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IB CHEMISTRY Option A Materials Higher level A.1 Materials science introduction OBJECTIVES • Materials are classified based on their uses, properties, or bonding and structure. • The properties of a material based on the degree of covalent, ionic or metallic character in a compound can be deduced from its position on a bonding triangle. • Composites are mixtures in which materials are composed of two distinct phases, a reinforcing phase that is embedded in a matrix phase. • Use of bond triangle diagrams for binary compounds from electronegativity data. • Evaluation of various ways of classifying materials. • Relating physical characteristics (melting point, permeability, conductivity, elasticity, brittleness) of a material to its bonding and structures (packing arrangements, electron mobility, ability of atoms to slide relative to one another). Classification of materials A. By bonding using the Van Arkel-Ketelaar Triangle of Bonding B. By physical properties: 1. Metallic 2. Ceramic 3. Polymer 4. Composite Van Arkel-Ketelaar Triangle of bonding One way to classify materials is based on their bonding. 100% ionic = CsF 100% covalent = F2 100% metallic = Cs All other materials have intermediate character of metallic, covalent, and ionic. The degree of ionic/covalent bonding is approximated by this formula: % ionic character = Δχ/3.2 x 100% Problem 1: Classify diamond, silicon dioxide, and bronze (Cu/Sn) according to their bonding. Substance χaverage Δχ % ionic character Diamond 2.6 0 Δχ/3.2 x 100% = 0 Silcon dioxide (1.9+3.4)/2=2.65 3.4-1.9=1.5 1.5/3.2 x 100% = 47 Cu/Zn (1.9+2.0)/2=1.95 2.0-1.9=0.1 Δχ/3.2 x 100% = 3 ∴ By using the bonding triangle: Diamond = covalent Silicon dioxide = polar covalent Bronze = metallic Problem 2: Tin(II) chloride and lead(II) chloride are both used in the production of artwork glass. One exists as molecules in the vapour phase. Determine which one this is likely to be. Δχ(SnCl2) = 3.2-2.0 = 1.2 % ionic character = Δχ/3.2 x 100% = 1.2/3.2 x 100% = 38% Δχ(PbCl2) = 3.2-1.8 = 1.4 % ionic character = Δχ/3.2 x 100% = 1.4/3.2 x 100% = 40% ∴ As SnCl2 has a more covalent character it is more likely to be the one that exists as discrete molecules in the vapour phase. 1. Metallic Melting point Permeability Electrical conductivity Elasticity Brittleness Ductility High as more electrons involved in bonding. Group 1 have the lowest melting points. Low High due to delocalized electrons. Low Low High as metal ions can be pulled apart and delocalized electrons can accommodate the change in the lattice structure. 2. Ceramic • Includes glass, semiconductors Melting point High due to strong ionic and covalent bonding Permeability Can be porous with gaps in structure Electrical conductivity Low as electrons held firmly in covalent bonds or ions Elasticity Brittleness Low High due to ionic and covalent bonding Ductility Low 3. Polymer Melting point Permeability Electrical conductivity Elasticity Brittleness Ductility Low to due weak intermolecular forces. Thermosetting plastics burn. None Low due to covalent bonding Low (except for some elastomers) Low (except for elastomers and thermoplastics) Low 4. Composite • Includes concrete, fibreglass Melting point High (depends on constituents) Permeability Can be gaps Low (can occur if metal or graphite is in the structure) Low (except for carbon fibre) Electrical conductivity Elasticity Composites generally consist of 2 phases: the reinforcing phase – the fibre, and the matrix phase. Brittleness Low (depends on constituents) Ductility Low A.2 Metals and inductively coupled plasma (ICP) spectroscopy OBJECTIVES • Reduction by coke (carbon), a more reactive metal, or electrolysis are means of obtaining some metals from their ores. • The relationship between charge and the number of moles of electrons is given by Faraday’s constant, F. • Alloys are homogeneous mixtures of metals with other metals or non-metals. • Diamagnetic and paramagnetic compounds differ in electron spin pairing and their behaviour in magnetic fields. • Trace amounts of metals can be identified and quantified by ionizing them with argon gas plasma in Inductively Coupled Plasma (ICP) Spectroscopy using Mass Spectroscopy ICP-MS and Optical Emission Spectroscopy ICP-OES. • Deduction of redox equations for the reduction of metals. • Relating the method of extraction to the position of a metal on the activity series. • Explanation of the production