2022 May/June · Paper 23 · Q3 (a(i))1 markDot-and-cross diagrams
2022 May/June · Paper 41 · Q4 (d)3 marksSigma & pi bonds / hybridisation
2022 May/June · Paper 41 · Q7 (a(ii))1 markSigma & pi bonds / hybridisation
2022 May/June · Paper 43 · Q4 (d)3 marksSigma & pi bonds / hybridisation
2022 May/June · Paper 43 · Q7 (a(ii))1 markSigma & pi bonds / hybridisation
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school. “Paper 42” means Paper 4, variant 2.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Sigma & pi bonds / hybridisation · Core · Paper 2 style · 6 marks
(a)State the type of hybridisation of each carbon atom in ethene, C₂H₄, and the H–C–H bond angle.[2]
(b)Describe how the σ bond and the π bond between the two carbon atoms in ethene are formed.[3]
(c)Explain why there is no free rotation about the C=C bond in ethene.[1]
Show answer
(a) sp²; 120°
B1 sp²; B1 120° (accept 117–121°).
(b) σ: end-on (head-on) overlap of sp² hybrid orbitals, electron density on the internuclear axis. π: sideways overlap of the unhybridised p orbitals, electron density above and below the plane of the molecule.
B1 σ by end-on/axial overlap (of sp² orbitals); B1 π by sideways overlap of p orbitals; B1 π electron density above and below the plane / not on the axis.
(c) Rotation would break the sideways overlap of the p orbitals (the π bond).
B1 rotation would destroy the p-orbital overlap / break the π bond.
Q2Sigma & pi bonds / hybridisation · Core · Paper 1 style · 1 mark
(a)In which molecule is a carbon atom sp hybridised?[1]
Show answer
(a) A CO₂
In CO₂ the carbon forms two double bonds and no lone pairs, two regions of electron density, linear → sp. CH₃CHO has sp² (C=O carbon) and sp³ carbons but no sp carbon; C₂H₄ is sp² only; CH₃OH is sp³.
Q3Sigma & pi bonds / hybridisation · Standard · Paper 1 style · 1 mark
(a)Propenenitrile, CH₂=CH–C≡N, is used to make acrylic fibres for winter shawls. How many σ bonds and how many π bonds are there in one molecule of propenenitrile?[1]
Show answer
(a) A 6 σ and 3 π
σ bonds: 3 C–H, 1 C–C and one σ in each of C=C and C≡N → 6. π bonds: 1 in C=C + 2 in C≡N → 3. Option C counts every line of every multiple bond as σ; a multiple bond contains only one σ bond.
Q4Sigma & pi bonds / hybridisation · Standard · Paper 2 style · 6 marks
(a)Give the hybridisation of each carbon atom, C1 to C5, in pent-3-en-1-yne, HC≡C–CH=CH–CH₃ (C1 is the carbon of HC≡, C5 is the CH₃ carbon).[3]
(b)Calculate the total number of σ bonds and of π bonds in one molecule.[2]
(c)State the bond angle around C2 and around C5.[1]
Show answer
(a) C1 sp, C2 sp, C3 sp², C4 sp², C5 sp³
B1 C1 and C2 sp; B1 C3 and C4 sp²; B1 C5 sp³.
(b) 10 σ bonds and 3 π bonds
6 C–H σ bonds + 4 C–C σ bonds (one in each carbon–carbon link) = 10 σ (B1). π: 2 in C≡C + 1 in C=C = 3 (B1).
(c) C2 180°; C5 about 109.5°
B1 both.
Q5Sigma & pi bonds / hybridisation · Challenge · Paper 2 style · 6 marks
Average bond energies (kJ/mol): C–C 350; C=C 610; C≡C 840.
(a)Use the data to estimate the energy of the π bond in C=C, and the average energy of each π bond in C≡C.[2]
(b)Explain, in terms of orbital overlap, why a π bond is weaker than a σ bond.[2]
(c)Suggest why alkenes are much more reactive than alkanes even though the C=C bond energy is greater than the C–C bond energy.[2]
Show answer
(a) π in C=C ≈ 260 kJ/mol; each π in C≡C ≈ 245 kJ/mol
(b) Sideways overlap of p orbitals is less effective than end-on overlap, so electron density between the nuclei is lower and the electrons are less strongly attracted by both nuclei.
