture and ethanoic acid itself; drag to rotate.

Weak acids still react with carbonate

Fluorine is very electronegative, so it pulls the C–F bonding pair towards itself, and the pull is passed along the σ bonds (a negative inductive effect). When H⁺ is lost, this spreads out the negative charge of the ion, so the ion is more stable, more of the acid dissociates and the pKa is lower: CH₃COOH 4.76, CH₂ClCOOH 2.87, CH₂FCOOH 2.59, CHF₂COOH 1.24, CF₃COOH 0.23 (data-book values, 298 K).

Addition, elimination and proton transfer

The carbonyl carbon of ethanoyl chloride is δ+, being bonded to both O and Cl. In the addition step, a lone pair on the O of water or ethanol, or on the N of ammonia or methylamine, bonds to that carbon as the C=O π pair moves to O, giving a tetrahedral intermediate with O⁻ and a positive O or N; in the elimination step, an O⁻ lone pair reforms C=O as the C–Cl pair leaves as Cl⁻. Finally H⁺ is removed from the positive O or N: Cl⁻ takes it (forming HCl) for water and ethanol, and a second ammonia or methylamine molecule takes it for the nitrogen nucleophiles, giving ethanoic acid, ethyl ethanoate, ethanamide or N-methylethanamide.

Fats and oils contain three ester links

A triglyceride is a triester of glycerol with three long-chain acids, which need not be identical. Each of the three ester links is cleaved separately at the acyl C–O bond, so complete hydrolysis gives one glycerol, HOCH₂CH(OH)CH₂OH, and three acid-derived fragments: three RCOOH with water and an acid catalyst (reversible), or three RCOO⁻Na⁺ (soap) with aqueous NaOH. Choose Methanol to compare transesterification to methyl esters (biodiesel).

Methanol exchanges the alcohol part of the ester

Transesterification: methanol, with an alkaline catalyst, exchanges the alcohol part of each ester link, so one triglyceride gives one glycerol and three methyl esters of long-chain acids, the mixture used as biodiesel. No water is a product of the ideal net reaction. Choose a hydrolysis reagent to compare the acid or soap products.

Develop the plate and reveal the spots

Mixture X and standards A, B and C are developed on one silica plate in a covered chamber, with the solvent starting below the pencil baseline. In the inset, molecules keep adsorbing onto the polar silica and desorbing into the moving solvent, so the component that spends a larger share of its time in the mobile phase travels further. The front is marked at 80 mm, the spots are located under UV, and Rf = distance to spot centre ÷ distance to solvent front gives 0.25, 0.60 and 0.80: matching A and B supports, but co-migration alone does not prove, their presence in X.

Column chromatography collects separated fractions

The mixture is applied as a narrow band and washed down the silica by the solvent. The component held less strongly by the silica spends more of its time dissolved in the moving solvent, so it travels faster and emerges first, while the more strongly retained band is still higher up the column. Successive fractions are collected and checked by TLC: the column separates the components but does not change them.

Gas chromatography uses a gaseous mobile phase

A small sample is injected into a heated inlet, where it vaporises, and an inert carrier gas sweeps it through a coiled column held in a temperature-controlled oven. Components that spend more time dissolved in the liquid film lining the column move more slowly, so each reaches the detector at its own time. Retention time is measured from injection to the maximum of each peak; the large early peak here is the solvent, not a sample component.

Keep the electron and ion pathways separate

Zinc atoms lose electrons and enter the solution as Zn²⁺; the electrons flow through the external circuit and Cu²⁺ ions gain them at the copper electrode, depositing copper. Ions, not electrons, carry the charge through the salt bridge: NO₃⁻ moves towards the zinc half-cell and K⁺ towards the copper half-cell. With the switch open, almost no current flows and the high-resistance voltmeter reads the EMF, +1.10 V; with the load switched in, it shows a slightly lower terminal p.d.

Read a cell from left to right

Each cell is written with the negative electrode on the left: | marks a phase boundary, || the salt bridge, and a comma separates species in the same phase. Platinum is included where the redox pair has no solid conductor of its own, as for Fe³⁺/Fe²⁺ and the hydrogen electrode. E°cell = E°right − E°left.

A potential needs a reference half-cell

The standard hydrogen electrode, platinum in contact with H₂ at 100 kPa and 1.00 mol dm⁻³ H⁺(aq) at 298 K, is assigned 0.00 V, so the cell EMF gives the other half-cell's standard electrode potential. Copper is the positive electrode against it, so E°(Cu²⁺/Cu) = +0.34 V; zinc is the negative electrode, so E°(Zn²⁺/Zn) = −0.76 V.

Learn the alkaline hydrogen–oxygen half-equations

Hydrogen is oxidised at the negative electrode, H₂ + 2OH⁻ → 2H₂O + 2e⁻, and oxygen is reduced at the positive electrode, O₂ + 2H₂O + 4e⁻ → 4OH⁻. Hydroxide made at the oxygen electrode moves through the electrolyte to the hydrogen electrode, electrons do work in the external load, and water and heat leave the cell. Doubling the hydrogen equation cancels the electrons, giving 2H₂ + O₂ → 2H₂O.

The specified simplified lithium model

In AQA's simplified lithium model, lithium atoms at the negative electrode lose electrons and the Li⁺ ions move through a non-aqueous electrolyte into cobalt oxide at the positive electrode, where Co(IV) gains the electron and becomes Co(III): Li + CoO₂ → LiCoO₂. Charging with an external supply reverses both electrode reactions. With illustrative potentials of +0.60 V and −3.00 V, the EMF is +3.60 V.

Reverse the discharge equations

On discharge, lead at the negative plate and lead(IV) oxide at the positive plate both become lead(II) sulfate, using up sulfate and H⁺ ions and forming water, while electrons flow through the load from the lead plate. Charging with an external supply reverses every change, regenerating Pb, PbO₂ and the acid. With the supplied potentials, E°cell = +1.69 − (−0.36) = +2.05 V.

Subtract reduction potentials once

Choose a cell to watch it discharge: oxidation at the negative electrode on the left releases electrons that travel through the external circuit to the positive electrode on the right, while K⁺ and NO₃⁻ ions carry the charge through the salt bridge. The panel gives the conventional cell representation and E°cell = E°right − E°left from the two standard reduction potentials. With the switch open, almost no current flows and the high-resistance voltmeter reads the EMF, equal to E°cell; with the load switched in, the cell discharges and the reading falls slightly to the terminal p.d.

Explain concentration changes for a specified cell

Changing a concentration changes an electrode potential and so the EMF: less Zn²⁺ makes E(Zn²⁺/Zn) more negative and raises Ecell, while less Cu²⁺ makes E(Cu²⁺/Cu) less positive and lowers it. The values are calculated for 298 K with the Nernst equation, which is not required here; the direction of the change is what matters.

Plan a concentration investigation

Changing a concentration changes an electrode potential and so the EMF: less Zn²⁺ makes E(Zn²⁺/Zn) more negative and raises Ecell, while less Cu²⁺ makes E(Cu²⁺/Cu) less positive and lowers it. The values are calculated for 298 K with the Nernst equation, which is not required here; the direction of the change is what matters.

Use a high-resistance voltmeter

Zinc atoms lose electrons and enter the solution as Zn²⁺; the electrons flow through the external circuit and Cu²⁺ ions gain them at the copper electrode, depositing copper. Ions, not electrons, carry the charge through the salt bridge: NO₃⁻ moves towards the zinc half-cell and K⁺ towards the copper half-cell. With the switch open, almost no current flows and the high-resistance voltmeter reads the EMF, +1.10 V; with the load switched in, it shows a slightly lower terminal p.d.

Standard enthalpy of combustion, ΔcH⦵

Standard enthalpy of combustion: 1 mol of methane burns completely in oxygen, so all its carbon ends up in CO₂(g) and all its hydrogen in H₂O(l), the standard state of water at 298 K. Heat is released: ΔcH⦵ = −890 kJ mol⁻¹. Use the selector for the other definitions.

Standard enthalpy of formation, ΔfH⦵

Standard enthalpy of formation: the elements in their standard states (graphite, H₂(g) and O₂(g)) rearrange into exactly 1 mol of C₂H₅OH(l), so the equation needs ½O₂, drawn as one O atom with its O₂ partner dashed. Heat is released, so ΔfH⦵ = −277 kJ mol⁻¹ is negative. Use the selector for the other definitions.

Enthalpy of neutralisation

Enthalpy of neutralisation: H⁺(aq) from the acid and OH⁻(aq) from the alkali join to form water, and the value is quoted per mole of water formed (about −57.6 kJ mol⁻¹ for a strong acid and strong alkali). Na⁺ and Cl⁻ are spectator ions and are unchanged. Use the selector for the other definitions.

Scaling and reversing equations

An enthalpy change belongs to the equation as written: for N₂(g) + 3H₂(g) → 2NH₃(g), 1 mol of N₂ and 3 mol of H₂ form 2 mol of NH₃ and ΔH = 2 × (−46) = −92 kJ mol⁻¹. Doubling the equation doubles ΔH; reversing it changes the sign. Use the selector for the other definitions.

ΔH from bond enthalpies

Bonds break one at a time, taking energy in, until only separate gaseous atoms remain; then the product bonds form, giving energy out. With the mean bond enthalpies quoted, ΔH = Σ(bonds broken) − Σ(bonds formed): for the hydrogenation of ethene 2700 − 2825 = −125 kJ mol⁻¹. The route through atoms is energy bookkeeping, not the reaction mechanism.

Gas-phase states and corrections

Enthalpy of atomisation: 1 mol of gaseous atoms is formed from the element in its standard state. For carbon, atoms must break free of the giant covalent layers of graphite, C(s, graphite) → C(g), ΔatH⦵ = +717 kJ mol⁻¹; bond-enthalpy cycles need this value because solid carbon has no bond enthalpy to use. Use the selector for the other definitions.

Mean bond enthalpy

Mean bond enthalpy: removing the four H atoms from gaseous methane one at a time takes a different energy each time (about +439, +462, +424 and +338 kJ mol⁻¹), because each step starts from a different fragment. Their total, about 1660 kJ mol⁻¹, divided by four gives a mean C–H value of about 415 kJ mol⁻¹ for methane; data-book values average over many compounds. Use the selector for the other definitions.

Required Practical 2: measuring an enthalpy change

Solution calorimetry in a polystyrene cup: the temperature is read every minute, the NaOH(aq) is added at a recorded time, and the readings continue as the mixture cools. Extrapolating the cooling line back to the mixing time gives ΔT, then q = mcΔT and ΔH per mole of water. The readings are a model, not measured data.

Required Practical 2: measuring an enthalpy change

Spirit-burner calorimetry: the burner is weighed with its cap on, heats 100.0 g of water in a copper can from 20.0 °C to 40.0 °C and is reweighed. q = mcΔT uses the mass of water, the mass of ethanol burned gives the moles, and ΔcH ≈ −8.4 × 10² kJ mol⁻¹ is less exothermic than the data-book −1367 kJ mol⁻¹ because of heat loss, heat absorbed by the can, incomplete combustion and evaporation.

Hydrated salt cycle

The hydration of anhydrous copper(II) sulfate cannot be measured reliably directly, so both solids are dissolved separately to the same final solution. Both routes start and finish in the same place, so ΔH₁ = ΔH(hydration) + ΔH₂, giving ΔH(hydration) = −66.0 − (+11.7) = −77.7 kJ mol⁻¹.

Dynamic equilibrium

In a sealed container at constant temperature, N₂O₄ molecules keep splitting into NO₂ and pairs of NO₂ keep joining back into N₂O₄, each time a suitable collision happens. The forward and reverse reactions continue at equal rates, so the amounts of N₂O₄ and NO₂ (and the brown colour) stay constant, but they are not equal: this is dynamic equilibrium.

Changing concentration

For A + B ⇌ C at equilibrium (0.40, 0.30 and 0.20 mol dm⁻³), extra A is added at fixed volume and temperature: only [A] jumps, to 0.60 mol dm⁻³. Net forward reaction then lowers [A] and [B] and raises [C] continuously until Q equals the unchanged Kc, with [A] settling above 0.40; the forward rate jumps, then forward and reverse rates become equal again. Use the buttons to add B or C, heat the mixture or add a catalyst.

Worked example: NO₂ and N₂O₄

Pushing the plunger of a sealed syringe in quickly squeezes the same mixture into half the volume, so [NO₂] doubles and the brown colour deepens at once. Then, at constant temperature, the position of equilibrium shifts left towards fewer gas moles (2 mol NO₂ → 1 mol N₂O₄): NO₂ molecules collide more often, so 2NO₂ → N₂O₄ outpaces N₂O₄ → 2NO₂ until the rates are equal again, and the colour partly fades. The new mixture is paler than just after the push but still darker than at the start, because [NO₂] stays higher.

Practical determination of Kc

Known amounts of ethanoic acid, ethanol, water and an acid catalyst are kept in a stoppered vessel at constant temperature until equilibrium. The whole mixture is transferred quantitatively, made up to 250.0 cm³ and a 25.0 cm³ aliquot is titrated with 0.200 mol dm⁻³ NaOH to the first permanent pale pink (22.00 cm³); a separate catalyst blank treated the same way needs 2.00 cm³. The titration measures acid equivalents, not ester or water, so the result is scaled to the whole mixture, the catalyst is subtracted and a mole table gives Kc = 2.75; the 250.0 cm³ is an analytical dilution, not the equilibrium volume.

Concentration and rate graphs

For A + B ⇌ C at equilibrium (0.40, 0.30 and 0.20 mol dm⁻³), extra A is added at fixed volume and temperature: only [A] jumps, to 0.60 mol dm⁻³. Net forward reaction then lowers [A] and [B] and raises [C] continuously until Q equals the unchanged Kc, with [A] settling above 0.40; the forward rate jumps, then forward and reverse rates become equal again. Use the buttons to add B or C, heat the mixture or add a catalyst.

Separation and recycling

Fresh N₂ and H₂ are compressed and join the recycled gas above the reactor, where part of the mixture reacts on the iron catalyst (N₂ + 3H₂ → 2NH₃). Downstream, a cooler condenses the ammonia, which collects in the separator and leaves as liquid, while unreacted N₂ and H₂ are recycled; a purge bleeds off some loop gas so inert argon cannot build up (close it to see the build-up). In the idealised model, 25% converts each pass, so overall conversion rises from 25% to 43.75% after two passes while the single-pass conversion stays 25%.

Yield graphs

The six illustrative values are plotted and a smooth curve drawn for each pressure. Comparing vertically at 450 °C changes only the pressure (30% → 43%), so the products have fewer gas moles; moving along the 100 atm curve changes only the temperature (48% → 30%), so the forward reaction is exothermic. A marker then trades yield against a schematic rate: low temperature gives a high yield too slowly, so a moderate temperature with a catalyst is used. Drag on the graph to move the marker.

What changes Kc?

Only temperature changes Kc for a given equation. Heating A + B ⇌ C (taking the forward reaction as exothermic) causes no jump in any concentration, but both rates rise, the reverse more, so Kc falls from 1.67 to 0.67 dm³ mol⁻¹ and the mixture shifts continuously towards A and B until Q equals the new Kc. Use the buttons to compare adding A, B or C (Kc unchanged, concentrations shift) and adding a catalyst (no shift at all).

A catalyst changes the journey, not the final ratio

Fresh N₂ and H₂ are compressed and join the recycled gas above the reactor, where part of the mixture reacts on the iron catalyst (N₂ + 3H₂ → 2NH₃). Downstream, a cooler condenses the ammonia, which collects in the separator and leaves as liquid, while unreacted N₂ and H₂ are recycled; a purge bleeds off some loop gas so inert argon cannot build up (close it to see the build-up). In the idealised model, 25% converts each pass, so overall conversion rises from 25% to 43.75% after two passes while the single-pass conversion stays 25%.

Only temperature changes Kp for a fixed equation

The six illustrative values are plotted and a smooth curve drawn for each pressure. Comparing vertically at 450 °C changes only the pressure (30% → 43%), so the products have fewer gas moles; moving along the 100 atm curve changes only the temperature (48% → 30%), so the forward reaction is exothermic. A marker then trades yield against a schematic rate: low temperature gives a high yield too slowly, so a moderate temperature with a catalyst is used. Drag on the graph to move the marker.

Metallic bonding and the non-smooth melting-point pattern

Down Group 2 the atomic radius increases and the first ionisation energy falls, but the melting points do not fall at every step: calcium (842 °C) melts above magnesium (650 °C). Each atom contributes two electrons to the delocalised electrons in the metallic lattice, and larger M²⁺ ions tend to attract them less strongly, which fits calcium to barium; crystal structure also affects melting. Use the Element buttons to compare the lattices.

Preparing and collecting an insoluble salt

Mixing barium chloride and sodium sulfate solutions precipitates white barium sulfate at once, while Na⁺ and Cl⁻ stay dissolved as spectator ions. Filtration keeps the insoluble solid on the paper and lets the solution through; washing with a little distilled water then removes the soluble ions still held in the wet residue. The residue is dried and weighed until its mass is constant.

OH⁻ and CN⁻: the two-arrow pattern

Bromoethane's polar Cδ+–Brδ− bond is attacked from the side opposite Br by a nucleophile, an electron-pair donor. One curly arrow goes from the nucleophile's lone pair to the δ+ carbon, and a second goes from the C–Br bond to Br, which leaves as Br⁻. Hydroxide (attacking through O) gives ethanol, and cyanide (attacking through C) gives propanenitrile, CH₃CH₂C≡N. The total charge is −1 on both sides; this one-step mechanism is known as SN2.

The three arrows of elimination

When 2-bromopropane is heated with ethanolic KOH, the hydroxide ion acts as a base: a curly arrow goes from a lone pair on its oxygen to an H on a carbon next to the C–Br carbon, a second goes from that C–H bond to the C–C bond to form C=C, and a third goes from the C–Br bond to Br. The products are propene, water and Br⁻, with a total charge of −1 on both sides.

Show that chlorine is regenerated

UV in the stratosphere breaks a C–Cl bond in CCl₂F₂ homolytically, giving •CClF₂ and Cl•; this initiation only supplies the radical and sits outside the cycle. Cl• + O₃ → ClO• + O₂ consumes Cl•, then ClO• + O₃ → 2O₂ + Cl• regenerates it, so adding the steps cancels Cl• and ClO• and leaves 2O₃ → 3O₂. Choose the second step to compare the supplied alternative ClO• + O → Cl• + O₂ (net O₃ + O → 2O₂).

