2017 H2 Chem P1 Q9 — Group 2 Solution

The answer is C (magnesium).
Polarising power increases with the charge density of the cation, \(\dfrac{q^+}{r_+}\): a higher charge and a smaller ionic radius give a more polarising cation. The more polarising the cation, the more it distorts the electron cloud of the large \(\text{O}_2^{2-}\) ion and the less stable the ionic peroxide.
The four cations are \(\text{Ba}^{2+}\), \(\text{Cs}^+\), \(\text{Mg}^{2+}\) and \(\text{Na}^+\). \(\text{Mg}^{2+}\) has the highest charge (2+) and the smallest radius of the four, so it has the highest charge density, is the most polarising and is least likely to form a stable peroxide.
• A is wrong: \(\text{Ba}^{2+}\) also has a 2+ charge, but it has many more electron shells than \(\text{Mg}^{2+}\), so it is much larger and has a much lower charge density.
• B is wrong: \(\text{Cs}^+\) has only a 1+ charge and is the largest of the four ions, so it has the lowest charge density and is the least polarising.
• D is wrong: \(\text{Na}^+\) has only a 1+ charge. It has the same number of electron shells as \(\text{Mg}^{2+}\) but one fewer proton, so it is also larger than \(\text{Mg}^{2+}\) and has a lower charge density.