Electron Configuration and Chemical Behaviour
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Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
Write your initial answer before reading on.
Key facts
- Main-group elements form ions to reach the nearest noble-gas (octet) configuration
- Group → typical ion charge: 1 → 1+, 2 → 2+, 13 → 3+, 15 → 3−, 16 → 2−, 17 → 1−, 18 → 0
- Transition metals show variable oxidation states because 3d and 4s electrons have similar energies
Concepts
- Why elements in the same group undergo the same type of reaction but with different vigour
- Why transition-metal compounds are often coloured (electron transitions within a partially-filled d subshell)
- Why 4s electrons are removed before 3d when transition metals form cations
Skills
- Predict the ion an element forms from its electron configuration
- Write the configuration of a main-group or transition-metal cation/anion
- Rank isoelectronic species by radius using nuclear charge
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 |
Na⁺ [Ne]: 1s²2s²2p⁶ (10e⁻)
Mg²⁺ [Ne]: 1s²2s²2p⁶ (10e⁻)
Al³⁺ [Ne]: 1s²2s²2p⁶ (10e⁻)
Si⁴⁺ [Ne]: 1s²2s²2p⁶ (10e⁻) — (rare, very high IE)
P³⁻ [Ar]: 1s²2s²2p⁶3s²3p⁶ (18e⁻)
S²⁻ [Ar]: 1s²2s²2p⁶3s²3p⁶ (18e⁻)
Cl⁻ [Ar]: 1s²2s²2p⁶3s²3p⁶ (18e⁻)
The cations tend toward [Ne]; the anions tend toward [Ar].
Octet rule: main-group atoms lose or gain electrons to match the nearest noble gas (full s²p⁶ outer shell). Cations: Group 1 → 1+, Group 2 → 2+, Group 13 → 3+. Anions: Group 17 → 1−, Group 16 → 2−, Group 15 → 3−. Isoelectronic species (e.g. N³⁻, O²⁻, F⁻, Ne, Na⁺, Mg²⁺, Al³⁺ — all 10 e⁻) have the same electron count but different nuclear charges; more protons → smaller ion.
Pause — copy the highlighted ion charge rules into your book before moving on.
Lock-in task: Aluminium (Al, Z=13) has 3 valence electrons. In one or two sentences, predict the charge of the ion it forms and state which noble gas configuration the ion adopts.
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.
We just saw how electron configuration determines ion charge and the octet rule. That raises a question: why do all elements in the same group behave similarly in reactions? This card answers it → elements in the same group have identical valence electron configurations, so they undergo the same type of reaction; their position down the group only affects the vigour.
All Group 1 metals have valence configuration ns¹ → same number and type of valence electrons → same kind of chemistry (react with water to give H₂ + metal hydroxide: 2M + 2H₂O → 2MOH + H₂). Reaction type depends on valence configuration; reaction vigour depends on position down the group (K more vigorous than Li because larger atom, lower IE).
Add the highlighted group chemistry rule to your notes before the check below.
Cloze: drag the terms into the right slots to complete the explanation.
Elements in the same group all have the same number of ___, which is why they share the same ___ of reaction. Going down a group, the ___ increases, so the outer electron is held more loosely — this increases the ___ of the reaction without changing what it is.
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.
We just saw how valence electron configuration explains group behaviour. That raises a question: why can transition metals like iron form more than one stable ion (Fe²⁺ and Fe³⁺), unlike sodium which only forms Na⁺? This card answers it → 3d and 4s energies are similar, so successive electron removals are feasible without breaking into a stable noble-gas core.
Transition metals form multiple oxidation states because 3d and 4s electrons have similar energies, so variable numbers can be removed. 4s electrons are lost first, then 3d: Fe = [Ar]3d⁶4s² → Fe²⁺ = [Ar]3d⁶ → Fe³⁺ = [Ar]3d⁵. Transition-metal compounds are usually coloured due to d–d electron transitions (partially filled d orbitals absorb visible light); empty (Sc³⁺, 3d⁰) or full (Zn²⁺, 3d¹⁰) d subshells give colourless compounds.
Pause — write the highlighted transition metal rules into your book.
True or false: when iron (Fe, [Ar]3d⁶4s²) forms the Fe²⁺ ion, the two 4s electrons are removed first, giving Fe²⁺ the configuration [Ar]3d⁶.
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
We just saw why transition metals can have multiple oxidation states and why their compounds are coloured. That raises a question: how do you write precise exam answers on ion charges, isoelectronic series, and transition metal properties? This card answers it → for each question type, cite the electron configuration as evidence and apply the specific rule (octet, isoelectronic ranking, 4s-first removal, d–d transition).
For exam answers on ion charge: state the octet rule, name the nearest noble gas, then give the electron change. For isoelectronic series: same e⁻, more protons → smaller; list smallest to largest by proton number. For transition metal oxidation states: 4s lost first, then 3d; colour arises from partially filled d orbitals absorbing visible light; colourless = fully empty or fully filled d subshell.
Pause — copy the highlighted exam strategies into your book before moving on.
Lock-in task: Na⁺, F⁻, and Ne are isoelectronic (all have 10 electrons). In one or two sentences, explain why Na⁺ is the smallest of the three.
Worked examples · reveal as you go
For each element, write the electron configuration, predict the likely ion formed, and explain the chemical behaviour: (a) Sulfur (Z=16), (b) Calcium (Z=20), (c) Iron (Z=26).
Copper has ground-state configuration [Ar]3d¹⁰4s¹. Predict the electron configurations of (i) Cu⁺ and (ii) Cu²⁺, and explain in one sentence why copper can form two oxidation states while sodium can only form Na⁺.
How close was your prediction?
Quick-fire practice · 5 reps +2 XP per reveal
Write the full sub-shell electron configuration of a chlorine atom (Z = 17).
How many valence electrons does sulfur (Z = 16) have, and what ion does it tend to form?
Why are the Group 1 metals (e.g. Na, K) so reactive?
Write the electron configuration of the calcium ion Ca²⁺ (Ca is Z = 20).
Element X has configuration 1s² 2s² 2p⁶ 3s². State its group, period and the ion it forms.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
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A student writes the electron configuration of the calcium ion (Ca²⁺, from Ca with Z=20). One line contains an error — click it.
- Neutral Ca has 20 electrons → 1s² 2s² 2p⁶ 3s² 3p⁶ 4s²
- Ca is in Group 2, so it loses its 2 valence electrons to reach a noble-gas configuration.
- Remove the 3s² electrons → Ca²⁺ = 1s² 2s² 2p⁶ 3p⁶ 4s²
- A correctly formed Ca²⁺ ion is isoelectronic with argon.
Q1. 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.
Q2. 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⁺.
📖 Comprehensive answers (click to reveal)
Activity 1
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⁺.
Activity 2
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).
❓ Multiple Choice
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.
Short Answer Model Answers
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).
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