The shape of the lesson
By the end of the lesson, learners can
- Across a period: Step through Period 3 in the simulator and watch the atom shrink from sodium to chlorine.
- Down a group: Each period down a group starts a new principal quantum shell.
- Ions and the isoelectronic series: Forming an ion changes the number of electrons but never the number of protons, so the same nucleus is pulling on a different-sized cloud.
- Nuclear charge: more protons, stronger pull, smaller radius.
- Shells: a new shell means further out and larger.
- Shielding: inner shells cancel part of the pull; electrons in the same shell shield poorly.
- Ions: protons never change; only the electron cloud does.
One picture every idea returns to
Across a period the radius decreases: each step adds one proton and one electron to the same shell, shielding barely changes, so the outer electrons are pulled in harder. Down a group the radius increases: each period adds a whole new shell, further out and more shielded. A cation is smaller than its atom (the outer shell is lost), an anion is larger (extra repulsion), and in an isoelectronic series the ion with the most protons is the smallest.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, How big is an atom?. Three things set the size of an atom: how strongly the nucleus pulls, how many shells the electrons occupy, and how much the inner electrons shield the outer ones. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Across a period | Step through Period 3 in the simulator and watch the atom shrink from sodium to chlorine. Project tab 8 of the lab (Atomic structure lab, radius trends) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Down a group | Each period down a group starts a new principal quantum shell. | Pairs |
| 21 to 30 min | Teach and check: Ions and the isoelectronic series | Forming an ion changes the number of electrons but never the number of protons, so the same nucleus is pulling on a different-sized cloud. | Anticipation / Reaction Guide |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 17 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Ions and the isoelectronic series, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| Ionic radius across Period 3 is not a smooth decrease. | It falls from Na + to Mg 2+ to Al 3+ (all 10 electrons, n = 2 outer shell), then jumps up at P 3−, S 2− and Cl −, which have 18 electrons and an n = 3 outer shell, and then falls again along that second isoelectronic set. |
Atomic and ionic radius: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Explain why atomic radius decreases across Period 3 from sodium to argon.
Explain why atomic radius increases down Group 2 from beryllium to barium.
Explain why a magnesium ion, Mg2+, is much smaller than a magnesium atom.
Explain why an oxide ion, O2−, is larger than an oxygen atom.
The ions S2−, Cl−, K+ and Ca2+ are isoelectronic. Arrange them in order of increasing ionic radius and explain the order.
Typical radii: Na 0.186 nm, Na+ 0.095 nm, Cl 0.099 nm, Cl− 0.181 nm. Use the data to explain why the sodium ion is smaller than its atom and the chloride ion larger than its atom.
Answer key and marking notes
- Nuclear charge increases by one proton for each successive element.
- The added electrons enter the same outer shell (n = 3), so the distance of the outer shell from the nucleus does not increase and shielding by the inner shells stays roughly the same.
- The outer electrons therefore experience a greater attraction and are pulled closer to the nucleus, so the radius decreases.
- Each element down the group has one more occupied principal quantum shell, so the outer electrons are further from the nucleus.
- Each extra inner shell adds shielding, which reduces the attraction felt by the outer electrons.
- Although nuclear charge also increases, the extra distance and shielding outweigh it, so the radius increases.
- Mg loses both of its 3s electrons, so the ion's outer shell is n = 2, which is closer to the nucleus than n = 3.
- The 12 protons now attract only 10 electrons, so the remaining electrons are held more tightly and the radius decreases further.
- The two extra electrons enter the same outer shell (n = 2) while the nuclear charge stays at 8 protons.
- More electrons means more electron-electron repulsion in the outer shell, so the electron cloud expands and the radius increases.
- All four ions have 18 electrons (1s22s22p63s23p6).
- The only difference is the nuclear charge: S 16, Cl 17, K 19, Ca 20 protons. More protons attract the same 18 electrons more strongly, giving a smaller ion.
- Increasing radius: Ca2+ < K+ < Cl− < S2−.
- Na to Na+: the radius roughly halves because the single 3s electron is removed, leaving n = 2 as the outer shell.
- The unchanged 11 protons now attract only 10 electrons, so those are held more tightly.
- Cl to Cl−: the radius nearly doubles because an extra electron is added to the same n = 3 shell.
