Describe the bonding in metals.

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

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

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

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

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

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

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

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

Outline why transition metals form coloured compounds.

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

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

Formulate an equation for this reaction.

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

electrostatic attraction

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

Accept “mobile electrons”.

Do not accept “metal atoms/nuclei”.

[2 marks]

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

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

Award [2] for correct final answer.

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

[2 marks]

Protons: 22 AND Neutrons: 26 AND Electrons: 22

[1 mark]

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

[1 mark]

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

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

Accept “atomic” for “ionic”.

[1 mark]

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

A log graph is acceptable.

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

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

[1 mark]

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

OR

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

prevent layers sliding over each other

Accept diagram showing different sizes of atoms/ions.

[2 marks]

pair of electrons provided by the ligand

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

[1 mark]

partially filled d-orbitals

«ligands cause» d-orbitals «to» split

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

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

colour observed is the complementary colour

[4 marks]

ionic

OR

«electrostatic» attraction between oppositely charged ions

[1 mark]

«simple» molecular structure

OR

weak«er» intermolecular bonds

OR

weak«er» bonds between molecules

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

Do not accept “covalent”.

[1 mark]

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

Accept ionic equation.

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

[2 marks]

HCl causes breathing/respiratory problems

OR

HCl is an irritant

OR

HCl is toxic

OR

HCl has acidic vapour

OR

HCl is corrosive

Accept TiO₂ causes breathing

problems/is an irritant.

Accept “harmful” for both HCl and TiO₂.

Accept “smoke is asphyxiant”.

[1 mark]

Discuss the bonding in the resonance structures of ozone.

Deduce one resonance structure of ozone and the corresponding formal charges on each oxygen atom.

The first six ionization energies, in kJ mol^–1, of an element are given below.

Explain the large increase in ionization energy from IE₃ to IE₄.

At a given time, the concentration of NO₂(g) and N₂O₄(g) were 0.52 and $0.10\text{ mol}\text{d}{\text{m}}^{-3}$ respectively.

Deduce, showing your reasoning, if the forward or the reverse reaction is favoured at this time.

Comment on the value of ΔG when the reaction quotient equals the equilibrium constant, Q = K.

lone pair on p orbital «of O atom» overlaps/delocalizes with pi electrons «from double bond»

both O–O bonds have equal bond length
OR
both O–O bonds have same/1.5 bond order
OR
both O–O are intermediate between O–O AND O=O 

both O–O bonds have equal bond energy

Accept “p/pi/$\pi$ electrons are delocalized/not localized”.

[3 marks]

ALTERNATIVE 1:

FC: –1 AND +1 AND 0

ALTERNATIVE 2:

FC: 0 AND +1 AND –1

Accept any combination of lines, dots or crosses to represent electrons.

Do not accept structure that represents 1.5 bonds.

Do not penalize missing lone pairs if already penalized in 3(b).

If resonance structure is incorrect, no ECF.

Any one of the structures with correct formal charges for [2 max].

[2 marks]

Any two of:
IE₄: electron in lower/inner shell/energy level
OR
IE₄: more stable/full electron shell

IE₄: electron closer to nucleus
OR
IE₄: electron more tightly held by nucleus

IE₄: less shielding by complete inner shells

Accept “increase in effective nuclear charge” for M2.

[2 marks]

«Q_c = $\frac{0.10}{{0.52}^2}$ =» 0.37
reaction proceeds to the left/NO₂(g) «until Q = K_c»
OR
reverse reaction «favoured»

Do not award M2 without a calculation for M1 but remember to apply ECF.

[2 marks]

ΔG = 0
reaction at equilibrium
OR
rate of forward and reverse reaction is the same
OR
constant macroscopic properties

[2 marks]

Magnesium ions produce no emission or absorption lines in the visible region of the electromagnetic spectrum. Suggest why most magnesium compounds tested in a school laboratory show traces of yellow in the flame.

(i) Explain the convergence of lines in a hydrogen emission spectrum.

(ii) State what can be determined from the frequency of the convergence limit.

Magnesium chloride can be electrolysed.

(i) 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 922K and 987K respectively.

