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.

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.

Suggest why many chemicals, including hydrogen peroxide, are kept in brown bottles instead of clear colourless bottles.

In a laboratory experiment solutions of potassium iodide and hydrogen peroxide were mixed and the volume of oxygen generated was recorded. The volume was adjusted to 0 at t = 0.

The data for the first trial is given below.

Plot a graph on the axes below and from it determine the average rate of formation of oxygen gas in cm³ O₂ (g) s⁻¹.

Average rate of reaction:

Additional experiments were carried out at an elevated temperature. On the axes below, sketch Maxwell–Boltzmann energy distribution curves at two temperatures T₁ and T₂, where T₂ > T₁.

Apart from a greater frequency of collisions, explain, by annotating your graphs in (b)(ii), why an increased temperature causes the rate of reaction to increase.

MnO₂ is another possible catalyst for the reaction. State the IUPAC name for MnO₂.

Comment on why peracetic acid, CH₃COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.

H₂O₂ (aq) + CH₃COOH (aq) $\rightleftharpoons$ CH₃COOOH (aq) + H₂O (l)

Sodium percarbonate, 2Na₂CO₃•3H₂O₂, is an adduct of sodium carbonate and hydrogen peroxide and is used as a cleaning agent.

M_r (2Na₂CO₃•3H₂O₂) = 314.04

Calculate the percentage by mass of hydrogen peroxide in sodium percarbonate, giving your answer to two decimal places.

decomposes in light  [✔]

Note: Accept “sensitive to light”.

points correctly plotted  [✔]

best fit line AND extended through (to) the origin  [✔]

Average rate of reaction:
«slope (gradient) of line =» 0.022 «cm³ O₂ (g) s⁻¹»  [✔]

Note: Accept range 0.020–0.024cm³ O₂ (g) s⁻¹.

peak of T₂ to right of AND lower than T₁  [✔]

lines begin at origin AND T₂ must finish above T₁  [✔]

E_a marked on graph  [✔]

explanation in terms of more “particles” with E ≥ E_a
OR
greater area under curve to the right of E_a in T₂  [✔]

manganese(IV) oxide
OR
manganese dioxide  [✔]

Note: Accept “manganese(IV) dioxide”.

move «position of» equilibrium to right/products  [✔]

Note: Accept “reactants are always present as the reaction is in equilibrium”.

M (H₂O₂) «= 2 × 1.01 + 2 × 16.00» = 34.02 «g»  [✔]

«% H₂O₂ = 3 × $\frac{34.02}{314.04}$ × 100 =» 32.50 «%»  [✔]

Note: Award [2] for correct final answer.

The explanation that the brown bottle prevented light causing a decomposition of the chemical was well answered but some incorrectly suggested it helped to stop mixing up of chemicals e.g. acid/water/peroxide.

The graphing was disappointing with a surprising number of students missing at least one mark for failing to draw a straight line or for failing to draw the line passing through the origin. Also some were unable to calculate the gradient.

The drawing of the two curves at T1 and T2 was generally poorly done.

Explaining why temperature increase caused an increase in reaction rate was generally incorrectly answered with most students failing to mention “activation energy” in their answer or failing to annotate the graph.

Many could correctly name manganese(IV)oxide, but there were answers of magnesium(IV) oxide or manganese(II) oxide.

Suggesting why peractic acid was sold in solution was very poorly answered and only a few students mentioned equilibrium and, if they did, they thought it would move to the left to restore equilibrium.

Calculating the % by mass was generally well answered although some candidates started by using rounded values of atomic masses which made their final answer unprecise.

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.

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⁻.