Cambridge International AS & A Level Chemistry 9701 · A Level topic 27

Group 2

What this chapter covers27.1.1–27.1.2

Limestone, calcium carbonate, is heated in lime kilns to about 900 °C to make quicklime, calcium oxide. Magnesium carbonate decomposes at a few hundred degrees; barium carbonate needs well over a thousand. At AS you learned this pattern — thermal stability increases down Group 2 — and a matching pair of patterns in solubility: the hydroxides become more soluble down the group and the sulfates less soluble. You described them. This chapter explains them.

Both explanations turn on the same property of the Group 2 cations. Every one carries a 2+ charge, but their radii increase steadily from Mg²⁺ to Ba²⁺. A small cation with a high charge concentrates that charge in a small volume: it has a high charge density. Such a cation distorts the electron clouds of the anions around it — which makes large anions such as carbonate and nitrate easier to break apart — and it attracts both the anions in a lattice and the water molecules around it in solution strongly. Whether a compound dissolves depends on the balance between those last two attractions, measured by the lattice energy and the enthalpy change of hydration.

The decomposition reactions and the trend27.1.1

Carbonates

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

All the compounds are white solids, so the change is observed by detecting the gas: carbon dioxide turns limewater milky. A carbonate that decomposes easily does so in a test-tube over a Bunsen flame; a very stable one shows no change.

Nitrates

The observations are brown fumes of nitrogen dioxide and a gas (oxygen) that relights a glowing splint. The white nitrate often melts before it decomposes. The same equation pattern applies to nitrates of other metals that form 2+ ions, such as Cu(NO₃)₂, Zn(NO₃)₂ and Pb(NO₃)₂: the products are always the metal oxide, NO₂ and O₂ — not the metal.

The trend

The thermal stability of both the carbonates and the nitrates increases down the group: a higher temperature is needed to decompose them. Figure 27.1 shows this quantitatively for the carbonates. The standard enthalpy change of decomposition becomes more endothermic from MgCO₃ to BaCO₃, while the entropy change is similar for all four (each forms one mole of gas). The temperature above which decomposition becomes feasible, where ΔG^⦵ = ΔH^⦵ − TΔS^⦵ falls to zero, therefore rises steeply down the group.

Explaining the trend: charge density and polarisation27.1.1

The carbonate ion, CO₃²⁻, and the nitrate ion, NO₃⁻, are large anions: several atoms, with the negative charge spread over them in a large, easily distorted electron cloud. In the solid, each anion is surrounded by cations. A cation attracts the electron cloud of a neighbouring anion towards itself, distorting it from its symmetrical shape. This distortion is called polarisation of the anion, and the ability of a cation to cause it is its polarising power.

Polarising power depends on charge density — the charge of the ion relative to its size. All Group 2 ions carry the same charge, 2+, but down the group the ionic radius increases because each ion has an extra complete shell of electrons. The same charge spread over a larger ion gives a lower charge density:

The explanation of the trend follows in three linked steps:

1. Down the group the cation radius increases, so the charge density of M²⁺ decreases.
2. A cation of lower charge density polarises (distorts) the electron cloud of the carbonate or nitrate ion less.
3. Polarisation pulls electron density out of the C–O (or N–O) bonds towards the cation, weakening the bond nearest the cation. The less the anion is polarised, the less the bond is weakened, so more energy (a higher temperature) is needed to break it and decompose the compound.

When a carbonate decomposes, one C–O bond breaks completely, leaving an oxide ion bonded to the metal ion and releasing a CO₂ molecule. Figure 27.3 shows this as a curly-arrow sequence: a lone pair on one O⁻ forms a second C=O bond, and the C–O bond to another oxygen breaks with both electrons going to that oxygen, which becomes O²⁻. A polarising cation beside that oxygen makes this step easier.

Using the model on unfamiliar compounds

Because the argument depends only on the charge and radius of the cation, it predicts the stability of carbonates and nitrates of metals outside Group 2 — the basis of many examination questions. Compare the charge ÷ radius of the new cation with the Group 2 values.

