Explain the general increase in trend in the first ionization energies of the period 3 elements, Na to Ar.

Sodium emits yellow light with a frequency of 5.09 × 10¹⁴Hz when electrons transition from 3p to 3s orbitals.

Calculate the energy difference, in J, between these two orbitals using sections 1 and 2 of the data booklet.

Darling, D, n.d. D lines (of sodium). [online] Available at <https://www.daviddarling.info/encyclopedia/D/D_lines.html> [Accessed 6 May 2020].

increasing number of protons
OR
increasing nuclear charge ✔

«atomic» radius/size decreases
OR
same number of shells/electrons occupy same shell
OR
similar shielding «by inner electrons» ✔

«ΔE = hν = 6.63 × 10^–34J s × 5.09 × 10¹⁴s^–1 =» 3.37 × 10^–19«J» ✔

Describe the bonding in metals.

Titanium exists as several isotopes. The mass spectrum of a sample of titanium gave the following data:

Calculate the relative atomic mass of titanium to two decimal places.

State the number of protons, neutrons and electrons in the ${}_{\text{22}}^{\text{48}}\text{Ti}$ atom.

State the full electron configuration of the ${}_{\text{22}}^{\text{48}}\text{T}{\text{i}}^{2+}$ ion.

Suggest why the melting point of vanadium is higher than that of titanium.

Sketch a graph of the first six successive ionization energies of vanadium on the axes provided.

Explain why an aluminium-titanium alloy is harder than pure aluminium.

Describe, in terms of the electrons involved, how the bond between a ligand and a central metal ion is formed.

Outline why transition metals form coloured compounds.

State the type of bonding in potassium chloride which melts at 1043 K.

A chloride of titanium, $\text{TiC}{\text{l}}_{\text{4}}$, melts at 248 K. Suggest why the melting point is so much lower than that of KCl.

Formulate an equation for this reaction.

Suggest one disadvantage of using this smoke in an enclosed space.

electrostatic attraction

between «a lattice of» metal/positive ions/cations AND «a sea of» delocalized electrons

Accept “mobile electrons”.

Do not accept “metal atoms/nuclei”.

[2 marks]

$\frac{(46\times 7.98)\text{ + }(47\times 7.32)\text{ + }(48\times 73.99)\text{ + }(49\times 5.46)\text{ + }(50\times 5.25)}{100}=47.93$

Answer must have two decimal places with a value from 47.90 to 48.00.

Award [2] for correct final answer.

Award [0] for 47.87 (data booklet value).

[2 marks]

Protons: 22 AND Neutrons: 26 AND Electrons: 22

[1 mark]

$\text{1}{\text{s}}^{\text{2}}\text{2}{\text{s}}^{\text{2}}\text{2}{\text{p}}^{\text{6}}\text{3}{\text{s}}^{\text{2}}\text{3}{\text{p}}^{\text{6}}\text{3}{\text{d}}^{\text{2}}$

[1 mark]

vanadium has smaller ionic radius «leading to stronger metallic bonding»

Accept vanadium has «one» more valence electron«s» «leading to stronger metallic bonding».

Accept “atomic” for “ionic”.

[1 mark]

regular increase for first five AND sharp increase to the 6th

A log graph is acceptable.

Accept log plot on given axes (without amendment of y-axis).

Award mark if gradient of 5 to 6 is greater than “best fit line” of 1 to 5.

[1 mark]

titanium atoms/ions distort the regular arrangement of atoms/ions

OR

titanium atoms/ions are a different size to aluminium «atoms/ions»

prevent layers sliding over each other

Accept diagram showing different sizes of atoms/ions.

[2 marks]

pair of electrons provided by the ligand

Do not accept “dative” or “coordinate bonding” alone.

[1 mark]

partially filled d-orbitals

«ligands cause» d-orbitals «to» split

light is absorbed as electrons transit to a higher energy level «in d–d transitions»
OR
light is absorbed as electrons are promoted

energy gap corresponds to light in the visible region of the spectrum

colour observed is the complementary colour

[4 marks]

ionic

OR

«electrostatic» attraction between oppositely charged ions

[1 mark]

«simple» molecular structure

OR

weak«er» intermolecular bonds

OR

weak«er» bonds between molecules

Accept specific examples of weak bonds such as London/dispersion and van der Waals.

Do not accept “covalent”.

[1 mark]

$\text{TiC}{\text{l}}_{\text{4}}\text{(l)}+\text{2}{\text{H}}_{\text{2}}\text{O(l)}\to \text{Ti}{\text{O}}_{\text{2}}\text{(s)}+\text{4HCl(aq)}$ correct products
correct balancing

Accept ionic equation.

Award M2 if products are HCl and a compound of Ti and O.

[2 marks]

HCl causes breathing/respiratory problems

OR

HCl is an irritant

OR

HCl is toxic

OR

HCl has acidic vapour

OR

HCl is corrosive

Accept TiO₂ causes breathing

problems/is an irritant.

Accept “harmful” for both HCl and TiO₂.

Accept “smoke is asphyxiant”.

[1 mark]

Outline why ozone in the stratosphere is important.

Dinitrogen monoxide in the stratosphere is converted to nitrogen monoxide, NO (g).

Write two equations to show how NO (g) catalyses the decomposition of ozone.

State one analytical technique that could be used to determine the ratio of ¹⁴N : ¹⁵N.

A sample of gas was enriched to contain 2 % by mass of ¹⁵N with the remainder being ¹⁴N.

Calculate the relative molecular mass of the resulting N₂O.

Predict, giving two reasons, how the first ionization energy of ¹⁵N compares with that of ¹⁴N.

Explain why the first ionization energy of nitrogen is greater than both carbon and oxygen.

Nitrogen and carbon:

Nitrogen and oxygen:

State what the presence of alternative Lewis structures shows about the nature of the bonding in the molecule.

State, giving a reason, the shape of the dinitrogen monoxide molecule.

Deduce the hybridization of the central nitrogen atom in the molecule.

absorbs UV/ultraviolet light «of longer wavelength than absorbed by O₂»     [✔]

NO (g) + O₃ (g) → NO₂ (g) + O₂ (g)       [✔]
NO₂ (g) + O₃ (g) → NO (g) + 2O₂ (g)     [✔]

Note: Ignore radical signs.

Accept equilibrium arrows.

Award [1 max] for NO₂ (g) + O (g) → NO (g) + O₂ (g).

mass spectrometry/MS     [✔]

« $\frac{(98\times 14)+(2\times 15)}{100}$ =» 14.02    [✔]

«M_r = (14.02 × 2) + 16.00 =» 44.04    [✔]

Any two:

same AND have same nuclear charge /number of protons/Z^_eff      [✔]

same AND neutrons do not affect attraction/ionization energy/Z^_eff
OR
same AND neutrons have no charge       [✔]

same AND same attraction for «outer» electrons     [✔]

same AND have same electronic configuration/shielding     [✔]

Note: Accept “almost the same”.

