Compounds, Amorphous Solids, Solid Solutions and the Structure of Polymers

This chapter belongs to the Engineering Sciences (XE) paper. It answers the middle headings of Section XE2.1 of XE2 Materials Science, one of the nine optional sections, any two of which a candidate chooses in the exam: “Crystalline structure of compounds: NaCl, CsCl, ZnS (Zinc blend and Wurtzite), Perovskite, Spinels; Pauling’s rule for structure of ionic compounds. Structure of amorphous materials: Crystalline and glassy silica; Fused silica and soda-lime glass. Solid solutions: Interstitial and substitutional; Hume-Rothery rules. Structure of polymers: Monomers and polymers … Natural and vulcanised rubber.” It teaches how radius ratio and Pauling’s rules fix the structure of ionic solids, how the common compound structures are built from the close-packed arrays and voids of the previous chapter, what makes silica crystalline or glassy, when two elements form a solid solution, and how a polymer chain is built, shaped and linked. The numericals are density of an ionic crystal, perovskite tolerance factor, degree of polymerisation, percentage crystallinity from density and the extent of sulfur crosslinking. Defects follow in the next chapter.

Ionic crystals: radius ratio and Pauling’s rules

An ionic crystal is built from spheres of two sizes, a small cation of radius r and a large anion of radius R, held together by electrostatic attraction that acts equally in all directions. The structure is then fixed by geometry and charge balance. Pauling’s first rule says each cation sits in a polyhedron of anions, and the coordination number is set by the radius ratio r/R: the cation should touch all its neighbours, and the anions should not touch each other. The limit comes when the anions just touch, which is the same void calculation as in the close-packed chapter.

Radius ratio and coordination
r/RCoordination numberAnion polyhedronLimit from geometry
0.155 to 0.2253Triangle0.155 = 2/√3 − 1
0.225 to 0.4144Tetrahedron0.225 = √6/2 − 1
0.414 to 0.7326Octahedron0.414 = √2 − 1
0.732 to 1.08Cube0.732 = √3 − 1

Second rule (electrostatic valence). The bond strength of a cation is s = z/CN, its charge divided by its coordination number. The strengths of the bonds reaching an anion add up to the magnitude of its charge. In MgO, s = 2/6 = ⅓ and each O²⁻ has six Mg²⁺ neighbours, so 6 × ⅓ = 2. In MgAl₂O₄, Mg²⁺ (CN 4) gives ½ and Al³⁺ (CN 6) gives ½; an O²⁻ with one Mg and three Al neighbours gets ½ + 3 × ½ = 2. Third rule: polyhedra prefer to share corners, less willingly edges and least willingly faces, because shared edges and faces bring the cations closer and raise their repulsion. Fourth rule: cations of high charge and low coordination avoid sharing polyhedron elements with each other. Fifth rule (parsimony): the number of different kinds of environments in a crystal tends to be small.

⚠️ A rule, not a law
The radius-ratio rule predicts the right coordination for many strongly ionic solids but fails for a good number of compounds, especially those with substantial covalent character; it says nothing when the ratio sits close to a limit, and it cannot separate structures of the same coordination, such as zinc blende and wurtzite. Use it to choose the most likely structure, and when a question gives a ratio and asks the coordination, read the table; do not claim the prediction is exact.

NaCl, CsCl, zinc blende and wurtzite

Each simple compound structure is a lattice plus a motif, and several are a close-packed array of anions with cations in a fraction of the voids. Rock salt (NaCl): the FCC lattice with Na⁺ at (0, 0, 0) and Cl⁻ at (½, 0, 0), that is, an FCC array of Cl⁻ with Na⁺ in all octahedral voids; CN 6 : 6, 4 formula units per cell. Examples: MgO, LiF, FeO. The ratio Na⁺/Cl⁻ ≈ 0.56 lies in the octahedral range. Caesium chloride: a simple cubic lattice with a two-ion motif, Cs⁺ at (0, 0, 0) and Cl⁻ at (½, ½, ½); CN 8 : 8, 1 formula unit per cell, and the ratio is close to 1. The ordered alloy β′-brass (CuZn) has the same structure.

