Draw the Lewis structures of oxygen, O₂, and ozone, O₃.

Outline why both bonds in the ozone molecule are the same length and predict the bond length in the ozone molecule. Refer to section 10 of the data booklet.

Reason: 

Length:

Distinguish ultraviolet light from visible light in terms of wavelength and energy.

Discuss how the different bond strengths between the oxygen atoms in O₂ and O₃ in the ozone layer affect radiation reaching the Earth’s surface.

NOTES: Coordinate bond may be represented by an arrow.

Do not accept delocalized structure for ozone.

resonance «structures»
OR
delocalization of «the double/pi bond» electrons ✔
121 «pm» < length < 148 «pm» ✔

NOTE: Accept any length between these two values.

«UV» shorter wavelength AND higher energy «than visible» ✔

«bond» in O₂ stronger than in O₃ ✔

ozone absorbs lower frequency/energy «radiation than oxygen»
OR
ozone absorbs longer wavelength «radiation than oxygen» ✔

NOTE: Accept ozone «layer» absorbs a range of frequencies.

State the electron configuration of a bromine atom.

Sketch the orbital diagram of the valence shell of a bromine atom (ground state) on the energy axis provided. Use boxes to represent orbitals and arrows to represent electrons.

Draw the Lewis (electron dot) structure for BrO₃⁻ that obeys the octet rule.

Predict, using the VSEPR theory, the geometry of the BrO₃⁻ ion and the O−Br−O bond angles.

Bromate(V) ions act as oxidizing agents in acidic conditions to form bromide ions.

Deduce the half-equation for this reduction reaction.

Bromate(V) ions oxidize iron(II) ions, Fe²⁺, to iron(III) ions, Fe³⁺.

Deduce the equation for this redox reaction.

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵

OR

[Ar] 4s² 3d¹⁰ 4p⁵ ✔

Accept 3d before 4s.

Accept double-headed arrows.

Accept dots, crosses or lines to represent electron pairs.

Geometry:
trigonal/pyramidal ✔

Reason:
three bonds AND one lone pair
OR
four electron domains ✔

O−Br−O angle:
107° ✔

Accept “charge centres” for “electron domains”.

Accept answers in the range 104–109°.

BrO₃⁻ (aq) + 6e⁻ + 6H⁺ (aq) → Br⁻ (aq) + 3H₂O (l)

correct reactants and products ✔

balanced equation ✔

Accept reversible arrows.

BrO₃⁻ (aq) + 6Fe²⁺ (aq) + 6H⁺ (aq) → Br⁻ (aq) + 3H₂O (l) + 6Fe³⁺ (aq) ✔

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 ${}_{22}^{48}\text{Ti}$ atom.

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

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

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

A chloride of titanium, TiCl₄, 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]

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

= 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]

1s²2s²2p⁶3s²3p⁶3d²

[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]

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]

TiCl₄(l) + 2H₂O(l) → TiO₂(s) + 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]

Suggest what can be concluded about the gold atom from this experiment.

Subsequent experiments showed electrons existing in energy levels occupying various orbital shapes.

Sketch diagrams of 1s, 2s and 2p.

State the electron configuration of copper.

Most ⁴He²⁺ passing straight through:

most of the atom is empty space
OR
the space between nuclei is much larger than ⁴He²⁺ particles
OR
nucleus/centre is «very» small «compared to the size of the atom» ✔

Very few ⁴He²⁺ deviating largely from their path:

nucleus/centre is positive «and repels ⁴He²⁺ particles»
OR
nucleus/centre is «more» dense/heavy «than ⁴He²⁺ particles and deflects them»
OR
nucleus/centre is «very» small «compared to the size of the atom» ✔

Do not accept the same reason for both M1 and M2.

Accept “most of the atom is an electron cloud” for M1.

Do not accept only “nucleus repels ⁴He²⁺ particles” for M2.

1s AND 2s as spheres ✔

one or more 2p orbital(s) as figure(s) of 8 shape(s) of any orientation (p_x, p_y p_z) ✔

1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰

OR

[Ar] 4s¹3d¹⁰ ✔

Accept configuration with 3d before 4s.

Explain why the first ionization energy of calcium is greater than that of potassium.

All models have limitations. Suggest two limitations to this model of the electron energy levels.

Draw an arrow, labelled X, to represent the electron transition for the ionization of a hydrogen atom in the ground state.

