This quiz covers Lessons 11–15: Polymers, Solubility, Atomic Structure and Models, Isotopes and Relative Atomic Mass, and the Periodic Table. 20 multiple choice + 3 short answer questions.
Lesson Summaries — Quick Review
Addition polymerisation: alkene monomers (C=C) join with no byproduct. Condensation polymerisation: bifunctional monomers form ester/amide links, releasing H₂O. Properties depend on chain length, branching, cross-linking, and IMF type. Thermoplastics soften on heating (IMFs); thermosets are permanently cross-linked (covalent bonds).
Polar and ionic solutes dissolve in polar solvents (water) via H-bonds and ion-dipole forces. Non-polar solutes dissolve in non-polar solvents (hexane) via dispersion forces. Ionic compounds may be insoluble if lattice energy exceeds hydration energy. Amphiphilic molecules (soaps) form micelles — non-polar tails in grease, polar heads in water.
Dalton (solid sphere) → Thomson (plum pudding, electrons embedded) → Rutherford (nuclear model: tiny dense positive nucleus, mostly empty space — from gold foil experiment) → Bohr (quantised electron shells — from discrete hydrogen emission spectrum). Each revision driven by new experimental evidence.
Isotopes: same Z, different A (same element, different neutron count). Same chemical properties (same electrons); different physical properties (different mass). Ar = Σ(isotopic mass × fractional abundance). Fractional abundance = % ÷ 100. All fractions sum to 1. Mass spectrometry gives isotopic masses and abundances.
Modern table: ordered by Z (atomic number). Periods = horizontal rows (same shell count). Groups = vertical columns (same valence electrons → similar chemistry). Key groups: 1 (alkali metals, 1 valence e⁻), 17 (halogens, 7 valence e⁻), 18 (noble gases, full shell). Metal reactivity ↑ down group; halogen reactivity ↓ down group.
Updates as you answer. SA marks are self-assessed from model answers.
1. Which correctly identifies the type of polymerisation used to produce PVC from vinyl chloride (CH₂=CHCl)? L11
2. Low density polyethylene (LDPE) is more flexible than high density polyethylene (HDPE). The structural reason is: L11
3. A polymer is described as thermosetting — it does not soften on heating and cannot be remoulded. The structural explanation is: L11
4. Nylon-6,6 is formed from hexamethylenediamine and adipic acid. What type of linkage connects the monomers and what small molecule is released? L11
5. Iodine (I₂) dissolves readily in hexane but is only sparingly soluble in water. The best explanation is: L12
6. AgCl is ionic but insoluble in water. The correct explanation involves: L12
7. Soaps are amphiphilic molecules that clean greasy surfaces. The mechanism involves: L12
8. As the carbon chain length of alcohols increases (methanol → ethanol → propanol → butanol), their solubility in water decreases. The explanation is: L12
9. What key observation from Rutherford's gold foil experiment showed that most of the atom is empty space? L13
10. Thomson's plum pudding model was accepted before Rutherford's experiment because it: L13
11. For the ion ⁵⁶₂₆Fe³⁺, the numbers of protons, neutrons, and electrons respectively are: L13
12. The Bohr model improved on Rutherford's by: L13
13. Naturally occurring chlorine consists of ³⁵Cl (75.77%) and ³⁷Cl (24.23%). The relative atomic mass of chlorine is approximately: L14
14. ¹²C and ¹⁴C are both carbon isotopes. Which correctly describes a difference and a similarity? L14
15. An element has two isotopes: mass 63 (69.2%) and mass 65 (30.8%). The relative atomic mass is: L14
16. A mass spectrum of an element shows a peak at m/z = 40 (99.6%) and a very small peak at m/z = 36 (0.3%) and m/z = 38 (0.06%). The relative atomic mass is closest to: L14
17. An element is in Group 17 (halogens), Period 3. Which correctly identifies this element and predicts its ion? L15
18. The modern periodic table orders elements by atomic number (Z) rather than atomic mass. One reason this is superior is: L15
19. Rubidium (Rb) is in Group 1, Period 5. Compared to potassium (K, Group 1, Period 4), rubidium is: L15
20. (HOT) A chemist discovers an unknown element with the following properties: it is a gas at room temperature, non-reactive with any common reagent, exists as single atoms rather than molecules, and has a very low boiling point. The element is most likely in: L15 HOT
Attempt all questions before checking answers below.
