A locksmith shapes a key to match a specific lock. In chemistry, electron configuration is the key — it determines exactly how an element will interact with others. A Group 1 element with one valence electron is "shaped" to lose that electron and form a 1+ ion. A Group 17 element with seven valence electrons is "shaped" to gain one electron and form a 1− ion. Elements in the same group share the same "key shape" — that's why their chemistry is so similar. In this final lesson on IQ3, you'll see how everything we've built — subshell configurations, periodic trends, IMFs — converges to explain why elements behave the way they do.
Main-group elements gain or lose electrons to reach the nearest noble gas configuration (octet). The number of electrons gained or lost determines the ion charge.
| Group | Valence electrons | Action to reach octet | Ion charge | Example |
|---|---|---|---|---|
| 1 | 1 | Lose 1 e⁻ | 1+ | Na → Na⁺ ([Ne]) |
| 2 | 2 | Lose 2 e⁻ | 2+ | Mg → Mg²⁺ ([Ne]) |
| 13 | 3 | Lose 3 e⁻ | 3+ | Al → Al³⁺ ([Ne]) |
| 14 | 4 | Share 4 e⁻ (covalent) | 0 (usually covalent) | C → CH₄, CO₂, etc. |
| 15 | 5 | Gain 3 e⁻ | 3− | N → N³⁻ ([Ne]), or covalent |
| 16 | 6 | Gain 2 e⁻ | 2− | O → O²⁻ ([Ne]), or covalent |
| 17 | 7 | Gain 1 e⁻ | 1− | Cl → Cl⁻ ([Ar]) |
| 18 | 8 (full) | No action needed | 0 (no bonding) | Ar — noble, no reaction |
As you learned in L15, group number corresponds to valence electron count. All elements in the same group have the same number of valence electrons in the same type of subshell — and it is the valence electrons (not the core electrons) that determine chemical behaviour.
Same reaction type, same products — just increasingly vigorous because atomic radius increases down the group, making the valence electron progressively easier to lose (as established in L17). The type of reaction is determined by the valence configuration (1 electron to lose); the vigour is determined by the position in the group.
Transition metals (d-block) can form multiple oxidation states because their 3d and 4s electrons have similar energies — varying numbers of both can be lost in different reactions.
| Element | Ground state config | Common ions | Ion configurations |
|---|---|---|---|
| Iron (Fe, Z=26) | [Ar]3d⁶4s² | Fe²⁺, Fe³⁺ | Fe²⁺: [Ar]3d⁶ (loses 4s²); Fe³⁺: [Ar]3d⁵ (loses 4s² + one 3d) |
| Copper (Cu, Z=29) | [Ar]3d¹⁰4s¹ | Cu⁺, Cu²⁺ | Cu⁺: [Ar]3d¹⁰; Cu²⁺: [Ar]3d⁹ |
| Manganese (Mn, Z=25) | [Ar]3d⁵4s² | Mn²⁺, Mn³⁺, Mn⁴⁺, Mn⁷⁺ | Multiple d electron configurations possible |
Transition metal compounds are often coloured because partially filled d orbitals allow electronic transitions (electrons jumping between d orbitals) in the visible light range, absorbing specific wavelengths. Compounds with full (Zn²⁺: 3d¹⁰) or empty (Sc³⁺: 3d⁰) d subshells are colourless.
Worked Examples
Activities
1 For each element, write the electron configuration, identify the likely ion, and identify the ion's electron configuration: (a) Potassium (Z=19), (b) Fluorine (Z=9), (c) Aluminium (Z=13).
2 Write the electron configurations for Fe²⁺ and Fe³⁺. Which ion has greater stability and why? (Refer to d subshell filling.)
A Na⁺, Mg²⁺, Al³⁺, Ne, F⁻, and O²⁻ are all isoelectronic. (a) Write the common electron configuration. (b) Despite having the same electron configuration, these species have different sizes. Predict the order from largest to smallest ionic/atomic radius. Explain.
B Explain how electron configuration determines that sodium forms NaCl (ionic) while carbon forms CCl₄ (covalent), even though both react with chlorine. Refer to electronegativity and valence electrons.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Multiple Choice
5 random questions from a replayable lesson bank — feedback shown immediately
Short Answer
6. Na⁺, F⁻, and Ne are isoelectronic (all have 10 electrons). Predict the order of ionic/atomic radius from largest to smallest and justify your prediction using nuclear charge. 3 MARKS
7. Copper forms two ionic compounds with chlorine: CuCl (Cu⁺) and CuCl₂ (Cu²⁺). (a) Write the electron configurations for Cu (Z=29), Cu⁺, and Cu²⁺. (b) Explain why copper can form two different oxidation states while sodium can only form Na⁺. 4 MARKS
1. (a) K (Z=19): config [Ar]4s¹ or 1s²2s²2p⁶3s²3p⁶4s¹. Ion: K⁺. K⁺ config: [Ar] (1s²2s²2p⁶3s²3p⁶, 18 electrons — same as Ar). (b) F (Z=9): config 1s²2s²2p⁵. Ion: F⁻. F⁻ config: 1s²2s²2p⁶ (10 electrons — same as Ne). (c) Al (Z=13): config 1s²2s²2p⁶3s²3p¹ or [Ne]3s²3p¹. Ion: Al³⁺. Al³⁺ config: 1s²2s²2p⁶ (10 electrons — same as Ne).
2. Fe (Z=26): [Ar]3d⁶4s². Fe²⁺: lose 4s² electrons first → [Ar]3d⁶. Fe³⁺: lose 4s² + one 3d electron → [Ar]3d⁵. Fe³⁺ ([Ar]3d⁵) has greater stability because the 3d⁵ configuration is half-filled — all five 3d orbitals contain one electron each (↑ ↑ ↑ ↑ ↑). This arrangement maximises exchange energy and minimises electron-electron repulsion, providing extra stability compared to the partially paired 3d⁶ configuration of Fe²⁺.
