(i) The diagram below shows the diffraction of two X-ray beams, y and z of wavelength λ, shining on a chromium crystal whose planes are a distance d nm apart.

Deduce the extra distance travelled by the second beam, z, compared to the first one, y.

(ii) State the Bragg’s condition for the observed diffraction to be at its strongest (constructive interference).

(i) The mass of one unit cell of chromium metal is 17.28 × 10⁻²³ g. Calculate the number of unit cells in one mole of chromium. A_r(Cr) = 52.00.

(ii) Deduce the number of atoms of chromium per unit cell.

i

2d sin θ
OR
2|AB| / 2|BC| / |AB| + |BC| / |AB| AND |BC|

Vertical lines indicating lengths not required. Answer may be conveyed in words also.
Do not accept |AC| – reference must be made to B.

ii

extra distance travelled/|AB| + |BC| = nλ/a whole number of wavelengths

Accept notations of extra distance as in (a)(i).

i
«$\frac{52.00\text{g}\text{mo}{\text{l}}^{-\text{1}}}{17.28\times {10}^{-23}\text{g}\text{unit}\text{cel}{\text{l}}^{-\text{1}}}=$» 3.009 × 10²³ «unit cells mol⁻¹»

ii
«$\frac{6.02\times {10}^{23}\text{atoms}\text{mo}{\text{l}}^{-\text{1}}}{3.01\times {10}^{23}\text{unit}\text{cells}\text{mo}{\text{l}}^{-\text{1}}}=$» 2 «atoms per unit cell»

Compare and contrast the Fenton and Haber–Weiss reaction mechanisms.

Adsorption and chelation are two methods of removing heavy metal ion pollution from the environment.

(i) Describe the process of adsorption.

(ii) Deduce the structure of the complex ion formed by the reaction of three H₂N−CH₂−CH₂−NH₂ chelating molecules with a mercury(II) ion.

One similarity:
both involve hydroxyl/•OH «radicals»

One difference:

Accept “hydroxy” for “hydroxyl”.
Do not penalize missing radical symbols if consistent throughout.
Accept “H₂O₂ → 2•OH” for the Fenton mechanism.

i
molecules/ions/substances are attracted to/form «non-covalent» interactions with the surface of the adsorbent

ii

Do not penalize missing charge or square brackets.
Bonds to Hg must be shown (in any format).

Describe the Meissner effect.

Outline one difference between type 1 and type 2 superconductors.

creation of a mirror image magnetic field of an external field «below the critical temperature/T_c of the superconductor»
OR
expulsion of a magnetic field from a superconductor «below its critical temperature/T_c»

Accept “Type 1: «most» metals AND Type 2: alloys/metal oxide ceramics/perovskites”.

Fermentation of sugars from corn starch produces propane-1,3-diol, which can be polymerized with benzene-1,4-dicarboxylic acid to produce the PTT polymer (polytrimethylene terephthalate).

(i) Draw the molecular structure of each monomer.

(ii) Deduce the name of the linkage formed on polymerization between the two monomers and the name of the inorganic product.

i

HO–CH₂–CH₂–CH₂–OH AND HOOC–C₆H₄–COOH

Accept full or condensed structural formulas. Labelling of monomers not required but penalize incorrect labels.

ii

Name of linkage: ester
AND
Name of inorganic product: water  

Do not accept “esterification”.
Do not accept formulas.

Precipitation is one method used to treat waste water.

Phosphates, ${\mathrm{PO}}_4^{3-}$, in waste water can be removed by precipitation with magnesium ions. $K_{sp}$ of magnesium phosphate is $1.04\times {10}^{-24}$.

$3{\mathrm{Mg}}^{2+}\left(\mathrm{aq}\right)+2{{\mathrm{PO}}_4}^{3-}\left(\mathrm{aq}\right)\to {\mathrm{Mg}}_3\left({\mathrm{PO}}_4{)}_2\right(\mathrm{s})$

Calculate the maximum solubility of phosphate ions in a solution containing $0.0100 \mathrm{mol} {\mathrm{dm}}^{-3}$ magnesium ions.

Precipitation is one method used to treat waste water.

Zinc, cadmium, nickel, and lead are metal ions which can be removed by precipitation. Explain why waste water is adjusted to a pH of 9−10 to remove these ions by referring to section 32 of the data booklet.

$\left[{{\mathrm{PO}}_4}^{3-}\right]=\sqrt{\frac{K_{\mathit{s}\mathit{p}}}{{\left[{\mathrm{Mg}}^{2+}\right]}^3}}$ ✔

$«\left[{{\mathrm{PO}}_4}^{3-}\right]=«\sqrt{\frac{1.04\times {10}^{-24}}{0.{0100}^3}}=»1.02\times {10}^{-9}«\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

Accept “$Ksp\mathit{=}{\left[\mathit{M}{\mathit{g}}^{\mathit{2}\mathit{+}}\right]}^{\mathit{3}}{\left[\mathit{P}{{\mathit{O}}_{\mathit{4}}}^{\mathit{3}\mathit{-}}\right]}^{\mathit{2}}$” for M1.
Award [2] for correct final answer.

Any two of:
precipitation occurs with a base/carbonate/${{\mathrm{CO}}_3}^{2-}$/hydroxide/${\mathrm{OH}}^{-}$ ✔

$\left[{\mathrm{OH}}^{-}\right]$ is high enough to cause metal hydroxide precipitation at that $\mathrm{pH}$ ✔

these ions are slightly acidic/more soluble in acidic conditions ✔

only small amounts of carbonate/hydroxides/anion needed at that $\mathrm{pH}$ ✔

solubility products of the hydroxides are very small ✔

Do not accept “hydroxyl” for “hydroxide”.

The Ksp problem was very well answered by the stronger candidates, who had no difficulty working out the maximum solubility of phosphate ions in solution as 1.02 x 10^-9 mol dm^-3.

This question although somewhat challenging was well answered by the stronger candidates.

Explain these properties of carbon nanotubes.

CNT can act as Type 2 superconductors. Outline why Type 2 superconductors are generally more useful than Type 1.

Explain the role of electrons in superconducting materials in terms of the Bardeen–Cooper–Schrieffer (BCS) theory.

Alloying metals changes their properties. Suggest one property of magnesium that could be improved by making a magnesium–CNT alloy.

Pure magnesium needed for making alloys can be obtained by electrolysis of molten magnesium chloride.

© International Baccalaureate Organization 2020

Calculate the theoretical mass of magnesium obtained if a current of 3.00 A is used for $10.0$ hours. Use charge :(Q) = current (I) × time (t)  and section 2 of the data booklet.

Suggest a gas which should be continuously passed over the molten magnesium in the electrolytic cell.

Zeolites can be used as catalysts in the manufacture of CNT. Explain, with reference to their structure, the high selectivity of zeolites.

Experiments have been done to explore the nematic liquid crystal behaviour of CNT. Justify how CNT molecules could be classified as nematic.

Excellent strength: defect-free AND rigid/regular 2D/3D ✔

Excellent conductivity: delocalized electrons ✔

Accept “carbons/atoms are all covalently bonded to each other” for M1.

