ALL my KS3 SCIENCE Revision Quizzes (~US K12 grades 6,7,8)

GCSE-IGCSE-KS4 Science-CHEMISTRY notes & quizzes (~US K12 grades 9-10)

Advanced Level CHEMISTRY GCE AS A2 IB notes and quizzes (~US K12 grades 11-12)

All my GCSE-IGCSE Science-CHEMISTRY etc. syllabus help links

 All my GCE-AS-A2-IB AQA, Edexcel, OCR etc. Advanced Level Chemistry syllabus-specification help links

 Doc Brown's Chemistry Clinic

My unofficial support for Salters AS Advanced Chemistry

Salters AS Chemistry - 'exam bashing' thoughts for

Unit M "From Minerals to Elements" - part of module 2848

M Unit map & learning objectives * other M backup material  * My Salters AS homepage * My Salters A2 homepage * My Salters revision index * EMAIL query?comment

PLEASE REMEMBER, THESE ARE NOT 'STAND ALONE' NOTES, and were designed for my classes for use alongside the Salters resources - Chemical Ideas, Chemical Storylines, Practical Activities-Investigations and the AS-A2 Revision guides all published by Heinemann Secondary Series, to reduce the reading workload and offer a study strategy. From your teacher (not me!), its handy to have the answers to the Chemical Ideas, Storylines Assignments and Activities Questions side by side with the texts and these strategy pages. You haven't time to redo the Q's but a quick read of the Q's and connecting with the official answers is valuable revision - there is too much hit and miss revision from doing past papers in my opinion.

Storylines M1 "CHEMICALS FROM THE SEA"

• Chemicals from the sea: sea water is a useful source of minerals, especially the highly concentrated Dead Sea (due to evaporation and no outflow), and it is particularly rich in bromides.

• Assignment 2 is a typical industrial scale calculation - important in the Salters course style!

• pH scale: revision reminder in green box p47

• Extracting bromine from the sea (eg from the Irish sea) which is done in 4 stages

• Stage 1 Oxidation of bromide ions Br^- to bromine Br₂:

  • filtered sea water is acidified with sulphuric acid to minimise reaction of halogens with water, which leads to loss of bromine and wasted chlorine X_2(aq) + H₂O_(aq)   HX_(aq) + HXO_(aq) (X = Br or Cl) increasing H⁺ moves equilibrium to the left, 'suppresses' H⁺XO^- acid formation

  • excess chlorine added to displace the bromine Cl_2(aq) + 2Br^-_(aq) ==> 2Cl^-_(aq) + Br_2(aq) 

  • be able to analyse the above equations in redox terms and oxidation state changes

    • eg Cl changes from 0 to -1, and Br changes from -1 to 0

• Stage 2 Removal of bromine vapour:

  • air blown through which extracts he easily vaporised bromine

• Stage 3 Reduction of bromine Br₂ to hydrobromic acid HBr:

  • sulphur dioxide gas is mixed with the air/bromine mixture and both are in contact with fresh water and the mixture condensed to form a concentrated solution of hydrobromic acid

  • this reduces the bromine to bromide (as hydrobromic acid) and also produces a much more concentrated solution: Br_2(aq) + SO_2(g) + 2H₂O_(l) ==> 2HBr_(aq) + H₂SO_4(aq) 

    • ox. state/number changes: S from +4 to +6, and Br from 0 to -1

• Stage 4 oxidation of hydrobromic acid HBr to bromine Br₂:

  • steam and chlorine are blown through the acid solution to reform bromine

  • be able to 'redox' analyse: Cl_2(aq) + 2HBr_(aq) ==> 2HCl_(aq) + Br_2(g) 

  • the hot vapour mixture is condensed forming an upper aqueous layer and a lower more dense wet bromine layer, the bromine is dried with concentrated sulphuric acid (powerful dehydrating agent)

  • the used sea water is treated with sulphur dioxide to remove traces of toxic chlorine or bromine and discharged - although quite acidic, it is rapidly diluted in the sea

• Assignment 3: asks you to consider +/- factors in make bromine from seawater or Dead sea water

  • eg no evaporation needed for sea water BUT more (eg chlorination) stages needed in the process

  • Dead Sea water is more concentrated in bromide and one chlorine treatment sufficient but there is still the energy costs for evaporation and crystallisation

