IB · CHEMISTRY HL

Chemistry HL

Structure 2: advanced bonding (HL) — Structure 2 HL

Name: ____________________Date: October 2, 2026

1.
Calculate the formal charge on each atom in the Lewis structure of the nitrate ion, NO₃⁻, drawn with one N=O double bond and two N−O single bonds.
[4 marks] · no calculator
Marking points
- States the formal charge formula: FC = (valence electrons) − (non-bonding electrons) − (1/2 bonding electrons).
- Calculates the formal charge on N: 5 − 0 − 4 = +1.
- Calculates the formal charge on the double-bonded O: 6 − 4 − 2 = 0.
- Calculates the formal charge on each single-bonded O: 6 − 6 − 1 = −1, confirming the structure's overall charge of +1 + 0 + (−1) + (−1) = −1, matching NO₃⁻.
Examiner tip: Formal charges on a correct Lewis structure must always sum to the overall charge of the species — use this as a check after calculating each individual formal charge.
2.
Marking analysis: A learner attempts the following task: “Calculate the formal charge on each atom in the Lewis structure of the nitrate ion, NO₃⁻, drawn with one N=O double bond and two N−O single bonds.” Their response addresses only this point: “States the formal charge formula: FC = (valence electrons) − (non-bonding electrons) − (1/2 bonding electrons).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States the formal charge formula: FC = (valence electrons) − (non-bonding electrons) − (1/2 bonding electrons).
- Identifies the missing requirement: Calculates the formal charge on N: 5 − 0 − 4 = +1.
- Identifies the missing requirement: Calculates the formal charge on the double-bonded O: 6 − 4 − 2 = 0.
- Identifies the missing requirement: Calculates the formal charge on each single-bonded O: 6 − 6 − 1 = −1, confirming the structure's overall charge of +1 + 0 + (−1) + (−1) = −1, matching NO₃⁻.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
3.
Explain, using the concept of resonance, why all three N−O bonds in the nitrate ion NO₃⁻ are experimentally found to have the same length, intermediate between a single and a double bond.
[3 marks] · no calculator
Marking points
- States that nitrate has three equivalent resonance structures, each differing only in which N−O bond is drawn as the double bond.
- Explains that the true structure is a resonance hybrid, an average of all three contributing structures, not any single one of them.
- Explains that since each bond has double-bond character one-third of the time on average, each N−O bond has a bond order of 4/3, giving a length between a single and double bond.
Examiner tip: A resonance hybrid is a single, real, averaged structure with delocalized electron density — it is not a molecule rapidly flipping between the drawn resonance forms.
4.
Marking analysis: A learner attempts the following task: “Explain, using the concept of resonance, why all three N−O bonds in the nitrate ion NO₃⁻ are experimentally found to have the same length, intermediate between a single and a double bond.” Their response addresses only this point: “States that nitrate has three equivalent resonance structures, each differing only in which N−O bond is drawn as the double bond.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that nitrate has three equivalent resonance structures, each differing only in which N−O bond is drawn as the double bond.
- Identifies the missing requirement: Explains that the true structure is a resonance hybrid, an average of all three contributing structures, not any single one of them.
- Identifies the missing requirement: Explains that since each bond has double-bond character one-third of the time on average, each N−O bond has a bond order of 4/3, giving a length between a single and double bond.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
5.
Determine the hybridization of the central atom and the molecular shape of sulfur hexafluoride, SF₆.
[3 marks] · no calculator
Marking points
- States that sulfur has 6 bonding electron domains around it (6 S−F bonds) and no lone pairs.
- States the hybridization as sp³d².
- States the molecular shape as octahedral, with bond angles of 90°.
Examiner tip: Count electron domains (bonds plus lone pairs) around the central atom first — this number directly tells you the hybridization (4 → sp³, 5 → sp³d, 6 → sp³d²) before you even consider the molecule's final shape.
6.
