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A2 · Practice questions · Group 2: solubility and thermal stability

Explain the trends.

Six original Cambridge-style questions on Group 2 thermal stability and solubility.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Charge density, then the energy tug of war.

01
Explain
[3 marks]

Explain why magnesium carbonate decomposes at a much lower temperature than barium carbonate.

  • 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. ✓
02
Recall
[2 marks]

Write the equation for the thermal decomposition of strontium nitrate and state one observation.

  • 2Sr(NO3)2 → 2SrO + 4NO2 + O2. ✓
  • Brown fumes of NO2 are seen (and a glowing splint relights in the O2). ✓
03
Calculation
[3 marks]

Use the data to calculate ΔHsol of MgSO4. ΔHhyd: Mg2+ −1920, SO42− −1160; lattice energy of MgSO4 −2833 (all kJ mol−1).

  • ΣΔHhyd = −1920 + (−1160) = −3080 kJ mol−1. ✓
  • ΔHsol = ΣΔHhyd − lattice energy. ✓
  • = −3080 − (−2833) = −247 kJ mol−1. ✓
04
Calculation
[2 marks]

ΔHsol of SrSO4 is −10 kJ mol−1. ΔHhyd: Sr2+ −1480, SO42− −1160 kJ mol−1. Calculate the lattice energy of SrSO4.

  • Lattice energy = ΣΔHhyd − ΔHsol. ✓
  • = (−1480 + −1160) − (−10) = −2640 + 10 = −2630 kJ mol−1. ✓
05
Explain
[4 marks]

Explain, in terms of lattice energy and hydration enthalpy, why the solubility of the Group 2 sulfates decreases down the group.

  • 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. ✓
06
Analysis
[3 marks]

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.

  • 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. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.