“Same” only needs to be stated once.

Nitrogen and carbon:

N has greater nuclear charge/«one» more proton «and electrons both lost from singly filled p-orbitals»    [✔]

Nitrogen and oxygen:

O has a doubly filled «p-»orbital
OR
N has only singly occupied «p-»orbitals     [✔]

Note: Accept “greater e– ^- e^- repulsion in O” or “lower e– ^- e^- repulsion in N”.

Accept box annotation of electrons for M2.

delocalization

OR

delocalized π-electrons    [✔]

Note: Accept “resonance”.

linear AND 2 electron domains

OR

linear AND 2 regions of electron density    [✔]

Note: Accept “two bonds AND no lone pairs” for reason.

sp     [✔]

Candidates sometimes failed to identify how ozone works in chemical terms, referring to protects/deflects, i.e., the consequence rather than the mechanism.

Many candidates recalled the first equation for NO catalyzed decomposition of ozone only. Some considered other radical species.

All candidates, with very few exceptions, answered this correctly.

Most candidates were able to calculate the accurate mass of N₂O, though quite a few candidates just calculated the mass of N and didn’t apply it to N₂O, losing an accessible mark.

Many students realized that neutrons had no charge and could not affect IE significantly, but many others struggled a lot with this question since they considered that ¹⁵N would have a higher IE because they considered the greater mass of the nucleus would result in an increase of attraction of the electrons.

Mixed responses here; the explanation of higher IE for N with respect to C was less well explained, though it should have been the easiest. It was good to see that most candidates could explain the difference in IE of N and O, either mentioning paired/unpaired electrons or drawing box diagrams.

Most candidates identified resonance for this given Lewis representation.

Though quite a number of candidates suggested a linear shape correctly, they often failed to give a complete correct explanation, just mentioning the absence of lone pairs but not two bonds, instead of referring to electron domains.

Hybridisation of the N atom was correct in most cases.

Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.

State the shape of the complex ion.

Deduce the charge on the complex ion and the oxidation state of cobalt.

Describe, in terms of acid-base theories, the type of reaction that takes place between the cobalt ion and water to form the complex ion.

Any three of:

Group 1:
atomic/ionic radius increases

smaller charge density

OR

force of attraction between metal ions and delocalised electrons decreases

Do not accept discussion of attraction between valence electrons and nucleus for M2.

Accept “weaker metallic bonds” for M2.

Group 17:
number of electrons/surface area/molar mass increase

London/dispersion/van der Waals’/vdw forces increase

Accept “atomic mass” for “molar mass”.

[Max 3 Marks]

«distorted» octahedral

Accept “square bipyramid”.

Charge on complex ion: 1+/+
Oxidation state of cobalt: +2

Lewis «acid-base reaction»

H2O: electron/e^– pair donor

OR

Co²⁺: electron/e^– pair acceptor

Write a balanced equation for the reaction that occurs.

Identify a metal, in the same period as magnesium, that does not form a basic oxide.

Calculate the amount of magnesium, in mol, that was used.

Determine the percentage uncertainty of the mass of product after heating.

Assume the reaction in (a)(i) is the only one occurring and it goes to completion, but some product has been lost from the crucible. Deduce the percentage yield of magnesium oxide in the crucible.

Evaluate whether this, rather than the loss of product, could explain the yield found in (b)(iii).

Suggest an explanation, other than product being lost from the crucible or reacting with nitrogen, that could explain the yield found in (b)(iii).

Calculate coefficients that balance the equation for the following reaction.

Ammonia is added to water that contains a few drops of an indicator. Identify an indicator that would change colour. Use sections 21 and 22 of the data booklet.

Determine the oxidation state of nitrogen in Mg₃N₂ and in NH₃.

Deduce, giving reasons, whether the reaction of magnesium nitride with water is an acid–base reaction, a redox reaction, neither or both.

State the number of subatomic particles in this ion.

Some nitride ions are ¹⁵N^3–. State the term that describes the relationship between ¹⁴N^3– and ¹⁵N^3–.

The nitride ion and the magnesium ion are isoelectronic (they have the same electron configuration). Determine, giving a reason, which has the greater ionic radius.

Suggest, giving a reason, whether magnesium or nitrogen would have the greater sixth ionization energy.

Suggest two reasons why atoms are no longer regarded as the indivisible units of matter.

State the types of bonding in magnesium, oxygen and magnesium oxide, and how the valence electrons produce these types of bonding.

2 Mg(s) + O₂(g) → 2 MgO(s) ✔

Do not accept equilibrium arrows. Ignore state symbols

aluminium/Al ✔

$⟨⟨\frac{53.726 \mathrm{g}-47.372 \mathrm{g}}{244.31 \mathrm{g} {\mathrm{mol}}^{-1}}=\frac{6.354 \mathrm{g}}{24.31 \mathrm{g} {\mathrm{mol}}^{-1}}⟩⟩=0.2614 «\mathrm{mol}»✔$

mass of product $«=56.941 \mathrm{g}-47.372 \mathrm{g}»=9.569 «\mathrm{g}»$ ✔

$\text{⟨⟨100 × }\frac{2\times 0.001 \mathrm{g}}{9.569 \mathrm{g}}\text{=0.0209⟩⟩ = 0.02 «\%»}$ ✔

Award [2] for correct final answer

Accept 0.021%

$⟨⟨ 0.2614 \mathrm{mol} \times (24.31 \mathrm{g} {\mathrm{mol}}^{-1}+16.00 \mathrm{g} {\mathrm{mol}}^{-1})=0.2614 \mathrm{mol}\times 40.31 \mathrm{g} {\mathrm{mol}}^{-1}⟩⟩=10.536 «\mathrm{g}»$ ✔

$⟨⟨100\times \frac{9.569 \mathrm{g}}{10.536 \mathrm{g}}= 90.822⟩⟩=91 «\%»$ ✔

Award «0.2614 mol x 40.31 g mol^–1»

Accept alternative methods to arrive at the correct answer.

Accept final answers in the range 90.5-91.5%

[2] for correct final answer.

yes
AND
«each Mg combines with $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ N, so» mass increase would be 14x$\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$ which is less than expected increase of 16x
OR
3 mol Mg would form 101g of Mg₃N₂ but would form 3 x MgO = 121 g of MgO
OR
0.2614 mol forms 10.536 g of MgO, but would form 8.796 g of Mg₃N₂ ✔

Accept Yes AND “the mass of N/N₂ that combines with each g/mole of Mg is lower than that of O/O₂”

Accept YES AND “molar mass of nitrogen less than of oxygen”.

incomplete reaction
OR
Mg was partially oxidised already
OR
impurity present that evaporated/did not react ✔

Accept “crucible weighed before fully cooled”.

Accept answers relating to a higher atomic mass impurity consuming less O/O₂.

