IB · CHEMISTRY HL

Chemistry HL

Structure 3: transition metals (HL) — Structure 3 HL

Name: ____________________Date: October 2, 2026

1.
Write the full electron configuration of a chromium atom, and explain why it is 1s²2s²2p⁶3s²3p⁶3d⁵4s¹ rather than the configuration 1s²2s²2p⁶3s²3p⁶3d⁴4s² predicted by the simple aufbau order.
[3 marks] · no calculator
Marking points
- States the correct electron configuration: 1s²2s²2p⁶3s²3p⁶3d⁵4s¹.
- Explains that a half-filled 3d subshell (3d⁵, with one electron in each of the five d-orbitals) has extra stability due to the symmetrical distribution of electrons and reduced electron-electron repulsion.
- Explains that this extra stability is enough to favour promoting one electron from 4s to 3d, despite 4s normally being filled first.
Examiner tip: Chromium and copper are the two classic exceptions to the simple aufbau filling order in the first transition series — both gain extra stability from a half-filled (Cr, 3d⁵) or fully-filled (Cu, 3d¹⁰) d subshell.
2.
Marking analysis: A learner attempts the following task: “Write the full electron configuration of a chromium atom, and explain why it is 1s²2s²2p⁶3s²3p⁶3d⁵4s¹ rather than the configuration 1s²2s²2p⁶3s²3p⁶3d⁴4s² predicted by the simple aufbau order.” Their response addresses only this point: “States the correct electron configuration: 1s²2s²2p⁶3s²3p⁶3d⁵4s¹.” 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 correct electron configuration: 1s²2s²2p⁶3s²3p⁶3d⁵4s¹.
- Identifies the missing requirement: Explains that a half-filled 3d subshell (3d⁵, with one electron in each of the five d-orbitals) has extra stability due to the symmetrical distribution of electrons and reduced electron-electron repulsion.
- Identifies the missing requirement: Explains that this extra stability is enough to favour promoting one electron from 4s to 3d, despite 4s normally being filled first.
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.
An iron atom has the electron configuration [Ar]3d⁶4s². Write the electron configuration of the Fe³⁺ ion, and state the number of unpaired electrons it contains.
[3 marks] · no calculator
Marking points
- States that electrons are removed from the 4s subshell before the 3d subshell when forming a transition metal cation.
- Removes both 4s electrons and one 3d electron to form Fe³⁺: [Ar]3d⁵.
- States that all 5 electrons in the 3d⁵ configuration are unpaired, one in each of the five d-orbitals (by Hund's rule).
Examiner tip: A common error is removing 3d electrons before 4s when forming a transition metal ion — always remove the 4s electrons first, even though 4s was filled before 3d in the neutral atom.
4.
Marking analysis: A learner attempts the following task: “An iron atom has the electron configuration [Ar]3d⁶4s². Write the electron configuration of the Fe³⁺ ion, and state the number of unpaired electrons it contains.” Their response addresses only this point: “States that electrons are removed from the 4s subshell before the 3d subshell when forming a transition metal cation.” 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 electrons are removed from the 4s subshell before the 3d subshell when forming a transition metal cation.
- Identifies the missing requirement: Removes both 4s electrons and one 3d electron to form Fe³⁺: [Ar]3d⁵.
- Identifies the missing requirement: States that all 5 electrons in the 3d⁵ configuration are unpaired, one in each of the five d-orbitals (by Hund's rule).
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.
Explain why zinc is not classified as a transition metal, even though it is a d-block element, while scandium is also excluded despite also being a d-block element.
[3 marks] · no calculator
Marking points
- States the definition: a transition metal must form at least one stable ion with a partially filled d subshell.
- Explains that zinc only ever forms the Zn²⁺ ion, which has a full 3d¹⁰ configuration, so zinc never has a partially filled d subshell in any of its compounds.
- Explains that scandium only ever forms the Sc³⁺ ion, which has an empty 3d⁰ configuration, so scandium never has a partially filled d subshell either.
Examiner tip: Being in the d-block is not the same as being a transition metal — always check the d-electron count of the element's actual ion(s), not just the neutral atom's position on the periodic table.
6.
Marking analysis: A learner attempts the following task: “Explain why zinc is not classified as a transition metal, even though it is a d-block element, while scandium is also excluded despite also being a d-block element.” Their response addresses only this point: “States the definition: a transition metal must form at least one stable ion with a partially filled d subshell.” 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 definition: a transition metal must form at least one stable ion with a partially filled d subshell.