ZnS has two polymorphs, both with cation and anion CN 4 : 4, both made of S²⁻ close-packed layers with Zn²⁺ in half of the tetrahedral voids (every second one, so that each Zn has four S neighbours and vice versa). Zinc blende (sphalerite) has the ABC stacking: an FCC lattice with a two-atom motif, 4 formula units per cubic cell, exactly the diamond structure with two kinds of atom (GaAs, CdTe, InP). Wurtzite has the AB stacking: a hexagonal lattice, 2 formula units per cell (ZnO, GaN, CdS, AlN). Their local environments are the same, and they differ only in the stacking of the layers, just as FCC differs from HCP.

Compound structures compared
StructureLatticeCN (cation : anion)Formula units per cell
NaClFCC6 : 64
CsClSimple cubic8 : 81
Zinc blendeFCC4 : 44
WurtziteHexagonal4 : 42

Density. ρ = (formula units per cell × M)/(V_c N_A), where M is the formula mass. NaCl, with a = 0.5640 nm, M = 22.99 + 35.45 = 58.44 g/mol: ρ = 4 × 58.44/[6.022 × 10²³ × (0.5640 × 10⁻⁷ cm)³] = 233.76/108.04 = 2.16 g/cm³. Counting 8 ions instead of 4 formula units is the usual factor-of-two slip.

Perovskite and spinel

Perovskite, ABO₃ (CaTiO₃, BaTiO₃, SrTiO₃): the large A cation (CN 12) sits at the cube corners, the small B cation (CN 6) at the body centre and O²⁻ at the three face-centre positions, so there is 1 formula unit per cell and the charges satisfy z_A + z_B = 6. Equivalently, A and O together form an FCC-like array (AO₃ layers) with B in one quarter of the octahedral voids, those surrounded only by oxygen. The ideal cubic cell has A–O distance a/√2 and B–O distance a/2, so touching ions would need (r_A + r_O) = √2 (r_B + r_O). The Goldschmidt tolerance factor t = (r_A + r_O)/[√2 (r_B + r_O)] measures the misfit; t near 1 favours the cubic structure and departures distort it. BaTiO₃ is cubic above about 120 °C and tetragonal and ferroelectric below it, since the small Ti⁴⁺ shifts off centre.

Spinel, AB₂O₄ (MgAl₂O₄, Fe₃O₄): an FCC array of O²⁻ with a cubic cell of 8 formula units, 32 O²⁻, 64 tetrahedral and 32 octahedral voids. In a normal spinel the 8 A²⁺ ions occupy 8 of the 64 tetrahedral voids (one eighth) and the 16 B³⁺ ions occupy 16 of the 32 octahedral voids (one half). In an inverse spinel the A²⁺ ions and half of the B³⁺ ions share the octahedral sites and the other half of the B³⁺ ions occupy tetrahedral sites, written B[AB]O₄. Magnetite is inverse, Fe³⁺[Fe²⁺Fe³⁺]O₄; ZnFe₂O₄ is normal.

⚠️ Count the sites, not the ions
In a spinel only 1/8 of the tetrahedral and 1/2 of the octahedral voids are filled, so the structure is mostly empty voids. Do not apply the NaCl idea (all octahedral voids filled) to it.

Amorphous solids: crystalline and glassy silica, fused silica, soda-lime glass

The unit of every silica structure is the SiO₄ tetrahedron, with Si–O about 0.16 nm. Tetrahedra join only at corners, each oxygen bridging two silicon atoms, which gives the formula SiO₂. In crystalline silica (quartz, cristobalite, tridymite) the Si–O–Si angles and the way tetrahedra rotate about each other repeat, so there is long-range order and a sharp melting point. In glassy silica the tetrahedra are the same and the local order is preserved, but the bridging angles and rotations vary randomly, so there is short-range order (neighbours, bond lengths) and no long-range order. Diffraction shows sharp peaks for the crystal and broad halos for the glass.

Fused silica is pure SiO₂ glass made by melting quartz: a fully connected network with very high softening temperature (above 1500 °C) and very low thermal expansion, hence thermal-shock resistance. Soda-lime glass (windows, bottles) is roughly 70 to 75 mass % SiO₂ with Na₂O and CaO making up most of the rest. SiO₂ is the network former. Na₂O is a modifier: each Na₂O breaks a Si–O–Si bridge and creates two non-bridging oxygens that carry the Na⁺, which loosens the network, lowers the softening temperature to roughly 700 °C and makes the glass easy to form. CaO makes the glass more durable against water, since the Na⁺ alone would let it dissolve. The price is a higher thermal expansion and poorer thermal-shock resistance than fused silica.

