We are asked to match the complex ions with their spin-only magnetic moments. The magnetic moment (\( \mu \)) for a complex ion can be calculated using the formula: \[ \mu = \sqrt{n(n+2)} \] where \( n \) is the number of unpaired electrons in the complex. Now, let’s determine the number of unpaired electrons for each complex and match them to the correct spin-only magnetic moment. Step 1: Determine the number of unpaired electrons for each complex. - (A) \( [\text{Cr}(\text{H}_2\text{O})_6]^{3+} \): - Chromium has an atomic number of 24. In the \( 3+ \) state, the electron configuration is \( \text{Cr}^{3+} : [Ar] 3d^3 \). - Number of unpaired electrons = 3. Therefore, \( \mu = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 \) BM. - Hence, A → Q. - (B) \( [\text{Cu}(\text{H}_2\text{O})_6]^{2+} \): - Copper has an atomic number of 29. In the \( 2+ \) state, the electron configuration is \( \text{Cu}^{2+} : [Ar] 3d^9 \). - Number of unpaired electrons = 1. Therefore, \( \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \) BM. - Hence, B → P. - (C) \( [\text{Mn}(\text{H}_2\text{O})_6]^{2+} \): - Manganese has an atomic number of 25. In the \( 2+ \) state, the electron configuration is \( \text{Mn}^{2+} : [Ar] 3d^5 \). - Number of unpaired electrons = 5. Therefore, \( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.93 \) BM. - Hence, C → S. - (D) \( [\text{Co}(\text{H}_2\text{O})_6]^{3+} \): - Cobalt has an atomic number of 27. In the \( 3+ \) state, the electron configuration is \( \text{Co}^{3+} : [Ar] 3d^6 \). - Number of unpaired electrons = 0. Therefore, \( \mu = \sqrt{0(0+2)} = 0 \) BM. - Hence, D → R. Step 2: Conclusion. The correct matches are: - A → Q - B → P - C → S - D → R Thus, the correct answer is (A) A → Q ; B → P ; C → S ; D → R. (A) A → Q ; B → P ; C → S ; D → R.
