The species that has a spin-only magnetic moment of 5.9 BM, is: (Td = tetrahedral)

🧠 Reverse-Engineer 5.9 BM

$\mu = \sqrt{n(n+2)} = 5.9$ BM ⟹ $n(n+2) = 34.81$ ⟹ $n = 5$ unpaired electrons.

So we need a metal centre with 5 unpaired electrons.

🗺️ Test Each Option

| Complex | Metal/OS | d-count | Geometry | Unpaired | |---|---|---|---|---| | $[\mathrm{Ni(CN)_4}]^{2-}$ | Ni²⁺ | d⁸ | sq. planar | 0 | | $[\mathrm{NiCl_4}]^{2-}$ | Ni²⁺ | d⁸ | Td | 2 | | $\mathrm{Ni(CO)_4}$ | Ni⁰ | d¹⁰ | Td | 0 | | $[\mathrm{MnBr_4}]^{2-}$ | Mn²⁺ | d⁵ | Td | 5 ✓ |

Mn²⁺ d⁵ in any tetrahedral field is automatically high-spin (Δ_t is too small to pair) → 5 unpaired → μ = 5.9 BM.

⚡ The d⁵ HS Signature

Mn²⁺, Fe³⁺ — both d⁵ — give μ ≈ 5.9 BM in any weak-to-moderate field. This is the most distinctive μ value in the question bank: see 5.9, think d⁵ HS.

⚠️ Td Δ_t Is Always Weak

For tetrahedral complexes, $\Delta_t \approx \frac{4}{9}\Delta_o$ — too small to ever overcome pairing energy in first-row metals. So all first-row Td complexes are HS by default. Never bother checking ligand strength for Td spin state.

$\boxed{\text{Answer: (4) } [\mathrm{MnBr_4}]^{2-}}$