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Chemistry  ·  Year 11  ·  Module 1  ·  Lesson 19

HSC Exam Practice

Electron Configuration and Chemical Behaviour

10 questions / 3 sections / 33 marks total
Section 1

Short answer

1.Short answer

1.1

Define isoelectronic. Give one example of an isoelectronic pair drawn from Period 3 of the periodic table.

2marks Band 3
1.2

Identify the ion formed by each of the following elements when they react to achieve a noble gas configuration, and write the electron configuration of each ion: (a) Potassium (Z = 19), (b) Sulfur (Z = 16), (c) Aluminium (Z = 13).

3marks Band 3
1.3

Explain why elements in the same group of the periodic table display similar chemical behaviour. In your answer, refer to valence electron configuration.

3marks Band 3–4
1.4

Distinguish between the ion formation behaviour of sodium (Group 1) and iron (Group 8, transition metal), with reference to the number of stable ions each element forms and the subshells involved.

4marks Band 3–4
1.5

Outline why a student who states that “anions are smaller than their parent atom because they have gained extra nuclear charge” is incorrect. Provide the correct explanation for how gaining electrons affects ionic radius.

3marks Band 4
1.6

Describe how the electron configuration of copper (Z = 29) explains why CuCl is a colourless solid while CuCl2 is a yellow-brown solid. Write the electron configurations of Cu+ and Cu²+ in your answer.

4marks Band 4
Section 2

Data response

2.Data response — reactivity of Group 2 metals with water

2.1

A student investigated the reactivity of Group 2 metals with water by measuring the volume of hydrogen gas produced in 60 seconds when 0.10 g of each metal was added to 50 mL of water at 25 °C. The table below shows the first ionisation energies and the experimental gas volumes.

Metal Z Atomic radius (pm) IE1 (kJ mol−¹) H2 volume (mL / 60 s)
Magnesium (Mg) 12 160 738 2
Calcium (Ca) 20 197 590 28
Strontium (Sr) 38 215 549 61
Table 2.1. Group 2 metal reactivity data. Illustrative values based on known trends. Reaction: M(s) + 2H2O(l) → M(OH)2(aq) + H2(g).

(a) Describe the trend in H2 gas production rate from Mg to Sr and link it to the trend in first ionisation energy. (3 marks)

(b) Account for the trend in first ionisation energy from Mg to Sr by referring to atomic radius and electron shielding. In your answer, write the valence electron configurations of Mg, Ca and Sr. (4 marks)

(c) Barium (Ba, Z = 56) is below strontium in Group 2. Predict the volume of H2 Ba would produce in 60 seconds under the same conditions, and justify your prediction using electron configuration principles. (2 marks)

9marks Band 4–5
Section 3

Extended response

3.Extended response

3.1

Evaluate the claim that “electron configuration completely determines how an element will behave chemically.” In your response, analyse how electron configuration explains (i) group behaviour and periodic trends, (ii) the distinction between ionic and covalent compound formation, and (iii) the behaviour of transition metals. Discuss at least one limitation of using electron configuration alone to predict chemical behaviour, and refer to specific named elements from the NSW HSC Chemistry curriculum.

7marks Band 5–6

Chemistry · Year 11 · Module 1 · Lesson 19

Answer Key & Marking Guidelines

1.1

Section 1 · Short answer · 2 marks · Band 3

Sample response. Isoelectronic describes two or more species that have the same number of electrons and the same electron configuration but different nuclear charges (1 mark). Example (Period 3): Na+ and Ne are isoelectronic — both have 10 electrons with configuration 1s²2s²2p&sup6; — or Cl− and Ar (18 electrons each, configuration 1s²2s²2p&sup6;3s²3p&sup6;) (1 mark).

Marking notes. 1 mark for a definition including same number of electrons and different nuclear charges. 1 mark for a valid Period 3 example (accept Na+/Ne/Mg²+/Al³+ for 10-electron group; Cl−/Ar/S²−/K+ for 18-electron group).

1.2

Section 1 · Short answer · 3 marks · Band 3

Sample response. (a) K (Z = 19): config [Ar]4s¹ → loses 1e− → K+, config [Ar] = 1s²2s²2p&sup6;3s²3p&sup6; (isoelectronic with Ar). (b) S (Z = 16): config [Ne]3s²3p&sup4; → gains 2e− → S²−, config [Ne]3s²3p&sup6; = [Ar] (isoelectronic with Ar). (c) Al (Z = 13): config [Ne]3s²3p¹ → loses 3e− → Al³+, config 1s²2s²2p&sup6; = [Ne] (isoelectronic with Ne).

Marking notes. 1 mark per element: ion correctly named + ion configuration correctly written. Accept [Ar] notation or full subshell notation.

1.3

Section 1 · Short answer · 3 marks · Band 3–4

Sample response. Elements in the same group have the same number of valence electrons in the same type of subshell (e.g. all Group 1 elements have the configuration ns¹; all Group 17 elements have ns²np&sup5;) (1 mark). It is the valence electrons — not the core electrons — that determine how an element bonds and reacts. Elements with the same valence configuration experience the same “drive” to gain, lose or share electrons to achieve a noble gas configuration, so they form the same types of ions and compounds (1 mark). The vigour or rate of reaction may differ down the group because atomic radius and ionisation energy change, but the type of reaction is fixed by the valence configuration (1 mark).

