$FeSO_4$ solution mixed with $(NH_4)_2SO_4$ 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?

(N/A) $(NH_4)_2SO_4 + FeSO_4 + 6H_2O \to FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ (Mohr's salt)
$CuSO_4 + 4NH_3 + 5H_2O \to [Cu(NH_3)_4]SO_4 \cdot 5H_2O$ (tetraamminocopper$(II)$ sulphate)
Both compounds are addition compounds,but they differ in nature. The former is a double salt,while the latter is a coordination compound.
$A$ double salt is stable in the solid state but dissociates into its constituent ions in the dissolved state. Thus,$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ dissociates into $Fe^{2+}$,$NH_4^+$,and $SO_4^{2-}$ ions,allowing it to give a positive test for $Fe^{2+}$.
$A$ coordination compound retains its identity in both solid and dissolved states. In $[Cu(NH_3)_4]SO_4 \cdot 5H_2O$,the $Cu^{2+}$ ion is part of the stable complex ion $[Cu(NH_3)_4]^{2+}$. Since the complex does not dissociate to release free $Cu^{2+}$ ions in the solution,it does not give the test for $Cu^{2+}$.