Group 2 (A Level)

0. What Examiners Want

This guide covers Topic 27 of Cambridge International AS & A Level Chemistry 9701 (2025–27) and follows the learning outcomes below, word for word. The right-hand column shows where each one is taught. The topic extends AS Topic 10 (Group 2) and uses the energy cycles of Topic 23.

1. Group 2 Recap and the Ions

From AS Topic 10 you know the elements magnesium to barium form M²⁺ ions and react more vigorously down the group, because the first and second ionisation energies fall as the atoms get larger. The reactions you must still know are summarised here.

Mg reacts with steam to give MgO and H₂. Calcium, strontium and barium react with dilute sulfuric acid only briefly, because an insoluble sulfate layer forms on the metal or carbonate. That is the first hint of the solubility trend explained in section 7.

The key to this topic is the size of the M²⁺ ion. Each step down the group adds a full electron shell.

The M²⁺ ion gets larger down the group. The charge is the same (2+), so the charge density falls and the hydration enthalpy becomes less exothermic.

Charge density is charge divided by size. Every Group 2 ion has the same 2+ charge, so a smaller ion has a higher charge density. The Mg²⁺ ion has the highest charge density in the group and Ba²⁺ the lowest. Every trend in this topic follows from that single fact.

2. Polarisation of the Anion

An ionic bond is not perfectly ionic. A positive ion attracts the electron cloud of a neighbouring negative ion. Polarisation is the distortion of the electron cloud of an anion by a cation. The more the cloud is pulled towards the cation, the more covalent character the bond has, and the weaker the bonds inside the anion become.

Polarisation of a large anion. A small, highly charged cation distorts the electron cloud strongly. A large cation distorts it only slightly.

So the polarising power of a Group 2 cation is greatest for Mg²⁺ and least for Ba²⁺. The anions that matter here, CO₃²⁻ and NO₃⁻, are large and have a spread-out electron cloud, so they are easily polarised.

3. Thermal Decomposition of the Carbonates

All the Group 2 carbonates decompose on heating to the oxide and carbon dioxide:

The temperature needed increases down the group: magnesium carbonate decomposes at a much lower temperature than barium carbonate. The values below are approximate and vary between sources; you will not be asked to recall them.

Approximate decomposition temperatures. The carbonates become more thermally stable down the group.

Explanation. A small cation with a high charge density (Mg²⁺) polarises the carbonate ion strongly. The electron cloud is pulled towards the cation, which weakens a C–O bond. When heated, the carbonate breaks up easily: CO₂ is lost and the very stable oxide ion is left next to the small cation. Going down the group the cation is larger, so its charge density is lower, it polarises the carbonate ion less, the C–O bonds are weakened less, and more heat is needed.

The cation pulls on an oxygen of the carbonate ion. The stronger the pull (smaller cation), the weaker the C–O bond and the lower the decomposition temperature.

Learn this chain. Every mark in a thermal-stability answer is one of these links.

Worked example. Calculate the volume of CO₂ produced at room conditions when 5.00 g of calcium carbonate is heated strongly. (A_r: Ca 40.1, C 12.0, O 16.0.)

M_r(CaCO₃) = 100.1. Moles = 5.00 ÷ 100.1 = 0.0500 mol. CaCO₃ → CaO + CO₂, so 0.0500 mol of CO₂. Volume = 0.0500 × 24.0 = 1.20 dm³.

4. Thermal Decomposition of the Nitrates

The Group 2 nitrates decompose on heating to the oxide, nitrogen dioxide and oxygen:

The brown gas is NO₂. The oxygen given off relights a glowing splint.

The trend is the same as for the carbonates: the nitrates become more stable to heat down the group. The explanation is also the same. A small, highly charged cation such as Mg²⁺ polarises the large nitrate ion strongly, weakening the N–O bonds, so the nitrate decomposes at a lower temperature. A large cation such as Ba²⁺ polarises it less.

The three marks. (1) Thermal stability increases down the group. (2) The radius of the cation increases (so the charge density decreases). (3) There is less polarisation (distortion) of the nitrate ion, so less weakening of the N–O bonds.

Contrast with Group 1. The Group 1 cations are larger and carry only a 1+ charge, so they polarise the anions far less. Group 1 carbonates (except lithium carbonate) do not decompose at Bunsen temperatures, and Group 1 nitrates (except lithium nitrate) give only the nitrite and oxygen: 2NaNO₃ → 2NaNO₂ + O₂. Lithium, with the smallest Group 1 ion, behaves like Group 2: 4LiNO₃ → 2Li₂O + 4NO₂ + O₂. Copper(II) nitrate also decomposes like Group 2 nitrates: 2Cu(NO₃)₂ → 2CuO + 4NO₂ + O₂.

Worked example. Calculate the total volume of gas, at room conditions, produced when 3.00 g of magnesium nitrate is heated until it has all decomposed. (A_r: Mg 24.3, N 14.0, O 16.0.)

M_r(Mg(NO₃)₂) = 24.3 + 2(14.0 + 48.0) = 148.3. Moles = 3.00 ÷ 148.3 = 0.02023 mol. From 2Mg(NO₃)₂ → 2MgO + 4NO₂ + O₂: each mole gives 2 mol NO₂ and 0.5 mol O₂, i.e. 2.5 mol of gas. Gas = 2.5 × 0.02023 = 0.05057 mol. Volume = 0.05057 × 24.0 = 1.21 dm³.