🎯 No melting point, only a glass transition
On cooling, a crystal shows a sudden drop in specific volume at its melting point. A glass-forming liquid becomes more viscous without crystallising, and its volume-temperature line only changes slope at the glass transition temperature T_g, which depends on cooling rate; below it the structure is frozen in.

Solid solutions and the Hume-Rothery rules

A solid solution is a single crystalline phase in which solute atoms are dissolved in the lattice of the solvent, so the structure of the solvent is kept. In a substitutional solution the solute atoms replace solvent atoms on lattice sites (Cu in Ni). In an interstitial solution the small solute atoms sit in the voids of the solvent lattice (C, N, H, O, B in iron); the void sizes of the previous chapter limit how much can dissolve, which is why carbon dissolves far more in FCC than in BCC iron. Substitutional solutions may be random or ordered.

  • Size factor: the atomic radii of solute and solvent differ by less than about 15 %, where the difference is (r_solute − r_solvent)/r_solvent.
  • Crystal structure: solute and solvent have the same crystal structure; this is required for complete solubility.
  • Electronegativity: the two are similar; a large difference favours compound formation instead of a solution.
  • Valence: other things equal, a metal dissolves more readily a metal of higher valence than one of lower valence.

Copper and nickel meet all four (radii 0.128 and 0.125 nm, a 2.3 % difference; both FCC; similar electronegativity) and are completely soluble in each other. Failing any rule limits solubility but does not forbid it: the rules are guides for substitutional systems, and an interstitial solute needs a small enough atom, not Hume-Rothery matching. A solid solution is not a mixture: the solute is dissolved at atomic scale in a single phase, and the lattice parameter changes smoothly with composition.

Polymer chains: monomers, bonding, degree of polymerisation and tacticity

A polymer is a molecule of many repeat units (mers) joined by covalent bonds; a monomer is the small molecule from which they come. In addition (chain-growth) polymerisation monomers with a C=C double bond add one after another by initiation, propagation and termination, and nothing is lost, so the repeat unit has the composition of the monomer: PE, PP, PVC, PS and PTFE. In condensation (step-growth) polymerisation bifunctional monomers react and release a small molecule, usually water or HCl, so the repeat unit has less mass than the monomers combined: nylon 6,6 from hexamethylenediamine and adipic acid (releasing water), and PET from ethylene glycol and terephthalic acid.

Bonding. Along the chain the atoms are joined by strong covalent bonds (C–C about 0.154 nm, about 350 kJ/mol, with a tetrahedral bond angle of 109.5° that makes the carbon backbone zigzag). Between chains only secondary bonds act, van der Waals forces and, where groups allow, dipole or hydrogen bonds, which are weaker by one to two orders of magnitude. This duality explains why polymers soften or melt long before any covalent bond breaks.

Degree of polymerisation. With the convention used here, DP = M̄_n/m, where M̄_n is the number-average molar mass of the polymer and m the molar mass of one repeat unit. For PVC, repeat unit –CH₂–CHCl–, m = 62.5 g/mol, so M̄_n = 50 000 g/mol gives DP = 800. Each repeat unit of a vinyl polymer adds two backbone carbons, so a chain of DP n has 2n C–C bonds; fully extended in the zigzag, each bond advances 0.154 × sin(109.5°/2) = 0.126 nm along the chain. For DP = 10 000 the extended length is 2 × 10 000 × 0.126 nm = 2.5 μm, although the real coil is far shorter.

Configuration vs conformation. Configuration is the arrangement fixed by the covalent bonds, such as the side-group placement, cis or trans geometry about a double bond; changing it requires breaking bonds. Conformation is the shape of the chain produced by rotation about single bonds, which changes freely with temperature or stress without breaking anything. For a vinyl chain –CH₂–CHR– the side groups R give three tacticities: isotactic (all R on the same side of the backbone), syndiotactic (R alternating from side to side) and atactic (R randomly placed). Regular chains (isotactic, syndiotactic) can pack in crystals; atactic chains cannot.

⚠️ Rotation cannot make an atactic chain isotactic
Rotation about C–C bonds changes only conformation. Tacticity is configuration, so no amount of bending, heating or stretching converts one into another without breaking bonds.