• BROMINE - dense, dark, dangerous BUT useful:

  • flame retardants like TBBA

  • 1,2-dibromoethane is used as a lead oxide remover in car engines because it forms 

  • agri-chemicals like the pest fumigant bromomethane

  • but bromine is an ozone 'destroyer' (see unit A next) so its use is hoped to be phased out

  • silver halide salts are used in photographic film - when visible light, uv, X-rays or gamma rays hit silver halide crystals metallic silver is formed. An electron is knocked off the halide ion and reduces the silver ion to silver: Ag⁺X^- + hv ==> Ag + X (Assignment 5 tests you on the ideas)

• Making Chlorine:

  • Chlorine is manufactured by the electrolysis of brine (salt solution of sodium chloride)

  • the salt is readily obtained from mining rock salt, which is almost pure NaCl

  • at the negative electrode (cathode) hydrogen ions are reduced by electron gain to form hydrogen gas:

    • 2H⁺_(aq) + 2e- ==> H_2(g)

  • at the positive electrode (anode) chloride ions are oxidised by electron loss to form chlorine gas:

    • 2Cl^-_(aq) ==> Cl_2(g) + 2e^-

  • left in solution is the useful alkali sodium hydroxide from the residual Na⁺ and OH^- ions.

  • Chlorine has many 'positive' uses eg manufacturing hydrochloric acid, the useful hard wearing plastic PVC, chlorinated organic solvents like trichloroethane for cleaning, bleaches, disinfectants (TCP, dettol etc.), water treatment to kill bacteria, BUT the use of  chlorinated organic pesticides is not so welcome due ecosystem damage increasing up the food chains.

  • The hydrogen formed can be used to make hydrochloric acid, or ...

    • hydrogenate unsaturated fats to more saturated margarine

      • >C=C< + H₂ => >CH-CH< (via Nickel Ni catalyst)

    • Haber Synthesis manufacture of ammonia: N_2(g) + 3H_2(g) ==> 2NH_3(g)

  • The sodium hydroxide is used in the manufacture of paper, soluble sodium salts of acidic materials (eg soluble aspirin), ceramics, soaps and detergents

Chemical Ideas CI 5.1 "Ions in solids and solutions"

• Ionic solids:

  • revise typical physical properties eg high mpt, insulator etc. based on the following model ...

  • ions held together by opposite electrical charges

  • each positive ion (cation) strongly attracts several negative ions (anions) and vice versa

  • ions held strongly in fixed positions to form the ionic lattice and this strong '3D' bonding accounts form the solids being hard and having high melting/boiling points

  • you should be able envisage the 6:6 in NaCl (Fig 1) and x:y in CsCl Q6

• Hydrated crystals:

  • when some salts crystallise, water molecules are incorporated into the crystal lattice, the so called water of crystallisation

  • these are known as hydrated crystals, but without them, they are referred to as the anhydrous form, often produced just by heating the hydrated crystals

  • eg blue copper(II) sulphate CuSO₄.5H₂O consists of a mole ratio lattice of 1 Cu²⁺ ion : 1 SO₄^2- ion : 5H₂O molecules) and anhydrous is just white CuSO₄

• Ionic substances in solution:

  • many ionic substances readily dissolve in water (particle picture Fig 2)

  • the ions become surrounded by water molecules and these +/- hydrated ions are no longer associated and are independently distributed at random throughout the solution

• Ionic equations:

  • If two solutions are mixed, and the combination contains the two ions that can form an insoluble substance, a precipitate forms.

  • The change can be concisely represented with an ionic equation, in which the spectator ions not involved are omitted from the equation representing the ionic precipitation.

  • Study the 1st example of an ionic equation showing the formation of insoluble silver chloride.

  • The 2nd example is the ionic equation for neutralisation. This simple ionic change happens when any acid (forming H⁺) is mixed with any alkali (forming OH^-) and the result is water formation. Again, note the exclusion of the spectator ions.

• Dissolving ionic solids:

  • Not all ionic solids dissolves and ultimately it is to do with energy and entropy changes.

  • Ions are strongly attracted in the solid, but they separate in the water when they become hydrated (solvated with water molecules, Fig 5).

  • The solvated ions are larger and the ionic charge is more spread out over the surface of the hydrated ions and considerably reduces the attractive force between them, and so they can randomly mix with the water molecules.