21. Explain why PVC (–[CH₂–CHCl]ₙ–) is stiffer than polyethylene (–[CH₂–CH₂]ₙ–), even though both are produced by addition polymerisation from alkene monomers. In your answer, refer to intermolecular forces. 3 MARKS
22. A mass spectrum of an unknown element shows three peaks: mass 69 (60.1%), mass 71 (39.9%). (a) Calculate the relative atomic mass. (b) Identify the element. (c) State whether isotopes of this element would have identical or different chemical properties. Justify. 4 MARKS
23. Evaluate the development of atomic models from Thomson to Bohr, explaining how each new model represented an improvement over the previous one. In your answer, identify the experimental evidence that led to each revision and the key feature each new model introduced. 5 MARKS
24. (L11 — 3 marks) Nylon-6,6 is formed from hexanediamine (H₂N–(CH₂)₆–NH₂) and adipic acid (HOOC–(CH₂)₄–COOH). (a) Identify the type of polymerisation and justify. (b) Name the functional group linkage in nylon-6,6. (c) Name the small molecule released during polymerisation. 3 MARKS
25. (L11 — 3 marks) High-density polyethylene (HDPE) is stiffer and has a higher melting point than low-density polyethylene (LDPE), even though both have the repeat unit –[CH₂–CH₂]ₙ–. Explain this difference using polymer chain structure and intermolecular forces. 3 MARKS
26. (L12 — 3 marks) Explain, using intermolecular force theory, why: (a) ethanol (CH₃CH₂OH) dissolves readily in water, and (b) hexane (C₆H₁₄) does not dissolve in water but dissolves readily in other non-polar solvents like diethyl ether. 3 MARKS
27. (L12 — 3 marks) A student adds iodine (I₂, non-polar) to a test tube containing water and hexane layers. Predict which layer I₂ will preferentially dissolve in, and explain using IMF theory. What would you observe? 3 MARKS
28. (L13 — 4 marks) Describe the Rutherford gold foil experiment. In your answer: (a) state the experimental setup and what was observed, (b) explain what each observation tells us about atomic structure, and (c) state the limitation of the Rutherford model that the Bohr model later addressed. 4 MARKS
29. (L14 — 3 marks) Chlorine has two stable isotopes: ³⁵Cl (75.77%) and ³⁷Cl (24.23%). (a) Define the term isotope. (b) Calculate the relative atomic mass of chlorine to 2 decimal places. (c) State one way in which ³⁵Cl and ³⁷Cl differ and one way in which they are identical in terms of chemical behaviour. 3 MARKS
30. (L15 — 3 marks) In Mendeleev's original periodic table (1869), he ordered elements by increasing atomic mass. This created two anomalies: tellurium (Te, mass 128) was placed before iodine (I, mass 127) despite having a higher mass, and cobalt (Co, mass 58.9) was placed before nickel (Ni, mass 58.7). Explain how the modern periodic table resolves these anomalies, and why the modern ordering produces a more scientifically consistent table. 3 MARKS
1. B — Addition polymerisation: C=C opens, no byproduct. PVC has no ester/amide links and Cl is a substituent, not released.
2. D — Branching prevents close chain packing → weaker dispersion forces → easier to deform (more flexible). Not about polarity or cross-linking.
3. A — Thermosetting = covalent cross-links. These bonds require far more energy to break than heating provides. Long chains alone still melt (UHMWPE does). H-bonds are strong but still overcome by heating.
4. C — –NH₂ + –COOH → amide linkage (–CONH–) + H₂O. This is condensation. Ester links come from –OH + –COOH reactions.
5. B — Like dissolves like: I₂ non-polar → compatible with non-polar hexane; incompatible with water's H-bonds. Not ionic, not mass-dependent.
6. C — Lattice energy vs hydration energy. AgCl's very high lattice energy (strong Ag⁺–Cl⁻ attraction) exceeds the hydration energy → insoluble despite being ionic.
7. D — Micelle mechanism: non-polar tails in grease, polar heads face water. No chemical reaction — purely physical.
8. A — Growing non-polar chain dominates over the fixed –OH group's hydrophilicity. –OH doesn't change; the hydrophobic portion increases.
9. C — Most particles pass through = mostly empty space. Back-scatter (A) shows the nucleus is small and dense. Large-angle deflections (B) show concentrated positive charge.
10. B — Thomson's model explained electrons (just discovered) within a neutral atom. No gold foil data existed yet; spectral lines weren't explained by any model at that time.
11. D — Z=26 → protons=26; neutrons=56−26=30; Fe³⁺ loses 3 electrons → electrons=26−3=23.
12. A — Bohr introduced quantised energy levels, explaining electron stability and discrete emission spectra. B was Rutherford. C was Chadwick. D was Dalton.
13. C — Ar = (35×0.7577) + (37×0.2423) = 26.52 + 8.97 = 35.49 ≈ 35.5.
14. B — Different neutrons: ¹²C has 6, ¹⁴C has 8. Same chemical properties because same Z → same electrons → same bonding behaviour → both form CO₂.