B1 sideways overlap is smaller/less effective than end-on; B1 electron density lies away from the internuclear axis, so weaker attraction to the nuclei.
(c) Only the weaker π bond needs to be broken in addition reactions; the exposed π electron density above and below the plane attracts electrophiles.
B1 only the π bond (≈260 kJ/mol) breaks; B1 π electrons are exposed/high electron density so electrophiles attack.
Q6Sigma & pi bonds / hybridisation · Challenge · Paper 2 style · 5 marks
(a)Describe the bonding between the two carbon atoms in ethyne, C₂H₂, in terms of hybrid and unhybridised orbitals. State the H–C–C bond angle.[3]
(b)The C–C bond lengths in ethane, ethene and ethyne are 0.154 nm, 0.134 nm and 0.120 nm. Explain this trend.[2]
Show answer
(a) Each C is sp hybridised; one σ bond from end-on overlap of sp orbitals; two π bonds from sideways overlap of the two pairs of unhybridised p orbitals, at right angles to each other; 180°.
B1 sp hybridisation and σ from sp–sp overlap; B1 two π bonds from two sets of p orbitals at 90° to each other; B1 180°.
(b) More shared electron pairs between the carbons (1, 2, 3) → greater electron density between the nuclei → nuclei pulled closer; also more s character in the hybrid orbitals (sp³ → sp² → sp).
B1 bond order increases / more electrons shared between nuclei; B1 stronger attraction pulls nuclei closer (credit increasing s character).
Q7Shapes & bond angles (VSEPR) · Core · Paper 1 style · 1 mark
(a)Which species has a bond angle closest to 107°?[1]
Show answer
(a) A H₃O⁺
H₃O⁺ has 3 bonding pairs and 1 lone pair around O → pyramidal, ≈107°. NH₄⁺ has 4 bonding pairs and no lone pair → tetrahedral 109.5°; H₂O has 2 lone pairs → 104.5°; BF₃ 120°.
Q8Shapes & bond angles (VSEPR) · Core · Paper 2 style · 4 marks
(a)The bond angles in CH₄, NH₃ and H₂O are 109.5°, 107° and 104.5°. Explain this sequence.[3]
(b)Name the shape of an NH₃ molecule.[1]
Show answer
(a) All have four electron pairs around the central atom. Number of lone pairs is 0, 1, 2. Lone pair–bonding pair repulsion is greater than bonding pair–bonding pair repulsion, so more lone pairs squeeze the bonds closer together.
B1 four electron pairs (tetrahedral arrangement) in each; B1 0, 1, 2 lone pairs; B1 lone pairs repel more strongly than bonding pairs.
(b) (Trigonal) pyramidal
B1.
Q9Shapes & bond angles (VSEPR) · Standard · Paper 2 style · 5 marks
(a)Complete the information for each molecule: number of bonding pairs and lone pairs around the central atom, shape, and bond angle(s).(i) BCl₃ (ii) PCl₅ (gas) (iii) SF₆ (iv) OF₂[4]
(b)State the principle used to predict these shapes.[1]
(b) Electron pairs around the central atom repel each other and arrange themselves as far apart as possible (lone pairs repel more than bonding pairs).
B1 valence-shell electron pairs repel to be as far apart as possible.
Q10Shapes & bond angles (VSEPR) · Standard · Paper 2 style · 4 marks
(a)BF₃ and NH₃ both have three atoms bonded to the central atom. Explain why BF₃ is trigonal planar but NH₃ is pyramidal.[2]
(b)BF₃ reacts with NH₃ to form the compound H₃N–BF₃. State the shape around the boron atom and around the nitrogen atom in this compound, and the F–B–F bond angle.[2]
Show answer
(a) B has 3 bonding pairs and no lone pair → 120° planar. N has 3 bonding pairs and 1 lone pair → four pairs, tetrahedral arrangement → pyramidal.
B1 BF₃: 3 bp, no lp; B1 NH₃: 3 bp + 1 lp, lone pair repels.
(b) Both tetrahedral; F–B–F about 109.5°
B1 both tetrahedral; B1 ≈109.5° (accept 107–110°).
