Let’s solve the hybridization of the coordination complex K₃[Al(C₂O₄)₃] step-by-step using Valence Bond Theory (VBT) in easy language and with proper reasoning on one page.

📘 What is the Hybridization of K₃[Al(C₂O₄)₃] According to VBT?

🔹 Step 1: Identify the Central Metal Ion

The central metal is Aluminium (Al).

K₃[Al(C₂O₄)₃] means:

• 3 potassium ions (K⁺)
• 1 complex ion: [Al(C₂O₄)₃]³⁻

So, Al³⁺ is the central ion in the complex.

🔹 Step 2: Write Electronic Configuration of Al³⁺

Atomic number of Al = 13
So, Al: 1s² 2s² 2p⁶ 3s² 3p¹
Al³⁺ means 3 electrons removed → from 3s and 3p

So, Al³⁺ = 1s² 2s² 2p⁶ → [Ne] → completely filled 2nd shell, no electrons in 3rd shell.

👉 It has no unpaired electrons but can still form coordination bonds by accepting lone pairs.

🔹 Step 3: Identify the Ligand and Its Type

Oxalate ion (C₂O₄²⁻) is a bidentate ligand (donates 2 lone pairs per ligand).
We have 3 oxalate ligands → total 6 donor atoms (6 bonds).

So, the coordination number = 6

🔹 Step 4: Hybridization Required

Coordination number 6 → needs 6 orbitals to bond.
According to VBT:

Coordination Number	Hybridization	Geometry
6	d²sp³	Octahedral

👉 So, Al³⁺ will use:

• 2 empty d-orbitals
• 1 s-orbital
• 3 p-orbitals

Total = 6 orbitals = d²sp³ hybridization

🔹 Step 5: Final Answer with Geometry

• Hybridization of Al³⁺ = d²sp³
• Geometry = Octahedral
• Magnetism = Diamagnetic (no unpaired electrons)

✅ Final Answer:

The hybridization of the complex ion [Al(C₂O₄)₃]³⁻ according to VBT is d²sp³, and the geometry is octahedral.