Match the LIST-I with LIST-II List-I Complex ion List-II Calculated...

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

A. $[Cr(H_2O)_6]^{2+}$: Chromium is in the $+2$ oxidation state. The electronic configuration of $Cr^{2+}$ is $[Ar] 3d^4$. Since $H_2O$ is a weak field ligand, no pairing occurs. The number of unpaired electrons is $n = 4$.
$\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM. (A $\rightarrow$ III)

B. $[Co(H_2O)_6]^{2+}$: Cobalt is in the $+2$ oxidation state. The electronic configuration of $Co^{2+}$ is $[Ar] 3d^7$. With $H_2O$ as a weak field ligand, the configuration is $t_{2g}^5 e_g^2$. The number of unpaired electrons is $n = 3$.
$\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM. (B $\rightarrow$ I)

C. $[Cu(H_2O)_6]^{2+}$: Copper is in the $+2$ oxidation state. The electronic configuration of $Cu^{2+}$ is $[Ar] 3d^9$. The number of unpaired electrons is $n = 1$.
$\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM. (C $\rightarrow$ IV)

D. $[Mn(H_2O)_6]^{2+}$: Manganese is in the $+2$ oxidation state. The electronic configuration of $Mn^{2+}$ is $[Ar] 3d^5$. With $H_2O$ as a weak field ligand, the configuration is $t_{2g}^3 e_g^2$. The number of unpaired electrons is $n = 5$.
$\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM. (D $\rightarrow$ II)

The correct matching is A-III, B-I, C-IV, D-II.

Answer: A-III, B-I, C-IV, D-II