The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 B.M. The suitable ligand for this complex is:

🧠 Decode μ = 5.9 BM for Octahedral Mn(II)

$\mu = 5.9$ BM ⟹ $n = 5$ unpaired electrons.

Mn²⁺ is d⁵. Five unpaired in octahedral means high-spin: $\mathrm{t_{2g}^3 e_g^2}$. So we need a weak-field ligand.

🗺️ Place Each Ligand on the Spectrochemical Series

Spectrochemical series (weak → strong): $\mathrm{I^-} < \mathrm{Br^-} < \mathrm{Cl^-} < \mathrm{F^-} < \mathrm{OH^-} < \mathrm{H_2O} < \mathrm{NCS^-} < \mathrm{NH_3} < \mathrm{en} < \mathrm{NO_2^-} < \mathrm{CN^-} \approx \mathrm{CO}$

| Ligand | Strength | |---|---| | CN⁻ | strong (LS) | | CO | strong (LS) | | en | strong (LS) | | NCS⁻ | weak/borderline (HS) ✓ |

CN⁻, CO, en would all force Mn²⁺ d⁵ to low-spin ($\mathrm{t_{2g}^5}$, 1 unpaired) → μ ≈ 1.73 BM. Only NCS⁻ keeps the complex high-spin → μ ≈ 5.9 BM.

⚡ The Mn²⁺ HS d⁵ Default

Mn²⁺ with weak/moderate ligands (Cl⁻, H₂O, NCS⁻, F⁻) is overwhelmingly HS d⁵ — five unpaired, μ ≈ 5.9 BM. It takes a true strong-field ligand (CN⁻, CO) to flip to LS.

⚠️ NCS⁻ vs SCN⁻ Linkage

The N-bonded form (isothiocyanate, NCS⁻) is borderline-weak; the S-bonded form (thiocyanate, SCN⁻) is even weaker. Either way, weaker than NH₃.

$\boxed{\text{Answer: (4) } \mathrm{NCS^-}}$