The hydroxides, and an energetic view of the trend27.1.1

The Group 2 hydroxides show the same trend. On heating, M(OH)₂(s) → MO(s) + H₂O(g); magnesium hydroxide decomposes readily and barium hydroxide only at a much higher temperature. The explanation is identical: down the group the cation radius increases, so the hydroxide ion is polarised less and the O–H bond is weakened less. (Do not confuse this with the solubility of the hydroxides, which is the subject of the next part.)

Why ΔH of decomposition becomes more endothermic

Polarisation is a qualitative, particle-level explanation. The same trend can also be seen in the energetics, and questions sometimes ask for this view. Decomposition can be broken into three imaginary stages (a Hess cycle): separate the carbonate lattice into gaseous ions; convert CO₃²⁻(g) into O²⁻(g) + CO₂(g); build the oxide lattice from gaseous ions.

The last term does not depend on the metal. Both lattice energies become less exothermic down the group as the cation gets bigger — but not equally. Lattice energy depends on the sum of the cation and anion radii. The oxide ion is much smaller than the carbonate ion, so a given increase in cation radius changes the oxide's sum by a larger proportion: ΔH_latt(MO) becomes less exothermic faster than ΔH_latt(MCO₃). The product lattice loses more of its stabilisation than the reactant lattice, so the decomposition becomes more endothermic down the group, as Figure 27.1 shows. For magnesium, the very exothermic lattice energy of MgO (small Mg²⁺, small O²⁻) is what makes MgCO₃ easy to decompose.

Past-paper practice · Set 27A · Thermal stability of the nitrates and carbonates

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 27A.1[4]

Answer and marking guidance

(a) thermal stability increases down the group ✔; the radius of the cation (M²⁺) increases ✔; so the nitrate ion is polarised / distorted less, weakening the N–O bonds less ✔. (b) Cu(NO₃)₂ → CuO + 2NO₂ + ½O₂ ✔ (or doubled). Examiner insight: both parts were answered well.

Question 27A.2[4]

Part (c) only. Parts (a) and (b) of the same question are on other topics.

Answer and marking guidance

(i) Zn(NO₃)₂ → ZnO + 2NO₂ + ½O₂ ✔ (or 2Zn(NO₃)₂ → 2ZnO + 4NO₂ + O₂). (ii) stability increases down the group AND the cation radius increases ✔; less polarisation / distortion of the nitrate ion, or less weakening of the N–O bond ✔. (iii) Mg(NO₃)₂ only ✔ — Mg²⁺ (65 pm) is the only Group 2 ion smaller than Zn²⁺ (74 pm), so only it polarises nitrate more. Examiner insight: the most common equation error was Zn(NO₃)₂ → Zn + 2NO₂ + O₂; the explanation was well known and clearly described; most gave the correct answer to (iii).

Question 27A.3[5]

Answer and marking guidance

(a) stability increases down the group ✔; the radius of the cation increases (or its charge density decreases) ✔; so there is less polarisation / distortion of the nitrate ion, or less weakening of the N–O bond ✔. (b)(i) Pb(NO₃)₂ → PbO + 2NO₂ + ½O₂ ✔. (ii) lead(II) nitrate decomposes more easily, because Pb²⁺ (120 pm) is smaller than Ba²⁺ (135 pm) — it has a higher charge density, so is more polarising ✔. Examiner insight: many gave the trend and the increasing cation radius, but decreasing polarisation of the nitrate anion was less frequently seen; some wrote of polarisation of the cation, or polarisation caused by the anion. In (b)(i) a common error was Pb as the product; in (b)(ii) many did not state clearly that Pb²⁺ has the smaller ionic radius, and using atomic radii was a common error.

Question 27A.4[8]

Parts (a) and (b); part (b)(i) revises the definition of lattice energy from topic 23.

