In Cr^{2+}, Mn^{2+}, Fe^{2+}, and Ni^{2+},the | vedclass

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(B) $H_2O$ is a weak field ligand. The magnetic moment is determined by the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)}

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$[Ni(H_2O)_6]^{2+}$

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(B) $H_2O$ is a weak field ligand. The magnetic moment is determined by the number of unpaired electrons $(n)$ using the formula $\mu = \sqrt{n(n+2)} \ BM$.For $[Cr(H_2O)_6]^{2+}$ $(3d^4)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.For $[Mn(H_2O)_6]^{2+}$ $(3d^5)$: $n = 5$,$\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92 \ BM$.For $[Fe(H_2O)_6]^{2+}$ $(3d^6)$: $n = 4$,$\mu = \sqrt{4(6)} = \sqrt{24} \approx 4.90 \ BM$.For $[Ni(H_2O)_6]^{2+}$ $(3d^8)$: $n = 2$,$\mu = \sqrt{2(4)} = \sqrt{8} \approx 2.83 \ BM$.Since $[Ni(H_2O)_6]^{2+}$ has the lowest number of unpaired electrons $(n=2)$,it has the lowest magnetic moment.

In $Cr^{2+}, Mn^{2+}, Fe^{2+},$ and $Ni^{2+}$,the number of electrons in the $3d$ orbital are $3d^4, 3d^5, 3d^6,$ and $3d^8$ respectively. Which of the following aqueous complexes will have the lowest magnetic moment?

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