Marking analysis: A learner attempts the following task: “Determine the hybridization of the central atom and the molecular shape of sulfur hexafluoride, SF₆.” Their response addresses only this point: “States that sulfur has 6 bonding electron domains around it (6 S−F bonds) and no lone pairs.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that sulfur has 6 bonding electron domains around it (6 S−F bonds) and no lone pairs.
- Identifies the missing requirement: States the hybridization as sp³d².
- Identifies the missing requirement: States the molecular shape as octahedral, with bond angles of 90°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
7.
Sulfur tetrafluoride, SF₄, has one lone pair on the central sulfur atom in addition to four bonding pairs. Predict the molecular shape of SF₄ and explain how the lone pair affects the bond angles.
[3 marks] · no calculator
Marking points
- States that with 5 electron domains (4 bonding, 1 lone pair), the electron domain geometry is trigonal bipyramidal, but the molecular shape is see-saw.
- Explains that the lone pair occupies an equatorial position, where it experiences less repulsion than it would in an axial position.
- Explains that lone pair–bonding pair repulsion is stronger than bonding pair–bonding pair repulsion, pushing the bonding pairs slightly closer together and distorting the bond angles from the ideal trigonal bipyramidal values.
Examiner tip: In a trigonal bipyramidal electron domain geometry, a lone pair always occupies an equatorial position first, since equatorial positions have fewer close (90°) neighbours than axial positions.
8.
Marking analysis: A learner attempts the following task: “Sulfur tetrafluoride, SF₄, has one lone pair on the central sulfur atom in addition to four bonding pairs. Predict the molecular shape of SF₄ and explain how the lone pair affects the bond angles.” Their response addresses only this point: “States that with 5 electron domains (4 bonding, 1 lone pair), the electron domain geometry is trigonal bipyramidal, but the molecular shape is see-saw.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that with 5 electron domains (4 bonding, 1 lone pair), the electron domain geometry is trigonal bipyramidal, but the molecular shape is see-saw.
- Identifies the missing requirement: Explains that the lone pair occupies an equatorial position, where it experiences less repulsion than it would in an axial position.
- Identifies the missing requirement: Explains that lone pair–bonding pair repulsion is stronger than bonding pair–bonding pair repulsion, pushing the bonding pairs slightly closer together and distorting the bond angles from the ideal trigonal bipyramidal values.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
9.
Calculate the enthalpy change for the complete combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), using the average bond enthalpies: C−H = 414, O=O = 498, C=O = 804, O−H = 463 (all in kJ mol⁻¹).
[5 marks]
Marking points
- Identifies the bonds broken in the reactants: 4 × C−H and 2 × O=O.
- Calculates the energy to break these bonds: 4(414) + 2(498) = 2652 kJ mol⁻¹.
- Identifies the bonds formed in the products: 2 × C=O (in CO₂) and 4 × O−H (in 2 H₂O).
- Calculates the energy released forming these bonds: 2(804) + 4(463) = 3460 kJ mol⁻¹.
- Uses ΔH = (energy to break bonds) − (energy released forming bonds) = 2652 − 3460 = −808 kJ mol⁻¹.
Examiner tip: Bond enthalpy calculations always follow 'break then make': sum the energy needed to break all reactant bonds, subtract the energy released forming all product bonds — a negative result confirms an exothermic reaction, consistent with combustion.
10.
Marking analysis: A learner attempts the following task: “Calculate the enthalpy change for the complete combustion of methane, CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g), using the average bond enthalpies: C−H = 414, O=O = 498, C=O = 804, O−H = 463 (all in kJ mol⁻¹).” Their response addresses only this point: “Identifies the bonds broken in the reactants: 4 × C−H and 2 × O=O.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Identifies the bonds broken in the reactants: 4 × C−H and 2 × O=O.
- Identifies the missing requirement: Calculates the energy to break these bonds: 4(414) + 2(498) = 2652 kJ mol⁻¹.
- Identifies the missing requirement: Identifies the bonds formed in the products: 2 × C=O (in CO₂) and 4 × O−H (in 2 H₂O).
