Practice exam questions SA1 to SA5, with answers built from the lectures

Intuition

Treat these like a driving test with the examiner’s notes pinned up beside you. First try each question cold. Then read the answer and see which lecture facts it leans on. The goal is to see the shape of a good answer, not to memorise this one.

Lecture 39 ends with five short-answer practice questions, labelled SA1 to SA5, on slides 16 to 20. Each slide also carries a marking-style sample answer from the lecturer. The final exam format is not stated in the slides, so this page makes no claim about it. The slides name the lectures each question draws on: SA1 Lecture 6 and the worksheet, SA2 Lecture 7 and the week 4 tutorial, SA3 Lectures 8 and 9, SA4 Lectures 10 and 11, SA5 Lectures 37 and 38.

Each question below has two parts. First, a step-by-step working in plain words, built from the linked topic pages. Second, the lecturer’s own model answer, copied exactly from the slide. The model answer is what earns the marks, so compare your attempt with it. Text in curly brackets {…} is the lecturer’s “by the way” note, not needed for marks.

SA1: a doubled unit cell halves the spacing of the spots

SA1: The material TiSe₂ is metallic at room temperature but on cooling converts to an insulator in which the structure distorts very slightly. This distortion causes a doubling of the size of the hexagonal unit cell along both the a and b axes (see figure: 1 = high temperature; 2 = low temperature).

Describe how the X-ray diffraction pattern is expected to change on cooling through the transition to the larger unit cell. (3 marks)

The slide shows two drawings of the hexagonal cell with Se and Ti atoms, 1 at high temperature and 2 at low temperature. Part of the question’s last line is hidden behind the answer box on the slide, so the words “through the transition to” above are reconstructed from the sense of the sentence.

Answer, from the lectures. Bragg diffraction says a set of planes with spacing $d$ reflects at an angle given by $n\lambda = 2d\sin\theta$. Bigger $d$ means a smaller angle, so the spots sit closer together. The lecture also says the observed pattern is a slice of the reciprocal lattice. So the distances in the pattern run opposite to the distances in the crystal.

1. The cell doubles along $a$ and $b$, so the plane spacings along those directions double.
2. The spots along those directions therefore move to half their old separation. In reciprocal-lattice language, $a^*$ and $b^*$ each halve.
3. The $c$ axis does not change, so $c^*$ does not change.

Lecturer’s model answer (L39 slide 16).

This question is mainly testing your understanding that spacing of diffraction peaks is inversely related to the spacing of Miller planes – there are multiple ways to answer it. Related important concepts here include the directions of the diffraction spots and what information is present in their intensities.

Since the diffraction pattern is inversely related to the structure (it is the reciprocal), the doubling along a and b will cause a halving of the distance between diffraction spots along the associated diffraction directions.

(Or more technically: the reciprocal lattice vectors a* and b* will each halve and c* will remain unchanged, causing … etc.)

Very good answer: Since there is only very small movement of the atoms away from the high symmetry structure, the extra diffraction peaks caused by the distortion will be very weak, i.e., the diffraction pattern will look very similar, but with extra very weak peaks midway between the main peaks along a* and b*.

{By the way, this type of metal-insulator transition is called a ‘charge density wave’ (CDW), and arises for the same reason as Jahn-Teller distortions – the delocalised electrons move to lower energy with the opening of a band gap associated with slight deformation of the structure.}

Why the “very good answer” is true, in plain words. Spot positions come from the cell size. Spot brightness (intensity) comes from where the atoms sit inside the cell. The new, bigger cell gives new spot positions halfway between the old ones. But the atoms have only shifted a tiny bit from the old pattern, so the new spots are very faint. Picture a fence where every second post leans very slightly: from far away it still looks like the old fence.