15. D — Ar = (63×0.692) + (65×0.308) = 43.60 + 20.02 = 63.62. (This is copper, Cu.)
16. A — Ar ≈ 40 because the dominant isotope is mass 40 (99.6%). The other isotopes contribute negligibly. (This is argon, Ar.)
17. B — Group 17, Period 3 = Cl. Halogens (7 valence e⁻) gain 1 electron → Cl⁻. Fluorine is Period 2. S²⁻ is Group 16. Cl does not form Cl⁷⁺ (would require removing 7 inner electrons).
18. C — Z determines electron configuration → chemical properties. Te/I is the classic example. Math convenience (A) is not a chemical reason.
19. D — Rb is below K in Group 1 → atomic radius larger → valence electron further from nucleus, more shielded → easier to remove → more reactive. Same valence electron count, but position matters.
20. B — Noble gases: monatomic, chemically inert, very low BP (dispersion forces only, smallest possible for their size). Halogens (C) are diatomic and reactive. Group 1 metals are not gases.
Q24 (3 marks): (a) Condensation polymerisation — hexanediamine has two –NH₂ groups (bifunctional) and adipic acid has two –COOH groups (bifunctional). Each monomer has reactive groups at both ends, so chains grow from both ends of each monomer. A small molecule byproduct is released at each bond-forming step (characteristic of condensation, unlike addition which has no byproduct). (b) Amide linkage: –CO–NH– (or –CONH–). Formed when the –COOH of adipic acid reacts with the –NH₂ of hexanediamine — the OH from –COOH and H from –NH₂ leave as water. (c) Water (H₂O) — one molecule released per amide bond formed.
Q25 (3 marks): Both LDPE and HDPE have the same repeat unit (–CH₂–CH₂–) and the same type of IMF between chains (London dispersion forces — non-polar polymer). LDPE (high-pressure synthesis): chains are highly branched — branches physically prevent chains from packing closely together. Gaps between chains → fewer and weaker dispersion force contacts between chains → less cohesive → softer, more flexible, lower melting point, lower density. HDPE (Ziegler-Natta catalyst, low pressure): chains are linear and unbranched. Linear chains align and pack closely together in a more crystalline arrangement → maximum contact area between adjacent chains → more and stronger dispersion forces → harder to separate → stiffer, higher melting point, higher density. The structural difference is chain branching, not monomer type or bond type.
Q26 (3 marks): (a) Ethanol–water: ethanol's –OH group participates in H-bonding with water (O is δ−, H is δ+, Δχ=1.2). New H-bonds form between ethanol and water molecules in place of water–water H-bonds — the energy trade-off is favourable. The non-polar ethyl tail is short (2 carbons) — insufficient to disrupt the H-bond network → ethanol is completely miscible with water. (b) Hexane–water: hexane (C₆H₁₄) has only non-polar C–H and C–C bonds (Δχ ≈ 0.4 or less, non-polar). Only dispersion forces between hexane molecules. Dissolving in water would require breaking strong H₂O–H₂O H-bonds to accommodate hexane molecules without providing any comparable interaction → very unfavourable → insoluble. Hexane–ether: diethyl ether is also non-polar (alkyl chains dominate). Hexane–ether dispersion forces are very similar in type and strength to hexane–hexane and ether–ether dispersion forces → dissolving causes no significant energy penalty → miscible ('like dissolves like').
Q27 (3 marks): Prediction: I₂ dissolves preferentially in the hexane (upper) layer. I₂ is non-polar (Δχ=0, identical atoms) → only London dispersion forces. Water is polar (H-bonded network). Dissolving I₂ in water would disrupt H-bonds without compensating interaction → energetically unfavourable → I₂ barely dissolves in water. Hexane is non-polar → dispersion forces between I₂ and hexane are compatible → 'like dissolves like' → I₂ dissolves readily in hexane. Observation: the hexane layer (top, less dense) turns pink/violet-purple — the characteristic colour of dissolved I₂ in non-polar solvents. The water layer (bottom) remains nearly colourless or shows only a faint yellow-brown tinge (trace aqueous I₂).
Q28 (4 marks): (a) Setup: alpha particles (α, helium-4 nuclei, charge 2+) from a radioactive source fired at a thin gold foil (~0.0004 mm). ZnS scintillation screen surrounds the foil to detect alpha particles at any angle. Observations: ~99% pass straight through or are deflected by tiny angles (< 1°). A small fraction (< 1%) deflect at large angles (up to 90°). Very rarely (1 in 20,000), an alpha particle is reflected back through nearly 180° ("as if you fired artillery shells at tissue paper and they came back and hit you" — Rutherford). (b) Interpretations: most pass straight through → atom is mostly empty space. Large deflections → the positive charge and most mass is concentrated in an extremely small, dense nucleus at the centre. Back-scatter → the nucleus is very hard and dense — incoming positive alpha particles are repelled by the highly concentrated positive nuclear charge. (c) Limitation: electrons in circular orbits are accelerating charges — classical electromagnetism requires they continuously emit radiation and lose energy → should spiral into the nucleus in ~10⁻⁸ s. Atoms would be unstable. Bohr proposed electrons occupy only specific, quantised orbits where they do not radiate — explaining atomic stability and discrete emission spectra.