Any two of:
have higher critical temperatures/T_c «than Type 1»
OR
can act at higher temperatures ✔

have higher critical magnetic fields/B_c «than Type 1» ✔

less time needed to cool to operating temperature ✔

less energy required to cool down/maintain low temperature ✔

Any three of:

passing electrons «slightly» deform lattice/displace positive ions/cations ✔

electrons couple/form Cooper pairs/condense with other electrons ✔

energy propagates along the lattice in wave-like manner/as phonons ✔

Cooper pair/electron condensate/pair of electrons moves through lattice freely
OR
phonons are «perfectly» elastic/cause no energy loss ✔

Any of:
ductility ✔
strength/resistance to deformation ✔
malleability ✔
hardness ✔
resistance to corrosion/chemical resistance ✔
range of working temperatures ✔
density ✔

Do not accept “conductivity”.

$«Q=I\times t=3.00\times 10.0\times 3600=»108 000 \mathrm{C}$ ✔

$«\frac{Q}{F}=\frac{108 000 C}{96 500 C {\mathrm{mol}}^{-1}}=»1.12«\mathrm{mol} {\mathrm{e}}^{-}»$ ✔

$«\frac{1.12 \mathrm{mol}}{2}=0.560 \mathrm{mol} \mathrm{Mg}»$
$«m=0.560 \mathrm{mol}\times 24.31 \mathrm{g} {\mathrm{mol}}^{-1}=»13.6«\mathrm{g}»$ ✔

Award [3] for correct final answer.

argon/$\mathrm{Ar}$/helium/$\mathrm{He}$ ✔

Accept any identified noble/inert gas.
Accept name OR formula.

Do not accept “nitrogen/$N_2$“.

pores/cavities/channels/holes/cage-like structures ✔

«only» reactants with appropriate/specific size/geometry/structure fit inside/go through/are activated/can react ✔

Accept “molecules/ions” for “reactants” in M2.

rod-shaped molecules
OR
«randomly distributed but» generally align
OR
no positional order AND have «some» directional order/pattern ✔

Accept “linear” for “rod-shaped”.

The stronger candidates knew that the excellent conductivity associated with CNTs is associated with delocalised electrons but few scored the mark for citing the property associated with excellent strength, which can be attributed to being defect-free and having a rigid/regular 2D/3D structure.

Most gained at least one mark here for stating that Type 2 superconductors have higher critical temperatures than Type 1.

The role of electrons in superconducting materials in terms of the Bardeen-Cooper-Schrieffer (BCS) theory was very well understood and many scored all three marks. 

This question proved to be difficult and few could suggest a suitable property (such as ductility) of magnesium that could be improved by making a magnesium-CNT alloy.

The better candidates scored all three marks for the electrolysis calculation. Even the weaker candidates managed to score at least one mark for calculating Q = 108,000 C.

The most common error here was "nitrogen" as the gas that should be continuously passed over the molten magnesium in the electrolytic cell. Magnesium can react with nitrogen forming magnesium nitride, which makes this choice of gas unsuitable (unlike argon for example).

The explanation of the high selectivity of zeolites, in terms of their structure, was very well answered and many scored both marks. A thorough understanding of zeolites was much better conveyed in N20 compared to previous sessions.

Most gained the one mark here, justifying how CNT molecules can be classified as nematic, by stating that they are "rod-shaped molecules".

Outline the two distinct phases of this composite.

Thermoplastic composites are increasingly replacing thermosets.

Suggest one advantage of thermoplastic polymers over thermosets.

Explain how thermoplastics, such as polyvinylchloride, PVC, can be made more flexible by the addition of phthalate ester plasticizers.

Explain why phthalates are replaced by other plasticizers in the production of plastics.

Classify PVC and polyethene terephthalate, PET, as addition or condensation polymers and deduce the structural formulas.

carbon fibre reinforcing phase ✔

«in a» matrix phase of epoxy ✔

Award [1 max] for “reinforcing phase «embedded» in a matrix”.

can be recycled
OR
can be reformed when hot
OR
high impact/chemical/abrasion resistance ✔

Any three of:

plasticizers embed/fit between «polymer» chains ✔

keep polymer strands/chains/molecules separated/apart ✔

weaken intermolecular/London/dispersion/attractive/forces/instantaneous induced dipole-induced dipole/forces «between chains» ✔

prevent chains from packing closely/forming regular packing/structure ✔

Accept “van der Waals/vdW” for “London”.

Any two of:
readily released into environment
OR
have weak intermolecular forces «rather than covalent bonds between chains» ✔

get into biological systems by ingestion/inhalation ✔

interrupt endocrine systems
OR
affect release of hormones
OR
effect development of male reproductive system ✔

considered carcinogenic
OR
can cause cellular damage ✔

can cause early puberty in females ✔

can cause thyroid effects ✔

can cause asthma ✔

Do not accept just “are a health concern”.

PVC: addition AND PET: condensation ✔

structure of PVC monomer ✔

structure of PET monomers ✔

Accept full OR condensed structural formulas.

Most had some idea of a composite and tended to gain at least one mark for stating "reinforcing phase embedded in a matrix".

This was well answered and most stated that one advantage of thermoplastic polymers over thermosets is that they can be recycled.

This also was well done and some managed to score all three marks.

This question required an explanation of why phthalates are replaced by other plasticizers in the production of plastics. This question was found to be very challenging and even the better candidates failed to score both marks. The weaker candidates typically stated that they are just a health concern, which was deemed insufficient to warrant even a salvage mark.

This question on PVC and PET was very well answered. All three marks for their correct classification and the corresponding structures of their monomers was frequently scored.

Identify the type of bonding in lithium hydride, using sections 8 and 29 of the data booklet.

Identify the colour of the emission spectrum of lithium using section 17 of the data booklet.

Suggest why ICP-OES does not give good quantitative results for distinguishing ⁶Li from naturally occurring lithium.

Suggest a better method.

Lithium is obtained by electrolysis of molten lithium chloride. Calculate the time, in seconds, taken to deposit 0.694 g Li using a current of 2.00 A.

Q (charge) = I (current) × t (time)

Lithium has shown some superconductive properties when doped into graphene or when under high pressure. Under high pressure, however, the Meissner effect is absent.

Describe the Meissner effect.

At very low temperatures, lithium atoms enhance the phonon binding of electrons in graphene suggesting the formation of Cooper pairs.

Explain how Cooper pairs are formed.

Lithium forms a crystalline lattice with the unit cell structure shown below.

X-ray diffraction shows that the length of the edge of the unit cell is 3.51 × 10⁻⁸ cm.

Determine the density of lithium, in g cm⁻³, using sections 2 and 6 of the data booklet.

ionic   [✔]

red   [✔]

emission spectra of both «⁶Li and natural Li» give same colour/produce same «range of» wavelengths
OR
they have same electron transitions/same nuclear charge    [✔]

Note: Accept “the spectra are almost identical”.

