A2 27 Group 2 | A Level Chemistry 9701

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At AS (topic 10) you learned two sets of trends in Group 2: the carbonates and nitrates get more thermally stable down the group, and the hydroxides get more soluble while the sulfates get less soluble. This A Level topic asks you to explain them — the first with the idea of polarisation, the second with the lattice and hydration enthalpies of topic 23.

🎯What you need to be able to do

Describe and explain the trend in thermal stability of the Group 2 nitrates and carbonates, including how the radius of the cation affects polarisation of the large anion.
Describe and explain the variation in solubility, and in enthalpy change of solution, of the Group 2 hydroxides and sulfates, in terms of the relative sizes of the lattice energy and the enthalpy change of hydration.

📚The chemistry

Thermal stability of the carbonates and nitrates

The trend: thermal stability increases down the group. Approximate decomposition temperatures for the carbonates:

MgCO₃ about 350 °C

CaCO₃ about 840 °C

SrCO₃ about 1290 °C

BaCO₃ about 1360 °C

The explanation is polarisation:

The carbonate ion, CO₃²⁻, and the nitrate ion, NO₃⁻, are large anions whose electron clouds are easily distorted.
A Group 2 cation attracts the anion’s electron cloud towards itself — it polarises the anion. The smaller the cation, the higher its charge density (same 2+ charge on a smaller ion), and the more strongly it polarises the anion.
Drawing electron density towards the cation weakens the C–O bonds in the carbonate ion (or the N–O bonds in nitrate), so less energy is needed to break the ion up into CO₂ (or NO₂ and O₂) and the oxide ion.
Down the group the cation radius increases, so charge density falls, the anion is polarised less, its bonds are weakened less, and a higher temperature is needed to decompose it.

Mg²⁺ is the smallest and most polarising, so MgCO₃ decomposes most easily; Ba²⁺ is the largest and least polarising, so BaCO₃ is the most stable. The same argument explains why Group 1 carbonates (1+ cations, lower charge density) are more stable than Group 2 ones.

Solubility of the hydroxides and sulfates

From topic 23, dissolving an ionic solid is breaking its lattice and then hydrating the ions:

\[ \Delta H_\mathrm{sol} = -\Delta H_\mathrm{latt} + \Delta H_\mathrm{hyd}(\mathrm{M^{2+}}) + \Delta H_\mathrm{hyd}(\text{anion}) \]

A more exothermic (or less endothermic) ΔH_sol generally goes with higher solubility. Down Group 2, the cation gets larger, so both terms involving it become less exothermic:

the lattice energy becomes less exothermic (larger ions, further apart);
the hydration enthalpy of M²⁺ becomes less exothermic (lower charge density, weaker attraction for water).

Which one falls faster decides the trend — and that depends on the size of the anion.

Sulfates: less soluble down the group

The sulfate ion is large. The lattice energy depends on the distance between ion centres, r₊ + r₋, and when r₋ is big, a change in the cation radius makes only a small difference to the sum. So the lattice energy decreases only slightly down the group. The hydration enthalpy of the cation, however, depends on the cation alone, and decreases considerably. The energy released by hydration falls faster than the energy needed to break the lattice, so ΔH_sol becomes less exothermic, then endothermic, and the sulfates become less soluble: MgSO₄ is soluble, BaSO₄ insoluble.

Hydroxides: more soluble down the group

The hydroxide ion is small. Now the cation radius is a large part of r₊ + r₋, so as it increases the lattice energy falls considerably — faster than the hydration enthalpy of the cation. Less energy is needed to break the lattice relative to what hydration returns, so ΔH_sol becomes more exothermic and the hydroxides become more soluble.

The single idea to carry into the exam: with a large anion, hydration enthalpy dominates the change; with a small anion, lattice energy dominates.

✏️Worked example

Data (kJ mol⁻¹): lattice energies MgSO₄ −2833, BaSO₄ −2374; hydration enthalpies Mg²⁺ −1921, Ba²⁺ −1305, SO₄²⁻ −1004. (a) Calculate ΔH_sol for MgSO₄ and BaSO₄. (b) Use the data to explain why the sulfates become less soluble down the group. (c) Explain why barium carbonate needs a higher temperature to decompose than magnesium carbonate.