- Identifies the missing requirement: Calculates the energy released forming these bonds: 2(804) + 4(463) = 3460 kJ mol⁻¹.
- Identifies the missing requirement: Uses ΔH = (energy to break bonds) − (energy released forming bonds) = 2652 − 3460 = −808 kJ mol⁻¹.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
11.
Explain, using Fajans' rules, why aluminium chloride, AlCl₃, shows significant covalent character despite being formed from a metal and a non-metal.
[4 marks] · no calculator
Marking points
- States that covalent character in an ionic compound increases when the cation is small and highly charged, giving it a high charge density (high polarizing power).
- Explains that Al³⁺ is small and has a 3+ charge, giving it an unusually high charge density for a metal cation.
- Explains that this high charge density distorts (polarizes) the electron cloud of the large, polarizable chloride anion, pulling shared electron density toward the aluminium and giving the bond significant covalent character.
- Concludes that the greater the degree of polarization, the more covalent (and less purely ionic) the bond becomes.
Examiner tip: Apply Fajans' rules by checking both ions: a small, highly-charged cation and a large, easily-polarizable anion together maximise covalent character — either factor alone is a weaker predictor.
12.
Marking analysis: A learner attempts the following task: “Explain, using Fajans' rules, why aluminium chloride, AlCl₃, shows significant covalent character despite being formed from a metal and a non-metal.” Their response addresses only this point: “States that covalent character in an ionic compound increases when the cation is small and highly charged, giving it a high charge density (high polarizing power).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that covalent character in an ionic compound increases when the cation is small and highly charged, giving it a high charge density (high polarizing power).
- Identifies the missing requirement: Explains that Al³⁺ is small and has a 3+ charge, giving it an unusually high charge density for a metal cation.
- Identifies the missing requirement: Explains that this high charge density distorts (polarizes) the electron cloud of the large, polarizable chloride anion, pulling shared electron density toward the aluminium and giving the bond significant covalent character.
- Identifies the missing requirement: Concludes that the greater the degree of polarization, the more covalent (and less purely ionic) the bond becomes.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
13.
Benzene, C₆H₆, has sp² hybridized carbon atoms arranged in a ring. Explain how this hybridization leads to a delocalized system of pi electrons above and below the ring.
[4 marks] · no calculator
Marking points
- States that each carbon atom is sp² hybridized, forming three sigma bonds (to two adjacent carbons and one hydrogen) that lie in the plane of the ring.
- States that each carbon has one unhybridized p-orbital remaining, oriented perpendicular to the plane of the ring.
- Explains that these six parallel p-orbitals overlap sideways with each other all the way around the ring, rather than forming three isolated localized double bonds.
- Concludes that this gives a delocalized ring of pi electron density above and below the plane of the carbon ring, explaining benzene's equal C−C bond lengths and extra stability.
Examiner tip: Delocalization always requires parallel, unhybridized p-orbitals on adjacent atoms — this is why only the sp² (not sp³) carbons in a structure can participate in an extended pi system.
14.
Marking analysis: A learner attempts the following task: “Benzene, C₆H₆, has sp² hybridized carbon atoms arranged in a ring. Explain how this hybridization leads to a delocalized system of pi electrons above and below the ring.” Their response addresses only this point: “States that each carbon atom is sp² hybridized, forming three sigma bonds (to two adjacent carbons and one hydrogen) that lie in the plane of the ring.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that each carbon atom is sp² hybridized, forming three sigma bonds (to two adjacent carbons and one hydrogen) that lie in the plane of the ring.
- Identifies the missing requirement: States that each carbon has one unhybridized p-orbital remaining, oriented perpendicular to the plane of the ring.
- Identifies the missing requirement: Explains that these six parallel p-orbitals overlap sideways with each other all the way around the ring, rather than forming three isolated localized double bonds.