ICP-MS   [✔]

Note: Accept “MS/mass spectrometry”.

n «=  $\frac{\text{m}}{{\text{M}}_{\text{r}}}=\frac{0.694}{6.94}$» =0.100«mol» 

« $t=\frac{0.100\text{mol}\times \text{96 500 C mo}{\text{l}}^{-1}}{2.00\text{ C }{\text{s}}^{-1}}$ »

4830 «s»   [✔]

creation of mirror image/opposing magnetic field of external field «below critical temperature/T of superconductor»
OR
expulsion of magnetic field from superconductor «below critical temperature/T»    [✔]

Any three of:
positive ions/cations in lattice are attracted to passing electron    [✔]

lattice is distorted «by this passing electron»     [✔]

creates «local» regions of increased positive charge     [✔]

second electron is attracted to deformation AND a coupling occurs     [✔]

mass of Li in unit cell = « $2\times \frac{6.94\text{ g mo}{\text{l}}^{-1}}{6.02\times {10}^{23}\text{ mo}{\text{l}}^{-1}}$ » 2.31 × 10^–23 g    [✔]

volume of unit cell = «(3.51 × 10^–8 cm)³ =» 4.32 × 10^–23 cm³    [✔]

«density = $\frac{2.31\times {10}^{-23}\text{ g}}{4.32\times {10}^{-23}\text{ c}{\text{m}}^3}$ =» 0.535 «g cm^–3»    [✔]

Note: Award [3] for correct final answer.

Most candidates correctly identified the type of bonding and the colour of the emission spectrum of lithium.

Most candidates correctly identified the type of bonding and the colour of the emission spectrum of lithium.

Most candidates correctly identified the type of bonding and the colour of the emission spectrum of lithium, but frequently referred to the ICP-OES spectra of 6Li and naturally occurring lithium as being the same, rather than similar and thus failed to score the mark in (b)(ii). 

A better method was selected by most candidates.

The calculation in was done well.

Candidates had some difficulty describing the Meissner effect, with several responses using the terms repelling or repulsion instead of opposing and expulsion. Correct terminology is required.

Poor expression was also evident in responses explaining the formation of Cooper pairs, with very few candidates scoring full marks.

Most candidates had difficulty determining the number of atoms in lithium in a unit cell, even with a diagram provided. However, ECF marks were frequently scored.

Outline the nature of the plasma state and how it is produced in ICP-MS.

Hydrogen sulfide could be used to remove antimony(III) ions from a solution.

Determine the concentration of antimony(III) ions that would be required to precipitate antimony(III) sulfide in a solution saturated with hydrogen sulfide.

[S²⁻] in water saturated with hydrogen sulfide = 1.0 × 10⁻¹⁴ mol dm⁻³ 

K_sp (Sb₂S₃) = 1.6 × 10⁻⁹³

Identify a ligand that could be used to chelate antimony(III) ions in solution.

electrons AND positive ions «in gaseous state»

high frequency/alternating current passed through argon
OR
«oscillating» electromagnetic/magnetic field
OR
high frequency radio waves

Accept “gas” instead of “argon”.

[2 marks]

K_sp = [Sb³⁺]² $\bullet$ [S²⁻]³

[Sb³⁺]² $\bullet$ (10⁻¹⁴)³ = 1.6 x 10⁻⁹³

[Sb³⁺] «$=\sqrt{1.6\times {10}^{-51}}$» = 4.0 x 10⁻²⁶ «mol dm⁻³»

Award [3] for correct final answer.

[3 marks]

EDTA/ethylenediaminetetraacetic aci
OR
H₂N–CH₂–CH₂–HN₂/ethane-1,2-diamine

Accept “EDTA^4–”.

Accept other chelating agents.

[1 mark]

Deduce the repeating unit of the polymer and the other product of the reaction.

State the class of polymer to which PETE belongs.

Repeating unit:

Other product: water/H₂O

Continuation bonds necessary for the mark.

Accept alternative repeating unit with O at other end.

Do not penalize square brackets or n.

[2 marks]

condensation

Accept polyester or thermoplastic.

[1 mark]

Lanthanum has a hexagonal close packed (hcp) crystal structure. State the coordination number of each lanthanum atom.

Lanthanum becomes superconducting below 5 K. Explain, in terms of Bardeen–Cooper–Schrieffer (BCS) theory, how superconductivity occurs.

Outline why superconductivity only occurs at low temperatures.

twelve/12

[1 mark]

«moving» electron attracts «nearby» positive charges/ions/cations

creates «local» regions of increased positive charge

positive charge/field attracts second electron «with opposite spin»

two electrons form a Cooper pair

«all» Cooper pairs «in sample» interact/form «electron» condensate

«electron» condensate/Cooper pairs move/flow «through sample» freely/without resistance

[3 marks]

reduces the band gap to zero
OR
«at high temperatures» thermal motion disrupts the formation of Cooper pairs

[1 mark]

Identify the other product formed.

Explain why EDTA, a chelating agent, is more effective in removing heavy metal ions from solution than monodentate ligands.

HCl/hydrogen chloride

Accept “hydrochloric acid”.

[1 mark]

forms four/six/several/multiple coordinate/coordination bonds «to a central metal ion»
OR
is a polydentate/tetradentate/hexadentate ligand

forms more stable complex/stronger bonds with central metal ion
OR
increases entropy/S by releasing smaller «monodentate ligand» molecules previously complexed

complex ions are much larger «and can be removed easily due to large size of chelate complexes»
OR
heavy metal ions trapped inside the ligand/become «biologically» inactive/nontoxic/harmless

Accept “dative «covalent»” for “coordinate/coordination”.

Do not accept just “chelates”.

[3 marks]

Label with an asterisk, *, the chiral carbon atom.

Explain the effects of very low and high temperatures on the liquid-crystal behaviour of this molecule.

Low temperature:

High temperature:

  [✔]

Low temperature:
intermolecular forces prevent molecules moving AND solid/«normal» crystal formation    [✔]

High temperature:
«above a critical temperature» disrupts alignment of molecules AND behaves as fluid/liquid    [✔]

Note: Accept “weak intermolecular forces break AND behaves as fluid/liquid”.

Identifying a chiral carbon atom was answered reasonably well.

Explaining effects of very low and very high temperatures on liquid-crystal behaviour proved difficult for most candidates. Responses lacked the required detail about intermolecular forces between molecules.

Calculate the total number of cobalt atoms within its unit cell.

The atomic radius, r, of cobalt is 1.18 × 10^–8 cm. Determine the edge length, in cm, of the unit cell, a, using the second diagram.

Determine a value for the density of cobalt, in g cm^–3, using data from sections 2 and 6 of the data booklet and your answers from (a) and (b) (i).

If you did not obtain an answer to (b) (i), use 3.00 × 10^–8 cm but this is not the correct answer.

«$8\times \frac{1}{8}+6\times \frac{1}{2}=$» 4

face diagonal $=\sqrt{2a}=4r$

«$a=\frac{\left(4\times 1.18\times {10}^{-8}\text{cm}\right)}{\sqrt{2}}=$» 3.34 x 10^–8 «cm»

mass of 4 atoms = $\frac{4\times 58.93gmo{l}^{-1}}{6.02\times {10}^{23}mo{l}^{-1}}=3.916\times {10}^{-22}$ «g»

«density $=\frac{3.916\times {10}^{-22}g}{{\left(3.34\times {10}^{-8}cm\right)}^3}=$» 10.5 «g cm^–3»

Answer using 3.00 x 10^–8 cm:

mass of 4 atoms = $\frac{4\times 58.93gmo{l}^{-1}}{6.02\times {10}^{23}mo{l}^{-1}}=3.916\times {10}^{-22}$ «g»

«density $=\frac{3.916\times {10}^{-22}g}{{\left(3.00\times {10}^{-8}cm\right)}^3}=$» 14.5 «g cm^–3»

Award [2] for correct final answer.