Boiling point: compare forces between molecules

Down Group 17 the X₂ molecules have more electrons (F₂ 18, Cl₂ 34, Br₂ 70, I₂ 106) and larger, more polarisable electron clouds. An instantaneous dipole in one molecule induces an opposite dipole in a neighbour, and the two attract: a London force between molecules, which is stronger for the larger molecules. More energy is needed to separate them, so boiling points rise (85, 239, 332 and 457 K). Boiling separates whole molecules; the covalent X–X bond inside each is not broken.

Boiling point: compare forces between molecules

Drawn to one scale, the halogen molecules from F₂ (18 electrons) to I₂ (106) have larger outer electron clouds that are more easily distorted. In each pair an instantaneous dipole in one molecule induces a dipole in its neighbour, and both the shifts and the attraction grow down the group, so more energy is needed to separate the molecules: boiling points rise from 85 K to 457 K. At 298 K fluorine and chlorine are gases, bromine is a liquid and iodine is a solid.

Why ammonia is a useful second test

Acidify a fresh portion with dilute nitric acid to remove carbonate, then add silver nitrate: chloride gives a white AgCl precipitate, bromide a cream AgBr precipitate and iodide a yellow AgI precipitate. Because these colours are hard to tell apart, ammonia confirms the identity: AgCl dissolves in dilute ammonia to form a colourless solution (as [Ag(NH₃)₂]⁺), AgBr dissolves only in concentrated ammonia, and AgI does not dissolve. AgF is soluble, so fluoride gives no precipitate.

Counting atoms in a skeletal formula

Butan-2-ol, CH₃CH(OH)CH₂CH₃, changes between three representations. The displayed formula shows every atom and bond, so all ten hydrogens can be counted: 3 + 1 + 2 + 3 on carbon plus the one on oxygen. The structural formula groups each carbon with its hydrogens, and the skeletal formula keeps only the carbon framework: every line end and corner is a carbon, hydrogens on carbon are implied, and the OH, including its H, is still drawn.

CIP: choose one priority group at each end

At each carbon of the C=C, compare the atomic numbers of the two atoms directly attached: in CH₃C(Br)=C(Cl)CH₂CH₃, Br (35) beats C (6) on the left and Cl (17) beats C (6) on the right. If the first atoms tie, compare the atoms attached to them, so ethyl (C,H,H) beats methyl (H,H,H). Priority groups on the same side of the C=C axis give Z and on opposite sides give E; twisting one carbon about the double bond would break the π bond, so rotation is restricted. Choose an example, or drag the molecule to try the twist.

Isomerism can change physical properties

In 1,2-dichloroethene each carbon of the C=C carries Cl (atomic number 17) and H (1), so Cl is the priority group at both ends. In the Z isomer the two C–Cl bond dipoles combine to give a net molecular dipole, adding permanent dipole–dipole attractions; in the E isomer the symmetric arrangement makes them cancel. This one example does not show that every Z isomer has the higher boiling point.

Recognise the electron-pair donor and acceptor

Bromoethane's polar Cδ+–Brδ− bond is attacked from the side opposite Br by a nucleophile, an electron-pair donor. One curly arrow goes from the nucleophile's lone pair to the δ+ carbon, and a second goes from the C–Br bond to Br, which leaves as Br⁻. Hydroxide (attacking through O) gives ethanol, and cyanide (attacking through C) gives propanenitrile, CH₃CH₂C≡N. The total charge is −1 on both sides; this one-step mechanism is known as SN2.

Work backwards from a name

To draw 3-methylpentan-2-ol, read the name in parts: “pent” gives a five-carbon parent chain, “an” means single C–C bonds, “-2-ol” puts OH on carbon 2 and “3-methyl” puts a CH₃ branch on carbon 3. Each carbon is then filled with hydrogens until it has four bonds: C1 and C5 are CH₃, C2 and C3 are CH and C4 is CH₂. The check gives six carbons, fourteen hydrogens and one oxygen, C₆H₁₄O, or CH₃CH(OH)CH(CH₃)CH₂CH₃; the skeletal version keeps the OH label and draws the methyl as a short branch.

Gas volume and mass loss

Calcium carbonate chips are tipped into excess hydrochloric acid only after the flask is sealed, and the stopwatch starts as they mix. Reading the syringe every 5 s gives a volume–time curve: the gradient of the tangent at t = 0 is the initial rate, and the curve levels off when the limiting calcium carbonate is used up. Temperature and pressure are kept constant because gas volume depends on both.

Required Practical 3: thiosulfate and acid

Sodium thiosulfate and hydrochloric acid are warmed separately in a water bath, mixed in a conical flask standing on a paper cross, and timed from the moment the acid is added until the cross can no longer be seen from above. The thermometer reads the mixture itself at the start and the end, and the mean is recorded; volumes, concentrations, flask, cross, lighting and viewing position stay the same each run. The clouding is time-lapsed (the stopwatch shows real seconds); use the selector to vary the temperature or the thiosulfate concentration at a fixed total volume.

Gas volume and mass loss

Calcium carbonate chips are tipped into excess dilute hydrochloric acid on a top-pan balance, the neck is plugged loosely with cotton wool and the stopwatch is started. CO₂ escapes through the cotton wool, which holds back acid spray, so the mass falls; reading the balance every 5 s gives a mass-lost curve that levels off at 0.22 g, the mass of CO₂ from 0.500 g of CaCO₃. The method is unsuitable for very soluble gases, or for very light gases such as H₂ whose mass loss is tiny compared with the balance resolution.

Processing RP3 results

Four runs at 20, 30, 40 and 50 °C are shown side by side; each stops when the same amount of sulfur hides the cross, so 1000/t is a relative rate. The values are plotted against temperature and a smooth curve of best fit is drawn, rising more steeply at higher temperature rather than in a straight line. Reading across from 1000/t = 35 meets the curve between 30 °C and 40 °C, and t = 1000 ÷ 35 = 28.6 s; the rate at 50 °C is 64 ÷ 13 = 4.9 times that at 20 °C.

Count environments, not just atoms

Equivalent carbons have the same chemical surroundings and give one signal in a proton-decoupled ¹³C spectrum. Turning propanone over about its mirror plane swaps its two CH₃ carbons, so it has three carbons but two environments; the two methyls of butan-2-one differ, giving four, 2-methylpropanal has three and benzene's six ring carbons are all equivalent. Shifts are illustrative values within the data-booklet ranges, TMS is the δ = 0 reference, and line heights are not carbon counts.

The n + 1 rule and its limits

In the simple n + 1 model, protons with n equivalent neighbouring H on the adjacent carbon give n + 1 lines. Each neighbouring H splits every line into two equal halves, so overlapping branches add up as Pascal's triangle: a singlet, a 1 : 1 doublet, a 1 : 2 : 1 triplet and a 1 : 3 : 3 : 1 quartet. In the schematic spectrum (shifts arbitrary) each whole multiplet is one environment, and its integration is the total area of the multiplet, not the number of lines.

Opposite rotations under the same conditions

In a polarimeter, plane-polarised monochromatic light passes through a tube of the sample, and the observer turns the analyser until the light is brightest: the angle turned is the observed rotation. A pure enantiomer turns the plane of polarisation progressively along the tube, and its mirror image turns it by the same angle in the opposite direction under the same conditions. In a racemate the equal and opposite rotations cancel, and an achiral solvent does not rotate the plane, so both read 0° (the 40° shown is illustrative).

Opposite rotations under the same conditions

Light is a transverse wave: in unpolarised light the electric field vibrates in all planes at right angles to the direction of travel. A polariser lets through only vibrations in one plane, giving plane-polarised light. A second filter, the analyser, passes all of this light when its axis is parallel to the polariser's and none when it is crossed at 90°; in between, the brightness falls smoothly.

Two equally accessible faces can make a racemate

The cyanide ion from KCN attacks ethanal through its carbon: one curly arrow goes from the lone pair on the C of :C≡N⁻ to the δ+ carbonyl carbon and another from the C=O π bond to O, giving CH₃CH(O⁻)CN, which H⁺ from dilute acid protonates to 2-hydroxypropanenitrile. Because the carbonyl group is planar, a red CN⁻ attacking from above and a blue CN⁻ attacking from below are equally likely (50% each). The two products, with their CN groups kept red and blue, are non-superimposable mirror images, so equal amounts of the two optical isomers form: a racemate.

Cancellation does not make each molecule achiral

A racemate contains equal amounts of the two enantiomers. Their equal and opposite rotations cancel, so the plane of polarisation leaves the tube unchanged and the analyser gives maximum brightness at 0°, although every molecule is still chiral. An achiral solvent also reads 0°, so a zero rotation alone does not identify a racemate; switch samples to compare the pure enantiomers.

Worked example: three labelled candidates

Three samples are known to be propanal, propan-2-ol and propanoic acid. Fresh portions of all three are tested with sodium carbonate: only C effervesces, and the gas turns limewater milky, so C is the acid. Fresh portions of A and B are warmed with Tollens’ reagent: A gives a silver mirror, so A is propanal, and B then reduces warm acidified dichromate from orange to green. Testing with dichromate first would not separate A from B, and the conclusions hold only for this candidate list.

Related analysis: a covalent halogen is not already a halide ion

A halogenoalkane must release X⁻ by hydrolysis before silver ions can precipitate AgX. Warmed with aqueous silver nitrate and ethanol, 1-chlorobutane gives white AgCl, 1-bromobutane cream AgBr and 1-iodobutane yellow AgI; the iodo compound responds first and the chloro compound last, so the delay reflects the organic hydrolysis as well as the detection step.

The core RP6 tests

Each RP6 test is shown with a positive and a negative tube. Shaken with bromine water, an alkene decolourises the orange-brown colour; warm acidified dichromate turns from orange to green with a primary or secondary alcohol but not with a tertiary one; warm Tollens’ reagent gives a silver mirror and warm Fehling’s a brick-red Cu₂O precipitate with an aldehyde but not a ketone; and sodium carbonate effervesces with a carboxylic acid, the gas turning limewater milky. Each card also states the limit of the conclusion.

Worked four-step route: ethanol to propylamine

A map of the AQA functional-group interconversions, with the four-step route from ethanol to propylamine traced across it: dehydration with concentrated phosphoric acid and heat gives ethene, HBr adds to give bromoethane, KCN in aqueous ethanol under reflux gives propanenitrile, and H₂ with a nickel catalyst reduces it to propylamine. The two original carbons stay orange and the cyanide carbon is blue, so the extra carbon can be followed into CH₂NH₂. Hover over or tap any arrow on the map for its reagents, conditions and planning note.

Reflux keeps volatile material in the reaction

In reflux, the vapour from the boiling mixture condenses in a vertical condenser and runs back into the flask. This lets the mixture be heated for a long time without losing volatile reactants or product. Water enters the condenser jacket at the bottom and leaves at the top. The top of the condenser is left open, because a sealed system would build up pressure when heated.

Washing and separating remove different impurities

A sodium carbonate wash releases CO₂ as it neutralises acidic impurities, so the inverted funnel is vented often through the tap, pointing away from people. On standing, droplets rise or sink and coalesce until the layers separate; the layer order shown applies to ethyl ethanoate over aqueous carbonate and depends on the densities of the liquids used. Removing the stopper before draining lets air in so the lower layer runs off freely.

Recrystallisation: exploit a change in solubility

Hot solvent is added a little at a time until the crude solid just dissolves, so the hot solution is close to saturated. As it cools slowly the product's solubility falls and crystals grow, and an ice bath lowers it further; the small amount of soluble impurity stays in the mother liquor. The graph uses illustrative solubility values: with too much solvent the solution only becomes saturated at a lower temperature and more product stays dissolved, so recovery is lower.

Why suction filtration needs a seal

A Büchner funnel with a flat perforated plate sits in a rubber collar in the neck of a thick-walled side-arm flask, whose side arm leads to the vacuum source; the stem opens inside the flask. The filter paper lies flat over every hole and is wetted so that it seals, then reduced pressure draws the mother liquor through while the crystals stay on the paper. The crystals are washed with a little ice-cold solvent, air is drawn through, and they are dried fully before weighing, because retained solvent would make the yield too high.

Explain the broad pattern without inventing a smooth trend

Na₂O and MgO are giant ionic lattices: melting must overcome strong attractions between oppositely charged ions throughout the solid, and MgO melts higher because Mg²⁺ is smaller and more highly charged than Na⁺. SiO₂ is a continuing giant covalent network (each Si bonded to four O, each O to two Si), so many strong Si–O bonds must break. P₄O₁₀, SO₂ and SO₃ are treated as simple molecular: melting overcomes only the weaker attractions between molecules, not the covalent bonds inside them, so they melt at much lower temperatures.

All the acidic hydrogens are bonded to oxygen

In H₃PO₄ (O=P(OH)₃), H₂SO₃ (O=S(OH)₂, with a lone pair on S) and H₂SO₄ (O=S(=O)(OH)₂) every acidic hydrogen is bonded to oxygen. Each dissociation removes H⁺ from an O–H group, leaving O⁻, so the charge falls by one at each step; a whole OH group is never removed. In the fully deprotonated phosphate, sulfite and sulfate ions the double bond is not fixed on one oxygen: the equivalent resonance contributors give equal X–O bonds, with the charge shared by the O atoms. PO₄³⁻ and SO₄²⁻ are tetrahedral; SO₃²⁻ is trigonal pyramidal.

Explain the two dips using the electron removed

First ionisation energy rises across Period 3 because the nuclear charge increases while the electron is removed from the same shell with similar shielding. It dips at Al, whose electron is removed from the higher-energy 3p sub-shell rather than 3s, and at S, where repulsion between the two electrons paired in one 3p orbital makes one of them easier to remove. The Period selector shows the matching Be → B and N → O dips in Period 2.

First ionisation energy: the general trend

First ionisation energy: each atom in one mole of gaseous atoms loses one electron, pulled away from the attraction of the nucleus, so the change is endothermic (Na: +496 kJ mol⁻¹). The second ionisation energy starts from gaseous 1+ ions and is larger, because the electron is removed from a positive ion (Mg: +1451 kJ mol⁻¹, against +738 for IE₁).

Why P₄, S₈, Cl₂ and Ar are much lower

Across Period 3 the structure changes, so no single explanation fits the melting points. Na, Mg and Al are giant metallic lattices, held by attraction between positive ions and delocalised electrons; silicon is a giant covalent network in which many strong covalent bonds must break, giving the peak; P₄, S₈ and Cl₂ melt when London forces between molecules are overcome (more electrons, stronger forces: S₈ > P₄ > Cl₂), and argon's separate atoms melt lowest. The covalent bonds inside each molecule stay intact.

Draw the smallest correct repeating section

Each monomer has a reactive group at both ends. At every new link the OH of a –COOH group and an H from –OH or –NH₂ leave as H₂O, forming an ester link –C(=O)–O– (Terylene) or an amide link –C(=O)–NH– (nylon 6,6 and Kevlar), and every amide N keeps one H. The repeat unit is bracketed with bonds passing through both brackets and n outside; the nylon 6,6 diacid has six carbons, four CH₂ plus two carbonyl carbons.

Amino acids can polymerise without a second monomer type

A molecule with a reactive group at each end can polymerise on its own. Alanine's –NH₂ of one molecule reacts with –COOH of the next to give amide links and the repeat unit –NH–CH(CH₃)–C(=O)–; 2-hydroxypropanoic acid gives ester links and –O–CH(CH₃)–C(=O)–. The methyl group stays as a side chain, and H₂O is lost at each link.

Separate forces between chains from bonds within chains

In nylon 6,6 each amide C=O on one chain faces an N–H on the next, so hydrogen bonds form between a lone pair on the carbonyl O and the δ+ H on N, near-linear at H; the solid lines are covalent bonds within each chain. Kevlar's rigid, flat chains form hydrogen-bonded sheets that stack closely. A polyester such as PET has polar C=O groups but no N–H or O–H in its repeat unit, so its chains attract by permanent dipole–dipole and London forces rather than hydrogen bonds.

Use the axis scale when finding Ea

Each rate constant is converted to 1/T and ln k and plotted; the best-fit line is straight, with gradient −Ea/R and intercept ln A. A large triangle between two points read from the line gives a gradient of −2.88 ÷ (4.00 × 10⁻⁴ K⁻¹) = −7200 K, so Ea = 7200 × 8.31 = 59 832 J mol⁻¹ = 59.8 kJ mol⁻¹. Zooming out to 1/T = 0 shows the intercept far from the data, and warmer runs lie further left. Switch the horizontal axis to 1000/T to see the gradient fall by a factor of 1000.

An iodine clock measures a fixed small change

The main reaction makes iodine steadily, but a small fixed amount of thiosulfate turns each I₂ straight back into iodide, so the starch stays colourless; only when the thiosulfate runs out does free iodine turn it blue-black and the clock stop. 2.00 × 10⁻⁵ mol of thiosulfate sets the threshold at 1.00 × 10⁻⁵ mol I₂, which is 2.00 × 10⁻⁴ mol dm⁻³ in 0.0500 dm³, so a 40.0 s run gives a mean rate of 5.00 × 10⁻⁶ mol dm⁻³ s⁻¹. A run with halved [I⁻] reaches the same threshold at half the slope, taking 80.0 s (illustrative times).

Instantaneous and initial rate

As one reactant A is used up (other concentrations and temperature fixed), its concentration–time curve is drawn and a tangent rides along it; the tangent's gradient magnitude is the rate, plotted against [A] on the second graph. Zero order gives a straight concentration–time line and a horizontal rate line; first order a curve with constant half-lives (30, 30, 30 s) and a straight rate line through the origin; second order a curve whose half-lives double and an upward-curving rate line. All three share the same first half-life so the shapes can be compared.

Oxidising and reducing agents

Separating the two half-reactions into a cell makes the electron transfer visible. Each zinc atom loses two electrons (0 → +2), so zinc is the reducing agent; each Cu²⁺ ion gains two electrons (+2 → 0), so Cu²⁺ is the oxidising agent. Every electron lost by zinc is gained by a copper(II) ion.

Construct a cycle that conserves matter

A Born–Haber cycle built one arrow at a time: the enthalpy of formation (the direct route) first, then atomisation, ionisation energies and electron affinities, with every particle and electron listed on each level, and finally the lattice formation enthalpy, found by Hess’s law. Up arrows are endothermic and down arrows exothermic; MgCl₂ needs IE₁ + IE₂ and two first electron affinities of chlorine, while MgO needs both electron affinities of oxygen. Values are typical data-book figures in kJ mol⁻¹, rounded.