Crystallinity, copolymers, common polymers, crosslinking and rubber

Long chains can crystallise only by folding into thin lamellae, and a large fraction of chains remains disordered, so most crystallisable polymers are semi-crystalline (crystalline lamellae in an amorphous matrix, grouped in spherulites); a polymer whose chains are irregular is wholly amorphous. Crystallinity raises density, stiffness, melting point and opacity. It needs linear, regular chains with small side groups, and it is helped by slow cooling. Mass-fraction crystallinity from density: X_c = ρ_c(ρ_s − ρ_a)/[ρ_s(ρ_c − ρ_a)]. For polyethylene with ρ_c = 1.000 and ρ_a = 0.870 g/cm³ and a sample of ρ_s = 0.960 g/cm³: X_c = 1.000 × 0.090/(0.960 × 0.130) = 0.721, or 72.1 %.

Common addition polymers
PolymerRepeat unitm (g/mol)Structural point
PE–CH₂–CH₂–28.05No side group; linear HDPE packs well, branched LDPE less
PP–CH₂–CH(CH₃)–42.08Isotactic crystallises; atactic is soft and amorphous
PVC–CH₂–CHCl–62.50Polar Cl; mostly atactic, largely amorphous
PTFE–CF₂–CF₂–100.02Fluorine sheath; highly crystalline, inert, low friction
PS–CH₂–CH(C₆H₅)–104.15Bulky phenyl; commercial atactic PS is amorphous and glassy

Copolymers contain two or more kinds of mer: random (A and B in no order), alternating (ABAB…) and block (long runs of A joined to long runs of B). Random copolymerisation disturbs the regularity and usually lowers crystallinity. Crosslinking joins chains by covalent bridges into a network that can swell but cannot dissolve or flow. Natural rubber is cis-1,4-polyisoprene (repeat unit C₅H₈, 68.12 g/mol), a soft, weak material of low T_g with coiled chains; its trans isomer is the hard, crystalline gutta-percha. Vulcanisation heats rubber with sulfur, which forms bridges of one or more S atoms between chains near the C=C double bonds. A few mass percent of sulfur gives an elastic rubber that snaps back, because crosslinks stop chains from flowing past each other; much more sulfur gives the rigid ebonite.

Worked example (crosslinking). 100 g of natural rubber is vulcanised with 3.0 g S, each crosslink bridge being two S atoms joining two repeat units. Repeat units: 100/68.12 = 1.468 mol. S: 3.0/32.06 = 0.0936 mol, giving 0.0468 mol of bridges and 2 × 0.0468 = 0.0936 mol of crosslinked units. Fraction crosslinked = 0.0936/1.468 = 0.064, or 6.4 % of the repeat units.

Key takeaways

  • Radius ratio r/R sets CN: 0.155–0.225 gives 3, 0.225–0.414 gives 4, 0.414–0.732 gives 6, 0.732–1 gives 8. Pauling’s second rule makes the bond strengths z/CN reaching an anion sum to its charge.
  • NaCl is FCC with 4 formula units and CN 6 : 6; CsCl is simple cubic with 1 unit and CN 8 : 8; zinc blende (FCC, 4 units) and wurtzite (hexagonal, 2 units) have CN 4 : 4 and differ only in stacking.
  • Perovskite ABO₃ has A at corners, B at the body centre, O at face centres, tolerance factor t = (r_A + r_O)/[√2 (r_B + r_O)]. Spinel AB₂O₄ fills 1/8 of tetrahedral and 1/2 of octahedral voids of an FCC oxygen array; inverse spinel puts half the B ions in tetrahedral sites.
  • Silica is corner-sharing SiO₄ tetrahedra, ordered over long range in the crystal and only over short range in the glass. Na₂O modifies the network and lowers the softening temperature; CaO restores durability.
  • Hume-Rothery: size within about 15 %, same crystal structure, similar electronegativity, favourable valence. Small atoms (C, N, H) dissolve interstitially in voids; larger ones substitute.
  • Polymers: strong covalent bonds along the chain, weak secondary bonds between chains; DP = M̄_n/m; tacticity is configuration, bond rotation changes only conformation; crosslinks turn chains into a network, as sulfur does in vulcanised rubber.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. In a strongly ionic compound the cation-to-anion radius ratio is r/R = 0.30. Pauling’s first rule (radius-ratio rule) predicts the cation to be surrounded by