- Identifies the missing requirement: Explains that zinc only ever forms the Zn²⁺ ion, which has a full 3d¹⁰ configuration, so zinc never has a partially filled d subshell in any of its compounds.
- Identifies the missing requirement: Explains that scandium only ever forms the Sc³⁺ ion, which has an empty 3d⁰ configuration, so scandium never has a partially filled d subshell either.
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.
Explain why transition metals commonly exhibit variable oxidation states, unlike most main group metals.
[3 marks] · no calculator
Marking points
- States that the 3d and 4s subshells of transition metals have similar energies.
- Explains that this allows successive electrons to be removed from both subshells at broadly similar energy costs, rather than there being one clearly preferred number of electrons to lose.
- Contrasts this with main group metals, where there is usually a large energy jump after removing the valence electrons, strongly favouring one particular ion charge.
Examiner tip: Compare the ionization energy graph of a transition metal (a gradual, steady increase) with that of a main group metal (a sharp jump after the valence electrons) to visualise why variable oxidation states arise.
8.
Marking analysis: A learner attempts the following task: “Explain why transition metals commonly exhibit variable oxidation states, unlike most main group metals.” Their response addresses only this point: “States that the 3d and 4s subshells of transition metals have similar energies.” 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 the 3d and 4s subshells of transition metals have similar energies.
- Identifies the missing requirement: Explains that this allows successive electrons to be removed from both subshells at broadly similar energy costs, rather than there being one clearly preferred number of electrons to lose.
- Identifies the missing requirement: Contrasts this with main group metals, where there is usually a large energy jump after removing the valence electrons, strongly favouring one particular ion charge.
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.
Define the term 'ligand', and state the coordination number of copper in the complex ion [Cu(H₂O)₆]²⁺.
[2 marks] · no calculator
Marking points
- Defines a ligand as a molecule or ion that donates a lone pair of electrons to a central metal ion, forming a coordinate (dative covalent) bond.
- States the coordination number of copper in [Cu(H₂O)₆]²⁺ as 6.
Examiner tip: The coordination number counts the number of coordinate bonds (dative bonds) to the central metal ion, not the number of separate ligand molecules — a bidentate ligand contributes 2 to the coordination number from a single molecule.
10.
Marking analysis: A learner attempts the following task: “Define the term 'ligand', and state the coordination number of copper in the complex ion [Cu(H₂O)₆]²⁺.” Their response addresses only this point: “Defines a ligand as a molecule or ion that donates a lone pair of electrons to a central metal ion, forming a coordinate (dative covalent) bond.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Marking points
- Recognises credit for the stated point: Defines a ligand as a molecule or ion that donates a lone pair of electrons to a central metal ion, forming a coordinate (dative covalent) bond.
- Identifies the missing requirement: States the coordination number of copper in [Cu(H₂O)₆]²⁺ as 6.
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, in terms of d-orbital splitting, why solutions of transition metal complex ions, such as [Cu(H₂O)₆]²⁺, are typically colored.
[4 marks] · no calculator
Marking points
- States that in a free transition metal ion, the five d-orbitals are degenerate (equal in energy).
- States that when ligands approach the metal ion in a complex, the d-orbitals split into two sets of different energies (in an octahedral field, three lower-energy and two higher-energy orbitals).
- Explains that an electron can absorb a photon of visible light and be excited from a lower-energy d-orbital to a higher-energy one, provided the energy gap corresponds to a visible wavelength.
- Explains that the observed color of the solution is the complementary color to the wavelength of light absorbed, since the remaining (unabsorbed) wavelengths are transmitted or reflected.
Examiner tip: Color always requires both a partially filled d subshell (so an electron has somewhere to be excited to) and ligands present to split the d-orbitals in the first place — this is why color is a signature property of transition metal complexes specifically.
12.
Marking analysis: A learner attempts the following task: “Explain, in terms of d-orbital splitting, why solutions of transition metal complex ions, such as [Cu(H₂O)₆]²⁺, are typically colored.” Their response addresses only this point: “States that in a free transition metal ion, the five d-orbitals are degenerate (equal in energy).” 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 in a free transition metal ion, the five d-orbitals are degenerate (equal in energy).
- Identifies the missing requirement: States that when ligands approach the metal ion in a complex, the d-orbitals split into two sets of different energies (in an octahedral field, three lower-energy and two higher-energy orbitals).
