2025 H2 Chem P1 Q4 — Periodic table Solution

The answer is A.
B is \([\text{He}]2\text{s}^2 2\text{p}^1\) and Be is \([\text{He}]2\text{s}^2\). The 2p electron removed from B is of higher energy than the 2s electron removed from Be, because it is shielded from the nuclear charge by the pair of 2s electrons. Less energy is needed to remove it, so B has a lower first ionisation energy than Be. Both the statement and the reason are correct.
• B is wrong: Cu is \([\text{Ar}]3\text{d}^{10}4\text{s}^1\) and Ni is \([\text{Ar}]3\text{d}^8 4\text{s}^2\). Both lose a 4s electron first, not a 3d electron, so the reason is false.
• C is wrong: first ionisation energy does increase from O to Ne, but the reason is that each successive element has one more proton in the nucleus while the shielding stays roughly constant (the electrons are added to the same shell). Having one more outer electron is not the cause.
• D is wrong: the second ionisation energy of P forms \(\text{P}^{2+}\), \([\text{Ne}]3\text{s}^2 3\text{p}^1\) (13 electrons), while the first ionisation energy of S forms \(\text{S}^{+}\), \([\text{Ne}]3\text{s}^2 3\text{p}^3\) (15 electrons). The ions formed do not have the same electronic configuration, so the reason is false.