Q7(b). Tetragonal distortion and the square-planar limit
Why tetragonal distortions occur in octahedral complexes? Illustrate it with suitable example. What will be the result of the extreme distortion to the octahedral complex? Present the d-orbital splitting scheme in this case.
Answer in brief
An unevenly filled e_g (or t_(2g)) set is orbitally degenerate; by the Jahn–Teller theorem the complex distorts, usually by lengthening the two axial bonds (d⁹ [Cu(H₂O)₆]²+). Removing the axial ligands altogether gives a square-planar complex with d_(x²-y²) far above the other four d orbitals.
Key points
- a non-linear molecule in an orbitally degenerate electronic state distorts to remove the degeneracy and lower its energy.
- the third e_g electron can sit in d_(z²) or d_(x²-y²) (degenerate state). Lengthening the two axial (z) bonds lowers d_(z²) and raises d_(x²-y²), each by δ₁/2.
- Two electrons go into the lowered d_(z²) and one into the raised d_(x²-y²): net gain 2(δ₁/2)-δ₁/2=δ₁/2. The t_(2g) set splits too (d_(xz),d_(yz) down, d_(xy) up) but, being full, gains nothing.
- The effect is large for uneven e_g occupancy (d⁹, high-spin d⁴, low-spin d⁷), because e_g orbitals point at the ligands; it is small for uneven t_(2g) (d¹, d²).
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Why the distortion occurs
- Jahn–Teller theorem: a non-linear molecule in an orbitally degenerate electronic state distorts to remove the degeneracy and lower its energy.
- In Oₕ, d⁹ is t_(2g)⁶e_g³: the third e_g electron can sit in d_(z²) or d_(x²-y²) (degenerate state). Lengthening the two axial (z) bonds lowers d_(z²) and raises d_(x²-y²), each by δ₁/2.
- Two electrons go into the lowered d_(z²) and one into the raised d_(x²-y²): net gain 2(δ₁/2)-δ₁/2=δ₁/2. The t_(2g) set splits too (d_(xz),d_(yz) down, d_(xy) up) but, being full, gains nothing.
- The effect is large for uneven e_g occupancy (d⁹, high-spin d⁴, low-spin d⁷), because e_g orbitals point at the ligands; it is small for uneven t_(2g) (d¹, d²).
Example
- [Cu(H₂O)₆]²+ and most Cu(II) complexes have four short and two long bonds (elongated octahedron); Cr(II) (d⁴ high spin) behaves the same way. The single broad visible band of [Cu(H₂O)₆]²+ is broadened by the split levels.
Extreme distortion
- As the axial ligands move away completely the complex becomes square planar (ML₄, D₄ₕ).
- Splitting: d_(x²-y²) (b_(1g)) points at the four ligands: highest. Then d_(xy) (b_(2g), in the plane between ligands), then d_(z²) (a_(1g), only its torus in the plane), and lowest d_(xz),d_(yz) (e_g). Order: x²-y²≫ xy>z²>xz,yz
- The large gap below d_(x²-y²) makes square planar the rule for low-spin d⁸: [Ni(CN)₄]²-, [PdCl₄]²-, [PtCl₄]²-, [AuCl₄]⁻. All eight electrons fill the four lower orbitals and the complexes are diamagnetic.
Application. the Jahn–Teller elongation explains why Cu(II) stands at the top of the Irving–Williams stability order and favours four short bonds, and why the fifth and sixth ligands of Cu(II) proteins and of CuSO₄.5H₂O are weakly held.
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All parts of Question 7
- 7(a) Maximum number of coexisting phases from the phase rule15 marks
- 7(b) Tetragonal distortion and the square-planar limit15 marks
- 7(c) Why N₂⁺ is weaker but O₂⁺ stronger than the neutral molecule10 marks
- 7(d) A cell for AgBr(s) → Ag⁺ + Br⁻10 marks