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What are the spin-only magnetic moments (in Bohr Magnetons, BM) of the complexes [Cr(NH3)6]³+ and [Ni(H2O)6]²+, respectively?
- 3.87 and 2.84
- 4.90 and 1.73
- 3.87 and 1.73
- 4.90 and 2.84
Correct answer: 3.87 and 2.84
Solution
Cr³+ has 3d³ configuration: 3 unpaired electrons, mu = sqrt(3*5) = sqrt(15) = 3.87 BM. Ni²+ has 3d⁸ configuration (high spin with weak field H2O): 2 unpaired electrons, mu = sqrt(2*4) = sqrt(8) = 2.83 ≈ 2.84 BM.
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