- Nuclear charge stays at 17, but the extra electron increases repulsion between the outer electrons, so the cloud expands.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
Which of these atoms has the smallest atomic radius?
- ACl
- BSi
- CNa
- DK
Which list shows the atoms in order of increasing atomic radius?
- AF < O < N
- BNa < Mg < Al
- CCl < F < Br
- DK < Na < Li
Which of these species has the largest radius?
- AI−
- BI
- CBr−
- DCs+
The ionic radius decreases from Na+ to Mg2+ to Al3+. What is the reason?
- AThe ions all have 10 electrons, but the nuclear charge increases from 11 to 13
- BEach ion has one fewer occupied shell than the one before
- CEach ion has fewer electrons than the one before
- DShielding by inner electrons increases from Na+ to Al3+
Ionic radius increases from F− to Cl− to Br−. What is the main reason?
- AEach ion has one more occupied shell of electrons
- BThe nuclear charge decreases down the group
- CThe charge on the ion becomes more negative down the group
- DEach ion gains more electrons from the metal it reacts with
The P3− ion is larger than the Cl− ion. Which statement gives the best explanation?
- ABoth ions have 18 electrons, but phosphorus has two fewer protons to attract them
- BP3− has one more occupied shell than Cl−
- CA phosphorus atom is larger than a chlorine atom, so its ion must also be larger
- DP3− contains more electrons than Cl−
Which of these ions has the smallest radius?
- ABe2+
- BLi+
- CMg2+
- DNa+
An Fe3+ ion is smaller than an Fe2+ ion. Which statement explains this?
- AFe3+ has one fewer electron held by the same 26 protons, so there is less repulsion and a stronger pull on each electron
- BFe3+ has lost an electron from the n = 4 shell, so it has one fewer occupied shell than Fe2+
- CFe3+ has more protons in its nucleus than Fe2+
- DFe3+ has more shielding from its inner electrons than Fe2+
Quiz key
Chlorine is furthest right in Period 3, so it has the greatest nuclear charge acting on outer electrons in the same n = 3 shell and the smallest radius. Silicon and sodium are further left with smaller nuclear charges, and potassium has an extra occupied shell (n = 4), making it the largest of the four.
Across Period 2 radius falls from N to O to F, so read backwards F < O < N is increasing. Na, Mg, Al runs across a period, where radius decreases, so that list is in decreasing order. Fluorine is smaller than chlorine (one fewer shell), so Cl < F is wrong. K, Na, Li goes up a group, so the radius decreases along that list.
I− has one more electron than the iodine atom, increasing repulsion so it is larger than I. Br− has one fewer occupied shell than I−. Cs+ and I− both have 54 electrons, but Cs+ has 55 protons against iodine's 53, so it pulls the same electrons in more tightly and is smaller.
All three ions are 1s22s22p6, so they have the same electrons in the same shells; only the proton number rises (11, 12, 13), pulling the electron cloud in more strongly. They have the same number of shells and the same number of electrons, and the shielding is essentially identical, so none of the other options explains the change.
Going down the group each halide ion has an extra principal shell, so its outer electrons are further from the nucleus and more shielded. Nuclear charge actually increases down the group. All three ions carry the same 1− charge, and each halogen atom gains exactly one electron to form its ion.
P3− (15 + 3) and Cl− (17 + 1) both have 18 electrons in the same shells, so the difference comes from nuclear charge: 15 protons hold the cloud less tightly than 17. Both ions have their outer electrons in n = 3, so there is no extra shell, and they have equal electron counts. The size of the atom does not by itself decide the size of the ion; the explanation must compare the ions.
Be2+ and Li+ both have only the 1s shell, and Be2+ has 4 protons holding 2 electrons compared with 3 protons in Li+, so Be2+ is smaller. Mg2+ and Na+ keep a full n = 2 shell, so both are larger than either Period 2 ion; a high charge alone does not make Mg2+ the smallest.
Fe2+ is [Ar] 3d6 and Fe3+ is [Ar] 3d5: the extra electron comes from 3d, not from a separate shell, since both 4s electrons were already gone in Fe2+. The nuclear charge stays at 26, so fewer electrons means less repulsion and each is pulled closer. Neither ion has more protons, and removing an outer electron does not increase shielding.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.