Q11Shapes & bond angles (VSEPR) · Challenge · Paper 2 style · 5 marks
(a)For each ion, give the number of bonding and lone pairs around the central atom, the shape and the bond angle.(i) NH₂⁻ (ii) PCl₄⁺ (iii) PCl₆⁻ (iv) AlCl₄⁻[4]
(b)Solid phosphorus(V) chloride is ionic and contains PCl₄⁺ and PCl₆⁻ ions. Suggest one way in which the shape of the phosphorus species changes when gaseous PCl₅ forms the solid.[1]
Q12Bonding types & properties · Core · Paper 1 style · 1 mark
(a)Which substance is most likely to have a giant ionic lattice?[1]
Show answer
(a) A melting point 770 °C; conducts when molten but not as a solid; dissolves in water
Ions are fixed in the solid but mobile when molten, so conduction only when molten; high melting point; soluble in water → giant ionic. B is giant covalent (like silicon), C is metallic, D is simple molecular.
Q13Bonding types & properties · Standard · Paper 2 style · 4 marks
(a)Define electronegativity.[1]
(b)Use the Pauling electronegativity values H 2.2, C 2.6, O 3.4, F 4.0, Cl 3.2 to place these bonds in order of increasing polarity: C–Cl, O–H, C–H, C–F.[1]
(c)Tetrachloromethane, CCl₄, contains polar bonds but is a non-polar molecule, whereas trichloromethane, CHCl₃, is polar. Explain.[2]
Show answer
(a) The power of an atom to attract the bonding pair of electrons in a covalent bond (towards itself).
B1.
(b) C–H < C–Cl < O–H < C–F
B1 differences 0.4 < 0.6 < 1.2 < 1.4.
(c) CCl₄ is a symmetrical tetrahedron, so the four equal bond dipoles cancel. In CHCl₃ one C–H bond replaces a C–Cl, so the dipoles do not cancel and there is a net dipole.
B1 CCl₄ symmetrical, dipoles cancel; B1 CHCl₃ dipoles unequal/do not cancel → net dipole.
Q14Bonding types & properties · Standard · Paper 2 style · 6 marks
A student in a Faisalabad college tests four solids.W: melting point 1710 °C; does not conduct as solid or liquid; insoluble in water and in hexane.X: melting point 660 °C; conducts as solid and liquid; insoluble in water.Y: melting point 114 °C; does not conduct as solid or liquid; insoluble in water, soluble in hexane.Z: melting point 996 °C; does not conduct as solid, conducts as liquid and in aqueous solution.
(a)Identify the type of structure of each of W, X, Y and Z.[4]
(b)Suggest a possible identity for W and for Y.[2]
Show answer
(a) W giant covalent; X giant metallic; Y simple molecular; Z giant ionic
B1 each.
(b) W: e.g. silicon(IV) oxide (sand) or diamond; Y: e.g. iodine or naphthalene (a non-polar simple molecular solid)
B1 each, any reasonable example consistent with the data.
Q15Bonding types & properties · Challenge · Paper 2 style · 5 marks
(a)Aluminium fluoride melts at 1290 °C, but aluminium chloride sublimes at about 180 °C. Suggest the type of bonding in each compound and explain the difference.[4]
(b)Suggest, with a reason, whether magnesium chloride or aluminium chloride would show more covalent character.[1]
Show answer
(a) AlF₃ is (mainly) ionic; AlCl₃ is (mainly) covalent/simple molecular. Al³⁺ is small with a high charge (high charge density) and polarises anions. Cl⁻ is larger and more easily polarised than F⁻, so its electron cloud is distorted towards Al³⁺ giving covalent character. Covalent AlCl₃ forms small molecules (Al₂Cl₆) with weak intermolecular forces, so low sublimation temperature; AlF₃ has a giant lattice with strong ionic bonds.
B1 AlF₃ ionic, AlCl₃ covalent; B1 Al³⁺ high charge density polarises anion; B1 Cl⁻ larger/more polarisable than F⁻; B1 weak intermolecular forces between molecules vs strong electrostatic attraction in giant lattice.
(b) Aluminium chloride: Al³⁺ has higher charge and smaller radius than Mg²⁺, so it is more polarising.
