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For octahedral [Mn(H2O)6]²⁺, d-electron count and unpaired e⁻ count:
Ad³, 3 unpaired
Bd⁵, 5 unpaired (high-spin)
Cd⁵, 1 unpaired
Dd⁷, 1 unpaired
Answer & Solution
Correct answer: B. d⁵, 5 unpaired (high-spin)
Mn²⁺: 25-23 = 3d⁵. H2O is weak/moderate field. High-spin d⁵: all 5 unpaired (each d-orbital singly occupied). Very paramagnetic. Color: nearly colorless because no allowed d-d transitions for d⁵ HS.
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