Draw an arrow, labelled Z, to represent the lowest energy electron transition in the visible spectrum.

increasing number of protons/nuclear charge/Z_eff ✔

«atomic» radius/size decreases
OR
same number of energy levels
OR
similar shielding «by inner electrons» ✔

Any two of:

does not represent sub-levels/orbitals ✔

only applies to atoms with one electron/hydrogen ✔

does not explain why only certain energy levels are allowed ✔

the atom is considered to be isolated ✔

does not take into account the interactions between atoms/molecules/external fields ✔

does not consider the number of electrons the energy level can fit ✔

does not consider probability of finding electron at different positions/OWTTE ✔

Do not accept “does not represent distance «from nucleus»”.

upward arrow X AND starting at n = 1 extending to n = ∞ ✔

downward or upward arrow between n = 3 and n = 2 ✔

It was surprising that this question that appears regularly in IB chemistry papers was not better answered. Many candidates only obtained one of the two marks for identifying one factor (often the larger nuclear charge of calcium or that the number of shells was the same for Ca and K). However, a few candidates did write thorough answers reflecting a good understanding of the factors affecting ionization energy. This question had a strong correlation between candidates who scored well and those who had a high score overall. Some candidates did not score any marks by focusing on trends in the Periodic Table without offering an explanation, or by discussing the number of electrons in Ca and K instead of the number of protons.

Only 30% of the candidates drew the correct arrow on the diagram representing the ionization of hydrogen. A few candidates missed the mark by having the arrow pointing downwards. The most common incorrect answer was a transition between n=1 and n=2.

Outline why metals, like iron, can conduct electricity.

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

Describe the bonding in this type of solid.

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.

electrostatic attraction ✔

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

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 ✔

Describe the nature of ionic bonding.

State the electron configuration of the Ca²⁺ ion.

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.

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.

electrostatic attraction AND oppositely charged ions

[1 mark]

1s²2s²2p⁶3s²3p⁶

OR

[Ar]

[1 mark]

«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]

pH > 7

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

[1 mark]

Deduce the ratio of Fe²⁺:Fe³⁺ in Fe₃O₄.

State the type of spectroscopy that could be used to determine their relative abundances.

State the number of protons, neutrons and electrons in each species.

Iron has a relatively small specific heat capacity; the temperature of a 50 g sample rises by 44.4°C when it absorbs 1 kJ of heat energy.

Determine the specific heat capacity of iron, in J g⁻¹K⁻¹. Use section 1 of the data booklet.

Write the half-equation for the reduction of hydrogen peroxide to water in acidic solution.

Deduce a balanced equation for the oxidation of Fe2+ by acidified hydrogen peroxide.

1:2 ✔

Accept 2 Fe³⁺: 1 Fe²⁺
Do not accept 2:1 only

mass «spectroscopy»/MS ✔

Award [1 max] for 4 correct values.

specific heat capacity « = $\frac{q}{m\times ∆T}/\frac{1000 \mathrm{J}}{50 \mathrm{g}\times 44 \mathrm{K}}$» = 0.45 «J g⁻¹K⁻¹» ✔

H₂O₂(aq) + 2H⁺(aq) + 2e⁻→ 2H₂O(l) ✔

H₂O₂(aq) + 2H⁺(aq) + 2Fe²⁺(aq) → 2H₂O(l) + 2Fe³⁺(aq) ✔

State the nuclear symbol notation, ${}_{\text{Z}}^{\text{A}}\text{X}$, for iron-54.

Mass spectrometry analysis of a sample of iron gave the following results:

Calculate the relative atomic mass, A_r, of this sample of iron to two decimal places.

An iron nail and a copper nail are inserted into a lemon.

Explain why a potential is detected when the nails are connected through a voltmeter.

${}_{\text{26}}^{\text{54}}\text{Fe}$  [✔]

«A_r =» 54 × 0.0584 + 56 × 0.9168 + 57 × 0.0217 + 58 × 0.0031
OR
«A_r =» 55.9111  [✔]

«A_r =» 55.91 [✔]

Notes:

Award [2] for correct final answer.
Do not accept data booklet value (55.85).

lemon juice is the electrolyte
OR
lemon juice allows flow of ions
OR
each nail/metal forms a half-cell with the lemon juice  [✔]

Note: Accept “lemon juice acts as a salt bridge”.

Any one of:
iron is higher than copper in the activity series
OR
each half-cell/metal has a different redox/electrode potential  [✔]

Note: Accept “iron is more reactive than copper”.

iron is oxidized
OR
Fe → Fe²⁺ + 2e^–
OR
Fe → Fe³⁺ + 3e⁻
OR
iron is anode/negative electrode of cell  [✔]

copper is cathode/positive electrode of cell
OR
reduction occurs at the cathode
OR
2H⁺ + 2e⁻ → H₂  [✔]

electrons flow from iron to copper  [✔]

Notes:
Accept “lemon juice acts as a salt bridge”.
Accept “iron is more reactive than copper”.