- Identifies the missing requirement: Explains that an electron can absorb a photon of visible light and be excited from a lower-energy d-orbital to a higher-energy one, provided the energy gap corresponds to a visible wavelength.
- Identifies the missing requirement: Explains that the observed color of the solution is the complementary color to the wavelength of light absorbed, since the remaining (unabsorbed) wavelengths are transmitted or reflected.
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.
Explain why the Zn²⁺ ion and its complexes are colorless, unlike most other transition metal ions.
[3 marks] · no calculator
Marking points
- States that Zn²⁺ has a 3d¹⁰ electron configuration, a completely filled d subshell.
- Explains that with no vacant d-orbital available, no d-to-d electron transition is possible, even though ligands still split the orbitals into two energy levels.
- Concludes that without any d-d transition to absorb visible light, no particular wavelength is absorbed, so the complex appears colorless (white or transparent in solution).
Examiner tip: A completely full (d¹⁰) or completely empty (d⁰) d subshell always produces a colorless ion, since both cases leave no partially filled set of orbitals for an electron to transition between.
14.
Marking analysis: A learner attempts the following task: “Explain why the Zn²⁺ ion and its complexes are colorless, unlike most other transition metal ions.” Their response addresses only this point: “States that Zn²⁺ has a 3d¹⁰ electron configuration, a completely filled d subshell.” 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 Zn²⁺ has a 3d¹⁰ electron configuration, a completely filled d subshell.
- Identifies the missing requirement: Explains that with no vacant d-orbital available, no d-to-d electron transition is possible, even though ligands still split the orbitals into two energy levels.
- Identifies the missing requirement: Concludes that without any d-d transition to absorb visible light, no particular wavelength is absorbed, so the complex appears colorless (white or transparent in solution).
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.
When excess ammonia solution is added to a solution of [Cu(H₂O)₆]²⁺ (pale blue), a ligand substitution reaction occurs, forming [Cu(NH₃)₄(H₂O)₂]²⁺ (deep blue). Write the equation for this reaction, and explain why a color change is observed.
[4 marks] · no calculator
Marking points
- Writes the equation: [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O.
- States that replacing water ligands with ammonia ligands changes the strength of the ligand field around the copper ion.
- Explains that a different ligand field strength changes the size of the energy gap between the split d-orbitals.
- Concludes that a different energy gap means a different wavelength of light is absorbed, producing the observed color change from pale blue to deep blue.
Examiner tip: A ligand substitution reaction almost always produces a visible color change, because different ligands create different crystal field splitting energies, even for the exact same metal ion and oxidation state.
16.
Marking analysis: A learner attempts the following task: “When excess ammonia solution is added to a solution of [Cu(H₂O)₆]²⁺ (pale blue), a ligand substitution reaction occurs, forming [Cu(NH₃)₄(H₂O)₂]²⁺ (deep blue). Write the equation for this reaction, and explain why a color change is observed.” Their response addresses only this point: “Writes the equation: [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O.” 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: Writes the equation: [Cu(H₂O)₆]²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄(H₂O)₂]²⁺ + 4H₂O.
- Identifies the missing requirement: States that replacing water ligands with ammonia ligands changes the strength of the ligand field around the copper ion.
- Identifies the missing requirement: Explains that a different ligand field strength changes the size of the energy gap between the split d-orbitals.
- Identifies the missing requirement: Concludes that a different energy gap means a different wavelength of light is absorbed, producing the observed color change from pale blue to deep blue.
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.
Explain, with reference to variable oxidation states, how a transition metal ion such as Fe²⁺/Fe³⁺ can act as a homogeneous catalyst in a redox reaction between two species that would otherwise react very slowly.
[4 marks] · no calculator
Marking points
- States that the transition metal ion provides an alternative reaction pathway with a lower activation energy than the direct, uncatalyzed reaction.
- Explains that the catalyst is first oxidized or reduced by one reactant (for example, Fe²⁺ is oxidized to Fe³⁺ by one species).
- Explains that the catalyst is then reduced or oxidized back to its original oxidation state by the second reactant (for example, Fe³⁺ is reduced back to Fe²⁺).
- Concludes that the catalyst is regenerated unchanged at the end of the overall reaction, having enabled two separate, faster steps instead of one slow direct step.
Examiner tip: The ability to readily switch between two oxidation states is exactly what makes a transition metal ion useful as a homogeneous redox catalyst — each oxidation state change corresponds to one of the two steps in the catalytic cycle.