(ii) Identify the type of reaction occurring at the cathode (negative electrode).

(iii) State the products when a very dilute aqueous solution of magnesium chloride is electrolysed.

Standard electrode potentials are measured relative to the standard hydrogen electrode. Describe a standard hydrogen electrode.

A magnesium half-cell, Mg(s)/Mg²⁺(aq), can be connected to a copper half-cell, Cu(s)/Cu²⁺(aq).

(i) Formulate an equation for the spontaneous reaction that occurs when the circuit is completed.

(ii) Determine the standard cell potential, in V, for the cell. Refer to section 24 of the data booklet.

(iii) Predict, giving a reason, the change in cell potential when the concentration of copper ions increases.

contamination with sodium/other «compounds»

i
energy levels are closer together at high energy / high frequency / short wavelength

ii
ionisation energy

i)

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).

ii)

reduction

iii)

Anode (positive electrode):
oxygen/O₂
OR
hydogen ion/proton/H⁺ AND oxygen/O₂
Cathode (negative electrode):
hydrogen/H₂
OR
hydroxide «ion»/OH^– AND hydrogen/H₂

Award [1 max] if correct products given at wrong electrodes.

Any two of:

«inert» Pt electrode
OR
platinum black conductor

1 mol dm^–3 H⁺ (aq)

H₂ (g) at 100 kPa

Accept 1 atm H₂ (g).
Accept 1 bar H₂ (g)
Accept a labelled diagram.
Ignore temperature if it is specified.

i

Mg(s) + Cu²⁺ (aq) → Mg²⁺ (aq) + Cu(s)

ii

«+0.34V – (–2.37V) = +»2.71 «V»

iii

cell potential increases

reaction «in Q4(k)(i)» moves to the right
OR
potential of the copper half-cell increases/becomes more positive

Accept correct answers based on the Nernst equation

Hydrogen spectral data give the frequency of 3.28 × 10¹⁵ s⁻¹ for its convergence limit.

Calculate the ionization energy, in J, for a single atom of hydrogen using sections 1 and 2 of the data booklet.

Calculate the wavelength, in m, for the electron transition corresponding to the frequency in (a)(iii) using section 1 of the data booklet.

Deduce any change in the colour of the electrolyte during electrolysis.

Deduce the gas formed at the anode (positive electrode) when graphite is used in place of copper.

Explain why transition metals exhibit variable oxidation states in contrast to alkali metals.

IE «= ΔE = hν = 6.63 × 10^–34 J s × 3.28 × 10¹⁵ s^–1» = 2.17 × 10^–18 «J»

[1 mark]

«$\lambda =\frac{C}{\text{v}}=\frac{3.00\times {10}^8\text{ m}{\text{s}}^{-1}}{3.28\times {10}^{15}{\text{s}}^{-1}}=$» 9.15 × 10^–8 «m»

[1 mark]

no change «in colour»

Do not accept “solution around cathode will become paler and solution around the anode will become darker”.

[1 mark]

oxygen/O₂

Accept “carbon dioxide/CO2”.

[1 mark]

Transition metals:

«contain» d and s orbitals «which are close in energy»

OR

«successive» ionization energies increase gradually

Alkali metals:

second electron removed from «much» lower energy level

OR

removal of second electron requires large increase in ionization energy

[2 marks]

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:

Predict the bond angle in the ozone molecule.

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.

Identify the steps which absorb ultraviolet light.

Determine, showing your working, the wavelength, in m, of ultraviolet light absorbed by a single molecule in one of these steps. Use sections 1, 2 and 11 of the data booklet.

Ozone depletion is catalysed by nitrogen monoxide, NO, which is produced in aircraft and motor vehicle engines, and has the following Lewis structure.

Show how nitrogen monoxide catalyses the decomposition of ozone, including equations in your answer.

