Ch 1 · Similarities and trends in the properties of the Group 2 metals and their compounds · AS & A Level Chemistry 9701
Topic 1 of 1 in Group 2 — AS & A Level Chemistry 9701 — 48 lessons.
Introduction
At AS Level you met the Group 2 metals mainly through their reactions with water and oxygen. At A2, the focus shifts to explaining two quantitative trends in their compounds — the thermal stability of nitrates and carbonates, and the solubility of hydroxides and sulfates — using the ionic model of polarisation and Hess's law energy cycles you met earlier in A2.
Key ideas — Thermal decomposition of nitrates and carbonates: the trend
Group 2 carbonates decompose on heating to the oxide and carbon dioxide, and the nitrates decompose to the oxide, nitrogen dioxide and oxygen:
🧑🏫 Teacher's explanation
Imagine you are holding a large, soft sponge (the nitrate or carbonate anion) with your fingers. If you use your thumb and forefinger to squeeze it very tightly, you distort its shape. This is what a small, highly charged ion does to a large anion. In chemistry, we call this 'polarisation'.
📖 Definitions
🔍 Worked example — Easy
Predict the relative thermal stability of MgCO_3 and BaCO_3 and justify your answer.
🔍 Worked example — Standard
Write the balanced chemical equation for the thermal decomposition of calcium nitrate, Ca(NO_3)_2(s).
🔍 Worked example — Exam-style
A Group 2 carbonate MC0_3 decomposes to form MO and CO_2. If the enthalpy change for the decomposition of MgCO_3 is Δ H = +118 kJ/mol, explain why the decomposition temperature increases down the group.
⚠️ Common errors
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🎯 Exam question
Key ideas — Explaining the trend: ionic radius and polarising power
Down Group 2, the ionic radius of the cation increases (extra electron shells), so its charge density, or polarising power, decreases. A more polarising (smaller, more charge-dense) cation distorts the electron cloud of the large NO_3^- or mathrmCO_3^2- anion more strongly, pulling electron density away from one of the anion's oxygen atoms and weakening a bond within the anion (a C-O or N-O bond) so that it breaks more readily on heating. Because mathrmMg^2+ is the smallest, most polarising Group 2 cation, magnesium salts distort and destabilise the anion most, so they decompose at the lowest temperature; mathrmBa^2+, the largest and least polarising, distorts the anion least and gives the most thermally stable compounds.
🧑🏫 Teacher's explanation
Imagine you are holding a large, soft sponge (the large nitrate or carbonate anion) and you try to press your finger into it. If you use a tiny, sharp needle (a small, highly charged cation like Mg^2+), you will create a deep, intense dent that disrupts the sponge's structure. If you use a large, flat palm (a large cation like Ba^2+), the pressure is spread out, and the sponge's shape remains mostly intact.
📖 Definitions
🔍 Worked example — Easy
Explain why MgCO_3 decomposes at a lower temperature than CaCO_3.
🔍 Worked example — Standard
Predict the trend in the thermal stability of Group 2 nitrates (Mg(NO_3)_2 to Ba(NO_3)_2) and justify it.
🔍 Worked example — Exam-style
A student claims that BaSO_4 is more thermally stable than MgSO_4 because the Ba^2+ ion is larger. Evaluate this statement.
⚠️ Common errors
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🎯 Exam question
Key ideas — Solubility trends of Group 2 hydroxides and sulfates
Down Group 2, the hydroxides become more soluble: Mg(OH)_2 is only very sparingly soluble (giving milky "milk of magnesia"), while Ba(OH)_2 is appreciably soluble. In contrast, the sulfates become less soluble down the group: MgSO_4 is freely soluble, while BaSO_4 is essentially insoluble and is used as the standard test for sulfate ions.
🧑🏫 Teacher's explanation
Imagine you are trying to pull two magnets apart. To dissolve a solid, the water molecules (the 'pullers') must overcome the attraction between the metal cation and the anion (the 'glue' holding the crystal together). This struggle is a competition between Lattice Energy (how hard the ions stick together) and Hydration Enthalpy (how much the water molecules want to pull the ions away).
📖 Definitions
🔍 Worked example — Easy
Explain why Mg(OH)_2 is less soluble than Ba(OH)_2 using enthalpy terms.
🔍 Worked example — Standard
Predict the solubility trend of Group 2 sulfates (MgSO_4 to BaSO_4) and justify it.
🔍 Worked example — Exam-style
A student observes that BaSO_4 forms a white precipitate when BaCl_2 is added to Na_2SO_4. Explain this in terms of solubility.
⚠️ Common errors
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🎯 Exam question
Key ideas — Explaining solubility: lattice energy versus hydration enthalpy
Dissolving depends on an enthalpy cycle in which the ionic lattice is first separated into gaseous ions (the reverse of lattice energy) and those ions are then hydrated:
🧑🏫 Teacher's explanation
Imagine you are trying to pull two strong magnets apart (the lattice) and then placing them into a crowd of people who want to hug them (the water molecules/hydration). To dissolve, the 'hug' from the water must be strong enough to overcome the 'magnetic pull' between the ions.
📖 Definitions
🔍 Worked example — Easy
A hypothetical Group 2 hydroxide has a lattice energy of 800 kJ mol^-1 and a total hydration enthalpy of 750 kJ mol^-1. Calculate the enthalpy of solution Δ H_sol^ominus.
🔍 Worked example — Standard
Compare the solubility trends of Group 2 hydroxides and sulfates. For hydroxides, as the cation radius increases, which term decreases more rapidly: Lattice Energy or Hydration Enthalpy?
🔍 Worked example — Exam-style
Explain why BaSO_4 is much less soluble than MgSO_4 using the relationship between lattice energy and hydration enthalpy.
⚠️ Common errors
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🎯 Exam question
Worked example
Question. Use the following data to calculate the standard enthalpy change of solution of magnesium hydroxide, and comment on what the sign of your answer suggests about its solubility: Δ H_latt^ominus[Mg(OH)_2] = -2993\ mathrmkJ\ mol^-1; Δ H_hyd^ominus[mathrmMg^2+] = -1920\ mathrmkJ\ mol^-1; Δ H_hyd^ominus[OH^-] = -460\ mathrmkJ\ mol^-1. [6]
🔍 Worked example — Easy
Calculate the enthalpy change of solution for a hypothetical salt where Δ H_latt^ominus = -2000 kJ mol^-1, Δ H_hyd^ominus(cation) = -1500 kJ mol^-1, and Δ H_hyd^ominus(anion) = -300 kJ mol^-1.
🔍 Worked example — Standard
A Group 2 sulfate has Δ H_latt^ominus = -2500 kJ mol^-1. The hydration enthalpies are Δ H_hyd^cation = -1800 kJ mol^-1 and Δ H_hyd^anion = -400 kJ mol^-1. Calculate Δ H_sol^ominus.
🔍 Worked example — Exam-style
Use the following data to calculate the standard enthalpy change of solution of magnesium hydroxide, and comment on what the sign of your answer suggests about its solubility: Δ H_latt^\ ominus[Mg(OH)_2] = -2993 kJ mol^-1; Δ H_hyd^\ ominus[Mg^2+] = -1920 kJ mol^-1; Δ H_hyd^\ ominus[OH^-] = -460 kJ mol^-1.
⚠️ Common errors
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