The shape of the lesson
By the end of the lesson, learners can
- Polarising the anion: Compare the cations next to a carbonate ion in the simulator and watch how much each one distorts the anion; then heat each carbonate and nitrate on the next tab and compare how hot it must get before gas is given off.
- Lattice versus hydration: Switch to the lattice versus hydration view and watch the two energy terms change as you move down the group for the sulfates.
- Hydroxides go the other way: Add sodium hydroxide to the four solutions on the bench: the precipitates run the opposite way to the sulfates, because the numbers in the cycle move in the opposite direction.
- Stability: small cation, high charge density, polarises the anion, decomposes easily.
- Cycle: ΔHsol = ΣΔHhyd − lattice energy.
- Sulfates: big anion, lattice falls slowly, hydration falls fast, less soluble.
- Hydroxides: small anion, lattice falls fast, more soluble.
One picture every idea returns to
Group 2 carbonates and nitrates become more thermally stable down the group. A small 2+ cation such as Mg2+ has a high charge density, so it has high polarising power: it distorts the electron cloud of the large CO32− or NO3− ion and weakens its C–O or N–O bonds, so less heat is needed to release CO2 or NO2. Larger cations polarise less. Solubility is linked to the enthalpy change of solution: ΔHsol = ΣΔHhyd − lattice energy. Down the group both the lattice energy and the hydration enthalpies become less exothermic, but at different rates. For sulfates the large anion means the lattice energy changes little while the cation hydration enthalpy falls a lot, so ΔHsol becomes less exothermic and solubility decreases. For hydroxides the small OH− means the lattice energy falls faster, so solubility increases.
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, Why the trends run that way.. At AS you learned the trends; now explain them. 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: Polarising the anion | Compare the cations next to a carbonate ion in the simulator and watch how much each one distorts the anion; then heat each carbonate and nitrate on the next tab and compare how hot it must get before gas is given off. Project tab 1 of the lab (Group 2 trends lab, polarising the anion) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Lattice versus hydration | Switch to the lattice versus hydration view and watch the two energy terms change as you move down the group for the sulfates. Project tab 3 of the lab (Group 2 trends lab, lattice versus hydration) and let a student drive it while the class predicts. | Pairs |
| 21 to 30 min | Teach and check: Hydroxides go the other way | Add sodium hydroxide to the four solutions on the bench: the precipitates run the opposite way to the sulfates, because the numbers in the cycle move in the opposite direction. Project tab 4 of the lab (Group 2 trends lab, solubility bench) and let a student drive it while the class predicts. | Cooperative Concept Mapping |
| 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 Hydroxides go the other way, 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.
Cooperative Concept Mapping
Teams receive the key terms on cards and arrange them on a large sheet, drawing and labelling the links. Compare maps between teams before showing the model map.
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 |
|---|---|
| Both terms get less exothermic down the group; the trend depends on which one falls faster, so never explain solubility with just one of them. | Polarisation is about the cation distorting the anion, not the anion being attracted to the cation more strongly. And watch the sign in the cycle: with lattice energy defined as formation (exothermic), Δ H sol = ΣΔ H hyd minus lattice energy, so −3080 − (−2833) = −247, not −5913. |
Group 2: solubility and thermal stability: worksheet
6 questions, 17 marks. Written in the style of the papers; no past paper question is reproduced.
Explain why magnesium carbonate decomposes at a much lower temperature than barium carbonate.
Write the equation for the thermal decomposition of strontium nitrate and state one observation.
Use the data to calculate ΔHsol of MgSO4. ΔHhyd: Mg2+ −1920, SO42− −1160; lattice energy of MgSO4 −2833 (all kJ mol−1).
ΔHsol of SrSO4 is −10 kJ mol−1. ΔHhyd: Sr2+ −1480, SO42− −1160 kJ mol−1. Calculate the lattice energy of SrSO4.
Explain, in terms of lattice energy and hydration enthalpy, why the solubility of the Group 2 sulfates decreases down the group.