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.

any value from 110°–119° ✔

«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.

steps 1 AND 3 ✔

ALTERNATIVE 1:
for oxygen:

$E=«\frac{498 000\text{\hspace{0.05em}J\hspace{0.05em}mo}{\text{l}}^{-1}}{6.02\times {10}^{23} \text{mo}{\text{l}}^{-1}}=» 8.27\times {10}^{-19} «\text{J}»$ ✔

$\mathrm{\lambda }\text{ = }«\frac{6.63\times {10}^{-34}\text{\hspace{0.05em}J\hspace{0.05em}s}\times \text{3}\text{.00}\times \text{1}{\text{0}}^8{\text{\hspace{0.05em}m\hspace{0.05em}s}}^{-1}}{8.27\times {10}^{-19} \text{J}}=» 2.40\times {10}^{-7} «\text{m}»$✔

ALTERNATIVE 2:
for ozone:

similar calculation using 200 < bond enthalpy < 400 for ozone, such as

$E=«\frac{300 000\text{\hspace{0.05em}J\hspace{0.05em}mo}{\text{l}}^{-1}}{6.02\times {10}^{23} \text{mo}{\text{l}}^{-1}}=» 4.98\times {10}^{-19}«\text{J}»$ ✔

$\mathrm{\lambda }\text{ = }«\frac{6.63\times {10}^{-34}\text{\hspace{0.05em}J\hspace{0.05em}s}\times \text{3}\text{.00}\times \text{1}{\text{0}}^8{\text{\hspace{0.05em}m\hspace{0.05em}s}}^{-1}}{4.98\times {10}^{-19} \text{J}}=» 3.99\times {10}^{-7} «\text{m}»$✔

NOTE: Award [2] for correct final answer.

•NO + O₃ → •NO₂ + O₂ ✔

•NO₂ + O₃ → •NO + 2O₂ ✔

NOTE: Accept •NO₂ → •NO + •O AND •O + O₃ → 2O₂ for M2.

Describe the nature of ionic bonding.

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

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

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

Outline why solid calcium is a good conductor of electricity.

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

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

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

Estimate the pH of the resultant solution.

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

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

electrostatic attraction AND oppositely charged ions

[1 mark]

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

OR

find the frequency of each isotope

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

OR

find/calculate the weighted average

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

[2 marks]

«promoted» electrons fall back to lower energy level

energy difference between levels is different

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

[2 marks]

Any two of:

stronger metallic bonding

smaller ionic/atomic radius

two electrons per atom are delocalized

OR

greater ionic charge

greater atomic mass

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

[2 marks]

delocalized/mobile electrons «free to move»

[1 mark]

general increase

only one discontinuity between “IE2” and “IE3”

[2 marks]

pH > 7

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

[1 mark]

sigma (σ):

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

OR

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

pi ($\pi$):

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

OR

sideways overlap «of p-orbitals»

Award marks for suitable diagrams.

[2 marks]

sigma (σ): 3

AND

pi ($\pi$): 2

[1 mark]

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

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

Describe the structure and bonding in solid sodium oxide.

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

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

$\frac{1}{2}$O₂(g) → O^2- (g):

Na (s) → Na⁺ (g):

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

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

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

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

Sodium oxide, Na₂O:

Phosphorus(V) oxide, P₄O₁₀:

Differentiation:

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

2Na₂O (s) + O₂ (g) → 2Na₂O₂ (s)

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

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

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

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

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

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

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

     [✔]

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

same number of electrons in outer shell

OR

all are s¹ [✔]

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

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

Note: Do not accept “ionic” without description.

$\frac{1}{2}$O₂(g) → O^2- (g)

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

Na (s) → Na⁺ (g)

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

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

«= +» 2481 «kJ mol⁻¹»    [✔]

Note: Award [2] for correct final answer.

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

K⁺ ion is larger than Na⁺

OR

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

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

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

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

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

OR

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

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

Note: Award [2] for correct final answer.

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

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

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

Note: Award [3] for correct final answer.

Award [2 max] for “+ 452.6 «kJ»”.

only valid for covalent bonds

OR

only valid in gaseous state     [✔]

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

Note: Accept “bond in ozone is longer”.

Any one of:

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

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

NaOH    [✔]

IV    [✔]

Generally well done with a correct plot of ionization energies.

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

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

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

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

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

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

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

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

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

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

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

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

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

The electron configuration of copper makes it a useful metal.

Determine the frequency of a photon that will cause the first ionization of copper. Use sections 1, 2 and 8 of the data booklet.

