2018 H2 Chem P1 Q4 — Atomic structure Solution

The answer is B (\(\Delta H_1 > \Delta H_3 > \Delta H_2\)).
Each reaction removes one electron from a species with 54 electrons: \(\text{Cs}^+\), \(\text{I}^-\) and Xe are isoelectronic, all with the configuration \([\text{Kr}]4\text{d}^{10}5\text{s}^2 5\text{p}^6\).
With the same number of electrons, the shielding is the same, so the effective nuclear charge depends only on the number of protons: \(\text{Cs}^+\) has 55, Xe has 54 and \(\text{I}^-\) has 53.
The greater the nuclear charge, the more strongly the outermost 5p electron is held and the more energy is needed to remove it. So the order of decreasing enthalpy change is \(\text{Cs}^+ (\Delta H_1) > \text{Xe} (\Delta H_3) > \text{I}^- (\Delta H_2)\).
• A is wrong: it puts \(\Delta H_2\) above \(\Delta H_3\), but \(\text{I}^-\) has one proton fewer than Xe, so its electron is held less strongly and \(\Delta H_2\) is the smallest.
• C is wrong: it puts \(\Delta H_2\) first, but \(\text{I}^-\) has the lowest nuclear charge of the three, so removing its electron needs the least energy.
• D is wrong: it is the reverse of the correct order; \(\text{Cs}^+\), with the highest nuclear charge (55 protons), needs the most energy, not the least.