The nuclear symbol notation was generally correct. However, some students swapped atomic and mass numbers and hence lost the mark.

Calculation of RAM was generally correctly calculated, but some candidates did not give their answer to two decimal places while they should use the provided periodic table.

Very few students gained the 2 marks available for explaining the potential generated in the lemon as they didn’t realise it was the lemon that acted as the electrolyte and allowed ions to flow. Some were able to gain a mark for explaining that electrons moved from iron to copper as iron is more reactive.

Explain why Si has a smaller atomic radius than Al.

Explain the decrease in radius from Na to Na⁺.

State the condensed electron configurations for Cr and Cr3⁺.

Describe metallic bonding and how it contributes to electrical conductivity.

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

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

Consider the following equilibrium reaction:

2SO₂(g) + O₂(g) $\rightleftharpoons$ 2SO₃(g)

State and explain how the equilibrium would be affected by increasing the volume of the reaction container at a constant temperature.

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 ✓

Na⁺ has one less energy level/shell
OR
Na⁺ has 2 energy levels/shells AND Na has 3 ✓

less shielding «in Na⁺ so valence electrons attracted more strongly to nucleus»
OR
effective nuclear charge/Z_eff greater «in Na⁺ so valence electrons attracted more strongly to nucleus» ✓

Accept “more protons than electrons «in Na⁺»” OR “less electron-electron repulsion «in Na⁺»” 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.

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.

pressure decrease «due to larger volume» ✓

reactant side has more moles/molecules «of gas» ✓

reaction shifts left/towards reactants ✓

Award M3 only if M1 OR M2 is awarded.

State the nuclear symbol notation, ${}_Z^{A}X$, for magnesium-26.

Mass spectroscopic analysis of a sample of magnesium gave the following results:

Calculate the relative atomic mass, A_r, of this sample of magnesium to two decimal places.

Magnesium burns in air to form a white compound, magnesium oxide. Formulate an equation for the reaction of magnesium oxide with water.

Describe the trend in acid-base properties of the oxides of period 3, sodium to chlorine.

In addition to magnesium oxide, magnesium forms another compound when burned in air. Suggest the formula of this compound

Describe the structure and bonding in solid magnesium oxide.

Magnesium chloride can be electrolysed.

Deduce the half-equations for the reactions at each electrode when molten magnesium chloride is electrolysed, showing the state symbols of the products. The melting points of magnesium and magnesium chloride are 922 K and 987 K respectively.

Anode (positive electrode):

Cathode (negative electrode):

${}_{12}^{26}\mathrm{M}\mathrm{g}$

«Ar =»$\frac{24\times 78.60+25\times 10.11+26\times 11.29}{100}$

«= 24.3269 =» 24.33

Award [2] for correct final answer.
Do not accept data booklet value (24.31).

MgO(s) + H₂O(l) → Mg(OH)₂(s)

OR

MgO(s) + H₂O(l) → Mg^₂+(aq) + 2OH^–(aq)

Accept $\rightleftharpoons$.

from basic to acidic

through amphoteric

Accept “alkali/alkaline” for “basic”.
Accept “oxides of Na and Mg: basic AND oxide of Al: amphoteric” for M1.
Accept “oxides of non-metals/Si to Cl acidic” for M2.
Do not accept just “become more acidic”

Mg₃N₂

Accept MgO₂, Mg(OH)₂, Mg(NOx)₂, MgCO₃.

«3-D/giant» regularly repeating arrangement «of ions»
OR
lattice «of ions»
Accept “giant” for M1, unless “giant covalent” stated.

electrostatic attraction between oppositely charged ions
OR
electrostatic attraction between Mg²⁺ and O^2– ions
Do not accept “ionic” without description.

Anode (positive electrode):
2Cl^– → Cl₂(g) + 2e^–

Cathode (negative electrode):
Mg²⁺ + 2e^– → Mg(l)

Penalize missing/incorrect state symbols at Cl₂ and Mg once only.
Award [1 max] if equations are at wrong electrodes.
Accept Mg (g).

Write a balanced equation for the reaction that occurs.

State the block of the periodic table in which magnesium is located.

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.

