`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?

Explanation

(a). `FeSO_4` and `(NH_4)_2SO_4` in `1:1` molar ratio form a double salt, called Mohr's salt.
`FeSO_4.(NH_4)_2SO_4.6H_2O` which ionises in aqueous solution to give `Fe^(2+)` ions. Hence it gives the test of `Fe^(2+)` ions.
Test of `Fe^(2+)` ions:
(i). `Fe^(2+)+underset("Ferrocyanide ion")([Fe^(2+)(CN)_6]^(4-))tounderset("Ferro-ferro cyanide")(Fe_3^(2+)[Fe(CN)_6]^(4-))`
(ii). `Fe^(2+)+underset("Ferricyanide ion")([Fe^(3+)(CN)_6]^(3-))tounderset("Ferro-ferri cyanide")(Fe_3^(2+)[Fe^(3+)(CN)_6]^(3-))`
(b). `CuSO_3` and aqueous `NH_3` in `1:4` molar ratio form a complex salt, `[Cu(NH_3)_4]SO_4`. The complex ion `[Cu(NH_3)_4]^(2+)` does not ionise to give `Cu^(2+)` ions. Hence it does not give the test of `Cu^(2+)` ions.