of aluminium by the electrolysis of alumina in molten cryolite • Explanation of how alloying alters properties of metals. • Solving stoichiometric problems using Faraday’s constant based on mass deposits in electrolysis. • Discussion of paramagnetism and diamagnetism in relation to electron structure of metals. • Explanation of the plasma state and its production in ICP- MS/OES. • Identify metals and abundances from simple data and calibration curves provided from ICP-MS and ICP-OES. • Explanation of the separation and quantification of metallic ions by MS and OES. • Uses of ICP-MS and ICP-OES. Extraction methods Reduction by carbon Coke is heated to make CO which is a better reducing agent: C(s) + O2(g) CO2(g) CO2(g) + C(s) 2CO(g) At high heat coke also acts as a reducing agent: Fe2O3(s) + 3C(s) 2Fe(l) + 3CO(g) Fe2O3(s) + 3CO(g) 2Fe(l) + 3CO2(g) Reduction by other metals Zn(s) + CuSO4(aq) Cu(s) + ZnSO4(aq) Reduction by electrolysis Bauxite (Al2O3) is dissolved in molten cryolite (Na3AlF6) which lowers the melting point decreasing energy costs. 2O2-(l) O2(g) + 4eC(s) + O2(g) CO2(g) Al3+(l) + 3e- Al(l) Quantitative electrolysis Q = It From data booklet Faraday’s constant: F = Q/n = 96 500 C/mol This is the charge in coulombs of one mole of electrons. F = 1 mole x charge e-1 ∴ charge e-1 = F/1 mole = 96500C/mol ÷ 6.02 x 1023 e-1/mol = 1.6 x 10-19 C Problem 1: Calculate the mass of aluminium that can be produced from an electrolytic cell in one year operating with an average current of 1.20 x 105 A. Al3+(l) + 3e- Al(l) One year = 365day/yr x 24hr/day x 60min/hr x 60s/min = 31536000s Q = It = 1.20 x 105 A x 31536000s = 3.78 x 1012 C F = Q/n = 96 500 C/mol n = Q/F = 3.78 x 1012 C ÷ 96 500 C/mol = 39215751 mol e-1 1 mol Al : 3 mol e-1 X : 39215751 mol e-1 X = 1.3 x 107 mol Al m(Al) = nM = 1.3 x 107 mol x 26.96g/mol = 3.5 x 108 g = 3.5 x 105 kg Alloys An alloy is a homogenous mixture containing at least one metal formed when liquid metals are added together and allowed to form a solid of uniform composition eg. steel = iron + carbon Heating and cooling an alloy can create crystals also changing it’s properties, eg. high carbon steel: Magnetism • Ferromagnetism – materials that retain magnetism in regions called domains eg. Co, Ni, Fe, and some rare earth metals • Paramagnetism – materials that are attracted to an external magnetic field due to unpaired electrons but do not retain their magnetic fields • Diamagnetism – materials that are repelled by and external magnetic field Problem 1: Determine the magnetic properties of H2O and [Fe(H2O)6]3+. All the electrons in water are paired so it is diamagnetic. The electron configuration of Fe3+ is [Ar]3d5 and so has 5 unpaired electrons and is paramagnetic. ICP Inductively coupled plasma A plasma of Ar+ and e- at 10 000K is created to excite sample materials for analysis. ICP – OES Inductively coupled plasma optical emission spectroscopy US EPA Environmental standards Heavy metal Maximum contaminant level (mg/dm3) Health effect Cadmium 0.005 Kidney damage Chromium 0.1 Dermatitis Copper 1.3 Liver and kidney damage Lead 0.015 Mental retardation ICP – MS Inductively coupled plasma mass spectroscopy Elements are ionized with the ICP and measured with MS. It is more effective with metals as they more easily form ions. ICP-OES and ICP-MS Advantages • Wavelength and intensity allow for qualitatively measuring a range of elements in a single sample at one time but also quantitatively their concentrations at values of µg/dm3 (ICP-OES) and ng/dm3 (ICP-MS). Disadvantages • It is difficult to create calibration curves at such low concentrations (large uncertainties) • Ar and CO2 which are part of the set up cannot be analysed, neither can the solvent usually containing C, O and/or H. A.3 Catalysts OBJECTIVES • Reactants adsorb onto heterogeneous catalysts at active sites and the products desorb. • Homogeneous catalysts chemically combine with the reactants to form a temporary activated complex or a reaction intermediate. • Transition metal catalytic properties depend on the adsorption/absorption properties of the metal and the variable oxidation states. • Zeolites act as selective catalysts because of their cage structure. • Catalytic particles are nearly always nanoparticles that have large surface areas per unit mass. • Explanation of factors involved in choosing a catalyst for a process. • Description of how metals work as heterogeneous catalysts. • Description of the benefits of