Mr Kidman's Classroom
Revise the key ideas
Group 1 and Group 2 patterns
- Explaining reactivity — Down Group 2, increased shielding and outer-electron distance outweigh the increased nuclear charge, so the first two electrons are generally removed more easily. First and second ionisation energies decrease overall and reactions become more vigorous. Do not explain this by a weaker nucleus or by each atom having fewer protons.
- Oxygen and chlorine — Magnesium to barium form predominantly ionic +2 compounds in the specified reactions: 2Mg + O2 → 2MgO and Mg + Cl2 → MgCl2 illustrate the stoichiometry. State the actual reagent and conditions; observations such as a bright magnesium flame support the reaction but do not themselves establish the balanced equation.
- Water and steam — Calcium, strontium and barium react with cold water to form hydroxides and hydrogen, with increasing vigour down the group: Ca + 2H2O → Ca(OH)2 + H2. Magnesium reacts very slowly with cold water, while heated magnesium reacts with steam to form MgO and H2. The oxide versus hydroxide product depends on conditions.
- Basic oxides and hydroxides — Group 2 oxides are basic: MgO + 2H+ → Mg2+ + H2O. An oxide that reacts with water produces the corresponding hydroxide, for example CaO + H2O → Ca(OH)2. Dissolved hydroxide supplies OH− for neutralisation. Low solubility does not mean a solid oxide or hydroxide cannot react with acid.
- Hydroxide solubility — Group 2 hydroxides become more soluble down the group. Magnesium hydroxide is sparingly soluble; calcium hydroxide gives an alkaline solution and barium hydroxide is more soluble. Compare concentrations only under comparable preparation conditions: an equal mass added to water need not all dissolve or give equal hydroxide concentrations.
- Sulfate solubility — Group 2 sulfates become less soluble down the group, with BaSO4 very sparingly soluble. A precipitate forms when dissolved ions exceed the relevant solubility conditions. This trend is opposite to the hydroxide trend; do not use a single rule that every Group 2 compound becomes more soluble down the group.
- Polarisation and thermal stability — A smaller, more highly charged cation has greater polarising power and distorts an anion's electron cloud more strongly. Down Group 2, larger cations polarise carbonate and nitrate less, so these salts generally become more thermally stable. Compare like anions and identify the cation before applying this explanation.
- Carbonate decomposition — Group 2 carbonates decompose on sufficient heating: MCO3 → MO + CO2. Lithium carbonate behaves similarly, while other Group 1 carbonates are much more resistant to ordinary laboratory heating. Test evolved CO2 with limewater rather than assuming every colourless gas is carbon dioxide.
- Nitrate decomposition — Group 2 nitrates and lithium nitrate form oxide, nitrogen dioxide and oxygen: 2M(NO3)2 → 2MO + 4NO2 + O2 for a +2 metal. Other Group 1 nitrates give nitrites and oxygen, for example 2KNO3 → 2KNO2 + O2. Match coefficients to the cation charge; nitrate and nitrite are different ions.
The lithium exception can be written 4LiNO3 → 2Li2O + 4NO2 + O2; do not apply the Group 2 coefficient pattern unchanged to a +1 metal.
- Flame-test method — Use an approved small-scale method with a clean wire and suitable acid, checking that the cleaned wire adds no misleading flame colour before testing a sample. Observe the non-luminous flame and compare known controls. Strong sodium contamination can mask other emissions; a mixed or weak colour is not a secure identification by itself.
- Flame colours and emission — Lithium gives crimson red, sodium yellow, potassium lilac, calcium orange-red, strontium red and barium pale apple-green under suitable tests. Magnesium has no characteristic visible flame-test colour. Excitation followed by electron transitions produces characteristic wavelengths; a flame colour is not the same as a solution's colour or a metal burning intensely.
Halogens, halides and inorganic analysis
- Halogen physical patterns — At room conditions chlorine is a pale green gas, bromine a red-brown liquid and iodine a grey-black solid with violet vapour on warming. Melting and boiling temperatures rise down the group because larger electron clouds give stronger London attractions between X2 molecules. The intermolecular explanation does not require stronger covalent X–X bonds.
- Oxidising ability — Halogens gain electrons in X2 + 2e− → 2X−. Oxidising ability generally decreases down Group 7 because greater shielding and distance reduce attraction for an incoming electron despite increased nuclear charge. Electronegativity also decreases. These trends concern different properties, so define each rather than using reactivity as an unexplained substitute.
- Displacement equations — Chlorine oxidises bromide and iodide; bromine oxidises iodide. For example Cl2 + 2Br− → 2Cl− + Br2. Iodine does not displace bromine from bromide under these ordinary conditions. Write the halogen as X2 and the halide as X−, and show both atom and charge balance.
- Observing halogen displacement — Released bromine gives orange/brown colour and iodine gives brown aqueous colour; a suitable non-reacting organic solvent can separate a coloured layer, with iodine violet/purple and bromine orange. Compare starting colours and controls before inferring reaction. Identify layers from actual solvent density rather than assuming the organic layer is always above water.
- Halogens and metals — A metal loses electrons while a halogen gains them, producing a metal halide, for example 2Na + Cl2 → 2NaCl. The metal's charge determines its formula. Visible reaction vigour alone does not prove the oxidation state or purity of the product, particularly for metals with several oxidation states.
- Chlorine and water — Cl2 + H2O ⇌ HCl + HClO describes chlorine's disproportionation in water: chlorine changes from 0 to −1 and +1. HClO is chloric(I) acid. Equilibrium and subsequent reactions depend on conditions; this is not simply chlorine dissolving as unchanged molecules or a reaction in which only one chlorine atom changes oxidation number.
- Cold and hot alkali — With cold dilute alkali, Cl2 + 2OH− → Cl− + ClO− + H2O. With hot concentrated alkali, 3Cl2 + 6OH− → 5Cl− + ClO3− + 3H2O. Distinguish chlorate(I) and chlorate(V), and use their different oxidation numbers and coefficients rather than treating bleach and hot-alkali products as identical.
- Hydrogen halides — Hydrogen chloride dissolves readily in water to produce an acidic solution through proton transfer: HCl + H2O → H3O+ + Cl−. With ammonia, HCl forms white NH4Cl fumes. Dry HCl gas and hydrochloric acid are not identical descriptions; the presence of water matters when explaining electrical conduction and aqueous acidity.
- Chloride with sulfuric acid — Solid sodium chloride with concentrated sulfuric acid initially undergoes acid displacement, giving steamy hydrogen chloride fumes: NaCl + H2SO4 → NaHSO4 + HCl. Chloride is not a strong enough reducing agent to reduce sulfuric acid under the stated test. This reaction is not a safe source of gas for unsupervised use.
- Bromide and iodide reduction — Bromide and especially iodide are stronger reducing agents than chloride. After acid displacement, HBr can reduce sulfuric acid to SO2 while forming Br2; HI can give products including SO2, sulfur and H2S while forming I2. Multiple products and conditions matter. Do not infer a unique balanced overall equation without specifying the sulfur reduction product.
- Balancing sulfur reduction — For the SO2 route, 2HBr + H2SO4 → Br2 + SO2 + 2H2O. For the H2S route, 8HI + H2SO4 → 4I2 + H2S + 4H2O. Sulfur changes from +6 to +4 or −2 respectively, requiring different electron amounts. These equations support observations without recommending smelling toxic products.
- Predicting unfamiliar halogens — Use periodic trends to make qualified predictions for fluorine or astatine, such as the direction of oxidising power or physical-state trend. A prediction is not a direct observation, and unusual reaction chemistry, radioactive scarcity and conditions limit simple extrapolation. State the trend and the property being predicted explicitly.
- Silver halide testing — Acidify a fresh sample with dilute nitric acid, then add silver nitrate: Ag+ + X− → AgX. Chloride gives white AgCl, bromide cream AgBr and iodide yellow AgI. Nitric acid removes interfering carbonate without introducing chloride; using hydrochloric acid would contaminate a chloride test.
- Ammonia confirmation — AgCl dissolves in dilute ammonia; AgBr needs concentrated ammonia; AgI remains insoluble under these tests. Dissolution reflects complex formation and equilibrium rather than disappearance of the halide element. Compare the initial precipitate and its response on a separate controlled sample; colour alone can be ambiguous.
- Carbonate and sulfate tests — Acid with carbonate or hydrogencarbonate releases CO2, which turns limewater milky through CaCO3 precipitation. For sulfate, acidify a separate sample with dilute hydrochloric acid and add barium chloride: Ba2+ + SO4^2− → BaSO4, a white precipitate. Acid removes carbonate interference; do not use sulfuric acid to test for sulfate.
- Ammonium and an analysis sequence — Warm a separate sample with aqueous NaOH; NH4+ + OH− → NH3 + H2O. Ammonia turns damp red litmus blue. Design a sequence using fresh portions, blanks and suitable confirmation tests so an earlier reagent does not introduce the ion sought later. Record observations before interpreting them, including negative results and mixture limitations.
Electrode potentials and electrochemical cells
- Half-cell equilibrium — An electrode potential describes the tendency for a specified reduction half-equilibrium relative to a reference. Tables conventionally write reduction, for example Cu2+ + 2e− ⇌ Cu. A more positive standard potential indicates a greater tendency for reduction under standard conditions; it is not an absolute voltage of an isolated electrode.
- Standard hydrogen electrode — The reference is assigned E° = 0.00 V: hydrogen gas at 100 kPa, aqueous H+ at 1 mol dm−3 and a platinised platinum electrode, platinum coated with finely divided platinum to provide a large catalytic surface, conventionally at 298 K. Platinum provides a conducting catalytic surface without being the species oxidised in the written half-equation. The reference conditions must be stated with the comparison.
- Measuring a potential — Connect the investigated half-cell to the hydrogen reference using an external high-resistance voltmeter and an appropriate salt bridge. Maintain the stated concentrations, temperature and gas pressure and note meter polarity. A high-resistance meter minimises current and disturbance; a reading without known terminal connections does not identify the reduction potential's sign.
- Metal and inert electrodes — A Zn2+/Zn half-cell uses zinc metal in contact with its aqueous ions. A Fe3+/Fe2+ half-cell needs an inert platinum electrode because both redox species are dissolved; platinum is not inserted into the redox equation as a reagent. Include every species required by the stated half-equilibrium and avoid unwanted electrode reactions.
For a non-metal gas couple such as Cl2/Cl−, bubble the specified gas at its standard pressure over an inert conducting electrode in contact with the specified halide solution, rather than using a nonexistent solid chlorine electrode. An ion/ion couple such as Fe3+/Fe2+ instead requires both oxidation-state solutions at their stated concentrations and an inert conductor.
- Salt bridge and charge balance — The bridge provides an ionic route and maintains bulk electrical neutrality while limiting direct mixing. Electrons move through the wire, not through the salt bridge. Choose bridge ions that do not precipitate or react with half-cell species; chloride is unsuitable in some silver-ion cells, for example.
- Cell notation — Write phase boundaries with a single vertical line and the junction with a double line. Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s) represents separated zinc and copper half-cells; add an inert terminal where no conducting redox solid exists. Use the question's notation convention and include concentrations when conditions are specified as non-standard.
- EMF and direction — For a cell written left to right, E°cell = E°right − E°left using reduction potentials. With Cu2+/Cu = +0.34 V and Zn2+/Zn = −0.76 V, the zinc-left/copper-right cell has +1.10 V: zinc oxidises and copper ions reduce for spontaneous discharge under standard conditions. Reversing the cell reverses the sign.
Schematic spontaneous zinc–copper discharge: electrons pass through the external wire from zinc to copper, while ions move through the salt bridge to maintain charge balance. At the stated standard reduction potentials, copper-right minus zinc-left is +1.10 V. Relative volumes and electrode sizes are not measured quantities. - Electrons and stoichiometry — Multiply half-equations to balance electron transfer before adding them, but do not multiply electrode potentials by stoichiometric coefficients. Potential is energy per charge, not an amount of energy. Two electrons in the Zn/Cu overall reaction do not make the 1.10 V value become 2.20 V.
- Feasibility and equilibrium — A positive standard cell EMF supports thermodynamic feasibility in the written direction under standard conditions. It corresponds to a favourable total entropy change and an equilibrium constant greater than one at that temperature for the reaction as written. A potential difference predicts neither the speed nor completion of a reaction by itself.
For a fixed balanced electron transfer, standard cell EMF is directly proportional to the corresponding total entropy change and to ln K. Changing the equation scale changes the proportionality through the transferred charge, while leaving the voltage itself unchanged.
- Limits of predictions — A feasible reaction may be very slow because of a high activation barrier, surface film or poor contact. Actual concentrations, pressures and complex formation can also change the potential from its standard value. A negative standard value does not rule out every possible non-standard condition; identify the assumptions used in the prediction.
- Selecting oxidants and reductants — Pair a relatively positive reduction half-equation with the reversed, less positive one to test a proposed reaction. The species reduced is the oxidising agent; the species oxidised is the reducing agent. Do not select a reagent simply because its element has the largest positive oxidation number: potential data and actual species are needed.
- Disproportionation from potentials — For Cu+ forming Cu2+ and Cu, use E°(Cu+/Cu) = +0.52 V and E°(Cu2+/Cu+) = +0.15 V. E°cell = 0.52 − 0.15 = +0.37 V for 2Cu+ → Cu2+ + Cu under the stated standard model. Compare the two adjacent couples with the intermediate species in both; reversing one subtraction reverses the conclusion.
- Core Practical 10 evidence — Construct specified electrochemical cells using clean electrodes, suitable solutions, bridge and a high-resistance voltmeter; record temperature, polarity and stable readings. Compare measured EMF with predictions and investigate contamination, surface oxidation, solution composition and meter limitations. A matching numerical answer does not establish competence in assembling and measuring a real cell.
Storage cells, fuel cells and redox titration
- Primary and rechargeable cells — A primary cell supplies energy through a reaction not conveniently reversed in normal use. In a rechargeable cell, an external source drives the reverse chemical changes within operating limits. Compare useful voltage, capacity, mass, cycling, materials and disposal using evidence; rechargeable does not mean indefinitely reversible or environmentally impact-free.
- Hydrogen fuel cells — A fuel cell continuously receives fuel and oxidant, unlike a closed store of a fixed initial reagent supply. For an acidic hydrogen–oxygen cell: anode H2 → 2H+ + 2e−; cathode O2 + 4H+ + 4e− → 2H2O. Oxidation occurs at the anode and reduction at the cathode: use these reaction roles to identify them. After multiplying the hydrogen half-equation, the overall reaction is 2H2 + O2 → 2H2O.
- Alkaline fuel cells — In an alkaline hydrogen–oxygen cell, anode 2H2 + 4OH− → 4H2O + 4e− and cathode O2 + 2H2O + 4e− → 4OH−. Adding cancels electrons, hydroxide and excess water to give the same overall reaction as in acid. Mixing an acidic half-equation with an alkaline one without conversion produces a misleading balance.
- Fuel choices and impacts — Hydrogen-rich fuels or methanol can be used in suitable systems, with different catalysts, reforming needs and carbon-containing products. Evaluate fuel production, storage, transport, operating efficiency and lifecycle emissions. Water at a hydrogen cell's point of use does not prove zero upstream emissions or solve hydrogen's storage and infrastructure problems.
- Manganate(VII) and iron(II) — In acid, MnO4− + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+. The purple titrant acts as its own indicator; the endpoint is a persistent faint pink from slight excess MnO4− under the method's conditions. Use suitable dilute sulfuric acid; chloride or oxidising acid can introduce unwanted redox chemistry.
- Titration amount calculation — If 20.00 cm3 of 0.0200 mol dm−3 MnO4− reacts with an iron(II) aliquot, manganate amount is 0.000400 mol and iron(II) amount is five times this, 0.00200 mol. Apply aliquot and dilution factors to infer the original sample. A 1:1 assumption would underestimate iron by a factor of five.
- Iodine and thiosulfate — I2 + 2S2O3^2− → 2I− + S4O6^2− gives the 1:2 amount ratio. Add starch near the endpoint when iodine is pale, then continue until the blue-black colour disappears. Early starch addition at high iodine concentration can make the endpoint less reliable; a disappearance endpoint differs from manganate's first persistent pink.
- Indirect iodine analysis — An oxidising analyte may liberate iodine from excess iodide; titrate that iodine with standard thiosulfate and link both equations. For example one mole of ClO− liberates one mole I2 in acid and corresponds to two moles thiosulfate. Account for dilution and aliquots, and avoid losing iodine or allowing unrelated oxidants to change the result.
- Core Practical 11 judgement — Use the specified redox-titration method, conditioned apparatus and concordant suitable titres to determine an unknown amount. Calculate the complete stoichiometric chain and explain the endpoint, acid and indicator choices. Evaluate air oxidation of Fe2+, iodine loss or reagent ageing where relevant; do not invent observations for an experiment not performed.
Transition elements and complex structures
- Definition and configurations — A transition element forms at least one stable ion with an incomplete d subshell. Sc forms Sc3+ with d0 and Zn forms Zn2+ with d10, so they are d-block elements but not transition elements under this course definition. For first-row cations, remove 4s electrons before 3d; Fe2+ is [Ar]3d6 and Fe3+ is [Ar]3d5.
- Variable oxidation states — Similar energies of 4s and 3d electrons allow several oxidation states, but their stability depends on ligands, solution conditions and redox partners. A list of possible Roman numerals is not a prediction that every state is equally stable. Use charge balance and potential evidence for the species actually present.
- Ligands and coordination — A ligand donates a lone pair to a central metal atom or ion, forming a coordinate bond. The coordination number counts donor atoms directly bonded to the centre, not the number of ligand molecules or the overall charge. Water and ammonia are neutral ligands; hydroxide and chloride carry −1 charge.
- Writing complex charges — For [Fe(H2O)6]3+, neutral waters leave iron at +3. In [CoCl4]2−, four −1 chlorides and total −2 require cobalt at +2. Square brackets identify the coordination entity. Do not infer a metal oxidation state by dividing the complex charge by the coordination number.
- Octahedral and tetrahedral — Six donor bonds commonly give octahedral geometry with 90° adjacent and 180° opposite angles. Four chloride ligands can give tetrahedral complexes such as [CoCl4]2− with ideal angles about 109.5°. Use perspective diagrams and actual coordination number; four-coordinate does not automatically mean square planar.
- Square-planar isomers — Square-planar [Pt(NH3)2Cl2] has cis and trans arrangements, with the two chloride ligands next to each other in cis and opposite each other in trans, with identical formula but different relative ligand positions. Cisplatin refers to the cis form and its biological activity depends on molecular arrangement and interactions. This structural example does not mean every square-planar isomer has the same effect or provides advice about medical treatment.
- Multidentate ligands — Ethane-1,2-diamine, often abbreviated en, is bidentate, meaning it bonds through two donor atoms: two nitrogen lone pairs can bind the same metal. EDTA4− is commonly hexadentate, providing six donor atoms. Three bidentate en ligands can therefore give coordination number six, while one hexadentate ligand can occupy six sites; ligand count and donor count differ.
Connectivity schematic of one ethane-1,2-diamine ligand donating from both nitrogen atoms to one metal. Each nitrogen also has two hydrogens, omitted for clarity; the connector shows the N–CH2–CH2–N chain, with no direct nitrogen–nitrogen bond. Three such ligands provide six donor atoms. This is not a full three-dimensional complex geometry. - Chelation and entropy — Chelation occurs when one ligand binds to the same metal through two or more donor atoms, forming a ring that includes the metal. A multidentate ligand can replace several separate ligands, often increasing the number of free particles in solution and favouring complex formation through entropy. For example [M(H2O)6]2+ + 3en ⇌ [M(en)3]2+ + 6H2O illustrates the particle argument in a simplified model. Overall stability also depends on enthalpy, protonation and solution conditions.
- Colour and d-level splitting — Ligands split the energies of d orbitals. Absorption of suitable visible light can promote an electron between these levels, leaving transmitted or reflected light with the complementary colour. Different oxidation states, ligands and geometries change the splitting. A substance's observed colour is not the colour of the light it absorbs.
- Limits of the colour model — d0 and d10 species cannot undergo ordinary d–d transitions, but they are not necessarily colourless by every mechanism: intensely coloured manganate(VII) and dichromate(VI) involve other electronic transitions. Colour also depends on concentration and mixture composition, so it is evidence to combine with chemistry rather than a unique metal identifier.
- Haem and ligand exchange — The iron(II) centre in haem is held by a surrounding ligand system and can bind oxygen reversibly, enabling oxygen transport in haemoglobin. Carbon monoxide binds strongly and interferes with oxygen carriage. Explain competing ligands and reversible binding without claiming that all iron in blood is a free hexaaqua ion or drawing an invented haem structure.
Transition-ion reactions and observations
- Vanadium oxidation states — In acidic aqueous chemistry, V(V) is represented by yellow VO₂⁺ with two oxygen atoms and total +1 charge, V(IV) by blue VO²⁺ with one oxygen atom and total +2 charge, V(III) by green V3+ and V(II) by violet V2+. Write these unambiguously as VO₂⁺ and VO²⁺: similar plain-text formulas can otherwise hide different species.
- Reduction sequence — A suitable reducing agent such as zinc in acid can reduce vanadium through several states. Follow electron transfers and actual potential data; mixed successive colours can occur while conversion is incomplete. Exposure to oxygen can reverse some changes, so an observed green mixture need not by itself establish pure V3+.
The successive acidic reduction half-equations are VO₂⁺ + 2H+ + e− → VO²⁺ + H2O, then VO²⁺ + 2H+ + e− → V3+ + H2O, then V3+ + e− → V2+. Their different oxygen contents explain why equal-looking plain text would be misleading.
- Chromium redox colours — Acidified orange dichromate(VI) can be reduced to green Cr3+; stronger reduction with zinc in acid can form blue Cr2+. These are changes in oxidation state, not merely ligand exchange. Chromium(III) solution colour depends on its ligand environment and preparation, so green is a useful specified observation rather than a universal colour for every Cr(III) complex.
The first acidic reduction half-equation is Cr2O7^2− + 14H+ + 6e− → 2Cr3+ + 7H2O; the further step is Cr3+ + e− → Cr2+. A reducing reagent must supply the corresponding electrons and the actual medium determines the feasible species.
- Chromate and dichromate — 2CrO4^2− + 2H+ ⇌ Cr2O7^2− + H2O links yellow chromate and orange dichromate. Acid favours dichromate and alkali favours chromate. Chromium remains +6 on both sides, so this equilibrium is not redox; changes of colour do not automatically establish electron transfer.
Two different chromium transformations: acid shifts yellow chromate(VI) toward orange dichromate(VI) without changing chromium +6; reduction of dichromate to chromium(III) requires electron transfer. Green is the specified aqueous Cr(III) observation, with actual shade dependent on coordination environment. Arrows are a conceptual sequence, not balanced equations. - Alkaline peroxide oxidation — In alkaline solution, hydrogen peroxide can oxidise chromium(III) to yellow chromate(VI). Acidification then shifts chromate toward orange dichromate without changing +6. Separate the redox step from the acid–base equilibrium and balance each in its own medium rather than treating peroxide as a reducing agent in every reaction.
A balanced simplified hydroxide-form equation is 2Cr(OH)3 + 3H2O2 + 4OH− → 2CrO4^2− + 8H2O. Chromium is oxidised from +3 to +6 while peroxide is reduced; subsequent acidification changes chromate/dichromate balance without another chromium oxidation-state change.
- Hydroxide precipitation — A hexaaqua 2+ ion can lose two protons to hydroxide: [M(H2O)6]2+ + 2OH− → [M(H2O)4(OH)2](s) + 2H2O. For a 3+ ion, three deprotonations give [M(H2O)3(OH)3](s). Simplified M(OH)n equations can also show precipitation, but do not label proton removal as replacement of water by neutral ammonia.
- Ammonia as a base — Limited aqueous ammonia accepts protons from coordinated water, giving hydroxide precipitates and NH4+. For example [Cu(H2O)6]2+ + 2NH3 → [Cu(H2O)4(OH)2](s) + 2NH4+. In excess it may also act as a ligand. Distinguish the initial acid–base reaction from a subsequent ligand-exchange reaction.
- Iron observations — Fe2+ solutions give green hydroxide precipitates with NaOH or ammonia, insoluble in excess under these tests and turning brown as air oxidises iron(II). Fe3+ gives red-brown hydroxide, also insoluble in excess. Observe promptly and on standing; an old brown sample cannot be assumed to have started as Fe3+.
- Copper with hydroxide and ammonia — Cu2+ gives a blue hydroxide precipitate with limited NaOH or ammonia. It remains insoluble in excess NaOH but dissolves in excess ammonia to a deep-blue solution, commonly represented by [Cu(NH3)4(H2O)2]2+. The change in excess ammonia is ligand exchange; copper remains +2 rather than being oxidised to a darker-blue state.
The aqua-ion exchange is [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O. Starting from the hydroxide precipitate, the simplified dissolution balance is [Cu(H2O)4(OH)2](s) + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 2OH− + 2H2O. Check atom and charge balance rather than treating ammonia as a source of extra OH groups in the final complex.
- Copper and chloride exchange — [Cu(H2O)6]2+ + 4Cl− ⇌ [CuCl4]2− + 6H2O illustrates exchange to a tetrahedral chloride complex. A blue-to-yellow change can appear green because both complexes coexist. Adding water can shift the equilibrium back; do not describe every green solution as a single green complex.
- Cobalt observations and oxidation — Cobalt(II) aqua solutions are pink. NaOH gives a blue hydroxide precipitate that may turn pink on standing and does not dissolve in excess NaOH. Limited ammonia also gives a hydroxide precipitate; excess ammonia can give a straw/brown ammine solution that darkens through air oxidation. Record reagent concentration, time and air exposure rather than treating one shade as independent of conditions.
A simplified aqua-ion exchange is [Co(H2O)6]2+ + 6NH3 ⇌ [Co(NH3)6]2+ + 6H2O before subsequent oxidation chemistry. The +2 metal state is unchanged in that exchange itself; air oxidation is a separate event.
- Cobalt ligand exchange — [Co(H2O)6]2+ + 4Cl− ⇌ [CoCl4]2− + 6H2O links pink octahedral and blue tetrahedral species without changing cobalt's +2 oxidation state. Chloride concentration and temperature influence the equilibrium. A blue cobalt complex here is a solution, distinct from a blue hydroxide precipitate in a base test.