B1 AlCl₃ with reason.
Q16Intermolecular forces & boiling points · Core · Paper 1 style · 1 mark
(a)Which compound has the highest boiling point?[1]
Show answer
(a) A pentane
The three C₅H₁₂ isomers have the same number of electrons, but unbranched pentane has the largest surface contact between molecules, so the strongest instantaneous dipole–induced dipole forces. Branching (B, C) reduces contact; butane has fewer electrons.
Q17Intermolecular forces & boiling points · Core · Paper 2 style · 4 marks
(a)The boiling points of the halogens are: F₂ −188 °C, Cl₂ −34 °C, Br₂ 59 °C, I₂ 184 °C. Name the intermolecular forces between halogen molecules and explain the trend.[3]
(b)State the physical state of bromine and of iodine at 25 °C.[1]
Show answer
(a) Instantaneous dipole–induced dipole (London/dispersion) forces. Down the group the number of electrons per molecule increases, so larger, more easily formed instantaneous dipoles; stronger forces need more energy to overcome.
B1 id–id forces; B1 more electrons down the group; B1 stronger forces → more energy to separate molecules.
(b) Bromine liquid; iodine solid
B1 both.
Q18Intermolecular forces & boiling points · Standard · Paper 2 style · 4 marks
Boiling points: butane −0.5 °C (Mr 58); propanone 56 °C (Mr 58); propan-1-ol 97 °C (Mr 60).
(a)Explain why propanone boils at a higher temperature than butane, even though both have Mr = 58.[2]
(b)Explain why propan-1-ol has an even higher boiling point than propanone.[2]
Show answer
(a) Propanone is polar (C=O dipole), so it has permanent dipole–permanent dipole forces as well as id–id forces; butane is non-polar with id–id forces only.
B1 propanone has pd–pd forces due to polar C=O; B1 butane only id–id, similar size so similar id–id.
(b) Propan-1-ol has an O–H group and forms hydrogen bonds between molecules (H bonded to O, lone pair on O); hydrogen bonds are stronger than pd–pd forces.
B1 hydrogen bonding via O–H; B1 hydrogen bonds stronger than pd–pd.
Q19Intermolecular forces & boiling points · Challenge · Paper 2 style · 5 marks
Boiling points: HCl −85 °C; HBr −67 °C. Electronegativities: H 2.2, Cl 3.2, Br 3.0.
(a)Calculate the electronegativity difference in H–Cl and in H–Br.[1]
(b)HCl is more polar than HBr, yet HBr has the higher boiling point. Explain this.[3]
(c)Predict, with a reason, whether HI boils above or below HBr.[1]
Show answer
(a) HCl 1.0; HBr 0.8
B1 3.2 − 2.2 = 1.0; 3.0 − 2.2 = 0.8.
(b) HBr has more electrons (36 vs 18), so stronger instantaneous dipole–induced dipole forces. The increase in id–id forces outweighs the decrease in permanent dipole–dipole forces, so more energy is needed to separate HBr molecules.
B1 HBr more electrons; B1 stronger id–id forces; B1 id–id contribution dominates over pd–pd difference.
(c) Above (−35 °C), because it has even more electrons → stronger id–id forces.
B1.
Q20Metallic bonding & properties · Core · Paper 1 style · 1 mark
(a)Why does magnesium have a higher melting point than sodium?[1]
Show answer
(a) A Each magnesium ion has a higher charge and is smaller, and there are more delocalised electrons per ion.
Mg²⁺ (2+, smaller) and two delocalised electrons per atom give stronger electrostatic attraction between cations and the sea of electrons. B is wrong: metals have metallic bonding, not intermolecular forces.
Q21Metallic bonding & properties · Core · Paper 2 style · 5 marks
(a)Describe metallic bonding.[2]
(b)Copper is used for electrical wiring in homes in Lahore. Explain why copper conducts electricity and why it can be drawn into wires.[3]
Show answer
(a) A lattice of positive metal ions in a sea of delocalised electrons; electrostatic attraction between the positive ions and the delocalised electrons.
B1 lattice of positive ions + delocalised electrons; B1 electrostatic attraction between them.