nanocatalysts in industry. Catalyst A catalyst increases the rate of a reaction without being used up in the reaction. It does this by lowering the activation energy of the reaction either by making the activated complex more stable, or by the creation of reaction intermediates. The area where these interactions take place is called the active site. Types of catalysts Heterogeneous • Is in a separate state to the reactants (adsorb) Homogeneous • The catalyst is in the same phase as the reactants, and is reformed back to its original state (absorb) Comparison of different catalysts Heterogeneous • Often work at stabilizing the activated complex • Less effective catalyst • Can become poisoned – other chemicals react with its surface • Easily removed at end Eg. Finely divided iron in the Haber process N2(g) + 3H2(g) ⇌ 2NH3(g) Homogeneous • Often work by creating intermediates • Have close contact with the reactants (eg. enzymes) • Can work under mild conditions with good selectivity • Catalyst needs to be removed which may destroy it eg. distillation Eg. TiCl4 catalyst converting ethane to polyethene AlR3 acts as a promoter – substances that act to increase catalytic activity (cf. inhibitors – which reduce activity by reacting with and removing reaction intermediates) Other classifications Transition metals Most common inorganic catalysts. Effective due to variable oxidation states and ability to temporarily absorb reactants. Nanocatalysts Large surface area creating a large number of active sites. Good heterogenous catalysts – eg. catalytic converters. Zeolites Minerals of aluminum silicate. Have channels allowing for shape selectivity, and increased surface area. Choosing a catalyst • Cost – how expensive is it to produce/maintain the catalyst? • Selectivity – is a high yield produced? • Efficiency – how much faster is the reaction? • Life expectancy – how long before the catalyst is poisoned? • Environmental impact – most transition metals are toxic • Range of temperature and pressure conditions A.4 Liquid crystals OBJECTIVES • Liquid crystals are fluids that have physical properties (electrical, optical and elasticity) that are dependent on molecular orientation to some fixed axis in the material. • Thermotropic liquid-crystal materials are pure substances that show liquidcrystal behaviour over a temperature range. • Lyotropic liquid crystals are solutions that show the liquid-crystal state over a (certain) range of concentrations. • Nematic liquid crystal phase is characterized by rod shaped molecules which are randomly distributed but on average align in the same direction. • Discussion of the properties needed for a substance to be used in liquid-crystal displays (LCD). • Explanation of liquid-crystal behaviour on a molecular level. the IB Chemistry student For God so loved the world that he gave his one life and only Son, that whoever believes in him shall not perish but have eternal life. (John 3:16) television, cell phones, and the internet Liquid crystals Liquid crystals are an intermediate state between a crystalline solid and a liquid. This intermediate stage is also called the nematic phase of a liquid. Molecules are randomly positioned as a liquid and can flow, but in directional order like a crystal. The intermolecular forces hold molecules in orientations called regions or domains. Domains visible under polarized microscopy. Thermotropic liquid crytals Thermotropic liquid crystals are substances that show the nematic state over a range of temperatures. Biphenyl nitriles as thermotropic liquid crystals Key features of liquid crystals: 1. Long and thin molecule allows for orientation 2. Biphenyl groups make the molecule rigid 3. Nitrile polar group allows influence by an electric field 4. Benzyl and alkyl groups make it chemically stable Lyotropic liquid crystals Lyotropic liquid crystals are substances that show the nematic state over a range of concentrations. Soap as a lyotropic liquid crystal Soap contains a polar head and a non-polar tail. At low concentrations if forms a spherical micelle. At higher concentrations these form rigid hexagonal rod-shaped liquid crystal structures and then bilayered sheets. Kevlar as a lyotropic liquid crystal Molecules line up because of hydrogen bonding between benzene rings. LCD displays Glass plates are scratched causing liquid crystals molecules to align in a set direction. These crystals twist allowing light through. Applying an electric current prevents the