Answer and marking guidance

(a)(i) arrow 1 from the O⁻ (lone pair) to the C–O bond; arrow 2 from the C–O bond (to the other O⁻) onto that oxygen atom ✔ — producing CO₂ and O²⁻. (ii) stability increases down the group ✔; the cationic radius increases (charge density of M²⁺ decreases) ✔; less polarisation / distortion of the carbonate ion ✔. (b)(i) the energy released when one mole of an ionic compound is formed ✔ from its gaseous ions (under standard conditions) ✔. (ii) any two of three for ✔, all three ✔✔: ΔH of decomposition becomes more positive (less negative) down the group; the oxide ion is smaller than the carbonate ion; so the lattice energy of the oxides becomes less exothermic faster (changes more) than that of the carbonates. Examiner insight: in (a)(i) many arrows ended at the C atom or broke the C=O bond; in (a)(ii) decreasing polarisation of the carbonate anion was less often seen and some wrote of polarisation of the cation; in (b)(i) common errors were omitting "one mole" and starting from gaseous ions as reactants; (b)(ii) was challenging — few answered in terms of the relative sizes of the anions and relative changes in lattice energy.

Question 27A.5[3]

Answer and marking guidance

thermal stability of the hydroxides increases down the group ✔; the radius of the cation increases ✔; so there is less polarisation / distortion of the hydroxide ion ✔. Examiner insight: answered well, though some candidates mistakenly discussed the solubility, instead of the thermal stability, of the hydroxides.

The solubility trends27.1.2

At AS you stated two trends in solubility down Group 2, from magnesium to barium:

• the hydroxides become more soluble: Mg(OH)₂ is only sparingly soluble, while Ba(OH)₂ dissolves readily to give a strongly alkaline solution;
• the sulfates become less soluble: MgSO₄ is soluble, CaSO₄ sparingly soluble and BaSO₄ almost insoluble.

These trends are used every day. Magnesium hydroxide is safe to swallow as an antacid ("milk of magnesia") because so little dissolves that the solution is only mildly alkaline. Calcium hydroxide is soluble enough to make limewater and to neutralise acidic soil. Barium sulfate is so insoluble that a suspension of it can be swallowed as a "barium meal" for X-ray imaging even though soluble barium compounds are toxic, and the precipitation of BaSO₄ from acidified barium chloride is the test for sulfate ions.

At A Level the question is why the two trends run in opposite directions. The answer lies in the energy changes that occur when an ionic solid dissolves.

Lattice energy, hydration and the enthalpy change of solution27.1.2

Dissolving can be pictured as two imaginary stages (Figure 27.4): the lattice is broken up into separate gaseous ions — the reverse of lattice formation, requiring −ΔH_latt — and the gaseous ions are then hydrated, releasing ΣΔH_hyd. By Hess's law,

Because both ΔH_latt and ΣΔH_hyd are large negative quantities, ΔH_sol is the small difference between two large numbers. If the hydration enthalpy is more exothermic than the lattice energy, dissolving is exothermic; if less, it is endothermic. Small changes in either term, from one compound to the next, can therefore change ΔH_sol considerably — which is exactly what happens down Group 2.

What controls lattice energy and hydration enthalpy27.1.2

Both quantities measure electrostatic attractions, and both become more exothermic for ions of higher charge and smaller radius:

• Lattice energy reflects the attraction between cations and anions in the solid. It depends on the product of the ionic charges and on the distance between the centres of neighbouring ions, which is roughly the sum of the cation and anion radii, r₊ + r₋. A larger sum means weaker attraction and a less exothermic lattice energy.
• Enthalpy change of hydration reflects the ion–dipole attraction between one ion and the water molecules around it — the δ⁻ oxygen atoms point towards a cation, the δ⁺ hydrogen atoms towards an anion. For the cation it depends on the cation's own charge and radius only. A small, highly charged cation such as Mg²⁺ attracts water very strongly; Ba²⁺ much less so.

Down the group, therefore, both terms in the cycle become less exothermic. Whether ΔH_sol becomes more or less exothermic depends on which of the two changes more — and that depends on the size of the anion.