Worked example: a fictional MX₂ dataset

The fictional MX₂ dataset from the worked example, built as a cycle. The X–X bond dissociation (+220) is used once because it makes both X atoms, M²⁺ needs IE₁ + IE₂, and the two first electron affinities act on two separate X atoms (2 × −330). The known steps total +1280, so the lattice formation enthalpy is −680 − 1280 = −1960 kJ mol⁻¹ (lattice dissociation +1960 kJ mol⁻¹).

Define the change, not just its name

Choose a definition to see the change it describes, particle by particle: a gaseous sample beside one magnified particle for the ionisation energies and electron affinities, and ions assembling into or leaving a lattice for lattice formation and dissociation. Each shows the equation with state symbols, the one-mole basis and the sign on an enthalpy level diagram with its reason. Values are typical data-book figures in kJ mol⁻¹.

Define the change, not just its name

Atomisation forms one mole of gaseous atoms from the element in its standard state, shown for graphite (+717 kJ mol⁻¹); for chlorine that means ½Cl₂(g) → Cl(g). Bond dissociation breaks one mole of a specified gaseous bond, and formation makes one mole of a compound from its elements: use the selector to compare them.

Count arrangements, then use data

A substance heated steadily: particles vibrate about fixed positions in the solid, move past each other in the liquid and move freely through the whole container as a gas. Entropy rises gradually within each state and jumps at melting and, by more, at boiling, as the number of ways of arranging the particles and their energy increases; for the same substance S(solid) < S(liquid) < S(gas). The graph shows the qualitative shape only.

Worked entropy calculation

CaCO₃(s) → CaO(s) + CO₂(g): a gas is released from a solid, so entropy is expected to increase. With the page’s rounded values, ΔS = (40 + 214) − 93 = +161 J K⁻¹ mol⁻¹, positive as predicted.

Extract ΔH and ΔS from a graph

ΔG = ΔH − TΔS against T is a straight line with intercept ΔH and gradient −ΔS: the page’s line through (300, −24.0) and (500, −12.0) has ΔH = −42.0 kJ mol⁻¹ and ΔS = −60.0 J K⁻¹ mol⁻¹ and crosses ΔG = 0 at 700 K, so the reaction is feasible (ΔG ≤ 0) below 700 K. Drag the marker to set the temperature and see ΔH and −TΔS add up to ΔG, or choose another reaction. The model assumes constant ΔH and ΔS; the intercept is an extrapolation, not a measurement at 0 K.

How covalent character develops

In the perfect ionic model both ions are spheres; in real compounds a small, highly charged cation attracts the anion’s electrons and distorts its cloud towards itself, increasing electron density between the ions without changing the number of electrons. Choose ions to compare: the distortion is greatest for a small, highly charged cation with a large anion. Born–Haber lattice enthalpies more negative than the perfect-ionic-model values are evidence of this covalent character (illustrative salts P and Q).

Two hypothetical steps give the actual change

Enthalpy of solution at the particle level: an NH₄NO₃ lattice breaks up in water and each ion becomes surrounded by water molecules. For ammonium nitrate the overall change is endothermic, so the solution cools; other salts dissolve exothermically.

Gas ions become aqueous ions

Enthalpy of hydration: a gaseous ion enters water and polar water molecules turn to face it, δ− oxygen towards the cation (Mg²⁺) and a δ+ hydrogen towards the anion (Cl⁻). Forming these attractions releases energy, so hydration enthalpies are negative; each ion has its own value, more negative for a smaller or more highly charged ion.

Worked example: keep the coefficients

A Hess cycle for the enthalpy of solution with the page’s illustrative data: separate the lattice into gaseous ions (lattice dissociation +2100), hydrate the M²⁺ ion (−1500) and both X⁻ ions (2 × −340), so ΔsolH = −80 kJ mol⁻¹. The gaseous-ion route is an accounting device, not what happens in the beaker; the second example (MX: +850, −410, −390) gives +50 kJ mol⁻¹.

Count donor atoms on each ligand

Each ethanedioate (C₂O₄²⁻) and each en (H₂NCH₂CH₂NH₂) donates two lone pairs to the same metal ion and closes a five-membered chelate ring; EDTA⁴⁻ wraps around and donates six (two N, four O). Compare with six monodentate waters: all four complexes have coordination number 6. Drag to rotate; the wedge-and-dash view adds brackets and the overall charge.

Coordination number narrows the possibilities

Each ligand donates a lone pair from its donor atom into a vacant site on the metal ion. Six coordinate bonds give an octahedral complex (90° and 180°; distorted for [Cu(NH₃)₄(H₂O)₂]²⁺, with longer axial Cu–O bonds), four give tetrahedral [CoCl₄]²⁻ (109.5°) or square planar Pt(NH₃)₂Cl₂ (90°), and two give linear [Ag(NH₃)₂]⁺ (180°).

Distinguish geometrical and optical isomers

Cis and trans [Co(NH₃)₄Cl₂]⁺ have the chloride ligands at 90° and 180°; cisplatin and its trans isomer differ in the same way in a square plane. The two forms of [Cr(en)₃]³⁺ are mirror images, each en drawn as one connected ligand. Choose “Try to superimpose” to see that no rotation makes either pair identical: at most one Cl lines up, and when all six N atoms of the en pair line up the chelate rings twist the other way.

Read the curve in the right direction

Two matched runs of acidified manganate(VII) with oxalate: A with no Mn²⁺ at the start, B with a little Mn²⁺ added. A's [MnO₄⁻] falls slowly at first, then steeply as the product Mn²⁺ catalyses the reaction, then levels off as MnO₄⁻ is used up; the steepest (most negative) gradient is the fastest rate. Added Mn²⁺ shortens the slow start. The curves are schematic predictions, not measured data.

Adsorb, react, desorb

Iron catalysing N₂(g) + 3H₂(g) ⇌ 2NH₃(g): the gases adsorb onto active sites, bonding to the surface weakens the N≡N and H–H bonds until they break, N–H bonds form between atoms held on the surface, and NH₃ desorbs, freeing the sites. A sulfur atom bonded strongly to one site poisons it, so molecules cannot adsorb there. Schematic: a simplified picture of the surface steps.

Copper and cobalt do not give identical formulae

Copper(II) keeps two water ligands: four NH₃ donate N lone pairs and two H₂O donate O lone pairs, so [Cu(NH₃)₄(H₂O)₂]²⁺ still has coordination number 6 (distorted octahedral, longer axial Cu–O bonds). Cobalt(II) can take six NH₃. Neutral ligands leave the overall charge equal to the metal ion's charge.

The same ligand idea operates in haemoglobin

A schematic of the iron site in haemoglobin. Fe²⁺ sits in the plane of a porphyrin ring, one multidentate ligand donating four N lone pairs; a histidine N from the globin protein bonds below, and O₂ bonds reversibly at the sixth site, arriving and leaving in turn. Choose CO to see it take the same site through its C atom: it binds more strongly, so passing O₂ molecules cannot bind there.

Use the appropriate ionic equation

Nine neutralisations, from 1 : 1 to 2 : 3 and 3 : 2. Each H⁺ from the acid joins one OH⁻ from the base to make one water molecule, so the H⁺ donated, the OH⁻ used and the water formed always match, and the ions left over make the salt. Matching the H⁺ per acid molecule with the OH⁻ per base formula unit gives the coefficients in the balanced equation; H₃PO₄ is a weak acid and Al(OH)₃ an insoluble base, so those steps show the reacting ratios for complete neutralisation.

Make a known solution quantitatively

One acid sample followed through the whole method: weigh by difference, dissolve in deionised water, transfer through a funnel and rinse the beaker, rod and funnel so the clinging HCl reaches the 250 cm³ volumetric flask, make up to the mark and invert, then pipette a 25.0 cm³ aliquot and titrate it with NaOH. The 7 HCl found in the aliquot scale up by 250.0 / 25.0 = 10 to the 70 in the original sample. Use the numbered steps to jump to any stage.

Titrate with a consistent end point

Hydrochloric acid runs from the burette into a 25.0 cm³ aliquot of sodium hydroxide solution containing methyl orange, while the flask is swirled on a white tile. Each addition turns the indicator red where the acid lands and the colour fades as it mixes in. Near the end point the acid is added dropwise, and the end point is the first permanent orange; the bottom of the meniscus is read at eye level to 2 d.p.

Titrate with a consistent end point

A particle model of the end point. Hydrochloric acid drips from the burette one drop at a time into 12 NaOH formula units with phenolphthalein; each H⁺ reacts with an OH⁻ to form water while Na⁺ and Cl⁻ stay as spectator ions. The solution stays pink while any OH⁻ remains and turns colourless once all 12 have reacted: 12 HCl : 12 NaOH = 1 : 1. In this model the colour changes exactly at equivalence; in a real titration a suitable indicator changes within the steep pH change near equivalence, so the end point is very close to it.

Ka measures equilibrium dissociation at a given temperature

Fluorine is very electronegative, so it pulls the C–F bonding pair towards itself, and the pull is passed along the σ bonds (a negative inductive effect). When H⁺ is lost, this spreads out the negative charge of the ion, so the ion is more stable, more of the acid dissociates and the pKa is lower: CH₃COOH 4.76, CH₂ClCOOH 2.87, CH₂FCOOH 2.59, CHF₂COOH 1.24, CF₃COOH 0.23 (data-book values, 298 K).

Molecular size and branching

Each extra CH₂ in an unbranched alkane adds 8 electrons and more contact with neighbouring chains, so at any instant more induced-dipole attractions (London forces) act between the molecules: boiling points rise from 112 K for methane to 447 K for decane and volatility falls. 2,2-Dimethylpropane has the same 42 electrons as pentane, but its compact shape touches its neighbours over a smaller area, so it boils lower (283 K against 309 K).

Different positions and further substitution

The same chain reaction under the same UV lamp, with chlorine in excess (CH₄ : Cl₂ = 1 : 3). Radicals are still only a small fraction of the molecules, but more Cl₂ absorbs more UV, so more Cl• are made over the run, and as CH₃Cl builds up a Cl• increasingly takes its H from an already-chlorinated molecule. Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄, so when the Cl₂ is used up the mixture is mainly polysubstituted.

Different positions and further substitution

Methane and chlorine (CH₄ : Cl₂ = 5 : 1) are sealed together under UV light. UV splits Cl₂ into Cl• radicals; in the two propagation steps a Cl• takes an H atom to form HCl and a carbon radical, which then takes a Cl atom from Cl₂ and regenerates Cl•. Because nearly every Cl• meets CH₄ rather than a chlorinated molecule, most of the product is CH₃Cl: excess methane favours monosubstitution but does not guarantee a pure product.

Consume one radical and regenerate another

UV light supplies the energy for homolytic fission of Cl–Cl: one fish-hook arrow goes to each atom, giving two chlorine radicals, Cl•, each with an unpaired electron and no charge. In propagation, Cl• removes a hydrogen atom from methane to give HCl and a methyl radical, CH₃•, which then reacts with Cl₂ to form chloromethane and regenerate Cl•, so the chain continues; adding the two steps cancels CH₃• and Cl• to give CH₄ + Cl₂ → CH₃Cl + HCl. In termination, two radicals combine (giving Cl₂, CH₃Cl or ethane), which removes chain carriers and explains why ethane can appear as an impurity.

Consume one radical and regenerate another

Abstraction from the middle carbon of propane gives CH₃CH•CH₃, with the dot on the carbon missing H; reaction with Br₂ then gives 2-bromopropane and regenerates Br•. Choose C1 to see abstraction from an end carbon, which gives CH₃CH₂CH₂• and 1-bromopropane: two positional isomers from one alkane.

Two different groups at each end

At each carbon of the C=C, compare the atomic numbers of the two atoms directly attached: in CH₃C(Br)=C(Cl)CH₂CH₃, Br (35) beats C (6) on the left and Cl (17) beats C (6) on the right. If the first atoms tie, compare the atoms attached to them, so ethyl (C,H,H) beats methyl (H,H,H). Priority groups on the same side of the C=C axis give Z and on opposite sides give E; twisting one carbon about the double bond would break the π bond, so rotation is restricted. Choose an example, or drag the molecule to try the twist.

One sigma bond and one pi bond

A C=C double bond is one σ bond plus one π bond. Use the View buttons to see the planar σ framework (about 120° at each carbon), the parallel p orbitals, and their sideways overlap into one π bond with electron density above and below the plane. Restricted rotation shows that twisting one CH₂ group removes the overlap. Drag to rotate the molecule.

Electrophilic addition of HBr

The π bond in propene's C=C is a region of high electron density, and it donates an electron pair to the Hδ+ end of polar HBr while the H–Br bonding pair moves to Br. Adding H to the terminal carbon gives the secondary carbocation CH₃CH⁺CH₃, and a lone pair on Br⁻ then forms the C–Br bond. This major pathway gives 2-bromopropane; the primary carbocation route gives 1-bromopropane as the minor product.

The double-bond carbons form the backbone

Nine monomers join one at a time: each C=C becomes a C–C single bond and continuation bonds join neighbouring units, with every substituent kept on its own carbon. Past a certain length the whole chain is no longer drawn: one repeat unit goes in brackets with n outside, for example [–CH₂–CHCl–]n for poly(chloroethene). Choose ethene, propene, chloroethene or 2-methylpropene above the figure.

Worked unfamiliar repeat unit: retain every side group

2-Methylpropene, CH₂=C(CH₃)₂, adds to the chain through its two C=C carbons only, so both methyl groups stay on the same backbone carbon and every second carbon carries two CH₃ groups and no H. The chain section joins successive units without internal brackets; the repeat unit is then written [–CH₂–C(CH₃)₂–]n.

The double-bond carbons form the backbone

Beyond the specification, this shows how PVC chains grow by radical addition polymerisation: heat splits a peroxide initiator, ROOR, homolytically into RO• radicals, and the chain-end radical then adds chloroethene molecules one at a time, regenerating a radical at the chain end after each addition. Termination joins two radicals; each fish-hook arrow shows the movement of one electron.

Excess ammonia favours the primary amine

Bromoethane is heated with excess ammonia (NH₃ : CH₃CH₂Br = 6 : 1) in ethanol in a sealed vessel. When a nitrogen lone pair meets the δ+ carbon from the side opposite Br, SN2 substitution forms the C–N bond and releases Br⁻, and another NH₃ removes H⁺ from the nitrogen, forming NH₄⁺. Bromoethane nearly always meets NH₃ before an amine, so ethylamine is the main product, although further substitution is not impossible.

Prepare a secondary or tertiary amine by following nitrogen’s bonds

The same reaction with bromoethane in excess (NH₃ : CH₃CH₂Br = 1 : 3). Ethylamine still has a lone pair on nitrogen, so, surrounded by bromoethane, it is attacked again, giving diethylamine, then triethylamine and some tetraethylammonium bromide, a quaternary salt with no N–H to lose. Each substitution also releases H⁺, which ends up on a nitrogen, so the reaction stops when no free lone pair remains, leaving a mixture of amines, mostly as their ammonium salts.

Recognise the C(=O)–N link

Joining the –COOH of one amino acid to the –NH₂ of another removes OH and H as H₂O and forms a –C(=O)–NH– amide (peptide) link, with N bonded directly to the carbonyl carbon. Glycine and alanine can give two different dipeptides, Gly–Ala and Ala–Gly, which have the same formula but different sequences; free terminal groups remain at each end.

Worked example: explain the new centre after addition

The cyanide ion from KCN attacks ethanal through its carbon: one curly arrow goes from the lone pair on the C of :C≡N⁻ to the δ+ carbonyl carbon and another from the C=O π bond to O, giving CH₃CH(O⁻)CN, which H⁺ from dilute acid protonates to 2-hydroxypropanenitrile. Because the carbonyl group is planar, a red CN⁻ attacking from above and a blue CN⁻ attacking from below are equally likely (50% each). The two products, with their CN groups kept red and blue, are non-superimposable mirror images, so equal amounts of the two optical isomers form: a racemate.

A racemic mixture can contain chiral molecules

In a polarimeter, plane-polarised monochromatic light passes through a tube of the sample, and the observer turns the analyser until the light is brightest: the angle turned is the observed rotation. A pure enantiomer turns the plane of polarisation progressively along the tube, and its mirror image turns it by the same angle in the opposite direction under the same conditions. In a racemate the equal and opposite rotations cancel, and an achiral solvent does not rotate the plane, so both read 0° (the 40° shown is illustrative).

A racemic mixture can contain chiral molecules

Light is a transverse wave: in unpolarised light the electric field vibrates in all planes at right angles to the direction of travel. A polariser lets through only vibrations in one plane, giving plane-polarised light. A second filter, the analyser, passes all of this light when its axis is parallel to the polariser's and none when it is crossed at 90°; in between, the brightness falls smoothly.

Water of crystallisation and constant mass

Hydrated magnesium sulfate is heated in a weighed crucible with the lid ajar so water vapour escapes while little solid is lost; the glassy crystals become an opaque white powder. It is cooled in a desiccator before each weighing, and heating, cooling and weighing are repeated until successive readings agree; the masses are illustrative. 2.46 g leaves 1.20 g of MgSO₄, so x = 0.0700 ÷ 0.0100 = 7; constant mass alone does not identify the product or rule out decomposition.

Water of crystallisation and constant mass

Magnesium oxidation: the crucible sits in a pipe-clay triangle on a tripod above the Bunsen burner, and the lid is raised briefly with tongs now and then to admit oxygen while limiting the loss of magnesium oxide smoke. The crucible is cooled and reweighed until the mass is constant; here 0.486 g of Mg gains 0.320 g of oxygen, a 1 : 1 amount ratio, so the formula is MgO (illustrative masses). Wear eye protection and do not stare at burning magnesium.

Three different observations test the model

Every route ends at cyclohexane. Hydrogenating one C=C releases about 120 kJ mol⁻¹, so a localised cyclohexa-1,3,5-triene would be expected to release 3 × 120 = 360 kJ mol⁻¹, but benzene releases only about 208 kJ mol⁻¹. Its hydrogenation is less exothermic, so benzene starts lower in enthalpy: about 360 − 208 = 152 kJ mol⁻¹ of stabilisation from delocalisation, compared with the hypothetical model.

Follow the electron pair, then restore delocalisation

Nitric and sulfuric acids generate the electrophile NO₂⁺, with its positive charge on N. A full-headed arrow from the benzene π system to N forms a new C–N σ bond; the intermediate is not aromatic, has an overall positive charge spread over the other five ring carbons and keeps H on the attacked carbon. The C–H bond pair, not the H, returns into the ring as H⁺ is removed, restoring delocalisation in nitrobenzene and regenerating sulfuric acid.