__ Mg₃N₂(s) + __ H₂O (l) → __ Mg(OH)₂(s) + __ NH₃(aq)

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 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

s ✔

Do not allow group 2

aluminium/Al ✔

$⟨⟨\frac{53.726 \mathrm{g}-47.372 \mathrm{g}}{244.31 \mathrm{g} {\mathrm{mol}}^{-1}}=\frac{6.354 \mathrm{g}}{24.31 \mathrm{g} {\mathrm{mol}}^{-1}}⟩⟩=0.2614$ «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 91-92%

[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)

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 ✔

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.

This was not as well done as one might have expected with the most common errors being O instead of O₂ oxygen and MgO rather than MgO₂.

Many students did not know what "block" meant, and often guessed group 2 etc.

Many students confused "period" and "group" and also many did not read metal, so aluminium was not chosen by the majority.

A number of students were not able to interpret the results and hence find the gain in mass and calculate the moles correctly.

Only a handful could work out the correct answer. Most had no real idea and quite a lot of blank responses. There also seems to be significant confusion between "percent uncertainty" and "percent error".

This was not well answered, but definitely better than the previous question with quite a few gaining some credit for correctly determining the theoretical yield.

This proved to be a very difficult question to answer in the quantitative manner required, with hardly any correct responses.

Quite a few students realised that incomplete reaction would lead to this, but only 30% of students gave a correct answer rather than a non-specific guess, such as "misread balance" or "impurities".

This was generally very well done with almost all candidates being able to determine the correct coefficients.

About 40% of students managed to correctly determine both the oxidation states, as -3, with errors being about equally divided between the two compounds.

Probably only about 10% could explain why this was an acid-base reaction. Rather more made valid deductions about redox, based on their answer to the previous question.

Most candidates could answer the question about subatomic particles correctly.

Identification of isotopes was answered correctly by most students.

In spite of being given the meaning of "isoelectronic", many candidates talked about the differing number of electrons and only about 30% could correctly analyse the situation in terms of nuclear charge.

The question was marked quite leniently so that the majority of candidates gained at least one of the marks by mentioning a subatomic particle. A significant number read "indivisible" as "invisible" however.

About a quarter of the students gained full marks and probably a similar number gained no marks. Metallic bonding was the type that seemed least easily recognised and least easily described. Another common error was to explain ionic bonding in terms of attraction of ions rather than describing electron transfer.

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.

Write the equation for the reaction of chloroethane with a dilute aqueous solution of sodium hydroxide.

Deduce the nucleophile for the reaction in d(iii).

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

Deduce the number of signals and their chemical shifts in the ${}_1\mathrm{H} \mathrm{NMR}$ 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.

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

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

Accept “chloroethane AND polar”.

${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\left(\mathrm{l}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\to {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$
OR
${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Cl}\left(\mathrm{l}\right)+\mathrm{NaOH}\left(\mathrm{aq}\right)\to {\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}\left(\mathrm{aq}\right)+\mathrm{NaCl}\left(\mathrm{aq}\right)$ ✔

Accept use of $C_{\mathit{2}}{H}_{\mathit{5}}Cl$ and $C_{\mathit{2}}{H}_{\mathit{5}}OH\mathrm{/}{C}_{\mathit{2}}{H}_{\mathit{6}}O$ in the equation.

hydroxide «ion»/${\mathrm{OH}}^{-}$ ✔

Do not accept $NaOH$.

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

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

$2$ «signals» ✔

$0.9–1.0 «\mathrm{ppm}»$ AND $3.3–3.7 «\mathrm{ppm}»$✔

Accept any values in the ranges.

Award [1 max] for two incorrect chemical shifts.

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

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

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.

Most candidates wrote the electron configuration of chlorine correctly.

Only half of the candidates deduced that the chloride ion has a larger radius than the chlorine atom with a valid reason. Many candidates struggled with this question and decided that the extra electron in the chloride ion caused a greater attraction between the nucleus and the outer electrons.

Only about a third of the candidates identified the extra proton in the chlorine nucleus as the cause of the smaller atomic radius when compared to the sulfur atom, and only the stronger candidates also compared the shielding or the number of shells in the two atoms. Many candidates had a poor understanding of factors affecting atomic radius and could not explain the difference.

About 60% of the candidates recognized that the peaks at m/z 35 and 37 in the mass spectrum of chlorine are due to its isotopes. A few students wrote 'isomers' instead of 'isotopes'.

This was the lowest scoring question on the paper, that was also left blank by 10% of the candidates. About 20% of the candidates identified the peak at m/z = 74 to be due to a molecule made up of two 37Cl atoms. And only very few candidates commented that the low abundance of the peak was due to the low abundance of the 37Cl isotope. A common incorrect answer was that chlorine has an isotope of mass number 74.

