Q.On the basis of VB theory explain the nature of bonding in [Co(C₂O₄)₃]³⁻.

Step 1. [Co(C₂O₄)₃]³⁻ has 3 K⁺ (or similar) counter ions implied by the -3 charge; with oxalate (ox²⁻) charge -2 each: x + 3(-2) = -3 gives x = +3, so cobalt is Co³⁺, outer configuration 3d⁶.

Step 2. Oxalate is bidentate, giving a coordination number of 6 (3 ligands × 2 donor O atoms each) -- octahedral geometry.

Step 3. In the spectrochemical series, ox²⁻ sits ABOVE F⁻ (a weak-field ligand) but below H₂O, NH₃ and CN⁻ -- meaning it is a moderately strong field ligand, unlike F⁻.

Step 4. Co³⁺ complexes are characteristically low-spin except with the very weakest ligands (like F⁻, as in [CoF₆]³⁻, VBT Illustration 4); since oxalate is a stronger field than F⁻, [Co(C₂O₄)₃]³⁻ is expected to be LOW spin, following the same pattern as [Fe(CN)₆]³⁻ (VBT Illustration 3). …

Work out cobalt's oxidation state and d-electron count, place oxalate on the spectrochemical series relative to the chapter's own worked F⁻ (high-spin, outer-orbital) and CN⁻ (low-spin, inner-o …

• Assuming oxalate must give a high-spin, outer-orbital complex just because [CoF₆]³⁻ (also weak-ish field, also octahedral Co³⁺) is high spin -- oxalate is actually a stronger field ligand than F⁻ in the spectrochemical series, which flips the expected spin state. …