SA2: charge first, then size

SA2: The materials Li^IF, Ca^IIO, Al^IIIN, Ni^IIO and Cs^II all adopt the rock-salt structure type. On the basis of the Born-Mayer equation below, place these in order of increasing $\Delta H^0_L$. Briefly justify your answer. (4 marks)

$\Delta H^0_L = \frac{N_A\,|z_A z_B|\,e^2}{4\pi\varepsilon_0 d}\left(1 - \frac{d^*}{d}\right)A$

Answer, from the lectures. The Born-Mayer equation has these ingredients: $N_A$, $e$ and $\varepsilon_0$ are constants. $d^*$ is a repulsion constant, 0.345 Å. $A$ is the Madelung constant, which is the same for every compound here because they all share the rock-salt structure. So only $z_A$, $z_B$ and $d$ differ. The lecture says higher charge gives higher lattice enthalpy, and compares melting points to show the charge effect is larger than the size effect: CaO (2572 °C) against NaCl (800 °C) for charge, CaO against BaO (1923 °C) for size.

Compound	$\lvert z_A z_B \rvert$	Note
CsI	1	larger $d$ than LiF
LiF	1	smaller $d$
CaO	4	larger $d$ than NiO
NiO	4	smaller $d$
AlN	9	highest charge product

1. Sort by $\lvert z_A z_B \rvert$: 1, 1, 4, 4, 9.
2. Break each tie by $d$. A larger $d$ lowers the enthalpy. Cs⁺ and I⁻ are both larger than Li⁺ and F⁻, so CsI has the larger $d$ and the lower enthalpy.
3. For CaO and NiO, the lecture on ionic radii and the MOF lecture both say radii decrease across a period. Ni²⁺ is further across the period than Ca²⁺, so it is smaller. NiO has the smaller $d$ and the higher enthalpy.

Order of increasing $\Delta H^0_L$: CsI < LiF < CaO < NiO < AlN.

The lecture’s Shannon radii table lets you check the sizes if you want numbers, but the periodic trends are enough here.

Lecturer’s model answer (L39 slide 17).

(This is similar to a week 4 tute question)

CsI < LiF < CaO < NiO < AlN

The only things varying across these compounds are zA, zB and d. The charges z have a larger influence on the enthalpy than changes in d and so are considered first. The values of zA x zB are 1 (CsI & LiF; CsI has larger d), 4 (CaO & NiO; CaO has larger d) and 9 (AlN).

SA3: one extra electron partly fills the band

SA3: Azafullerene, NC₅₉, has the same soccer-ball structure as C₆₀ but has one carbon atom replaced (‘doped’) by a nitrogen atom. (6 marks)

(i) Assuming that the solid state packing is the same as that of C₆₀, predict the electron conducting behaviour of azafullerene. Include in your answer a description and explanation of how the conductivity might be expected to change upon cooling. (3 marks)

(ii) Predict how the electrical conductivity of this material would be expected to change if Na⁺ ions were intercalated into the octahedral holes of the structure. Give a justification for this prediction. (3 marks)

(i), from the lectures. The conductors lecture counts C₆₀ as $4 \times 60 = 240$ valence electrons and gives C₆₀ as a semiconductor with a 1.6 eV band gap. Alkali-metal doping (A₃C₆₀) donates electrons and gives a metal with a partly filled $t_{1u}$ band. For Rb₂CsC₆₀ the lecture says the $t_{1u}$ band is half filled and the compound is a superconductor with $T_c = 31$ K.

1. Nitrogen has 5 valence electrons and carbon has 4. NC₅₉ has $4 \times 59 + 5 = 241$ valence electrons, one more than C₆₀.
2. The extra electron has to go into the next empty band, the $t_{1u}$ band. Three electrons make A₃C₆₀ half filled, so the band holds six. One electron fills one sixth of it.
3. A partly filled band means a metal.
4. For metals, the conductivity falls as temperature rises because phonon scattering shortens the relaxation time while the carrier number barely changes. Cooling reverses this, so the conductivity rises on cooling as vibration amplitudes drop and electrons scatter less.
5. The lecture’s doped fullerenes include a superconductor, so low-temperature superconductivity is possible. Whether it happens would depend on the orbital overlap, which the lecture does not let you settle.

