For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following statements: (I) Both the complexes can…

🧠 Walk Each Statement

(I) Both can be high-spin: ✓ correct.

• Octahedral Mn(II) d⁵ with weak ligands (Cl⁻, H₂O) → HS → 5 unpaired.
• Tetrahedral Ni(II) d⁸ → always HS (Δ_t too small to cause LS in 3d).

(II) Ni(II) tetrahedral can very rarely be low-spin: ✓ correct.

• Td $\Delta_t \approx \frac{4}{9}\Delta_o$ — almost always too small to overcome pairing energy. So Td 3d complexes are virtually never LS — Ni(II) Td included.

(III) With strong-field ligands, Mn(II) octahedral can be low-spin: ✓ correct.

• Mn(II) d⁵ with strong ligands (CN⁻) → LS → $t_{2g}^5$ → 1 unpaired. Real example: $[\mathrm{Mn(CN)_6}]^{4-}$, μ ≈ 1.73 BM.

(IV) Aqueous Mn(II) is yellow: ✗ incorrect.

• $[\mathrm{Mn(H_2O)_6}]^{2+}$ is pale pink (almost colourless), not yellow. The HS d⁵ configuration has all transitions spin-forbidden + Laporte-forbidden → very weak colour.

🗺️ Why HS Mn²⁺ Is So Pale

For HS d⁵, every d–d transition requires changing one electron's spin (since all five t₂g + e_g electrons are unpaired with parallel spins). Spin-forbidden transitions are extraordinarily weak (ε ~ 0.01–1 M⁻¹cm⁻¹). The result is the famously pale pink Mn²⁺(aq).

⚡ Quick Mn²⁺ Identifiers

• Aqueous Mn²⁺: pale pink, paramagnetic (5 unpaired, μ ≈ 5.92 BM).
• HS d⁵ in any weak/moderate field: same colour pattern.

⚠️ The Yellow Trap

A common error is associating "Mn²⁺" with "yellow", probably confused with $\mathrm{MnO_4^{2-}}$ (manganate, deep green) or $\mathrm{MnO_4^-}$ (permanganate, intense purple). Aqueous Mn(II) is pale pink only.

$\boxed{\text{Answer: (3) (I), (II) and (III) only}}$