Question

Match the LIST-I with LIST-II

• A. A-I, B-III, C-IV, D-II
• B. A-II, B-I, C-III, D-IV
• C. A-IV, B-II, C-I, D-III
• D. A-III, B-I, C-IV, D-II

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The "spin-only" magnetic moment ($\mu_s$) is calculated using the number of unpaired electrons ($n$) in the metal ion's d-orbitals. Water ($H_2O$) is a weak field ligand, so these complexes follow high-spin configurations.

Step 2: Key Formula or Approach:
1. Formula: $\mu_s = \sqrt{n(n+2)}$ Bohr Magnetons (BM). 2. Determine oxidation states (all are +2) and d-electron counts.

Step 3: Detailed Explanation:
1. [Cr(H₂O)₆]²⁺: $Cr^{2+}$ is $d^4$. Since $H_2O$ is weak, $n = 4$. $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90$ BM. (A matches III) 2. [Co(H₂O)₆]²⁺: $Co^{2+}$ is $d^7$. High spin means $t_{2g}^5 e_g^2$. $n = 3$. $\mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87$ BM. (B matches I) 3. [Cu(H₂O)₆]²⁺: $Cu^{2+}$ is $d^9$. $n = 1$. $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73$ BM. (C matches IV) 4. [Mn(H₂O)₆]²⁺: $Mn^{2+}$ is $d^5$. High spin means $n = 5$. $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92$ BM. (D matches II)

Step 4: Final Answer:
Matching sequence: A-III, B-I, C-IV, D-II.