Chemistry • Year 11 • Module 1 • Lesson 19
Electron Configuration and Chemical Behaviour
Apply your understanding of config–to–ion charge, isoelectronic series, and group reactivity to data, scenarios and a reasoning challenge on transition metals.
1. Interpret experimental data — isoelectronic series and ionic radius
The table below shows data for six species that are all isoelectronic (they all have 10 electrons). 9 marks
| Species | Z (nuclear charge) | Electron configuration | Ionic / atomic radius (pm) | Larger or smaller than Ne? |
|---|---|---|---|---|
| O²− | 8 | 1s²2s²2p&sup6; | 140 | |
| F− | 9 | 1s²2s²2p&sup6; | 133 | |
| Ne | 10 | 1s²2s²2p&sup6; | 112 | — (reference) |
| Na+ | 11 | 1s²2s²2p&sup6; | 102 | |
| Mg²+ | 12 | 1s²2s²2p&sup6; | 72 | |
| Al³+ | 13 | 1s²2s²2p&sup6; | 53 |
1.1 Complete the “Larger or smaller than Ne?” column for O²−, F−, Na+, Mg²+ and Al³+. 2 marks
1.2 Describe the trend in ionic radius across the isoelectronic series from O²− to Al³+. 2 marks
1.3 Explain the trend using the concept of nuclear charge and its effect on the same 10-electron cloud. 3 marks
1.4 Predict the radius order (largest to smallest) for the isoelectronic series P³−, S²−, Cl−, Ar, K+, Ca²+ (all have 18 electrons). Justify briefly. 2 marks
2. Interpret graph — successive ionisation energies of sodium
The graph below shows the successive ionisation energies (IE1 to IE11) of sodium (Z = 11). 7 marks
Figure 2. Successive ionisation energies of sodium (Z = 11) on a log scale. Data: NIST Webbook / illustrative values.
2.1 Identify where the largest jump in ionisation energy occurs and state what this indicates about the electron configuration of sodium. 2 marks
2.2 Explain why there is a smaller but noticeable second jump between IE9 and IE10. 2 marks
2.3 Use the graph to justify why sodium forms only Na+ ions in normal chemical conditions, and not Na²+ or Na³+. 3 marks
3. Compare ionic and covalent compound formation from electron configuration
Complete the two-column table to contrast how electron configuration leads to ionic versus covalent compound formation. 10 marks (1 per cell)
| Feature | Ionic compound (e.g. NaCl) | Covalent compound (e.g. CCl4) |
|---|---|---|
| Valence electron count of central atom | ||
| Electronegativity difference (Δχ) | ||
| How octet is satisfied | ||
| Ion formed? | ||
| Type of bonding |
4. Predict and justify — a Queensland copper mine scenario
Mount Isa in Queensland is one of the world’s largest copper mines. The copper extracted exists as Cu+ in CuCl (white solid) and as Cu²+ in CuCl2 (yellow-brown solid). A student claims: “Copper must be a main-group metal because it forms two different ions just like other metals do.” 5 marks
4.1 Write the ground-state electron configurations for Cu (Z = 29), Cu+, and Cu²+. Note that Cu has an anomalous ground-state configuration. 3 marks
4.2 Identify the error in the student’s claim and explain what feature of copper’s electron configuration actually allows it to form two stable ions. 2 marks
5. Cause-and-effect chain — reactivity of Group 1 metals down the group
The boxes on the left are causes; write the effect (consequence) in the right-hand boxes. 5 marks (1 per effect + 1 overall outcome)
Overall outcome (so…): The reactivity of Group 1 metals with water _______________
Q1.1 — Larger/smaller column
O²−: Larger (Z = 8 < Z(Ne) = 10). F−: Larger (Z = 9 < 10). Na+: Smaller (Z = 11 > 10). Mg²+: Smaller. Al³+: Smaller.
Q1.2 — Trend description
Ionic radius decreases continuously from O²− (140 pm) to Al³+ (53 pm) across the isoelectronic series [1]. The decrease is consistent and substantial — a reduction of 87 pm from the largest to the smallest species [1].
