2024 H2 Chem P1 Q5 — Atomic structure Solution

The answer is B (10).
Write the valence-shell configuration of each element and fill the three 3p orbitals singly before pairing (Hund's rule):
Na: [Ne]\(3\text{s}^1\), 1 unpaired
Mg: [Ne]\(3\text{s}^2\), 0 unpaired
Al: [Ne]\(3\text{s}^2 3\text{p}^1\), 1 unpaired
Si: [Ne]\(3\text{s}^2 3\text{p}_x^1 3\text{p}_y^1\), 2 unpaired
P: [Ne]\(3\text{s}^2 3\text{p}_x^1 3\text{p}_y^1 3\text{p}_z^1\), 3 unpaired
S: [Ne]\(3\text{s}^2 3\text{p}_x^2 3\text{p}_y^1 3\text{p}_z^1\), 2 unpaired
Cl: [Ne]\(3\text{s}^2 3\text{p}_x^2 3\text{p}_y^2 3\text{p}_z^1\), 1 unpaired
Ar: [Ne]\(3\text{s}^2 3\text{p}^6\), 0 unpaired
Total \(= 1 + 0 + 1 + 2 + 3 + 2 + 1 + 0 = 10\)
• A is wrong: 8 is the number of elements, as if each element had exactly one unpaired electron; Mg and Ar have none, while Si, P and S have more than one.
• C is wrong: 12 is more than the total from the orbital diagrams; S and Cl have paired p electrons (only 2 and 1 unpaired respectively), and Mg and Ar have none.
• D is wrong: 14 overcounts even further; no configuration above gives more than 10 unpaired electrons in total.