(b) Delocalised electrons are free to move through the structure and carry charge when a voltage is applied. Layers of ions can slide over each other without breaking the metallic bonding, because the delocalised electrons still hold the ions together (ductile).
B1 delocalised electrons mobile; B1 layers of ions slide; B1 bonding is non-directional / attraction maintained after sliding.
Q22Metallic bonding & properties · Challenge · Paper 2 style · 6 marks
(a)Explain why potassium has a lower melting point than sodium.[2]
(b)Explain why magnesium has a much higher melting point than sodium.[2]
(c)Aircraft bodies use an alloy of aluminium containing small amounts of copper and magnesium. Explain why the alloy is harder than pure aluminium.[2]
Show answer
(a) K⁺ is larger than Na⁺ (same charge, one more shell), so the attraction between the cations and the delocalised electrons is weaker and less energy is needed to overcome it.
B1 K⁺ ion larger / outer electron further from nucleus; B1 weaker attraction between cations and delocalised electrons → less energy needed.
(b) Mg²⁺ has a higher charge and smaller radius than Na⁺, and each atom provides two delocalised electrons instead of one, so the metallic bonding is stronger.
B1 Mg²⁺ higher charge / smaller; B1 two delocalised electrons per atom → stronger attraction.
(c) Atoms of different sizes disrupt the regular layers of ions, so the layers cannot slide over each other easily.
B1 different-sized atoms distort the lattice; B1 layers less able to slide.
Q23Dot-and-cross diagrams · Core · Paper 2 style · 5 marks
(a)Describe, in words, a dot-and-cross diagram for carbon dioxide, CO₂. State the number of bonding pairs between each pair of atoms and the number of lone pairs on each atom. Show outer electrons only.[3]
(b)Describe a dot-and-cross diagram for nitrogen, N₂.[2]
Show answer
(a) O=C=O: two shared pairs (a double bond) between C and each O; carbon has no lone pairs; each oxygen has two lone pairs. Carbon has 8 outer electrons around it; each O has 8.
B1 two double bonds (2 shared pairs each, one electron of each pair from C, one from O); B1 no lone pairs on C; B1 two lone pairs on each O.
(b) Three shared pairs (triple bond) between the two N atoms, each N contributing 3 electrons; one lone pair on each N.
B1 triple bond; B1 one lone pair on each N.
Q24Dot-and-cross diagrams · Challenge · Paper 2 style · 6 marks
(a)Calculate the total number of outer-shell electrons in a molecule of carbon monoxide, CO.[1]
(b)In CO both atoms have a full outer shell. Describe the dot-and-cross diagram, showing which atom provides the electrons in each bond.[3]
(c)The cyanide ion, CN⁻, is isoelectronic with CO. State the number of outer electrons in CN⁻ and describe its bonding.[2]
Show answer
(a) 10
B1 4 + 6 = 10.
(b) Triple bond: two shared pairs, each with one electron from C and one from O, plus one dative pair with both electrons from O; one lone pair on C and one lone pair on O.
B1 three bonding pairs; B1 one of them dative from O to C; B1 one lone pair on each atom.
(c) 10 outer electrons; C≡N triple bond (three shared pairs) with one lone pair on C and one on N; the extra electron (charge) is on carbon.
B1 4 + 5 + 1 = 10; B1 triple bond + one lone pair on each atom.
Q25Dative bonds & electron pairs · Core · Paper 1 style · 1 mark
(a)Which species contains a dative covalent (co-ordinate) bond?[1]
Show answer
(a) A H₃O⁺
In H₃O⁺ an oxygen lone pair is donated to H⁺, forming a dative bond. BF₃ is electron-deficient and could accept a pair, but in BF₃ itself all three bonds are ordinary covalent bonds.
Q26Dative bonds & electron pairs · Standard · Paper 2 style · 6 marks
(a)At about 180 °C aluminium chloride vapour contains Al₂Cl₆ molecules. Explain why an AlCl₃ unit can form a dative bond and describe how two AlCl₃ units join.[3]
(b)State the arrangement of the chlorine atoms around each aluminium atom in Al₂Cl₆ and the approximate Cl–Al–Cl bond angle.[2]
(c)State how many dative bonds there are in one Al₂Cl₆ molecule.[1]
Show answer
(a) In AlCl₃ aluminium has only 6 outer electrons (electron-deficient, an empty orbital). A lone pair on a Cl atom of one AlCl₃ is donated into the empty orbital of the Al in the other unit; this happens twice, so two bridging Cl atoms each form a dative bond.
