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

Solution

Chemistry ma’am · 4 steps

Every option is octahedral, so the real test is zero unpaired electrons and a strong-field ligand.

1. Shape decides nothing. Each complex has six ligands around the metal, so all four are octahedral. Magnetism and stability decide the answer.
2. Find the d count. Work out the metal's oxidation state and its d electrons, then fill t_(2g) and e_g. A weak-field ligand (Cl⁻, H₂O) gives high spin. A strong-field ligand (CN⁻) forces the electrons to pair in t_(2g).

   Complex	Metal, d count	Ligand field	Configuration	Unpaired
   Na₃[CoCl₆]	Co³⁺, d⁶	weak	t_(2g)⁴ e_g²	4, paramagnetic
   [Co(H₂O)₆]Cl₂	Co²⁺, d⁷	weak	t_(2g)⁵ e_g²	3, paramagnetic
   K₃[Co(CN)₆]	Co³⁺, d⁶	strong	t_(2g)⁶ e_g⁰	0, diamagnetic
   [Ni(NH₃)₆]Cl₂	Ni²⁺, d⁸	strong, but no effect	t_(2g)⁶ e_g²	2, paramagnetic

3. Why most stable. In [Co(CN)₆]³⁻ all six electrons sit in the lower t_(2g) set, giving the largest CFSE, -2.4Δₒ. CN⁻ produces a very large Δₒ, and the +3 charge on cobalt holds the ligands strongly, so this complex is the most stable of the four.
4. So. Only (c), K₃[Co(CN)₆], is octahedral, diamagnetic and the most stable.

The idea to keep. For octahedral d⁶, the ligand decides everything: weak field gives 4 unpaired, strong field gives t_(2g)⁶, which is diamagnetic. And d⁸ always keeps 2 unpaired in e_g, whatever the ligand.

Answer from the NTA final answer key. Solution by PrepWiser.