Worked formula translation: count connections before hydrogens

To draw 3-methylbutan-1-ol, “but” gives a four-carbon parent chain and “-1-ol” puts the OH on carbon 1, so the chain is numbered from the OH end; “3-methyl” adds a CH₃ branch on carbon 3, giving five carbons in total although the stem has four. Filling each carbon to four bonds gives 12 H including the one on oxygen, C₅H₁₂O or CH₃CH(CH₃)CH₂CH₂OH. In the skeletal version the last line ends at a labelled OH, which is not a hidden extra carbon.

Molecular, structural, displayed and skeletal formulae

Butan-2-ol, CH₃CH(OH)CH₂CH₃, changes between three representations. The displayed formula shows every atom and bond, so all ten hydrogens can be counted: 3 + 1 + 2 + 3 on carbon plus the one on oxygen. The structural formula groups each carbon with its hydrogens, and the skeletal formula keeps only the carbon framework: every line end and corner is a carbon, hydrogens on carbon are implied, and the OH, including its H, is still drawn.

Infer a structure from several properties together

Structures compared in 3D: giant ionic sodium chloride, a metal (magnesium), giant covalent diamond and graphite, and simple molecular iodine and ice. Each panel links the particles and forces to the melting point and to what can carry charge; the Bonding view shows those forces. Drag to rotate.

Draw a covalent structure from a valence-electron inventory

Coordinate bonds in 3D: NH₃ donates its nitrogen lone pair to H⁺, and H₂O donates one of oxygen's lone pairs. NH₄⁺ has four bonding pairs and no lone pair, so it is tetrahedral, while H₃O⁺ keeps one lone pair and is pyramidal; the Electron origins view shows the donated pair as two of the donor's electrons.

A shared pair attracts both nuclei

Dot-and-cross diagrams built step by step: dots are electrons from O or N, crosses from H. In H₂O each shared pair is one dot and one cross; in NH₄⁺ and H₃O⁺ the pair donated to H⁺ is two of the central atom's dots, and the brackets and + charge go round the whole ion (5 + 4(1) − 1 = 8 and 6 + 3(1) − 1 = 8 outer electrons).

Ionic bonding extends throughout a lattice

Giant ionic lattices of sodium chloride and magnesium oxide: each ion is surrounded by six ions of opposite charge, and the attraction acts in all directions throughout the lattice. Mg²⁺ and O²⁻ have larger charges and are smaller (all ions are drawn at about 40% of their ionic radii so the lattice shows), so the attractions in MgO are stronger and its melting point is much higher.

Choose the explanation that distinguishes the pair

Each extra CH₂ in an unbranched alkane adds 8 electrons and more contact with neighbouring chains, so at any instant more induced-dipole attractions (London forces) act between the molecules: boiling points rise from 112 K for methane to 447 K for decane and volatility falls. 2,2-Dimethylpropane has the same 42 electrons as pentane, but its compact shape touches its neighbours over a smaller area, so it boils lower (283 K against 309 K).

London forces, permanent dipoles and hydrogen bonds

At an instant the electron density in one I₂ molecule piles up at one end, an instantaneous dipole (δ− where it builds up, δ+ where it thins), although the molecule stays neutral. This induces a dipole in the neighbouring molecule, lined up δ+ to δ−, and the two attract: an induced dipole–dipole interaction (London force). Pulling back to six molecules and then to solid iodine shows such correlated, short-lived dipoles forming and vanishing everywhere; they are strong because each I₂ has 106 electrons in an easily polarised cloud.

Bond polarity and molecular polarity are different

Each bonding pair is drawn as dots of electron density. When the two atoms differ in electronegativity, the dots stream along the bond towards the more electronegative atom, which becomes δ− (H–Cl, and more so H–F); in H–H the pair stays evenly shared. Choose H₂O to see both O–H pairs drawn towards O in a bent molecule, giving an overall dipole, or CO₂, where two equal bond dipoles cancel.

Add bond dipoles using the geometry

Oxygen (3.44) is more electronegative than hydrogen (2.20), so the dots of both O–H bonding pairs stream towards O: O becomes δ− and each H δ+. Water is bent, so the two bond dipoles add up to an overall dipole; choose CO₂ to see two equal bond dipoles cancel in a linear molecule.

Draw the hydrogen bond to the electron pair

Each O–H bonding pair is drawn towards O, leaving the H strongly δ+ with little electron density around it. A hydrogen bond (dotted line) forms when this δ+ H lines up with a lone pair on the O of another molecule, O–H···O close to 180°; each water molecule can donate two and accept two, giving the open network of ice. Choose Why N, O or F? to see that a C–H bond is almost non-polar, so it cannot form a hydrogen bond.

Add bond dipoles using the geometry

Bond dipoles add as vectors in three dimensions. The two C=O dipoles of linear CO₂ and the four C–Cl dipoles of tetrahedral CCl₄ cancel, but bent H₂O, CHCl₃ and CH₃Cl keep a net dipole, even though the two carbon compounds are still tetrahedral. The C–H bonds are treated as almost non-polar.

Water’s unusual properties and molecular solubility

Hydrogen bonding in ice: each water molecule donates two hydrogen bonds through its δ+ H atoms and accepts two at the lone pairs on its O, each O–H···O close to 180°, so every molecule has four neighbours arranged tetrahedrally in an open hexagonal network. On melting some hydrogen bonds break and the network partly collapses, so liquid water packs more closely and is denser than ice.

Apply the same counting to unfamiliar species

Choose a shape to see its example molecule, its bonding pairs and lone pairs, and its bond angles. The faint outline shows the arrangement of all the electron regions (linear, trigonal planar, tetrahedral, trigonal bipyramidal or octahedral); lone pairs take some of its positions (equatorial with five regions, opposite each other with six), so the shape name describes only where the atoms are. Drag to rotate the molecule, or switch to Wedge and dash to see the same orientation drawn with in-plane lines, wedges and dashed wedges.

Two tests answer two different questions

Each RP6 test is shown with a positive and a negative tube. Shaken with bromine water, an alkene decolourises the orange-brown colour; warm acidified dichromate turns from orange to green with a primary or secondary alcohol but not with a tertiary one; warm Tollens’ reagent gives a silver mirror and warm Fehling’s a brick-red Cu₂O precipitate with an aldehyde but not a ketone; and sodium carbonate effervesces with a carboxylic acid, the gas turning limewater milky. Each card also states the limit of the conclusion.

Cyanide forms a new carbon–carbon bond

Cyanide from KCN attacks propanal through its carbon lone pair while the C=O π pair moves to O, giving CH₃CH₂CH(O⁻)CN; an O⁻ lone pair then takes the H of H–CN, forming 2-hydroxybutanenitrile and regenerating CN⁻. A red CN⁻ attacking the planar carbonyl from above and a blue CN⁻ attacking from below are equally likely, so the two enantiomers (shown with red and blue CN groups either side of a mirror line) form in equal amounts and the mixture has no net optical rotation.

Build the hydride mechanism in two stages

Aqueous NaBH₄ supplies the hydride ion, H⁻, which uses its lone pair to attack the δ+ carbonyl carbon while the C=O π pair moves onto oxygen; the planar carbonyl carbon becomes tetrahedral in an alkoxide intermediate with O⁻. An O⁻ lone pair then takes an H from water as that O–H bond pair returns to the water oxygen, giving the alcohol and OH⁻. Butan-2-one gives butan-2-ol, a secondary alcohol, and propanal gives propan-1-ol, a primary alcohol.

Hydrogen bonding explains two different trends

Carboxylic acid molecules hydrogen-bond using O–H as the donor and an oxygen lone pair as the acceptor, so two molecules can pair up through two hydrogen bonds; these strong intermolecular attractions contribute to relatively high boiling points. Short-chain acids also hydrogen-bond with water, which is why they dissolve. Choose Carboxylate ion to see why the conjugate base is stabilised.

Treat COOH as one functional group

When a carboxylic acid such as ethanoic acid donates H⁺, it forms a carboxylate ion, here CH₃COO⁻. In the ion the p orbitals on the carbon and both oxygens overlap sideways into one delocalised π system above and below the O–C–O plane, so the negative charge is spread over both oxygens and the two C–O bonds are the same length (0.126 nm). Use the View buttons to watch the separate p orbitals merge, or compare the localised picture and ethanoic acid itself; drag to rotate.

Equal rates, constant concentrations

In a sealed container at constant temperature, N₂O₄ molecules keep splitting into NO₂ and pairs of NO₂ keep joining back into N₂O₄, each time a suitable collision happens. The forward and reverse reactions continue at equal rates, so the amounts of N₂O₄ and NO₂ (and the brown colour) stay constant, but they are not equal: this is dynamic equilibrium.

Distinguish the imposed change from the later response

For A + B ⇌ C at equilibrium (0.40, 0.30 and 0.20 mol dm⁻³), extra A is added at fixed volume and temperature: only [A] jumps, to 0.60 mol dm⁻³. Net forward reaction then lowers [A] and [B] and raises [C] continuously until Q equals the unchanged Kc, with [A] settling above 0.40; the forward rate jumps, then forward and reverse rates become equal again. Use the buttons to add B or C, heat the mixture or add a catalyst.

Choose conditions for useful output

Fresh N₂ and H₂ are compressed and join the recycled gas above the reactor, where part of the mixture reacts on the iron catalyst (N₂ + 3H₂ → 2NH₃). Downstream, a cooler condenses the ammonia, which collects in the separator and leaves as liquid, while unreacted N₂ and H₂ are recycled; a purge bleeds off some loop gas so inert argon cannot build up (close it to see the build-up). In the idealised model, 25% converts each pass, so overall conversion rises from 25% to 43.75% after two passes while the single-pass conversion stays 25%.

Choose conditions for useful output

The six illustrative values are plotted and a smooth curve drawn for each pressure. Comparing vertically at 450 °C changes only the pressure (30% → 43%), so the products have fewer gas moles; moving along the 100 atm curve changes only the temperature (48% → 30%), so the forward reaction is exothermic. A marker then trades yield against a schematic rate: low temperature gives a high yield too slowly, so a moderate temperature with a catalyst is used. Drag on the graph to move the marker.

Retention time and peak area answer different questions

A small sample is injected into a heated inlet, where it vaporises, and an inert carrier gas sweeps it through a coiled column held in a temperature-controlled oven. Components that spend more time dissolved in the liquid film lining the column move more slowly, so each reaches the detector at its own time. Retention time is measured from injection to the maximum of each peak; the large early peak here is the solvent, not a sample component.

The baseline and solvent front define the measurement

Mixture X and standards A, B and C are developed on one silica plate in a covered chamber, with the solvent starting below the pencil baseline. In the inset, molecules keep adsorbing onto the polar silica and desorbing into the moving solvent, so the component that spends a larger share of its time in the mobile phase travels further. The front is marked at 80 mm, the spots are located under UV, and Rf = distance to spot centre ÷ distance to solvent front gives 0.25, 0.60 and 0.80: matching A and B supports, but co-migration alone does not prove, their presence in X.

Choose the reduction and oxidation, then check the stated direction

For Ag⁺ + Fe²⁺ → Ag + Fe³⁺, Fe²⁺ is oxidised at a platinum electrode and Ag⁺ is reduced to silver: E°cell = +0.80 − (+0.77) = +0.03 V. This small positive value supports feasibility under standard conditions, but it says nothing about rate, and changes in concentration could alter the favoured direction.

Worked example: sum the electrode reactions

Hydrogen supplied to the negative electrode is oxidised and oxygen supplied to the positive electrode is reduced; electrons flow through the external load while ions carry the charge through the electrolyte. In acid, H⁺ moves towards the oxygen side and water forms there; in alkali, OH⁻ moves towards the hydrogen side. Either way, adding the half-equations gives 2H₂ + O₂ → 2H₂O.

Apply the same principles to unfamiliar batteries

On discharge, lead at the negative plate and lead(IV) oxide at the positive plate both become lead(II) sulfate, using up sulfate and H⁺ ions and forming water, while electrons flow through the load from the lead plate. Charging with an external supply reverses every change, regenerating Pb, PbO₂ and the acid. With the supplied potentials, E°cell = +1.69 − (−0.36) = +2.05 V.

Why the circuit needs both electron and ion movement

Electrons travel through the external wire from the zinc electrode, where zinc is oxidised, to the copper electrode, where Cu²⁺ is reduced. The salt bridge completes the circuit by ion movement: NO₃⁻ moves towards the zinc half-cell as Zn²⁺ forms, and K⁺ towards the copper half-cell as Cu²⁺ is removed. With the switch open, the high-resistance voltmeter draws almost no current and reads the e.m.f., +1.10 V; with the load switched in, it shows a slightly lower terminal p.d.

An isolated half-cell has no directly measurable absolute potential

A half-cell's standard electrode potential is measured against the standard hydrogen electrode, assigned 0.00 V. When both members of the redox pair are dissolved ions, as in Fe³⁺/Fe²⁺, platinum provides the conducting surface without appearing in the net reaction. Choose other cells to compare metal/metal-ion half-cells and their cell potentials.

Construct the standard hydrogen electrode

The hydrogen electrode has H₂ at 100 kPa in contact with platinum and 1.00 mol dm⁻³ H⁺(aq), at 298 K for tabulated values, and is assigned 0.00 V. The polarity gives the sign: copper is the positive terminal against it, so E°(Cu²⁺/Cu) = +0.34 V, while zinc is negative, so E°(Zn²⁺/Zn) = −0.76 V.

Shells contain sub-shells; sub-shells contain orbitals

Each dot cloud shows where an electron is most likely to be found. The 1s and 2s orbitals are spherical, with 2s larger than 1s. Each 2p orbital is a dumbbell with a nodal plane through the nucleus, and the three 2p orbitals lie along x, y and z at right angles. Each orbital holds a maximum of 2 electrons with opposite spins. Drag the diagram to rotate it.

Fill lower energies first

Sub-shells fill in the order 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p, with up to two electrons per orbital. Equal-energy orbitals fill singly with parallel spins before pairing, and a pair has opposite spins, so oxygen's 2p boxes hold ↑↓, ↑, ↑. Chromium and copper are shown as observed ([Ar] 3d⁵ 4s¹ and [Ar] 3d¹⁰ 4s¹), but these exceptions are outside the OCR AS configuration assessment.

Write an equation that matches the enthalpy definition

The formation equation for ethanol forms exactly one mole of C₂H₅OH(l) from graphite, H₂(g) and O₂(g), the elements in their standard states, so ½O₂ appears: one O atom, half an O₂ molecule, shown dashed. ΔfH⦵ = −277 kJ mol⁻¹. Use the selector for the other definitions.

Use the correct one-mole definition

Enthalpy change of reaction belongs to the reacting amounts in the balanced equation: for N₂(g) + 3H₂(g) → 2NH₃(g) it is −92 kJ mol⁻¹, for 2 mol of NH₃. Formation, combustion and neutralisation each fix a different one-mole basis, shown in the other selector options with standard conditions of 100 kPa and a stated temperature, usually 298 K.

Reduce and estimate heat exchange

Insulated-cup calorimetry for 25.0 cm³ of 1.00 mol dm⁻³ HCl and 25.0 cm³ of 1.00 mol dm⁻³ NaOH: readings every minute, the alkali added at 3.0 min, and the cooling trend extrapolated back to the mixing time. With the 50.0 g of solution, q = 50.0 × 4.18 × 6.5 = 1358.5 J and ΔneutH = −54.3 kJ mol⁻¹. The readings are a model, not measured data.

Fuel calorimetry

Spirit-burner calorimetry: the burner is weighed with its cap on, heats 100.0 g of water in a copper can from 20.0 °C to 40.0 °C and is reweighed. q = mcΔT uses the mass of water, the mass of ethanol burned gives the moles, and ΔcH ≈ −8.4 × 10² kJ mol⁻¹ is less exothermic than the data-book −1367 kJ mol⁻¹ because of heat loss, heat absorbed by the can, incomplete combustion and evaporation.

Break minus make

Break minus make: for H₂(g) + Cl₂(g) → 2HCl(g), breaking H–H (436) and Cl–Cl (242) takes in 678 kJ mol⁻¹, forming two H–Cl bonds (2 × 431) gives out 862 kJ mol⁻¹, so ΔH = −184 kJ mol⁻¹. Average values assume gases; the route through separate atoms is energy accounting, not the mechanism. The selector adds N₂ + 3H₂ → 2NH₃.

Break minus make

Average bond enthalpy: even within methane, removing the four H atoms one at a time takes four different energies (about +439, +462, +424 and +338 kJ mol⁻¹), so tables quote an average. The total, about 1660 kJ mol⁻¹, gives about 415 kJ mol⁻¹ per C–H bond for methane; bond breaking is always endothermic. Use the selector for the other definitions.

Make the final solutions chemically identical

The direct route makes the hydrate, then dissolves it; the other dissolves the anhydrous salt straight to the same final solution. Same start and finish, so ΔH(hydration) + 8.0 = −52.0 and ΔH(hydration) = −60.0 kJ mol⁻¹. Both dissolving arrows point towards the common solution, even though one value is positive.

Energy can be distributed in different ways

A substance heated steadily: particles vibrate about fixed positions in the solid, move past each other in the liquid and move freely through the whole container as a gas. Entropy rises gradually within each state and jumps at melting and, by more, at vaporisation, because gas particles can occupy far more positions and spread their energy more widely. The graph shows the qualitative shape only.

Worked example: coefficients and units matter

N₂(g) + 3H₂(g) → 2NH₃(g): four moles of gas become two, so ΔS is expected to be negative; with the page’s values, ΔS = 2(195) − [190 + 3(130)] = −190 J K⁻¹ mol⁻¹ for the reaction as written. Switch to CaCO₃ decomposition, where gas forms from solids and ΔS is positive.

Worked example: explain the temperature threshold

ΔG = ΔH − TΔS against T is a straight line with intercept ΔH and gradient −ΔS. For the worked example (ΔH = +48.0 kJ mol⁻¹, ΔS = +120 J K⁻¹ mol⁻¹) the line crosses zero at 400 K and the forward change is favourable above it; with ΔH and ΔS both negative the same 400 K is an upper limit. Drag the marker to set the temperature and see ΔH and −TΔS add up to ΔG.

