MHTCET 26 Apr 2025 Shift 2 Paper

Concept:The spin-only magnetic moment is calculated using the number of unpaired electrons with the formula $\mu = \sqrt{n(n+2)}$ BM.Explanation:The atomic number of copper (Cu) is 29.The ground state electronic configuration of Cu is $[\mathrm{Ar}]\,3d^{10}4s^1$.For $\mathrm{Cu}^{2+}$, two electrons are removed from the neutral atom.Electrons are removed first from the 4s orbital, then from the 3d orbital.Therefore, the electronic configuration of $\mathrm{Cu}^{2+}$ is $[\mathrm{Ar}]\,3d^9$.In the $3d^9$ configuration, nine electrons occupy the five d-orbitals.Since a fully filled $3d^{10}$ has zero unpaired electrons, $3d^9$ will have exactly one unpaired electron.Thus, the number of unpaired electrons is $n = 1$.Now, substitute $n = 1$ into the spin-only magnetic moment formula:$$\mu = \sqrt{n(n+2)} = \sqrt{1(1+2)} = \sqrt{3}$$$$\mu \approx 1.73 \, \text{BM}$$Answer:Option C: $1.73$ BM