A transition metal complex is experimentally observed to have a spin-only magnetic moment of approxi — Atomic Structure Chemistry Question

💡 Solution & Explanation

The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \text{ BM}$, where $n$ is the number of unpaired electrons. For $\mu \approx 2.83 \text{ BM}$, we have $\sqrt{n(n+2)} \approx 2.83$, yielding $n(n+2) \approx 8$, which implies $n = 2$ unpaired electrons. $Ni^{2+}$ ($Z=28$) has configuration $[Ar] 3d^8$ ($2$ unpaired electrons). $Ti^{3+}$ ($Z=22$) is $[Ar] 3d^1$ ($1$ unpaired electron). $Cr^{3+}$ ($Z=24$) is $[Ar] 3d^3$ ($3$ unpaired electrons). $V^{3+}$ ($Z=23$) is $[Ar] 3d^2$ ($2$ unpaired electrons). Therefore, $Ni^{2+}$ and $V^{3+}$ are the correct matches.