For Mg(OH)2, ΣΔHhyd = −2840 and lattice energy = −2993; for Ba(OH)2, ΣΔHhyd = −2280 and lattice energy = −2228 (all kJ mol−1). Calculate both values of ΔHsol and use them to explain the solubility trend.
Answer key and marking notes
- Mg2+ and Ba2+ have the same charge, but Mg2+ is much smaller (0.065 nm against 0.135 nm), so it has a higher charge density.
- Mg2+ therefore polarises (distorts) the electron cloud of the large CO32− ion more.
- This weakens a C–O bond, so less energy is needed to break up the carbonate into MgO and CO2.
- 2Sr(NO3)2 → 2SrO + 4NO2 + O2.
- Brown fumes of NO2 are seen (and a glowing splint relights in the O2).
- ΣΔHhyd = −1920 + (−1160) = −3080 kJ mol−1.
- ΔHsol = ΣΔHhyd − lattice energy.
- = −3080 − (−2833) = −247 kJ mol−1.
- Lattice energy = ΣΔHhyd − ΔHsol.
- = (−1480 + −1160) − (−10) = −2640 + 10 = −2630 kJ mol−1.
- Down the group the cation radius increases, so both the lattice energy and the hydration enthalpy of the cation become less exothermic.
- The hydration enthalpy of the cation depends on its own radius, so it falls a lot (−1920 for Mg2+ to −1360 for Ba2+).
- The SO42− ion is large, so r+ + r− changes relatively little and the lattice energy falls much less.
- So ΔHsol becomes less exothermic (more endothermic), dissolving is less favourable and solubility decreases.
- Mg(OH)2: −2840 − (−2993) = +153 kJ mol−1.
- Ba(OH)2: −2280 − (−2228) = −52 kJ mol−1.
- The lattice energy falls by 765 but the hydration enthalpies by only 560, because OH− is small; ΔHsol becomes more exothermic, so the hydroxides become more soluble down the group.
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.
The chart shows the enthalpy change for the thermal decomposition of the Group 2 carbonates. Which statement best explains the trend?
- Athe cation radius increases down the group, so its charge density falls and it polarises the carbonate ion less, leaving the C-O bonds less weakened
- Bthe cation charge increases down the group, so the carbonate ion is held more strongly
- Cthe carbonate ion becomes smaller down the group, so it is harder to polarise
- Dthe oxides formed become more stable down the group, so less energy is needed
Which products form when magnesium nitrate is heated strongly?
- AMgO, NO2 and O2
- BMg(NO2)2 and O2
- CMgO and NO2 only
- DMg, NO2 and O2
Why does the solubility of the Group 2 sulfates decrease down the group?
- Athe hydration enthalpy of the cation becomes much less exothermic, while the lattice energy changes only a little because SO42− is large, so ΔHsol becomes less exothermic
- Bthe lattice energy becomes much more exothermic down the group
- Cthe hydration enthalpy of the cation becomes more exothermic down the group
- Dlarger cations polarise SO42− more, so the lattice becomes covalent
Use the data to calculate the enthalpy change of solution of CaSO4. ΔHhyd: Ca2+ −1650, SO42− −1160; lattice energy of CaSO4 −2780 (all kJ mol−1).
- A−30 kJ mol−1
- B+30 kJ mol−1
- C−5590 kJ mol−1
- D+1130 kJ mol−1
Which Group 2 hydroxide is the most soluble in water, and why?
- ABa(OH)2, because with the small OH− ion the lattice energy becomes less exothermic faster than the hydration enthalpy
- BMg(OH)2, because Mg2+ has the most exothermic hydration enthalpy
- CBa(OH)2, because Ba2+ has the most exothermic hydration enthalpy
- DMg(OH)2, because its lattice energy is the least exothermic
The energy cycle is for dissolving strontium sulfate, which has ΔHsol = −10 kJ mol−1. Using the hydration enthalpies shown, what is the lattice energy of SrSO4?