(ii), from the lectures. The lecture puts alkali atoms in the holes of the C₆₀ structure and says each donates its electron, as in A₃C₆₀.

1. Each Na⁺ in a hole comes with one electron donated to the fullerene. NaNC₅₉ adds one more electron per molecule.

2. The $t_{1u}$ band goes from 1/6 filled (one electron) to 1/3 filled (two electrons).

3. The band is still partly filled, so the material stays metallic.

4. Conductivity goes up. Why: a band’s density of states (how many electron states sit at each energy) is small at the bottom edge and rises towards the middle. At 1/6 filling the Fermi level sits low in the band, where there are few states. At 1/3 filling it sits higher, where there are more states. More states at the Fermi level means more electrons that can carry current. The lecture’s band picture shows this rising shape.

Lecturer’s model answer (L39 slide 18).

(i) NC59 has 1 more valence electron than C60 (same as C60(−)). It would therefore lead to a partially-filled valence band (1/6-filled t(1u)), and so be metallic. For metals, the conductivity increases on cooling due to the lower vibrational amplitudes leading to less electron scattering. The material might be expected to superconduct at very low temperature, but this could depend on the extent of the overlap of the orbitals.

(ii) NaNC59 would add 1 extra electron per NC59, and so would increase the band filling from 1/6 up to 1/3. Given the general shape of density of states curves, this might be expected to increase the density of states at the Fermi level, leading to an increase in the conductivity.

{Note: if you research azafullerene you’ll see there are some additional complexities, all of which can’t be readily foreseen and so certainly not expected if this was an actual exam question}

SA4: hysteresis in common, structure versus spin as the difference

SA4: There are a number of similarities and differences in the ways that electric dipoles (e.g., polar materials) behave in electric fields and the way that magnetic dipoles (e.g., unpaired electron spins) behave in magnetic fields. List one similarity and one difference, and briefly describe and explain each. (4 marks)

Similarity, from the lectures. Ferroelectrics and ferromagnets both show hysteresis loops. In the ferroelectric loop the polarisation saturates in a strong field ($P_s$), some remains at zero field (remnant polarisation) and a reverse field, the coercive field $E_c$, brings it back to zero. In the ferromagnet the magnetisation follows a hysteresis loop as well, and its shape sets its use. In both cases the dipoles sit in domains that line up with the field and stay partly aligned when it is removed.

Difference, from the lectures. The origin of the dipole differs. A ferroelectric needs a structure with no inversion symmetry: in BaTiO₃ the polar form (P4mm) is ferroelectric and the non-polar form with inversion symmetry (Pm-3m) is paraelectric. So the electric dipole comes from the arrangement of the charges in the structure. A magnetic dipole comes from unpaired electron spins, with the order held by exchange coupling, and needs no change in the structure.

Any one similarity and one difference, each with a reason, gets the marks. The lecturer lists several you can pick from.

Lecturer’s model answer (L39 slide 19).

Examples of similarities:

• Ferroelectric & ferromagnetic – both involve spontaneous dipoles in zero field and exhibit hysteresis loops when the field is cycled in the positive and negative directions, and require negative fields (coercive fields) to reverse the dipole alignment. Both can occur in domains, and transition to randomised states on heating.
• Paraelectric & paramagnetic – both involve gradual alignment of thermally-randomised dipoles with increasing electric/magnetic field strength, reaching saturation at high field and/or low temperature.
• Electric and magnetic susceptibilities are analogous in giving the relative ease with which materials become polarised/magnetised in an electric/magnetic field. Polarisation and magnetisation are analogous in giving the extent to which electric and magnetic dipoles are induced in the material.
• {Not mentioned in L10, but can have antiferroelectric materials, like antiferromagnetic.}

Both a similarity and a difference:

• Dipole ordering occurs due to electrostatic interactions, and/but in the case of electric dipoles this is through the electric field of the dipoles (and results in an attractive interaction – as seen in intermolecular forces) whereas in magnetic dipoles it is through exchange coupling (e.g., superexchange, double exchange).

