Deduce the equation for the decomposition of guanidinium nitrate.

Calculate the total number of moles of gas produced from the decomposition of 10.0 g of guanidinium nitrate.

Calculate the pressure, in kPa, of this gas in a 10.0 dm³ air bag at 127°C, assuming no gas escapes.

Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled, while nitrogen does not.

Another airbag reactant produces nitrogen gas and sodium.

Suggest, including an equation, why the products of this reactant present a safety hazard.

C(NH₂)₃NO₃(s) → 2N₂(g) + 3H₂O (g) + C (s) ✔

moles of gas = « $5\times \frac{10.0 \text{g}}{122.11 \text{g mo}{\text{l}}^{-1}}=$» 0.409 «mol» ✔

«$p=\frac{0.409 \text{mol}\times \text{8}{\text{.31\hspace{0.05em}J\hspace{0.05em}K}}^{-1}\text{mo}{\text{l}}^{-1}\times (127+273) \text{K}}{\text{10}\text{.0\hspace{0.05em}d}{\text{m}}^3}$» = 136 «kPa» ✔

Any two of:
nitrogen non-polar/London/dispersion forces AND water polar/H-bonding ✔
water has «much» stronger intermolecular forces ✔
water molecules attract/condense/occupy smaller volume «and therefore deviate from ideal behaviour» ✔

2Na (s) + 2H₂O (l) → 2NaOH (aq) + H₂(g) ✔

hydrogen explosive
OR
highly exothermic reaction
OR
sodium reacts violently with water
OR
forms strong alkali ✔

NOTE: Accept the equation of combustion of hydrogen.
Do not accept just “sodium is reactive/dangerous”.

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

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» ✔

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] 

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).

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 ✔

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.

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]

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]

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.

Before its isolation, scientists predicted the existence of rhenium and some of its properties.

Suggest the basis of these predictions.

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.

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»  [✔]

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 «%»  [✔]

This nature of science question generated a lot of discussion among teachers. Some in support of such questions and others concerned that it takes a lot of time for candidates to know how to answer. Some teachers thought it was unclear what the question was asking. It is pleasing that about a quarter of the candidates answered both parts successfully and many candidates gained one mark usually for “periodic trends”. However, some candidates only focused on one part of the question. Quite a few candidates discussed isotopes, probably thrown off by the stem. A teacher was concerned that since transition metals are not part of the SL syllabus that Re was a bad choice, however, the question did not really require any transition metal chemistry to be answered.

This question was a good discriminator between high-scoring and low-scoring candidates. It was well answered by more than half of the candidates who had obviously carried out such displacement reactions and interpreted the outcomes during the course. Some candidates did not state the obvious of dipping the metal into the sulfates.

More than half of the candidates named ReCl₃ correctly. Common mistakes included “rhenium chloride” and “trichlororhenium”.

The majority of candidates calculated the percentage, by mass, of rhenium in ReCl₃ correctly. Some rounding errors were seen that students should be more careful with.

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.

Magnesium can be produced by the electrolysis of molten magnesium chloride.

Write the half-equation for the formation of magnesium.

Suggest an experiment that shows that magnesium is more reactive than zinc, giving the observation that would confirm this.

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

Identify the strongest force between the molecules of Compound B.

Draw the structural formula of the alkene required.

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.

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.

Mg²⁺ + 2 e^- → Mg ✔

Do not penalize missing charge on electron.

Accept equation with equilibrium arrows.

Alternative 1

put Mg in Zn²⁺(aq) ✔

Zn/«black» layer forms «on surface of Mg» ✔

Award [1 max] for “no reaction when Zn placed in Mg²⁺(aq)”.

Alternative 2

place both metals in acid ✔

bubbles evolve more rapidly from Mg
OR
Mg dissolves faster ✔

Alternative 3

construct a cell with Mg and Zn electrodes ✔

bulb lights up
OR
shows (+) voltage
OR
size/mass of Mg(s) decreases «over time»
OR
size/mass of Zn increases «over time»

Accept “electrons flow from Mg to Zn”.

Accept Mg is negative electrode/anode
OR
Zn is positive electrode/cathode

Accept other correct methods.

propanone ✔

Accept 2-propanone and propan-2-one.

hydrogen bonds ✔

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» ✔

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”.

Unfortunately, only 40% of the students could write this quite straightforward half equation.

Many candidates gained some credit by suggesting voltaic cell or a displacement reaction, but most could not gain the second mark and the reason was often a failure to be able to differentiate between "what occurs" and "what is observed".

Even though superfluous numbers (2-propanone, propan-2-one) were overlooked, only about half of the students could correctly name this simple molecule.

Probably just over half the students correctly identified hydrogen bonding, with dipole-dipole being the most common wrong answer, though a significant number identified an intramolecular bond.

Few candidates could correctly eliminate water to deduce the identity of the required reactant.

Correct answers to this were very scarce and even when candidates had an incorrect alkene for the previous part, they were unable to score an ECF mark, by deducing the formula of the polymer it would produce.