Q29 (3 marks): (a) Isotopes are atoms of the same element (same atomic number Z — same number of protons) that have different numbers of neutrons, giving different mass numbers (A = protons + neutrons). They are chemically identical but differ in mass. (b) Ar = (35 × 0.7577) + (37 × 0.2423) = 26.5195 + 8.9651 = 35.4846 ≈ 35.48. (c) Difference: ³⁷Cl has 20 neutrons; ³⁵Cl has 18 neutrons — they differ in neutron count and therefore mass. ³⁷Cl is heavier (mass number 37 vs 35). Instruments like mass spectrometers can distinguish them. Identical in chemical behaviour: both have Z=17 with electron configuration 1s²2s²2p⁶3s²3p⁵ (7 valence electrons). Chemical reactivity is determined solely by electron configuration — both react with the same substances, in the same ratios, at essentially the same rates (minor kinetic isotope effects exist but are negligible at HSC level).
Q30 (3 marks): Mendeleev ordered by atomic mass → Te (128) placed after I (127) in mass sequence, but Te has group-16 chemical behaviour (forms TeO₂, like SO₂ and SeO₂) while I has group-17 behaviour (forms HI and reacts like F and Cl). Mass ordering placed them in the wrong chemical groups. Modern table: ordered by atomic number Z (number of protons). Te has Z=52, I has Z=53 → Te correctly precedes I. This is more scientifically consistent because Z (not mass) determines electron configuration (via the Aufbau principle) — electron configuration determines valence electrons — valence electrons determine all chemical behaviour. Ordering by Z automatically groups elements with the same valence electron configuration into the same vertical group, making the table predictive: any element's properties can be inferred from its Z and position, including elements not yet discovered (as Mendeleev demonstrated by predicting germanium from the pattern — but the modern table provides the mechanistic justification via quantum mechanics).
Q21 (3 marks): PVC has –Cl substituents on alternate carbons of the polymer chain (1 mark). The C–Cl bond is polar because Cl is electronegative (χ = 3.0) → each PVC repeat unit has a permanent dipole → dipole-dipole forces exist between adjacent PVC chains (1 mark). Polyethylene has only –H substituents; the C–H bond is essentially non-polar → only dispersion forces act between PE chains. Dipole-dipole forces are stronger than dispersion forces for molecules of similar size → PVC chains are held more tightly → greater resistance to chain sliding → PVC is stiffer than PE (1 mark).
Q22 (4 marks): (a) Fractions: 69 → 0.601; 71 → 0.399. Sum = 1.000 ✓. Ar = (69 × 0.601) + (71 × 0.399) = 41.469 + 28.329 = 69.80 ≈ 69.8 (1 mark for working; 1 mark for correct answer). (b) Ar ≈ 69.8 → this is gallium (Ga), Ar = 69.72 on the periodic table (1 mark). (c) Both isotopes (⁶⁹Ga and ⁷¹Ga) have identical chemical properties because they have the same atomic number (Z = 31) and therefore the same number and arrangement of electrons. Chemical behaviour is determined by electron configuration, which is the same for all gallium isotopes (1 mark).
Q23 (5 marks): Thomson's plum pudding model (1904) explained the existence of electrons (discovered by Thomson 1897) within an overall neutral atom — electrons embedded in a diffuse positive sphere. This was consistent with cathode ray tube evidence (1 mark). Rutherford's gold foil experiment (1909–11): most alpha particles passed through gold foil (consistent with mostly empty space) but a small fraction deflected at large angles and a tiny fraction reflected back — impossible in the plum pudding model where positive charge is diffuse (1 mark). Rutherford introduced the nuclear model: tiny, dense, positive nucleus surrounded by mostly empty space, electrons orbiting at large distances. Improvement: explains large-angle and back-scatter deflections (1 mark). However, Rutherford's model couldn't explain why orbiting electrons didn't spiral into the nucleus (classical physics problem) nor hydrogen's discrete emission spectrum. Hydrogen's spectrum shows specific coloured lines, not a continuous rainbow, indicating quantised energy emissions (1 mark). Bohr introduced quantised energy levels: electrons occupy fixed orbits at specific energies; photons of specific energy (frequency) are emitted when electrons fall between levels. This explained electron stability (no energy loss in a fixed level) and exactly matched the wavelengths of hydrogen's spectral lines. Improvement: resolves the instability problem and explains spectra (1 mark).
Tick when you have checked all answers and noted areas to review.