A) A deep blue solution is obtained.
B) A blue precipitate of\[Cu{{(OH)}_{2}}\]is obtained.
C) A black precipitate of\[CuO\] is obtained.
D) No change takes place.
Correct Answer: A
Solution :
With ammonium solution, \[\text{CuS}{{\text{O}}_{\text{4}}}\]forms a stable blue complex. First it forms a precipitate of \[\text{Cu(OH}{{\text{)}}_{\text{2}}}\] which dissolves in excess of ammonia solution forming a deep blue coloured complex. \[CuS{{O}_{4}}+2N{{H}_{4}}OH\to Cu{{(OH)}_{2}}+{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\] \[Cu{{(OH)}_{2}}+2N{{H}_{4}}OH+{{(N{{H}_{4}})}_{2}}S{{O}_{4}}\to \] \[\underset{\text{Tetraammine}\,\text{cupric}\,\text{sulphate}}{\mathop{Cu{{(N{{H}_{3}})}_{4}}S{{O}_{4}}}}\,+4{{H}_{2}}O\] Hence, the correct option is [a].
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