Explaining the two trends27.1.2

The hydroxides: more soluble down the group

The hydroxide ion is small — similar in size to the cations. When the cation grows from Mg²⁺ to Ba²⁺, the sum r₊ + r₋ increases by a large proportion, so the lattice energy becomes much less exothermic. The hydration enthalpy of the cation also becomes less exothermic, but the hydration of the two OH⁻ ions does not change, so the total hydration term changes by a smaller amount. The lattice energy is the dominant change:

The sulfates: less soluble down the group

The sulfate ion is large — about twice the radius of hydroxide. Now the anion dominates the sum r₊ + r₋, so the same increase in cation radius changes the sum by only a small proportion, and the lattice energy changes only a little. The hydration enthalpy of the cation, which depends on the cation alone, still becomes much less exothermic. This time the hydration term changes more:

Past-paper practice · Set 27B · Solubility and enthalpy change of solution

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 27B.1[3]

Answer and marking guidance

solubility decreases down the group ✔; because both the lattice energy and the hydration energy become less exothermic ✔; and the hydration energy decreases (becomes less exothermic) by a greater extent than the lattice energy ✔ — so ΔH_sol becomes more endothermic, as the question states. Examiner insight: the trend and its explanation were well understood.

Question 27B.2[4]

Answer and marking guidance

solubility increases down the group ✔; ΔH_latt and ΔH_hyd both become less exothermic ✔; ΔH_latt changes more than ΔH_hyd (OH⁻ is small compared with M²⁺) ✔; so ΔH_sol becomes more exothermic / less endothermic ✔. Examiner insight: for a quantity that can be exothermic or endothermic, such as ΔH_sol, "increases" or "decreases" is ambiguous; describe it as becoming more or less exothermic (or endothermic).

Question 27B.3[6]

Answer and marking guidance

(i) ΔH_latt and ΔH_hyd both become less exothermic ✔; ΔH_latt decreases more than ΔH_hyd (OH⁻ is small) ✔; ΔH_sol becomes more exothermic ✔. (ii) for MCO₃ the decrease in ΔH_hyd is larger than the decrease in ΔH_latt (CO₃²⁻ is large) ✔. (iii) Sr and Ba hydroxides could be used, Mg(OH)₂ could not ✔; because SrCO₃ and BaCO₃ are much less soluble than their hydroxides, so the carbonate precipitates, whereas MgCO₃ is more soluble than Mg(OH)₂ ✔. Examiner insight: in (i) many explained why each energy decreased but did not relate the relative decreases; (ii) was challenging, most answers discussing decomposition instead of solubility; in (iii) many identified Sr and Ba but few explained using the solubilities of both hydroxides and carbonates.

Question 27B.4[6]

Parts (a)–(c); the question continues with definitions and entropy (topic 23).

Answer and marking guidance

(a) cycle with K⁺(g) + Cl⁻(g) at the top and K⁺(aq) + Cl⁻(aq) (or KCl(aq)) ✔; three correctly directed, labelled arrows ✔. (b) ΔH_hyd(Mg²⁺) = ΔH_sol + ΔH_latt − 2ΔH_hyd(Cl⁻) = −155 + (−2493) − 2(−364) ✔ = −1920 kJ mol⁻¹ ✔. (c) Mg²⁺ is smaller than K⁺ and has a greater charge ✔ (any two points), so there is greater attraction between the ions in MgCl₂ ✔. Examiner insight: species were generally identified, but completing the cycle was harder; in (b) common errors were −2284 (only one Cl⁻) and −1610 (one sign error), and omitting −364 scored nothing; in (c) many identified the greater charge and smaller radius but did not state that this leads to greater attraction between the ions.

Question 27B.5[7]

Part (b) combines the solubility of Be(OH)₂ with K_sp from topic 25.

Answer and marking guidance

(a) ΔH_latt and ΔH_hyd both become less exothermic ✔; ΔH_hyd changes less (ΔH_latt is the dominant factor / changes more) ✔; ΔH_sol becomes more exothermic ✔. (b)(i) K_sp = [Be²⁺][OH⁻]² ✔; units mol³ dm⁻⁹ ✔. (ii) solubility = 2.40 × 10⁻⁶ ÷ 43.0 = 5.58 × 10⁻⁸ mol dm⁻³ ✔; K_sp = 4s³ = 4 × (5.58 × 10⁻⁸)³ = 6.95 × 10⁻²² ✔. Examiner insight: the trend was well understood, but ΔH_latt being the dominant factor was less frequently seen; the commonest error in (b)(ii) was not converting the solubility from g dm⁻³ to mol dm⁻³ (giving 5.53 × 10⁻¹⁷).