  • The energy is needed to separate the ions comes from the hydration process. The water molecule has a permanent dipole H^d+-O^d--H^d+ and the ^d- polar end is attracted to the cation and the ^d+ polar end is attracted to the anion.

  • The water-ion bonds are weak, but if strong enough (usually with the cation), the water molecules cling to the cation when the water is evaporated to form the salt eg hydrated copper(II) sulphate.

  • Table 1 shows that (i) cations are much more strongly attracting than anions (ii) the smaller the cation, and (iii) the more highly charged it is, the more water molecules are held in the hydrated ion. 

Chemical Ideas CI 9.1 "Oxidation and reduction"

• Oxidation and reduction:

  • oxygen loss (reduction) and oxygen gain (oxidation) are far too limited for A level!

  • electron definitions of oxidation or reduction (electron loss or gain)

  • conceiving of 'redox' reactions as two half-reactions or half-equations, one an oxidation and the other a reduction

  • an electron acceptor is an oxidising agent (and gets reduced in its action)

  • an electron donor is a reducing agent (and gets oxidised in the process)

  • in the book examples:

    • the magnesium atoms are oxidised (electron loss) by the electron accepting oxidising agents oxygen or chlorine

    • the oxygen and chlorine molecules are reduced (electron gain) by the electron donating reducing agent magnesium

• Displacement reactions:

  • displacement reactions (for metals or non-metals) are often redox reactions

  • they can be 'split' into two half-reactions, note spectator ions should not be included (these are ions which do not take part in the reaction ie do not change their chemical 'status'!

  • the equations for them can be expressed in full formula style

    • eg Cl_2(aq) + KI_(aq) ==> etc.

  • OR more fundamentally as redox ionic equations with spectator ions (eg K⁺) excluded

    • eg Cl_2(aq) + I^-_(aq) ==> etc.

  • note on the balancing: electrons are shown in the half-cell reactions BUT not in the ionic equation, getting the right ratio of one half-cell reaction to another should ensure their transfer is 'hidden'!

• Oxidation state:

  • when dealing with atoms and simple ions, redox analysis is usually quite straight forward BUT what about when dealing with electrically neutral molecules eg H₂O and ions like PO₄^3- ?

  • these situations can redox analysed by using the concept of oxidation number or oxidation state

  • some basic rules (and learn them as quickly as possible):

    1. atoms in elements, ie not combined with atoms of other different elements, are considered to have an oxidation state of zero or 0

    2. in simple ions the oxidation state is the same as the charge on the ion eg Al³⁺ has an oxidation state of +3, oxide ion O^2- has an oxidation state of -2 (note in chemical symbols the charge sign is after the number, the charge sign is placed first for oxidation states and exam boards are strict about this sign convention)

    3. since compounds have no overall charge, the sum of the oxidation states is zero, this is easy to follow in simple ionic compounds but not so easy at first in covalent molecules or molecular ions like PO₄^3-

    4. For covalent molecules or 'molecule ions' the most electronegative element or elements will have the negative oxidation state.

    5. irrespective of being in an ionic or covalent compound: F is always -1, H is +1 except in the hydride ion H^- (eg in NaH),  O is -2 except in peroxides H₂O₂ or O₂^2-,were it is -1 and O is +2 in F₂O. Cl is -1 except when combined with O or F (see examples later eg Cl is +1 in Cl₂O)

    6. for ions like PO₄^3- the charge on the ion is equal to the sum of the oxidation numbers, so P is +5, O is -2, so +5 + (4 x -2) = -3 = 3- for overall charge on ion.

       • Note the convention:

         • for oxidation state, the sign comes before the number

         • for charge on an ion, the sign comes after the number

    7. using these rules you can work out the oxidation states of the other elements in the molecule or ion

• Oxidation states in names:

  • different compounds or ions can be formed from the same elements, so at least one of the elements is in a different oxidation state

  • so there is a need to indicate this when writing the name, but note:

    • Roman numerals are used

    • the number shows the oxidation state of the element (usually +ve)

    • the number is placed in brackets after the name, but with no space between it.

• Using oxidation states:

  • you can do a detailed redox analysis of any equation ie what is oxidised and what is reduced

  • it helps you to balance equations when given two half-reactions

  • and note:

    • oxidation is increase in oxidation state/number (e.g. -2 to -1, -1 to 0, 0 to +1, +2 to +4)

    • reduction is decrease in oxidation state/number (e.g. -1 to -3, 0 to -1, +2 to 0, +3 to +2) 

Chemical Ideas CI 2.4 "Electronic Structure: sub-shells and orbitals" tricky!

1. The principal quantum electronic energy levels are split into sub-levels denoted by s, p, d and f (Figs 10 and 11, don't worry too much about f electrons!).

2. for n = 1, just one sub-shell (s), for n = 2 there are two sub-shells (s and p), for n = 3 there are three sub-shells (s, p and d).

3. The maximum electrons permitted in each sub-shell is s (2 e^-s), p (6 e^-s), d (10 e^-s) and f (14 e^-s).

4. The 'space' in which the electron exists with its particular quantum level energy is called the atomic orbital.

5. The s, p, d and f sub-shells are further split into these atomic orbitals, each orbital can contain a maximum of two electrons of opposite spin, so the maximum number of orbitals per sub-shell is s (1 orbital), p (3 orbitals), d (5 orbitals) and f (7 orbitals).

6. Check out electron number connection between points 2, 3 and 5, it should all add up!

7. To accurately describe an electron requires four quantum numbers. They are for principal shell, its sub-shell, orbital type and spin (don't worry too much about these numbers, the features described just come out in the 'descriptive process'.

8. The arrangement of electrons in the shells and orbitals is called the electronic configuration.

9. The orbital electrons are denoted in the form of eg 2p³, means there are three electrons in the p sub-shell in the second principal quantum shell.

10. The orbitals are filled in a definite order to produce the system of lowest energy.

11. The order of filling (up to atomic number Z = 36, H to Kr)  is 1s 2s 2p 3s 3p 4s 3d 4p (total 36 electrons), note the 'quirk' in order for 3d.

12. Table 8 shows how they are written out up to Z = 20 and note the orbital order when writing out (but see Q3 p33, for higher atomic numbers they are written out in strict order of 1, 2, 3 etc. and each number followed by s, p or d order sub-levels  etc., and this is irrespective of the order of filling, ie ignores the 3d 'quirk' etc. 

13. On page 20 some are written out in box diagram format, each box represents an orbital with a maximum of two electrons of opposite spin (shown by the opposing arrows). Note the electrons only pair up when all sub-orbitals are filled separately (this minimises electron pair repulsion).

14. Elements with one or two outer s electrons, and no outer p electrons, are called s-block elements (Groups 1 and 2); elements with at least one outer p electron are called p-block elements (Groups 3 to 8/0); elements where the d sub-shell is being filled are called d-block elements (*Transition Metals). See Figs 14 and 15, note the outer electron arrangements shown in Fig 14.

    • * Sc-Zn is the 3d block, BUT true transition elements form at least one ion with a partly filled sub-shell of d electrons

      • (Sc only forms Sc³⁺ [Ar]3d⁰, Zn only forms Zn²⁺ [Ar]3d¹⁰4s² 

15. To work out the electron configuration of ions ...

    • for negative ions, use the same rule as in point 11. eg chloride is chlorine + 1 electron so it becomes 1s²2s²2p⁶3s²3p⁶, ie it has the electronic structure of argon.

    • for positive ions, the electrons are removed in the strict reverse order of how the electron arrangement is written.

    • eg for Z=20, Ca is 1s²2s²2p⁶3s²3p⁶4s²; Ca²⁺ is 1s²2s²2p⁶3s²3p⁶

    • for Z=26;  Fe is 1s²2s²2p⁶3s²3p⁶3d⁶4s²; Fe³⁺ is 1s²2s²2p⁶3s²3p⁶3d⁵ (remove 4s first and then one of the 3d electrons)

    • see Fig 20 p260 for box diagrams (also can you spot the other electronic 'quirks' for chromium and copper!).

16.  

Chemical Ideas CI 11.4 "The p-block: Group 7 Halogens"

• When studying Group 7 be prepared to 'predict' the properties of radioactive astatine (At) in true Mendeleev style! (and F too, since too reactive to do expts. in school labs.)

• They exist as diatomic molecules, X₂, and know the appearance (state/colour) of fluorine, chlorine, bromine and iodine (s, l or g and in solution) - particularly at room temperature, see tables 4 and 5.

• Be able to deduce ox electronic diagrams for simple covalent molecules (element or compound)  and ionic compounds (p256 and p34), possibly more awkward ones like Fig 19 p257.

• Be able to deduce the oxidation state of the halogen in a variety of situations (see Q2, +5 in Fig 19, 0 in Fig 17, -1 in fig 18 etc.).

• They react with metals, usually to form ionic compounds. (revise typical physical properties)

• They react with non-metallic elements to form covalent compounds (including other halogens!).

• Use the reactivity order to deduce which halogen will displace another ..

  • eg X_2(aq) + 2KY_(aq) ==> 2KX_(aq) + Y_2(aq) (where halogen X is more reactive than halogen Y)

  • and write out the corresponding ionic redox equation: X_2(aq) + 2Y^-_(aq) ==> 2X^-_(aq) + Y_2(aq) 

  • and the reactivity decreases down the group, to give order F > Cl > Br > I

  • these are examples of non-metallic displacement reactions and be able to 'analyse' them in terms of oxidation state/number changes.

• Halide ions, X^-, can be converted into the free halogen using an oxidising agent (eg potassium manganate(VII) or anodic electrolysis (see Act M 1.5 and CS M1).

• The precipitation of silver salts is important and know the observations eg the colours of the precipitate for halide ion identification (AgX salts are used in photography, see CS M1)

  • eg AgNO_3(aq) + NaX_(aq) ==> AgX_(s) + NaNO_3(aq) where X is the halogen

  • better expressed as the ionic (NOT redox) equation: Ag⁺_(aq) + X^-_(aq) ==> AgX_(s) 

  • when silver salts are exposed to light, silver is formed.

• Halide salts are formed by neutralising the corresponding acid and alkali ...

  • eg NaOH_(aq) + HCl_(aq) ==> NaCl_(aq) + H₂O_(l)

  • but this is a ridiculously expensive method of production, evaporation of seawater is much more simple and cost effective!

• Make sure you can clearly distinguish between a redox, precipitation or acid-base neutralisation (and not impossible for two of the terms to coincide for the same reaction!).

Activity M1.1 "Solutions of ions"

• Know the meaning and use of the words: mixing, diluting, dissolving, precipitating, solution, insoluble and soluble salt, acid-base reaction.

• The relative molarity ideas are important ie distinguish between the molarity of the salt and its constituent ions free in solution.

• Should mull over the full and ionic equations for the reactions observed.  

Activity M1.2 "Bromine production"

• The questions are of a useful interpretation style.

Activity M1.3 "Halogens and their compounds"

• This activity supports CI 11.4 but adds little to the exam study of the chapter in terms of observations BUT the Questions a to h are important and of exam style. 

Activity M1.4 "This liquid is dangerous"

• Re-read to revise ideas on chemical industry safety when dealing with hazardous chemicals and some exam style interpretation questions. 

Activity M1.5 "Manufacturing chlorine"

• Plenty of exam style Q's to study from basic principles of moles to interpretation and concept Q's eg about the design and function of the membrane. 

Storylines M2 "COPPER FROM DEEP UNDERGROUND"

• Copper, discovered by accident, useful and attractive properties eg easily shaped, beautiful appearance, cast in moulds etc.

• It can be alloyed with tin to make bronze an even harder and more useful material.

• You do not need to know the detailed origin of copper minerals, but you should realise the mineral must have enough copper in it to be economically viable.

• You also need to be able to interpret ore formulae (p54) eg in calculating % copper and writing balanced ore reduction equations (see assignment 6).

  • (can be tricky, and be able to deduce the oxidation state of the metal in a given ore)

• You are expected to be able to interpret or produce a flow diagram Fig 21 (input/output style is important) and answer specific questions on the various stages of the extraction processes but only when given information. (Rates of reaction factors from CI 10.1 may be included in interpretation questions)

• So read p53-59 once carefully, so that you have an overview of how a pure metal is obtained from its ore and any resulting environmental impact (Assignment 7). More specific knowledge is pointed out below.

  • Mining the ore - major environmental impact, mined ore is not very concentrated in terms of % copper, lots of waste material involved.

  • Concentration - the idea is to concentrate the high % copper particles using froth flotation, which proves a very effective process (Assignment 9).

  • Smelting - the concentrated copper sulphide ore is now reduced to copper (using oxygen, not carbon!), redox interpretation of the process (Assignment 10). (oxidation state changes)

  • Electrolytic refining - the impure copper from the smelter is made the anode (+) in an electrolytic cell, pure copper is deposited on the cathode (-). You need to be able to write the electrode equations and interpret them as redox changes (p58). Note valuable 'waste' material - the residual 'anode sludge' contains silver and gold! contains

• Rock eating bacteria - two themes to appreciate (i) redox chemistry (Assignment 10) and (ii) an economically viable way of getting more copper from the waste material in the dump! Bacterial leaching is cheaper, quieter and less polluting BUT it is much slower.

• Copper: A vital element - Study the 'green box' thoroughly as an example of the typical properties and related uses of a Transition Metal and its alloys and compounds.

  • Physical properties: the metal is high melting, high density, good conductor of heat and electricity, malleable, can be cast - use in pans, electrical wiring, bronze, brass  etc.

  • Chemical properties: Variable oxidation state (eg +1 and +2 in oxides and chlorides), catalytic properties (in enzymes and industrial processes), coloured compounds (pottery glazes, stained glass) and electronic structure of 3d block and Transition elements

Chemical Ideas CI "1.5 Concentration of solutions"

• Concentration can be quoted as g dm^-3 but it is usually more useful to use molarity, especially when its important to monitor or predict the relative amounts of reactants.

• Molarity is defined as moles of solute per dm³ of solvent (1 dm³ = 1 litre = 1000 cm³).

• Moles of solute = mass in g/ relative formula mass

• To go from g dm^-3 to molarity you divide by the formula mass in g (molar mass)

• To go from molarity to g dm^-3 you multiply by the relative formula mass in g (molar mass)

• mol = molarity x volume in dm³, or mol = molarity x volume in cm³ /1000 (since V in cm³/1000 = dm³).

• Volumetric problem solving is not adequately covered in CI 1.5 but if you can do Q's 11-15 you have cracked the more difficult ones.

• The only way to really learn calculations is to work through as many as possible, and not just the CI 1.5 Q's, eg some odd scribbles of help on CI 1.5 Q's and the  extra volumetric problem sheet

• but please note other calculations will crop up eg % composition, empirical formula, reacting mass in g or kg, % of a metal in a rock - based on % of its compound in the rock

Chemical Ideas CI 8.1 "Acid-base reactions"

• A proton is the H⁺ ion. In water it exists as the oxonium ion H₃O⁺_(aq), where the proton forms a dative covalent bond with the donated lone pair from the oxygen atom.

• The Bronsted-Lowry theory of acid states that acids are proton donors (eg HCl, H₂O, H₃O⁺), and bases are proton acceptors (eg NH₃, H₂O, OH^-), see the two examples on p191-192, and note water like the HCO₃^- ion, can act as an acid or a base (amphoteric behaviour).

• A reaction is which a proton is donated from one species to another is called an acid-base reaction.

• An acid will form the H⁺ or H₃O⁺ in water and an alkali will form the OH^- ion in water.

• An alkali is a water soluble base (Fig 4), and don't assume it will have OH in the formula!, ammonia and the carbonate ion react with water slightly to form the hydroxide ion. Similarly don't assume all acids have a 'H' in the formula, carbon dioxide forms an acidic solution.

• When an acid donates a proton, the anion formed is a base, because it could attract the proton back.

• Similarly, when a base accepts a proton, it could donate it back again, and so acts as an acid.

• This is known as a conjugate acid-base pair situation (see Tables 1 and 2).

• A strong acid is a powerful proton donor eg HCl, HNO₃ or H₂SO₄, ie these are highly ionised in solution, on proton donation they form a very weak conjugate base eg the Cl^-, NO₃^- or SO₄^2- ions.

• A weak acid eg CH₃COOH, is a weak proton donor, ie it only ionises to a small extent to form the H⁺_(aq) ion and the corresponding anion CH₃COO^- is a strong conjugate base.

• So in terms of conjugate acid-base pairs: a strong acid forms a weak conjugate base and a weak acid forms a strong conjugate base.

• Many acid-base indicators like litmus, methyl orange or phenolphthalein are weak acids. The acid and its conjugate base are different colours and so the relative concentrations of the two species is pH dependant (ie H⁺ ion concentration dependant). The interchange between the two coloured forms is an acid-base reaction.  

