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9.2  FeSO_{4} solution mixed with (NH_{4})_{2}SO_{4} solution in 1:1 molar ratio gives the test of Fe^{2+} ion but  CuSO_{4} solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu^{2+} ion. Explain why?

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FeSO4.(NH4)2SO4.6H2O →Fe2+ + SO42- + 2NH4+ + 6H2O

Fe2+ ions are free and form a green precipitate with potassium ferricyanide.

While Cu2++ 4NH→ [Cu(NH3)4]2+

So Cu2+ ions are not free and will not form blue precipitate with potassium ferrocyanide.

The major difference between both the compounds is that the first compound is a salt and the other one is a coordination compound. In case of double salt compounds (Mohr's salt), the compound breaks into its constituent ions when dissolved in water, therefore it gives a positive test for the presence of Fe2+. But in case of coordinate compounds, they maintain their identity in both solid and dissolved state. Thus the individual property of each constituent is lost. And therefore it doesn't give a positive test for Cu2+.

Posted by

Devendra Khairwa

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