Step 1: Understanding the molecular composition:
The complex is given as Co.5NH\(_3\).Cl.SO\(_4\). It consists of a central Co\(^3+\) ion surrounded by 5 ammonia molecules (NH\(_3\)) and one chloride (Cl) ion as well as a sulfate (SO\(_4^{2-}\)) ion.
Step 2: Identifying the isomers:
Isomer A: The solution of A gives a white precipitate with AgNO\(_3\) solution. This indicates the presence of chloride ions (Cl\(^-\)) in the solution because AgNO\(_3\) reacts with chloride ions to form a white precipitate of AgCl.
Isomer B: The solution of B gives a white precipitate with BaCl\(_2\) solution. This indicates the presence of sulfate ions (SO\(_4^{2-}\)) in the solution because BaCl\(_2\) reacts with sulfate ions to form a white precipitate of BaSO\(_4\).
Step 3: Analyzing the isomerism:
The two isomers A and B are ionisation isomers.
In ionisation isomerism, two or more compounds have the same molecular formula but release different ions in solution.
In this case, the only difference between the two isomers is that in A, chloride (Cl\(^-\)) is the counter-ion, while in B, sulfate (SO\(_4^{2-}\)) is the counter-ion. These two isomers give different ions in solution, which is the defining characteristic of ionisation isomerism.
Step 4: Conclusion:
Since the two isomers differ in the ions that are released into the solution (chloride in A and sulfate in B), this is a classic example of ionisation isomerism.