Question

The CORRECT order of \( \Delta_0 \) (the octahedral crystal field splitting of d orbitals) values for the following anionic metal complexes is

• A. \([ \text{Ir(CN)}_6 ]^{3-} < [ \text{Rh(CN)}_6 ]^{3-} < [ \text{RhI}_6 ]^{3-} < [ \text{CoI}_6 ]^{3-} \)
• B. \([ \text{CoI}_6 ]^{3-} < [ \text{RhI}_6 ]^{3-} < [ \text{Rh(CN)}_6 ]^{3-} < [ \text{Ir(CN)}_6 ]^{3-} \)
• C. \([ \text{CoI}_6 ]^{3-} < [ \text{Rh(CN)}_6 ]^{3-} < [ \text{RhI}_6 ]^{3-} < [ \text{Ir(CN)}_6 ]^{3-} \)
• D. \([ \text{Ir(CN)}_6 ]^{3-} < [ \text{CoI}_6 ]^{3-} < [ \text{Rh(CN)}_6 ]^{3-} < [ \text{RhI}_6 ]^{3-} \)

Hint:

For octahedral complexes, the metal with a higher oxidation state or larger charge will generally have a larger \( \Delta_0 \) value.

Answer 1

The Correct Option is B

Crystal field splitting (Δₒ) in octahedral complexes:

Δₒ depends on:
1. Metal identity: 3d < 4d < 5d (increases down the group)
2. Ligand strength (spectrochemical series): Weak field < Strong field
3. Oxidation state: Higher oxidation state → larger Δₒ

Spectrochemical series (relevant ligands):
I⁻ < Cl⁻ < CN⁻

Analyzing the complexes:

[Ir(CN)₆]³⁻:
- Ir³⁺ (5d⁶, third row transition metal)
- CN⁻ (strongest field ligand)
- Very large Δₒ

[Rh(CN)₆]³⁻:
- Rh³⁺ (4d⁶, second row transition metal)
- CN⁻ (strongest field ligand)
- Large Δₒ (but less than Ir³⁺)

[RhI₆]³⁻:
- Rh³⁺ (4d⁶, second row)
- I⁻ (weakest field ligand among these)
- Small Δₒ

[CoI₆]³⁻:
- Co³⁺ (3d⁶, first row transition metal)
- I⁻ (weakest field ligand)
- Smallest Δₒ

[RhCl₆]³⁻:
- Rh³⁺ (4d⁶, second row)
- Cl⁻ (intermediate field ligand)
- Moderate Δₒ (larger than I⁻, smaller than CN⁻)

Ordering from smallest to largest Δₒ:

1. [CoI₆]³⁻ - 3d metal, weakest ligand
2. [RhI₆]³⁻ - 4d metal, weakest ligand (larger than Co³⁺)
3. [RhCl₆]³⁻ - 4d metal, intermediate ligand (Cl⁻ > I⁻)
4. [Rh(CN)₆]³⁻ - 4d metal, strongest ligand
5. [Ir(CN)₆]³⁻ - 5d metal, strongest ligand (largest)

Answer: (B)