A chloride of fourth group cation in qualitative analysis gives a green coloured complex [A] in aqueous solution which when treated with ethane –1, 2 – diamine (en) gives pale - yellow solution [B] which on subsequent addition of ethane –1, 2 – diamine turns to blue/purple [C] and finally to violet [D]. Write the structures of complexes [A], [B], [C] and [D].

The cations present in the group IV include Co⁺², Zn⁺², Ni⁺², Mn⁺². The chloride of fourth group cation which gives a green coloured complex is NiCl₂. When NiCl₂ is dissolved in water, it gives green coloured complex due to formation of complex ion [Ni (H₂O) ₆]⁺². Therefore the compounds A is as follows:

A = [Ni (H₂O) ₆] ⁺²

The reaction of Nickel [II] Chloride with water is as follows:

NiCl₂ + 2H₂O → [Ni (H₂O) ₆] ⁺²

Therefore when the compound A that is Hexaaquanickel [II] ion, [Ni (H₂O) ₆] ⁺² reacts with ethane-1, 2-diamine, a pale yellow coloured complex is formed initially. The reaction is as follows:

[Ni (H₂O) ₆] ⁺² + Ethane-1, 2-diamine → [Ni (H₂O) ₄ en] ⁺²

Therefore the compound B = [Ni (H₂O) ₄ en] ⁺²

On further addition of ethane-1, 2-diamine to the complex again an intermediate compound of blue or purple colour is formed. The reaction is as follows:

[Ni (H₂O) ₄ en] ⁺² + Ethane-1, 2-diamine → [Ni (H₂O) ₂ (en) ₂] ⁺²

On further addition of ethane-1, 2-diamine to the complex, all the water molecules present in the complex gets replaced by the Ethane-1, 2-diamine molecules and the final complex formed is of violet colour. The reaction is as follows:

[Ni (H₂O) ₄ (en) ₂] ⁺² + Ethane-1, 2-diamine → [Ni (en) ₃] ⁺²

Therefore the compounds A, B, C, D are as follows: