IB DP Chemistry Structure 2.4: From models to materials – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions test Structure 2.4, From models to materials, understandings 2.4.1 to 2.4.6, from the IB Chemistry guide for the Diploma Programme (issued February 2023; first assessment 2025). Questions 1 to 8 and 11 are for SL and HL. Questions 9, 10 and 12 test Structure 2.4.6, which is additional higher level, so they are labelled HL only.

Work through the study guide and the revision notes first. Questions 11 and 12 combine several understandings. Other units sit on the course page; log progress on the syllabus checklist.

Electronegativities are Pauling values given in each question. For masses, take C = 12.01, H = 1.01, O = 16.00, N = 14.01 and Cl = 35.45.

Questions

1. Aluminium oxide is used to make the insulating body of spark plugs. Using a bonding model, explain why this oxide melts only at a very high temperature and is a non-conductor in the solid state. [3]

2. Use these electronegativity values: Ca 1.00, O 3.44, C 2.55, S 2.58, Ni 1.91, Al 1.61.

(a) Calculate the electronegativity difference and the average electronegativity for CaO, CS₂ and NiAl. [3] (b) State the region of the bonding triangle in which each compound lies. [2]

3. Use these electronegativity values: Si 1.90, Cl 3.16, Mg 1.31, F 3.98.

(a) Calculate the electronegativity difference and the average electronegativity for SiCl₄ and MgF₂. [2] (b) Use their positions in the bonding triangle to predict which has the higher melting point and which conducts when molten. [3]

4. Cupronickel, an alloy of copper and nickel, is used for coins.

(a) Define the term alloy. [1] (b) Explain why cupronickel is described as a mixture rather than a compound. [1] (c) Explain, in terms of non-directional bonding, why cupronickel is harder than pure copper but still conducts electricity. [3]

5. Poly(ethene) is a plastic made of long hydrocarbon chains. Explain, in terms of its structure, why poly(ethene):

(a) is an electrical insulator [1] (b) softens when heated [2] (c) persists for a long time in the environment [1]

6. Classify each of these polymers as natural or synthetic: cellulose, nylon, DNA, poly(chloroethene), natural rubber, a polyester. [2]

7. 1-Chloropropene, CH₃CH=CHCl, undergoes addition polymerization.

(a) Draw the repeating unit. [2] (b) State the atom economy of the polymerization and explain your answer. [2] (c) A sample of the polymer has an average molar mass of 2.68 × 10⁵ g mol⁻¹. Calculate the average number of monomer units per chain. [2]

8. A section of an addition polymer is shown.

–CHCl–CCl₂–CHCl–CCl₂–CHCl–CCl₂–

(a) Draw the repeating unit. [1] (b) Deduce the structural formula of the monomer and name it. [2]

9. (HL only) 8-Aminooctanoic acid, H₂N(CH₂)₇COOH, polymerizes by condensation with itself.

(a) Draw the repeating unit formed. [2] (b) Name the type of link formed and the small molecule released. [2] (c) Calculate the atom economy for the formation of the polymer. [2]

10. (HL only)

(a) Hydroxyethanoic acid, HOCH₂COOH, forms a polyester. Draw its repeating unit. [2] (b) Another polyester has the repeating unit –[–O(CH₂)₃O–CO(CH₂)₂CO–]–ₙ. Deduce the structural formulas of its two monomers. [2] (c) Explain why polyesters can be broken down by hydrolysis but poly(ethene) cannot. [2]

11. A kettle has a stainless steel body, a plastic handle and an aluminium oxide insulator round its heating element. Use these electronegativity values: Al 1.61, O 3.44, Ti 1.54, Cl 3.16.

(a) Calculate the electronegativity difference and the average electronegativity for Al₂O₃. [2] (b) Describe the position of Al₂O₃ in the bonding triangle and predict two of its properties that suit it to this use. [2] (c) Explain why stainless steel is chosen for the body rather than pure iron, and why it still conducts heat. [2] (d) Explain why the handle is an electrical insulator, and why it could soften if it became too hot. [2] (e) TiCl₄ has an electronegativity difference only a little smaller than that of Al₂O₃, yet it is a liquid at room temperature. Suggest what this shows about discrete bonding categories. [1]

12. (HL only)

(a) State the type of reaction by which all biological macromolecules form. [1] (b) A polypeptide contains 85 amino acid residues. Deduce the number of water molecules released when it forms, and calculate the mass of water released when 1 mol of these chains forms. [2] (c) Calculate the mass of glucose formed by complete hydrolysis of 4.86 g of cellulose, taking each unit in the chain as C₆H₁₀O₅. [3] (d) State two differences between addition and condensation polymerization. [2]

Answers

1. Ionic lattice of Al³⁺ and O²⁻ ions [1]. Highly charged ions attract strongly, so much energy is needed to separate them: very high melting point [1]. Ions are fixed in the lattice with no mobile charges: no conduction as a solid [1]. [3] Examiner insight: “Strong bonds” alone earns little; credit comes from naming the ions and their electrostatic attraction.

2. (a) CaO: Δχ = 2.44, average = 2.22 [1]. CS₂: Δχ = 0.03, average = 2.565 [1]. NiAl: Δχ = 0.30, average = 1.76 [1]. (b) CaO ionic; CS₂ covalent [1]; NiAl metallic [1]. [5] Examiner insight: Use one value per element. Multiplying the sulfur value by two for CS₂ gives a wrong average and loses the mark.

3. (a) SiCl₄: Δχ = 1.26, average = 2.53 [1]. MgF₂: Δχ = 2.67, average = 2.645 [1]. (b) MgF₂ lies much higher, in the ionic region: a lattice of ions with strong attraction [1]. MgF₂ has the higher melting point [1]. SiCl₄ is lower, polar covalent and molecular; only MgF₂ conducts when molten, as its ions become mobile [1]. [5] Examiner insight: Tie each prediction to the triangle position; “MgF₂ is ionic” with no reference to Δχ is incomplete.

4. (a) A metal mixed with one or more further elements, metallic or non-metallic [1]. (b) Its composition can vary; there is no fixed formula [1]. (c) Non-directional metallic bonding lets nickel atoms take places in the copper lattice [1]. Being a different size, they distort the layers, which slide less easily: harder [1]. Mobile delocalized electrons remain: still conducts [1]. [5] Examiner insight: “The alloy has stronger bonds” does not explain hardness; credit goes to the idea that layers of cations can no longer slide easily.

5. (a) All electrons are in localized covalent bonds; there are no mobile ions or delocalized electrons [1]. (b) Chains are held together only by London dispersion forces [1]. Heating overcomes these so chains move past each other; covalent bonds in the chains do not break [1]. (c) The C–C and C–H bonds are strong and non-polar, so the chains are unreactive and break down very slowly [1]. [4] Examiner insight: Saying the plastic softens because “bonds break” is wrong unless you specify intermolecular forces, not covalent bonds.

6. Natural: cellulose, DNA, natural rubber. Synthetic: nylon, poly(chloroethene), polyester. [1] for four or five correct; [1] for all six correct. [2] Examiner insight: Where a mark is shared across a list, every item usually needs to be correct for full credit, so check each one.

7. (a) Two-carbon backbone, single bond, each carbon keeping its groups: –CH(CH₃)–CHCl– [1]; brackets, continuation bonds and n: –[–CH(CH₃)–CHCl–]–ₙ [1]. (b) 100% [1]. The polymer is the only product; every atom of the monomer is in the polymer [1]. (c) M(C₃H₅Cl) = 3(12.01) + 5(1.01) + 35.45 = 76.53 g mol⁻¹ [1]. n = 2.68 × 10⁵ / 76.53 = 3.50 × 10³ [1]. [6] Examiner insight: Both methyl and chlorine are side groups here; drawing CH₃ as part of the backbone gives a chain that does not match the monomer.

