Explain the formation of [CoF6]3– complex ion with respect to(i) Oxidation State of Cobalt(ii) Type of hybridisation(iii) Magnetic property(iv) Geometrical shape(b) Write the name of ligand used to estimate hardness of water.

Answer:

(a) (i) Oxidation state: \(x + 6(-1) = -3\), so Co is +3. Co (Z = 27): [Ar]3d⁷4s²; Co³⁺: [Ar]3d⁶ (four unpaired electrons).

(ii) Hybridisation: F^– is a weak field ligand, so the 3d electrons are not paired. The six F^– lone pairs occupy one 4s, three 4p and two 4d orbitals: sp³d² (outer orbital, high spin complex).

(iii) Magnetic property: four unpaired electrons, hence paramagnetic (μ ≈ \(\sqrt{4(4+2)} = 4.9\) BM).

(iv) Shape: sp³d² hybridisation gives an octahedral complex.

(b) EDTA (ethylenediaminetetraacetate), a hexadentate ligand, used in complexometric titration of Ca²⁺ and Mg²⁺.