Examples of differences:

• Electric dipoles arise from symmetry/structural arrangements of ionic charges; magnetic dipoles aren’t related to structural change. As such, the structures of electric dipole materials change in response to an electric field, whereas the structures of magnetic materials don’t change (at least, at most only very slightly … beyond examinable since I didn’t mention this in the lectures, but there can be a very weak effect called magnetostriction.)
• There is no electric dipole equivalent of diamagnetism. All atoms are slightly repelled by a magnetic field (superconductors are very strongly repelled), whereas there is no similar effect for electric fields.
• Electric dipoles can exist across a very wide range of molecule and materials since they just require a net imbalance in the direction/displacement of positive and negative charges, whereas magnetic materials require unpaired electrons and so generally (but not solely) involve metal ions.

Where each point is taught on this site: hysteresis, coercive field and domains on ferroelectric loops and magnetic order; paramagnetism, susceptibility and diamagnetism on magnetic moments; polarisation and $\chi_e$ on dielectrics; superexchange and double exchange on magnetic order.

SA5: methane storage, or negative thermal expansion

SA5: Answer one of the following two questions (i.e., choose A or B) for a total of 8 marks:

A) Metal-organic frameworks (MOFs) show potential for the storage of methane gas. Briefly describe the features of these materials that make them suitable for this application. List and briefly discuss one challenge that needs to be overcome to enable their technological use. (8 marks)

B) Prussian Blue analogues of the form M^IIPt^IV(CN)₆ (M^II = transition metal ion; see picture) display negative thermal expansion (NTE) behaviour.

(i) Explain why most materials expand upon warming. (2 marks)

(ii) Explain the origin of the NTE behaviour of these Prussian Blue analogues. (3 marks)

(iii) Explain how can the coefficient of thermal expansion be varied. (3 marks)

The slide’s picture shows a cubic framework of linked octahedral nodes with cyanide bridges.

Option A, from the lectures

Features, from MOFs for hydrogen, CO₂ and methane:

• High surface area. The lecture seeks a high surface area, and says MOFs reach up to 6000 m² g⁻¹.
• A favourable interaction energy and local structure. In HKUST-1 (Cu₃(btc)₂), CH₄ binds to the bare metal site, creating an ion-induced dipole on the CH₄ that strengthens the binding. Neighbouring CH₄ molecules interact with each other and strengthen it further. The binding enthalpies are $-21.8$ kJ mol⁻¹ in the windows, $-9.4$ kJ mol⁻¹ at the metal only and about $-15$ kJ mol⁻¹ with CH₄–CH₄ interactions.
• Performance. Cu₃(btc)₂ has the highest known room-temperature volumetric methane uptake: 230 cm³(STP) cm⁻³ at 35 bar and 270 cm³(STP) cm⁻³ at 65 bar, meeting the volumetric target set by DOE.

One challenge the lecture supports: getting the interaction energy right. The lecture’s Langmuir analysis, worked for H₂, shows that binding which is too weak gives little uptake at practical pressure, and binding which is too strong needs pressures below 1 bar to release the gas. The lecture works this numbers game only for hydrogen. Applying the same argument to methane is a step the lecture does not take, so treat it as the lecture’s logic carried over, not a result on a slide.

Lecturer’s model answer, option A (L39 slide 20).

Very high surface areas achieved through construction of porous frameworks, leading to very high methane gas uptake both volumetrically and gravimetrically. Ability to optimise the interaction energy of the methane molecules with the surface of the framework, thereby leading to gas uptake and release at ideal conditions (i.e., room temperature, pressures up to 250 atm). Design approaches include the use of bare metal sites to bind the methane molecules.

