School / IB / CHEMISTRY HL / Structure 2: advanced bonding (HL) Exam-style + marking analysis
Structure 2: advanced bonding (HL) Formal charge, resonance, expanded octets, hybridization, and bond enthalpy calculations.
Chemistry HL Structure 2: advanced bonding (HL)
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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. Medium 4 marks No calculator + 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. Marking analysis Medium 4 marks No calculator + 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. Easy 3 marks No calculator + 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. Marking analysis Easy 3 marks No calculator + 5 Determine the hybridization of the central atom and the molecular shape of sulfur hexafluoride, SF₆. Easy 3 marks No calculator + 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. Marking analysis Easy 3 marks No calculator + 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. Easy 3 marks No calculator + 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. Marking analysis Easy 3 marks No calculator + 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). Medium 5 marks Calculator + 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. Marking analysis Medium 5 marks Calculator + 11 Explain, using Fajans' rules, why aluminium chloride, AlCl₃, shows significant covalent character despite being formed from a metal and a non-metal. Medium 4 marks No calculator + 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. Marking analysis Medium 4 marks No calculator + 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. Medium 4 marks No calculator + 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. Marking analysis Medium 4 marks No calculator + 15 Carbon dioxide, CO₂, contains two polar C=O bonds, yet the molecule as a whole is non-polar. Explain why. Easy 3 marks No calculator + 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. Marking analysis Easy 3 marks No calculator + 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. Medium 5 marks No calculator + 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. Marking analysis Medium 5 marks No calculator + 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. Easy 3 marks No calculator + 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. Marking analysis Easy 3 marks No calculator + 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. Medium 4 marks No calculator + 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. Marking analysis Medium 4 marks No calculator + 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. Medium 4 marks No calculator + 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. Marking analysis Medium 4 marks No calculator + 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. Easy 3 marks No calculator + 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. Marking analysis Easy 3 marks No calculator + 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. Medium 5 marks Calculator + 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. Marking analysis Medium 5 marks Calculator + 29 During the addition reaction of ethene (C₂H₄) with hydrogen to form ethane (C₂H₆), describe the change in hybridization of each carbon atom. Easy 3 marks No calculator + 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. Marking analysis Easy 3 marks No calculator + 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. Easy 3 marks No calculator + 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. Marking analysis Easy 3 marks No calculator + 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. Easy 3 marks No calculator + 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. Marking analysis Easy 3 marks No calculator + 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. Medium 4 marks No calculator + 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. Marking analysis Medium 4 marks No calculator + Self-assessed 0 / 0
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