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(B) To determine the paramagnetism,we calculate the number of unpaired electrons in each complex:$1$. In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ is $3d^3$,which

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

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(B) To determine the paramagnetism,we calculate the number of unpaired electrons in each complex:$1$. In $[Cr(H_2O)_6]^{3+}$,$Cr^{3+}$ is $3d^3$,which has $3$ unpaired electrons.$2$. 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 $4$ unpaired electrons.$3$. In $[Cu(H_2O)_6]^{2+}$,$Cu^{2+}$ is $3d^9$,which has $1$ unpaired electron.$4$. In $[Zn(H_2O)_6]^{2+}$,$Zn^{2+}$ is $3d^{10}$,which has $0$ unpaired electrons.Since $[Fe(H_2O)_6]^{2+}$ has the highest number of unpaired electrons $(4)$,it exhibits the highest paramagnetism.

Among the following ions,which one has the highest paramagnetism?

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