Examples of challenges:

• Optimise the MOF-methane interaction energy to allow uptake and release in the right temperature and pressure range. This should not be too low, since high pressures/low temperature would be needed, but also not too high, since this would cause the uptake/release to be highly exo/endothermic.
• Maximise the surface area – this involves the construction of highly open framework, which requires the use of large, light building units. A particular challenge here is making such frameworks sufficiently chemically and mechanically stable to withstand gas sorption/desorption cycles.
• Densification – ie. work out how to pack as much MOF into a given volume as possible, thereby maximising the gas storage capabilities. Ideas here include processing of the crystals (which has been shown to decrease their uptake ability) and sol-gel processing.

So the “interaction energy” challenge worked above is exactly the lecturer’s first challenge. You only need one.

Option B, from the lectures

(i) Atoms vibrate along the bond with greater amplitude at higher temperature. The interaction potential is asymmetric: steep on the squeezed side and gentle on the stretched side. So the average bond length increases.

(ii) In these cyanide frameworks the links are long and thin. Low-energy transverse vibrations, at 90° to the bond, bend the M–C≡N–Pt link sideways like a skipping rope. When a link bows, its two ends move closer, so the metal-to-platinum distance shrinks. The amplitude of this sideways motion grows with temperature, and it outweighs the small lengthening of each bond. The framework contracts: NTE.

(iii) The topic on tuning NTE gives two routes.

1. Change the metal M. The strength of the M–N bond sets the stiffness of the bridge. Stronger bonding gives smaller sideways amplitude and smaller NTE. The ionic radius falls across the period and the crystal field stabilisation energy changes: Ni(II) has $\mathrm{CFSE} = -1.2\,\Delta_o$ and the strongest bonds, while Zn(II) has $\mathrm{CFSE} = 0$ and weaker bonds.
2. Change the guests in the pores. The slide’s cadmium and zinc-platinum examples show the expansion changing sign when guest molecules enter or leave. The slide’s model answer mentions this as an alternative route.

Lecturer’s model answer, option B (L39 slide 20).

(i) Longitudinal vibration at higher temperatures causes an average lengthening of bonding interactions due to the asymmetry of the interaction potential.

(ii) Transverse vibration (like a skipping rope) of the Pt-CN-M bridges leads to the Pt…M distance getting shorter at higher temperatures due to the tranverse vibrational amplitude increasing.

(iii) Changing the metal ion M^II leads to a change in the strength of the M^II-N bond, which changes the ‘stiffness’ of the Pt-CN-M bridges. Strong bonding (eg. associated with large crystal field stabilisation energies, e.g. Ni(II) > Zn(II)) leads to decreased NTE since the amplitude of the transverse vibration is less. (Could alternatively talk about the role of adsorbed guest molecules in the pores.)

Exam detail

What the five model answers show about marking:

• Short and causal. Each mark is one idea plus its reason (“doubling along a and b, so spots halve”, “because the charges matter more than d”).
• Use the given equation. SA2 asks you to rank “on the basis of the Born-Mayer equation”, so name which terms change ($z_A$, $z_B$, $d$) and which stay fixed.
• Count electrons, then fill bands. SA3 is electron counting → band filling → metal or not → temperature trend.
• Pick one and explain it. SA4 and SA5 accept any valid point from a list, as long as it is explained.
• A “very good answer” goes one step further, like the faint extra peaks in SA1.

Pitfall

Do not copy one of these model answers into a different question. Questions like SA2 ask you to rank with the equation, and a different set of compounds changes the charge products and sizes. Redo the steps each time.

Recall

For SA2, why do you sort by charge product before looking at distance d?

The charge product $\lvert z_A z_B \rvert$ changes the enthalpy by much more than the plausible change in $d$ between compounds. The lecture shows this with melting points: moving from charge 1 to charge 2 (NaCl to CaO) changes more than swapping to a bigger ion of the same charge (CaO to BaO). Use $d$ only to break ties.