Describe how the monomers of addition polymers and of condensation polymers differ.

Identify the type of intermolecular bonding that is responsible for Kevlar^®’s strength.

addition: C=C

AND

condensation: two functional groups needed on each monomer

Accept "alkene/alkenyl" OR "double bond" OR "multiple bond".

hydrogen bonds

Accept “$\pi -\pi$ stacking/interactions”.

The diagram illustrates the crystal structure of aluminium metal with the unit cell indicated. Outline the significance of the unit cell.

When X-rays of wavelength 0.154 nm are directed at a crystal of aluminium, the first order diffraction pattern is observed at 18°. Determine the separation of layers of aluminium atoms in the crystal, in m, using section 1 of the data booklet.

Deduce what the shape of the graph indicates about aluminium.

Outline why the resistance of aluminium increases above 1.2 K.

The concentration of aluminium in drinking water can be reduced by precipitating aluminium hydroxide. Calculate the maximum concentration of aluminium ions in water of pH 7 at 298 K. Solubility product of aluminium hydroxide = 3.3 × 10⁻³⁴ at 298 K.

the smallest repeating unit «from which the crystal structure can be derived»

Accept “building block that the structure is made from”.

[1 mark]

«nλ = 2d sin θ»

1 × 1.54 × 10^–10 = 2 × d × sin 18

d «$\frac{1.54\times {10}^{-10}\text{ m}}{2\times 0.309}$» = 2.49 × 10^–10 «m»

Award [2] for correct final answer.

[2 marks]

type 1

superconductor

[2 marks]

collisions between electrons and «lattice of metal» ions become more frequent

OR

thermal oscillations/vibrations disrupt the Cooper electron pairs

[1 mark]

K_sp = [Al³⁺][OH^–]³ «= 3.3 × 10^–34»

[Al³] = «$\frac{3.3\times {10}^{-34}}{{(1\times {10}^{-7})}^3}=$» 3.3 × 10^–13 «mol dm^–3»

Award [2] for correct final answer.

[2 marks]

State the name of one method, other than precipitation, of removing heavy metal ions from solution in water.

The solubility product, K_sp , of cadmium sulfide, CdS, is 8.0 × 10^–27. Determine the concentration of cadmium ions in 1.0 dm³ of a saturated solution of cadmium sulfide to which 0.10 mol of solid sodium sulfide has been added, stating any assumption you make.

adsorption

OR

chelation

OR

ion exchange

Accept other valid methods such as “phytoremediation” OR “Fenton reaction” OR “electrolysis”.

Calculation:

K_sp = [Cd²⁺] x [S^2–] ✔

[Cd²⁺] = 8.0 x 10^–26 «mol dm^–3» ✔

Assumption:

volume of solution remains 1.0 dm³

OR

concentration of sulfide ions in original solution is negligible

OR

hydrolysis of sulfide ions is negligible

Award [2] for correct numerical answer of [Cd²⁺] for M1 and M2.

Accept “0.10 + x ∼ 0.10 «mol dm^–3»”.

Nickel(II) ions are least soluble at pH 10.5. Calculate the molar solubility of nickel(II) hydroxide at this pH. K_spNi(OH)₂ = 5.48 × 10^–16.

Rhodium is paramagnetic with an electron configuration of [Kr] 5s¹4d⁸.

Explain, in terms of electron spin pairing, why paramagnetic substances are attracted to a magnetic field and diamagnetic substances are not.

Rhodium is a type 1 superconductor.

Sketch graphs of resistance against temperature for a conductor and superconductor.

Contrast type 1 and type 2 superconductors by referring to three differences between them.

K_sp= [Ni²⁺][OH⁻]²
OR
5.48 x 10⁻¹⁶ = [Ni²⁺][10^−3.5]²

«[Ni²⁺] =» 5.48 x 10⁻⁹ «mol dm⁻³»

Award [2] for correct final answer.

[2 marks]

paramagnetic materials have unpaired electrons
OR
diamagnetic materials have all electrons «spin-»paired

unpaired electrons align with an external magnetic field
OR
paired electrons are not influenced by magnetic field

Accept “diamagnetic materials have no unpaired electrons" for M1.

[2 marks]

Conductor:
Accept any concave upwards curve or line showing resistance increasing with temperature. There should be a
y-axis intercept. Do not accept x-axis intercept for conductor.

Superconductor:
Sharp transition with vertical line to x-axis. Greater than T_c, accept any concave upwards curve or line showing resistance increasing with temperature.

[2 marks]

Any three of:

type 1 have lower critical temperature/T_c «than type 2»
OR
type 2 can superconduct at higher temperatures «than type 1»

type 1 are «elemental» metals AND type 2 can be alloys/composites/metal oxide ceramics/perovskites

type 1 have sharp transition to superconductivity AND type 2 have more gradual transition

type 1 have all «magnetic» flux expelled to normal state AND type 2 have partial penetration of flux in mixed state

type 1 typically work via Cooper pairs AND type 2 may not necessarily use this mechanism

magnetic fields can penetrate type 2 in the mixed state «in a type of Vortex» AND type 1 has no mixed state

type 1 have one critical magnetic field/B_c AND type 2 have two/B_c1 and B_c2

Award [1 max] if three correct pieces of information are given for one type only without contrasting with the other type.

Marks may also be awarded from suitable sketch(es).

Accept “H” for “B”.

[3 marks]

Deduce the number of atoms per unit cell in vanadium.

Calculate the expected first order diffraction pattern angle, in degrees, if x-rays of wavelength 150 pm are directed at a crystal of vanadium. Assume the edge length of the crystal to be the same as separation of layers of vanadium atoms found by x-ray diffraction. Use section 1 of the data booklet.

Calculate the average mass, in g, of a vanadium atom by using sections 2 and 6 of the data booklet.

Determine the volume, in cm³, of a vanadium unit cell.

Determine the density, in g cm⁻³, of vanadium by using your answers to (a)(i), (a)(iii) and (a)(iv).

Vanadium and other transition metals can interfere with cell metabolism.

State and explain one process, other than by creating free radicals, by which transition metals interfere with cell metabolism.

Vanadium(IV) ions can create free radicals by a Fenton reaction.

Deduce the equation for the reaction of V⁴⁺ with hydrogen peroxide.

2

[1 mark]

nλ = 2dsinθ

OR

$\theta ={\sin }^{-1}\left(\frac{n\lambda }{2d}\right)$

θ = «${\sin }^{-1}\left(\frac{150}{2\times 303}\right)=$» 14.3 «°»

Award [2] for correct final answer.

[2 marks]

m = «$\frac{50.94}{6.02\times {10}^{23}}=$ » 8.46 × 10^–23 «g»

[1 mark]

«303 pm = 303 × 10^–10 cm»

V = «(303 × 10^–10)³ =» 2.78 × 10^–23 «cm³ »

[1 mark]

«8.46 × 10^–23 g × 2 =» 1.69 × 10^–22 «g»

d = «$\frac{1.69\times {10}^{-22}\text{ g}}{2.78\times {10}^{-23}\text{ c}{\text{m}}^3}=$» 6.08 «g cm^–3»

Accept any value in the range 6.07–6.09 «g cm^–3».