plane polarized light from rotating to pass through the second plate as the molecules form a liquid crystal. A.5 Polymers OBJECTIVES • Thermoplastics soften when heated and harden when cooled. • A thermosetting polymer is a prepolymer in a soft solid or viscous state that changes irreversibly into a hardened thermoset by curing. • Elastomers are flexible and can be deformed under force but will return to nearly their original shape once the stress is released. • High density polyethene (HDPE) has no branching allowing chains to be packed together. • Low density polyethene (LDPE) has some branching and is more flexible. • Plasticizers added to a polymer increase the flexibility by weakening the intermolecular forces between the polymer chains. • Atom economy is a measure of efficiency applied in green chemistry. • Isotactic addition polymers have substituents on the same side. • Atactic addition polymers have the substituents randomly placed. • Description of the use of plasticizers in polyvinyl chloride and volatile hydrocarbons in the formation of expanded polystyrene. • Solving problems and evaluating atom economy in synthesis reactions. • Description of how the properties of polymers depend on their structural features. • Description of ways of modifying the properties of polymers, including LDPE and HDPE. • Deduction of structures of polymers formed from polymerizing 2-methylpropene. Addition Polymerization Addition polymers are formed when the double bond of a monomer breaks to form a long repeating carbon chain. Monomer Polymer Ethene Polyethene Chloroethene (vinyl chloride) Polychloroethene (PVC) Propene Polypropene Styrene Polystyrene Problem 1: Deduce the structure of an addition polymer made from methylpropene. Include three repeating units. LDPE and HDPE LDPE – branching in polyethene prevents interaction and alignment of the chains making intermolecular forces weak resulting in low density, low melting point, and flexibility. HDPE – with a catalyst and low temperature, branching is decreased creating a higher density plastic. Side group orientation Isotatic addition polymers – side groups on one side making it harder, more rigid and a higher melting point due to increased intermolecular forces. Atactic addition polymers – side groups are randomly placed Syndiotatic addition polymers – side groups alternate sides allowing more compactness than atactic. Plasticizers Plasticizers (usually diesters) are chemicals that separate chains to make them softer and more flexible. eg. Plasticizers are added to PVC which by itself is hard and brittle due to the strong permanent dipoles of the Cl groups. Plasticers can evaporate out, resulting in plastics becoming more brittle and the ‘new car smell’. ester Thermoplastics vs thermoset plastics Thermoplastics or thermosoftening plastics have weak intermolecular forces allowing heat reshaping. Thermoset plastics cannot be reshaped with heat due to covalent crosslinks. Elastomers Elastomers can be made from either thermoplastics or thermoset polymers and exhibit higher strength. They return to their original shape when deformed due to covalent crosslinking bonds. Eg. In vulcanization, sulphur is added to create covalent disulphide links making rubber Expansion moulding Expansion moulding is mixing a plastic with a volatile hydrocarbon and heating it in a mould. The volatile hydrocarbon evaporates and causes expansion due to gas bubbles. Eg. Expanded polystyrene is formed by the addition of 5% pentane. Green Chemistry – Atom Economy % Atom Economy is different to % Yield, in that it takes into account the amount of waste produced, not just the efficiency of product made. (from Data Booklet) Problem 1: Chloroethene is the monomer used in the manufacture of PVC. It can be produced from 1,2-dichloroethane by the following reaction: CH2ClCH2Cl CH2=CHCl + HCl Calculate the atom economy for the reaction. Molar mass of reactant = (2 x 12.01) + (4 x 1.01) + (2 x 35.45) = 98.96 Molar mass desired product = (2 x 12.01) + (3 x 1.01) + (1 x 35.45) = 62.50 % atom economy = molar mass of desired product/molar mass of reactants x 100% = 62.50/98.96 x 100% = 63.2% A.6 Nanotechnology OBJECTIVES • Molecular self-assembly is the bottom-up assembly of nanoparticles and can occur by selectively attaching molecules to specific surfaces. Self-assembly can also occur spontaneously in solution. • Possible methods of producing nanotubes are arc discharge, chemical vapour deposition (CVD) and high