Some students deduced that, as it was a tertiary alcohol, there would be no reaction, but almost all were lucky that this was accepted as well as the correct observation - "it would remain orange".

About a quarter of the students identified this as a substitution reaction, though quite a number then lost the mark by incorrectly stating it was either "free radical" or "electrophilic". A very common wrong answer was "displacement" or "single displacement" and this makes one wonder whether this terminology is being taught instead of substitution

Generally well done with the vast majority of students correctly citing "correct orientation" and many only failed to gain the second mark through failing to equate the energy required to the activation energy.

Another question that was not well answered with probably only a quarter of candidates stating that the polarity would decrease because of decreasing electronegativity down the group.

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.

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⁻.

State a physical property of sodium oxide.

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⁺ ✔

Any one of:
brittle ✔
high melting point/crystalline/solid «at room temperature» ✔
low volatility ✔
conducts electricity when molten ✔
does not conduct electricity at room temperature ✔

Do not accept soluble in water.

Ignore any chemical properties.

Determine the coefficients that balance the equation for the reaction of lithium with water.

Calculate the molar concentration of the resulting solution of lithium hydroxide.

Calculate the volume of hydrogen gas produced, in cm³, if the temperature was 22.5 °C and the pressure was 103 kPa. Use sections 1 and 2 of the data booklet.

Suggest a reason why the volume of hydrogen gas collected was smaller than predicted.

The reaction of lithium with water is a redox reaction. Identify the oxidizing agent in the reaction giving a reason.

Describe two observations that indicate the reaction of lithium with water is exothermic.

2 Li (s) + 2 H₂O (l) → 2 LiOH (aq) + H₂(g) ✔

$n_{Li}«\frac{0200 g}{6.94 g}=»0.0288«mol»$✔

«n_LiOH = n_Li»

$\left[\mathrm{LiOH}\right] «=\frac{0.0288 mol}{0.5000 d{m}^3}=»0.0576 «mol d{m}^{-3}»$ ✔

Award [2] for correct final answer.

$«n_{H_2}=\frac{1}{2}\times 0.0288 mol=0.0144 mol»$

$«V=\frac{nRT}{P}=»\left(\frac{0.0144 mol\times 8.31 J{K}^{-1}mo{l}^{-1}\times \left(22.5+273\right)K}{103 kPa}\right) «\times {10}^3»$✔

$V=343 «{\mathrm{cm}}^3»$✔

Award [2] for correct final answer.

Accept answers in the range 334 – 344 cm³.

Award [1 max] for 0.343 «cm³/dm³/m³».

Award [1 max] for 26.1 cm³ obtained by using 22.5 K.

Award [1 max] for 687 cm³ obtained by using 0.0288 mol.

lithium was impure/«partially» oxidized

OR

gas leaked/ignited ✔

Accept “gas dissolved”.

H₂O AND hydrogen gains electrons «to form H₂»

OR

H₂O AND H oxidation state changed from +1 to 0 ✔

Accept “H₂O AND H/H₂O is reduced”.

Any two:

temperature of the water increases ✔

lithium melts ✔

pop sound is heard ✔

Accept “lithium/hydrogen catches fire”.

Do not accept “smoke is observed”.

This part-question was better answered than part (ii). 50% of the candidates drew a correct arrow between n=2 and n=3. Both absorption and emission transitions were accepted since the question did not specify which type of spectrum was required. Some teachers commented on this in their feedback. Mistakes often included transitions between higher energy levels.

Describe the structure and bonding in solid sodium oxide.

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 5.00 g of sodium oxide produces 5.50 g 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 (c)(i).

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.

«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.

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 «= $\frac{5.00\text{ g}}{61.98\text{ mo}{\text{l}}^{-1}}$» = 0.0807/8.07 × 10⁻² «mol»
OR
mass Na₂O₂ theoretical yield «= $\frac{5.00\text{ g}}{61.98\text{ mo}{\text{l}}^{-1}}$ × 77.98 gmol⁻¹» = 6.291 «g»  [✔]

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

Note: Award [2] for correct final answer.

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

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

ΔH = «ΣΔH_f products − ΣΔH_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  [✔]

NaOH  [✔]

Note: Accept correct equation showing NaOH as a product.

IV  [✔]

Disappointingly many students did not realise that sodium oxide was held by ionic bonds, many said it was covalent or metallic bonding. The ones that knew it was ionic failed to describe it adequately to earn the 2 marks.

Very few students could correctly write out the two equations and so often were unable to realise it was acid/base behaviour that would differentiate the oxides.

Many candidates were able to correctly calculate the % yield but some weaker candidates just used 5.0/5.5 to find %.

The calculation of the enthalpy change using enthalpies of formation was generally answered well but common mistakes were students forgetting to multiply by 2 or adding extra terms for oxygen.

Most students didn’t gain a mark and “values are average” was the most common incorrect answer. The fact this was an ionic compound did not register with them. Some students did gain a mark for stating that the substances were not in a gaseous state.

Some students correctly identified sodium hydroxide as the correct product, but hydrogen, oxygen and sodium oxide were common answers.

Oxidation number of +4 was often correctly identified.

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.

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]