Award [2] for correct final answer.

[2 marks]

Any one of these alternatives:

ALTERNATIVE 1

disrupt enzyme binding sites

which can inhibit/over-stimulate enzymes

ALTERNATIVE 2

disrupt endocrine system

because they compete for active sites of enzymes/cellular receptors

ALTERNATIVE 3

form complexes/coordination compounds

which can bind to enzymes

ALTERNATIVE 4

act as oxidizing/reducing agents

OR

act as catalysts

which can initiate unwanted reactions

Accept “can undergo oxidation–reduction reactions” for M1 in Alternative 4.

[2 marks]

V⁴⁺(aq) + H₂O₂(aq) → V⁵⁺(aq) + OH^–(aq) + •OH(aq)

Do not accept • on H.

Accept answer without •

[1 mark]

Distinguish between the manufacture of polyester and polyethene.

Any one of these alternatives:

ALTERNATIVE 1

Polyester: produced by condensation/esterification polymerization

Polyethene: produced by addition polymerization

ALTERNATIVE 2

Polyester: reaction between monomers/molecules containing two functional groups per molecule

Polyethene: reaction between monomers/molecules containing a carbon–carbon double bond/C=C

ALTERNATIVE 3

polyester polymerization forms a by-product/H₂O

polyethene has no by-products/100% atom economy

Accept the names of different catalysts used for each polymerization as an alternative answer.

[2 marks]

Estimate the atom economy of this first step.

Suggest, giving one reason, whether this is an addition or condensation reaction.

Subsequent steps proceed under differing conditions, forming the dendrimer polymer with the following repeating unit.

State the name of one functional group in this repeating unit.

100%

Accept “almost 100%” if a catalyst is referred to.

[1 mark]

addition AND no atoms removed/all atoms accounted for/no loss of water/ammonia/inorganic by-product/small molecules
OR
addition AND there is only one «reaction» product

[1 mark]

amido
OR
amino

Accept “amide/carboxamide/carbamoyl” for “amido”.

Accept “amine“ for “amino”.

Accept “carbonyl”.

[1 mark]

State the type of particle present in the plasma formed.

Calculate the concentration of lead ions in the sample in mol dm^‒3.

Lead ions are toxic and can be precipitated using hydroxide ions.

Pb²⁺ (aq) + 2OH^‒ (aq) $\rightleftharpoons$ Pb(OH)₂ (s)

Sufficient sodium hydroxide solid is added to the antacid sample to produce a 1.0 × 10^‒2 mol dm^‒3 hydroxide ion solution at 298 K.

Deduce if a precipitate will be formed, using section 32 of the data booklet.

If you did not calculate the concentration of lead ions in (b)(i), use the value of 2.4 × 10⁻⁴ mol dm^‒3, but this is not the correct value.

Electrolysis is used to obtain lead from Pb²⁺ (aq) solution.

Determine the time, in hours, required to produce 0.0500 mol lead using a current (I) of 1.34 A. Use section 2 of the data booklet and the equation, charge (Q) = current (I) × time (t, in seconds).

State one feature of a chelating agent.

An aqueous lead(II) ion reacts with three ethane-1,2-diamine molecules to form an octahedral chelate ion.

Outline why the chelate ion is more stable than the reactants.

positive ions/cations/Pb²⁺

OR

free electrons ✔

Accept “ions” OR “charged species/particle”.

[Pb²⁺] = 0.50 × 10^‒6/5.0 × 10^–7 «g dm^–3» ✔

[Pb²⁺] «$=\frac{0.50\times {10}^{-6}\text{g}\text{d}{\text{m}}^{-3}}{207.20\text{g}\text{mo}{\text{l}}^{-1}}$» = 2.4 × 10^‒9 «mol dm^‒3» ✔

Award [2] for correct final answer.

«K_sp = 1.43 × 10^–20»

ALTERNATIVE 1:

«Q = [Pb²⁺] [OH^–]² = 2.4 × 10^–9 × (1.0 × 10^–2)²» = 2.4 × 10^–13 ✔

Q > K_sp AND precipitate will form

OR

2.4 × 10^–13 > 1.43 × 10^–20 AND precipitate will form ✔

ALTERNATIVE 2:

critical [Pb²⁺] for hydroxide solution «$=\frac{K_{sp}}{{\left[\text{O}{\text{H}}^{-}\right]}^2}=\frac{1.43\times {10}^{-20}}{{\left(1.0\times {10}^{-2}\right)}^2}$» = 1.4 × 10^–16 ✔

initial concentration > critical concentration AND precipitate will form

OR

2.4 × 10^–9 > 1.4 × 10^–16 AND precipitate will form ✔

If value given is used:

ALTERNATIVE 3:

«Q = [Pb²⁺] [OH^–]² = 2.4 × 10^–4 × (1.0 × 10^–2)²» = 2.4 × 10^–8 ✔

Q > K_sp AND precipitate will form

OR

2.4 × 10^–8 > 1.43 × 10^–20 AND precipitate will form ✔

«Faraday’s constant, F = 9.65 × 10⁴ C mol^‒1 and 1 A = 1 C s^–1»
Q «= 0.0500 mol × 2 × 96500 C mol^‒1» = 9650 «C» ✔

t «$=\frac{Q}{I}=\frac{9650\text{C}}{1.34\text{C}{\text{s}}^{-1}}\approx 7200\text{s}$ so $\frac{7200\text{s}}{60\times 60\text{s}{\text{h}}^{-1}}$» = 2.00 «hours» ✔

Award [2] for correct final answer.

Any one of:

two «or more» lone/non-bonding pairs on different atoms

OR

two «or more» atoms/centres that act as Lewis bases ✔

form «at least» two coordination/coordinate bonds

OR

«at least» two atoms can form coordination/coordinate bonds ✔

Reference to “on DIFFERENT atoms” required.

Accept “dative «covalent» bond” for “coordination/coordinate bond”.

increase in entropy

OR

ΔS > 0/ΔS positive ✔

Accept “ΔG < 0” but not “ΔH < 0”.

Outline why this type of classification is not entirely satisfactory by using magnesium diboride, MgB₂, as an example. Refer to sections 8 and 29 of the data booklet.

Structures of poly(methyl acrylate), PMA, and Bakelite^® are shown.

Suggest, giving reasons, which is the thermoplastic polymer and which is the thermosetting polymer.

A zeolite is an alternative catalyst for this reaction.

Explain how zeolites act as selective catalysts.

State the names of the two terminal functional groups in X.

Deduce the repeating unit of the polymer of X.

Repeating units of several polymers are listed.

The infrared (IR) spectrum of one of these polymers is shown.

Deduce, giving a reason, the name of this polymer and its Resin Identification Code (RIC), using sections 26 and 30 in the data booklet.

$\mathrm{\Delta }\chi$ = 0.7 AND average $\mathrm{\Delta }\chi$ = 1.7 ✔

bonding between metallic and ionic

OR

more than one type of bonding present

OR

bond type difficult to determine as close to several regions/several types/named bonding types «eg ionic and covalent etc.»

OR

bond is mostly covalent «based on % covalent scale on diagram»

OR

bond has « $\frac{0.7}{3.2}\times 100=$» 22% ionic character ✔

Accept “EN” for “$\chi$".

