Chemistry • Year 11 • Module 1 • Lesson 19
Electron Configuration and Chemical Behaviour
Build HSC Band 5–6 extended-response technique on evaluating evidence, designing investigations, and synthesising the links between electron configuration, periodic trends, and real-world chemical behaviour.
1. Data + scenario: First row transition metals in the Broken Hill ore body (Band 5–6)
8 marks Band 5–6
Scenario. The Broken Hill ore body in New South Wales is one of Australia’s most significant mineral deposits. It contains large concentrations of zinc (Zn), lead (Pb), and silver (Ag), as well as smaller amounts of iron (Fe), manganese (Mn), and copper (Cu). The table below summarises some key data for three of these elements.
| Element | Z | Ground-state configuration | Common ions | Colour of common compound |
|---|---|---|---|---|
| Iron (Fe) | 26 | [Ar]3d&sup6;4s² | Fe²+, Fe³+ | Fe²+ compounds: pale green; Fe³+ compounds: orange-brown |
| Copper (Cu) | 29 | [Ar]3d¹&sup0;4s¹ | Cu+, Cu²+ | Cu+ compounds: colourless or white; Cu²+ compounds: blue or green |
| Zinc (Zn) | 30 | [Ar]3d¹&sup0;4s² | Zn²+ only | All Zn²+ compounds: colourless or white |
Data adapted from standard inorganic chemistry references.
Q1. Analyse and evaluate the data above to explain the differences in the number of oxidation states and the colour of compounds for Fe, Cu and Zn. In your response you must:
- Write the electron configurations for Fe²+, Fe³+, Cu+, Cu²+ and Zn²+, clearly showing which electrons are removed first and why.
- Explain why Fe and Cu form more than one stable oxidation state while Zn forms only Zn²+, using the concept of similar 3d/4s energies.
- Explain the pattern in compound colour (coloured vs. colourless) with reference to d-subshell occupancy.
- Evaluate why Cu+ compounds are colourless even though Cu is a transition metal, using specific electron configuration evidence.
- Identify one limitation of using colour alone to determine the oxidation state of a transition metal in a solution sample.
2. Experimental design — testing group reactivity and electron configuration (Band 5–6)
7 marks Band 5–6
Research question. A student hypothesises that the reactivity of Group 2 metals with dilute hydrochloric acid increases down the group (Mg → Ca → Sr) because the first ionisation energy decreases as atomic radius increases.
Constraints: You have access to standard Year 11 laboratory equipment. Small samples of Mg ribbon, Ca turnings and Sr metal are available (warning: Ca and Sr react vigorously with water; use dilute HCl instead). You may use a gas syringe, a digital balance and a stopwatch. Your investigation must be safe and completed within a single laboratory session.
Q2. Design the investigation and present it in the format below.
- State your hypothesis (a testable prediction linking electron configuration, ionisation energy, and reactivity).
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe the procedure in at least four numbered steps, including how you will measure reactivity quantitatively.
- State a prediction for the order of gas production rates (Mg, Ca, Sr) and link it to ionisation energy trends.
- Identify two limitations of your design and one improvement for reliability.
Q1 — Sample Band 6 response (8 marks), annotated
Ion configurations: Fe (Z = 26): [Ar]3d&sup6;4s². When forming ions, 4s electrons are removed before 3d because in the cation the 3d subshell is lower in energy. Fe²+: [Ar]3d&sup6; (removed 4s²). Fe³+: [Ar]3d&sup5; (removed 4s² + one 3d) [1 mark]. Cu (Z = 29, anomalous): [Ar]3d¹&sup0;4s¹. Cu+: remove 4s¹ → [Ar]3d¹&sup0;. Cu²+: remove 4s¹ + one 3d → [Ar]3d&sup9; [1 mark]. Zn (Z = 30): [Ar]3d¹&sup0;4s². Zn²+: remove 4s² → [Ar]3d¹&sup0; [1 mark].
Multiple oxidation states: Fe and Cu can form more than one stable oxidation state because their 3d and 4s subshells have similar energies; varying numbers of electrons from both subshells can be removed under different conditions. Fe³+ is stabilised by the half-filled 3d&sup5; configuration (maximum exchange energy). Zn can form only Zn²+ because after removing the 4s² electrons, the 3d¹&sup0; subshell is fully filled and substantially more stable — removing a 3d electron requires much more energy, so a Zn³+ ion is not formed under normal chemical conditions [1 mark].
