2019 H2 Chem P1 Q3 — Atomic structure Solution

The answer is B.

Sb and As are in Group 15 (valence shell \(n\text{s}^2 n\text{p}^3\)); Se and Te are in Group 16 (\(n\text{s}^2 n\text{p}^4\)).
• Group 15: the 2nd and 3rd IE remove the last two p electrons, but the 4th IE removes an electron from the inner s subshell, which is lower in energy and closer to the nucleus. So there is a large jump from IE\(_3\) to IE\(_4\).
• Group 16: the 2nd, 3rd and 4th IE all remove p electrons, so the values rise steadily with no large jump.
Jumps from IE\(_3\) to IE\(_4\) (kJ mol\(^{-1}\)): A 4270 - 2440 = 1830; B 3680 - 3010 = 670; C 4850 - 2730 = 2120; D 4140 - 3090 = 1050.
So A and C (large jumps) are the Group 15 elements, and B and D (smaller jumps) are the Group 16 elements, Se and Te.
Te is below Se in Group 16: it has one more electron shell, so its outer electrons are further from the nucleus and more shielded, and its IEs are lower. B has lower values than D at every stage (1800 against 2080, 3010 against 3090, 3680 against 4140), so B is tellurium and D is selenium.

• A is wrong: its large IE\(_3\) to IE\(_4\) jump (1830) marks a Group 15 element; it has the lower IEs of the two Group 15 elements, so it is antimony.
• C is wrong: its jump (2120) also marks Group 15, and its higher IEs make it arsenic, the element above Sb.
• D is wrong: it is a Group 16 element, but its IEs are all higher than those of B, so it is selenium, the element above Te.