Accept “non-stoichiometric compounds formed”.

Do not accept "human error", "wrongly calibrated balance" or other non-chemical reasons.

If answer to (b)(iii) is >100%, accept appropriate reasons, such as product absorbed moisture before being weighed.

«1» Mg₃N₂ (s) + 6 H₂O (l) → 3 Mg(OH)₂ (s) + 2 NH₃ (aq) ✔

phenol red ✔

Accept bromothymol blue or phenolphthalein.

Mg₃N₂: -3
AND
NH₃: -3 ✔

Do not accept 3 or 3-

Acid–base:
yes AND N^3- accepts H⁺/donates electron pair«s»
OR
yes AND H₂O loses H⁺ «to form OH^-»/accepts electron pair«s» ✔

Redox:
no AND no oxidation states change ✔

Accept “yes AND proton transfer takes place”

Accept reference to the oxidation state of specific elements not changing.

Accept “not redox as no electrons gained/lost”.

Award [1 max] for Acid–base: yes AND Redox: no without correct reasons, if no other mark has been awarded

Protons: 7 AND Neutrons: 7 AND Electrons: 10 ✔

isotope«s» ✔

nitride AND smaller nuclear charge/number of protons/atomic number ✔

nitrogen AND electron lost from first «energy» level/s sub-level/s-orbital AND magnesium from p sub-level/p-orbital/second «energy» level
OR
nitrogen AND electron lost from lower level «than magnesium» ✔

Accept “nitrogen AND electron lost closer to the nucleus «than magnesium»”.

Any two of:

subatomic particles «discovered»
OR
particles smaller/with masses less than atoms «discovered»
OR
«existence of» isotopes «same number of protons, different number of neutrons» ✔

charged particles obtained from «neutral» atoms
OR
atoms can gain or lose electrons «and become charged» ✔

atom «discovered» to have structure ✔

fission
OR
atoms can be split ✔

Accept atoms can undergo fusion «to produce heavier atoms»

Accept specific examples of particles.

Award [2] for “atom shown to have a nucleus with electrons around it” as both M1 and M3.

Award [1] for all bonding types correct.

Award [1] for each correct description.

Apply ECF for M2 only once.

Done very well. However, it was disappointing to see the formula of oxygen molecule as O and the oxide as Mg₂O and MgO₂ at HL level.

Average performance; the question asked to identify a metal; however, answers included S, Si, P and even noble gases besides Be and Na. The only choice of aluminium; however, since its oxide is amphoteric, it could not be the answer in the minds of some.

Very good performance; some calculated the mass of oxygen instead of magnesium for the calculation of the amount, in mol, of magnesium. Others calculated the mass, but not the amount in mol as required.

Mediocre performance; instead of calculating percentage uncertainty, some calculated percentage difference.

Satisfactory performance; however, a good number could not answer the question correctly on determining the percentage yield.

Poorly done. The question asked to evaluate and explain but instead many answers simply agreed with the information provided instead of assessing its strength and limitation.

Mediocre performance; explaining the yield found was often a challenge by not recognizing that incomplete reaction or Mg partially oxidized or impurities present that evaporated or did not react would explain the yield.

Calculating coefficients that balance the given equation was done very well.

Well done; some chose bromocresol green or methyl red as the indicator that would change colour, instead of phenol red, bromothymol blue or phenolphthalein.

Good performance; however, surprising number of candidates could not determine one or both oxidation states correctly or wrote it as 3 or 3−, instead of −3.

Average performance; choosing the given reaction as an acid-base or redox reaction was not done well. Often answers were contradictory and the reasoning incorrect.

Stating the number of subatomic particles in a ¹⁴N^3- was done very well. However, some answers showed a lack of understanding of how to calculate the number of relevant subatomic particles given formula of an ion with charge and mass number.

Exceptionally well done; A few candidates referred to isomers, rather than isotopes.

There was reference to nitrogen and magnesium, rather than nitride and magnesium ions. Also, instead identifying smaller nuclear charge in nitride ion, some referred to core electrons, Z_eff, increased electron-electron repulsion or shielding.

Common error in suggesting nitrogen would have the greater sixth ionization energy was that for nitrogen, electron is lost from first energy level without making reference to magnesium losing it from second energy level.

Good performance; some teachers were concerned about the expected answers. However, generally, students were able to suggest two reasons why matter is divisible.

One teacher commented that not asking to describe bonding in terms of electrostatic attractions as in earlier papers would have been confusing and some did answer in terms of electrostatic forces of attractions involved. However, the question was clear in its expectation that the answer had to be in terms of how the valence electrons produce the three types of bonds and the overall performance was good. Some had difficulty identifying the bond type for Mg, O₂ and MgO.

Explain why Si has a smaller atomic radius than Al.

Explain why the first ionization energy of sulfur is lower than that of phosphorus.

State the condensed electron configurations for Cr and Cr3⁺.

Describe metallic bonding and how it contributes to electrical conductivity.

Deduce, giving a reason, which complex ion [Cr(CN)₆]³⁻ or [Cr(OH)₆]³⁻ absorbs higher energy light. Use section 15 of the data booklet.

[Cr(OH)₆]³⁻ forms a green solution. Estimate a wavelength of light absorbed by this complex, using section 17 of the data booklet.

Deduce the Lewis (electron dot) structure and molecular geometry of sulfur tetrafluoride, SF₄, and sulfur dichloride, SCl₂.

Suggest, giving reasons, the relative volatilities of SCl₂ and H₂O.

nuclear charge/number of protons/Z/Z_eff increases «causing a stronger pull on the outer electrons» ✓

same number of shells/«outer» energy level/shielding ✓

P has «three» unpaired electrons in 3p sub-level AND S has one full 3p orbital «and two 3p orbitals with unpaired electrons»
OR
P: [Ne]3s²3p_x¹3p_y¹3p_z¹ AND S: [Ne]3s²3p_x²3p_y¹3p_z¹ ✓

Accept orbital diagrams for 3p sub-level for M1. Ignore other orbitals or sub-levels.

repulsion between paired electrons in sulfur «and therefore easier to remove» ✓

Accept “removing electron from S gives more stable half-filled sub-level" for M2.

Cr:
[Ar] 4s¹3d⁵ ✓

Cr³⁺:
[Ar] 3d³ ✓

Accept “[Ar] 3d⁵4s¹”.

Accept “[Ar] 3d³4s⁰”.

Award [1 max] for two correct full electron configurations “1s²2s²2p⁶3s²3p⁶4s¹3d⁵ AND 1s²2s²2p⁶3s²3p⁶3d³”.

Award [1 max] for 4s¹3d⁵ AND 3d³.

electrostatic attraction ✓

between «a lattice of» cations/positive «metal» ions AND «a sea of» delocalized electrons ✓

mobile electrons responsible for conductivity
OR
electrons move when a voltage/potential difference/electric field is applied ✓

Do not accept “nuclei” for “cations/positive ions” in M2.

