Q. 6 B5.0( 1 Vote )

A chloride of fou

Answer :

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:

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