Chemical Ideas CI 5.2 "Molecules and networks"

• The oxides of carbon and silicon, XO₂, in group 4 show strikingly contrasting physical properties and you need to be able to explain in words and diagrams why this is so.

• Carbon dioxide O=C=O is a small covalent molecular substance* with a low melting point due to the weak inter-molecular forces being easily overcome by mild thermal agitation. It is moderately soluble in solvents like water, again, because the inter-molecular forces can be overcome in the solvation process.

  • * note 2 polar bonds, C^d+=O^d-, but cancel out to give overall a non-polar molecule

• In carbon dioxide, the smaller C atom can form a double bond with oxygen, but in silicon dioxide (silicon(IV) oxide, silica, quartz) the larger Si atom can only form single Si-O bonds.

• The result is SiO₂ has a giant 3D covalent lattice structure or network in which each silicon atom forms 4 bonds to an oxygen in a 'tetrahedral' spatial arrangement.

• The structure is therefore held together by strong covalent bonds (not weak intermolecular forces) and so it is far more thermally stable giving a high melting point, and insoluble in any solvent, because solvation energies are much lower than covalent bond energies.

• A summary of the contrasting sets of properties is given on page 92.

• Elements which form small molecules, ie non-network bonding arrangements are shown in Fig 7 p92. These have low melting/boiling points and usually soluble in at least one solvent.

• Carbon and silicon are two elements which form giant covalent structures ie they are high melting and insoluble solids.

• Carbon (diamond) and silicon form networks based on tetrahedral arrangements of C-C or Si-Si bonds around each atom (Figs 10a/11a). Both are very hard substances because of the strong bonding, diamond is harder because the smaller C atoms give shorter stronger bonds. Both are poor conductors of electricity because the outer electrons are strongly held and localised between the two atoms of any bond.

• However, carbon in the form of graphite, forms hexagonal ring layers in which the three C-C single bonds are supplemented by delocalised electron bonding from the 4th out electron of carbon. This makes graphite a moderately good electrical conductor as the electrons can move freely through a layer. The layers are held together by weak inter-molecular forces and easily slip over each other making graphite a 'slippery' brittle solid. But as a giant covalent structure it is still high melting and insoluble.

• Relatively recently (and another case of serendipity!) a 3rd form of carbon has been discovered in the form of the 'ball shaped' fullerenes (Fig 13).

Activity M2.1 "Mineral spotting"

• Just makes the point of how complex any mineral mixture is, and so poses the problems of separation, concentration of the ore to finally reduce it to the metal. 

Activity M2.2 "Getting at the minerals"

• No use for module exam except different chemical extraction methods work, choose the most efficient one. 

Activity M2.3 "Extracting copper"

• Illustrates complex separation and extraction techniques and in Q a,b,c you have to consider maximum yield possible versus time/efficiency of eg multiple leaching/washing ie the need for an optimum economic option.

• you need to be able to sketch the apparatus and describe the procedure for vacuum filtration

  • (this reduced pressure considerably speeds up the filtration rate, offer a few details like wash filtered solid with a little clean solvent etc.).

• You need to be able to write equations asked for in Q d and appreciate the displacement chemistry used to free the copper. In industry scrap iron is used instead of zinc.

Activity M2.4 "Finding out how much acid there is in solution"

• Both the method and calculations must be known in detail and a nice little industrial scale calculation as a finale.

• A simple working knowledge error calculations will do for the exam (I don't think they have asked for this yet?). 

• Be able to describe the basic volumetric procedure, name/describe apparatus and its use, indicator/end-point, calculations)

• also need to be able to do calculations involving sodium thiosulphate and iodine

Activity M2.5 "The philosopher's microbe"

• Read once for ideas? 

Activity M2.6 "Molecules and networks"

• This activity supports and summarises many of the important points of Chemical Ideas 5.2. The Q's are worth mulling over briefly. 

Storylines M3 SUMMARY and Activity M3 Check your notes on From Minerals to Elements, Learning Objectives List and M UNIT TEST all for Module 2851 (2nd module)

GENERAL REVISION NOTES

 * Salters Advanced Level Chemistry * Salters Advanced Level Chemistry * Salters Advanced Level Chemistry * Salters Advanced Level Chemistry * Salters Advanced Level Chemistry *