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Number of unpaired electrons in [Mn(CN)₆]³⁻ (Mn(III), d⁴, strong field CN⁻):
A5
B4
C2 (low spin d⁴: t₂g⁴ e_g⁰, 2 unpaired)
D0
Answer & Solution
Correct answer: C. 2 (low spin d⁴: t₂g⁴ e_g⁰, 2 unpaired)
Mn(III) = d⁴. CN⁻ strong field → low spin: t₂g⁴ e_g⁰. Two paired electrons + 2 unpaired in t₂g. μ = sqrt(8) ≈ 2.83 BM.
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