- Chromium with bases — Cr3+ gives a green hydroxide precipitate with limited NaOH. Excess NaOH dissolves it to a green hydroxo complex, often written [Cr(OH)6]3−, showing amphoteric behaviour: the hydroxide can react with acid and with a sufficiently strong base. Ammonia initially gives green hydroxide; ligand exchange in sufficiently concentrated excess can be slow and condition-dependent. Do not equate an immediate dilute-ammonia test with every long-standing concentrated-ammonia result.
A simplified amphoteric dissolution equation is Cr(OH)3(s) + 3OH− ⇌ [Cr(OH)6]3−. This differs from ammonia acting as a ligand. Pearson's practical-guide excess-ammonia observations include a green solution after the initial green precipitate under its stated test conditions; the speed and exact coordinated species depend on preparation and conditions, so colour alone must not be used to assign an exact ammine formula.
- Comparing reaction types — Classify a change by species and conservation: precipitation removes an insoluble phase; deprotonation transfers H+; ligand exchange changes donors; redox changes oxidation number. A single sequence may contain more than one type. Equations, phases and conditions provide stronger evidence than a rule that colour change always means a particular reaction.
Catalysis and preparing a complex
- Catalyst and pathway — A catalyst provides an alternative pathway with a lower activation barrier and is regenerated overall. It can change reaction rate without changing the equilibrium constant or equilibrium composition at fixed temperature. Faster approach to equilibrium is not a higher equilibrium yield; follow temporary catalyst species through the mechanism before cancelling them.
- Heterogeneous surfaces — A catalyst in a different phase can adsorb reactants onto active surface sites, weaken suitable bonds, allow reaction and desorb products. Increased accessible surface can increase rate, but binding that is too strong can block sites and poisons can deactivate them. Adsorption is surface attachment, distinct from absorption into a bulk material.
- Vanadium oxide cycle — In the Contact process, a simplified cycle is V2O5 + SO2 → V2O4 + SO3, followed by V2O4 + 1/2 O2 → V2O5. Adding gives SO2 + 1/2 O2 → SO3. Vanadium changes between +5 and +4 and the catalyst is regenerated; these equations summarise redox participation rather than specifying every surface intermediate.
- Vehicle exhaust catalysis — Catalytic conversion can combine oxidation of CO and hydrocarbons with reduction of nitrogen oxides. For example 2CO + 2NO → 2CO2 + N2 is balanced redox at a suitable catalytic surface. Performance depends on temperature, reactant mixture and catalyst condition; a converter does not remove all pollutants or all carbon dioxide production.
- Homogeneous iron cycle — For oxidation of iodide by peroxodisulfate, 2Fe2+ + S2O8^2− → 2Fe3+ + 2SO4^2−, then 2Fe3+ + 2I− → 2Fe2+ + I2. Adding cancels the iron species, giving S2O8^2− + 2I− → 2SO4^2− + I2. The dissolved catalyst changes oxidation state temporarily and is regenerated.
- Autocatalysis — In the acidified manganate(VII)–ethanedioate reaction, Mn2+ formed can catalyse further reaction. A slow initial rate followed by acceleration can reflect increasing catalyst concentration, before reactant depletion eventually slows the reaction. Autocatalysis does not mean a reaction needs no activation energy or that products always speed reactions.
- Core Practical 12 preparation — Prepare the specified transition-metal complex using the approved reagent amounts, reaction conditions and isolation method. Separate, wash and dry the product appropriately, record mass and assess yield or purity using relevant evidence. Colour alone cannot confirm coordination structure or purity, and a wet product can give an artificially high apparent yield.
- Synthesis evaluation — Trace the limiting reagent, formula and hydration state into a theoretical yield, then compare actual dry mass. Explain losses during transfer, solubility or filtration and risks from oxidation or contamination. Preserve actual experimental records; an independently answered quiz cannot certify the practical preparation, identification or safe technique.