Identify the diamagnetic octahedral complex ions from below: A. [Mn(CN)6]3- B. [Co(NH3)6]3+ C. [Fe(CN)6]4- D.…

🧠 Diamagnetic ⇒ All Electrons Paired

For octahedral complexes, diamagnetic = LS d⁶ (filled $t_{2g}$, empty $e_g$) is the canonical case. Otherwise, check d-config and ligand field.

🗺️ Evaluate Each Complex

| # | Complex | Metal/d-config | Ligand field | Configuration | Unpaired | Diamagnetic? | |---|---|---|---|---|---|---| | A | $[\mathrm{Mn(CN)_6}]^{3-}$ | Mn(III) d⁴ | Strong (CN⁻) → LS | $t_{2g}^4 e_g^0$ | 2 | ✗ | | B | $[\mathrm{Co(NH_3)_6}]^{3+}$ | Co(III) d⁶ | Strong → LS | $t_{2g}^6 e_g^0$ | 0 | ✓ | | C | $[\mathrm{Fe(CN)_6}]^{4-}$ | Fe(II) d⁶ | Strong (CN⁻) → LS | $t_{2g}^6 e_g^0$ | 0 | ✓ | | D | $[\mathrm{Co(H_2O)_3 F_3}]$ | Co(III) d⁶ | Weak (H₂O + F⁻) → HS | $t_{2g}^4 e_g^2$ | 4 | ✗ |

B and C are diamagnetic.

⚡ Co(III) Strong-Field Special

Co(III) with NH₃ (and stronger ligands) is universally LS d⁶ → diamagnetic. With weak-field ligands (F⁻, H₂O), Co(III) goes HS d⁶ → 4 unpaired.

⚠️ CN⁻ Doesn't Always Force LS

For d⁴ Mn(III), CN⁻ gives LS but $t_{2g}^4$ still has 2 unpaired (only 4 electrons in 3 $t_{2g}$ orbitals). Diamagnetism requires all electrons paired.

$\boxed{\text{Answer: (4) B and C only}}$