- Identifies the missing requirement: Concludes that this gives a delocalized ring of pi electron density above and below the plane of the carbon ring, explaining benzene's equal C−C bond lengths and extra stability.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
15.
Carbon dioxide, CO₂, contains two polar C=O bonds, yet the molecule as a whole is non-polar. Explain why.
[3 marks] · no calculator
Marking points
- States that each individual C=O bond is polar, since oxygen is more electronegative than carbon.
- States that CO₂ has a linear molecular shape, with the two C=O bond dipoles pointing in exactly opposite directions.
- Explains that these two equal and opposite bond dipoles cancel out, giving a net molecular dipole moment of zero, so the molecule is non-polar overall.
Examiner tip: Molecular polarity depends on both bond polarity and molecular shape — symmetric arrangements of identical polar bonds (linear, trigonal planar, tetrahedral with identical substituents) always cancel to give a non-polar molecule.
16.
Marking analysis: A learner attempts the following task: “Carbon dioxide, CO₂, contains two polar C=O bonds, yet the molecule as a whole is non-polar. Explain why.” Their response addresses only this point: “States that each individual C=O bond is polar, since oxygen is more electronegative than carbon.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that each individual C=O bond is polar, since oxygen is more electronegative than carbon.
- Identifies the missing requirement: States that CO₂ has a linear molecular shape, with the two C=O bond dipoles pointing in exactly opposite directions.
- Identifies the missing requirement: Explains that these two equal and opposite bond dipoles cancel out, giving a net molecular dipole moment of zero, so the molecule is non-polar overall.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
17.
Two possible Lewis structures can be drawn for the thiocyanate ion, SCN⁻: one with S=C=N (cumulated double bonds) and one with S≡C−N (a triple and a single bond). Use formal charges to determine which structure is the more stable, major contributor.
[5 marks] · no calculator
Marking points
- Calculates formal charges for S=C=N: S: 6−4−2=0, C: 4−0−4=0, N: 5−4−2=−1.
- Calculates formal charges for S≡C−N: S: 6−2−3=+1, C: 4−0−4=0, N: 5−6−1=−2.
- States that the most stable Lewis structure minimises formal charge magnitudes and places any negative formal charge on the most electronegative atom.
- Compares the two structures: S=C=N has smaller formal charge magnitudes (0, 0, −1) than S≡C−N (+1, 0, −2).
- Concludes that S=C=N is the more stable, major contributing structure.
Examiner tip: When comparing candidate Lewis structures, prefer the one with formal charges closest to zero overall, and when a negative formal charge is unavoidable, prefer it on the most electronegative atom present.
18.
Marking analysis: A learner attempts the following task: “Two possible Lewis structures can be drawn for the thiocyanate ion, SCN⁻: one with S=C=N (cumulated double bonds) and one with S≡C−N (a triple and a single bond). Use formal charges to determine which structure is the more stable, major contributor.” Their response addresses only this point: “Calculates formal charges for S=C=N: S: 6−4−2=0, C: 4−0−4=0, N: 5−4−2=−1.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks] · no calculator
Marking points
- Recognises credit for the stated point: Calculates formal charges for S=C=N: S: 6−4−2=0, C: 4−0−4=0, N: 5−4−2=−1.
- Identifies the missing requirement: Calculates formal charges for S≡C−N: S: 6−2−3=+1, C: 4−0−4=0, N: 5−6−1=−2.
- Identifies the missing requirement: States that the most stable Lewis structure minimises formal charge magnitudes and places any negative formal charge on the most electronegative atom.
- Identifies the missing requirement: Compares the two structures: S=C=N has smaller formal charge magnitudes (0, 0, −1) than S≡C−N (+1, 0, −2).
- Identifies the missing requirement: Concludes that S=C=N is the more stable, major contributing structure.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
19.
Count the number of sigma (σ) and pi (π) bonds in a molecule of ethyne (acetylene), C₂H₂, which contains a carbon-carbon triple bond.
[3 marks] · no calculator
Marking points
- States that each C−H bond is a single sigma bond, giving 2 sigma bonds from the two C−H bonds.