B1 Al has only 6 electrons / incomplete octet; B1 Cl lone pair donated to Al; B1 two dative bonds / two bridging Cl.
(b) Tetrahedral, about 109.5°
B1 tetrahedral; B1 ≈109.5°.
(c) 2
B1.
Q27Giant covalent structures (graphite, diamond) · Core · Paper 2 style · 6 marks
(a)Describe the structures of diamond and graphite.[4]
(b)Explain why graphite is used for electrodes but diamond is used on drill tips and for cutting glass.[2]
Show answer
(a) Diamond: each C covalently bonded to 4 others, tetrahedral, giant 3-D lattice. Graphite: each C bonded to 3 others in flat hexagonal layers; one electron per carbon delocalised; weak (id–id) forces between layers.
(b) Graphite conducts because delocalised electrons move along the layers; diamond is very hard because every atom is held by four strong covalent bonds in all directions.
Q28Giant covalent structures (graphite, diamond) · Challenge · Paper 2 style · 5 marks
(a)Describe the structure of silicon(IV) oxide and explain its high melting point (1710 °C).[3]
(b)A sample of pure quartz sand has a mass of 6.00 g. Calculate the amount, in mol, of Si–O bonds in the sample. (Mr SiO₂ = 60.1)[2]
Show answer
(a) Giant covalent lattice: each Si bonded to 4 O tetrahedrally, each O bonded to 2 Si. Melting requires breaking many strong covalent Si–O bonds, which needs a lot of energy.
B1 each Si to 4 O / each O to 2 Si; B1 giant 3-D covalent lattice; B1 many strong covalent bonds must break.
(b) 0.399 mol
M1 n(SiO₂) = 6.00/60.1 = 0.0998 mol; A1 × 4 = 0.399 mol (each Si forms four Si–O bonds).
Q29Hydrogen bonding · Core · Paper 2 style · 5 marks
(a)Describe a hydrogen bond between two water molecules. Include the partial charges, the atoms involved and the bond angle at the hydrogen atom.[3]
(b)Explain why ice floats on water, for example in the frozen lakes of Gilgit-Baltistan.[2]
Show answer
(a) δ+ H of one molecule (bonded to the very electronegative O) is attracted to a lone pair on the δ− O of another molecule; O–H···O approximately linear, 180°.
B1 δ+ on H and δ− on O; B1 lone pair on O involved; B1 O–H···O angle 180°.
(b) In ice each molecule forms hydrogen bonds to four others in an open tetrahedral lattice, so molecules are further apart than in liquid water; ice is less dense.
B1 open (rigid) lattice held by hydrogen bonds; B1 molecules further apart → lower density.
Q30Hydrogen bonding · Challenge · Paper 2 style · 7 marks
0.250 g of ethanoic acid vapour occupies 82.3 cm³ at 400 K and 101 kPa. R = 8.31 J K⁻¹ mol⁻¹.
(a)Calculate the apparent relative molecular mass of ethanoic acid in the vapour. Give your answer to 3 significant figures.[3]
(b)The formula mass of CH₃COOH is 60.0. Explain why the value in (a) is larger, and describe the species responsible.[2]
(c)Assume the vapour contains only monomers and dimers. Calculate the percentage of CH₃COOH units that are present as dimers. Give your answer to 2 significant figures.[2]
(b) Ethanoic acid molecules form dimers held by two hydrogen bonds (C=O···H–O between two molecules), so many particles have Mr 120 and the average is between 60 and 120.
B1 dimers via hydrogen bonds; B1 two hydrogen bonds between C=O of one molecule and O–H of the other.
(c) 80 %
M1 apparent M = 60.0/(1 − x/2) where x = fraction of units in dimers; A1 x = 2(1 − 60.0/100) = 0.80 → 80 %.
Similar questions
Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.
References only: find each question in its paper (official link where available, otherwise through your school).
Sigma & pi bonds / hybridisationBanker45 questions · 9/9 sessions
