Question

For a given metal ion and set of ligands, is the crystal-field splitting energy larger for a tetrahedral or an octahedral geometry?

Short answer

In conclusion, for a given metal ion and set of ligands, the crystal-field splitting energy is larger for an octahedral geometry than for a tetrahedral geometry. This is because the ligand interactions are stronger in octahedral geometry due to the ligand-metal ion bond angles. The relationship between the crystal-field splitting energies in octahedral (Δ_oct) and tetrahedral (Δ_tet) geometries can be defined as: Δ_tet = \( \frac{4}{9} \) * Δ_oct The factor \( \frac{4}{9} \) indicates that the tetrahedral geometry has a smaller crystal-field splitting energy compared to the octahedral geometry.

Step-by-step solution

Understanding Tetrahedral and Octahedral Geometries

Tetrahedral geometry consists of a metal ion surrounded by four ligands, while octahedral geometry consists of a metal ion surrounded by six ligands. In both cases, the ligands interact differently with the metal ion's d-orbitals, leading to different crystal-field splitting energies.

Crystal-Field Splitting in Octahedral Geometry

In octahedral geometry, the d-orbitals of the metal ion are split into two groups: the t₂g orbitals (d(xy), d(xz), and d(yz)) and the e_g orbitals (d(x²-y²) and d(z²)). The t₂g orbitals are lower in energy, while the e_g orbitals are higher in energy. The crystal-field splitting energy in octahedral geometry (Δ_oct) is the energy difference between the t₂g and e_g orbitals.

Crystal-Field Splitting in Tetrahedral Geometry

In tetrahedral geometry, the d-orbitals are also split into two groups, but this time the e (d(xy), d(xz), and d(yz)) orbitals are lower in energy, and the t₂ orbitals (d(x²-y²) and d(z²)) are higher in energy. The crystal-field splitting energy in tetrahedral geometry (Δ_tet) is the energy difference between the e and t₂ orbitals.

Comparing Crystal-Field Splitting Energies

In general, the crystal-field splitting energy in octahedral geometry is larger than that in tetrahedral geometry. This is because the ligand interactions are stronger in octahedral geometry due to the ligand-metal ion bond angles (90° for octahedral vs 109.5° for tetrahedral). To make a comparison, we can define a relationship between Δ_oct and Δ_tet: Δ_tet = 4/9 * Δ_oct The factor 4/9 indicates that the crystal-field splitting energy is smaller for tetrahedral geometry than for octahedral geometry.

Conclusion

In conclusion, for a given metal ion and set of ligands, the crystal-field splitting energy is larger for an octahedral geometry than for a tetrahedral geometry.