    1. six anions at the corners of an octahedron, since the ratio exceeds the 0.225 limit of tetrahedral sites
    2. four anions at the corners of a tetrahedron, since the ratio lies between 0.225 and 0.414
    3. eight anions at the corners of a cube, since the ratio is well above the 0.155 limit
    4. three anions at the corners of a triangle, since the ratio lies between 0.155 and 0.225, a planar coordination
    Show answer

    Answer: B — four anions at the corners of a tetrahedron, since the ratio lies between 0.225 and 0.414

    The coordination windows are 0.155–0.225 (CN 3), 0.225–0.414 (CN 4), 0.414–0.732 (CN 6) and 0.732–1 (CN 8). A ratio of 0.30 lies in the tetrahedral window. CN 6 would need r/R of at least 0.414, so that reading, the tempting one, is wrong; CN 8 needs at least 0.732 and CN 3 applies only below 0.225.

  2. In silica each Si⁴⁺ has coordination number 4 towards O²⁻. Pauling’s electrostatic valence rule then requires every O²⁻ to be bonded to how many Si⁴⁺ ions?

    1. One, so that every oxygen is terminal and carries one unshared bond
    2. Four, so that each oxygen sits at the centre of its own tetrahedron of silicon
    3. Three, so that each oxygen receives bond strength 3 from silicon in total
    4. Two, so that each oxygen bridges two tetrahedra and receives 2 × 1 = 2
    Show answer

    Answer: D — Two, so that each oxygen bridges two tetrahedra and receives 2 × 1 = 2

    The bond strength of Si⁴⁺ is z/CN = 4/4 = 1. The bonds arriving at O²⁻ must sum to 2, so two Si⁴⁺ neighbours are needed: every oxygen is a corner shared by two SiO₄ tetrahedra, which gives SiO₂. One neighbour would give 1 and leave the oxygen under-bonded; three or four would over-bond it.

  3. Zinc blende and wurtzite are two forms of ZnS. Which statement about them is correct?

    1. Both have CN 4 : 4 with Zn²⁺ in half of the tetrahedral voids of close-packed S²⁻ layers; they differ in the layer stacking, ABC… in blende and AB… in wurtzite
    2. Blende has CN 4 : 4 and wurtzite has CN 6 : 6, since the hexagonal stacking brings extra sulfur neighbours closer to each zinc ion, as it does in the hexagonal metals
    3. Wurtzite is the cubic form with 4 formula units per cell and blende is hexagonal with 2 formula units per cell, both with Zn²⁺ in all tetrahedral voids
    4. Blende is FCC in its S²⁻ sublattice, wurtzite is BCC, and the two have different numbers of Zn²⁺ neighbours around each sulfur ion because of the different packing
    Show answer

    Answer: A — Both have CN 4 : 4 with Zn²⁺ in half of the tetrahedral voids of close-packed S²⁻ layers; they differ in the layer stacking, ABC… in blende and AB… in wurtzite

    Blende is the ABC (FCC) form, 4 formula units per cubic cell; wurtzite is the AB (HCP) form, 2 per hexagonal cell. In both, Zn²⁺ fills every second tetrahedral void, so the coordination is 4 : 4. The cubic/hexagonal labels are interchanged in one distractor, and the BCC claim is false because neither is a body-centred array.

  4. Which description of the rock-salt (NaCl) structure is correct?

    1. A simple cubic lattice with a two-ion motif, one formula unit per cell and CN 8 : 8, as found in CsCl and ordered β′-brass
    2. An FCC array of Cl⁻ with Na⁺ filling all the tetrahedral voids, eight Na⁺ per cell, each surrounded by four Cl⁻ ions
    3. An FCC lattice with Na⁺ at (0, 0, 0) and Cl⁻ at (½, 0, 0): Na⁺ fills all octahedral voids, CN 6 : 6, four formula units per cell
    4. An FCC array of Cl⁻ with Na⁺ in half the tetrahedral voids, CN 4 : 4, four formula units per cell, the arrangement of zinc blende
    Show answer

    Answer: C — An FCC lattice with Na⁺ at (0, 0, 0) and Cl⁻ at (½, 0, 0): Na⁺ fills all octahedral voids, CN 6 : 6, four formula units per cell

    NaCl is an FCC lattice with a two-ion motif. Equivalently the Cl⁻ ions form an FCC array and Na⁺ occupies all four octahedral voids per cell, giving CN 6 : 6. The simple cubic description is CsCl, filling all the tetrahedral voids would give an anti-fluorite-type arrangement, and the half-filled tetrahedral description is zinc blende.