- States that the carbon-carbon triple bond consists of one sigma bond and two pi bonds.
- Totals 3 sigma bonds (2 C−H + 1 C−C) and 2 pi bonds.
Examiner tip: Every single bond is always exactly one sigma bond; a double bond is always one sigma plus one pi; a triple bond is always one sigma plus two pi — only the first bond between two atoms can ever be a sigma bond.
20.
Marking analysis: A learner attempts the following task: “Count the number of sigma (σ) and pi (π) bonds in a molecule of ethyne (acetylene), C₂H₂, which contains a carbon-carbon triple bond.” Their response addresses only this point: “States that each C−H bond is a single sigma bond, giving 2 sigma bonds from the two C−H bonds.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that each C−H bond is a single sigma bond, giving 2 sigma bonds from the two C−H bonds.
- Identifies the missing requirement: States that the carbon-carbon triple bond consists of one sigma bond and two pi bonds.
- Identifies the missing requirement: Totals 3 sigma bonds (2 C−H + 1 C−C) and 2 pi bonds.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
21.
Predict the molecular shape of chlorine trifluoride, ClF₃, which has 3 bonding pairs and 2 lone pairs on the central chlorine atom, and explain the effect of the lone pairs on the shape.
[4 marks] · no calculator
Marking points
- States that with 5 electron domains (3 bonding, 2 lone pairs), the electron domain geometry is trigonal bipyramidal.
- States that both lone pairs occupy equatorial positions, to minimise the stronger lone pair–lone pair and lone pair–bonding pair repulsions.
- States that the resulting molecular shape is T-shaped.
- Explains that the lone pairs compress the F−Cl−F bond angles to slightly less than the ideal 90°.
Examiner tip: With a trigonal bipyramidal electron domain geometry, lone pairs always fill equatorial positions before axial ones, since this minimises the number of close 90° repulsions they experience.
22.
Marking analysis: A learner attempts the following task: “Predict the molecular shape of chlorine trifluoride, ClF₃, which has 3 bonding pairs and 2 lone pairs on the central chlorine atom, and explain the effect of the lone pairs on the shape.” Their response addresses only this point: “States that with 5 electron domains (3 bonding, 2 lone pairs), the electron domain geometry is trigonal bipyramidal.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that with 5 electron domains (3 bonding, 2 lone pairs), the electron domain geometry is trigonal bipyramidal.
- Identifies the missing requirement: States that both lone pairs occupy equatorial positions, to minimise the stronger lone pair–lone pair and lone pair–bonding pair repulsions.
- Identifies the missing requirement: States that the resulting molecular shape is T-shaped.
- Identifies the missing requirement: Explains that the lone pairs compress the F−Cl−F bond angles to slightly less than the ideal 90°.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
23.
Compare the structure and bonding of diamond and carbon dioxide, and explain why diamond has an extremely high melting point while carbon dioxide sublimes at a low temperature.
[4 marks] · no calculator
Marking points
- States that diamond is a giant covalent (network) structure, in which every carbon atom is covalently bonded to four others in a continuous three-dimensional lattice.
- States that carbon dioxide is a simple molecular structure, consisting of discrete CO₂ molecules held together only by weak London (dispersion) forces.
- Explains that melting diamond requires breaking strong covalent bonds throughout the lattice, which needs a very large amount of energy.
- Explains that subliming carbon dioxide only requires overcoming the weak intermolecular forces between molecules, not breaking any covalent bonds, which needs far less energy.
Examiner tip: When comparing melting or boiling points, always identify what must actually be broken: strong covalent (or ionic) bonds throughout a giant structure, or only weak intermolecular forces between separate molecules.
24.
