Identify the metal ions among Co^{2+}, Ni^{2+ | vedclass

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(A) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.For $\mu > 3.0 \ BM$, we need

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$7$

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(A) The spin-only magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$, where $n$ is the number of unpaired electrons.For $\mu > 3.0 \ BM$, we need $\sqrt{n(n+2)} > 3.0$, which implies $n(n+2) > 9$. This holds for $n \geq 3$.Electronic configurations:$V^{3+} (3d^2): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$$Ti^{2+} (3d^2): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$$Ni^{2+} (3d^8): n = 2, \mu = \sqrt{2(4)} = 2.83 \ BM$$Fe^{2+} (3d^6): n = 4, \mu = \sqrt{4(6)} = 4.90 \ BM$$Co^{2+} (3d^7): n = 3, \mu = \sqrt{3(5)} = 3.87 \ BM$Thus, $Fe^{2+}$ and $Co^{2+}$ have $\mu > 3.0 \ BM$.In high spin octahedral complexes:$Fe^{2+} (3d^6)$ has $4$ unpaired electrons.$Co^{2+} (3d^7)$ has $3$ unpaired electrons.Sum of unpaired electrons = $4 + 3 = 7$.

Identify the metal ions among $Co^{2+}, Ni^{2+}, Fe^{2+}, V^{3+}$ and $Ti^{2+}$ having a spin-only magnetic moment value more than $3.0 \ BM$. The sum of unpaired electrons present in the high spin octahedral complexes formed by those metal ions is . . . . . . .

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