Download File - Mr Weng`s IB Chemistry

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