'Spin only' magnetic moment is same for which of the following ions? A. $\mathrm{Ti}{}^{3+}$ B. $\mathrm{Cr}{}^{2+}$ C. $\mathrm{Mn}{}^{2+}$ D. $\mathrm{Fe}{}^{2+}$ E. $\mathrm{Sc}{}^{3+}$ Choose the most appropriate answer from the options given below:

Solution & Explanation

\textbf{Answer:} (D) B and D only \textbf{Solution:} The spin-only moment is $\mu =\sqrt{n(n+2)}$ BM, so two ions have equal $\mu$ only if they have the same number of unpaired electrons $n$. A. $\mathrm{Ti}{}^{3+}$ ($3d^1$): $n=1$. B. $\mathrm{Cr}{}^{2+}$ ($3d^4$): $n=4$. C. $\mathrm{Mn}{}^{2+}$ ($3d^5$): $n=5$. D. $\mathrm{Fe}{}^{2+}$ ($3d^6$): $n=4$. E. $\mathrm{Sc}{}^{3+}$ ($3d^0$): $n=0$. Ions B ($\mathrm{Cr}{}^{2+}$) and D ($\mathrm{Fe}{}^{2+}$) both have $n=4$, giving the same $\mu =\sqrt{24}=4.90$ BM. Hence option D.