5. Enthalpy of Solution: Hydration versus Lattice Energy

When an ionic solid dissolves in water, the ions leave the lattice and become surrounded by water molecules. Two energy changes are involved (Topic 23):

• Lattice energy, ΔH_latt: the enthalpy change when one mole of an ionic solid forms from its gaseous ions. It is always exothermic (negative). A higher charge and smaller ions give a more exothermic lattice energy.
• Enthalpy change of hydration, ΔH_hyd: the enthalpy change when one mole of a gaseous ion dissolves in water to form an aqueous ion. It is always exothermic. A higher charge and a smaller ion give a more exothermic value, because water molecules are attracted more strongly.

The enthalpy change of solution, ΔH_sol, is the enthalpy change when one mole of a solid dissolves in enough water that no further change occurs on dilution. It follows from a cycle.

The solid is broken into gaseous ions (the reverse of the lattice energy, endothermic). The ions are then hydrated (exothermic).

Worked example. Use the data to find ΔH_sol for magnesium sulfate and for barium sulfate. ΔH_hyd(SO₄²⁻) = −1000 kJ mol⁻¹ for both.

MgSO₄: ΔH_sol = +2829 − 1920 − 1000 = −91 kJ mol⁻¹. BaSO₄: ΔH_sol = +2379 − 1360 − 1000 = +19 kJ mol⁻¹. So dissolving becomes less favourable down the group, because the hydration enthalpy of the cation falls by more than the lattice energy does.

The general argument. For a given anion, both ΔH_latt and ΔH_hyd(M²⁺) become less exothermic down the group as the cation gets larger. ΔH_sol depends on which falls more:

6. Solubility of the Hydroxides

The Group 2 hydroxides become more soluble down the group. This is why a saturated solution of barium hydroxide has a higher pH than limewater, and why Mg(OH)₂ (milk of magnesia) is barely soluble.

Explanation. The hydroxide ion is small. The sum of the two radii in the lattice, r(M²⁺) + r(OH⁻), changes a lot when the cation changes size, so the lattice energy becomes much less exothermic down the group. The hydration enthalpy of the cation also becomes less exothermic, but the lattice energy falls by more. So ΔH_sol becomes more exothermic (less endothermic), and solubility increases.

Illustrative data chosen to show the pattern. The lattice energy falls by more than the hydration enthalpy, so ΔH_sol becomes more exothermic. Note that the vertical axis does not start at zero.

7. Solubility of the Sulfates

The Group 2 sulfates become less soluble down the group. Magnesium sulfate is very soluble, calcium sulfate is slightly soluble, strontium sulfate is only sparingly soluble and barium sulfate is insoluble.

Opposite trends for two anions. Note the log scale: each gridline is a factor of 10.

Explanation. The sulfate ion is large. In the lattice the distance between the ion centres is r(M²⁺) + r(SO₄²⁻), and the sulfate radius is so much larger than the cation radius that changing the cation barely changes this sum. The lattice energy therefore changes only a little. The hydration enthalpy of the cation, however, depends on its own size, so it becomes much less exothermic from Mg²⁺ to Ba²⁺. The fall in hydration enthalpy is larger than the fall in lattice energy, so ΔH_sol becomes less exothermic and then endothermic, and the solubility decreases.

Illustrative data chosen to show the pattern. The hydration enthalpy falls by more than the lattice energy, so ΔH_sol moves from exothermic towards endothermic. Note that the vertical axis does not start at zero.

8. Predicting and Using the Trends

Use the trends for compounds you have not met. Radium, below barium, follows the pattern: the Ra²⁺ ion is the largest, so radium is the most reactive metal, RaCO₃ and Ra(NO₃)₂ are the most stable to heat, Ra(OH)₂ is the most soluble hydroxide (a strongly alkaline solution) and RaSO₄ is the least soluble sulfate (a white precipitate when Ra²⁺ meets sulfate ions).

Uses that rely on the trends

• Barium sulfate is so insoluble that the toxic Ba²⁺ does not enter the body; it is used in the ‘barium meal’ to show the gut in X-rays.
• Test for sulfate ions: add acidified barium chloride or nitrate. A white precipitate of BaSO₄ forms: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). The acid removes carbonate ions that would also give a precipitate.
• Calcium hydroxide (slaked lime) is used in agriculture to neutralise acidic soil, and limewater (Ca(OH)₂(aq)) is the test for CO₂.
• Magnesium hydroxide is used in antacids, and calcium carbonate is heated in a lime kiln to make calcium oxide (about 900 °C), which shows the high stability of the carbonate down the group.

Worked example. Strontium carbonate and magnesium carbonate are heated separately with the same Bunsen burner. Predict and explain what you would see.

MgCO₃ decomposes readily to the white MgO and CO₂, because the small Mg²⁺ polarises the carbonate ion strongly. SrCO₃ is much more stable, since Sr²⁺ is larger, has a lower charge density and polarises the carbonate ion less; little or no decomposition occurs at the Bunsen temperature.

9. Quick Sheet and Checklist

All the trends in one picture. The wide end of each wedge is the larger value.

Checklist. Before the exam you should be able to:

• [ ] explain what is meant by the charge density of an ion and how it changes down Group 2
• [ ] define polarisation and say which ion polarises which
• [ ] write equations for the thermal decomposition of Group 2 carbonates and nitrates, with state symbols
• [ ] describe the trend in thermal stability and explain it in terms of ionic radius and polarisation of the anion
• [ ] state the observations on heating a nitrate (brown gas, oxygen) and contrast Group 1 nitrates and carbonates
• [ ] define enthalpy change of solution, hydration and lattice energy, and draw a cycle linking them
• [ ] calculate ΔH_sol from lattice energy and hydration enthalpies, with the right number of anions
• [ ] explain the solubility trend of the hydroxides (small anion) and of the sulfates (large anion) in terms of the relative magnitudes of ΔH_latt and ΔH_hyd
• [ ] use the trends to predict the properties of radium compounds