K constrains the equilibrium ratio

For A ⇌ B with Kc = [B]/[A] = 4, extra A at fixed temperature raises [A] from 0.10 to 0.35 mol dm⁻³ at once, so the ratio [B]/[A] falls below Kc. Net A → B then lowers [A] and raises [B] continuously until the ratio is 4 again, at [A] = 0.15 and [B] = 0.60 mol dm⁻³: the concentrations change while the constant stays fixed. The buttons also show adding B, heating and adding a catalyst.

Lose two outer-shell electrons

Down Group 2 each atom has an extra occupied shell, so the outer electrons are farther from the nucleus and more shielded. Although the nuclear charge increases, the outer electrons are attracted less strongly, so the first and second ionisation energies fall and the 2+ ion forms more easily: reactivity generally increases from magnesium to barium. Use the Element buttons to compare the atoms.

Show both electron-pair movements

Bromoethane's polar Cδ+–Brδ− bond is attacked from the side opposite Br by a nucleophile, an electron-pair donor. One curly arrow goes from the nucleophile's lone pair to the δ+ carbon, and a second goes from the C–Br bond to Br, which leaves as Br⁻. Hydroxide (attacking through O) gives ethanol, and cyanide (attacking through C) gives propanenitrile, CH₃CH₂C≡N. The total charge is −1 on both sides; this one-step mechanism is known as SN2.

Chlorine and nitrogen monoxide catalytic cycles

UV breaks a C–Cl bond in CF₂Cl₂ homolytically to release Cl•, which then catalyses ozone breakdown: Cl• + O₃ → ClO• + O₂ followed by ClO• + O → Cl• + O₂. Cl• is consumed and regenerated (the catalyst) and ClO• is formed and used up (an intermediate), so both cancel to leave O₃ + O → 2O₂. Choose NO• to see the nitrogen monoxide cycle, which gives the same overall reaction.

Diatomic molecules and London forces

Cl₂, Br₂ and I₂ have more electrons down the group (34, 70 and 106) and larger, more polarisable electron clouds, so the instantaneous dipole–induced dipole attractions (London forces) between molecules get stronger. More energy is needed to separate the molecules, so melting and boiling points rise. The covalent X–X bond is inside each molecule and is not broken on boiling.

Diatomic molecules and London forces

Drawn to one scale, Cl₂ (34 electrons), Br₂ (70) and I₂ (106) have larger, more polarisable electron clouds down the group. In each pair an instantaneous dipole induces a dipole in the neighbouring molecule, and these induced dipole–dipole interactions (London forces) get stronger, so melting and boiling points rise (boiling points 239, 332 and 457 K): chlorine is a gas, bromine a liquid and iodine a solid at room temperature.

Precipitate the silver halide

Acidify with dilute nitric acid (not HCl, which adds the chloride being tested), then add aqueous silver nitrate: white AgCl, cream AgBr and yellow AgI precipitates form. Add dilute ammonia to the precipitate: only the silver chloride solid dissolves. If the solid remains, concentrated ammonia dissolves AgBr but not AgI.

Transfer the method to a 2+ metal or a 2− ion

For MgCl₂ the route needs the first and second ionisation energies of magnesium and two first electron affinities of chlorine, one for each Cl atom; for MgO one oxygen atom gains two electrons, so both EA₁ (−141) and the positive EA₂ (+798) appear. The strongly exothermic lattice enthalpy still makes formation exothermic overall. Values are typical data-book figures in kJ mol⁻¹, rounded.

Worked example: label the route before substituting

The page’s worked example built as a cycle: with the illustrative NaCl-like data, the known indirect steps (110 + 122 + 500 − 350) sum to +382 kJ mol⁻¹, so ΔLEH = −410 − 382 = −792 kJ mol⁻¹. Up arrows are endothermic and down arrows exothermic; switch to NaCl for the same cycle with typical data-book values (ΔLEH = −787 kJ mol⁻¹).

Build the gaseous ions from the elements

Ionisation energies remove one electron from each atom or 1+ ion in one mole of gaseous particles, and electron affinities add one electron to each, shown in a gaseous sample beside one magnified particle. Removing an electron is endothermic, the first electron affinity of chlorine is exothermic, and adding a second electron to O⁻ needs energy to overcome repulsion. Values are typical data-book figures in kJ mol⁻¹.

Build the gaseous ions from the elements

Atomisation forms one mole of gaseous atoms from the element in its standard state, shown for graphite (+717 kJ mol⁻¹); for chlorine that means ½Cl₂(g) → Cl(g). Switch to bond enthalpy to see H₂(g) → 2H(g): one mole of bonds gives two moles of atoms, so check which process a value describes before halving it.

Start with the direction and the physical states

Lattice enthalpy: gaseous Na⁺ and Cl⁻ ions attract and form one mole of solid NaCl, releasing energy, so ΔLEH is negative (−787 kJ mol⁻¹ for NaCl). It is per mole of solid formed, not per mole of bonds; lattice dissociation is the reverse, positive with the same magnitude.

Dissolution has an energy cost and an energy release

Enthalpy change of solution at the particle level: an NH₄NO₃ lattice breaks up in water and each ion becomes surrounded by water molecules. For ammonium nitrate the overall change is endothermic, so the solution cools; dissolving need not be exothermic.

Water stabilises ions through ion–dipole attraction

Enthalpy change of hydration: a gaseous ion enters water and polar water molecules turn to face it, δ− oxygen towards the cation (Na⁺) and a δ+ hydrogen towards the anion (Cl⁻). These ion–dipole attractions release energy, so hydration enthalpies are negative; the ion stays an ion, surrounded and stabilised by water.

Dissolution has an energy cost and an energy release

ΔsolH = −ΔLEH + ΣΔhydH, built as a Hess cycle with the page’s worked example: ΔLEH = −780, so separating the lattice is +780, and hydration enthalpies of −390 and −370 give ΔsolH = +20 kJ mol⁻¹ (endothermic). The second example (MX₂: ΔLEH −2200, −1600 and 2 × −320) gives −40 kJ mol⁻¹; the gaseous-ion route is an accounting device.

Worked environment count: compare both ends

A ¹³C signal represents a carbon environment, not one atom. Pentan-3-one turns over about the mirror plane through its C=O and lands on itself, so its five carbons give three signals; pentan-2-one has no such symmetry and gives five, which helps distinguish the two isomers. Shifts are illustrative values within the data-sheet ranges (ethyl ethanoate uses the notes' 14, 21, 60 and 171 ppm), and the decoupled carbon lines are neither split nor used as carbon counts.

The simple n + 1 rule has a defined setting

For simple first-order coupling, a set of protons with n equivalent neighbouring H on adjacent carbon gives n + 1 lines; each neighbouring H splits every line in two, so the intensities follow Pascal's triangle up to 1 : 3 : 3 : 1 for a quartet. The CH of an isopropyl group, next to six equivalent methyl H, gives a septet, 1 : 6 : 15 : 20 : 15 : 6 : 1, while those six methyl H give a 6H doublet. Each multiplet is one environment whose total area gives its relative number of H.

Worked two-step route

A map of the Year 12 OCR route connections with worked two-step routes traced across it. Bromoethane is heated under reflux with aqueous NaOH to give ethanol, which is then refluxed with excess acidified dichromate to give ethanoic acid, keeping two carbons throughout. The other routes work backwards from a ketone to a secondary alcohol (2-bromobutane to butan-2-one) and convert propene, through mainly propan-2-ol, into propanone. Hover over or tap any arrow for its reagents and conditions.

Worked route: bromoethane to methyl propanoate

Bromoethane has only two carbons, so the route to methyl propanoate first heats it with cyanide ions in ethanol to make propanenitrile, then hydrolyses the nitrile with dilute aqueous acid under reflux to propanoic acid, and finally heats the acid with methanol and an acid catalyst. The carbon colours track their origins: two from bromoethane, one from cyanide and one from methanol. Hover over or tap any arrow on the map for its reagents and conditions.

Exploit a temperature-dependent solubility difference

Hot solvent is added a little at a time until the crude solid just dissolves, so the hot solution is close to saturated. As it cools slowly the product's solubility falls and crystals grow, and an ice bath lowers it further; the small amount of soluble impurity stays in the mother liquor. The graph uses illustrative solubility values: with too much solvent the solution only becomes saturated at a lower temperature and more product stays dissolved, so recovery is lower.

Wash impurities away without redissolving the product

A Büchner funnel with a flat perforated plate sits in a rubber collar in the neck of a thick-walled side-arm flask, whose side arm leads to the vacuum source; the stem opens inside the flask. The filter paper lies flat over every hole and is wetted so that it seals, then reduced pressure draws the mother liquor through while the crystals stay on the paper. The crystals are washed with a little ice-cold solvent, air is drawn through, and they are dried fully before weighing, because retained solvent would make the yield too high.

Assemble the apparatus for the job

In reflux, vapour from the boiling mixture condenses in a vertical condenser and runs back into the flask, so the mixture can be heated for longer without losing volatile reactants or products. Cooling water enters the jacket at the bottom and leaves at the top, and the top of the condenser stays open: never heat a sealed system.

Separate immiscible liquid layers

A sodium carbonate wash releases CO₂ as it neutralises acidic impurities, so the inverted funnel is vented often through the tap, pointing away from people. On standing, droplets rise or sink and coalesce until the layers separate; the layer order shown applies to ethyl ethanoate over aqueous carbonate and depends on the densities of the liquids used. Removing the stopper before draining lets air in so the lower layer runs off freely.

Assemble the apparatus for the job

Simple distillation carries vapour into a separate receiver. The thermometer bulb sits at the side arm, so it measures the vapour entering the condenser; cooling water enters the jacket at the bottom and leaves at the top, and the receiver is open to the air. Collect only the fraction that boils over the range expected for the product.

Explain the trend and both exceptions

First ionisation energy rises across Period 3 because the nuclear charge increases while the electron is removed from the same shell with similar shielding. It dips at Al, whose electron is removed from the higher-energy 3p sub-shell rather than 3s, and at S, where repulsion between the two electrons paired in one 3p orbital makes one of them easier to remove. The Period selector shows the matching Be → B and N → O dips in Period 2.

Worked inference: what does a large jump actually tell you?

For the invented values 590, 1150, 4920, 6490 and 8150 kJ mol⁻¹, every step rises because each electron leaves a more positive ion, but the second-to-third step is over fourfold: the third electron comes from an inner shell, closer to the nucleus and less shielded. Two electrons are removed before the jump, so the atom has two outer-shell electrons (Group 2); the data alone do not say which Group 2 element it is.

First and successive ionisation energies

First ionisation energy: one electron is removed from each atom in one mole of gaseous atoms, against the attraction of the nucleus, so the change is endothermic (Na: +496 kJ mol⁻¹). The second ionisation energy removes an electron from each gaseous 1+ ion and is larger (Mg: +1451 kJ mol⁻¹, against +738 for the first).

Period 2 and Period 3 change structure

Across Period 3 the structure changes, so no single explanation fits the melting points. Na, Mg and Al are giant metallic lattices, held by attraction between positive ions and delocalised electrons; silicon is a giant covalent network in which many strong covalent bonds must break, giving the peak; P₄, S₈ and Cl₂ melt when London forces between molecules are overcome (more electrons, stronger forces: S₈ > P₄ > Cl₂), and argon's separate atoms melt lowest. The covalent bonds inside each molecule stay intact.

Networks of strong covalent bonds

Diamond, graphite and graphene are forms of carbon. In diamond every carbon forms four tetrahedral covalent bonds in a 3D network with no delocalised electrons; in graphite and graphene each carbon forms three bonds in flat hexagonal sheets and one electron per carbon is delocalised over the sheet, so they conduct. Graphite's layers slide because only weak London forces hold them together, while the strong bonds within the layers give it a very high melting point.

Metal ions and delocalised electrons

Metallic bonding: a giant lattice of positive ions in a sea of delocalised electrons, held together by the electrostatic attraction between the ions and those electrons. Magnesium gives two delocalised electrons per atom and forms smaller Mg²⁺ ions, so its attraction is stronger and it melts at a higher temperature than sodium. Drag to rotate.

Replace the oxygen linkage with the correct nitrogen linkage

A diamine and a dicarboxylic acid form amide links –C(=O)–NH– with H₂O lost at each link, and every N keeps one H. Glycine alone gives [–NH–CH₂–C(=O)–]ₙ because it has both groups. With a diacyl chloride the same amide links form but HCl is eliminated instead of water.

Worked construction: keep every atom in the backbone

Ethane-1,2-diol and butanedioic acid form a polyester: at each link the OH of –COOH and the H of –OH leave as H₂O and the alcohol O bonds to the carbonyl carbon, giving –C(=O)–O–. The repeat unit [–O–CH₂–CH₂–O–C(=O)–CH₂–CH₂–C(=O)–]ₙ has bonds passing through both brackets, and terminal H and OH are not put into every repeat.

Initial-rate comparisons and clock reactions

The main reaction makes iodine steadily, but a small fixed amount of thiosulfate turns each I₂ straight back into iodide, so the starch stays colourless; only when the thiosulfate runs out does free iodine turn it blue-black and the clock stop. 2.00 × 10⁻⁵ mol of thiosulfate sets the threshold at 1.00 × 10⁻⁵ mol I₂, which is 2.00 × 10⁻⁴ mol dm⁻³ in 0.0500 dm³, so a 40.0 s run gives a mean rate of 5.00 × 10⁻⁶ mol dm⁻³ s⁻¹. A run with halved [I⁻] reaches the same threshold at half the slope, taking 80.0 s (illustrative times).

Read the axes before deciding the order

A concentration–time graph follows one reaction mixture; the gradient of its tangent at any moment gives the rate, which is plotted against [A] to build the rate–concentration graph. Zero order: straight concentration–time line, horizontal rate line. First order: rate line through the origin with gradient k, and constant half-lives — 0.080 to 0.040 and 0.040 to 0.020 mol dm⁻³ each take 120 s, so k = ln 2/120 = 5.78 × 10⁻³ s⁻¹. Second order: upward-curving rate line and half-lives that double.

Worked example: read the straight-line form

Each rate constant is converted to 1/T and ln k and plotted, and a best-fit straight line is drawn: y = ln k, x = 1/T, gradient −Ea/R, intercept ln A. A large triangle on the line gives a gradient of −7200 K, so Ea = 7200 × 8.314 = 59 860.8 J mol⁻¹ = 59.9 kJ mol⁻¹, and the intercept 24.0 gives A = e²⁴ = 2.65 × 10¹⁰. The line gives ln k = 0 at 300 K and 1.50 at 320 K: warming moves left on the graph. Switch to 1000/T to see the gradient change by a factor of 1000.

Choose a method that measures a useful change

A gas syringe gives repeated volume readings as hydrogen is made from magnesium and excess hydrochloric acid. The joints are gas-tight, the plunger moves freely and the 100 cm³ capacity is above the expected final volume. Timing starts as the reactants mix, and temperature and pressure are recorded so the final volume can be converted to an amount; hydrogen's volume is easy to measure even though its mass loss would be tiny.

Rate is a gradient, not the total product

Sulfur from thiosulfate and acid gradually obscures a cross viewed from above, and the time is taken when the cross disappears. Here the thiosulfate is diluted with water so the total volume, and so the liquid depth the eye looks through, stays the same; the flask, cross, lighting, viewing position and temperature are kept the same too. Because every run stops at the same amount of sulfur, 1/t is a comparative rate (s⁻¹), not a concentration rate; the times are illustrative and the clouding is time-lapsed.

Choose a method that measures a useful change

Mass loss works when the gas escapes from a vessel on a balance: here CO₂ from calcium carbonate and excess hydrochloric acid leaves through a loose cotton-wool plug, which reduces spray loss without sealing the outlet. Readings every 5 s give a mass-lost curve whose gradient falls as the limiting carbonate is used up, levelling off at 0.22 g of CO₂. The same amount of H₂ would weigh only 0.010 g, too little for many balances even though its volume is easy to measure.

Oxidation and reduction happen together

Separating the two half-reactions into a cell makes the electron transfer visible. Each zinc atom loses two electrons (0 → +2), so zinc is the reducing agent; each Cu²⁺ ion gains two electrons (+2 → 0), so Cu²⁺ is the oxidising agent. Every electron lost by zinc is gained by a copper(II) ion.

A ligand supplies the electron pair

H₂O and NH₃ are monodentate: each donates one lone pair and forms one coordinate bond. Ethane-1,2-diamine (en) uses a lone pair on each N, so three en ligands give six coordinate bonds in [Co(en)₃]³⁺. Neutral ligands do not change the charge, so the complex keeps the 3+ of Co³⁺.

Work from coordination number to the appropriate geometry

Six coordinate bonds give an octahedral complex (90° and 180°). Four can give a tetrahedral complex, such as [CuCl₄]²⁻ (about 109.5°), or a square planar one, such as Pt(NH₃)₂Cl₂ (90°): coordination number 4 alone does not decide the shape.

Recognise characteristic behaviour without making it universal

Iron catalysing N₂(g) + 3H₂(g) ⇌ 2NH₃(g): the gases adsorb onto active sites, bonding to the surface weakens the N≡N and H–H bonds until they break, N–H bonds form between atoms held on the surface, and NH₃ desorbs, freeing the sites. A sulfur atom bonded strongly to one site poisons it, so molecules cannot adsorb there. Schematic: a simplified picture of the surface steps.

Ammonia and chloride can replace water ligands

Three copper(II) complexes: six water ligands (octahedral), four ammonia and two water ligands (deep-blue, still six coordinate bonds), and four larger chloride ligands (tetrahedral, coordination number 4). Copper stays +2 throughout: the overall charge changes only because chloride ligands are charged.

A ligand change can alter a biological function

A schematic of the iron site in haemoglobin. Fe²⁺ sits in the plane of a porphyrin ring, one multidentate ligand donating four N lone pairs; a histidine N from the globin protein bonds below, and O₂ bonds reversibly at the sixth site, arriving and leaving in turn. Choose CO to see it take the same site through its C atom: it binds more strongly, so passing O₂ molecules cannot bind there.

Geometric isomers have different ligand arrangements

The two forms of [Ni(en)₃]²⁺ are mirror images: six Ni–N bonds, with each pair of N donors joined as one en ligand. Choose “Try to superimpose” to see that a rotation can line up all six N atoms, but the CH₂–CH₂ links then twist the other way, so they are optical isomers. Pt(NH₃)₂Cl₂ and an octahedral MA₄B₂ complex show cis (90°) and trans (180°) isomers instead.

