An octahedral complex with molecular composition M.5NH$$_3$$.Cl.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 octahedral complex has the molecular composition M·5NH₃·Cl·SO₄, meaning it contains a central metal atom M, five ammonia ligands (NH₃), one chloride ion (Cl⁻), and one sulfate ion (SO₄²⁻). Since the complex is neutral, the oxidation state of M must be determined. Ammonia is neutral, chloride has a charge of -1, and sulfate has a charge of -2. Therefore, the metal M must have an oxidation state of +3 to balance the total charge: $$(+3) + 5 \times (0) + (-1) + (-2) = 0$$.

This complex exhibits two isomers, A and B. The solution of isomer A produces a white precipitate when treated with AgNO₃ solution. Silver nitrate (AgNO₃) tests for free chloride ions (Cl⁻) by forming insoluble silver chloride (AgCl): $$\text{AgNO}_3 + \text{Cl}^- \rightarrow \text{AgCl} \downarrow + \text{NO}_3^-$$. Thus, isomer A must have chloride ions outside the coordination sphere.

The solution of isomer B produces a white precipitate when treated with BaCl₂ solution. Barium chloride (BaCl₂) tests for free sulfate ions (SO₄²⁻) by forming insoluble barium sulfate (BaSO₄): $$\text{BaCl}_2 + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \downarrow + 2\text{Cl}^-$$. Thus, isomer B must have sulfate ions outside the coordination sphere.

Given that both isomers share the same molecular formula but differ in which ions are inside or outside the coordination sphere, we can write their structures as:

• Isomer A: Chloride is outside, so sulfate must be coordinated. The formula is [M(NH₃)₅SO₄]Cl. In aqueous solution, it dissociates to give free Cl⁻ ions: $$[\text{M(NH}_3\text{)}_5\text{SO}_4]^+ \text{Cl}^- \rightleftharpoons [\text{M(NH}_3\text{)}_5\text{SO}_4]^+ + \text{Cl}^-$$. This explains the positive test with AgNO₃.
• Isomer B: Sulfate is outside, so chloride must be coordinated. The formula is [M(NH₃)₅Cl]SO₄. In aqueous solution, it dissociates to give free SO₄²⁻ ions: $$[\text{M(NH}_3\text{)}_5\text{Cl}]^{2+} \text{SO}_4^{2-} \rightleftharpoons [\text{M(NH}_3\text{)}_5\text{Cl}]^{2+} + \text{SO}_4^{2-}$$. This explains the positive test with BaCl₂.

The isomers have identical atoms bonded but differ in the ionization of the counter ions, leading to different ions in solution. This is characteristic of ionization isomerism, where the ligands exchange roles between being coordinated and being counter ions.

Now, evaluating the options:

• Option A (Coordinate isomerism) involves complexes with multiple metal centers and different ligand distributions. Since this complex has only one metal center, it does not apply.
• Option B (Geometrical isomerism) arises from different spatial arrangements of the same ligands (e.g., cis-trans). Here, the ligands differ (Cl⁻ vs. SO₄²⁻ coordinated), so it does not apply.
• Option C (Ionisation isomerism) fits, as the isomers produce different ions in solution due to the exchange of ionic ligands between the coordination sphere and the counter ions.
• Option D (Linkage isomerism) occurs when a ligand bonds through different atoms (e.g., ambidentate ligands). Sulfate typically coordinates via oxygen, and there is no evidence of different bonding modes, so it does not apply.

Hence, the correct answer is Option C.