Correct Answer

[Cr(H₂O)₆]³⁺ exhibits perfect octahedral geometry due to its d³ electron configuration, which shows no Jahn–Teller distortion in an octahedral crystal field. Chromium (atomic number 24) in the +3 oxidation state has a d³ arrangement with three unpaired electrons symmetrically distributed in the t₂g orbitals. Water acts as a weak field ligand, but the high-spin configuration maintains ideal symmetry.

Option Analysis

• (A) Slightly distorted octahedral geometry: Incorrect, as no minor distortions occur; the electron setup ensures balance without perturbations from ligands or other factors.
• (B) Tetragonally elongated octahedral geometry: Incorrect; elongation happens in high-spin d⁹ or d⁴ cases (e.g., Cu²⁺, Cr²⁺) where eg electrons unevenly populate dz² and dx²-y² orbitals.
• (C) Tetragonally compressed octahedral geometry: Incorrect; compression is rare and typically seen in specific low-spin cases, not applicable to this d³ high-spin complex.
• (D) Perfect octahedral geometry: Correct; d³ configuration places one electron each in t₂g orbitals (dxy, dxz, dyz), keeping t₂g and eg sets degenerate and undistorted.

Crystal Field Splitting Basics

In octahedral fields, d orbitals split into t₂g (lower) and eg (higher) sets. For [Cr(H₂O)₆]³⁺, water ligands create moderate splitting (Δo), but as a weak field ligand for first-row transition metals, it yields high-spin d³: three electrons singly occupy t₂g, maintaining symmetry. No Jahn–Teller distortion occurs since t₂g³ leaves no uneven eg population.

Why Not Distorted?

Jahn–Teller theorem predicts distortion for degenerate, unequally filled orbitals. Here:

• d⁴ high-spin or d⁹ (e.g., [Mn(H₂O)₆]²⁺) elongate tetragonally.
• d³ like [Cr(H₂O)₆]³⁺ stays perfect, unlike [Cr(H₂O)₆]²⁺ (d⁴).

Valence bond theory supports d²sp³ hybridization for octahedral shape.

Key Comparisons