- A−2630 kJ mol−1
- B−2650 kJ mol−1
- C+2630 kJ mol−1
- D−1470 kJ mol−1
Calculate ΔHsol of Ca(OH)2. ΔHhyd: Ca2+ −1650, OH− −460; lattice energy of Ca(OH)2 −2643 (all kJ mol−1).
- A+73 kJ mol−1
- B−73 kJ mol−1
- C+533 kJ mol−1
- D−5213 kJ mol−1
A 4.00 g mixture of MgCO3 (Mr 84.3) and BaCO3 is heated at 700 °C, where only the MgCO3 decomposes. The mass falls by 0.990 g. What is the percentage by mass of MgCO3 in the mixture?
- A47.4%
- B52.6%
- C24.8%
- D23.7%
Quiz key
Mg2+ (0.065 nm) is much smaller than Ba2+ (0.135 nm) with the same 2+ charge, so it has a far higher charge density. It distorts the electron cloud of CO32− and weakens a C-O bond, so MgCO3 decomposes most easily (+101 kJ mol−1) and BaCO3 least easily (+275 kJ mol−1). Every Group 2 cation carries a 2+ charge, so charge does not increase. The carbonate ion is the same ion in every compound. A more stable oxide would make decomposition easier, the opposite of the trend shown, and the explanation required is polarisation of the anion.
2Mg(NO3)2 → 2MgO + 4NO2 + O2: brown NO2 fumes and oxygen, which relights a glowing splint. Forming the nitrite and oxygen is what Group 1 nitrates (except lithium) do. MgO and NO2 alone does not balance for oxygen, so O2 must also form. Magnesium metal is not produced; the cation keeps its +2 oxidation state in MgO.
The cation hydration enthalpy depends on the cation radius alone and falls from −1920 (Mg2+) to −1360 (Ba2+) kJ mol−1. The lattice energy depends on r+ + r−, and the large sulfate ion makes that sum change relatively little, so it falls less. ΔHsol therefore becomes less exothermic. Lattice energy becomes less exothermic, not more, as the ions get bigger. Hydration enthalpy also becomes less exothermic down the group. Larger cations polarise less, not more.
ΣΔHhyd = −1650 + (−1160) = −2810. ΔHsol = −2810 − (−2780) = −30 kJ mol−1. +30 subtracts the wrong way round (lattice energy − hydration). −5590 adds the lattice energy instead of subtracting it. +1130 uses only the calcium hydration enthalpy and leaves out the sulfate ion.
Hydroxide solubility increases down the group. OH− is small, so r+ + r− depends mostly on the cation; a bigger cation makes the lattice energy fall faster than the hydration enthalpy, so ΔHsol becomes more exothermic. Mg2+ does have the most exothermic hydration enthalpy, but Mg(OH)2 also has by far the most exothermic lattice energy, and it is only sparingly soluble. Ba2+ has the least exothermic hydration enthalpy, not the most. Mg(OH)2 has the most exothermic lattice energy, not the least.
Lattice energy = ΣΔHhyd − ΔHsol = (−1480 + −1160) − (−10) = −2640 + 10 = −2630 kJ mol−1. −2650 subtracts 10 instead of adding it. A lattice energy defined as formation from gaseous ions is always exothermic, so +2630 is wrong. −1470 uses only the strontium hydration enthalpy and leaves out the sulfate ion.
ΣΔHhyd = −1650 + 2(−460) = −2570. ΔHsol = −2570 − (−2643) = +73 kJ mol−1. −73 subtracts the wrong way round. +533 counts only one OH− ion. −5213 adds the lattice energy instead of subtracting it.
n(CO2) = 0.990 ÷ 44.0 = 0.0225 mol = n(MgCO3), so the mass of MgCO3 = 0.0225 × 84.3 = 1.90 g, which is 47.4% of 4.00 g. Only MgCO3 decomposes because the small Mg2+ ion polarises the carbonate ion strongly; BaCO3 needs a far higher temperature. 52.6% is the percentage of BaCO3. 24.8% is just the percentage mass loss. 23.7% halves the moles, as if two carbonates were needed per CO2.
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.