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(A) The magnetic moment depends on the number of unpaired electrons $(n)$.For $[Cr(H_2O)_6]^{2+}$,$Cr^{2+}$ is $3d^4$. Since $H_2O$ is a weak field li

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

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(A) The magnetic moment depends on the number of unpaired electrons $(n)$.For $[Cr(H_2O)_6]^{2+}$,$Cr^{2+}$ is $3d^4$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^3 e_g^1$,so $n = 4$.For $[Fe(H_2O)_6]^{2+}$,$Fe^{2+}$ is $3d^6$. Since $H_2O$ is a weak field ligand,the configuration is $t_{2g}^4 e_g^2$,so $n = 4$.Since both complexes have $n = 4$,they exhibit the same magnetic moment.

Which pair of complexes exhibits the same theoretical magnetic moment?

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Number of complexes from the following with an even number of unpaired $d$ electrons is . . . . . . . . . .
$[V(H_2O)_6]^{3+}, \quad [Cr(H_2O)_6]^{2+}, \quad [Fe(H_2O)_6]^{3+}, \quad [Ni(H_2O)_6]^{3+}, \quad [Cu(H_2O)_6]^{2+}$
[Given atomic numbers: $V=23, Cr=24, Fe=26, Ni=28, Cu=29$]

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