Compound colour: Transition metal compounds are coloured when the metal ion has a partially filled d subshell. Electrons absorb specific wavelengths of visible light to move between d orbitals of slightly different energy (crystal-field d–d transitions). Fe²+ ([Ar]3d&sup6;) has an incomplete d subshell → pale green colour. Fe³+ ([Ar]3d&sup5;) has an incomplete d subshell → orange-brown. Cu²+ ([Ar]3d&sup9;) has an incomplete d subshell → blue-green. Zn²+ ([Ar]3d¹&sup0;) has a full d subshell → no d–d transitions possible → colourless [1 mark].
Why Cu+ is colourless: Cu+ has the configuration [Ar]3d¹&sup0; — a completely filled d subshell. There are no empty d orbitals at a lower energy for electrons to jump into; d–d transitions are impossible. Therefore Cu+ compounds (e.g. CuCl) do not absorb visible light and appear colourless or white, despite copper being a transition metal in the ground state. This contrasts with Cu²+ ([Ar]3d&sup9;), which has one hole in the d subshell, enabling d–d transitions and the observed blue-green colour [1 mark].
Limitation of colour alone: The colour of a transition metal compound in solution depends not only on the d-subshell electron count but also on the identity and geometry of the surrounding ligands (which affect the size of the d-orbital energy splitting), the concentration of the solution, and the pH. For example, CuSO4(aq) is deep blue, but adding ammonia changes it to intense royal blue (tetraamminecopper(II)) without changing the oxidation state. Colour alone cannot confirm the oxidation state without additional analysis [1 mark]. (5 marks for the five dot-point requirements above; award up to 3 additional marks for precision of chemical language, correct formulae, and depth of reasoning: mark as 8 total.)
Marking criteria (8 marks): 1 = all five ion configurations written correctly with 4s removed before 3d; 1 = explains multiple oxidation states via similar 3d/4s energies with Zn as contrast; 1 = correct explanation of colour from partial d-filling with named examples; 1 = explains Cu+ colourlessness from d¹&sup0; configuration; 1 = valid limitation stated; 1 = supporting evidence used throughout; 1 = precise chemical language (d–d transitions, crystal field, exchange energy, partially/fully filled); 1 = evaluative judgement integrating all three metals as a comparative analysis.
Q2 — Sample Band 6 response (7 marks), annotated
Hypothesis: If first ionisation energy decreases down Group 2 (Mg → Ca → Sr) due to increasing atomic radius and electron shielding, then the rate of hydrogen gas production when each metal reacts with dilute HCl will increase in the order Mg < Ca < Sr. Independent variable: identity of Group 2 metal (Mg, Ca, Sr). Dependent variable: volume of H2 gas produced per minute (measured by gas syringe). Controlled variables: HCl concentration (0.5 mol/L), HCl volume (20 mL), mass of metal (0.10 g), temperature (room temperature) [1 mark].
Procedure: (1) Set up three identical conical flasks each fitted with a delivery tube connected to a calibrated gas syringe. Measure 20 mL of 0.5 mol/L HCl into each flask. (2) Weigh 0.10 g each of Mg ribbon, Ca turnings and Sr metal (handle Ca and Sr in a fume hood; avoid direct skin contact). (3) Add the metal to each flask simultaneously (or sequentially with careful timing), immediately seal the flask, and start the stopwatch. Record the volume of H2 gas produced in the syringe every 30 seconds for 5 minutes. (4) Plot volume of H2 (mL) vs time (s) for each metal; calculate the initial rate of gas production from the gradient of the steepest part of each curve. Compare rates [1 mark].
Predicted order: Sr > Ca > Mg. Sr has the largest atomic radius and lowest first ionisation energy (IE1: Mg 738, Ca 590, Sr 549 kJ mol−¹), so Sr loses its two valence electrons most easily and reacts fastest with HCl. The reaction is M(s) + 2HCl(aq) → MCl2(aq) + H2(g) [1 mark].
Limitations: (1) Ca and Sr react vigorously; the gas produced may escape before the syringe can measure it if the connection is imperfect, underestimating the rate [1 mark]. (2) Metal surface area is not controlled — Ca turnings and Mg ribbon have different surface areas per gram, which affects the observed reaction rate independently of the ionisation energy trend [1 mark].
Improvement: Repeat each trial a minimum of three times and average the initial rates to improve reliability. Use metal powder of a standardised mesh size for all three elements to control surface area [1 mark].
Falsification criterion: If Mg reacted faster than Ca or Sr, or if no consistent trend in rate was observed, the hypothesis would be falsified [1 mark].
Marking criteria (7 marks): 1 = testable hypothesis naming IV, DV and the ionisation energy–reactivity link; 1 = four clear procedure steps including a quantitative measurement method; 1 = correct predicted order with IE values or trend reasoning; 1 = first valid limitation (gas escape / measurement issue); 1 = second valid limitation (surface area not controlled); 1 = one specific improvement addressing reliability; 1 = correct equation for the reaction or precise chemical terminology throughout.