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(B) Paramagnetism is determined by the number of unpaired electrons $(n)$. The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$.$1$. In $[Cr(H_2

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

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(B) Paramagnetism is determined by the number of unpaired electrons $(n)$. The magnetic moment is given by $\mu = \sqrt{n(n+2)} \ BM$.$1$. In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ is $3d^3$,so $n = 3$.$2$. In $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $3d^6$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^4 e_g^2$,resulting in $n = 4$.$3$. In $[Cu(H_2O)_6]^{2+}$,$Cu^{2+}$ is $3d^9$,so $n = 1$.$4$. In $[Zn(H_2O)_6]^{2+}$,$Zn^{2+}$ is $3d^{10}$,so $n = 0$.Since $[Fe(H_2O)_6]^{2+}$ has the highest number of unpaired electrons $(n = 4)$,it exhibits the highest paramagnetism.

Which of the following aquated metal ions has the highest paramagnetism?

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