2020 H2 Chem P1 Q8 — Group 2 Solution

The answer is A (the charge density of the \(\text{Mg}^{2+}\) ion is higher than that of the \(\text{Ca}^{2+}\) ion).

On heating, \(\text{MCO}_3 \rightarrow \text{MO} + \text{CO}_2\). For this to happen, a C-O bond in the carbonate ion must break.
• \(\text{Mg}^{2+}\) and \(\text{Ca}^{2+}\) have the same charge, but \(\text{Mg}^{2+}\) is smaller (ionic radius 0.065 nm, against 0.099 nm for \(\text{Ca}^{2+}\)), so \(\text{Mg}^{2+}\) has a higher charge density and a greater polarising power.
• \(\text{Mg}^{2+}\) distorts the electron cloud of the large \(\text{CO}_3^{2-}\) anion more, which weakens a C-O bond more.
• So less energy is needed to break the C-O bond, and \(\text{MgCO}_3\) decomposes at a lower temperature than \(\text{CaCO}_3\).

• B is wrong: the enthalpy change of decomposition is \(\Delta H_\text{f}(\text{MO}) + \Delta H_\text{f}(\text{CO}_2) - \Delta H_\text{f}(\text{MCO}_3)\). A less negative \(\Delta H_\text{f}\) of MgO, on its own, would make the decomposition more endothermic, so it would make \(\text{MgCO}_3\) harder to decompose, not easier.
• C is wrong: a more negative lattice energy means the \(\text{MgCO}_3\) lattice is more stable, which on its own would raise its decomposition temperature, not lower it.
• D is wrong: melting is a physical change that separates the ions without breaking any C-O bond, so the melting point has no bearing on the thermal decomposition of the carbonate ion.