Accept “mobile/free” for “delocalized” electrons in M2.

Accept “electrons move when connected to a cell/battery/power supply” OR “electrons move when connected in a circuit” for M3.

[Cr(CN)₆]³⁻ AND CN⁻/ligand causes larger splitting «in d-orbitals compared to OH⁻»
OR
[Cr(CN)₆]³⁻ AND CN⁻/ligand associated with a higher Δ/«crystal field» splitting energy/energy difference «in the spectrochemical series compared to OH⁻ » ✓

Accept “[Cr(CN)₆]³⁻ AND «CN⁻» strong field ligand”.

any value or range between 647 and 700 nm ✓

SF₄/SCl₂ structure does not have to be 3-D for mark.

Penalize missing lone pairs of electrons on halogens once only.

Accept any combination of dots, lines or crosses for bonds/lone pairs.

Accept “non-linear” for SCl₂ molecular geometry.

Award [1] for two correct electron domain geometries, e.g. trigonal bipyramidal for SF₄ and tetrahedral for SCl₂.

H₂O forms hydrogen bonding «while SCl₂ does not» ✓

SCl₂ «much» stronger London/dispersion/«instantaneous» induced dipole-induced dipole forces ✓

Alternative 1:
H₂O less volatile AND hydrogen bonding stronger «than dipole–dipole and dispersion forces» ✓

Alternative 2:
SCl₂ less volatile AND effect of dispersion forces «could be» greater than hydrogen bonding ✓

Ignore reference to Van der Waals.

Accept “SCl₂ has «much» larger molar mass/electron density” for M2.

State the full electron configuration of the chlorine atom.

State, giving a reason, whether the chlorine atom or the chloride ion has a larger radius.

Outline why the chlorine atom has a smaller atomic radius than the sulfur atom.

The mass spectrum of chlorine is shown.

NIST Mass Spectrometry Data Center Collection © 2014 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved.

Outline the reason for the two peaks at $m/z=35$ and $37$.

Explain the presence and relative abundance of the peak at $m/z=74$.

Calculate the amount, in $\mathrm{mol}$, of manganese(IV) oxide added.

Determine the limiting reactant, showing your calculations.

Determine the excess amount, in $\mathrm{mol}$, of the other reactant.

Calculate the volume of chlorine, in ${\mathrm{dm}}^3$, produced if the reaction is conducted at standard temperature and pressure (STP). Use section 2 of the data booklet.

State the oxidation state of manganese in ${\mathrm{MnO}}_2$ and ${\mathrm{MnCl}}_2$.

Deduce, referring to oxidation states, whether ${\mathrm{MnO}}_2$ is an oxidizing or reducing agent.

Hypochlorous acid is considered a weak acid. Outline what is meant by the term weak acid.

State the formula of the conjugate base of hypochlorous acid.

Calculate the concentration of ${\mathrm{H}}^{+} \left(\mathrm{aq}\right)$ in a $\mathrm{HClO} \left(\mathrm{aq}\right)$ solution with a $\mathrm{pH}=3.61$.

State the type of reaction occurring when ethane reacts with chlorine to produce chloroethane.

Predict, giving a reason, whether ethane or chloroethane is more reactive.

Explain the mechanism of the reaction between chloroethane and aqueous sodium hydroxide, $\mathrm{NaOH} \left(\mathrm{aq}\right)$, using curly arrows to represent the movement of electron pairs.

Ethoxyethane (diethyl ether) can be used as a solvent for this conversion.
Draw the structural formula of ethoxyethane

Deduce the number of signals and chemical shifts with splitting patterns in the ¹H NMR spectrum of ethoxyethane. Use section 27 of the data booklet.

Calculate the percentage by mass of chlorine in ${\mathrm{CCl}}_2{\mathrm{F}}_2$.

Comment on how international cooperation has contributed to the lowering of $\mathrm{CFC}$ emissions responsible for ozone depletion.

$\mathrm{CFC}$s produce chlorine radicals. Write two successive propagation steps to show how chlorine radicals catalyse the depletion of ozone.

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^5$ ✔

Do not accept condensed electron configuration.

${\mathrm{Cl}}^{-}$ AND more «electron–electron» repulsion ✔

Accept $C{l}^{\mathit{-}}$ AND has an extra electron.

$\mathrm{Cl}$ has a greater nuclear charge/number of protons/${\mathrm{Z}}_{\mathrm{eff}}$ «causing a stronger pull on the outer electrons» ✔

same number of shells
OR
same «outer» energy level
OR
similar shielding ✔

«two major» isotopes «of atomic mass $35$ and $37$» ✔

«diatomic» molecule composed of «two» chlorine-37 atoms ✔

chlorine-37 is the least abundant «isotope»
OR
low probability of two ${}_{{}_{37}}\mathrm{Cl}$ «isotopes» occurring in a molecule ✔

$«\frac{2.67 \mathrm{g}}{86.94 \mathrm{g} {\mathrm{mol}}^{-1}}=»0.0307 «\mathrm{mol}»$ ✔

$«{\mathrm{n}}_{\mathrm{HCl}}=2.00 \mathrm{mol} {\mathrm{dm}}^{-3}\times 0.2000 {\mathrm{dm}}^3»=0.400 \mathrm{mol}$✔

$«\frac{0.400}{4}=» 0.100 \mathrm{mol}$ AND ${\mathrm{MnO}}_2$ is the limiting reactant ✔

Accept other valid methods of determining the limiting reactant in M2.

$«0.0307 \mathrm{mol}\times 4=0.123 \mathrm{mol}»$

$«0.400 \mathrm{mol}–0.123 \mathrm{mol}=»0.277 «\mathrm{mol}»$ ✔

$«0.0307 \mathrm{mol}\times 22.7 {\mathrm{dm}}^3 {\mathrm{mol}}^{-1}=» 0.697 «{\mathrm{dm}}^3»$ ✔

Accept methods employing $pV=nRT$.

${\mathrm{MnO}}_2: +4$ ✔

${\mathrm{MnCl}}_2: +2$ ✔

oxidizing agent AND oxidation state of $\mathrm{Mn}$ changes from $+4$ to $+2$/decreases ✔

partially dissociates/ionizes «in water» ✔

${\mathrm{ClO}}^{-}$ ✔

$«\left[{\mathrm{H}}^{+}\right]={10}^{-3.61}=» 2.5\times {10}^{-4} «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

«free radical» substitution/${\mathrm{S}}_{\mathrm{R}}$ ✔

Do not accept electrophilic or nucleophilic substitution.

chloroethane AND C–Cl bond is weaker/$324 \mathrm{kJ} {\mathrm{mol}}^{-1}$ than C–H bond/$414 \mathrm{kJ} {\mathrm{mol}}^{-1}$
OR
chloroethane AND contains a polar bond ✔

Accept “chloroethane AND polar”.

curly arrow going from lone pair/negative charge on $\mathrm{O}$ in ⁻OH to $\mathrm{C}$ ✔

curly arrow showing $\mathrm{Cl}$ leaving ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

Accept $O{H}^{\mathit{-}}$ with or without the lone pair.

