18/07/2026
9701/42/O/N/25
Question 1
(a) Thermal decomposition of magnesium nitrate:
2Mg(NO₃)₂(s) → 2MgO(s) + 4NO₂(g) + O₂(g))
(b) Compound that decomposes at a lower temperature: Mg(NO₃)₂
Mg²⁺ and Sr²⁺ have the same charge, but Mg²⁺ is smaller, so it has a higher charge density (greater polarising power). The Mg²⁺ ion polarises/distorts the large nitrate ion more strongly, weakening an N–O bond within it. Less heat energy is then needed to break it down, so magnesium nitrate decomposes at a lower temperature. (Polarising power decreases down the group as ionic radius increases.)
(c)(i) MgO + H₂SO₄ → MgSO₄ + H₂O, so the products are magnesium sulfate (MgSO₄) and water. Salt A is magnesium sulfate.
(c)(ii) Solubility depends on the balance between the lattice enthalpy (energy to break the ionic lattice) and the hydration enthalpy of the ions (energy released when ions are hydrated). The sulfate ion is large, so the lattice enthalpy is controlled mainly by the anion and changes little from MgSO₄ to SrSO₄. The hydration enthalpy, though, depends on the cation: Mg²⁺ is smaller with a higher charge density than Sr²⁺, so it is hydrated much more strongly and releases more energy. For MgSO₄ the hydration enthalpy is large enough to outweigh the lattice enthalpy, so it dissolves; for SrSO₄ the smaller hydration enthalpy no longer compensates, so it is much less soluble. Group 2 sulfate solubility therefore decreases down the group.
Key marking points for (ii): balance of lattice enthalpy vs hydration enthalpy (1); lattice enthalpy roughly constant because sulfate is large (1); Mg²⁺ smaller/higher charge density so greater hydration enthalpy, tipping the balance towards dissolving (1).