8. (a) –[–CHCl–CCl₂–]–ₙ [1]. (b) CHCl=CCl₂ [1]; trichloroethene [1]. [3] Examiner insight: A monomer is a molecule, so continuation bonds must not appear in its formula.

9. (a) Remove H from –NH₂ and OH from –COOH and join the N to the C=O [1]: –[–NH(CH₂)₇CO–]–ₙ [1]. (b) Amide link [1]; water [1]. (c) M(monomer, C₈H₁₇NO₂) = 159.26 g mol⁻¹; M(repeating unit) = 159.26 − 18.02 = 141.24 g mol⁻¹ [1]. Atom economy = 141.24 / 159.26 × 100 = 88.7% [1]. [6] Examiner insight: One monomer with both groups gives one residue per repeating unit; two residues is not the simplest unit.

10. (a) H from –OH and OH from –COOH lost as water; ester link formed [1]: –[–OCH₂CO–]–ₙ [1]. (b) HO(CH₂)₃OH (propane-1,3-diol) [1]; HOOC(CH₂)₂COOH (butanedioic acid) [1]. (c) Polyesters contain polar ester links that water can attack, splitting the chain [1]. Poly(ethene) offers water no polar link to attack: its bonds are all C–C or C–H [1]. [6] Examiner insight: When deducing monomers, add H₂O back at every link: each –O– gains H and each C=O gains OH.

11. (a) Δχ = 3.44 − 1.61 = 1.83 [1]; average = (1.61 + 3.44) / 2 = 2.525 [1]. (b) High in the triangle, in or near the ionic region, with some covalent character [1]. Predicted high melting point and no electrical conduction as a solid [1]. (c) Other atoms distort the iron layers so they slide less easily (stronger, harder), and it resists rusting [1]. Delocalized electrons remain, so it still conducts heat [1]. (d) No mobile charged particles in the plastic [1]. Heating overcomes the intermolecular forces between chains, so they can slide [1]. (e) TiCl₄ (Δχ = 1.62, average 2.35) sits close to Al₂O₃ yet is molecular, so bonding changes gradually and discrete categories cannot describe every compound [1]. [9] Examiner insight: Each property needs a structural reason; a bare list earns little.

12. (a) Condensation [1]. (b) Number of links = 85 − 1 = 84 water molecules [1]; mass = 84 × 18.02 = 1.51 × 10³ g [1]. (c) M(C₆H₁₀O₅) = 162.16 g mol⁻¹; moles of units = 4.86 / 162.16 = 0.02997 mol [1]. Each unit adds one H₂O to give glucose, M = 180.18 g mol⁻¹ [1]. Mass of glucose = 0.02997 × 180.18 = 5.40 g [1]. (d) Addition gives no other product; condensation releases a small molecule such as water [1]. Addition needs a C=C in the monomer; condensation needs two functional groups per monomer [1]. [8] Examiner insight: Hydrolysis adds water, so the glucose mass must exceed the cellulose mass; a smaller answer means the ratio is upside down.

Where marks are usually lost

• Multiplying an electronegativity by a subscript when finding Δχ or the average.
• Explaining alloy hardness through “stronger bonds” instead of layers that cannot slide.
• Saying covalent bonds break when a plastic softens.
• Drawing a side group of an addition polymer in the main chain.
• Leaving continuation bonds on a monomer, or leaving them off a repeating unit.
• (HL only) Keeping the H or OH that should have been lost at each condensation link.
• (HL only) Counting N water molecules instead of N − 1 for a chain of N units.

Next steps

• Review weak points in the unit revision notes
• Reread explanations in the unit study guide
• Course page and syllabus checklist
• Check other units with all free 10-minute diagnostics
• Learn with a Marlbridge teacher: book a free trial class

Official syllabus

Document followed: Chemistry guide, Diploma Programme, International Baccalaureate Organization (February 2023; first examined 2025). Scope here: Structure 2.4, From models to materials; 2.4.1 to 2.4.5 for SL and HL, 2.4.6 additional higher level.