Marking notes. 1 mark for identifying valence electrons as the basis of group behaviour. 1 mark for explaining that same valence configuration means same bonding tendency (gaining/losing/sharing). 1 mark for distinguishing reaction type (fixed by config) from reaction vigour (varies by position in group).

1.4

Section 1 · Short answer · 4 marks · Band 3–4

Sample response. Sodium (Group 1): config [Ne]3s¹. It forms only one stable ion, Na+ ([Ne]), by losing its single 3s valence electron. After Na+ is formed, the second ionisation energy is nearly 10 times greater (IE2 ≈ 4562 kJ mol−¹ vs IE1 ≈ 496 kJ mol−¹) because the next electron comes from the tightly held inner 2p shell. No second ion (Na²+) forms under normal conditions (2 marks: 1 for config + one ion; 1 for IE evidence/explanation). Iron (Z = 26): config [Ar]3d&sup6;4s². It forms two common ions, Fe²+ ([Ar]3d&sup6;) and Fe³+ ([Ar]3d&sup5;), by losing 4s electrons first, then a 3d electron. The 3d and 4s subshells have similar energies in the transition metals, so varying numbers of electrons from both can be removed under different oxidising conditions. Note Fe³+ is stabilised by the extra stability of the half-filled 3d&sup5; configuration (2 marks: 1 for two ions with correct configs; 1 for 3d/4s similar energy explanation).

Marking notes. 1 mark for Na+ config with correct explanation (only one valence electron; IE2 too large). 1 mark for explicitly contrasting that removing a 2p electron from Na would require far more energy than any lattice energy can compensate. 1 mark for Fe²+ and Fe³+ both correctly identified with configurations (4s removed before 3d). 1 mark for explaining variable oxidation states via similar 3d/4s energies.

1.5

Section 1 · Short answer · 3 marks · Band 4

Sample response. The student’s claim is incorrect in two ways: (1) gaining electrons does not add nuclear charge — the nuclear charge (number of protons) is unchanged when electrons are gained (1 mark). (2) Anions are larger than the parent atom, not smaller. When an atom gains electrons, the number of electrons increases while the nuclear charge stays the same. The fixed nuclear charge must now attract a larger electron cloud; the electron–electron repulsions increase and the effective nuclear charge per electron decreases, allowing the electron cloud to expand (1 mark). Example: Cl (radius 99 pm) → Cl− (radius 181 pm): adding one electron increases the radius by 82 pm because the 17 protons now hold 18 electrons less tightly than the original 17 (1 mark).

Marking notes. 1 mark for identifying that nuclear charge does NOT change when electrons are gained. 1 mark for correctly stating that anions are larger and explaining why (increased e−–e− repulsion / reduced effective nuclear charge per electron). 1 mark for a specific named example with numerical evidence or precise chemical language.

1.6

Section 1 · Short answer · 4 marks · Band 4

Sample response. Cu (Z = 29, anomalous ground state): [Ar]3d¹&sup0;4s¹. Cu+: remove the 4s¹ electron → [Ar]3d¹&sup0;. Cu²+: remove 4s¹ + one 3d electron → [Ar]3d&sup9; (1 mark for both configurations). CuCl contains Cu+ with a fully filled 3d¹&sup0; subshell. No d–d transitions are possible because all d orbitals are occupied and there is no empty orbital of lower energy within the d subshell for electrons to jump into. Cu+ compounds therefore do not absorb visible light and appear colourless or white (1 mark). CuCl2 contains Cu²+ with a 3d&sup9; configuration — one hole in the d subshell. Electrons can absorb visible light to jump between d orbitals of slightly different energy (crystal-field d–d transitions), and the complementary colour to the absorbed wavelength is observed as the yellow-brown/blue-green colour of the compound (1 mark). This demonstrates that it is the occupancy of the d subshell in the ion, not the metal itself, that determines whether a compound is coloured (1 mark).

Marking notes. 1 mark for correct Cu+ and Cu²+ electron configurations. 1 mark for explaining CuCl colourlessness via full 3d¹&sup0; (no d–d transitions). 1 mark for explaining CuCl2 colour via partial 3d&sup9; and d–d transitions. 1 mark for an integrative statement linking d-subshell occupancy to colour/colourlessness.

2.1

Section 2 · Data response · 9 marks · Band 4–5

Sample response (a). H2 gas production increases substantially down the group: Mg produces only 2 mL in 60 s, Ca produces 28 mL, and Sr produces 61 mL. This is an inverse relationship with first ionisation energy, which decreases from Mg (738) to Ca (590) to Sr (549 kJ mol−¹). As IE1 decreases, the metal gives up its two valence electrons more readily to water, increasing the reaction rate and gas production (3 marks: 1 describe trend; 1 link to IE trend; 1 explain the link).

