k3 [al(c2o4)3] hybridization state? According to. VBT theory Slove on the page plzzzz

Question:
Find the hybridization state of K₃[Al(C₂O₄)₃] according to VBT.

Solution:

Given complex:
K₃[Al(C₂O₄)₃]

Complex ion = [Al(C₂O₄)₃]³⁻

Central metal atom = Al
Ligand = Oxalate ion (C₂O₄)²⁻

Oxalate is a bidentate ligand.
One oxalate ligand donates 2 lone pairs.

Number of oxalate ligands = 3

Therefore,
Coordination number = 3 × 2
= 6

Now determine oxidation state of Al:

Let oxidation state of Al = x

x + 3(-2) = -3
x – 6 = -3
x = +3

So metal ion is:
Al³⁺

Electronic configuration of Al:
Al = 1s² 2s² 2p⁶ 3s² 3p¹

Electronic configuration of Al³⁺:
Al³⁺ = 1s² 2s² 2p⁶

For coordination number 6,
hybridization can be:
d²sp³ or sp³d²

Since Al³⁺ has no occupied 3d electrons,
outer orbitals are used:

3s + 3p + 2d

Therefore,
Hybridization = sp³d²

Geometry = Octahedral

Final Answer:
Hybridization of K₃[Al(C₂O₄)₃] = sp³d²
Geometry = Octahedral
“`