The pair of metal ions that can give a spin-o | vedclass

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(A) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.For $\mu = 3.9 \,

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$V^{2+}$ and $Co^{2+}$

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(A) The spin-only magnetic moment is given by the formula $\mu = \sqrt{n(n+2)} \, BM$,where $n$ is the number of unpaired electrons.For $\mu = 3.9 \, BM$,we have $\sqrt{n(n+2)} \approx 3.9$,which implies $n = 3$.Thus,the metal ion $M^{2+}$ must have $3$ unpaired electrons.Electronic configurations in the presence of the weak field ligand $H_2O$:$V^{2+} (d^3): t_{2g}^3 e_g^0$ (unpaired electrons $n = 3$)$Co^{2+} (d^7): t_{2g}^5 e_g^2$ (unpaired electrons $n = 3$)$Fe^{2+} (d^6): t_{2g}^4 e_g^2$ (unpaired electrons $n = 4$)$Cr^{2+} (d^4): t_{2g}^3 e_g^1$ (unpaired electrons $n = 4$)$Mn^{2+} (d^5): t_{2g}^3 e_g^2$ (unpaired electrons $n = 5$)Therefore,the pair of metal ions with $3$ unpaired electrons is $V^{2+}$ and $Co^{2+}$.

The pair of metal ions that can give a spin-only magnetic moment of $3.9 \, BM$ for the complex $[M(H_2O)_6]Cl_2$ is

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