(a)

\[ \Delta H_\mathrm{sol}(\mathrm{MgSO_4}) = +2833 + (-1921) + (-1004) = -92\ \mathrm{kJ\ mol^{-1}} \] \[ \Delta H_\mathrm{sol}(\mathrm{BaSO_4}) = +2374 + (-1305) + (-1004) = +65\ \mathrm{kJ\ mol^{-1}} \]

(b) From Mg to Ba the lattice energy becomes less exothermic by 2833 − 2374 = 459 kJ mol⁻¹, but the cation’s hydration enthalpy becomes less exothermic by 1921 − 1305 = 616 kJ mol⁻¹. The hydration term falls by more, because the sulfate ion is large and so the lattice energy is less sensitive to the cation’s size. So ΔH_sol changes from exothermic (−92) to endothermic (+65), and BaSO₄ is much less soluble.

(c) Mg²⁺ is much smaller than Ba²⁺, so it has a higher charge density and polarises the large carbonate ion more strongly, distorting its electron cloud and weakening the C–O bonds. Less energy is then needed to break the carbonate ion into CO₂ and O²⁻. Ba²⁺ polarises it far less, so a higher temperature is required.

Check it. The difference in ΔH_sol between the two, 65 − (−92) = 157, should equal 616 − 459 = 157 — the amount by which the hydration change outweighs the lattice change. It does, which confirms the arithmetic and the explanation at once.

“Barium sulfate is insoluble because its lattice energy is larger.” It is the opposite: BaSO₄ has the smaller lattice energy. The explanation is always about the balance between the two terms and which falls faster. In thermal stability, the matching error is to talk about the cation’s charge changing — it is 2+ throughout; only its radius, and so its charge density, changes.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Explain what is meant by the charge density of an ion, and how it changes down Group 2.

Charge density is the charge of an ion relative to its size (charge per unit volume). All Group 2 ions carry a 2+ charge, but their radius increases down the group (more electron shells), so the same charge is spread over a larger ion and the charge density decreases from Mg²⁺ to Ba²⁺.

2. Place the nitrates of Mg, Ca, Sr and Ba in order of the temperature at which they decompose, and write the equation for strontium nitrate.

Decomposition temperature increases down the group: Mg(NO₃)₂ < Ca(NO₃)₂ < Sr(NO₃)₂ < Ba(NO₃)₂. 2Sr(NO₃)₂ → 2SrO + 4NO₂ + O₂. The nitrate ion, like carbonate, is large and polarisable, so the same polarisation argument applies.

3. Explain why magnesium carbonate decomposes at a lower temperature than calcium carbonate.

Mg²⁺ has a smaller ionic radius than Ca²⁺ with the same charge, so it has a higher charge density. It attracts the electron cloud of the large carbonate ion more strongly, polarising it more. This distortion weakens the C–O bonds within the carbonate ion, so less energy (a lower temperature) is needed to break it down into CO₂ and the oxide.

4. Explain why the Group 2 hydroxides become more soluble down the group.

ΔH_sol = −ΔH_latt + ΔH_hyd(M²⁺) + 2ΔH_hyd(OH⁻). Down the group both the lattice energy and the cation’s hydration enthalpy become less exothermic. Because the OH⁻ ion is small, the distance between ions depends strongly on the cation’s radius, so the lattice energy decreases more than the hydration enthalpy. Breaking the lattice becomes relatively easier, ΔH_sol becomes more exothermic, and solubility increases.

5. Why does the lattice energy of the sulfates change less down Group 2 than the lattice energy of the hydroxides?

Lattice energy depends on the distance between the ion centres, roughly r_cation + r_anion. The sulfate ion is large, so this sum is dominated by r_anion; increasing the cation radius changes the total only a little, so the lattice energy falls only slightly. The hydroxide ion is small, so the cation radius is a large fraction of the sum and changes in it have a much bigger effect.

6. Predict, with a reason, whether lithium carbonate or potassium carbonate decomposes more easily.

Lithium carbonate. Li⁺ is much smaller than K⁺ (both 1+), so it has a much higher charge density and polarises the carbonate ion more, weakening its C–O bonds. (Indeed Li₂CO₃ decomposes on strong heating, while the other Group 1 carbonates do not at Bunsen temperatures — lithium behaves like a Group 2 metal here.)

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

Chemguide (Jim Clark) — the Group 2 pages on thermal stability and on solubility, which work through the lattice/hydration argument with data