Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code: Column I (a) $\mathrm{Co}{}^{3+}$ (b) $\mathrm{Cr}{}^{3+}$ (c) $\mathrm{Fe}{}^{3+}$ (d) $\mathrm{Ni}{}^{2+}$ Column II (i) $\sqrt{8}$ BM (ii) $\sqrt{35}$ BM (iii) $\sqrt{3}$ BM (iv) $\sqrt{24}$ BM (v) $\sqrt{15}$ BM Choose the correct answer:

Solution & Explanation

\textbf{Answer:} (c) a-iv, b-v, c-ii, d-i. \textbf{Solution:} Use the spin-only formula $\mu =\sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. (a) $\mathrm{Co}{}^{3+}$ is $\mathrm{d}{}^6$ with $n=4$ (free ion) → $\mu =\sqrt{4(4+2)}=\sqrt{24}$ BM → (iv). (b) $\mathrm{Cr}{}^{3+}$ is $\mathrm{d}{}^3$ with $n=3$ → $\mu =\sqrt{3(3+2)}=\sqrt{15}$ BM → (v). (c) $\mathrm{Fe}{}^{3+}$ is $\mathrm{d}{}^5$ with $n=5$ → $\mu =\sqrt{5(5+2)}=\sqrt{35}$ BM → (ii). (d) $\mathrm{Ni}{}^{2+}$ is $\mathrm{d}{}^8$ with $n=2$ → $\mu =\sqrt{2(2+2)}=\sqrt{8}$ BM → (i). Thus a-iv, b-v, c-ii, d-i, which is option (c).