Sample response (b). Valence configurations: Mg [Ne]3s²; Ca [Ar]4s²; Sr [Kr]5s². All three have ns² valence configurations (same group behaviour). Going down the group, the valence 4s (Ca) and 5s (Sr) electrons are progressively further from the nucleus and shielded by increasing numbers of inner-shell electrons. The effective nuclear charge experienced by the outer electrons decreases, reducing the energy needed to remove them (IE1 falls). This is why Mg (smaller radius, less shielding) holds its valence electrons more tightly and reacts most slowly with water (4 marks: 1 for all three valence configs; 1 for increasing radius explanation; 1 for increasing shielding explanation; 1 for linking these to decreasing IE).

Sample response (c). Ba (Z = 56, Group 2, Period 6): config [Xe]6s². Ba has a larger atomic radius and lower IE1 (~503 kJ mol−¹) than Sr, so it should be more reactive than Sr. Predicted H2 volume: greater than 61 mL in 60 s (accept any value reasonably extrapolating the trend, e.g. 80–100 mL). Justification: the 6s valence electrons are further from the nucleus and more heavily shielded; the effective nuclear charge is lower than for Sr, so Ba ionises even more easily and reacts faster with water (2 marks: 1 prediction above 61 mL with reasoning; 1 reference to larger radius/lower IE/more shielding).

Marking notes. Part (a): 1 mark for correctly describing the trend in gas volume from the table; 1 mark for identifying the inverse relationship with IE; 1 mark for mechanistic explanation (lower IE = more readily loses electrons = faster reaction). Part (b): 1 mark for all three correct valence configurations; 1 mark for atomic radius trend explanation; 1 mark for electron shielding explanation; 1 mark for explicitly linking both to decreasing IE. Part (c): 1 mark for a prediction above Sr (>61 mL) with justification; 1 mark for reference to larger radius, increased shielding, lower IE from electron configuration principles.

3.1

Section 3 · Extended response · 7 marks · Band 5–6

Sample response. Electron configuration is a highly powerful predictor of chemical behaviour for main-group elements, because the valence configuration determines the number of electrons available for bonding and the drive to achieve a noble gas configuration. For group behaviour: all Group 1 elements have an ns¹ valence configuration and react similarly with water to produce hydrogen gas and a metal hydroxide (e.g. 2Na + 2H2O → 2NaOH + H2; 2K + 2H2O → 2KOH + H2). The reaction type is identical; only the vigour differs, because atomic radius increases and ionisation energy decreases down the group, making the valence electron progressively easier to lose. For ionic vs covalent character: electron configuration, combined with electronegativity, predicts bond type. Na ([Ne]3s¹) has one valence electron and very low electronegativity (χ = 0.9). When Na reacts with Cl (χ = 3.2), Δχ = 2.3 (≥ 1.7), so electron transfer is favoured: NaCl is ionic. Carbon ([He]2s²2p²) has four valence electrons and moderate electronegativity (χ = 2.6). With Cl, Δχ = 0.6 (< 1.7), so electron sharing is preferred: CCl4 is covalent. For transition metals: the similar energies of the 3d and 4s subshells mean transition metals can form multiple oxidation states (e.g. Fe²+ [Ar]3d&sup6; and Fe³+ [Ar]3d&sup5;), variable compound colour (partial d-filling enables d–d transitions), and catalytic properties. These behaviours cannot be predicted from group number alone but are explained by d-subshell configuration. However, electron configuration alone has limitations. The same configuration does not guarantee identical behaviour in all contexts: the ligand environment (surrounding molecules or ions) strongly modifies transition metal colour and reactivity. Cu+ ([Ar]3d¹&sup0;) is colourless in CuCl but the anomalous ground state of Cu ([Ar]3d¹&sup0;4s¹ rather than [Ar]3d&sup9;4s²) illustrates that simple electron configuration rules can break down. Furthermore, the physical state, concentration, temperature, and reaction partner can all modify observed chemical behaviour independently of configuration. In conclusion, electron configuration is the fundamental basis of chemical behaviour for most elements, particularly for group trends and bond-type prediction, but it must be combined with electronegativity, ionisation energy data, and knowledge of reaction conditions for a complete and accurate prediction.

Marking criteria (7 marks). 1 = correctly explains group behaviour using valence electron configuration with a named example (Na or K with water). 1 = correctly explains the ionic vs covalent distinction using valence electron count and electronegativity difference (Δχ) with named elements (NaCl vs CCl4 or equivalent). 1 = correctly explains transition metal variable oxidation states and/or colour using d-subshell configuration with named ion configurations. 1 = identifies a specific limitation of electron configuration as a sole predictor (e.g. ligand effects on colour; anomalous ground states; role of reaction conditions). 1 = uses a second named element or compound from the HSC curriculum as supporting evidence. 1 = integrates the three aspects (group, bond type, transition metal) into a coherent evaluative argument rather than treating them as isolated points. 1 = reaches an explicit evaluative conclusion assessing the degree to which the claim is valid, acknowledging both its strength and limitations.