  5. In a normal spinel AB₂O₄ the O²⁻ ions form an FCC array. Which fraction of the voids is occupied by cations?

    1. All octahedral voids and none of the tetrahedral voids, as in rock salt
    2. One half of the tetrahedral voids and one eighth of the octahedral voids, by A and B ions respectively
    3. One half of both the tetrahedral and the octahedral voids, shared equally between the A and B ions
    4. One eighth of the tetrahedral voids (A ions) and one half of the octahedral voids (B ions)
    Show answer

    Answer: D — One eighth of the tetrahedral voids (A ions) and one half of the octahedral voids (B ions)

    Per cell there are 32 O²⁻, 64 tetrahedral and 32 octahedral voids, with 8 A and 16 B ions. The 8 A²⁺ ions fill 8/64 = 1/8 of the tetrahedral voids, and the 16 B³⁺ ions fill 16/32 = 1/2 of the octahedral voids. The reversed fractions in the distractor swap the roles of the sites, and rock-salt filling applies to NaCl, not spinel.

  6. Adding Na₂O to silica glass lowers its softening temperature mainly because the Na₂O

    1. raises the number of long-range ordered regions, so the glass flows like a crystal
    2. fills the voids of the network with Na⁺ and so increases the density of crosslinks between tetrahedra
    3. breaks Si–O–Si bridges into non-bridging oxygens, which loosens the network connectivity
    4. replaces silicon by sodium on the tetrahedral sites, so every tetrahedron becomes smaller and weaker
    Show answer

    Answer: C — breaks Si–O–Si bridges into non-bridging oxygens, which loosens the network connectivity

    Each Na₂O supplies an oxygen that terminates a Si–O–Si bridge, forming two non-bridging oxygens balanced by Na⁺. A less connected network softens at a lower temperature. Na⁺ does not substitute for Si⁴⁺ (the charges and sizes are wrong) and adding Na₂O reduces, not increases, the connectivity, which is what the crosslink distractor reverses.

  7. A pair of metals is expected to show complete solid solubility in each other under the Hume-Rothery rules. Which pair?

    1. Both FCC, radii 0.128 and 0.180 nm, similar electronegativity and valence
    2. One FCC and one BCC, radii 0.125 and 0.126 nm, similar electronegativity and valence
    3. Both FCC, radii 0.143 and 0.144 nm, electronegativities 0.9 and 2.4, same valence
    4. Both FCC, radii 0.128 and 0.125 nm, similar electronegativity and valence
    Show answer

    Answer: D — Both FCC, radii 0.128 and 0.125 nm, similar electronegativity and valence

    Only the pair with the close FCC radii 0.128 and 0.125 nm satisfies every rule: the same crystal structure, a size difference of 2.3 % (below 15 %), similar electronegativity and valence. The 0.128 and 0.180 nm pair differs in size by 41 %, the FCC and BCC pair in crystal structure, and the last in electronegativity, which would favour a compound rather than a solution.

  8. Which statement correctly contrasts addition and condensation polymerisation?

    1. Addition polymerisation releases water at every step, and condensation polymerisation releases nothing, so the repeat unit of a condensation polymer has the monomer’s composition
    2. Condensation polymerisation needs a C=C double bond in each monomer, whereas addition polymerisation needs two reactive end groups of different kinds
    3. Both release a small molecule, but addition polymerisation does it only at termination, so in either case the repeat-unit mass equals the sum of the monomer masses
    4. Addition polymerisation loses no atoms, so the repeat unit has the monomer’s composition, whereas condensation polymerisation releases a small molecule such as H₂O
    Show answer

    Answer: D — Addition polymerisation loses no atoms, so the repeat unit has the monomer’s composition, whereas condensation polymerisation releases a small molecule such as H₂O

    In addition polymerisation (PE, PVC, PS) monomers with C=C add without loss. In condensation polymerisation (nylon 6,6, PET) bifunctional monomers react and release water or HCl, so the repeat unit is lighter than the monomers combined. The other statements swap these features or invent a release step in addition polymerisation.

  9. A sample of isotactic polypropylene is heated, stretched, bent and allowed to relax, with no bond ever broken. It can then end up as which of the following?

