Coordination Compounds Part 4

Concept:The spin-only magnetic moment, $\mu = \sqrt{n(n+2)}$ Bohr Magneton, depends on the number of unpaired electrons $n$ in the metal ion.Explanation:First, write the electronic configurations of the given ions (considering the loss of $4\mathrm{s}$ electrons before $3\mathrm{d}$): $\mathrm{Co}^{2+}$: $[\mathrm{Ar}] 3\mathrm{d}^7$ $\mathrm{Ni}^{2+}$: $[\mathrm{Ar}] 3\mathrm{d}^8$ $\mathrm{Fe}^{2+}$: $[\mathrm{Ar}] 3\mathrm{d}^6$ $\mathrm{V}^{3+}$: $[\mathrm{Ar}] 3\mathrm{d}^2$ $\mathrm{Ti}^{2+}$: $[\mathrm{Ar}] 3\mathrm{d}^2$ For a spin-only moment $>3.0$ BM, we need $n(n+2) > 9$ i.e. $n > 3$ approximately. Thus only ions with more than 3 unpaired electrons qualify. In high-spin octahedral complexes: $\mathrm{Fe}^{2+}$ ($\mathrm{d}^6$) gives 4 unpaired electrons ($t_{2g}^4 e_g^2$). $\mathrm{Co}^{2+}$ ($\mathrm{d}^7$) gives 3 unpaired electrons ($t_{2g}^5 e_g^2$). The other ions ($\mathrm{Ni}^{2+}$, $\mathrm{V}^{3+}$, $\mathrm{Ti}^{2+}$) have 2 or fewer unpaired electrons, so their $\mu$ values are $\le 3.0$ BM. Therefore, the required ions are $\mathrm{Fe}^{2+}$ and $\mathrm{Co}^{2+}$. Number of unpaired electrons in high-spin complexes: $\mathrm{Fe}^{2+}=4$, $\mathrm{Co}^{2+}=3$. Sum $= 4 + 3 = 7$. Answer:7