Misconceptions and how the topic is assessed27.1

The misconceptions below recur in examiner reports on this topic. Each is set out as the incorrect idea, why it fails, the correct model and what it costs in an examination.

How the topic is assessed

Self-test

Ten questions across the whole unit, each with the reasoning behind the answer.

Definitions to learn

Data used in this chapter

Ionic radii of Mg²⁺, Ca²⁺, Sr²⁺, Ba²⁺ and Zn²⁺ are those printed in an examination question; other ionic radii are reference values of the kind printed in the data booklet. The values for KI and CaBr₂ in Worked examples 27.3 and 27.4 are printed in examination questions. The enthalpy changes of hydration in Table 27.3, and the ΔH_f^⦵ and S^⦵ values behind Figure 27.1, are reference values, not the official data booklet; use your own data booklet in an examination.

Figure 27.5 and the solubility model use the Kapustinskii equation, U = 1202.5 νz₊z₋/(r₊ + r₋) × (1 − 0.345/(r₊ + r₋)) kJ mol⁻¹ (r in 10⁻¹⁰ m), with thermochemical anion radii. It is an estimate — for Mg(OH)₂ it gives 3009 kJ mol⁻¹ against 2993 kJ mol⁻¹ quoted in a mark scheme — and is not required by the syllabus. The same equation applied to the oxides and carbonates predicts that ΔH of carbonate decomposition rises by about 180 kJ mol⁻¹ from Mg to Ba; the reference data give 174 kJ mol⁻¹. The model is not reliable when the two changes are close: for the carbonates (CO₃²⁻, 178 pm) it predicts the lattice term to change slightly more than the hydration term, i.e. the wrong direction for the observed decrease in solubility. The model on the page reports this failure rather than hiding it.

Summary

27.1.1 Thermal stability

• MCO₃ → MO + CO₂; 2M(NO₃)₂ → 2MO + 4NO₂ + O₂; M(OH)₂ → MO + H₂O.
• Thermal stability of carbonates, nitrates and hydroxides increases down the group.
• Down the group the cation radius increases and its charge density decreases, so it polarises (distorts) the large anion less; the C–O or N–O bonds are weakened less and more heat is needed.
• Energetically: ΔH_latt(MO) becomes less exothermic faster than ΔH_latt(MCO₃) because O²⁻ is smaller than CO₃²⁻, so ΔH of decomposition becomes more endothermic.
• The same reasoning predicts the behaviour of other M²⁺ compounds from their ionic radii.

27.1.2 Solubility

• Hydroxides become more soluble down the group; sulfates (and carbonates) less soluble.
• ΔH_sol = ΣΔH_hyd − ΔH_latt; both terms become less exothermic down the group.
• Small OH⁻: lattice energy changes more → ΔH_sol more exothermic → more soluble.
• Large SO₄²⁻: hydration enthalpy changes more → ΔH_sol more endothermic → less soluble.
• The explanation ignores entropy; it is a model that predicts the direction of the trends.

Examination checklist

• Can I write balanced decomposition equations for Group 2 carbonates, nitrates and hydroxides, and for Cu, Zn or Pb nitrates?
• Does my explanation of thermal stability contain all three ideas: trend, cation radius or charge density, polarisation of the named anion weakening its bonds?
• Do I say that the cation polarises the anion, and use ionic (not atomic) radii?
• Can I draw two curly arrows for CO₃²⁻ → CO₂ + O²⁻?
• Can I explain the trend in ΔH of decomposition using the relative sizes of O²⁻ and CO₃²⁻?
• Can I define lattice energy, ΔH_hyd and ΔH_sol, draw the cycle and calculate any one term, counting ions correctly?
• Do I describe enthalpy changes as "more/less exothermic" rather than "increase/decrease"?
• Can I explain both solubility trends by saying which term changes more, and why?

Knowledge organiser