pressure carbon monoxide (HIPCO). • Arc discharge involves either vaporizing the surface of one of the carbon electrodes, or discharging an arc through metal electrodes submersed in a hydrocarbon solvent, which forms a small rod-shaped deposit on the anode. • Distinguishing between physical and chemical techniques in manipulating atoms to form molecules. • Description of the structure and properties of carbon nanotubes. • Explanation of why an inert gas, and not oxygen, is necessary for CVD preparation of carbon nanotubes. • Explanation of the production of carbon from hydrocarbon solvents in arc discharge by oxidation at the anode. • Deduction of equations for the production of carbon atoms from HiPCO. • Discussion of some implications and applications of nanotechnology. • Explanation of why nanotubes are strong and good conductors of electricity. Nanotechnology Nanotechnology is research and technology development in the 1 to 100nm range. ( 1nm = x10-9) Approaches to nanotechnology Top down approach – starts with bulk material and breaks it down to smaller parts Bottom up approach – builds material from atomic or molecular species Chemical approach – these include arc discharge, laser ablation, chemical vapour deposition (CVD), and high pressure carbon monoxide disproportionation (HiPCO). Physical approach – these include atomic force microscopy (AFM), scanning tunneling microscopy (STM), and electronbeam-induced deposition. Carbon structures Buckminsterfullerene C60 Graphene – single sheet of graphite, strong, transparent, conducts electricity, impermeable even to H atoms. Nanotubes can be made of different diameters and can be capped or open. Their electrical conductivity can be varied due to size dimensions. They can be effective catalysts due to large surface areas and size selectivity. Strength is 50-100 times that of iron. A nanotube consists of repeating hexagons of carbon and can exist as single-walled carbon nanotubes (SWNT), double-walled carbon nanotubes (DWNT) or multi-walled carbon nanotubes (MWNT). Arc discharge An electric discharge between two graphite rods in low pressure inert gas (nanotubes at cathode) or hydrocarbon solvent (nanotubes at anode). 50% yield, 5g/day, mostly defect free but small and random and difficult to purify. Laser ablation Laser pulse strikes and vaporizes graphite under low pressure inert gas and heat. 70% yield, 1g/day, very high quality SWNT with fine control of diameter size but very expensive Chemical vapour deposition (CVD) Heat cracks hydrocarbon gases to carbon atoms with are deposited on transition metal catalyst. 50% yield, 1000kg/day, can be produced on industrial levels, but many defects and SWNT and MWNT difficult to separate. High pressure carbon monoxide disproportionation (HiPCO) A type of CVD which uses the following reaction: 2CO(g) C(s) + CO2(g) 95% yield, 1kg/day, but still some defects and random production. Atomic force microscopy (AFM) A 10nm Si3N4 tip is attracted or repelled by London dispersion forces. The reflected laser gives a picture of the surface. Y2O3 Scanning tunneling microscopy (STM) Uses a similar approach to AFM, however a voltage is applied to the tip. The change in electrical properties due to electron clouds gives an atomic picture. Electron-beam-induced deposition Like a 3D printer, an electron beam directs where the nanostructure is to be formed such as in techniques like HiPCO. Implications of nanotechnology Uses: • Zeolites allowing slow release of water and nutrients in agriculture • Biological diagnostic sensors • Energy efficient applications • Smaller and new types of electrical components • ICT flexible displays, increased data storage and CPU speed • Membranes for water treatment Concerns: • Unknown health effects and ability of the immune system to cope • Large scale production could lead to explosions • Disposal of nanoparticles • Control of nano-weapons (alien jokes aside - this is legitimate) A.7 Environmental impact—plastics OBJECTIVES • Plastics do not degrade easily because of their strong covalent bonds. • Burning of polyvinyl chloride releases dioxins, HCl gas and incomplete hydrocarbon combustion products. • Dioxins contain unsaturated six-member heterocyclic rings with two oxygen atoms, usually in positions 1 and 4. • Chlorinated dioxins are hormone disrupting, leading to cellular and genetic damage. • Plastics require more processing to be recycled than other materials. • Plastics are recycled based on different