The electron configuration of copper makes it a useful metal.

Explain why a copper(II) solution is blue, using section 17 of the data booklet.

The electron configuration of copper makes it a useful metal.

Copper plating can be used to improve the conductivity of an object.

State, giving your reason, at which electrode the object being electroplated should be placed.

$«E=\frac{745 000 \mathrm{J} {\mathrm{mol}}^{-1}}{6.02\times {10}^{23} {\mathrm{mol}}^{-1}}=»1.24\times {10}^{-18} \mathrm{J}$ ✔

$«E=h\nu »$
$«1.24\times {10}^{-18} \mathrm{J}=6.63\times {10}^{-34} \mathrm{J} \mathrm{s}\times \mathrm{\nu }»$
$\mathrm{\nu }=1.87\times {10}^{15} «{\mathrm{s}}^{-1}/\mathrm{Hz}»$ ✔

Award [2] for correct final answer.
Award [1] for 1.12 × 10³⁹ «Hz».

orange light is absorbed «and the complementary colour is observed» ✔

Any TWO from:
partially filled d-orbitals ✔
«ligands/water cause» d-orbitals «to» split ✔
light is absorbed as electrons move to a higher energy orbital «in d–d transitions»
OR
light is absorbed as electrons are promoted ✔
energy gap corresponds to «orange» light in the visible region of the spectrum ✔

cathode/negative «electrode» AND ${\mathrm{Cu}}^{2+}$ reduced «at that electrode» ✔

Accept cathode/negative «electrode» AND copper forms «at that electrode».

Determining the frequency of a photon that will cause the first ionization of copper was the most challenging question on the exam. Many could not do it all, although some came up with the answer that came from using the result that would arise from the ionization energy in J/mole (and frequently kJ/mole) rather than J/atom.

Many students were able to fully explain why solutions containing Cu²⁺ appear blue, however the misconception between absorption and emission spectra is still quite evident.

Surprisingly not that well answered. Most students identified the cathode as the electrode where electroplating occurs but few could adequately justify why.

Explain the decrease in atomic radius from Na to Cl.

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

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

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

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

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

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

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

more protons in Na⁺ ✔

Sketch showing:

largest increase between third and fourth ionization energies ✔

IE₁ < IE₂ < IE₃ < IE₄ < IE₅ ✔

Fe²⁺ AND smaller size/radius

OR

Fe²⁺ AND higher charge density ✔

stronger interaction with «polar» water molecules ✔

M1 not needed for M2.

Explain why Si has a smaller atomic radius than Al.

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

State the condensed electron configurations for Cr and Cr3⁺.

Describe metallic bonding and how it contributes to electrical conductivity.

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

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

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

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

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

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

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

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

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

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

Cr:
[Ar] 4s¹3d⁵ ✓

Cr³⁺:
[Ar] 3d³ ✓

Accept “[Ar] 3d⁵4s¹”.

Accept “[Ar] 3d³4s⁰”.

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

Award [1 max] for 4s¹3d⁵ AND 3d³.

electrostatic attraction ✓

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

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

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

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

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

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

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

any value or range between 647 and 700 nm ✓

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

Penalize missing lone pairs of electrons on halogens once only.

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

Accept “non-linear” for SCl₂ molecular geometry.

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

H₂O forms hydrogen bonding «while SCl₂ does not» ✓

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

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

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

Ignore reference to Van der Waals.

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

Outline why metals, like iron, can conduct electricity.

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

Sketch the first eight successive ionisation energies of sulfur.

Describe the bonding in this type of solid.

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

State the full electron configuration of the sulfide ion.

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

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

Write the equation for this reaction.

Deduce the change in the oxidation state of sulfur.

Suggest why this process might raise environmental concerns.

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

mobile/delocalized «sea of» electrons

Any two of:

forms acidic oxides «rather than basic oxides» ✔

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

forms anions «rather than cations» ✔

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

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

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

large increase from 6th to 7th ✔

Accept line/curve showing these trends.

electrostatic attraction ✔

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

X-ray crystallography ✔

1s² 2s² 2p⁶ 3s² 3p⁶ ✔

Do not accept “[Ne] 3s² 3p⁶”.