Marking analysis: A learner attempts the following task: “Compare the structure and bonding of diamond and carbon dioxide, and explain why diamond has an extremely high melting point while carbon dioxide sublimes at a low temperature.” Their response addresses only this point: “States that diamond is a giant covalent (network) structure, in which every carbon atom is covalently bonded to four others in a continuous three-dimensional lattice.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that diamond is a giant covalent (network) structure, in which every carbon atom is covalently bonded to four others in a continuous three-dimensional lattice.
- Identifies the missing requirement: States that carbon dioxide is a simple molecular structure, consisting of discrete CO₂ molecules held together only by weak London (dispersion) forces.
- Identifies the missing requirement: Explains that melting diamond requires breaking strong covalent bonds throughout the lattice, which needs a very large amount of energy.
- Identifies the missing requirement: Explains that subliming carbon dioxide only requires overcoming the weak intermolecular forces between molecules, not breaking any covalent bonds, which needs far less energy.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
25.
Determine the hybridization of bromine and the molecular shape of bromine pentafluoride, BrF₅, which has 5 bonding pairs and 1 lone pair on the central bromine atom.
[3 marks] · no calculator
Marking points
- States that with 6 electron domains (5 bonding, 1 lone pair), the hybridization is sp³d².
- States that the electron domain geometry is octahedral.
- States that the molecular shape, once the lone pair is accounted for, is square pyramidal.
Examiner tip: The hybridization only depends on the total number of electron domains, but the final molecular shape name depends on how many of those domains are lone pairs versus bonds — always state both separately.
26.
Marking analysis: A learner attempts the following task: “Determine the hybridization of bromine and the molecular shape of bromine pentafluoride, BrF₅, which has 5 bonding pairs and 1 lone pair on the central bromine atom.” Their response addresses only this point: “States that with 6 electron domains (5 bonding, 1 lone pair), the hybridization is sp³d².” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that with 6 electron domains (5 bonding, 1 lone pair), the hybridization is sp³d².
- Identifies the missing requirement: States that the electron domain geometry is octahedral.
- Identifies the missing requirement: States that the molecular shape, once the lone pair is accounted for, is square pyramidal.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
27.
The enthalpy of hydrogenation of cyclohexene (one C=C double bond) is −120 kJ mol⁻¹. If benzene behaved as a theoretical molecule with three isolated, non-interacting C=C double bonds ('Kekulé benzene'), predict its theoretical enthalpy of hydrogenation. Given that the experimental enthalpy of hydrogenation of benzene is actually −208 kJ mol⁻¹, calculate the resonance (delocalization) stabilization energy of benzene.
[5 marks]
Marking points
- Predicts the theoretical enthalpy of hydrogenation of 'Kekulé benzene' as 3 × (−120) = −360 kJ mol⁻¹.
- States the experimental value as −208 kJ mol⁻¹, which is less exothermic (less negative) than the theoretical prediction.
- Calculates the resonance stabilization energy as the difference: −360 − (−208) = −152 kJ mol⁻¹.
- Interprets this to mean benzene is 152 kJ mol⁻¹ more stable (lower in energy) than the hypothetical Kekulé structure would predict.
- Attributes this extra stability to the delocalization of the pi electrons around the whole ring, rather than their being localized in three separate double bonds.
Examiner tip: This enthalpy of hydrogenation comparison is the classic experimental evidence for benzene's delocalized structure — the molecule 'needs less energy released' than predicted because it started out more stable than three isolated double bonds.
28.
Marking analysis: A learner attempts the following task: “The enthalpy of hydrogenation of cyclohexene (one C=C double bond) is −120 kJ mol⁻¹. If benzene behaved as a theoretical molecule with three isolated, non-interacting C=C double bonds ('Kekulé benzene'), predict its theoretical enthalpy of hydrogenation. Given that the experimental enthalpy of hydrogenation of benzene is actually −208 kJ mol⁻¹, calculate the resonance (delocalization) stabilization energy of benzene.” Their response addresses only this point: “Predicts the theoretical enthalpy of hydrogenation of 'Kekulé benzene' as 3 × (−120) = −360 kJ mol⁻¹.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 marks]
Marking points
- Recognises credit for the stated point: Predicts the theoretical enthalpy of hydrogenation of 'Kekulé benzene' as 3 × (−120) = −360 kJ mol⁻¹.