Ka measures a particular acid equilibrium

Fluorine is very electronegative, so it pulls the C–F bonding pair towards itself, and the pull is passed along the σ bonds (a negative inductive effect). When H⁺ is lost, this spreads out the negative charge of the ion, so the ion is more stable, more of the acid dissociates and the pKa is lower: CH₃COOH 4.76, CH₂ClCOOH 2.87, CH₂FCOOH 2.59, CHF₂COOH 1.24, CF₃COOH 0.23 (data-book values, 298 K).

Water of crystallisation from mass loss

The hydrate is weighed in a clean, dry crucible and lid, heated gently at first to reduce spitting, then cooled in a desiccator and reweighed; heating, cooling and weighing are repeated until successive readings agree. The partly open lid lets water vapour escape while limiting loss of solid. 2.460 g leaves 1.200 g of MgSO₄, so x = 0.0700 ÷ 0.009967 = 7.02 (illustrative masses); constant mass alone does not identify the product or rule out decomposition.

Formula from a mass gain: magnesium oxide

Magnesium oxidation: the crucible sits in a pipe-clay triangle on a tripod above the Bunsen burner, and the lid is raised briefly with tongs now and then to admit oxygen while limiting the loss of magnesium oxide smoke. The crucible is cooled and reweighed until the mass is constant; here 0.486 g of Mg gains 0.320 g of oxygen, a 1 : 1 amount ratio, so the formula is MgO (illustrative masses). Wear eye protection and do not stare at burning magnesium.

Core Practical 1: measure, rather than assume, molar volume

Pearson's CP1 example collects the carbon dioxide over water; this shows the gas-syringe alternative, which reduces dissolving in collection water. Calcium carbonate is tipped into excess ethanoic acid only after sealing, and the final volume is read once the plunger stops. Repeating with 0.100, 0.200 and 0.300 g gives a straight line through the origin; gradient (cm³ g⁻¹) × 100.1 ÷ 1000 gives the molar volume, here 23.7 dm³ mol⁻¹ at the conditions shown.

Core Practical 2: prepare a known acid solution

One acid sample followed through the whole method: weigh by difference, dissolve in deionised water, transfer through a funnel and rinse the beaker, rod and funnel so the clinging HCl reaches the 250 cm³ volumetric flask, make up to the mark and invert, then pipette a 25.0 cm³ aliquot and titrate it with NaOH. The 7 HCl found in the aliquot scale up by 250.0 / 25.0 = 10 to the 70 in the original sample. Use the numbered steps to jump to any stage.

Acid–base does not always mean 1:1

Nine neutralisations, from 1 : 1 to 2 : 3 and 3 : 2. Each H⁺ from the acid joins one OH⁻ from the base to make one water molecule, so the H⁺ donated, the OH⁻ used and the water formed always match, and the ions left over make the salt. Matching the H⁺ per acid molecule with the OH⁻ per base formula unit gives the coefficients in the balanced equation; H₃PO₄ is a weak acid and Al(OH)₃ an insoluble base, so those steps show the reacting ratios for complete neutralisation.

Good technique protects the measured amount

Hydrochloric acid runs from the burette into a 25.0 cm³ aliquot of sodium hydroxide solution containing methyl orange, while the flask is swirled on a white tile. Each addition turns the indicator red where the acid lands and the colour fades as it mixes in. Near the end point the acid is added dropwise, and the end point is the first permanent orange; the bottom of the meniscus is read at eye level to 2 d.p.

Equivalence and end point are different ideas

A particle model of the end point. Hydrochloric acid drips from the burette one drop at a time into 12 NaOH formula units with phenolphthalein; each H⁺ reacts with an OH⁻ to form water while Na⁺ and Cl⁻ stay as spectator ions. The solution stays pink while any OH⁻ remains and turns colourless once all 12 have reacted: 12 HCl : 12 NaOH = 1 : 1. In this model the colour changes exactly at equivalence; in a real titration a suitable indicator changes within the steep pH change near equivalence, so the end point is very close to it.

Use the AS functional-group tests together

The CP7 organic tests on separate portions, each with a positive and a negative or comparison tube: bromine water is decolourised by C=C; warm acidified dichromate turns from orange to green with an oxidisable alcohol; warm Fehling’s solution gives a red Cu₂O precipitate with a simple aliphatic aldehyde; warming a halogenoalkane with NaOH(aq) and ethanol, acidifying with dilute HNO₃ and adding AgNO₃ gives white, cream or yellow precipitates; and carbonate effervesces with an acid, the gas turning limewater cloudy.

Worked pattern: do not force a more precise identity than the tests allow

Unknowns A, B and C are one each of an alkene, a primary alcohol and a bromoalkane. Bromine water identifies A by decolourising; fresh portions of B and C warmed with acidified dichromate show that B is oxidised (orange to green); C gives a cream precipitate only after warming with NaOH(aq) and ethanol, acidifying with dilute nitric acid and adding silver nitrate. The results identify a functional group within this candidate set, not the exact compound.

A ring can hide several equivalent carbons

Rotating benzene by 60° maps every ring carbon onto another, so all six give one signal. In methylbenzene the mirror plane through the CH₃ pairs the ortho and the meta carbons, leaving four ring environments (ipso, ortho, meta, para) plus the methyl carbon: five signals for seven carbons. The expansion separates the close aromatic signals; shifts are illustrative.

Count equivalent positions before counting peaks

In a proton-decoupled ¹³C spectrum each chemically distinct carbon environment normally gives one signal. Propan-2-ol's end methyls are related by symmetry, so it gives two signals, while propan-1-ol gives three; pentan-3-one gives three and pentan-2-one five. The shifts are illustrative, and the line heights are not interpreted as carbon counts.

Both use a column; their mobile phases differ

A small sample is injected into a heated inlet, where it vaporises, and an inert carrier gas sweeps it through a coiled column held in a temperature-controlled oven. Components that spend more time dissolved in the liquid film lining the column move more slowly, so each reaches the detector at its own time. Retention time is measured from injection to the maximum of each peak; the large early peak here is the solvent, not a sample component.

Place the start line above the solvent and preserve the front

Mixture X and standards A, B and C are developed on one silica plate in a covered chamber, with the solvent starting below the pencil baseline. In the inset, molecules keep adsorbing onto the polar silica and desorbing into the moving solvent, so the component that spends a larger share of its time in the mobile phase travels further. The front is marked at 80 mm, the spots are located under UV, and Rf = distance to spot centre ÷ distance to solvent front gives 0.25, 0.60 and 0.80: matching A and B supports, but co-migration alone does not prove, their presence in X.

Use n+1 only for an appropriate neighbouring set

For simple first-order coupling, a set of protons with n equivalent neighbouring H on adjacent carbon gives n + 1 lines; each neighbouring H splits every line in two, so the intensities follow Pascal's triangle up to 1 : 3 : 3 : 1 for a quartet. The CH of an isopropyl group, next to six equivalent methyl H, gives a septet, 1 : 6 : 15 : 20 : 15 : 6 : 1, while those six methyl H give a 6H doublet. Each multiplet is one environment whose total area gives its relative number of H.

An s orbital is a sphere; a p orbital has two lobes

Each dot cloud shows where an electron is most likely to be found. The 1s and 2s orbitals are spherical, with 2s larger than 1s. Each 2p orbital is a dumbbell with a nodal plane through the nucleus, and the three 2p orbitals lie along x, y and z at right angles. Each orbital holds a maximum of 2 electrons with opposite spins. Drag the diagram to rotate it.

Build configurations in a controlled order

Electrons fill the sub-shells in the order 1s 2s 2p 3s 3p 4s 3d 4p: lowest energy first, one electron in each equal-energy orbital with parallel spins before any pairing, and at most two electrons per orbital with opposite spins. 4s fills before 3d, but configurations are written with 3d before 4s, and 4s electrons are the first lost when transition-metal ions form. Chromium ([Ar] 3d⁵ 4s¹) and copper ([Ar] 3d¹⁰ 4s¹) are the two exceptions to learn.

Small dips reveal subshell energy and pairing

First ionisation energy across Period 2 rises generally but dips twice. Boron loses a 2p electron, from a higher-energy sub-shell than beryllium's 2s; oxygen loses one of a pair of electrons in one 2p orbital, and repulsion within the pair makes it easier to remove than nitrogen's unpaired 2p electrons. Switch to Period 3 to see the analogous Mg → Al and P → S dips.

A large jump identifies the change to an inner shell

Constructed data (600, 1200, 7500, 9800 and 12 000 kJ mol⁻¹) on a log₁₀ scale: log₁₀(600) = 2.778, log₁₀(1200) = 3.079 and log₁₀(7500) = 3.875. The rise from the second to the third value is a factor of 6.25, against a factor of two before it, because the third electron comes from an inner shell, closer to the nucleus and less shielded. Two electrons are removed before the jump, so this is a Group 2 element.

Ionisation energy specifies both the species and its state

First ionisation enthalpy: one electron is removed from each atom in one mole of gaseous atoms, against the attraction of the nucleus, so the change is endothermic (Na: +496 kJ mol⁻¹). The second removes an electron from each gaseous 1+ ion and is larger (Mg: +1451 kJ mol⁻¹, against +738 for the first), so Mg²⁺ needs IE₁ + IE₂.

Compare like quantities across a period

First ionisation energy rises across Period 3 because the nuclear charge increases while the electron is removed from the same shell with similar shielding. It dips at Al, whose electron is removed from the higher-energy 3p sub-shell rather than 3s, and at S, where repulsion between the two electrons paired in one 3p orbital makes one of them easier to remove. The Period selector shows the matching Be → B and N → O dips in Period 2.

A melting trend requires the actual structure

Across Period 3 the structure changes, so no single explanation fits the melting points. Na, Mg and Al are giant metallic lattices, held by attraction between positive ions and delocalised electrons; silicon is a giant covalent network in which many strong covalent bonds must break, giving the peak; P₄, S₈ and Cl₂ melt when London forces between molecules are overcome (more electrons, stronger forces: S₈ > P₄ > Cl₂), and argon's separate atoms melt lowest. The covalent bonds inside each molecule stay intact.

Ion charge and separation affect ionic attraction

Giant ionic lattices of sodium chloride and magnesium oxide: each ion is surrounded by six ions of opposite charge, and the attraction acts in all directions throughout the lattice. Mg²⁺ and O²⁻ have larger charges and are smaller (all ions are drawn at about 40% of their ionic radii so the lattice shows), so the attractions in MgO are stronger and its melting point is much higher.

Metallic bonding also involves electrostatic attraction

Metallic bonding: a giant lattice of positive ions in a sea of delocalised electrons, held together by the electrostatic attraction between the ions and those electrons. Magnesium gives two delocalised electrons per atom and forms smaller Mg²⁺ ions, so its attraction is stronger and it melts at a higher temperature than sodium. Drag to rotate.

Al₂Cl₆ contains two bridging chlorine atoms

Al₂Cl₆ forms when a lone pair on one Cl of each AlCl₃ is donated to the electron-deficient Al of the other: two Cl→Al coordinate bonds, from different bridging Cl atoms to different Al atoms. The Electron origins view marks each electron as a dot (from Al) or a cross (from Cl) and counts the 48 outer electrons; the 3D model view shows the dimer forming.

A coordinate bond uses a pair from one donor

NH₄⁺ as a dot-and-cross diagram: nitrogen's lone pair (two dots) becomes the fourth shared pair when H⁺, which has no electrons, bonds to it. The whole ion goes in brackets with the + charge outside; there are 5 + 4 − 1 = 8 outer electrons in four shared pairs.

Show three dimensions honestly

Three-dimensional shapes are drawn with in-plane lines, solid wedges (towards you) and dashed wedges (away from you). Choose a shape and drag to rotate it: the drawing updates to match the orientation, and switching to 3D model shows the same molecule as a ball-and-stick model inside a faint outline of its electron-pair arrangement, with lone pairs and bond angles.

Chain length and branching change London attractions

Each extra CH₂ in an unbranched alkane adds 8 electrons, a larger polarisable cloud and more contact with neighbouring chains, so more London attractions act between the molecules: boiling temperatures rise from 112 K for methane to 447 K for decane and volatility falls. 2,2-Dimethylpropane and pentane both have 42 electrons; the compact branched isomer touches its neighbours over a smaller area, so it boils lower (283 K against 309 K).

HF is anomalous within the hydrogen halides

Boiling points of the Group 14–17 hydrides. Down each group boiling points rise as London forces increase, but NH₃, H₂O and HF boil far above the trend (the hollow markers continue the period 3 to 4 trend back to period 2) because they also form hydrogen bonds; CH₄ cannot, so Group 14 rises smoothly.

Hydrogen bonding gives water unusual physical properties

Hydrogen bonding in ice: each water molecule donates two hydrogen bonds through its δ+ H atoms and accepts two at the lone pairs on its O, each O–H···O close to 180°, so every molecule has four neighbours arranged tetrahedrally in an open hexagonal network. On melting some hydrogen bonds break and the network partly collapses, so liquid water packs more closely and is denser than ice.

London forces arise from fluctuating electron distributions

A fluctuation piles the electron density in one I₂ molecule up at one end, an instantaneous dipole (δ− where it builds up, δ+ where it thins), although the molecule stays neutral. This changes the electron distribution in the neighbouring molecule, inducing a dipole lined up δ+ to δ−, and the correlated partial charges attract: a London force (instantaneous dipole–induced dipole attraction). Pulling back to six molecules and then to solid iodine shows such short-lived dipoles forming and vanishing everywhere.

Electronegativity describes attraction within a bond

Each bonding pair is drawn as dots of electron density. When the two atoms differ in electronegativity, the dots stream along the bond towards the more electronegative atom, which becomes δ− (H–Cl, and more so H–F); in H–H the pair stays evenly shared. Choose H₂O to see both O–H pairs drawn towards O in a bent molecule, giving an overall dipole, or CO₂, where two equal bond dipoles cancel.

Bond dipoles add as vectors

Oxygen (3.44) is more electronegative than hydrogen (2.20), so the dots of both O–H bonding pairs stream towards O: O becomes δ− and each H δ+. Water is bent, so the two bond dipoles add up to an overall dipole; choose CO₂ to see two equal bond dipoles cancel in a linear molecule.

Hydrogen bonding needs a donor and an acceptor

Each O–H bonding pair is drawn towards O, leaving the H strongly δ+ with little electron density around it. A hydrogen bond (dashed line) forms when this δ+ H lines up with a lone pair on the O of another molecule, O–H···O close to 180°; each water molecule can donate two and accept two, giving the open network of ice. Choose Why N, O or F? to see that a C–H bond is almost non-polar, so it cannot form a hydrogen bond.

Bond dipoles add as vectors

Bond dipoles add as vectors in three dimensions: they cancel in linear CO₂, tetrahedral CCl₄ and square planar XeF₄ (despite its lone pairs), but not in bent H₂O, pyramidal NH₃ or CH₃Cl, which are polar. The C–H bonds are treated as almost non-polar.

Graphite and graphene separate strength from slipperiness

Carbon allotropes in 3D: in diamond every carbon forms four tetrahedral bonds in a giant network, while graphite and graphene have three bonds per carbon in flat hexagonal sheets with one delocalised electron per carbon, and graphite's sheets slide because only weak London forces hold them together. C₆₀ is a simple molecular cage and a nanotube is a sheet of graphene rolled into a cylinder. Switch to the Bonding view to see the electrons and forces, and drag to rotate.

Bond type and structure answer different questions

Giant and molecular structures in 3D: ionic sodium chloride, metallic magnesium, giant covalent diamond, graphite and silicon dioxide, and iodine and ice, whose covalent bonds stay inside separate molecules. Each panel links bonding and structure to melting and conductivity; the Bonding view shows the forces involved. Drag to rotate.

Standard conditions are a comparison convention

Standard states at 298 K and 100 kPa: carbon as graphite, hydrogen and oxygen as diatomic gases, ethanol as a liquid. Forming exactly one mole of C₂H₅OH(l) from these elements is the standard enthalpy change of formation, −277 kJ mol⁻¹. Use the selector for the other definitions.

Breaking bonds consumes energy

For a strong acid and a strong alkali the main net change is H⁺(aq) + OH⁻(aq) → H₂O(l); the spectator ions are unchanged, so the values are similar, about −57 kJ mol⁻¹ per mole of water formed. Use the selector for the other definitions.

Four definitions with four different bases

Standard enthalpy change of reaction uses the amounts in the balanced equation: 2H₂(g) + O₂(g) → 2H₂O(l) forms two moles of liquid water, so ΔrH⦵ = 2 × (−286) = −572 kJ mol⁻¹. Formation, combustion and neutralisation each fix a different one-mole basis; use the selector to compare them.

Break reactant bonds, then form product bonds

Breaking the reactant bonds takes energy in and forming the product bonds gives energy out, so ΔH ≈ Σ(bonds broken) − Σ(bonds formed): for ethene + hydrogen, 2696 − 2820 = −124 kJ mol⁻¹ with the values quoted. The path through separate atoms is a Hess construction, not the actual mechanism. The selector adds the combustion of methane.

Bond enthalpy and mean bond enthalpy

Bond enthalpy: breaking one mole of H–H bonds in gaseous H₂ takes in +436 kJ mol⁻¹. The cleavage is homolytic, so each H atom keeps one electron of the shared pair. Use the selector for the other definitions.

Rearrange to find an unknown mean bond enthalpy

Mean bond enthalpy: the four successive C–H removals from methane need different energies (about +439, +462, +424 and +338 kJ mol⁻¹). Their total, about 1660 kJ mol⁻¹, is the atomisation of methane, and dividing by four gives about 415 kJ mol⁻¹ per C–H bond. Use the selector for the other definitions.

Design the insulated-container experiment

Insulated-cup method for 25.0 cm³ of 2.00 mol dm⁻³ HCl and 40.0 cm³ of 1.00 mol dm⁻³ NaOH: readings every minute, the alkali added at 3.0 min, and the cooling line extrapolated back to give the corrected ΔT. NaOH is limiting, so 0.0400 mol of water forms: Q = 65.0 × 4.18 × 8.0 = 2173.6 J and ΔH = −54.3 kJ mol⁻¹. The readings are a model, not measured data.

Spirit-burner calorimetry: water mass and fuel mass have different roles

Spirit-burner calorimetry: 0.920 g of ethanol heats 200.0 g of water by 22.0 K, so Q = 200.0 × 4.18 × 22.0 = 18 392 J, n = 0.0200 mol and ΔcH ≈ −920 kJ mol⁻¹. This is less negative than the data-book −1367 kJ mol⁻¹ because of heat loss, heating the can, incomplete combustion and evaporation; the burner masses are illustrative.