Most candidates were able to determine the number of moles of MnO₂ using the mass.

It was pleasing that the majority of the candidates were able to determine the limiting reactant by using the stoichiometric ratio.

Half of the candidates were able to determine the amount of excess reactant. Some candidates who determined the limiting reactant in the previous part correctly, forgot to use the stoichiometric ratio in this part, and ended up with incorrect answers.

60% of the candidates determined the volume of chlorine produced correctly. Some candidates made mistakes in the units when using PV = nRT and had a power of 10 error.

The majority of candidates were able to determine the oxidation states of Mn in the two compounds correctly.

Less than half of the candidates were awarded the mark. Some did identify MnO2 as the oxidizing agent but did not give the explanation in terms of oxidation state as required in the question. Other candidates did not have an understanding of oxidizing and reducing agents. 

A very well answered question - 80% of candidates understood what is meant by the term weak acid. Incorrect answers included 'acids that have high pH'.

Half of the candidates deduced the formula of the conjugate base of hypochlorous acid. Incorrect answers included H₂O and HCl.

A well answered question. It was pleasing to see that 70% of the candidates were able to calculate [H⁺] from the given pH.

More than half of the candidates identified the type of reaction between ethane and chlorine as a substitution reaction. A few candidates lost the marks for writing 'electrophilic substitution' or 'nucleophilic substitutions'.

This was a challenging question that was answered correctly by only 30% of the candidates. A variety of incorrect answers were seen such as 'chlorine is a halogen and hence it is reactive', and 'ethane is more reactive because it is an alkane'. For students who answered correctly, the polarity was the most frequently given reason.

Half of the candidates wrote the correct equation for the hydrolysis of chloroethane. Incorrect answers often included carbon dioxide and water as the products.

This was a highly discriminating question. Only 30% of the candidates were able to identify the hydroxide ion as the nucleophile in the hydrolysis of chloroethane. Incorrect answers included NaOH where the ion was not specified. 14% of the candidates left this question blank.

Half of the candidates were able to give the structural formula of ethoxyethane. Incorrect answers included methoxymethane, ketones and esters.

Nearly half of the candidates were able to identify the number of signals obtained in the 1H NMR spectrum of ethoxyethane, obtaining the first mark of this question. Many candidates were awarded the mark as 'error carried forward' from an incorrect structure of ethoxyethane. The second mark for this question required candidates to look up values of chemical shift from the data booklet. Nearly a third of the candidates were able to match the chemical environments of the hydrogen atoms in ethoxyethane to those listed in the data booklet successfully. 

This was the highest scoring question in the paper. The majority of candidates were able to calculate the percentage by mass of chlorine in CCl₂F₂. Mistakes included incorrect rounding and arithmetic errors.

This nature of science question was well answered by half of the candidates. Some teachers commented that the wording was rather vague. Incorrect answers were mainly assuming that CFCs were related to the combustion of fuels and greenhouse gas emissions.

Draw arrows in the boxes to represent the electron configuration of a nitrogen atom.

Draw the Lewis (electron dot) structure of the ammonia molecule.

Deduce the expression for the equilibrium constant, K_c, for this equation.

Explain why an increase in pressure shifts the position of equilibrium towards the products and how this affects the value of the equilibrium constant, K_c.

State how the use of a catalyst affects the position of the equilibrium.

Determine the enthalpy change, ΔH, for the Haber–Bosch process, in kJ. Use Section 11 of the data booklet.

Calculate the enthalpy change, ΔH^⦵, for the Haber–Bosch process, in kJ, using the following data.

$∆H_{ \mathrm{f}}^{⦵}\left({\mathrm{NH}}_3\right)=-46.2 \mathrm{kJ} {\mathrm{mol}}^{-1}$.

Suggest why the values obtained in (d)(i) and (d)(ii) differ.

State the relationship between NH₄⁺ and NH₃ in terms of the Brønsted–Lowry theory.

Determine the concentration, in mol dm^–3, of the solution formed when 900.0 dm³ of NH₃(g) at 300.0 K and 100.0 kPa, is dissolved in water to form 2.00 dm³ of solution. Use sections 1 and 2 of the data booklet.

Calculate the concentration of hydroxide ions in an ammonia solution with pH = 9.3. Use sections 1 and 2 of the data booklet.

Accept all 2p electrons pointing downwards.

Accept half arrows instead of full arrows.