Accept “bond is ionic but close to several regions/several types/other named bonding type(s) (eg covalent, metallic and covalent etc.)”.

Do not accept just “bond is ionic”.

Accept any value for % ionic character in range 15–24% or % covalent character in range 76–85%.

Thermoplastic polymer:

PMA AND «weak» intermolecular/IMFs/London/dispersion/van der Walls/vdW/dipole-dipole forces «between layers/chains»

OR

PMA AND no/few cross-links «between layers/chains» ✔

Thermosetting polymer:

Bakelite^® AND «strong» covalent bonds «between layers/chains»

OR

Bakelite^® AND extensive cross-links «between layers/chains» ✔

Do not accept “hydrogen bonding” for M1.

Award [1 max] for correct reasons for both polymer classes even if named polymers are incorrectly classified.

pores/cavities/channels/holes/cage-like structures «in zeolites» have specific shape/size ✔

only reactants «with appropriate size/geometry» fit inside/go through/are activated/can react ✔

amino AND carboxyl ✔

Do not accept “carbonyl”, “hydroxyl”.

Continuation bonds at NH and CO are required for mark.

Ignore any brackets and n.

Name and reason:

PET/PETE AND peak for C=O «at 1700–1750 cm^–1» ✔

RIC:

1 ✔

Accept “PET/PETE AND peak for C–O «at 1050–1410 cm^–1»” for M1.

Accept “PET/PETE AND peak(s) for COO” for M1.

Accept name or abbreviation for polymer.

No ECF for M2.

State the name of the crystal structure of gold.

Calculate the number of atoms per unit cell of gold, showing your working.

The edge length of the gold unit cell is 4.08 × 10^‒8 cm.

Determine the density of gold in g cm^‒3, using sections 2 and 6 of the data booklet.

face-centred cube/fcc

OR

cubic close packed/ccp ✔

$\frac{1}{2}$ «atom per face» × 6 «faces per cube» × 3 «atoms» AND $\frac{1}{8}$ «atom per corner» × 8 «corners per cube» = 1 «atom» ✔

«atoms per unit cell = 3 + 1 =» 4 ✔

Award [1 max] for “4” without working shown.

«4 atoms per unit cell»

mass of 4 atoms «$=4\times \frac{196.97\text{g}\text{mo}{\text{l}}^{-1}}{6.02\times {10}^{23}\text{mo}{\text{l}}^{-1}}=$» 1.31 × 10^–21 «g»

volume of unit cell «= (4.08 × 10^‒8)³ cm³» = 6.79 × 10^–23 «cm³»

density = «$\frac{1.31\times {10}^{-21}\text{g}}{6.79\times {10}^{-23}\text{c}{\text{m}}^3}$» = 1.93 × 10¹/19.3 «g cm^‒3»

Award [3] for correct final answer.

Describe the structure and bonding of a carbon nanotube.

Structure:

Bonding:

Suggest one application for carbon nanotubes.

Structure:
giant covalent/network covalent    [✔]

Bonding:
each carbon covalently bonded to 3 other carbons
OR
each bond has order of 1.5    [✔]

Note: Accept “cylindrical/tube shaped”.

Accept “has delocalized electrons” OR “has sp2 hybridization”.

Any one of:
3D electrodes   [✔]

catalysts    [✔]

biosensors    [✔]

molecular stents    [✔]

body armour    [✔]

synthetic muscles    [✔]

micro transistors/circuitry/capacitors/electrodes    [✔]

reinforcing phase in a matrix/composite material «such as concrete»    [✔]

micro antenna    [✔]

stealth technology    [✔]

water/air filtration    [✔]

solar cells    [✔]

tennis racquets    [✔]

microelectronic circuits    [✔]

Note: Do not accept just general answers such as “medicine” or “defence”.

Describing the structure and bonding of a carbon nanotube was generally answered satisfactorily, although some candidates simply said the bonding was covalent with no further detail.

There were some vague responses for applications of carbon nanotubes when specific details were needed to score the mark in (b).

X-ray crystallography of a metal crystal produces a diffraction pattern of bright spots.

Using X-rays of wavelength 1.54 × 10⁻¹⁰ m, the first bright spots were produced at an angle θ of 22.3° from the centre.

Calculate the separation between planes of atoms in the lattice, in meters, using section 1 of the data booklet.

«d = $\frac{n\lambda }{2\sin \text{θ}}$»

d = «$\frac{1\times 1.54\times {10}^{-10}\text{m}}{2\times \sin {22.3}^{\circ }}$» 2.03 × 10⁻¹⁰ «m» ✔

Outline two differences between heterogeneous and homogeneous catalysts.

Suggest, giving a reason, how elastomers used for the tyre tread can increase the traction between the tyre and the road.

Tyre fires emit trace quantities of polychlorinated dibenzofurans and polychlorinated dibenzo-p-dioxin.

Outline, using section 31 of the data booklet, why polychlorinated dibenzofuran is not classed chemically as a dioxin but considered “dioxin-like”.

Classify polybutadiene as either an addition or condensation polymer, giving a reason.

State one factor considered when making green chemistry polymers.

Any two of:
heterogeneous catalyst is in different phase than reactants AND homogeneous catalyst in same phase   [✔]

homogeneous catalysts chemically change/react and reformed at end of reaction
OR
reactants adsorb onto heterogenous catalyst and products desorb    [✔]

heterogeneous catalysts are more easily removed than homogenous catalysts     [✔]

heterogeneous catalysts can function at higher temperatures     [✔]

homogeneous catalysts are «generally» more selective     [✔]

homogeneous catalysts offer a broader range of reactions     [✔]

Note: Accept “state” for “phase”.

Accept “heterogeneous catalyst provides a surface to activate reaction”.

elastomers bend under force «and return to original form when force is released»
OR
elastomers make tyre more flexible   [✔]

allows greater contact with road    [✔]

does not contain heterocyclic ring with 2 oxygen atoms
OR
middle ring has only 1 oxygen atom    [✔]

produces similar toxic effects to dioxins    [✔]

addition AND not two different functional groups reacting
OR
addition AND formed by breaking one bond of the carbon–carbon double bonds
OR
addition AND empirical formula of monomer equals empirical formula of polymer
OR
addition AND no atoms removed/all atoms accounted for/no loss of water/ammonia/inorganic by-product/small molecules
OR
addition AND atom economy/efficiency is 100 %
OR
addition AND there is only one «reaction» product   [✔]

Any one of:
high content of raw materials in product/high atom economy    [✔]

use of low toxic chemicals/catalysts/materials/solvents     [✔]

renewable feedstock/raw materials     [✔]

use of renewable/clean/low carbon energy source     [✔]

high safety standards     [✔]

increase energy efficiency     [✔]

waste recycling     [✔]

Note: Accept other reasonable answers.

Most candidates correctly stated one difference between heterogeneous and homogeneous catalysts. Few gave a second difference even though the question is worth 2 marks.

Most explained well how elastomers increase tyre traction.

But had difficulty applying their knowledge to outline why polychlorinated dibenzofuran is considered dioxin-like but is not classified as a dioxin.

Some candidates failed to score the mark as they did not give a reason for classifying polybutadiene as an addition polymer.