Do not accept curly arrows originating on $H$ in $O{H}^{-}$.

Accept curly arrows in the transition state.

Do not penalize if $HO$ and $Cl$ are not at 180°.

Do not award M3 if $OH-C$ bond is represented. 

 / ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{OCH}}_2{\mathrm{CH}}_3$ ✔

Accept $\mathit{(}C{H}_{\mathit{3}}C{H}_{\mathit{2}}{\mathit{)}}_{\mathit{2}}O$.

2 «signals» ✔

0.9−1.0 AND triplet ✔

3.3−3.7 AND quartet ✔

Accept any values in the ranges.

Award [1] for two correct chemical shifts or two correct splitting patterns.

$«M\left({\mathrm{CCl}}_2{\mathrm{F}}_2\right) =» 120.91 «\mathrm{g} {\mathrm{mol}}^{-1}»$ ✔

$\frac{2\times 35.45 \mathrm{g} {\mathrm{mol}}^{-1}}{120.91 \mathrm{g} {\mathrm{mol}}^{-1}}\times 100\%=» 58.64 «\%»$ ✔

Award [2] for correct final answer.

Any of:

research «collaboration» for alternative technologies «to replace $\mathrm{CFC}$s»
OR
technologies «developed»/data could be shared
OR
political pressure/Montreal Protocol/governments passing legislations ✔

Do not accept just “collaboration”.

Do not accept any reference to $CFC$ as greenhouse gas or product of fossil fuel combustion.

Accept reference to specific measures, such as agreement on banning use/manufacture of $CFC$s.

${\mathrm{O}}_3+\mathrm{Cl}·\to \mathrm{O}2+\mathrm{ClO}·$ ✔

$\mathrm{ClO}·+\mathrm{O}·\to {\mathrm{O}}_2+\mathrm{Cl}·$
OR
$\mathrm{ClO}·+{\mathrm{O}}_3\to \mathrm{Cl}·+2{\mathrm{O}}_2$ ✔

Penalize missing/incorrect radical dot (∙) once only.

Well answered question with 90% of candidates correctly identifying the complete electron configuration for chlorine.

Most candidates could correctly explain the relative sizes of chlorine atom and chloride ion.

Fairly well answered though some candidates missed M2 for not recognizing the same number of shells affected.

More than 80% could identify that the two peaks in the MS of chlorine are due to different isotopes.

Not well answered. Some candidates were able to identify m/z 74 being due to the m/z of two Cl-37 atoms, however fewer candidates were able to explain the relative abundance of the isotope.

Stoichiometric calculations were generally well done and over 90% could calculate mol from a given mass.

90% of candidates earned full marks on this 2-mark question involving finding a limiting reactant.

Surprisingly, quite a number of candidates struggled with the quantity of excess reactant despite correctly identifying limiting reactant previously.

Most candidates could find the volume of gas produced in a reaction under standard conditions.

More than 90% could identify the oxidation number of manganese in both MnO₂ and MnCl₂.

Most candidates stated that MnO₂ is an oxidizing agent in the reaction but many did not get the mark because there was no reference to oxidation states.

Another well answered 1-mark question where candidates correctly identified a weak acid as an acid which partially dissociates in water. 

Roughly ⅓ of the candidates failed to identify the conjugate base, perhaps distracted by the fact it was not contained in the equation given.

Vast majority of candidates could calculate the concentration of H⁺ (aq) in a HClO (aq) solution with a pH =3.61.

Many identified the reaction of chlorine with ethane as free-radical substitution, or just substitution, with some erroneously stating nucleophilic or electrophilic substitution.

The underlying reasons for the relative reactivity of ethane and chloroethane were not very well known with a few giving erroneous reasons and some stating ethane more reactive.

Few earned full marks for the curly arrow mechanism of the reaction between sodium hydroxide and chloroethane. Mistakes being careless curly arrow drawing, inappropriate –OH notation, curly arrows from the hydrogen or from the carbon to the C–Cl bond, or a method that missed the transition state.

Approximately 60% could draw ethoxyethane however many demonstrated little knowledge of structure of an ether molecule.

A poorly answered question with some getting full marks on this 1HNMR spectrum of ethoxyethane question. Very few could identify all 3 of number of signals, chemical shift, and splitting pattern.

Another good example of candidates being well rehearsed in calculations with 90% earning 2/2 on this question of calculation percentage by mass composition. 

Somewhat disappointing answers on this question about how international cooperation has contributed to the lowering of CFC emissions. Many gave vague answers and some referred to carbon emissions and global warming.

Few could construct the propagation equations showing how CFCs affect ozone, and many lost marks by failing to identify ClO· as a radical.

Describe the nature of ionic bonding.

Describe how the relative atomic mass of a sample of calcium could be determined from its mass spectrum.

When calcium compounds are introduced into a gas flame a red colour is seen; sodium compounds give a yellow flame. Outline the source of the colours and why they are different.

Suggest two reasons why solid calcium has a greater density than solid potassium.

Outline why solid calcium is a good conductor of electricity.

Sketch a graph of the first six ionization energies of calcium.

Calcium carbide reacts with water to form ethyne and calcium hydroxide.

CaC₂(s) + H₂O(l) → C₂H₂(g) + Ca(OH)₂(aq)

Estimate the pH of the resultant solution.

Describe how sigma (σ) and pi ($\pi$) bonds are formed.

Deduce the number of σ and $\pi$ bonds in a molecule of ethyne.

electrostatic attraction AND oppositely charged ions

[1 mark]

multiply relative intensity by «m/z» value of isotope

OR

find the frequency of each isotope

sum of the values of products/multiplication «from each isotope»

OR

find/calculate the weighted average

Award [1 max] for stating “m/z values of isotopes AND relative abundance/intensity” but not stating these need to be multiplied.

[2 marks]

«promoted» electrons fall back to lower energy level

energy difference between levels is different

Accept “Na and Ca have different nuclear charge” for M2.

[2 marks]

Any two of:

stronger metallic bonding

smaller ionic/atomic radius

two electrons per atom are delocalized

OR

greater ionic charge

greater atomic mass

Do not accept just “heavier” or “more massive” without reference to atomic mass.

[2 marks]

delocalized/mobile electrons «free to move»

[1 mark]

general increase

only one discontinuity between “IE2” and “IE3”

[2 marks]

pH > 7

Accept any specific pH value or range of values above 7 and below 14.

[1 mark]

sigma (σ):

overlap «of atomic orbitals» along the axial/internuclear axis

OR

head-on/end-to-end overlap «of atomic orbitals»

pi ($\pi$):

overlap «of p-orbitals» above and below the internuclear axis

OR

sideways overlap «of p-orbitals»

Award marks for suitable diagrams.