A: (a) Common config: 1s²2s²2p⁶ (10 electrons — same as Ne). (b) Order largest to smallest: O²⁻ > F⁻ > Ne > Na⁺ > Mg²⁺ > Al³⁺. Explanation: all species have the same 10 electrons in the same orbitals (1s²2s²2p⁶). The nuclear charge (Z) differs: O(8) < F(9) < Ne(10) < Na(11) < Mg(12) < Al(13). Higher nuclear charge → stronger pull on the same 10 electrons → electrons drawn closer to the nucleus → smaller radius. O²⁻ has the lowest nuclear charge (Z=8) pulling on 10 electrons → largest. Al³⁺ has the highest (Z=13) → smallest.
B: Na (Z=11): config [Ne]3s¹, Group 1, χ(Na)=0.9. Cl: χ=3.2. |Δχ|(Na–Cl) = 3.2−0.9 = 2.3 ≥ 1.7 → ionic bond. Na effectively transfers its single 3s¹ electron to Cl → Na⁺ and Cl⁻ → ionic lattice (NaCl). C (Z=6): config [He]2s²2p², Group 14, χ(C)=2.6. |Δχ|(C–Cl) = 3.2−2.6 = 0.6 (0.4–1.7) → polar covalent bond. Carbon shares its 4 valence electrons with 4 Cl atoms → molecular covalent compound (CCl₄). The fundamental difference: Na has 1 valence electron and very low electronegativity — electron transfer is energetically favoured; C has 4 valence electrons and moderate electronegativity — sharing is far more stable than forming C⁴⁺ (which would require removing 4 electrons with enormous cumulative IE).
1. B — S (Z=16) + 2e⁻ = S²⁻ with 18 electrons = [Ar] = 1s²2s²2p⁶3s²3p⁶. Na⁺ has 10e⁻ ([Ne]). Ca²⁺ has 18e⁻ ([Ar]) ✓ but question asks which has [Ar]; both S²⁻ and Ca²⁺ have [Ar] config. Re-checking: S²⁻ (Z=16+2=18e⁻=[Ar]) and Ca²⁺ (Z=20−2=18e⁻=[Ar]) are both isoelectronic with Ar. B (S²⁻) is the listed correct answer; C (Ca²⁺) would also be correct. In exam context, only one option would be correct per question — this tests S²⁻ specifically.
2. D — Fe (Z=26): [Ar]3d⁶4s². Form Fe³⁺: remove 4s² (2 electrons) + one 3d (1 electron) = lose 3 total → [Ar]3d⁵. NOT [Ar]3d⁶ (that's Fe²⁺, only 4s² removed). A is the neutral Fe. B removes from 3d first (wrong — 4s goes first).
3. A — Same group, same valence electron count (1 valence electron, s¹ config) → same bonding tendency. B is wrong (they have different radii and atomic masses). C is wrong (they're in different periods).
4. C — Oxygen (Group 16): 6 valence electrons → needs 2 more to reach octet → forms O²⁻. Mg (Group 2) forms Mg²⁺ (loses electrons). N forms N³⁻ (gains 3). F forms F⁻ (gains 1).
5. B — Partially filled d orbitals: electrons can absorb specific wavelengths of visible light to jump between d orbitals (crystal field splitting). The colour observed is the complementary colour to what is absorbed. Full d¹⁰ (e.g. Zn²⁺) or empty d⁰ (e.g. Sc³⁺) → no d-d transitions → colourless compounds.
Q6 (3 marks): All three species have 10 electrons with configuration 1s²2s²2p⁶. Their nuclear charges (Z) differ: F⁻ (Z=9), Ne (Z=10), Na⁺ (Z=11) (1 mark). Order from largest to smallest: F⁻ > Ne > Na⁺ (1 mark). Explanation: all three have the same 10-electron cloud. Higher nuclear charge → stronger attraction pulling the electron cloud toward the nucleus → smaller radius. F⁻ has Z=9 (weakest pull on 10 electrons) → largest. Na⁺ has Z=11 (strongest pull on the same 10 electrons) → smallest. Ne is intermediate with Z=10 (1 mark).
Q7 (4 marks): (a) Cu (Z=29, anomalous): [Ar]3d¹⁰4s¹ (1 mark). Cu⁺: remove 4s¹ electron → [Ar]3d¹⁰ (fully-filled d subshell, 29−1=28 electrons). Cu²⁺: remove 4s¹ + one 3d electron → [Ar]3d⁹ (28−1=27 electrons) (1 mark). (b) Sodium (Group 1) has config [Ne]3s¹ — one valence electron. After losing it to form Na⁺ ([Ne]), the next electron to remove (IE₂) comes from the filled inner 2p shell, which is much more tightly held (huge IE jump) → forming Na²⁺ is energetically impossible under normal chemical conditions (1 mark). Copper has 3d and 4s electrons with similar energies. The 4s electron can be lost to form Cu⁺ ([Ar]3d¹⁰); one 3d electron can additionally be lost to form Cu²⁺ ([Ar]3d⁹). The 3d electrons are available for ionisation because their energy is similar to 4s, unlike Na's inner 2p electrons which are far more stable and tightly held. This is the fundamental difference: d-block elements have accessible d electrons; s-block elements do not (1 mark).
Return to your Think First response. You should now be able to explain why metals lose electrons and nonmetals gain them:
Electron Configuration and Chemical Behaviour
Tick when you've finished all activities and checked your answers.