    1. Atactic polypropylene, because heating randomises the side-group positions
    2. Syndiotactic polypropylene, because stretching alternates the side groups along the chain
    3. A chain of a different conformation but of the same configuration, still isotactic
    4. Polyethylene, because bending removes the CH₃ side groups from the backbone
    Show answer

    Answer: C — A chain of a different conformation but of the same configuration, still isotactic

    Tacticity is a configuration, defined by which side of the backbone each CH₃ group is attached to, and it can change only if bonds are broken and remade. Heating, stretching and bending change only conformation through rotation about C–C bonds, so the chain stays isotactic. The atactic and syndiotactic options wrongly treat tacticity as a conformation, and removing CH₃ groups needs chemical reaction.

  10. Which polymer is most likely to be semi-crystalline under normal processing?

    1. Commercial atactic polystyrene, because the bulky phenyl groups pack in an orderly way
    2. Linear polyethylene, because it has no side groups and its chains fold regularly
    3. A random copolymer of ethylene and propylene with equal amounts of both mers
    4. Atactic polypropylene, because the methyl groups are all on the same side of the chain
    Show answer

    Answer: B — Linear polyethylene, because it has no side groups and its chains fold regularly

    Linear polyethylene has the simplest, most regular chain, so it crystallises to a high degree. Atactic polystyrene and atactic polypropylene have irregular side-group placement and stay amorphous (and the methyl groups of atactic PP are not all on one side, that is isotactic). A random copolymer with a 1 : 1 mix has little regularity.

  11. Which of the following statements about Pauling’s rules and the radius-ratio rule are correct?

    1. The sum of the electrostatic bond strengths z/CN reaching an anion equals the magnitude of the anion’s charge
    2. For a strongly ionic compound with r/R = 0.50 the radius-ratio rule predicts octahedral coordination of the cation
    3. Sharing of polyhedron faces between neighbouring cation polyhedra stabilises a structure more than sharing of corners does
    4. The radius-ratio rule predicts the structure of ZnS exactly, since the bonding in ZnS is purely ionic
    Show answer

    Answer: A — The sum of the electrostatic bond strengths z/CN reaching an anion equals the magnitude of the anion’s charge; B — For a strongly ionic compound with r/R = 0.50 the radius-ratio rule predicts octahedral coordination of the cation

    The valence rule and the 0.414 to 0.732 window for CN 6 are both correct. The third rule says corner sharing is preferred, then edges, then faces, because shared faces bring cations too close and raise repulsion. ZnS has substantial covalent character, so the radius-ratio rule is not exact for it.

  12. Which of the following statements about wurtzite and spinel structures are correct?

    1. In wurtzite ZnS the S²⁻ ions are hexagonally close-packed and Zn²⁺ occupies half of the tetrahedral voids
    2. In a normal spinel AB₂O₄ the B³⁺ ions occupy tetrahedral voids and the A²⁺ ions occupy octahedral voids
    3. In zinc blende each Zn²⁺ has six S²⁻ neighbours, the same coordination as in rock salt
    4. In magnetite, an inverse spinel, the Fe³⁺ ions are split equally between tetrahedral and octahedral sites
    Show answer

    Answer: A — In wurtzite ZnS the S²⁻ ions are hexagonally close-packed and Zn²⁺ occupies half of the tetrahedral voids; D — In magnetite, an inverse spinel, the Fe³⁺ ions are split equally between tetrahedral and octahedral sites

    Wurtzite is the HCP form of ZnS with Zn in half the tetrahedral voids. In magnetite, Fe³⁺[Fe²⁺Fe³⁺]O₄, one Fe³⁺ per formula unit is tetrahedral and one is octahedral. In a normal spinel it is A²⁺ that is tetrahedral and B³⁺ that is octahedral, the reverse of the statement, and zinc blende is 4 : 4, not 6 : 6.

  13. Which of the following statements about polymer structure are correct?

    1. Vulcanised natural rubber cannot be dissolved or remelted into a flowing liquid, because its chains are linked into a covalent network
    2. In a polymer the secondary bonds between chains are stronger than the covalent bonds along the chain, which is why polymers are soft
    3. Isotactic and syndiotactic chains are regular enough to crystallise, whereas atactic chains of a bulky side-group polymer usually cannot
    4. The repeat unit of polyethylene is –CH₂–CH₂– and that of PTFE is –CF₂–CF₂–, so the molar mass of the PTFE unit is below that of the PVC unit
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    Answer: A — Vulcanised natural rubber cannot be dissolved or remelted into a flowing liquid, because its chains are linked into a covalent network; C — Isotactic and syndiotactic chains are regular enough to crystallise, whereas atactic chains of a bulky side-group polymer usually cannot

    Crosslinks make rubber a network, which swells but does not dissolve or flow. Regular chains crystallise and atactic bulky-side-group chains do not. Secondary bonds are weaker, not stronger, than covalent ones. The PTFE unit (100.02 g/mol) is heavier than the PVC unit (62.50 g/mol), so the statement about molar masses is false.