[2 marks]

sigma (σ): 3

AND

pi ($\pi$): 2

[1 mark]

Plot the relative values of the first four ionization energies of sodium.

Outline why the alkali metals (group 1) have similar chemical properties.

Describe the structure and bonding in solid sodium oxide.

Calculate values for the following changes using section 8 of the data booklet.

ΔH_atomisation (Na) = 107 kJ mol⁻¹
ΔH_atomisation (O) = 249 kJ mol⁻¹

$\frac{1}{2}$O₂(g) → O^2- (g):

Na (s) → Na⁺ (g):

The standard enthalpy of formation of sodium oxide is −414 kJ mol⁻¹. Determine the lattice enthalpy of sodium oxide, in kJ mol⁻¹, using section 8 of the data booklet and your answers to (d)(i).

(If you did not get answers to (d)(i), use +850 kJ mol⁻¹ and +600 kJ mol⁻¹ respectively, but these are not the correct answers.)

Justify why K₂O has a lower lattice enthalpy (absolute value) than Na₂O.

Write equations for the separate reactions of solid sodium oxide and solid phosphorus(V) oxide with excess water and differentiate between the solutions formed.

Sodium oxide, Na₂O:

Phosphorus(V) oxide, P₄O₁₀:

Differentiation:

Sodium peroxide, Na₂O₂, is formed by the reaction of sodium oxide with oxygen.

2Na₂O (s) + O₂ (g) → 2Na₂O₂ (s)

Calculate the percentage yield of sodium peroxide if 5.00g of sodium oxide produces 5.50g of sodium peroxide.

Determine the enthalpy change, ΔH, in kJ, for this reaction using data from the table and section 12 of the data booklet.

Outline why bond enthalpy values are not valid in calculations such as that in (g)(i).

An allotrope of molecular oxygen is ozone. Compare, giving a reason, the bond enthalpies of the O to O bonds in O₂ and O₃.

Outline why a real gas differs from ideal behaviour at low temperature and high pressure.

The reaction of sodium peroxide with excess water produces hydrogen peroxide and one other sodium compound. Suggest the formula of this compound.

State the oxidation number of carbon in sodium carbonate, Na₂CO₃.

     [✔]

Notes: Accept curve showing general trend.
Award mark only if the energy difference between the first two points is larger than that between points 2/3 and 3/4.

same number of electrons in outer shell

OR

all are s¹ [✔]

«3-D/giant» regularly repeating arrangement «of ions»
OR
lattice «of ions»    [✔]

electrostatic attraction between oppositely charged ions
OR
electrostatic attraction between Na⁺ and O²⁻ ions    [✔]

Note: Do not accept “ionic” without description.

$\frac{1}{2}$O₂(g) → O^2- (g)

«ΔH_atomisation (O) + 1st EA + 2nd EA = 249 k Jmol⁻¹ − 141 kJmol⁻¹ + 753 kJmol⁻¹ =» «+»861 «kJmol⁻¹»    [✔]

Na (s) → Na⁺ (g)

«ΔH_atomisation (Na) + 1st IE = 107 kJmol⁻¹ + 496 kJmol⁻¹ =» «+»603 «kJmol⁻¹»     [✔]

lattice enthalpy = 861 «kJ mol⁻¹» + 2 × 603 «kJ mol⁻¹» −(−414 «kJ mol⁻¹»)     [✔]

«= +» 2481 «kJ mol⁻¹»    [✔]

Note: Award [2] for correct final answer.

If given values are used:
M1: lattice enthalpy = 850 «kJ mol⁻¹» +
2 × 600 «kJ mol⁻¹» −(−414 «kJ mol⁻¹»)
M2: «= +» 2464 «kJ mol⁻¹»

K⁺ ion is larger than Na⁺

OR

smaller attractive force because of greater distance between ion «centres»      [✔]

Sodium oxide:
Na₂O(s) + H₂O(l) → 2NaOH (aq)      [✔]

Phosphorus(V) oxide:
P₄O₁₀ (s) + 6H₂O(l) → 4H₃PO₄ (aq)     [✔]

Differentiation:
NaOH/product of Na₂O is alkaline/basic/pH > 7 AND H₃PO₄/product of P₄O₁₀ is acidic/pH < 7     [✔]

n(Na₂O₂) theoretical yield «= $\frac{5.00\text{g}}{61.98\text{g mo}{\text{l}}^{-1}}$» = 0.0807/8.07 × 10⁻² «mol»

OR

mass of Na₂O₂ theoretical yield «= $\frac{5.00\text{g}}{61.98\text{g mo}{\text{l}}^{-1}}$ × 77.98 gmol⁻¹» = 6.291 «g»    [✔]

% yield «= $\frac{5.50\text{g}}{6.291\text{g}}$ × 100» OR « $\frac{0.0705}{0.0807}$ × 100» = 87.4 «%»     [✔]

Note: Award [2] for correct final answer.

∑ΔH_f products = 2 × (−1130.7) / −2261.4 «kJ»    [✔]

∑ΔH_f reactants = 2 × (−510.9) + 2 × (−393.5) / −1808.8 «kJ»     [✔]

ΔH = «∑ΔH_f products − ∑ΔH_f reactants = −2261.4 −(−1808.8) =» −452.6 «kJ»     [✔]

Note: Award [3] for correct final answer.

Award [2 max] for “+ 452.6 «kJ»”.

only valid for covalent bonds

OR

only valid in gaseous state     [✔]

bond in O₃ has lower enthalpy AND bond order is 1.5 «not 2»    [✔]

Note: Accept “bond in ozone is longer”.

Any one of:

finite volume of particles «requires adjustment to volume of gas»     [✔]

short-range attractive forces «overcomes low kinetic energy»    [✔]

NaOH    [✔]

IV    [✔]

Generally well done with a correct plot of ionization energies.

The majority answered correctly stating same number of valence electrons as the reason. Some candidates stated same size or similar ionization energy but the majority scored well.

Many candidates lost one or two marks for missing “electrostatic forces” between “oppositely charged ions”, or “lattice”. Some candidates’ answers referred to covalent bonds and shapes of molecules.

Good performance with typical error being in the calculation for the first equation, ½O2 (g) → O₂⁻ (g), where the value for the first electron affinity of oxygen was left out.

Many candidates earned some credit for ECF based on (d)(i).

Average performance with answers using atomic size rather than ionic size or making reference to electronegativities of K and Na.

An average of 1.1 out of 3 earned here. Many candidates could write a balanced equation for the reaction of sodium oxide with water but not phosphorus(V) oxide. Mediocre performance in identifying the acid/base nature of the solutions formed.

The majority earned one or two marks in finding a % yield.

The average was 2.2 out 3 for this question on enthalpy of formation. Enthalpy calculations were generally well done.

