Which of the following complex is octahedral, diamagnetic and the most stable?

🧠 Three Conditions: Octahedral + Diamagnetic + Most Stable

Need a complex that is:

• Octahedral (rules out tetrahedral / square-planar)
• Diamagnetic (zero unpaired electrons)
• Most stable (typically: largest CFSE, strong-field ligand)

🗺️ Test Each

| Complex | Geometry | Magnetic | |---|---|---| | (1) $\mathrm{Na_3[CoCl_6]}$ | octahedral | Co(III) d⁶ + weak Cl⁻ → high-spin → 4 unpaired (paramagnetic) ✗ | | (2) $[\mathrm{Ni(NH_3)_6}]\mathrm{Cl_2}$ | octahedral | Ni(II) d⁸ → 2 unpaired (paramagnetic) ✗ | | (3) $\mathrm{K_3[Co(CN)_6]}$ | octahedral | Co(III) d⁶ + strong CN⁻ → low-spin → 0 unpaired ✓ | | (4) $[\mathrm{Co(H_2O)_6}]\mathrm{Cl_2}$ | octahedral | Co(II) d⁷ + weak H₂O → high-spin → 3 unpaired ✗ |

Only (3) is diamagnetic. CN⁻ is the strongest spectrochemical-series ligand → also gives the largest CFSE → most stable.

⚡ The "Co(III) + CN⁻ = Diamagnetic + Stable" Trio

$[\mathrm{Co(CN)_6}]^{3-}$ is one of the most stable octahedral complexes in coordination chemistry — combining Co(III)'s preference for low-spin d⁶ with CN⁻'s strong-field, π-accepting character.

⚠️ Don't Pick Ni²⁺ Just Because of NH₃

Ni(II) $\mathrm{d^8}$ is always paramagnetic in octahedral geometry (2 unpaired in $\mathrm{e_g}$). It can never be made diamagnetic in octahedral form, regardless of ligand. Square-planar Ni(II) (like $[\mathrm{Ni(CN)_4}]^{2-}$) is diamagnetic, but that's a different geometry.

$\boxed{\text{Answer: (3) } \mathrm{K_3[Co(CN)_6]}}$