resin types. • Deduction of the equation for any given combustion reaction. • Discussion of why the recycling of polymers is an energy intensive process. • Discussion of the environmental impact of the use of plastics. • Comparison of the structures of polychlorinated biphenyls (PCBs) and dioxins. • Discussion of the health concerns of using volatile plasticizers in polymer production. • Distinguish possible Resin Identification Codes (RICs) of plastics from an IR spectrum. Plastic disposal A large proportion of plastic ends up in landfills or in the calm spots of rotating ocean currents called gyres. Incineration is difficult because incomplete combustion releases plastic toxins (dioxins and polychlorinated biphenyls) as well as carbon dioxide, carbon monoxide, and HCl gas causing acid rain from PVC. Problem 1: Deduce a balanced equation for the complete combustion of PVC (CH2CHCl)n. (CH2CHCl)n + O2 CO2 + H2O + HCl Balance for Cl: (CH2CHCl)n + O2 CO2 + H2O + nHCl Balance for C: (CH2CHCl)n + O2 2nCO2 + H2O + nHCl Balance for H: (CH2CHCl)n + O2 2nCO2 + nH2O + nHCl Balance for O: (CH2CHCl)n + (5n/2)O2 2nCO2 + nH2O + nHCl 2(CH2CHCl)n + 5nO2 4nCO2 + 2nH2O + 2nHCl Plastic toxins Phthalate plasticizers (especially in PVC) are present in many products including children’s toys and there is concern from animal testing that they disrupt the endocrine system and are carinogenic. Dioxins are unsaturated six-member heterocyclic rings with 2 oxygen atoms (1,4dioxin the most common). They disrupt the endocrine system and cause cellular and genetic damage. The are not destroyed in the environment (Persistant Organic Pollutant – POP) and bioaccumulate up the food chain. Other chemicals also act like dioxins and are carcinogenic such as polychlorinated biphenyls (PCBs). Plastic recycling Advantages: less use of raw material, less energy costs, less pollutants, less land for waste disposal. Methods: 1. Themoplastics are melted and remoulded. Pieces are separated by density from a floating tank. 2. Depolymerization by heating in the absence of air causes monomers to split – called pyrolysis. Fractional distillation allows for the polymer process to begin again. Plastic recycling Sorting: Chopped up pieces of plastic are read with an IR spectrophotometer to determine their correct resin identification code (RIC) – see data booklet. An air blower then shoots air accordingly to sort them. Problem 1: Deduce the structure of the polymer from the data booklet. Absorption 100-1400cm-1: C-F present No absorption 2850-3050cm-1: no C-H present Therefore the structure must be Teflon: OBJECTIVES • Superconductors are materials that offer no resistance to electric currents below a critical temperature. • The Meissner effect is the ability of a superconductor to create a mirror image magnetic field of an external field, thus expelling it. • Resistance in metallic conductors is caused by collisions between electrons and positive ions of the lattice. • The Bardeen–Cooper–Schrieffer (BCS) theory explains that below the critical temperature electrons in superconductors form Cooper pairs which move freely through the superconductor. • Type 1 superconductors have sharp transitions to superconductivity whereas Type 2 superconductors have more gradual transitions. • X-ray diffraction can be used to analyse structures of metallic and ionic compounds. • Crystal lattices contain simple repeating unit cells. • Atoms on faces and edges of unit cells are shared. • The number of nearest neighbours of an atom/ion is its coordination number. • Analysis of resistance versus temperature data for Type 1 and Type 2 superconductors. • Explanation of superconductivity in terms of Cooper pairs moving through a positive ion lattice. • Deduction or construction of unit cell structures from crystal structure information. • Application of the Bragg equation, , in metallic structures. • Determination of the density of a pure metal from its atomic radii and crystal packing structure. Higher level A.8 Superconducting metals and X-ray crystallography At low temperatures (the critical temperature) the kinetic energy of a lattice is sufficiently reduced to allow zero resistance electrical movement – superconduction. Resistance due to KE At or below critical temperature Higher level Superconductors Cooper pairs are thought to exist where an electron causes an distortion in the lattice, which creates a partial positive charge and attracts another electron as a matching pair