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 option:

Solution & Explanation

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