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27.1 Trends in Group 2 metals and compounds

Syllabus
9701–2028–2029
Topic
27.1
Level
A2

Learning objectives

27.1.1Trends: the trend in the thermal stability• Describe/explain trends: the trend in the thermal stability of the nitrates and carbonates, including the effect of ionic radius on the polarisation of the large anion27.1.2Trends: the variation in solubility• Describe/explain trends: the variation in solubility and of enthalpy change of solution, ΔH ⦵ sol, of the hydroxides and sulfates in terms of relative magnitudes of the enthalpy change of hydration and the lattice energy

Thermal stability of Group 2 carbonates and nitrates rises down the group

MCOX3(s)→MO(s)+COX2(g)\ce{MCO3(s) -> MO(s) + CO2(g)}

2 M(NOX3)X2(s)→2 MO(s)+4 NOX2(g)+OX2(g)\ce{2M(NO3)2(s) -> 2MO(s) + 4NO2(g) + O2(g)}

From Mg²⁺ to Ba²⁺, the carbonates and nitrates require progressively stronger heating to decompose: their thermal stability increases down Group 2.

All cations have charge +2, but ionic radius increases down the group. Charge density and polarising power therefore decrease. The larger cation distorts the electron cloud of the large CO₃²⁻ or NO₃⁻ anion less, so the anion's internal bonds are weakened less and decomposition is harder.

Cation Relative radius / charge density Anion polarisation Thermal stability
Mg²⁺ smaller / higher greater lower
Ba²⁺ larger / lower smaller higher

The explanation is not that every bond simply becomes stronger down the group. The key causal change is how strongly the fixed 2+ cation polarises the large polyatomic anion.

Hydroxide and sulfate solubilities follow different enthalpy balances

ΔHsol∘=−ΔHlatt∘(formation)+∑ΔHhyd∘\Delta H^\circ_{sol}=-\Delta H^\circ_{latt}(\mathrm{formation})+\sum\Delta H^\circ_{hyd}

Down Group 2, M²⁺ becomes larger. Its hydration enthalpy becomes less exothermic because ion–dipole attraction to water weakens. Lattice formation also becomes less exothermic because attraction between the larger cation and anion weakens.

Salt series down Mg → Ba Which magnitude changes faster? ΔH°sol trend Solubility trend
hydroxides, M(OH)₂ lattice-energy magnitude falls faster than hydration magnitude more exothermic increases
sulfates, MSO₄ hydration magnitude falls faster than lattice-energy magnitude more endothermic decreases

OH⁻ is relatively small, so changing M²⁺ radius strongly changes the lattice term. SO₄²⁻ is already large, so changing M²⁺ has a smaller relative effect on its lattice term; the weakening M²⁺ hydration then dominates.

Thus Mg(OH)₂ is sparingly soluble while Ba(OH)₂ is much more soluble; sulfate behaviour is opposite, with BaSO₄ effectively insoluble compared with MgSO₄.

Keep the lattice convention explicit: the cycle reverses negative lattice formation. ΔH°sol helps explain the trend but solubility is an equilibrium property involving entropy as well, so do not equate one enthalpy value with solubility in every system.