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

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

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

Accept other valid answers.

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

Accept any correct ratio.

+6
OR
−2 to +4 ✔

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

sulfur dioxide/SO₂ causes acid rain ✔

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

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

prevents layers/atoms sliding over each other ✔

Outline why ozone in the stratosphere is important.

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

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

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

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

Calculate the relative molecular mass of the resulting N₂O.

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

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

Nitrogen and carbon:

Nitrogen and oxygen:

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

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

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

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

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

Note: Ignore radical signs.

Accept equilibrium arrows.

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

mass spectrometry/MS     [✔]

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

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

Any two:

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

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

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

same AND have same electronic configuration/shielding     [✔]

Note: Accept “almost the same”.

“Same” only needs to be stated once.

Nitrogen and carbon:

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

Nitrogen and oxygen:

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

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

Accept box annotation of electrons for M2.

delocalization

OR

delocalized π-electrons    [✔]

Note: Accept “resonance”.

linear AND 2 electron domains

OR

linear AND 2 regions of electron density    [✔]

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

sp     [✔]

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

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

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

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

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

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

Most candidates identified resonance for this given Lewis representation.

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

Hybridisation of the N atom was correct in most cases.

State the full electronic configuration of Fe²⁺.

Explain why there is a large increase from the 8th to the 9th ionization energy of iron.

Calculate the oxidation state of sulfur in iron(II) disulfide, FeS₂.

Describe the bonding in iron, Fe (s).

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ ✔

Any two of:

IE₉: electron in lower energy level
OR
IE₉: more stable/full electron level ✔

IE₉: electron closer to nucleus
OR
IE₉: electron more tightly held by nucleus ✔

IE₉: less shielding by «complete» inner levels ✔

–1 ✔

Accept “– I”.

electrostatic attraction/hold between «lattice of» positive ions/cations AND delocalized «valence» electrons ✔

Mostly well done which was a pleasant surprise since this is not overly easy, predictably some gave [Ar] 4s² 3d⁴.

Despite some confusion regarding which sub-level the electrons were being removed from, many candidates were able to make at least one valid point, commonly in terms of lower energy/ full sub level/closer to nucleus.

This was an easy question, yet 30% of the candidates were unable to work it out; some wrote the oxidation state in the conventionally incorrect format, 1- and lost the mark.

Most candidates knew the bonding in Fe is metallic but some did not “describe” it or missed the type of attraction, a minor mistake; others referred to nuclei or protons instead of cations/positive ions. In some cases, candidates referred too ionic bonding, probably still thinking of FeS₂ (not reading the question well). Overall, only 30% answered satisfactorily.

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

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

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

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

increasing number of protons
OR
increasing nuclear charge ✔

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

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

Write a balanced equation for the reaction that occurs.

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

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

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

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

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

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

Calculate coefficients that balance the equation for the following reaction.

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

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

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

State the number of subatomic particles in this ion.

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

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

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

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

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

2 Mg(s) + O₂(g) → 2 MgO(s) ✔

Do not accept equilibrium arrows. Ignore state symbols

aluminium/Al ✔

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

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

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

Award [2] for correct final answer

Accept 0.021%

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

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

Award «0.2614 mol x 40.31 g mol^–1»

Accept alternative methods to arrive at the correct answer.

Accept final answers in the range 90.5-91.5%

[2] for correct final answer.

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

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

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

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

Accept “crucible weighed before fully cooled”.

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

Accept “non-stoichiometric compounds formed”.

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

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

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

phenol red ✔

Accept bromothymol blue or phenolphthalein.

Mg₃N₂: -3
AND
NH₃: -3 ✔

Do not accept 3 or 3-

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

Redox:
no AND no oxidation states change ✔

Accept “yes AND proton transfer takes place”

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

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

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

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

isotope«s» ✔

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

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

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

Any two of:

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

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

atom «discovered» to have structure ✔

fission
OR
atoms can be split ✔

Accept atoms can undergo fusion «to produce heavier atoms»

Accept specific examples of particles.

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

Award [1] for all bonding types correct.

Award [1] for each correct description.

Apply ECF for M2 only once.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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