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Question

Two coloured salt solutions of two metals A and B gave precipitates with NaOH and NH4OH. The oxidation states of salts metal A and B are same. On further addition of NH4OH, the precipitate disappears, in the case of B. There is no such change in the case of 'A' either with NaOH or NH4OH. Identify A and B. Give necessary equations :

A
Fe+2,Cu+2
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B
Cu+2,Fe+2
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C
Fe+,Fe+3
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D
Fe+3,Cu+2
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Solution

The correct option is A Fe+2,Cu+2
Both Fe2+ and Fe3+ gives precipitate of their hydroxides (Fe(OH)2) and (Fe(OH)3) with NH4OH or NaOH.
When NH4OH or NaOH is added to a solution of copper ions the copper ions will react with the few OH to make insoluble
Cu(OH)2. The solid exists in a suspension. As more NH4OH is added the insoluble Cu(OH)2 will dissolve as [Cu(NH3)4]2+ is formed. The copper-ammonia complex ion is a deep purple, and it will be difficult to see through the solution, but there won't be any solid precipitate remaining.
As the oxidation state of both salts is similar thus the salts contain Fe2+ and Cu2+ ions respectively.

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