An octahedral complex having molecular composition $$Co \cdot 5NH_3 \cdot Cl \cdot SO_4$$ has two isomers A and B. The solution of A gives a white precipitate with $$AgNO_3$$ solution and the solution of B gives white precipitate with $$BaCl_2$$ solution. The type of isomerism exhibited by the complex is,

The given molecular composition can produce two octahedral complexes:

$$[Co(NH_3)_5SO_4]Cl$$   and   $$[Co(NH_3)_5Cl]SO_4$$

In the first species, chloride is the counter-ion, whereas sulphate is inside the coordination sphere. In the second species, sulphate is the counter-ion, while chloride is inside the coordination sphere.

Identification of the two isomers by qualitative tests:

Case A: The solution gives a white precipitate with $$AgNO_3$$.
The reagent $$AgNO_3$$ detects free $$Cl^-$$ ions by forming white $$AgCl$$. Therefore the complex that produces this test must have chloride outside the coordination sphere: $$[Co(NH_3)_5SO_4]Cl \;+\; AgNO_3 \;\rightarrow\; AgCl\downarrow$$

Case B: The solution gives a white precipitate with $$BaCl_2$$.
The reagent $$BaCl_2$$ detects free $$SO_4^{2-}$$ ions by forming white $$BaSO_4$$. Hence the complex furnishing this test must have sulphate outside the coordination sphere: $$[Co(NH_3)_5Cl]SO_4 \;+\; BaCl_2 \;\rightarrow\; BaSO_4\downarrow$$

The two compounds differ only in the exchange of $$Cl^-$$ and $$SO_4^{2-}$$ between the coordination sphere and the counter-ion position. Such a pair of complexes, which produce different ions in solution because of this exchange, represents ionisation isomerism.

Therefore the type of isomerism exhibited is Ionisation isomerism  →  Option C.