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Magnetic moment for [Mn(H₂O)₆]²⁺ (high spin, d⁵):
A1.73
B0
C2.83
D5.92 BM (5 unpaired e⁻: μ = sqrt(5×7) = sqrt(35) ≈ 5.92)
Answer & Solution
Correct answer: D. 5.92 BM (5 unpaired e⁻: μ = sqrt(5×7) = sqrt(35) ≈ 5.92)
Mn²⁺ is d⁵. H₂O weak field, so high-spin: t₂g³ e_g² with 5 unpaired e⁻. μ = sqrt(n(n+2)) = sqrt(35) ≈ 5.92 BM.
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