18.
Marking analysis: A learner attempts the following task: “Explain, with reference to variable oxidation states, how a transition metal ion such as Fe²⁺/Fe³⁺ can act as a homogeneous catalyst in a redox reaction between two species that would otherwise react very slowly.” Their response addresses only this point: “States that the transition metal ion provides an alternative reaction pathway with a lower activation energy than the direct, uncatalyzed reaction.” 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 the transition metal ion provides an alternative reaction pathway with a lower activation energy than the direct, uncatalyzed reaction.
- Identifies the missing requirement: Explains that the catalyst is first oxidized or reduced by one reactant (for example, Fe²⁺ is oxidized to Fe³⁺ by one species).
- Identifies the missing requirement: Explains that the catalyst is then reduced or oxidized back to its original oxidation state by the second reactant (for example, Fe³⁺ is reduced back to Fe²⁺).
- Identifies the missing requirement: Concludes that the catalyst is regenerated unchanged at the end of the overall reaction, having enabled two separate, faster steps instead of one slow direct step.
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.
Determine the oxidation state of cobalt in the complex ion [Co(NH₃)₅Cl]²⁺.
[4 marks]
Marking points
- States that ammonia, NH₃, is a neutral ligand, contributing 0 to the overall charge.
- States that chloride, Cl⁻, contributes a charge of −1 as a ligand.
- Sets up the equation: x + 5(0) + (−1) = +2, where x is the oxidation state of cobalt.
- Solves to obtain x = +3, the oxidation state of cobalt.
Examiner tip: To find a metal's oxidation state in a complex ion, set the sum of the metal's oxidation state and all ligand charges equal to the overall charge of the complex ion — neutral ligands like NH₃ and H₂O always contribute zero.
20.
Marking analysis: A learner attempts the following task: “Determine the oxidation state of cobalt in the complex ion [Co(NH₃)₅Cl]²⁺.” Their response addresses only this point: “States that ammonia, NH₃, is a neutral ligand, contributing 0 to the overall charge.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Marking points
- Recognises credit for the stated point: States that ammonia, NH₃, is a neutral ligand, contributing 0 to the overall charge.
- Identifies the missing requirement: States that chloride, Cl⁻, contributes a charge of −1 as a ligand.
- Identifies the missing requirement: Sets up the equation: x + 5(0) + (−1) = +2, where x is the oxidation state of cobalt.
- Identifies the missing requirement: Solves to obtain x = +3, the oxidation state of cobalt.
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.
Explain why the same transition metal ion, such as Fe²⁺, can form complexes of different colors depending on the ligand attached, referring to the spectrochemical series.
[3 marks] · no calculator
Marking points
- States that different ligands produce different strengths of crystal field splitting, as ranked by the spectrochemical series.
- Explains that a stronger-field ligand produces a larger energy gap between the split d-orbitals, while a weaker-field ligand produces a smaller gap.
- Explains that since the energy gap determines which wavelength of visible light is absorbed, a different ligand field strength (even with the same metal ion) causes a different color to be observed.
Examiner tip: The spectrochemical series ranks ligands by field strength, not by charge — the small, strongly-donating CN⁻ ligand is a much stronger-field ligand than the larger, weaker-field I⁻ ion, so do not assume charge alone predicts field strength.
22.
Marking analysis: A learner attempts the following task: “Explain why the same transition metal ion, such as Fe²⁺, can form complexes of different colors depending on the ligand attached, referring to the spectrochemical series.” Their response addresses only this point: “States that different ligands produce different strengths of crystal field splitting, as ranked by the spectrochemical series.” 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 different ligands produce different strengths of crystal field splitting, as ranked by the spectrochemical series.
- Identifies the missing requirement: Explains that a stronger-field ligand produces a larger energy gap between the split d-orbitals, while a weaker-field ligand produces a smaller gap.
- Identifies the missing requirement: Explains that since the energy gap determines which wavelength of visible light is absorbed, a different ligand field strength (even with the same metal ion) causes a different color to be observed.
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.
State the shape of the complex ion [CuCl₄]²⁻, and explain why it adopts a different shape from the six-coordinate [Cu(H₂O)₆]²⁺.
[3 marks] · no calculator
Marking points
- States that [CuCl₄]²⁻ has a tetrahedral shape.
- States that [CuCl₄]²⁻ has a coordination number of 4, while [Cu(H₂O)₆]²⁺ has a coordination number of 6.