Most candidates were able to state a factor considered when making green chemistry polymers.

Determine the mass of aluminium, in g, that could be extracted from an appropriate solution by a charge of 48 250 C. Use sections 2 and 6 of the data booklet.

Once extracted, the purity of the metal can be assessed using ICP-MS. Suggest two advantages of using plasma technology rather than regular mass spectrometry.

Explain the action of metals as heterogeneous catalysts.

Outline how alloys conduct electricity and why they are often harder than pure metals.

Conduct electricity:

Harder than pure metals:

Carbon nanotubes are added to metals to increase tensile strength.

Write an equation for the formation of carbon nanotubes from carbon monoxide.

moles of electrons «= $\frac{48250\text{ C}}{96500\text{ C mo}{\text{l}}^{-1}}$» = 0.5000 «mol»   [✔]

moles of aluminium «= $\frac{0.5000\text{ mol}}{3}$» = 0.1667 «mol»  [✔]

mass of aluminium «= 26.98 g mol^–1 × 0.1667 mol» = 4.50 «g»  [✔]

Note: Award [3] for correct final answer.

Any two of:
larger linear calibration  [✔]

«accurate» detection of multiple elements/metals [✔]

«accurate» detection of elements in low concentration  [✔]

temperature around 10 000 K atomises/ionises every material  [✔]

Any two of:
reactant(s) adsorb onto active sites/surface  [✔]

bonds weakened/broken/stretched «in adsorbed reactants»
OR
activation energy lowered [✔]

products desorbed [✔]

Note: Accept “products released” for M3.

Conduct electricity:
«delocalized/valence» electrons free to move «under potential difference»  [✔]

Harder than pure metals:
atoms/ions of different sizes prevent layers «of atoms/ions» from sliding over one another  [✔]

2CO (g) → C (s) + CO₂ (g)  [✔]

Many candidates did reasonably well in this question but some struggled with the number of electrons required.

Most candidates did not seem to understand any advantages of using plasma technology rather than regular mass spectrometry.

This question was reasonably answered with many candidates receiving a mark for the action of a catalyst. The terms adsorbed and desorbed were often missing.

Most candidates were awarded one mark for how alloys conduct electricity. Some struggled with describing why they are harder than pure metals.

Carbon nanotubes proved to be difficult for the majority of the candidates. Hardly any candidates stated an equation for the formation of carbon nanotubes from carbon monoxide.

Describe the characteristics of the nematic liquid crystal phase.

Shape of molecules:

Distribution: 

Shape of molecules:
linear
OR
rod «shaped»  [✔]

Distribution:
no positional order AND «some» directional order  [✔]

Note: Accept “partly ordered”.

Most candidates were able to obtain at least one mark on this question but struggled with the distribution of the nematic liquid crystal phase.

State what is meant by a superconductor.

Outline the difference in behaviour of Type 1 and Type 2 superconductors when the temperature is lowered.

«material with» no electrical resistance  [✔]

Type 1 has sharper transition to superconductivity  [✔]

Note: Accept annotated plot of electrical resistance against temperature.

The question about superconductor was well answered by the candidates.

Some candidates struggled to outline the difference in the behaviour of Type 1 and Type 2 superconductors when the temperature is lowered.

Outline why heavy metals are toxic.

Determine the maximum concentration of lead(II) ions at 298 K in a solution in which the concentration of carbonate ions is maintained at 1.10 × 10⁻⁴ mol dm⁻³. Use section 32 of the data booklet.

State a method, other than precipitation, of removing heavy metal ions from solution.

Any one of:
disrupt endocrine system
OR
compete for active sites of enzymes/cellular receptors
OR
form complexes with/inhibit enzymes
OR
denature proteins
OR
change shape of active site  [✔]

participate in redox reactions
OR
disturb normal redox balance «in cells»  [✔]

initiate «free» radical reactions «in electron transfer»  [✔]

«K_sp = 7.40 × 10^–14»
K_sp = [Pb²⁺][CO₃^2–]  [✔]

[Pb²⁺] «= $\frac{7.40\times {10}^{-14}}{1.10\times {10}^{-4}}$» = 6.73 × 10^–10 «mol dm^–3»  [✔]

Note: Award [2] for correct final answer.

Any one of:
chelation «by EDTA/polydentate ligand anchored»  [✔]

ion exchange systems [✔]

adsorption by «water» plants  [✔]

Note: Accept “use of zeolites”.

The candidates seemed to have difficulty in outlining why heavy metals are toxic.

Majority of the candidates managed to get two marks in determining the maximum concentration of lead(II) ions using solubility product constant.

This was not well answered by most of the candidates.

Draw a section of isotactic polychloroethene (polyvinylchloride, PVC) showing all the atoms and all the bonds of four monomer units.

The infrared (IR) spectrum of polyethene is given.

Suggest how the IR spectrum of polychloroethene would diff er, using section 26 of the data booklet.

Explain how plasticizers affect the properties of plastics.

Suggest why the addition of plasticizers is controversial.

Outline, giving a reason, how addition and condensation polymerization compare with regard to green chemistry.

Draw the full structural formula of the organic functional group formed during the polymerization of the two reactants below.

correct bonding  [✔]

Cl atoms all on same side and alternate  [✔]

Note: Continuation bonds must be shown.

Award [1 max] if less than or more than four units shown.

Accept a stereo formula with all atoms and bonds shown.

«strong additional» absorption at 600–800 «cm^–1»  [✔]

Any two of:
embedded/fit between chains of polymers  [✔]

prevent chains from forming crystalline regions  [✔]

keep polymer strands/chains/molecules separated/apart [✔]

increase space/volume between chains  [✔]

weaken intermolecular/dipole-dipole/London/dispersion/instantaneous dipoleinduced dipole/van der Waals/vdW forces «between chains» [✔]

increase flexibility/durability/softness [✔]

make polymers less brittle [✔]

Note: Accept “lowers density/melting point”.

leach into foodstuffs/environment
OR
«unknown» health/environmental consequences  [✔]

Note: Accept “plasticizers cannot be recycled”.

addition produces only the polymer «AND more green»
OR
addition has no by/side-product/condensation produces by-product/small molecules/HCl/NH₃ «AND less green»
OR
addition has high atom economy/condensation has lower atom economy «AND less green»
OR
condensation polymers «often» more biodegradable than addition polymers «AND more green»  [✔]

Note: Accept “if water produced by condensation «AND condensation and addition equally green»”.

Accept for addition “all of reactants change into products”.

 [✔]

Note: Continuation bonds must be shown.

Do not accept condensed formula.

Few candidates scored at least one mark although most either scored both or none for this polymer structure. Some did not read that only four monomer units are required.

Almost all candidates received the mark for identifying the correct absorption band for polychloroethene.

This was a well-answered question; with most candidates identifying at least one method plasticizers affect the properties of plastic.

Several candidates wrote vague answers as to why the addition of plasticizers is controversial.

Candidates seemed to have difficulty in comparing addition and condensation polymerisation with regard to green chemistry.

Several candidates struggled to draw the full structural formula of the peptide linkage formed during the polymerisation of the two reactants.

MWCNT are very small in size and can greatly increase switching speeds in a liquid crystal allowing the liquid crystal to change orientation quickly.

