Calculate the theoretical spin-only magnetic moment () in Bohr Magnetons (BM) of a ion in its ground — Atomic Structure Chemistry Question

💡 Solution & Explanation

Cobalt (Co, $Z=27$) has the configuration $[Ar] 4s^2 3d^7$. In the $Co^{2+}$ cation, two $4s$ electrons are lost, leaving $[Ar] 3d^7$. Following Hund's rule, the five $3d$ orbitals are occupied as follows: two paired sets and three singly occupied orbitals. Thus, the number of unpaired electrons $n = 3$. The magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{3(3+2)} = \sqrt{15} \text{ BM}$. Squaring this yields $\mu^2 = 15$.