The majority of candidates referred to “bond enthalpy values are average”, rather than not valid for solids or only used for gases.

Some candidates recognized that ozone had a resonance structure but then only compared bond length between ozone and oxygen rather than bond enthalpy.

Few candidates could distinguish the cause for difference in behaviour between real and ideal gases at low temperature or high pressure. Many answers were based on increase in number of collisions or faster rate or movement of gas particles.

Na₂O was a common formula in many candidates’ answers for the product of the reaction of sodium peroxide with water.

The vast majority of candidates could correctly state the oxidation number of carbon in sodium carbonate.

State the name of Compound B, applying International Union of Pure and Applied Chemistry (IUPAC) rules.

Compound A and Compound B are both liquids at room temperature and pressure. Identify the strongest intermolecular force between molecules of Compound A.

State the number of $\text{σ}$ (sigma) and $\mathrm{\pi }$ (pi) bonds in Compound A.

Deduce the hybridization of the central carbon atom in Compound A.

Identify the isomer of Compound B that exists as optical isomers (enantiomers).

Draw the structural formula of the alkene required.

Explain why the reaction produces more (CH₃)₃COH than (CH₃)₂CHCH₂OH.

Deduce the structural formula of the repeating unit of the polymer formed from this alkene.

Deduce what would be observed when Compound B is warmed with acidified aqueous potassium dichromate (VI).

Identify the type of reaction.

Outline the requirements for a collision between reactants to yield products.

Explain the mechanism of the reaction using curly arrows to represent the movement of electron pairs.

The polarity of the carbon–halogen bond, C–X, facilitates attack by HO^–.

Outline, giving a reason, how the bond polarity changes going down group 17.

2-methylpropan-2-ol /2-methyl-2-propanol ✔

Accept methylpropan-2-ol/ methyl-2-propanol.

Do not accept 2-methylpropanol.

dipole-dipole ✔

Do not accept van der Waals’ forces.

$\text{σ}$: 9
AND
$\mathrm{\pi }$: 1 ✔

sp² ✔

butan-2-ol/CH₃CH(OH)C₂H₅ ✔

carbocation formed from (CH₃)₃COH is more stable / (CH₃)₃C⁺ is more stable than (CH₃)₂CHCH₂⁺ ✔

«because carbocation has» greater number of alkyl groups/lower charge on the atom/higher e^- density
OR
«greater number of alkyl groups» are more electron releasing
OR
«greater number of alkyl groups creates» greater inductive/+I effect ✔

Do not award any marks for simply quoting Markovnikov’s rule.

Do not penalize missing brackets or n.

Do not award mark if continuation bonds are not shown.

no change «in colour/appearance/solution» ✔

«nucleophilic» substitution
OR
SN2 ✔

Accept “hydrolysis”.

Accept SN1

energy/E ≥ activation energy/E_a ✔

correct orientation «of reacting particles»
OR
correct geometry «of reacting particles» ✔

curly arrow going from lone pair/negative charge on O in ^-OH to C ✔

curly arrow showing I leaving ✔

representation of transition state showing negative charge, square brackets and partial bonds ✔

Accept OH^- with or without the lone pair.

Do not allow curly arrows originating on H, rather than the -, in OH^-.

Accept curly arrows in the transition state.

Do not penalize if HO and I are not at 180°.

Do not award M3 if OH–C bond is represented.

Award [2 max] if S_N1 mechanism shown.

decreases/less polar AND electronegativity «of the halogen» decreases ✔

Accept “decreases” AND a correct comparison of the electronegativity of two halogens.

Accept “decreases” AND “attraction for valence electrons decreases”.

Naming the organic compound using IUPAC rules was generally done well.

Mediocre performance in stating the number of σ (sigma) and π (pi) bonds in propanone; the common answer was 3 σ and 1 π instead of 9 σ and 1 π, suggesting the three C-H σ bonds in each of the two methyl groups were ignored.

sp² hybridization of the central carbon atom in the ketone was very done well.

Mediocre performance; some identified 2-methylpropan-1-ol or -2-ol, instead butan-2-ol/CH₃CH(OH)C₂H₅ as the isomer that exists as an optical isomer.

Good performance; some had a H and CH₃ group on each C atom across double bond instead of having two H atoms on one C and two CH₃ groups on the other.

Poor performance, particularly in light of past feedback provided in similar questions since there was repeated reference simply to Markovnikov's rule, without any explanation.

Mediocre performance; deducing structural formula of repeating unit of the polymer was challenging in which continuation bonds were sometimes missing, or structure included a double bond or one of the CH₃ group was missing.

Mediocre performance; deducing whether the tertiary alcohol could be oxidized solicited mixed responses ranging from the correct one, namely no change (in colour, appearance or solution), to tertiary alcohol will be reduced, or oxidized, or colour will change will occur, and such.

Excellent performance on the type of reaction but with some incorrect answers such as alkane substitution, free radical substitution or electrophilic substitution.

Good performance. For the requirements for a collision between reactants to yield products, some suggested necessary, sufficient or enough energy or even enough activation energy instead of energy/E ≥ activation energy/E_a.

Mechanism for SN2 not done well. Often the negative charge on OH was missing, the curly arrow was not going from lone pair/negative charge on O in -OH to C, or the curly arrow showing I leaving placed incorrectly and specially the negative charge was missing in the transition state. Formation of a carbocation intermediate indicating SN1 mechanism could score a maximum of 2 marks.

Good performance on how the polarity of C-X bond changes going down group 17.

The stable isotope of rhenium contains 110 neutrons.

State the nuclear symbol notation ${}_{\text{Z}}^{\text{A}}\text{X}$ for this isotope.

Suggest the basis of these predictions.

A scientist wants to investigate the catalytic properties of a thin layer of rhenium metal on a graphite surface.

Describe an electrochemical process to produce a layer of rhenium on graphite.

Predict two other chemical properties you would expect rhenium to have, given its position in the periodic table.

Describe how the relative reactivity of rhenium, compared to silver, zinc, and copper, can be established using pieces of rhenium and solutions of these metal sulfates.

State the name of this compound, applying IUPAC rules.

Calculate the percentage, by mass, of rhenium in ReCl₃.

Suggest why the existence of salts containing an ion with this formula could be predicted. Refer to section 6 of the data booklet.

Deduce the coefficients required to complete the half-equation.

ReO₄⁻ (aq) + ____H⁺ (aq) + ____e⁻ ⇌ [Re(OH)₂]²⁺ (aq) + ____H₂O (l)        E^θ = +0.36 V

Predict, giving a reason, whether the reduction of ReO₄⁻ to [Re(OH)₂]²⁺ would oxidize Fe²⁺ to Fe³⁺ in aqueous solution. Use section 24 of the data booklet.

