Question

1 : 1 Molar mixture of FeSO$_4$ and (NH$_4$)$_2$SO$_4$ gives test of Fe$^{2+$ ions, but 1 : 4 molar mixture of CuSO$_4$ and NH$_3$ (aq) does not give test of Cu$^{2+}$ ions. Why?}

Hint:

Double salts ionize completely and show individual ion tests, but complex salts do not dissociate completely and thus fail to give usual ionic tests.

Answer 1

Step 1: Case of FeSO$_4$ + (NH$_4$)$_2$SO$_4$.
A 1:1 molar mixture of ferrous sulfate and ammonium sulfate forms
Mohr’s salt \((FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O)\). In this double salt, the \(Fe^{2+}\) ions are free and ionize completely in aqueous solution. Hence, Mohr’s salt gives the usual tests of \(Fe^{2+}\) ions. Step 2: Case of CuSO$_4$ + NH$_3$ (aq).
When CuSO$_4$ is treated with excess ammonia (1:4 ratio), a
complex compound \([Cu(NH_3)_4]^{2+}\) is formed. In this complex, copper ions are coordinated with ammonia molecules and are not free in solution. Hence, it does not give the normal tests of \(Cu^{2+}\) ions. Conclusion:
\[ \text{Mohr’s salt → Double salt (ions free)} \quad \text{but} \quad [Cu(NH_3)_4]SO_4 \to \text{Complex (ions not free)} \]