- Explains that the larger size of the chloride ligand, compared to water, means fewer chloride ions can fit around the central copper ion, resulting in a lower coordination number and a different geometry.
Examiner tip: Coordination number and shape are directly linked: 4-coordinate complexes are typically tetrahedral (or occasionally square planar), while 6-coordinate complexes are typically octahedral — ligand size is a key factor in determining which coordination number occurs.
24.
Marking analysis: A learner attempts the following task: “State the shape of the complex ion [CuCl₄]²⁻, and explain why it adopts a different shape from the six-coordinate [Cu(H₂O)₆]²⁺.” Their response addresses only this point: “States that [CuCl₄]²⁻ has a tetrahedral shape.” 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 [CuCl₄]²⁻ has a tetrahedral shape.
- Identifies the missing requirement: States that [CuCl₄]²⁻ has a coordination number of 4, while [Cu(H₂O)₆]²⁺ has a coordination number of 6.
- Identifies the missing requirement: Explains that the larger size of the chloride ligand, compared to water, means fewer chloride ions can fit around the central copper ion, resulting in a lower coordination number and a different geometry.
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.
The square planar complex [Pt(NH₃)₂Cl₂] exists as two distinct geometric isomers, cis and trans. Describe the difference between the cis and trans arrangements.
[3 marks] · no calculator
Marking points
- Describes the cis isomer as having the two identical ligands (the two NH₃ groups, or equivalently the two Cl ligands) positioned adjacent to each other, at 90° to one another.
- Describes the trans isomer as having the two identical ligands positioned directly opposite each other, at 180° to one another.
- States that cis and trans platin have measurably different chemical and physical properties, despite having the same molecular formula.
Examiner tip: Geometric (cis-trans) isomerism in square planar complexes requires two pairs of identical ligands — this is directly analogous to cis-trans isomerism around a C=C double bond, just in a different geometric arrangement.
26.
Marking analysis: A learner attempts the following task: “The square planar complex [Pt(NH₃)₂Cl₂] exists as two distinct geometric isomers, cis and trans. Describe the difference between the cis and trans arrangements.” Their response addresses only this point: “Describes the cis isomer as having the two identical ligands (the two NH₃ groups, or equivalently the two Cl ligands) positioned adjacent to each other, at 90° to one another.” 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: Describes the cis isomer as having the two identical ligands (the two NH₃ groups, or equivalently the two Cl ligands) positioned adjacent to each other, at 90° to one another.
- Identifies the missing requirement: Describes the trans isomer as having the two identical ligands positioned directly opposite each other, at 180° to one another.
- Identifies the missing requirement: States that cis and trans platin have measurably different chemical and physical properties, despite having the same molecular formula.
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 octahedral complex [Ni(en)₃]²⁺ (where 'en' is the bidentate ligand ethylenediamine) can exist as non-superimposable mirror images. Explain why this complex is chiral (optically active).
[3 marks] · no calculator
Marking points
- States that the complex has no plane of symmetry, due to the way the three bidentate ligands wrap around the central metal ion in a propeller-like arrangement.
- Explains that the mirror image of the complex cannot be rotated to superimpose exactly onto the original structure.
- Concludes that the complex and its non-superimposable mirror image are a pair of enantiomers, which rotate plane-polarized light in opposite directions.
Examiner tip: Chirality in coordination complexes with bidentate or polydentate ligands follows the same 'non-superimposable mirror image' test used for chiral carbon centres in organic chemistry — look for an absence of any internal symmetry plane.
28.
Marking analysis: A learner attempts the following task: “The octahedral complex [Ni(en)₃]²⁺ (where 'en' is the bidentate ligand ethylenediamine) can exist as non-superimposable mirror images. Explain why this complex is chiral (optically active).” Their response addresses only this point: “States that the complex has no plane of symmetry, due to the way the three bidentate ligands wrap around the central metal ion in a propeller-like arrangement.” 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 the complex has no plane of symmetry, due to the way the three bidentate ligands wrap around the central metal ion in a propeller-like arrangement.
- Identifies the missing requirement: Explains that the mirror image of the complex cannot be rotated to superimpose exactly onto the original structure.
- Identifies the missing requirement: Concludes that the complex and its non-superimposable mirror image are a pair of enantiomers, which rotate plane-polarized light in opposite directions.
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.