${}_{\text{75}}^{\text{185}}\text{Re}$    [✔]

gap in the periodic table
OR
element with atomic number «75» unknown
OR
break/irregularity in periodic trends     [✔]

«periodic table shows» regular/periodic trends «in properties»      [✔]

electrolyze «a solution of /molten» rhenium salt/Reⁿ⁺     [✔]

graphite as cathode/negative electrode
OR
rhenium forms at cathode/negative electrode     [✔]

Note: Accept “using rhenium anode” for M1.

Any two of:
variable oxidation states     [✔]

forms complex ions/compounds     [✔]

coloured compounds/ions     [✔]

«para»magnetic compounds/ions     [✔]

Note: Accept other valid responses related to its chemical metallic properties.

Do not accept “catalytic properties”.

place «pieces of» Re into each solution    [✔]

if Re reacts/is coated with metal, that metal is less reactive «than Re»    [✔]

Note: Accept other valid observations such as “colour of solution fades” or “solid/metal appears” for “reacts”.

rhenium(III) chloride
OR
rhenium trichloride    [✔]

«M_r ReCl₃ = 186.21 + (3 × 35.45) =» 292.56    [✔]
«100 × $\frac{186.21}{292.56}$ =» 63.648 «%»   [✔]

same group as Mn «which forms MnO₄^-»
OR
in group 7/has 7 valence electrons, so its «highest» oxidation state is +7    [✔]

ReO₄⁻ (aq) + 6H⁺ (aq) + 3e⁻ ⇌ [Re(OH)₂]²⁺ (aq) + 2H₂O (l)    [✔]

no AND ReO₄⁻ is a weaker oxidizing agent than Fe³⁺
OR
no AND Fe³⁺ is a stronger oxidizing agent than ReO₄⁻
OR
no AND Fe²⁺ is a weaker reducing agent than [Re(OH)₂]²⁺
OR
no AND [Re(OH)₂]²⁺ is a stronger reducing agent than Fe²⁺
OR
no AND cell emf would be negative/–0.41 V     [✔]

It was expected that this question would be answered correctly by all HL candidates. However, many confused the A-Z positions or calculated very unusual numbers for A, sometimes even with decimals.

This is a NOS question which required some reflection on the full meaning of the periodic table and the wealth of information contained in it. But very few candidates understood that they were being asked to explain periodicity and the concept behind the periodic table, which they actually apply all the time. Some were able to explain the “gap” idea and other based predictions on properties of nearby elements instead of thinking of periodic trends. A fair number of students listed properties of transition metals in general.

Generally well done; most described the cell identifying the two electrodes correctly and a few did mention the need for Re salt/ion electrolyte.

Generally well answered though some students suggested physical properties rather than chemical ones.

Many candidates chose to set up voltaic cells and in other cases failed to explain the actual experimental set up of Re being placed in solutions of other metal salts or the reaction they could expect to see.

Almost all candidates were able to name the compound according to IUPAC.

Most candidates were able to answer this stoichiometric question correctly.

This should have been a relatively easy question but many candidates sometimes failed to see the connection with Mn or the amount of electrons in its outer shell.

Surprisingly, a great number of students were unable to balance this simple half-equation that was given to them to avoid difficulties in recall of reactants/products.

Many students understood that the oxidation of Fe²⁺ was not viable but were unable to explain why in terms of oxidizing and reducing power; many students simply gave numerical values for E^Θ often failing to realise that the oxidation of Fe²⁺ would have the inverse sign to the reduction reaction.

Outline why metals, like iron, can conduct electricity.

Justify why sulfur is classified as a non-metal by giving two of its chemical properties.

Sketch the first eight successive ionisation energies of sulfur.

Describe the bonding in this type of solid.

State a technique that could be used to determine the crystal structure of the solid compound.

State the full electron configuration of the sulfide ion.

Outline, in terms of their electronic structures, why the ionic radius of the sulfide ion is greater than that of the oxide ion.

Suggest why chemists find it convenient to classify bonding into ionic, covalent and metallic.

Write the equation for this reaction.

Deduce the change in the oxidation state of sulfur.

Suggest why this process might raise environmental concerns.

Explain why the addition of small amounts of carbon to iron makes the metal harder.

mobile/delocalized «sea of» electrons

Any two of:

forms acidic oxides «rather than basic oxides» ✔

forms covalent/bonds compounds «with other non-metals» ✔

forms anions «rather than cations» ✔

behaves as an oxidizing agent «rather than a reducing agent» ✔

Award [1 max] for 2 correct non-chemical properties such as non-conductor, high ionisation energy, high electronegativity, low electron affinity if no marks for chemical properties are awarded.

two regions of small increases AND a large increase between them✔

large increase from 6th to 7th ✔

Accept line/curve showing these trends.

electrostatic attraction ✔

between oppositely charged ions/between Fe²⁺ and S²⁻ ✔

X-ray crystallography ✔

1s² 2s² 2p⁶ 3s² 3p⁶ ✔

Do not accept “[Ne] 3s² 3p⁶”.

«valence» electrons further from nucleus/extra electron shell/ electrons in third/3s/3p level «not second/2s/2p»✔

Accept 2,8 (for O^2–) and 2,8,8 (for S^2–)

allows them to explain the properties of different compounds/substances
OR
enables them to generalise about substances
OR
enables them to make predictions ✔

Accept other valid answers.

4FeS(s) + 7O₂(g) → 2Fe₂O₃(s) + 4SO₂(g) ✔

Accept any correct ratio.

+6
OR
−2 to +4 ✔

Accept “6/VI”.
Accept “−II, 4//IV”.
Do not accept 2- to 4+.

sulfur dioxide/SO₂ causes acid rain ✔

Accept sulfur dioxide/SO₂/dust causes respiratory problems
Do not accept just “causes respiratory problems” or “causes acid rain”.

disrupts the regular arrangement «of iron atoms/ions»
OR
carbon different size «to iron atoms/ions» ✔

prevents layers/atoms sliding over each other ✔

Explain the decrease in atomic radius from Na to Cl.

Explain why the radius of the sodium ion, Na⁺, is smaller than the radius of the oxide ion, O²⁻.

Sketch a graph to show the relative values of the successive ionization energies of boron.

Predict, giving your reasons, whether Mn²⁺ or Fe²⁺ is likely to have a more exothermic enthalpy of hydration.

nuclear charge/number of protons/Z_eff increases «causing a stronger pull on the outer electrons» ✔

same number of shells/«outer» energy level/shielding ✔

Accept “atomic number” for “number of protons”.

isoelectronic/same electronic configuration/«both» have 2.8 ✔

more protons in Na⁺ ✔

Sketch showing:

largest increase between third and fourth ionization energies ✔

IE₁ < IE₂ < IE₃ < IE₄ < IE₅ ✔

Fe²⁺ AND smaller size/radius

OR

Fe²⁺ AND higher charge density ✔

stronger interaction with «polar» water molecules ✔

M1 not needed for M2.