Which of the following are the example of double salt?
(A) $$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$$
(B) $$CuSO_4 \cdot 4NH_3 \cdot H_2O$$
(C) $$K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$$
(D) $$Fe(CN)_2 \cdot 4KCN$$
Choose the correct answer.

We need to identify which of the given compounds are double salts.

What is a double salt?

A double salt is a salt that contains two different cations or two different anions (or both) in the same crystal lattice. The key property of double salts is that they dissociate completely into their constituent ions when dissolved in water. In solution, they behave as a mixture of two simple salts and give tests for all the individual ions.

What is a coordination compound?

A coordination compound contains a central metal ion bonded to ligands through coordinate bonds, forming a complex ion. When dissolved in water, the complex ion remains intact and does not release all individual ions separately.

Analyzing each compound:

(A) $$FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$$ (Mohr's salt):

This is a double salt. When dissolved in water, it completely dissociates into $$Fe^{2+}$$, $$NH_4^+$$, and $$SO_4^{2-}$$ ions. It gives individual tests for both $$Fe^{2+}$$ and $$NH_4^+$$. It is simply a combination of ferrous sulfate and ammonium sulfate crystallized together.

(B) $$CuSO_4 \cdot 4NH_3 \cdot H_2O$$:

This is actually the coordination compound $$[Cu(NH_3)_4]SO_4 \cdot H_2O$$ (tetraamminecopper(II) sulfate). The four $$NH_3$$ molecules are coordinated to $$Cu^{2+}$$ through coordinate bonds, forming the complex ion $$[Cu(NH_3)_4]^{2+}$$. In solution, $$NH_3$$ is not released as free molecules. This is not a double salt.

(C) $$K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O$$ (Potash alum):

This is a double salt. When dissolved in water, it completely dissociates into $$K^+$$, $$Al^{3+}$$, and $$SO_4^{2-}$$ ions. It gives individual tests for all ions. It is simply a combination of potassium sulfate and aluminium sulfate crystallized together.

(D) $$Fe(CN)_2 \cdot 4KCN$$:

This is the coordination compound $$K_4[Fe(CN)_6]$$ (potassium ferrocyanide). The six $$CN^-$$ groups are coordinated to $$Fe^{2+}$$, forming the complex ion $$[Fe(CN)_6]^{4-}$$. In solution, it does not give the test for free $$Fe^{2+}$$ or $$CN^-$$ ions. This is not a double salt.

The correct answer is Option A: A and C only.