Give the systematic (IUPAC) name of the complex ion [Cu(NH₃)₄(H₂O)₂]²⁺.
[3 marks] · no calculator
Marking points
- Names the ammonia ligands using the prefix 'tetraammine' (4 ammonia ligands).
- Names the water ligands using the prefix 'diaqua' (2 water ligands), listing ligands in alphabetical order (ammine before aqua).
- States the full name as tetraamminediaquacopper(II), including the oxidation state of copper in Roman numerals.
Examiner tip: IUPAC naming of complex ions always lists ligands alphabetically by ligand name (ignoring the multiplying prefixes), then names the metal with its oxidation state in Roman numerals in brackets immediately after.
30.
Marking analysis: A learner attempts the following task: “Give the systematic (IUPAC) name of the complex ion [Cu(NH₃)₄(H₂O)₂]²⁺.” Their response addresses only this point: “Names the ammonia ligands using the prefix 'tetraammine' (4 ammonia ligands).” 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: Names the ammonia ligands using the prefix 'tetraammine' (4 ammonia ligands).
- Identifies the missing requirement: Names the water ligands using the prefix 'diaqua' (2 water ligands), listing ligands in alphabetical order (ammine before aqua).
- Identifies the missing requirement: States the full name as tetraamminediaquacopper(II), including the oxidation state of copper in Roman numerals.
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.
State the number of unpaired electrons in the Fe³⁺ ion ([Ar]3d⁵), and explain how this relates to the magnetic behaviour of its compounds.
[2 marks] · no calculator
Marking points
- States that by Hund's rule, the five electrons in the 3d⁵ configuration occupy all five d-orbitals singly before any pairing occurs, giving 5 unpaired electrons.
- States that unpaired electrons each act as tiny magnets, making compounds containing Fe³⁺ paramagnetic (weakly attracted into a magnetic field).
Examiner tip: The number of unpaired electrons directly determines magnetic behaviour: any unpaired electrons make a species paramagnetic, while a species with all electrons paired is diamagnetic (weakly repelled by a magnetic field).
32.
Marking analysis: A learner attempts the following task: “State the number of unpaired electrons in the Fe³⁺ ion ([Ar]3d⁵), and explain how this relates to the magnetic behaviour of its compounds.” Their response addresses only this point: “States that by Hund's rule, the five electrons in the 3d⁵ configuration occupy all five d-orbitals singly before any pairing occurs, giving 5 unpaired electrons.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Marking points
- Recognises credit for the stated point: States that by Hund's rule, the five electrons in the 3d⁵ configuration occupy all five d-orbitals singly before any pairing occurs, giving 5 unpaired electrons.
- Identifies the missing requirement: States that unpaired electrons each act as tiny magnets, making compounds containing Fe³⁺ paramagnetic (weakly attracted into a magnetic field).
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.
Describe the general trend in first ionization energy across the first row of transition metals (Sc to Zn), and explain why this trend is much less pronounced than the trend across a main group period.
[3 marks] · no calculator
Marking points
- Describes the general trend as a relatively small, somewhat irregular increase in first ionization energy from Sc to Zn, rather than a large, steady increase.
- Explains that across the transition series, additional electrons are added to the inner 3d subshell rather than to the outer (valence) shell.
- Explains that because the outermost 4s electrons being removed experience only a gradually increasing effective nuclear charge (due to d-electron shielding), the ionization energy increases only slightly and somewhat irregularly across the series.
Examiner tip: The key contrast is where new electrons are added: across a main group period they enter the valence shell directly (causing a steep ionization energy increase), while across a transition series they enter an inner d subshell (causing only a gentle increase).
34.
Marking analysis: A learner attempts the following task: “Describe the general trend in first ionization energy across the first row of transition metals (Sc to Zn), and explain why this trend is much less pronounced than the trend across a main group period.” Their response addresses only this point: “Describes the general trend as a relatively small, somewhat irregular increase in first ionization energy from Sc to Zn, rather than a large, steady increase.” 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: Describes the general trend as a relatively small, somewhat irregular increase in first ionization energy from Sc to Zn, rather than a large, steady increase.
- Identifies the missing requirement: Explains that across the transition series, additional electrons are added to the inner 3d subshell rather than to the outer (valence) shell.
- Identifies the missing requirement: Explains that because the outermost 4s electrons being removed experience only a gradually increasing effective nuclear charge (due to d-electron shielding), the ionization energy increases only slightly and somewhat irregularly across the series.
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.