Octahedral complexes of copper(II) undergo structural distortion (Jahn-Teller). Which one of the given copper(II)…

🧠 Jahn-Teller Distortion in d⁹ Cu(II)

Cu(II) is d⁹: $t_{2g}^6 e_g^3$. The unequal $e_g$ population causes Jahn-Teller distortion (typically axial elongation).

The magnitude of distortion depends on the ligand field. Stronger field ligands enhance the distortion (larger $\Delta_o$ → more $e_g$ splitting visible). Weaker field ligands → less distortion magnitude.

Wait — that's not quite right. Let me reconsider.

🗺️ Reasoning Through Each Option

For octahedral Cu(II) complexes, J-T distortion magnitude correlates with the symmetry of the ligand environment. With identical ligands all around (homoleptic), the distortion is unambiguous and large. Mixed-ligand complexes with strongly bound ligands in cis/trans positions can lock geometry and reduce J-T magnitude.

| Complex | Ligand environment | J-T extent | |---|---|---| | $[\mathrm{Cu(H_2O)_6}]\mathrm{SO_4}$ | All 6 H₂O — homoleptic, weak field, free axial elongation | Maximum | | $[\mathrm{Cu(en)(H_2O)_4}]\mathrm{SO_4}$ | en + 4 H₂O — en restricts axial movement | Less | | cis-$[\mathrm{Cu(en)_2 Cl_2}]$ | 2 en + 2 Cl cis | Even less (en holds geometry) | | trans-$[\mathrm{Cu(en)_2 Cl_2}]$ | 2 en + 2 Cl trans | Less still |

🗺️ Why Aqua Complex Distorts Most

In $[\mathrm{Cu(H_2O)_6}]^{2+}$, all six H₂O are weakly bound — the molecule freely elongates two axial Cu–O bonds (from ~2.0 Å to ~2.4 Å) under J-T. With chelating ligands like en, the bidentate constraint resists axial elongation.

⚡ J-T Distortion Magnitude Rule

For octahedral Cu(II):

• Homoleptic with weak-field monodentate (H₂O, F⁻) → largest distortion.
• Strong-field ligands or chelates → reduce distortion magnitude (constrained geometry).

⚠️ J-T Always Present, Magnitude Varies

All Cu(II) octahedral complexes have some J-T distortion (since d⁹ is unsymmetrical). The question is which has the most.

$\boxed{\text{Answer: (1) } [\mathrm{Cu(H_2O)_6}]\mathrm{SO_4}}$