Discuss two other properties a substance should have to be suitable for use in liquid crystal displays.

Any two from:

chemically stable AND does not «chemically» degrade over time

stable over range of temperatures AND to avoid «voltage/random shift» fluctuations

polar AND influenced by an electric field

strong intermolecular forces AND allow molecule to align in specific orientations

Award [1 max] for identifying two correct properties without any discussion given or incorrect interpretation of suitability.

Accept “voltage” for “electric field”.

[2 marks]

Draw the structure of the monomers of Kevlar^® if the by-product of the condensation polymerization is hydrogen chloride.

State and explain why plasticizers are added to polymers.

Discuss why the recycling of plastics is an energy intensive process.

OR
H₂NC₆H₄NH₂ ✔

OR
Cl(O)CC₆H₄C(O)Cl ✔

increases flexibility/softness/plasticity ✔
break/weaken intermolecular forces/IMF/H-bonds «between chains» ✔

Any two of:
collection/transportation of plastic waste ✔

separation/sorting of different types «of plastic»
OR
separation/sorting of plastic from other materials ✔

melting plastic ✔

processing/washing/cleaning/drying/manufacture of recycled plastic ✔

Outline how resistance to electric currents occurs in metals.

Suggest why the resistance of metals increases with temperature.

State two differences between Type I and Type II superconductors.

electrons collide with cations/positive ions ✔

increased vibrations of «lattice» ions ✔
increased «probability of» collisions «between electrons and cations» ✔

NOTE: Accept “increases lattice vibrations” for M1.

Any two of:
Type I have sharper transitions to superconductivity «than Type II» ✔
Type I have lower critical/operating temperatures «than Type II» ✔
Type I have lower critical magnetic field «strength than Type II» ✔
Type I carry lower currents «than Type II» ✔
Type I are «pure» metals/metalloids AND Type II are alloys/metal oxide ceramics/perovskites/metallic compounds ✔
Type II exist in a mixed state/are partly permeable to the magnetic field AND Type I do not/are not ✔

1.40 × 10⁻³ g of NaOH (s) are dissolved in 250.0 cm³ of 1.00 × 10⁻¹¹ mol dm⁻³ Pb(OH)₂ (aq) solution.

Determine the change in lead ion concentration in the solution, using section 32 of the data booklet.

«[OH⁻] =$\frac{1.40\times {10}^{-3}\text{g}}{40.00\text{ g mo}{\text{l}}^{-1}\times 0.2500\text{ d}{\text{m}}^3}$ =» 1.40 × 10⁻⁴ «mol dm⁻³» ✔

«[OH⁻] from dissolved Pb(OH)₂ is negligible»

NOTE: Accept «ratio $\frac{{\left[P{b}^{2+}\right]}_{initial}}{{\left[P{b}^{2+}\right]}_{final}}$ =» 13.7 OR  «ratio $\frac{{\left[P{b}^{2+}\right]}_{final}}{{\left[P{b}^{2+}\right]}_{initial}}$ =» 0.0730 for M4.

K_sp = [Pb²⁺][OH⁻]²
OR
1.43 × 10⁻²⁰ = [Pb²⁺] × (1.40 × 10⁻⁴)² ✔

[Pb²⁺]_final = 7.30 × 10⁻¹³ «mol dm⁻³» ✔

NOTE: Award [4] for correct final answer.

«change in [Pb²⁺] = 1.00 × 10⁻¹¹ − 7.30 × 10⁻¹³ =» 9.27 × 10⁻¹² «mol dm⁻³» ✔

NOTE: Award [3] for correct [Pb²⁺]_final.

Explain why Type 2 superconductors are generally more useful than Type 1.

Any two of:

have higher critical temperatures/T_c «than Type 1»

OR

can act at higher temperatures

have higher critical magnetic fields/B_c «than Type 1»

less time needed to cool to operating temperature

less energy required to cool down/maintain low temperature

Do not accept “Type 2 has a gradual transition to a superconducting state but in Type 1 it is a sharp transition”.

[Max 2 Marks]

Draw the structure of the monomer from which nylon-6 is produced by a condensation reaction.

Deduce, giving a reason, whether the atom economy of a condensation polymerization, such as this, would be greater or less than an addition polymerization, such as the formation of HDPE.

–NH₂ AND –COOH

six C-atoms

Accept –COCl instead of –COOH.

[2 marks]

less AND a second molecule/product formed

Accept “not all the reactant molecules «in the equation» are converted «to product molecules»”.

[1 mark]

State the number of atoms in the unit cell.

Determine the density of calcium, in g cm⁻³, using section 2 of the data booklet.

Ar = 40.08; metallic radius (r) = 1.97 × 10⁻¹⁰ m

« $8\times \frac{1}{8}+6\times \frac{1}{2}=$ » 4  [✔]

a = « $\frac{4r}{\sqrt{2}}=\frac{4\times 1.97\times {10}^{-10}\text{m}}{\sqrt{2}}$ =» 5.572 × 10^–10 «m»
OR
volume of unit cell = «(5.572 × 10^–10 m)³ × 10⁶ =» 1.73 × 10^–22 «cm³»  [✔]

mass of unit cell =« $\frac{40.08\text{ g mo}{\text{l}}^{-1}\times 4}{6.02\times {10}^{23}\text{mo}{\text{l}}^{-1}}$ =» 2.66 × 10^–22 «g»  [✔]

density = « $\frac{2.66\times {10}^{-22}\text{g}}{{(5.572\times {10}^{-10})}^3\times {10}^6}$ » 1.54 «g cm^–3»  [✔]

Note: Award [3] for correct final answer.

The number of atoms in the unit cell was correctly calculated by most of the candidates.

Majority of the candidates managed to get three marks in determining the density of the calcium.

Explain how entropy affects this equilibrium.

State the number of coordinate covalent bonds EDTA forms with Ni²⁺.

entropy increases «and the reaction proceeds to the right»   [✔]

more species / free molecules are formed
OR
more ways of distributing energy   [✔]

six    [✔]

There were several incorrect responses that products were more ordered than the reactants.

Proved very challenging with very few candidates knowing the number of coordinate covalent bonds EDTA forms with a nickel ion.

Describe the effect of infrared (IR) radiation on carbon dioxide molecules.

Outline one approach to controlling industrial emissions of carbon dioxide.

bond length/C=O distance changes
OR
«asymmetric» stretching «of bonds»
OR
bond angle/OCO changes [✔]

polarity/dipole «moment» changes
OR
dipole «moment» created «when molecule absorbs IR» [✔]

Note: Accept appropriate diagrams.

Any one of:
capture where produced «and store» [✔]

use scrubbers to remove [✔]

use as feedstock for synthesising other chemicals [✔]

carbon credit/tax/economic incentive/fines/country specific action [✔]

use alternative energy
OR
stop/reduce use of fossil fuels for producing energy [✔]

use carbon reduced fuels «such as methane» [✔]

increase efficiency and reduce energy use [✔]

Note: Do not accept “planting more trees”.

Accept specific correct examples.

This part was fairly well answered with most candidates receiving one of the two marks. There were many candidates who stated asymmetric stretching and bonds vibrate but missed writing polarity and dipole changes, which deprived them of the second mark.

This part was reasonably answered although there were many candidates who gave vague answers that did not receive marks.