2022 H2 Chem P1 Q8 — Periodic table Solution

The answer is B.

The four elements are P and S (Period 3) and As and Se (Period 4). The sixth ionisation energy removes an electron from the \(\text{X}^{5+}\) ion:
• P (Group 15, Period 3): \(\text{P}^{5+}\) is [He]\(2s^2 2p^6\), so a 2p electron is removed from a stable closed shell.
• S (Group 16, Period 3): \(\text{S}^{5+}\) is [Ne]\(3s^1\), so a 3s electron is removed, leaving a stable closed shell.
• As (Group 15, Period 4): \(\text{As}^{5+}\) is [Ar]\(3d^{10}\), so a 3d electron is removed from a filled subshell.
• Se (Group 16, Period 4): \(\text{Se}^{5+}\) is [Ar]\(3d^{10}4s^1\), so a 4s electron is removed, leaving a stable \(3d^{10}\) configuration.

For the Group 16 elements the sixth electron comes from the outermost shell, so it is removed more easily than for the Group 15 elements, where it comes from an inner shell. Hence G and H are the Group 16 elements and J and M are the Group 15 elements.
Within each group, the electron removed is further from the nucleus in Period 4 than in Period 3, so the Period 4 element has the lower sixth ionisation energy.

Order of increasing sixth ionisation energy: Se (G) < S (H) < As (J) < P (M). The Group 16, Period 3 element is S, which is H.

• A is wrong: G has the lower sixth ionisation energy of the Group 16 pair, so it is Se (Group 16, Period 4).
• C is wrong: J is As (Group 15, Period 4).
• D is wrong: M is P (Group 15, Period 3).