Step 1: Understand ionisation isomers.
The two complexes \([Co(NH_3)_5SO_4]Cl\) and \([Co(NH_3)_5Cl]SO_4\) have the same formula, but they swap which group is bonded inside (coordination sphere) and which stays outside as a free ion. So in water they release different free ions.
Step 2: Find the free ion in each.
In \([Co(NH_3)_5SO_4]Cl\), the free ion outside is chloride \(Cl^-\). In \([Co(NH_3)_5Cl]SO_4\), the free ion outside is sulphate \(SO_4^{2-}\).
Step 3: Test for the free sulphate.
Add barium chloride \((BaCl_2)\) solution. Only \([Co(NH_3)_5Cl]SO_4\) gives a white precipitate of barium sulphate \((BaSO_4)\), because it has free \(SO_4^{2-}\). The other complex gives no such precipitate.
Step 4: Test for the free chloride.
Add silver nitrate \((AgNO_3)\) solution. Only \([Co(NH_3)_5SO_4]Cl\) gives a white precipitate of silver chloride \((AgCl)\), because it has free \(Cl^-\). This confirms the two are ionisation isomers.
Step 5: Chelate effect.
When a single ligand uses two or more donor atoms to grip the metal and form a ring (a chelate ring), the complex becomes much more stable than a similar complex made from separate single-donor (monodentate) ligands. This extra stability is called the chelate effect.
Step 6: Example.
Ethylenediamine (en) is a bidentate ligand. \([Co(en)_3]^{3+}\) is far more stable than \([Co(NH_3)_6]^{3+}\), even though both use six Co-N bonds, because en forms stable five-membered chelate rings.
Answer: Use \(BaCl_2\) (detects free \(SO_4^{2-}\)) and \(AgNO_3\) (detects free \(Cl^-\)) to tell the ionisation isomers apart. The chelate effect is the extra stability of a complex containing chelate rings, for example \([Co(en)_3]^{3+}\).