The sum of the spin-only magnetic moment values (in BM) of [Mn(Br)₆]³⁻ and [Mn(CN)₆]³⁻ is

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Question

The sum of the spin-only magnetic moment values (in BM) of [Mn(Br)₆]³⁻ and [Mn(CN)₆]³⁻ is

Accepted Answer

Correct Answer: 7.72 BMStep-by-step Solution:Both complexes contain Mn³⁺ ion (Mn has configuration [Ar]3d⁵4s² → Mn³⁺ has lost 3 electrons → [Ar]3d⁴).[Mn(Br)₆]³⁻:Br⁻ is a weak field ligand (does not cause pairing of electrons).Mn³⁺ (3d⁴): In weak field, electrons remain unpaired as much as possible.Configuration: ↑|↑|↑|↑| → 4 unpaired electrons (n = 4).μ = √(n(n+2)) = √(4×6) = √24 = 2√6 ≈ 4.90 BM[Mn(CN)₆]³⁻:CN⁻ is a strong field ligand (causes maximum electron pairing in t₂g set).Mn³⁺ (3d⁴): In strong field (octahedral), fill t₂g first before eg:t₂g: ↑↓|↑|↑ (all 4 electrons in t₂g: 1 paired pair + 2 unpaired → n = 2 unpaired).Actually: 3d⁴ in strong field → t₂g⁴ e_g⁰: ↑↓|↑|↑ → 2 unpaired electrons (n = 2).μ = √(2×4) = √8 = 2√2 ≈ 2.83 BMSum = 4.90 + 2.83 = 7.73 ≈ 7.72 BM(More precisely: √24 + √8 = 2√6 + 2√2 = 2(√6 + √2) = 2(2.449 + 1.414) = 2(3.863) = 7.726 ≈ 7.72 BM)Key Concepts:Spin-only magnetic moment formula: μ = √(n(n+2)) BM, where n = number of unpaired electrons. This formula is valid for first-row transition metal complexes where orbital contribution to magnetic moment is quenched.Crystal Field Theory (CFT) for octahedral complexes: Ligands split the 3d orbitals into t₂g (lower, triply degenerate) and e_g (upper, doubly degenerate) sets. Strong field ligands cause large Δ_oct (crystal field splitting), favouring pairing; weak field ligands cause small Δ_oct, favouring unpaired electrons (high spin).Spectrochemical series: I⁻ Mn³⁺ (d⁴) configurations:High spin (weak field ligand): t₂g³ e_g¹ → 4 unpaired e⁻ → μ = √24 ≈ 4.90 BMLow spin (strong field ligand): t₂g⁴ e_g⁰ → 2 unpaired e⁻ → μ = √8 ≈ 2.83 BM

The sum of the spin-only magnetic moment values (in BM) of [Mn(Br)₆]³⁻ and [Mn(CN)₆]³⁻ is

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The sum of the spin-only magnetic moment values (in BM) of [Mn(Br)₆]³⁻ and [Mn(CN)₆]³⁻ is

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