Combustion data: reactants minus products

The hydration of anhydrous copper(II) sulfate cannot be measured reliably directly, so both solids are dissolved separately to the same final solution. Both routes start and finish in the same place, so ΔH₁ = ΔH(hydration) + ΔH₂, giving ΔH(hydration) = −66.0 − (+11.7) = −77.7 kJ mol⁻¹.

Construct the cycle before substituting

A Born–Haber cycle built one arrow at a time from the elements in their standard states: atomisation, ionisation enthalpies and electron affinities lead to the separated gaseous ions, and the lattice energy (formation) closes the cycle, found by Hess’s law. MgCl₂ needs IE₁ + IE₂ and 2EA₁(Cl); in MgO the endothermic EA₂(O) raises the separated ions above every other level. Values are typical data-book figures in kJ mol⁻¹, rounded.

Worked example: keep the formation convention

The page’s MCl₂ worked example as a cycle: the steps before lattice formation sum to 150 + 600 + 1200 + 2(120) + 2(−350) = +1490 kJ mol⁻¹, so −650 = +1490 + ΔlattH and the lattice energy is −2140 kJ mol⁻¹. The two released electrons stay on each level until both chlorine atoms have gained one.

Name the species and state in every definition

The first ionisation enthalpy removes one electron from each atom in one mole of gaseous atoms and the second removes one from each gaseous 1+ ion, so Mg²⁺ needs IE₁ + IE₂. The first electron affinity of chlorine is exothermic because the electron is attracted by the nucleus; the second electron affinity of oxygen is endothermic because O⁻ repels the incoming electron. Values are typical data-book figures in kJ mol⁻¹.

Name the species and state in every definition

Atomisation forms one mole of gaseous atoms from an element in its standard state, shown for graphite (+717 kJ mol⁻¹); for chlorine that means ½Cl₂(g) → Cl(g). Switch to bond enthalpy to see H₂(g) → 2H(g): a diatomic bond enthalpy makes two moles of atoms, so it is twice the atomisation enthalpy.

Lattice energy has a direction

Lattice energy: gaseous Na⁺ and Cl⁻ ions attract and form one mole of solid NaCl, releasing energy, so the lattice energy is negative (−787 kJ mol⁻¹ for NaCl). Some data sources quote lattice dissociation instead, with the same magnitude and the opposite sign, so write the defining equation before using a value.

Use states and gas amounts, then consider competing effects

A substance heated steadily: particles vibrate about fixed positions in the solid, move past each other in the liquid and move freely through the whole container as a gas. Entropy rises gradually within each state; melting increases it and vaporisation causes a larger increase, because gas particles have much greater freedom of arrangement. The graph shows the qualitative shape only.

Worked example: keep the mole-of-reaction basis

N₂(g) + 3H₂(g) → 2NH₃(g): four moles of gas become two, so ΔSsystem is expected to be negative; with the page’s values, 2(193) − [192 + 3(131)] = −199 J mol⁻¹ K⁻¹ per mole of reaction as written. Switch to CaCO₃ decomposition, where gas forms from solids and ΔSsystem is positive.

Read a Gibbs-energy graph as a straight-line equation

For approximately constant ΔH and ΔS, ΔG against T follows y = mx + c with intercept ΔH and gradient −ΔSsystem. With ΔH = +48.0 kJ mol⁻¹ and ΔS = +120 J mol⁻¹ K⁻¹ the reaction is favourable above 400 K; with −36.0 and −90.0 the same 400 K boundary is an upper limit. Drag the marker to set the temperature and see ΔH and −TΔS add up to ΔG.

A dissolving lattice has two competing enthalpy changes

Enthalpy change of solution at the particle level: an NH₄NO₃ lattice breaks up in water and each ion becomes surrounded by water molecules. Dissolving ammonium nitrate is endothermic, so the solution cools, yet it still happens: enthalpy alone does not decide solubility.

Water stabilises ions through ion–dipole attractions

Enthalpy change of hydration: a gaseous ion enters water and polar water molecules orient their dipoles, δ− oxygen towards the cation (Mg²⁺) and a δ+ hydrogen towards the anion (Cl⁻). Attractions form, so hydration is exothermic; no O–H bonds are broken.

Separate polarising power from polarisability

The cation polarises the anion: a small, highly charged cation has high polarising power and distorts a large, polarisable anion’s electron cloud towards itself, increasing electron density between the ions. Choose ions to compare, from Na⁺ (little distortion) to Al³⁺ with I⁻ (much more). An experimental lattice energy 60 kJ mol⁻¹ more exothermic than the ionic-model value (illustrative data) is evidence of this extra stabilisation.

A dissolving lattice has two competing enthalpy changes

ΔsolH = −ΔlattH(formation) + ΣΔhydH, built as a Hess cycle with the page’s example: −(−2100) + (−1500) + 2(−320) = −40 kJ mol⁻¹, exothermic because hydration releases slightly more energy than separating the lattice requires. The second example (MX: −780, −400, −350) gives +30 kJ mol⁻¹; dissolving crystals do not actually release gaseous ions.

Equal rates do not mean equal concentrations

In a sealed container at constant temperature, N₂O₄ molecules keep splitting into NO₂ and pairs of NO₂ keep joining back into N₂O₄, each time a suitable collision happens. The forward and reverse reactions continue at equal rates, so the amounts of N₂O₄ and NO₂ (and the brown colour) stay constant, but they are not equal: this is dynamic equilibrium.

Changing concentration disturbs the balance

For A + B ⇌ C at equilibrium (0.40, 0.30 and 0.20 mol dm⁻³), extra A is added at fixed volume and temperature: only [A] jumps, to 0.60 mol dm⁻³. Net forward reaction then lowers [A] and [B] and raises [C] continuously until Q equals the unchanged Kc, with [A] settling above 0.40; the forward rate jumps, then forward and reverse rates become equal again. Use the buttons to add B or C, heat the mixture or add a catalyst.

Separate product and recycle unreacted material

Fresh N₂ and H₂ are compressed and join the recycled gas above the reactor, where part of the mixture reacts on the iron catalyst (N₂ + 3H₂ → 2NH₃). Downstream, a cooler condenses the ammonia, which collects in the separator and leaves as liquid, while unreacted N₂ and H₂ are recycled; a purge bleeds off some loop gas so inert argon cannot build up (close it to see the build-up). In the idealised model, 25% converts each pass, so overall conversion rises from 25% to 43.75% after two passes while the single-pass conversion stays 25%.

High equilibrium yield and fast production are different objectives

The six illustrative values are plotted and a smooth curve drawn for each pressure. Comparing vertically at 450 °C changes only the pressure (30% → 43%), so the products have fewer gas moles; moving along the 100 atm curve changes only the temperature (48% → 30%), so the forward reaction is exothermic. A marker then trades yield against a schematic rate: low temperature gives a high yield too slowly, so a moderate temperature with a catalyst is used. Drag on the graph to move the marker.

Temperature changes K itself

For an endothermic A ⇌ B with [A] + [B] = 0.600 mol dm⁻³, heating raises Kc = [B]/[A] from 2.0 to 5.0. Nothing jumps: the old mixture ([A] = 0.200, [B] = 0.400) now has Q below the new Kc, so net forward reaction continues until [A] = 0.100 and [B] = 0.500 mol dm⁻³. Both rates rise on heating, the forward (endothermic) more, and they become equal again. The buttons compare adding A or B and adding a catalyst.

A constant can stay fixed while composition changes

For H₂ + I₂ ⇌ 2HI with Kc = 8.00, adding H₂ at fixed volume and temperature raises [H₂] from 0.100 to 0.400 mol dm⁻³, making [HI]²/([H₂][I₂]) = 2.00. More HI forms until the ratio is 8.00 again (H₂ 0.351, I₂ 0.051, HI 0.380 mol dm⁻³): the disturbance changes the quotient, not the equilibrium constant. The violet tint follows [I₂]; the buttons show other changes.

Catalysts alter the approach, not the target

Fresh N₂ and H₂ are compressed and join the recycled gas above the reactor, where part of the mixture reacts on the iron catalyst (N₂ + 3H₂ → 2NH₃). Downstream, a cooler condenses the ammonia, which collects in the separator and leaves as liquid, while unreacted N₂ and H₂ are recycled; a purge bleeds off some loop gas so inert argon cannot build up (close it to see the build-up). In the idealised model, 25% converts each pass, so overall conversion rises from 25% to 43.75% after two passes while the single-pass conversion stays 25%.

Catalysts alter the approach, not the target

The six illustrative values are plotted and a smooth curve drawn for each pressure. Comparing vertically at 450 °C changes only the pressure (30% → 43%), so the products have fewer gas moles; moving along the 100 atm curve changes only the temperature (48% → 30%), so the forward reaction is exothermic. A marker then trades yield against a schematic rate: low temperature gives a high yield too slowly, so a moderate temperature with a catalyst is used. Drag on the graph to move the marker.

Why magnesium to barium becomes more reactive

Down Group 2 each atom has an extra occupied shell, so the outer electrons are farther from the nucleus and more shielded. Although the nuclear charge increases, the outer electrons are attracted less strongly, so the first and second ionisation energies fall and the 2+ ion forms more easily: reactivity generally increases from magnesium to barium. Use the Element buttons to compare the atoms.

Physical state depends on intermolecular attraction

Drawn to one scale, the halogen molecules from F₂ (18 electrons) to I₂ (106) have larger electron clouds that are more readily polarised. In each pair an instantaneous dipole in one molecule induces a dipole in its neighbour, and the London attraction grows down the group, so more energy is needed to separate the molecules: boiling temperatures rise from 85 K to 457 K. At room temperature fluorine and chlorine are gases, bromine is a liquid and iodine is a solid.

Continuous monitoring: gas volume or mass loss

Calcium carbonate is tipped into excess hydrochloric acid after the flask is sealed, so no gas is lost before timing starts at mixing. The syringe is read every 5 s: the tangent at t = 0 gives the initial rate in cm³ s⁻¹, and the curve flattens to its plateau, the final volume of CO₂, as the carbonate is used up. The joints must not leak, the plunger must move freely and the syringe must hold the final volume.

Fixed-change timing: when is 1/t meaningful?

Sulfur from thiosulfate and acid gradually obscures a cross viewed from above, and the time is taken when the cross disappears. Here the thiosulfate is diluted with water so the total volume, and so the liquid depth the eye looks through, stays the same; the flask, cross, lighting, viewing position and temperature are kept the same too. Because every run stops at the same amount of sulfur, 1/t is a comparative rate (s⁻¹), not a concentration rate; the times are illustrative and the clouding is time-lapsed.

Continuous monitoring: gas volume or mass loss

The flask stands on a balance and the mass is recorded as CO₂ leaves. A loose cotton-wool plug reduces spray while letting the gas escape; a tightly sealed bung would prevent the mass loss and could build up pressure. The mass-lost curve is steepest at the start and levels off at 0.22 g when the limiting calcium carbonate has reacted; evaporation or ejected liquid would also lower the mass, so control these and consider a blank.

Recognise order from rate and concentration graphs

As one reactant A is used up (other concentrations and temperature fixed), its concentration–time curve is drawn and a tangent rides along it; the tangent's gradient magnitude is the rate, plotted against [A] on the second graph. Zero order gives a straight concentration–time line and a horizontal rate line; first order a curve with constant half-lives (30, 30, 30 s) and a straight rate line through the origin; second order a curve whose half-lives double and an upward-curving rate line. All three share the same first half-life so the shapes can be compared.

Original data and a transparent calculation route

The CP14-style clock times are converted to 1/T and ln t and plotted. At fixed concentrations a clock time is inversely proportional to k, so ln t against 1/T is a straight line with a positive gradient, +5800 K, giving Ea = 5800 × 8.314 = 48 221 J mol⁻¹ = 48.2 kJ mol⁻¹. Switch the plot to ln(1/t) to see the gradient become −5800 K and the same positive Ea come from −gradient × R.

Turn an exponential relationship into a straight line

Rate constants at several temperatures are converted to 1/T and ln k and plotted; ln k = ln A − (Ea/R)(1/T), so the best-fit line has gradient −Ea/R and intercept ln A. Two points on the line, (0.00310, −3.20) and (0.00340, −5.00), give a gradient of −1.80 ÷ 0.00030 = −6000 K, so Ea = 6000 × 8.314 = 49 884 J mol⁻¹ = 49.9 kJ mol⁻¹. Switch the axis to 1000/T to see how its scale factor must be restored.

Activation energy is a kinetic barrier

Iron catalysing N₂(g) + 3H₂(g) ⇌ 2NH₃(g): the gases adsorb onto active sites, bonding to the surface weakens the N≡N and H–H bonds until they break, N–H bonds form between atoms held on the surface, and NH₃ desorbs, freeing the sites. A sulfur atom bonded strongly to one site poisons it, so molecules cannot adsorb there. Schematic: a simplified picture of the surface steps.

Worked clock calculation and rate-law comparison

Peroxodisulfate oxidises iodide to iodine, but 5.00 × 10⁻⁵ mol of thiosulfate turns each I₂ straight back into iodide until it runs out; then free iodine turns the starch blue-black and the clock stops. Every run therefore times the same change, 2.50 × 10⁻⁵ mol I₂ or Δ[I₂] = 5.00 × 10⁻⁴ mol dm⁻³ in 50.0 cm³, so the 40.0 s run has a mean rate of 1.25 × 10⁻⁵ mol dm⁻³ s⁻¹. Halving [I⁻] doubles the time to 80.0 s and doubling [S₂O₈²⁻] halves it to 20.0 s: rate = k[S₂O₈²⁻][I⁻].

Restricted rotation produces E/Z stereoisomerism

At each carbon of the C=C, compare the atomic numbers of the two atoms directly attached: in CH₃C(Br)=C(Cl)CH₂CH₃, Br (35) beats C (6) on the left and Cl (17) beats C (6) on the right. If the first atoms tie, compare the atoms attached to them, so ethyl (C,H,H) beats methyl (H,H,H). Priority groups on the same side of the C=C axis give Z and on opposite sides give E; twisting one carbon about the double bond would break the π bond, so rotation is restricted. Choose an example, or drag the molecule to try the twist.

Build an unambiguous IUPAC name

To draw 3-methylpentan-2-ol, read the name in parts: “pent” gives a five-carbon parent chain, “an” means single C–C bonds, “-2-ol” puts OH on carbon 2 and “3-methyl” puts a CH₃ branch on carbon 3. Each carbon is then filled with hydrogens until it has four bonds: C1 and C5 are CH₃, C2 and C3 are CH and C4 is CH₂. The check gives six carbons, fourteen hydrogens and one oxygen, C₆H₁₄O, or CH₃CH(OH)CH(CH₃)CH₂CH₃; the skeletal version keeps the OH label and draws the methyl as a short branch.

Six formula types answer different questions

Butan-2-ol, CH₃CH(OH)CH₂CH₃, changes between three representations. The displayed formula shows every atom and bond, so all ten hydrogens can be counted: 3 + 1 + 2 + 3 on carbon plus the one on oxygen. The structural formula groups each carbon with its hydrogens, and the skeletal formula keeps only the carbon framework: every line end and corner is a carbon, hydrogens on carbon are implied, and the OH, including its H, is still drawn.

Polymerisation joins molecules into long chains

Nine monomers join one at a time: the π bond of each C=C is used to make new bonds to neighbouring units, so the two former C=C carbons become adjacent atoms in the saturated backbone and no small molecule is lost. The chain is then written as one repeat unit with bonds through both brackets and n outside, such as [–CH₂–CH(Cl)–]ₙ for poly(chloroethene). Choose ethene, propene or chloroethene above the figure.

Keep side groups out of the backbone

Only the two C=C carbons of propene enter the backbone, so its CH₃ stays a side group on every second carbon and the repeat unit is –CH₂–CH(CH₃)–, not a three-carbon chain. With 2-methylpropene both CH₃ groups stay on the same backbone carbon, giving –CH₂–C(CH₃)₂–.

Polymerisation joins molecules into long chains

Beyond the specification, this shows how the π bond is used to join monomer units when PVC forms by radical addition polymerisation: RO• radicals from a peroxide initiator start a chain, the chain-end radical adds chloroethene molecules one at a time, and two radicals combine to end it. Each fish-hook arrow shows the movement of one electron.

Core Practical 5: oxidise ethanol, reflux, then distil

In Core Practical 5, ethanol is heated under reflux with excess acidified dichromate(VI). Vapour condenses in the vertical condenser and runs back, so volatile ethanol and the intermediate aldehyde stay in the flask and are oxidised fully to ethanoic acid. Cooling water enters at the bottom and leaves at the top, and the top stays open to the atmosphere.

Core Practical 5: oxidise ethanol, reflux, then distil

After refluxing, the cooled apparatus is rearranged for distillation to separate the volatile products from the chromium-containing mixture. The thermometer bulb sits at the side arm, cooling water enters at the bottom and the receiver is open to the air; the distillate contains water as well as ethanoic acid, so it is not a pure product.

Track both electrons when HBr adds

The π bond in propene's C=C is a region of high electron density, and it donates an electron pair to the Hδ+ end of polar HBr while the H–Br bonding pair moves to Br. Adding H to the terminal carbon gives the secondary carbocation CH₃CH⁺CH₃, and a lone pair on Br⁻ then forms the C–Br bond. This major pathway gives 2-bromopropane; the primary carbocation route gives 1-bromopropane as the minor product.

The double bond contains one σ bond and one π bond

A C=C double bond is one σ bond plus one π bond. Use the View buttons to see the planar σ framework (about 120° at each carbon), the parallel p orbitals, and their sideways overlap into one π bond with electron density above and below the plane. Restricted rotation shows that twisting one CH₂ group removes the overlap. Drag to rotate the molecule.

Why radical substitution rarely makes one pure product

The same chain reaction under the same UV lamp, with chlorine in excess (CH₄ : Cl₂ = 1 : 3). Radicals are still only a small fraction of the molecules, but more Cl₂ absorbs more UV, so more Cl• are made over the run, and as CH₃Cl builds up a Cl• increasingly takes its H from an already-chlorinated molecule. Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄, so when the Cl₂ is used up the mixture is mainly polysubstituted.

Why radical substitution rarely makes one pure product

Methane and chlorine (CH₄ : Cl₂ = 5 : 1) are sealed together under UV light. UV splits Cl₂ into Cl• radicals; in the two propagation steps a Cl• takes an H atom to form HCl and a carbon radical, which then takes a Cl atom from Cl₂ and regenerates Cl•. Because nearly every Cl• meets CH₄ rather than a chlorinated molecule, most of the product is CH₃Cl: excess methane favours monosubstitution but does not guarantee a pure product.