Accept lines or dots or crosses for electrons, or a mixture of these

$K_c=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$ ✔

shifts to the side with fewer moles «of gas»
OR
shifts to right as there is a reduction in volume✔

«value of » K_c unchanged ✔

Accept “K_c only affected by changes in temperature”.

same/unaffected/unchanged ✔

bonds broken: N≡N + 3(H–H) / «1 mol×»945 «kJ mol^–1» + 3«mol»×436 «kJ mol^–1» / 945 «kJ» + 1308 «kJ» / 2253 «kJ» ✔

bonds formed: 6(N–H) / 6«mol»×391 «kJ mol^–1» / 2346 «kJ» ✔

ΔH = «2253 kJ – 2346 kJ = » –93 «kJ» ✔

Award [2 max] for (+)93 «kJ»

–92.4 «kJ» ✔

«N-H» bond enthalpy is an average «and may not be the precise value in NH₃» ✔

Accept it relies on average values not specific to NH₃

conjugate «acid and base» ✔

amount of ammonia $⟨⟨=\frac{P.V}{R.T}=\frac{100.0 kPa\times 900.0 d{m}^3}{8.31 J K^{-1}mo{l}^{-1}\times 300.0 K}⟩⟩ = 36.1 «\mathrm{mol}»$ ✔

concentration $⟨⟨=\frac{n}{V}=\frac{36.1}{2.00}⟩⟩=18.1 «\mathrm{mol} {\mathrm{dm}}^{-3}»$ ✔

Award [2] for correct final answer.

[OH⁻] $⟨⟨=\frac{{\mathrm{K}}_{\mathrm{W}}}{\left[{\mathrm{H}}^{+}\right]}=\frac{{10}^{-14}}{{10}^{-9.3}}={10}^{-4.7}⟩⟩=2.0 \times {10}^{-5} ⟨⟨\mathrm{mol} {\mathrm{dm}}^{-3}⟩⟩$ ✔

Most students realised that the three p-orbitals were all singly filled.

Even more candidates could draw the correct Lewis structure of ammonia, with omission of the lone pair being the most common error.

Most students could deduce the equilibrium constant expression from the equilibrium equation.

Many students realised that increasing pressure shifts an equilibrium to the side with the most moles of gas (though the "of gas" was frequently omitted!) but probably less than half realised that, even though the equilibrium position changes, the value of the equilibrium constant remains constant.

It was pleasing to see that about a third of students gaining full marks and an equal number only lost a single mark because they failed to locate the correct bond enthalpy for molecular nitrogen.

Very few students could determine the enthalpy change from enthalpy of formation data, with many being baffled by the absence of values for the elemental reactants and more than half who overcame this obstacle failed to note that 2 moles of ammonia are produced.

About half the candidates recognised the species as a conjugate acid-base pair, though some lost the mark by confusing the acid and base, even though this information was not asked for.

About 40% of candidates gained full marks for the calculation and a significant number of others gained the second mark to calculate the concentration as an ECF.

This question was very poorly answered with many candidates calculating the [H⁺] instead of [OH^-].

Outline why ozone in the stratosphere is important.

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.

Suggest why it is surprising that dinitrogen monoxide dissolves in water to give a neutral solution.

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

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.

oxides of nitrogen/non-metals are «usually» acidic  [✔]

60 % of the candidates were aware that ozone in the atmosphere absorbs UV light. Some candidates did not gain the mark for not specifying the type of radiation absorbed.

Well answered. More than half of the candidates stated mass spectrometry is used to determine the ratio of the isotopes.

Many candidates successfully calculated the relative atomic mass of nitrogen in the sample. M2 was awarded independently of M1, so candidates who calculated the relative molecular mass using the A_r of nitrogen in the data booklet (14.01) were awarded M2. Many candidates scored both marks.

This was a challenging question for many candidates, while stronger candidates often showed clarity of thinking and were able to conclude that the ionization energies of the two isotopes must be the same and to provide two different reasons for this. Some candidates did realize that the ionization energies are similar but did not give the best reasons to support their answer. Many candidates thought the ionization energies would be different because the size of the nucleus was different. Some teachers commented that the question was difficult while others liked it because it made students apply their knowledge in an unfamiliar situation. The question had a good discrimination index.

Only a quarter of the candidates answered correctly. Some simply stated that N₂O forms HNO₃ with water which did not gain the mark.

Draw the first four energy levels of a hydrogen atom on the axis, labelling n = 1, 2, 3 and 4.

Draw the lines, on your diagram, that represent the electron transitions to n = 2 in the emission spectrum.

Outline why atomic radius decreases across period 3, sodium to chlorine.

Outline why the ionic radius of K⁺ is smaller than that of Cl⁻.

Copper is widely used as an electrical conductor.