placing both in a lower energy state, moving throughout the lattice as a unit unimpeded. This is also known as the Bardeen-Cooper-Schrieffer (BCS) theory. Higher level Cooper pairs The Meissner effect is when the external magnetic field creates eddy currents the superconductor which then causes and equal and opposite magnetic field. Superconductors work as perfect diamagnets. Higher level Meissner effect Mainly metals or metalloids that show some conductivity at room temperature. Below a certain temperature and not too strong a magnetic field, the Cooper pairs can form. Critical temperatures are too low to be practical. Higher level Type 1 superconductors Mainly metallic compounds and alloys. Do not sharply lose their conductivity. Can operate at higher temperatures. Higher level Type 2 superconductors A unit cell is the simplest repeating unit of a crystal lattice. The most compact is the FCC with a coordination number of 12 eg. Au, Ag The second most compact is the BCC with a coordination number of 8 eg. Na, K, Fe. The simple cubic cell has a coordination number of 6 eg. NaCl. Higher level Unit cells Position of atom in unit cell Atom contribution Centre 1 Corner 1/8 Face 1/2 Edge 1/4 Higher level Atom contribution to unit cell Center atoms = 1 Contribution = 1 x 1 = 1 Corner atoms = 8 Contribution = 8 x 1/8 = 1 ∴ There are a total of 2 atoms per unit cell Higher level Problem 1: Calculate how many atoms there are in a BCC unit cell. BCC FCC Higher level Atomic radii and unit cell length a. Data booklet states r(Cu) = 122 x 10-12m Pythagoras: d2 + d2 = (4r)2 d = 4r/√2 = 3.45 x 10-10m b. Volume of unit cell = (3.45 x 10-10m)3 Number of atoms = 4 Mass of copper atom = M/n = 63.55g/mol / 6.02 x 1023 Density = m/V = 4 x (63.55g/mol / 6.02 x 1023) / (3.45 x 10-10)3 = 10.3kg/m3 Higher level Problem 2: Calculate a. the unit cell length and b. density of copper knowing it as a FCC structure. The distance between 2 layers in a crystal gives us the unit cell edge length. When two X-rays (X-rays are small enough for atoms to diffract) diffract at an integer difference in length, the form constructive interference and occur as a bright dot. We can then use the Bragg equation to determine this distance. Higher level X-ray crystallography nλ=2dsinθ 154pm = 2dsin13.49⁰ d = 154/(2 x 0.2333) = 330pm r = 330/2 = 165pm Higher level Problem 3: A Tantalum Ta superconductor is analysed by X-ray crystallography of 154pm and found to give a first order (n=1) diffraction pattern at 13.49⁰. Determine the atomic radius. OBJECTIVES • Condensation polymers require two functional groups on each monomer. • NH 3 , HCl and H 2 O are possible products of condensation reactions. • Kevlar® is a polyamide with a strong and ordered structure. The hydrogen bonds between O and N can be broken with the use of concentrated sulfuric acid. • Distinguishing between addition and condensation polymers. • Completion and descriptions of equations to show how condensation polymers are formed. • Deduction of the structures of polyamides and polyesters from their respective monomers. • Explanation of Kevlar®’s strength and its solubility in concentrated sulfuric acid. Higher level A.9 Condensation polymers Condensation polymerization requires monomers to have two functional groups. Small molecules are produced such as H2O, NH3, or HCl. Higher level Condensation polymerization Polyesters are polymers with ester linkages formed from hydroxyl (alcohol) and carboxyl (carboxylic acid) groups. Higher level Polyesters PETE or PET is the most commonly used plastic. It can be reused as it has no plasticizers, it is unreactive, nontoxic, impermeable to gases and has resembles glass with good optical properties. phthalate terephthalate Higher level Polyethylene terephthalate (PETE) benzene-1,4-dicarboxylic acid (terephthalic acid) ester ethane-1,2-diol Higher level Polyethylene terephthalate (PETE) Polyamides are polymers with amide linkages formed from amino (amine) and carboxyl (carboxylic acid) groups. Higher level Polyamides OR Higher level Nylon Although this plastic is strong enough to stop bullets due to the aligning of the hydrogen bonds, it will dissolve in concentrated sulphuric acid which breaks up the hydrogen bonds between the amino and carboxyl groups. Higher level Kevlar Due to the condensation product the atom economy of condensation polymerization is not as good as addition polymerization. Green polymers should: • Have high atom economy and high resource