- Identifies the missing requirement: States the experimental value as −208 kJ mol⁻¹, which is less exothermic (less negative) than the theoretical prediction.
- Identifies the missing requirement: Calculates the resonance stabilization energy as the difference: −360 − (−208) = −152 kJ mol⁻¹.
- Identifies the missing requirement: Interprets this to mean benzene is 152 kJ mol⁻¹ more stable (lower in energy) than the hypothetical Kekulé structure would predict.
- Identifies the missing requirement: Attributes this extra stability to the delocalization of the pi electrons around the whole ring, rather than their being localized in three separate double bonds.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
29.
During the addition reaction of ethene (C₂H₄) with hydrogen to form ethane (C₂H₆), describe the change in hybridization of each carbon atom.
[3 marks] · no calculator
Marking points
- States that each carbon atom in ethene is sp² hybridized, part of a C=C double bond with a trigonal planar arrangement around each carbon.
- States that each carbon atom in ethane is sp³ hybridized, part of a C−C single bond with a tetrahedral arrangement around each carbon.
- Explains that the addition of hydrogen breaks the pi bond of the double bond, converting each carbon from sp² to sp³ as it gains a new sigma bond to hydrogen.
Examiner tip: Whenever a carbon atom gains an extra sigma bond during a reaction, its hybridization always increases in s-character complexity in the sequence sp → sp² → sp³ — breaking a pi bond is what enables this change.
30.
Marking analysis: A learner attempts the following task: “During the addition reaction of ethene (C₂H₄) with hydrogen to form ethane (C₂H₆), describe the change in hybridization of each carbon atom.” Their response addresses only this point: “States that each carbon atom in ethene is sp² hybridized, part of a C=C double bond with a trigonal planar arrangement around each carbon.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that each carbon atom in ethene is sp² hybridized, part of a C=C double bond with a trigonal planar arrangement around each carbon.
- Identifies the missing requirement: States that each carbon atom in ethane is sp³ hybridized, part of a C−C single bond with a tetrahedral arrangement around each carbon.
- Identifies the missing requirement: Explains that the addition of hydrogen breaks the pi bond of the double bond, converting each carbon from sp² to sp³ as it gains a new sigma bond to hydrogen.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
31.
The nitrite ion, NO₂⁻, has two equivalent resonance structures. Draw both resonance structures, state the formal charge on each atom in one of them, and state the bond order of each N−O bond in the resonance hybrid.
[3 marks] · no calculator
Marking points
- Draws two resonance structures of NO₂⁻, each with one N=O double bond and one N−O single bond, differing in which oxygen carries the double bond.
- States the formal charges in one structure: N (one double bond, one single bond, one lone pair): 5 − 2 − 3 = 0; the double-bonded O: 6 − 4 − 2 = 0; the single-bonded O: 6 − 6 − 1 = −1.
- States the bond order of each N−O bond in the resonance hybrid as 3/2 (1.5), since each bond is a double bond in one structure and a single bond in the other, averaged equally.
Examiner tip: Bond order in a resonance hybrid is found by averaging the bond order of that same bond across all equivalent contributing structures — with two structures and bonds that are double in one and single in the other, the average is always 1.5.
32.
Marking analysis: A learner attempts the following task: “The nitrite ion, NO₂⁻, has two equivalent resonance structures. Draw both resonance structures, state the formal charge on each atom in one of them, and state the bond order of each N−O bond in the resonance hybrid.” Their response addresses only this point: “Draws two resonance structures of NO₂⁻, each with one N=O double bond and one N−O single bond, differing in which oxygen carries the double bond.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: Draws two resonance structures of NO₂⁻, each with one N=O double bond and one N−O single bond, differing in which oxygen carries the double bond.
