Which of the following complex will show largest splitting of d-orbitals?

Solution

Chemistry ma’am · 3 steps

The metal and its charge are the same in all four, so only the ligand decides the splitting.

1. Same metal, same charge. Every complex here is octahedral with Fe³⁺ at the centre. The metal's contribution to Δₒ is therefore identical, and only the ligand's field strength can make a difference.
2. Spectrochemical series. Ligands ranked by how strongly they split the d-orbitals: F⁻ < C₂O₄²⁻ < H₂O < NH₃ < CN⁻. CN⁻ is near the strong end because it donates through carbon and also accepts electron density back from the metal into its empty π^* orbitals (back bonding).

   Complex	Ligand	Field strength
   [FeF₆]³⁻	F⁻	weak, smallest Δₒ
   [Fe(C₂O₄)₃]³⁻	C₂O₄²⁻	weak, O donor
   [Fe(NH₃)₆]³⁺	NH₃	moderate, N donor
   [Fe(CN)₆]³⁻	CN⁻	strong, largest Δₒ

3. So. CN⁻ is the strongest field ligand of the four, so [Fe(CN)₆]³⁻ shows the largest d-orbital splitting. The answer is (b).

The idea to keep. When the metal ion and geometry are fixed, compare ligands using the spectrochemical series. Halides sit at the weak end and CN⁻ and CO at the strong end.

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