A radical chain has three distinct stages

UV light supplies the energy for homolytic fission of Cl–Cl: one fish-hook arrow goes to each atom, giving two chlorine radicals, Cl•, each with an unpaired electron and no charge. In propagation, Cl• removes a hydrogen atom from methane to give HCl and a methyl radical, CH₃•, which then reacts with Cl₂ to form chloromethane and regenerate Cl•, so the chain continues; adding the two steps cancels CH₃• and Cl• to give CH₄ + Cl₂ → CH₃Cl + HCl. In termination, two radicals combine (giving Cl₂, CH₃Cl or ethane), which removes chain carriers and explains why ethane can appear as an impurity.

Distillation, cracking and reforming solve different problems

Each extra CH₂ in an unbranched alkane adds 8 electrons, a larger polarisable cloud and more contact with neighbouring chains, so more London attractions act between the molecules: boiling temperatures rise from 112 K for methane to 447 K for decane and volatility falls. 2,2-Dimethylpropane and pentane both have 42 electrons; the compact branched isomer touches its neighbours over a smaller area, so it boils lower (283 K against 309 K).

Why radical substitution rarely makes one pure product

To form 2-bromopropane, Br• removes the H on the central carbon, giving HBr and CH₃C•HCH₃; this radical reacts with Br₂ to give CH₃CHBrCH₃ and regenerate Br•. Choose C1 to see that the terminal radical, CH₃CH₂CH₂•, describes the other product, 1-bromopropane. Fish-hook arrows are shown for interest; the radical dots and species are what matter here.

Compare the whole fuel life cycle

Transesterification: methanol, with an alkaline catalyst, exchanges the alcohol part of each ester link, so one triglyceride gives one glycerol and three methyl esters of long-chain acids, the mixture used as biodiesel. No water is a product of the ideal net reaction. Choose a hydrolysis reagent to compare the acid or soap products.

Ammonia gives a charged intermediate before the amine

Bromoethane is heated with excess ammonia (NH₃ : CH₃CH₂Br = 6 : 1) in ethanol in a sealed vessel. When a nitrogen lone pair meets the δ+ carbon from the side opposite Br, SN2 substitution forms the C–N bond and releases Br⁻, and another NH₃ removes H⁺ from the nitrogen, forming NH₄⁺. Bromoethane nearly always meets NH₃ before an amine, so ethylamine is the main product, although further substitution is not impossible.

Ammonia gives a charged intermediate before the amine

The same reaction with bromoethane in excess (NH₃ : CH₃CH₂Br = 1 : 3). Ethylamine still has a lone pair on nitrogen, so, surrounded by bromoethane, it is attacked again, giving diethylamine, then triethylamine and some tetraethylammonium bromide, a quaternary salt with no N–H to lose. Each substitution also releases H⁺, which ends up on a nitrogen, so the reaction stops when no free lone pair remains, leaving a mixture of amines, mostly as their ammonium salts.

Ammonia gives a charged intermediate before the amine

With excess NH₃ in ethanol, heated in suitable sealed apparatus, the N lone pair of ammonia attacks the carbon bearing Br as the C–Br pair moves to Br, giving CH₃CH₂NH₃⁺ (three N–H bonds, charge +1) and Br⁻. Another NH₃ accepts a proton from it as the N–H bond pair returns to the original N, giving the primary amine CH₃CH₂NH₂ and NH₄⁺, which forms ammonium bromide with Br⁻; two ammonia molecules are used overall.

Primary substitution: bond making and breaking together

Bromoethane's polar Cδ+–Brδ− bond is attacked from the side opposite Br by a nucleophile, an electron-pair donor. One curly arrow goes from the nucleophile's lone pair to the δ+ carbon, and a second goes from the C–Br bond to Br, which leaves as Br⁻. Hydroxide (attacking through O) gives ethanol, and cyanide (attacking through C) gives propanenitrile, CH₃CH₂C≡N. The total charge is −1 on both sides; this one-step mechanism is known as SN2.

Neutralisation and drying remove different impurities

Washing the crude 2-chloro-2-methylpropane with sodium hydrogencarbonate solution removes residual HCl and releases CO₂, so the inverted funnel is vented often through the tap, pointing away from people. In this preparation the organic product is the upper layer once the layers settle; remove the stopper, then run off the lower aqueous layer.

SN2: backside attack inverts the reacting centre

Bromoethane's polar Cδ+–Brδ− bond is attacked from the side opposite Br by a nucleophile, an electron-pair donor. One curly arrow goes from the nucleophile's lone pair to the δ+ carbon, and a second goes from the C–Br bond to Br, which leaves as Br⁻. Hydroxide (attacking through O) gives ethanol, and cyanide (attacking through C) gives propanenitrile, CH₃CH₂C≡N. The total charge is −1 on both sides; this one-step mechanism is known as SN2.

A racemate contains chiral molecules but has no net rotation

In a polarimeter, plane-polarised monochromatic light passes through a tube of the sample, and the observer turns the analyser until the light is brightest: the angle turned is the observed rotation. A pure enantiomer turns the plane of polarisation progressively along the tube, and its mirror image turns it by the same angle in the opposite direction under the same conditions. In a racemate the equal and opposite rotations cancel, and an achiral solvent does not rotate the plane, so both read 0° (the 40° shown is illustrative).

A racemate contains chiral molecules but has no net rotation

Light is a transverse wave: in unpolarised light the electric field vibrates in all planes at right angles to the direction of travel. A polariser lets through only vibrations in one plane, giving plane-polarised light. A second filter, the analyser, passes all of this light when its axis is parallel to the polariser's and none when it is crossed at 90°; in between, the brightness falls smoothly.

Use observations to answer a defined question

Each RP6 test is shown with a positive and a negative tube. Shaken with bromine water, an alkene decolourises the orange-brown colour; warm acidified dichromate turns from orange to green with a primary or secondary alcohol but not with a tertiary one; warm Tollens’ reagent gives a silver mirror and warm Fehling’s a brick-red Cu₂O precipitate with an aldehyde but not a ketone; and sodium carbonate effervesces with a carboxylic acid, the gas turning limewater milky. Each card also states the limit of the conclusion.

Hydrogen bonding explains boiling and water solubility

Carboxylic acid molecules hydrogen-bond using O–H as the donor and an oxygen lone pair as the acceptor, so two molecules can pair up through two hydrogen bonds; these strong intermolecular attractions contribute to relatively high boiling points. Short-chain acids also hydrogen-bond with water, which is why they dissolve. Choose Carboxylate ion to see why the conjugate base is stabilised.

COOH is one functional group

Ethanoic acid only partly dissociates in water, giving the ethanoate ion and H₃O⁺. In this conjugate base the p orbitals on the carbon and both oxygens overlap sideways into one π system above and below the O–C–O plane, so the negative charge is delocalised over the two oxygen atoms and the two C–O bonds are the same length (0.126 nm). Use the View buttons to watch the separate p orbitals merge, or compare the localised picture and ethanoic acid itself; drag to rotate.

Two steps with explicit arrow origins and charge bookkeeping

Cyanide from KCN attacks propanal through its carbon lone pair while the C=O π pair moves to O, giving CH₃CH₂CH(O⁻)CN; an O⁻ lone pair then takes the H of H–CN, forming 2-hydroxybutanenitrile and regenerating CN⁻. A red CN⁻ attacking the planar carbonyl from above and a blue CN⁻ attacking from below are equally likely, so the two enantiomers (shown with red and blue CN groups either side of a mirror line) form in equal amounts and the mixture has no net optical rotation.

Worked energy comparison uses a common final product

Every route ends at cyclohexane. Hydrogenating one C=C releases about 120 kJ mol⁻¹, so a localised cyclohexa-1,3,5-triene would be expected to release 3 × 120 = 360 kJ mol⁻¹, but benzene releases only about 208 kJ mol⁻¹. Its hydrogenation is less exothermic, so benzene starts lower in enthalpy: about 360 − 208 = 152 kJ mol⁻¹ of stabilisation from delocalisation, compared with the hypothetical model.

From a useful formula to an evidence-based bonding model

Benzene is a flat, regular hexagon: each carbon forms three σ bonds at 120°, and all six C–C bonds are the same length (0.140 nm, between C=C and C–C). Use the View buttons to see the six parallel p orbitals, then watch them overlap sideways with their neighbours all round the ring and merge into two continuous rings of delocalised π electron density, one above and one below the plane, and see how the circle in the skeletal formula represents them. Drag to rotate the molecule.

Turn a single link into repeated links

Condensation polymerisation joins monomers with a reactive group at each end. A diacid with a diol gives ester links –C(=O)–O–; with a diamine it gives amide links –C(=O)–NH–, keeping one N–H at each link. Water is lost from a diacid route and HCl from a diacyl chloride route, and the repeat unit is drawn with bonds through both brackets.

Track protonation as the polymer is broken down

Polyamide chains with N–H groups form hydrogen bonds from the N–H hydrogen of one chain to a lone pair on a carbonyl O of the next, alongside London and permanent dipole forces. A typical polyester repeat has no N–H or O–H donor, so its chains attract by permanent dipole–dipole and London forces instead.

Compare availability of the nitrogen lone pair

The nitrogen lone pair accepts H⁺. An alkyl group donates electron density towards N, so butylamine's lone pair is more available than ammonia's; in phenylamine the lone pair is partly delocalised into the aromatic ring, so it is less available, although phenylamine is still a weak base. The illustrative conjugate-acid pKa values 10.6, 9.3 and 4.6 give the order alkylamine > ammonia > phenylamine.

Make the C(=O)–NH bond and keep the end groups

Joining the –COOH of one amino acid to the –NH₂ of another removes OH and H as H₂O and forms a –C(=O)–NH– amide (peptide) link, with N bonded directly to the carbonyl carbon. Glycine and alanine can give two different dipeptides, Gly–Ala and Ala–Gly, which have the same formula but different sequences; free terminal groups remain at each end.

Purify by solubility, then dry before measuring

Hot solvent is added a little at a time until the crude solid just dissolves, so the hot solution is close to saturated. As it cools slowly the product's solubility falls and crystals grow, and an ice bath lowers it further; the small amount of soluble impurity stays in the mother liquor. The graph uses illustrative solubility values: with too much solvent the solution only becomes saturated at a lower temperature and more product stays dissolved, so recovery is lower.

Purify by solubility, then dry before measuring

A Büchner funnel with a flat perforated plate sits in a rubber collar in the neck of a thick-walled side-arm flask, whose side arm leads to the vacuum source; the stem opens inside the flask. The filter paper lies flat over every hole and is wetted so that it seals, then reduced pressure draws the mother liquor through while the crystals stay on the paper. The crystals are washed with a little ice-cold solvent, air is drawn through, and they are dried fully before weighing, because retained solvent would make the yield too high.

Three stages organise every aromatic substitution

The three stages of electrophilic substitution, shown with E⁺ = NO₂⁺: the acid mixture generates the electrophile, a ring π pair forms the C–E bond, and loss of H⁺ restores delocalisation. In the intermediate the attacked carbon carries both H and E, and the delocalisation arc and + charge exclude that tetrahedral carbon. The second arrow starts at the C–H bond, and H₂SO₄ is regenerated separately from HSO₄⁻ + H⁺.

Use the change in solubility with temperature

Hot solvent is added a little at a time until the crude solid just dissolves, so the hot solution is close to saturated. As it cools slowly the product's solubility falls and crystals grow, and an ice bath lowers it further; the small amount of soluble impurity stays in the mother liquor. The graph uses illustrative solubility values: with too much solvent the solution only becomes saturated at a lower temperature and more product stays dissolved, so recovery is lower.

Use the change in solubility with temperature

A Büchner funnel with a flat perforated plate sits in a rubber collar in the neck of a thick-walled side-arm flask, whose side arm leads to the vacuum source; the stem opens inside the flask. The filter paper lies flat over every hole and is wetted so that it seals, then reduced pressure draws the mother liquor through while the crystals stay on the paper. The crystals are washed with a little ice-cold solvent, air is drawn through, and they are dried fully before weighing, because retained solvent would make the yield too high.

Core Practical 10: construct, measure and evaluate cells

Core Practical 10 compares combinations of metal half-cells joined by a potassium nitrate salt bridge, with a high-resistance voltmeter so that very little current flows and the reading approaches the emf. The values shown are for 1.00 mol dm⁻³ solutions at 298 K; with the 0.10 mol dm⁻³ silver nitrate used in the practical, the silver cells are not standard cells.

Separate the two electron-transfer processes

Separating Zn/Zn²⁺ and Cu²⁺/Cu into half-cells forces the electron transfer through the external wire: zinc is oxidised at the negative anode and Cu²⁺ is reduced at the positive cathode. Nitrate ions move towards the zinc compartment and potassium ions towards the copper compartment through the salt bridge; no electrons pass through it. With the switch open, the high-resistance voltmeter draws almost no current and reads the emf, +1.10 V; with the load switched in, it shows a slightly lower terminal p.d.

Write a conventional cell diagram and calculate emf

The cell diagram has the oxidation half-cell on the left and the reduction half-cell on the right, with Pt where the redox pair has no conductor of its own. For copper with Fe³⁺/Fe²⁺, E°cell = 0.77 − 0.34 = +0.43 V: the iron half-equation is doubled to balance electrons, but its E° is not.

A reference electrode makes comparisons possible

The standard hydrogen electrode has hydrogen gas at 100 kPa over platinum coated with finely divided platinum, in 1.00 mol dm⁻³ H⁺(aq) at 298 K, and is assigned E° = 0.00 V. Measured against it with negligible current, copper is the positive electrode (E° = +0.34 V) and zinc the negative electrode (E° = −0.76 V).

Non-standard conditions and kinetic barriers are different limitations

Lowering [Cu²⁺] reduces the forward driving force of Zn + Cu²⁺ → Zn²⁺ + Cu and makes E(Cu²⁺/Cu) less positive, so the emf falls; lowering [Zn²⁺], a product, makes E(Zn²⁺/Zn) more negative and the emf rises. The values are calculated for 298 K; the direction of the change is the key idea.

Hydrogen–oxygen fuel cell with an acidic electrolyte

Hydrogen loses electrons at the negative anode, and oxygen gains them at the positive cathode, combining with H⁺ to form water. Electrons travel through the external load while H⁺ crosses the proton-exchange membrane, which is not an electron path. Twice the hydrogen half-equation supplies the four electrons one O₂ needs, and E°cell = +1.23 V.

Hydrogen–oxygen fuel cell with an alkaline electrolyte

In alkaline conditions the half-equations use OH⁻ and water: oxygen reduction makes OH⁻ at the cathode and hydrogen oxidation uses it at the anode, so hydroxide moves from the oxygen side to the hydrogen side, opposite to proton transport in the acidic cell. With E° values of +0.40 V and −0.83 V, E°cell = +1.23 V, and the overall reaction is again 2H₂ + O₂ → 2H₂O.

Hydrogen-rich fuels and an evidence-based comparison

In an idealised acidic direct methanol fuel cell, methanol and water are oxidised at the anode, CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻, and oxygen is reduced at the cathode as in the hydrogen cell. The overall reaction, CH₃OH + 1½O₂ → CO₂ + 2H₂O, releases carbon dioxide at the point of use, unlike hydrogen.

Interpret a rechargeable cell from supplied data

On discharge, lead at the negative plate and lead(IV) oxide at the positive plate both become lead(II) sulfate, using up sulfate and H⁺ ions and forming water, while electrons flow through the load from the lead plate. Charging with an external supply reverses every change, regenerating Pb, PbO₂ and the acid. With the supplied potentials, E°cell = +1.69 − (−0.36) = +2.05 V.

Excess ammonia can switch from base to ligand

Copper(II) keeps two water ligands: four NH₃ donate N lone pairs and two H₂O donate O lone pairs, so [Cu(NH₃)₄(H₂O)₂]²⁺ still has coordination number 6 (distorted octahedral, longer axial Cu–O bonds). Cobalt(II) can take six NH₃. Neutral ligands leave the overall charge equal to the metal ion's charge.

A reaction product can accelerate its own formation

Two matched runs of acidified manganate(VII) with ethanedioate: A with no Mn²⁺ at the start, B with a little Mn²⁺ added. A's [MnO₄⁻] falls slowly at first, then steeply as the product Mn²⁺ catalyses the reaction, then levels off as MnO₄⁻ is used up; the magnitude of the gradient is the rate of manganate consumption. Added Mn²⁺ shortens the slow start. The curves are schematic, not measured data.

Adsorb, react, then desorb

A catalytic converter surface: CO and NO adsorb onto active sites, the N–O bond weakens and breaks, each CO combines with an adsorbed O atom to form CO₂ and two N atoms join as N₂, and the products desorb, freeing the sites. A lead atom bonded strongly to one site poisons it. Schematic: a simplified picture of the surface steps.

One ligand can occupy several coordination positions

Each en (H₂NCH₂CH₂NH₂) donates the lone pair on both N atoms to the same ion, closing a chelate ring, so three en give six coordinate bonds in [Co(en)₃]³⁺. EDTA⁴⁻ wraps round and donates six pairs, two from N and four from carboxylate O, as one ligand. Six water ligands also give coordination number 6.

Connect coordination number and geometry

Each ligand donates an electron pair to form a dative (coordinate) bond. Six small ligands such as H₂O, NH₃ or OH⁻ give octahedral complexes; larger chloride ligands give four-coordinate tetrahedral [CoCl₄]²⁻ and [CuCl₄]²⁻; Pt(NH₃)₂Cl₂ is four-coordinate but square planar. The panel checks each overall charge, for example +3 + 6 × (−1) = 3− for [Cr(OH)₆]³⁻.

Ligand exchange can change oxygen transport

A schematic of the iron site in haemoglobin. Fe²⁺ sits in the plane of a porphyrin ring, one multidentate ligand donating four N lone pairs; a histidine N from the globin protein bonds below, and O₂ bonds reversibly at the sixth site, arriving and leaving in turn. Choose CO to see it take the same site through its C atom: it binds more strongly, so passing O₂ molecules cannot bind there.

Adjacent and opposite ligands give different biological behaviour

Square-planar Pt(NH₃)₂Cl₂: in cis-platin the two chloride ligands are adjacent (Cl–Pt–Cl = 90°); in the trans form they are opposite (180°). Choose “Try to superimpose” to turn the trans isomer over the cis one: at most one Cl ever lines up, because rotating a whole molecule cannot turn adjacent positions into opposite ones.