Draw arrows in the boxes to represent the electronic configuration of copper in the 4s and 3d orbitals.

Impure copper can be purified by electrolysis. In the electrolytic cell, impure copper is the anode (positive electrode), pure copper is the cathode (negative electrode) and the electrolyte is copper(II) sulfate solution.

Formulate the half-equation at each electrode.

Outline where and in which direction the electrons flow during electrolysis.

4 levels showing convergence at higher energy

[1 mark]

arrows (pointing down) from n = 3 to n = 2 AND n = 4 to n = 2

[1 mark]

same number of shells/«outer» energy level/shielding AND nuclear charge/number of protons/Z_eff increases «causing a stronger pull on the outer electrons»

[1 mark]

K⁺ 19 protons AND Cl^– 17 protons

OR

K⁺ has «two» more protons

same number of electrons/isoelectronic «thus pulled closer together»

[2 marks]

[1 mark]

Anode (positive electrode):

Cu(s) → Cu²⁺(aq) + 2e^–

Cathode (negative electrode):

Cu²⁺(aq) + 2e^– → Cu(s)

Accept Cu(s) – 2e^– → Cu²⁺(aq).

Accept $\rightleftharpoons$ for →

Award [1 max] if the equations are at the wrong electrodes.

[2 marks]

«external» circuit/wire AND from positive/anode to negative/cathode electrode

Accept “through power supply/battery” instead of “circuit”.

[1 mark]

State the electron configuration of the Cu⁺ ion.

Copper(II) chloride is used as a catalyst in the production of chlorine from hydrogen chloride.

4HCl (g) + O₂ (g) → 2Cl₂ (g) + 2H₂O (g)

Calculate the standard enthalpy change, ΔH^θ, in kJ, for this reaction, using section 12 of the data booklet.

The diagram shows the Maxwell–Boltzmann distribution and potential energy profile for the reaction without a catalyst.

Annotate both charts to show the activation energy for the catalysed reaction, using the label E_{a (cat)}.

Explain how the catalyst increases the rate of the reaction.

Solid copper(II) chloride absorbs moisture from the atmosphere to form a hydrate of formula CuCl₂•$x$H₂O.

A student heated a sample of hydrated copper(II) chloride, in order to determine the value of $x$. The following results were obtained:

Mass of crucible = 16.221 g
Initial mass of crucible and hydrated copper(II) chloride = 18.360 g
Final mass of crucible and anhydrous copper(II) chloride = 17.917 g

Determine the value of $x$.

State how current is conducted through the wires and through the electrolyte.

Wires: 

Electrolyte:

Write the half-equation for the formation of gas bubbles at electrode 1.

[Ar] 3d¹⁰
OR
1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ ✔

ΔH^θ = ΣΔH^θ_f (products) − ΣΔH^θ_f (reactants) ✔
ΔH^θ = 2(−241.8 «kJ mol⁻¹») − 4(−92.3 «kJ mol⁻¹») = −114.4 «kJ» ✔

NOTE: Award [2] for correct final answer.

E_{a (cat)} to the left of E_a ✔                        

peak lower AND E_{a (cat)} smaller ✔

«catalyst provides an» alternative pathway ✔

«with» lower E_a
OR
higher proportion of/more particles with «kinetic» E ≥ E_a(cat) «than E_a» ✔

mass of H₂O = «18.360 g – 17.917 g =» 0.443 «g» AND mass of CuCl₂ = «17.917 g – 16.221 g =» 1.696 «g» ✔

moles of H₂O = «$\frac{0.443 \text{g}}{18.02 \text{g mo}{\text{l}}^{-1}}$=» 0.0246 «mol»
OR
moles of CuCl₂ =«$\frac{1.696 \text{g}}{134.45 \text{g mo}{\text{l}}^{-1}}$= » 0.0126 «mol» ✔

«water : copper(II) chloride = 1.95 : 1»
«$x$ =» 2 ✔

NOTE: Accept «$x$ =» 1.95.

NOTE: Award [3] for correct final answer.

Wires:
«delocalized» electrons «flow» ✔

Electrolyte:
«mobile» ions «flow» ✔

2Cl⁻ → Cl₂ (g) + 2e⁻
OR
Cl⁻ → $\frac{1}{2}$Cl₂ (g) + e⁻ ✔

NOTE: Accept e for e⁻.

After heating 3.760 g of a silver oxide 3.275 g of silver remained. Determine the empirical formula of Ag_xO_y.

Suggest why the final mass of solid obtained by heating 3.760 g of Ag_xO_y may be greater than 3.275 g giving one design improvement for your proposed suggestion. Ignore any possible errors in the weighing procedure.