Q1.3 — Nuclear charge explanation
All six species have exactly 10 electrons in the same orbitals (1s²2s²2p&sup6;). The only variable is nuclear charge (Z) [1]. A higher Z means a larger positive charge pulling on the same 10-electron cloud [1]. This stronger attractive force draws the electrons closer to the nucleus, resulting in a smaller radius. O²− has the lowest Z (8) pulling on 10 electrons — electrons are held loosely and the radius is largest; Al³+ has Z = 13 — the nuclear pull is much stronger, giving the smallest radius [1].
Q1.4 — [Ar]-isoelectronic series prediction
Largest to smallest: P³− > S²− > Cl− > Ar > K+ > Ca²+ [1]. Same reasoning: all have 18 electrons, so the one with the lowest Z (P, Z = 15) has the weakest nuclear pull on 18 electrons = largest; Ca (Z = 20) has the strongest pull = smallest [1].
Q2.1 — Largest IE jump
The largest jump occurs between IE1 (496 kJ mol−¹) and IE2 (4562 kJ mol−¹) [1]. This indicates that sodium has only one valence electron in its outermost (3s) shell. After removing it, the second electron must come from the filled 2p inner shell, which is much more tightly held (higher effective nuclear charge, closer to nucleus), requiring far more energy [1].
Q2.2 — Jump between IE9 and IE10
IE2 through IE9 correspond to removing the eight electrons of the second shell (2s²2p&sup6;). IE10 and IE11 involve removing electrons from the innermost first shell (1s²), which is closer to the nucleus and shielded by far fewer electrons [1]. The much greater effective nuclear charge experienced by 1s electrons results in another large jump in ionisation energy at IE10 [1].
Q2.3 — Why only Na+ forms
IE1 (496 kJ mol−¹) is relatively low — removing the single 3s valence electron requires modest energy that is more than compensated by lattice energy in ionic compounds [1]. IE2 (4562 kJ mol−¹) is nearly 10 times higher because the second electron comes from an inner shell that is much more tightly held [1]. The energy required to form Na²+ would never be recouped by any ionic lattice formation or other bonding process under normal chemical conditions; therefore Na forms only Na+ [1].
Q3 — Compare and contrast table
Valence electron count: Ionic — 1 (Na has 1 valence e−); Covalent — 4 (C has 4 valence e−). Δχ: Ionic — large (≥ 1.7, e.g. Na–Cl: 3.2−0.9=2.3); Covalent — small (< 1.7, e.g. C–Cl: 3.2−2.6=0.6). Octet satisfied by: Ionic — electron transfer (Na gives 1 e− to Cl); Covalent — electron sharing (C shares 4 pairs with 4 Cl atoms). Ion formed? Ionic — Yes (Na+, Cl−); Covalent — No (no ions formed). Bond type: Ionic — electrostatic attraction between oppositely charged ions; Covalent — shared electron pair between atoms.
Q4.1 — Cu, Cu+, Cu²+ configurations
Cu (Z = 29): [Ar]3d¹&sup0;4s¹ (anomalous; full 3d is more stable than [Ar]3d&sup9;4s²) [1 mark]. Cu+: remove the 4s¹ electron → [Ar]3d¹&sup0; (28 electrons) [1 mark]. Cu²+: remove 4s¹ + one 3d electron → [Ar]3d&sup9; (27 electrons) [1 mark].
Q4.2 — Error in student’s claim
The student’s claim is wrong because the ability to form two different ions is a feature of transition metals, not main-group metals [1]. Copper is a d-block transition metal; its 3d and 4s subshells have similar energies, so varying numbers of both can be lost. This gives Cu+ ([Ar]3d¹&sup0;) and Cu²+ ([Ar]3d&sup9;). Main-group metals (like Na, Mg) form only one characteristic ion because there is a very large jump in IE after the valence electrons are removed [1].
Q5 — Cause-and-effect chain
Effect 1: The valence electron is held less tightly by the nucleus (nuclear attraction is weaker and shielding is greater). Effect 2: The effective nuclear charge experienced by the valence electron decreases. Effect 3: The metal loses its valence electron more readily (higher reactivity). Effect 4: The reaction with water and other reagents becomes more vigorous. Overall outcome: The reactivity of Group 1 metals with water increases down the group (Li reacts gently, K ignites, Rb/Cs react explosively) because the valence electron is progressively easier to lose.