  14. Sodium chloride has the rock-salt structure with lattice parameter a = 0.5640 nm. Take M(Na) = 22.99 g/mol, M(Cl) = 35.45 g/mol and N_A = 6.022 × 10²³ mol⁻¹. Calculate its theoretical density in g/cm³, to two decimal places.

    Numerical answer — type the value.

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    Answer: 2.16

    The cell holds 4 formula units. V_c = (0.5640 × 10⁻⁷ cm)³ = 1.7940 × 10⁻²² cm³. ρ = 4 × 58.44/(1.7940 × 10⁻²² × 6.022 × 10²³) = 233.76/108.04 = 2.16 g/cm³. Counting 8 formula units would double it to 4.32.

  15. For the cubic perovskite BaTiO₃ use the ionic radii r(Ba²⁺, coordination 12) = 0.161 nm, r(Ti⁴⁺, coordination 6) = 0.0605 nm and r(O²⁻) = 0.140 nm. Compute the Goldschmidt tolerance factor t = (r_A + r_O)/[√2 (r_B + r_O)] to two decimal places.

    Numerical answer — type the value.

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    Answer: 1.06

    Numerator: 0.161 + 0.140 = 0.301 nm. Denominator: √2 × (0.0605 + 0.140) = 1.4142 × 0.2005 = 0.28355 nm. t = 0.301/0.28355 = 1.0615, that is 1.06. A value above 1 means the A–O distance is too large for the B–O distance, which is why the small Ti⁴⁺ can shift off centre.

  16. Polyvinyl chloride has the repeat unit –CH₂–CHCl–. Take C = 12.01, H = 1.008 and Cl = 35.45 g/mol. A sample has a number-average molar mass M̄_n = 75 000 g/mol. Using DP = M̄_n/m, with m the molar mass of one repeat unit, find the number-average degree of polymerisation, rounded to the nearest integer.

    Numerical answer — type the value.

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    Answer: 1200

    The repeat unit is C₂H₃Cl: m = 2 × 12.01 + 3 × 1.008 + 35.45 = 24.02 + 3.024 + 35.45 = 62.494 g/mol. DP = 75 000/62.494 = 1200.1, so 1200. Using the mass of the vinyl chloride molecule is the same here; using the two-carbon backbone count 2n instead would double the answer.

  17. A polyethylene sample has density ρ_s = 0.945 g/cm³. Take the density of the fully crystalline polymer ρ_c = 1.000 g/cm³ and of the fully amorphous polymer ρ_a = 0.870 g/cm³. Use the mass-fraction relation X_c = ρ_c(ρ_s − ρ_a)/[ρ_s(ρ_c − ρ_a)]. Find the percentage crystallinity, rounded to the nearest integer.

    Numerical answer — type the value.

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    Answer: 61

    X_c = 1.000 × (0.945 − 0.870)/[0.945 × (1.000 − 0.870)] = 0.0750/0.12285 = 0.6105, that is 61 %. Using the simple linear (volume-fraction) interpolation (0.945 − 0.870)/0.130 = 57.7 % ignores that the crystalline mass fraction and volume fraction differ.

  18. 100 g of natural rubber (repeat unit C₅H₈, 68.12 g/mol) is vulcanised with 5.0 g of sulfur (32.06 g/mol). Assume all the sulfur forms crosslink bridges, each bridge consisting of exactly four S atoms and joining exactly two repeat units, with no repeat unit in more than one bridge. What percentage of the repeat units is crosslinked, to one decimal place?

    Numerical answer — type the value.

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    Answer: 5.3

    Repeat units = 100/68.12 = 1.468 mol. S = 5.0/32.06 = 0.1560 mol, so bridges = 0.1560/4 = 0.0390 mol. Crosslinked units = 2 × 0.0390 = 0.0780 mol. Fraction = 0.0780/1.468 = 0.0531, that is 5.3 %. Forgetting that each bridge joins two units would give 2.7 %.