- Identifies the missing requirement: States the formal charges in one structure: N (one double bond, one single bond, one lone pair): 5 − 2 − 3 = 0; the double-bonded O: 6 − 4 − 2 = 0; the single-bonded O: 6 − 6 − 1 = −1.
- Identifies the missing requirement: States the bond order of each N−O bond in the resonance hybrid as 3/2 (1.5), since each bond is a double bond in one structure and a single bond in the other, averaged equally.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
33.
Phosphorus pentachloride, PCl₅, has 5 bonding pairs and no lone pairs on the central phosphorus atom. State the hybridization of phosphorus and the molecular shape of PCl₅, including the bond angles.
[3 marks] · no calculator
Marking points
- States the hybridization as sp³d.
- States the molecular shape as trigonal bipyramidal.
- States that the bond angles are 90° (axial-equatorial), 120° (equatorial-equatorial), and 180° (axial-axial), noting that unlike a tetrahedral or octahedral shape, a trigonal bipyramidal shape has more than one distinct bond angle.
Examiner tip: Trigonal bipyramidal is the only common VSEPR shape with more than one distinct ideal bond angle — always state all of them (90°, 120°, and 180°) rather than giving a single value.
34.
Marking analysis: A learner attempts the following task: “Phosphorus pentachloride, PCl₅, has 5 bonding pairs and no lone pairs on the central phosphorus atom. State the hybridization of phosphorus and the molecular shape of PCl₅, including the bond angles.” Their response addresses only this point: “States the hybridization as sp³d.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Marking points
- Recognises credit for the stated point: States the hybridization as sp³d.
- Identifies the missing requirement: States the molecular shape as trigonal bipyramidal.
- Identifies the missing requirement: States that the bond angles are 90° (axial-equatorial), 120° (equatorial-equatorial), and 180° (axial-axial), noting that unlike a tetrahedral or octahedral shape, a trigonal bipyramidal shape has more than one distinct bond angle.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
35.
Explain why the bond angle in ammonia, NH₃ (107°), is slightly smaller than the ideal tetrahedral bond angle of 109.5°, while the bond angle in water, H₂O (104.5°), is smaller still.
[4 marks] · no calculator
Marking points
- States that lone pair–lone pair repulsion is greater than lone pair–bonding pair repulsion, which in turn is greater than bonding pair–bonding pair repulsion.
- Explains that ammonia has one lone pair, which compresses the H−N−H bond angles slightly below the tetrahedral ideal through extra lone pair–bonding pair repulsion.
- Explains that water has two lone pairs, which exert even greater repulsion (including lone pair–lone pair repulsion) on the two bonding pairs, compressing the H−O−H angle further still.
- Concludes that more lone pairs on the central atom produce progressively greater compression of the bond angle below the ideal tetrahedral value.
Examiner tip: Remember the repulsion strength order: lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair — this single ranking explains the entire trend in bond angle compression across methane, ammonia, and water.
36.
Marking analysis: A learner attempts the following task: “Explain why the bond angle in ammonia, NH₃ (107°), is slightly smaller than the ideal tetrahedral bond angle of 109.5°, while the bond angle in water, H₂O (104.5°), is smaller still.” Their response addresses only this point: “States that lone pair–lone pair repulsion is greater than lone pair–bonding pair repulsion, which in turn is greater than bonding pair–bonding pair repulsion.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that lone pair–lone pair repulsion is greater than lone pair–bonding pair repulsion, which in turn is greater than bonding pair–bonding pair repulsion.
- Identifies the missing requirement: Explains that ammonia has one lone pair, which compresses the H−N−H bond angles slightly below the tetrahedral ideal through extra lone pair–bonding pair repulsion.
- Identifies the missing requirement: Explains that water has two lone pairs, which exert even greater repulsion (including lone pair–lone pair repulsion) on the two bonding pairs, compressing the H−O−H angle further still.
- Identifies the missing requirement: Concludes that more lone pairs on the central atom produce progressively greater compression of the bond angle below the ideal tetrahedral value.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.