2024 H2 Chem P1 Q29 — Transition elements Solution

The answer is C (nickel(II), 2).
In an octahedral complex the five 3d orbitals split into three lower-energy orbitals and two higher-energy orbitals. Filling each orbital singly before pairing (as the question states), the first 5 electrons go one into each of the five orbitals (3 lower, 2 higher); electrons 6, 7 and 8 then pair up in the three lower orbitals, and electrons 9 and 10 pair up in the two higher orbitals.
Nickel(II): Ni is [Ar]\(3\text{d}^8 4\text{s}^2\), so \(\text{Ni}^{2+}\) is [Ar]\(3\text{d}^8\). Six electrons fill the three lower orbitals (paired) and 2 go singly into the higher orbitals, as shown in the explanation figure. The row value 2 is correct.
• A is wrong: Mn is [Ar]\(3\text{d}^5 4\text{s}^2\), so \(\text{Mn}^{2+}\) is \(3\text{d}^5\): 3 electrons in the lower orbitals and 2 in the higher orbitals, not 1.
• B is wrong: Fe is [Ar]\(3\text{d}^6 4\text{s}^2\), so \(\text{Fe}^{3+}\) is \(3\text{d}^5\): again 2 electrons in the higher orbitals, not 3.
• D is wrong: Cu is [Ar]\(3\text{d}^{10} 4\text{s}^1\), so \(\text{Cu}^+\) is \(3\text{d}^{10}\): all five orbitals are full, so 4 electrons are in the higher orbitals, not 0.