effectiveness • Have a clean production process preventing waste and greenhouse emissions • Use renewable resources and renewable energy • Have a low carbon footprint Higher level Green chemistry Find the ester link and break it to form and acid and an alcohol. Higher level Problem 1: Poly(ethylene furanoate) (PEF) is a green polymer made from bio-based monomers. Deduce the structure of the two monomers from which it is made. OBJECTIVES • Toxic doses of transition metals can disturb the normal oxidation/reduction balance in cells through various mechanisms. • Some methods of removing heavy metals are precipitation, adsorption, and chelation. • Polydentate ligands form more stable complexes than similar monodentate ligands due to the chelate effect, which can be explained by consideringentropy changes. • Explanation of how chelating substances can be used to remove heavy metals. • Deduction of the number of coordinate bonds a ligand can form with a central metal ion. • Calculations involving K sp as an application of removing metals in solution. • Compare and contrast the Fenton and Haber–Weiss reaction mechanism. Higher level A.10 Environmental impact—heavy metals • Interfer with enzymes by bonding instead of Ca2+, Mg2+, or Zn2+ ions. • Disrupt the oxidation/reduction balance in a cell. Most dangerous: Mercury (Hg) Lead (Pb) Cadmium (Cd) Mode of action Attracted to sulfur atoms thereby denaturing enzymes. Particularly disrupts the sodium pump in the nervous system. Deactivates enzymes that make haemaglobin. Mimics zinc making enzymes ineffective. Effects Blindness, insanity, Minamata disease, death. Kidney, liver, heart Brittle bones, failure. Brain kidney and lung damage and cancer, death. coma/death. Higher level Toxicity Solid surfaces can remove heavy metals. These include: • Activated carbon (expensive) • Charcoal filters • Clays • Ion-exchange mechanisms (resins or zeolites) • Natural materials (cheap eg. coconut shells, rice husks, sugar cane) Higher level Removal by absorption Most heavy metals have poor solubility as sulfide or hydroxide or phosphate salts and can be used to precipitate out the metal from the solution. Higher level Removal by precipitation The Ksp is similar to the Keq but the solid salt is removed from the equation as it is heterogenous and not a part of the solution: MpNq(s) ⇌ pMm+(aq) + qNn-(aq) Ksp = [Mm+]p[Nn-]q (values in data booklet) If ionic concentrations exceed the Ksp then a precipitate will form. Addition of either M or N ions will decrease the solubility of the compound pushing the reaction to the left due to Le Chatelier’s Principle – known as the common ion effect. Higher level Solubility product constant Ksp Ksp(Pb2+) = 1.43 x 10-20 (data booklet) Pb(OH)2(s) ⇌ Pb2+(aq) + 2OH-(aq) Ksp = [Pb2+][OH-]2 1.43 x 10-20 = (x)(2x)2 R 4x3 = 1.43 x 10-20 I x =1.53 x 10-7 1 mol Pb2+ : 1 mol Pb(OH)2 Pb2+ OH- 1 2 0 0 C +x +2x E x 2x ∴ The molar solubility is 1.53 x 10-7 mol/dm3 Higher level Problem 1: Calculate the maximum solubility of lead(II) hydroxide in solution. At pH = 11, pOH = 3 [OH-] = 10-pOH = 10-3 Cd(OH)2(s) ⇌ Cd2+(aq) + 2OH-(aq) Addition of hydroxide ions pushes the equilibrium to the left taking cadmium out of solution. Ksp(Cd2+) 10-15 = 7.2 x (data booklet) Ksp = [Cd2+][OH-]2 7.2 x 10-15 = (x)(10-3 +2x)2 7.2 x 10-15 = (x)(10-3)2 (as x is small) x(10-6) = 7.2 x 10-15 x =7.2 x 10-9 mol/dm3 Cd2+ OH- R 1 2 I 0 10-3 C +x 10-3 +2x E x 10-3 +2x (at normal pH the cadmium level is 10 000 times higher) Higher level Problem 2: Calculate the molar solubility of Cd2+ ion after lime (adding calcium hydroxide) precipitation to a pH of 11. Chelation is when a metal is surrounded by ligands. This large complex prevents it entering cells and can be then excreted from the body as a water soluble ion. The hexadentate ligand EDTA4- works effectively as it releases water ligands and therefore increases entropy. Higher level Removal by chelation Heavy metals participate in (Fenton) or catalyse (Haber-Weiss) the creation of destructive hydroxyl free radicals. Fenton: Fe2+ + H2O2 Fe3+ + ·OH + OH(·OH will oxidise organic pollutants, OH- will precipitate heavy metals) Haber-Weiss: (catalyzed Fenton) The slow natural reaction: (·O2- superoxide ion occurring in living systems ) ·O2- + H2O2 O2 + ·OH + OHHowever, Fe3+ ions are reduced: ·O2- + Fe3+ O2 + Fe2+ This drives the rapid creation of hydroxyl ions by the Fenton reaction: Fe2+ + H2O2 Fe3+ + ·OH + OH- Higher level Fenton and Haber-Weiss reactions