Match Complex Ions with Calculated Spin Only Magnetic Moments

To determine the spin-only magnetic moment ($\mu_{\text{s}}$) of each complex ion, we use the formula: $\mu_{\text{s}} = \sqrt{n(n+2)} \text{ BM}$ where $n$ is the number of unpaired electrons in the central metal ion.

Since $\text{H}_2\text{O}$ is a weak field ligand, these octahedral complexes will form high-spin states:

1. A. $[\text{Cr}(\text{H}_2\text{O})_6]^{2+}$:

   • Oxidation state of $\text{Cr} = +2$
   • Electronic configuration of $\text{Cr}^{2+} = [\text{Ar}] 3d^4$
   • Distribution of electrons: $t_{2g}^3 e_g^1$
   • Number of unpaired electrons ($n$) = $4$
   • Magnetic moment: $\mu_{\text{s}} = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$
   • A matches with III.

2. B. $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$:

   • Oxidation state of $\text{Co} = +2$
   • Electronic configuration of $\text{Co}^{2+} = [\text{Ar}] 3d^7$
   • Distribution of electrons: $t_{2g}^5 e_g^2$
   • Number of unpaired electrons ($n$) = $3$
   • Magnetic moment: $\mu_{\text{s}} = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \text{ BM}$
   • B matches with I.

3. C. $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$:

   • Oxidation state of $\text{Cu} = +2$
   • Electronic configuration of $\text{Cu}^{2+} = [\text{Ar}] 3d^9$
   • Distribution of electrons: $t_{2g}^6 e_g^3$
   • Number of unpaired electrons ($n$) = $1$
   • Magnetic moment: $\mu_{\text{s}} = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \text{ BM}$
   • C matches with IV.

4. D. $[\text{Mn}(\text{H}_2\text{O})_6]^{2+}$:

   • Oxidation state of $\text{Mn} = +2$
   • Electronic configuration of $\text{Mn}^{2+} = [\text{Ar}] 3d^5$
   • Distribution of electrons: $t_{2g}^3 e_g^2$
   • Number of unpaired electrons ($n$) = $5$
   • Magnetic moment: $\mu_{\text{s}} = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \text{ BM}$
   • D matches with II.

Thus, the correct matching is: $\text{A-III, B-I, C-IV, D-II}$

Correct Option: D