Naturally occurring silver is composed of two stable isotopes, ¹⁰⁷Ag and ¹⁰⁹Ag.

The relative atomic mass of silver is 107.87. Show that isotope ¹⁰⁷Ag is more abundant.

Some oxides of period 3, such as Na₂O and P₄O₁₀, react with water. A spatula measure of each oxide was added to a separate 100 cm³ flask containing distilled water and a few drops of bromothymol blue indicator.

The indicator is listed in section 22 of the data booklet.

Deduce the colour of the resulting solution and the chemical formula of the product formed after reaction with water for each oxide.

Explain the electrical conductivity of molten Na₂O and P₄O₁₀.

Outline the model of electron configuration deduced from the hydrogen line emission spectrum (Bohr’s model).

n(Ag) = «$\frac{3.275\text{ g}}{107.87\text{ g}\text{mol}}=$» 0.03036 «mol»

AND

n(O) = «$\frac{3.760\text{ g}-3.275\text{ g}}{16.00\text{ g}\text{mo}{\text{l}}^{-1}}=\frac{0.485}{16.00}=$» 0.03031 «mol»

«$\frac{0.03036}{0.03031}\approx 1$ / ratio of Ag to O approximately 1 : 1, so»

AgO

Accept other valid methods for M1.

Award [1 max] for correct empirical formula if method not shown.

[2 marks]

temperature too low
OR
heating time too short
OR
oxide not decomposed completely

heat sample to constant mass «for three or more trials»

Accept “not heated strongly enough”.

If M1 as per markscheme, M2 can only be awarded for constant mass technique.

Accept "soot deposition" (M1) and any suitable way to reduce it (for M2).

Accept "absorbs moisture from atmosphere" (M1) and "cool in dessicator" (M2).

Award [1 max] for reference to impurity AND design improvement.

[2 marks]

A_r closer to 107/less than 108 «so more ¹⁰⁷Ag»
OR
A_r less than the average of (107 + 109) «so more ¹⁰⁷Ag»

Accept calculations that gives greater than 50% ¹⁰⁷Ag.

[1 mark]

Do not accept name for the products.

Accept “Na⁺ + OH^–” for NaOH.

Ignore coefficients in front of formula.

[3 marks]

«molten» Na₂O has mobile ions/charged particles AND conducts electricity

«molten» P₄O₁₀ does not have mobile ions/charged particles AND does not conduct electricity/is poor conductor of electricity

Do not award marks without concept of mobile charges being present.

Award [1 max] if type of bonding or electrical conductivity correctly identified in each compound.

Do not accept answers based on electrons.

Award [1 max] if reference made to solution.

[2 marks]

electrons in discrete/specific/certain/different shells/energy levels

energy levels converge/get closer together at higher energies
OR
energy levels converge with distance from the nucleus

Accept appropriate diagram for M1, M2 or both.

Do not give marks for answers that refer to the lines in the spectrum.

[2 marks] 

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

Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.

State an equation for the reaction of phosphorus (V) oxide, P₄O₁₀ (s), with water.

Describe the emission spectrum of hydrogen.

Identify the strongest reducing agent in the given list.

A voltaic cell is made up of a Mn²⁺/Mn half-cell and a Ni²⁺/Ni half-cell.

Deduce the equation for the cell reaction.

The voltaic cell stated in part (ii) is partially shown below.

Draw and label the connections needed to show the direction of electron movement and ion flow between the two half-cells.

increasing number of protons

OR

increasing nuclear charge

«atomic» radius/size decreases

OR

same number of shells

OR

similar shielding «by inner electrons»

«greater energy needed to overcome increased attraction between nucleus and electrons»

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.

P₄O₁₀ (s) + 6H₂O (l) → 4H₃PO₄ (aq)

Accept “P₄O₁₀ (s) + 2H₂O (l) → 4HPO₃ (aq)” (initial reaction).

«series of» lines

OR

only certain frequencies/wavelengths

convergence at high«er» frequency/energy/short«er» wavelength

M1 and/or M2 may be shown on a diagram.

Mn

Mn (s) + Ni²⁺ (aq) → Ni (s) + Mn²⁺ (aq)

wire between electrodes AND labelled salt bridge in contact with both electrolytes

anions to right (salt bridge)
OR
cations to left (salt bridge)
OR
arrow from Mn to Ni (on wire or next to it)

Electrodes can be connected directly or through voltmeter/ammeter/light bulb, but not a battery/power supply.

Accept ions or a specific salt as the label of the salt bridge.