Q.$CoSO_4Cl \cdot 5NH_3$ exists in two isomeric forms 'A' and 'B'. Isomer 'A' reacts with $AgNO_3$ to give white precipitate, but does not react with $BaCl_2$. Isomer 'B' gives white precipitate with $BaCl_2$ but does not react with $AgNO_3$. Answer the following questions.

Concept: Coordination Compound Nomenclature & Ionisation Isomerism

Reasoning:

1. Analyse the precipitates:

   $AgNO_3$ tests for free $Cl^-$ ions (white $AgCl$ precipitate). $BaCl_2$ tests for free $SO_4^{2-}$ ions (white $BaSO_4$ precipitate).

2. Interpret the data for A:

   A gives $AgCl$ with $AgNO_3$ → $Cl^-$ is outside the coordination sphere.

   A does not react with $BaCl_2$ → $SO_4^{2-}$ is inside the coordination sphere.

   So A is $[Co(NH_3)_5SO_4]Cl$.

3. Interpret the data for B:

   B gives $BaSO_4$ with $BaCl_2$ → $SO_4^{2-}$ is outside the sphere.

   B does not react with $AgNO_3$ → $Cl^-$ is inside the sphere.

   So B is $[Co(NH_3)_5Cl]SO_4$.

4. Isomerism type: …

Why this approach works

In coordination compounds, the central metal ion and the ligands directly attached to it form the coordination sphere (written inside square brackets). Ions outside the sphere are free in solution and behave like simple ions — they can be detected by precipitation reactions.

$AgNO_3$ tests for free chloride ions ($Cl^-$), giving a white precipitate of $AgCl$.

$BaCl_2$ tests for free sulfate ions ($SO_4^{2-}$), giving a white precipitate of $BaSO_4$.

So if an isomer gives a precipitate with one reagent but not the other, it tells us exactly which ion is outside the coordination sphere — and therefore which ion must be inside as a ligand.

Step-by-step reasoning

1. Write the molecular formula clearly

   The compound is $CoSO_4Cl \cdot 5NH_3$. This means one cobalt, one sulfate, one chloride, and five ammonia molecules. Total charge: $Co^{3+}$ (common oxidation state in such complexes), $SO_4^{2-}$, $Cl^-$, and $5NH_3$ (neutral) — so the complex is neutral overall.

2. Interpret the test results for isomer A

   • A reacts with $AgNO_3$ → white precipitate → free $Cl^-$ ions present.
   • A does not react with $BaCl_2$ → no free $SO_4^{2-}$ ions. Therefore, chloride is outside the coordination sphere, and sulfate must be inside as a ligand. Structure of A: $[Co(NH_3)_5SO_4]Cl$

3. Interpret the test results for isomer B

   • B gives white precipitate with $BaCl_2$ → free $SO_4^{2-}$ ions present.
   • B does not react with $AgNO_3$ → no free $Cl^-$ ions. Therefore, sulfate is outside, and chloride must be inside as a ligand. Structure of B: $[Co(NH_3)_5Cl]SO_4$

4. Identify the type of isomerism

   Both isomers have the same molecular formula but differ in which anion is coordinated and which is free. This is ionisation isomerism — a type of structural isomerism where the isomers give different ions in solution. …

Method: Deductive Reasoning from Conductometric / Precipitation Data

This method uses the precipitation behaviour of the complex with specific reagents to determine which ions are free (outside the coordination sphere) and which are coordinated (inside the sphere).

Steps

Step 1: Identify the total composition

The formula is $CoSO_4Cl \cdot 5NH_3$.

• Cobalt(III) is the metal centre (common oxidation state +3 in ammine complexes).
• Ligands: 5 ammonia ($NH_3$) molecules.
• Anionic species: one sulphate ($SO_4^{2-}$) and one chloride ($Cl^-$).

Step 2: Use precipitation tests to locate free ions

• Isomer A: reacts with $AgNO_3$ → free $Cl^-$ present.

  Does not react with $BaCl_2$ → $SO_4^{2-}$ is coordinated (inside sphere).

• Isomer B: reacts with $BaCl_2$ → free $SO_4^{2-}$ present.

  Does not react with $AgNO_3$ → $Cl^-$ is coordinated (inside sphere).

Step 3: Write structural formulas

• Isomer A:

  Free $Cl^-$ outside, $SO_4^{2-}$ inside.

  Coordination sphere: $[Co(NH_3)_5SO_4]^+$

  Counter ion: $Cl^-$

  Formula: $\mathbf{[Co(NH_3)_5SO_4]Cl}$

• Isomer B:

  Free $SO_4^{2-}$ outside, $Cl^-$ inside. …

🚩 Common Mistakes & How to Avoid Them

1. Mistaking the formula as a simple double salt

• Mistake: Treating $CoSO_4Cl \cdot 5NH_3$ as a mixture of separate salts (like $CoSO_4$ and $CoCl_2$ with ammonia).
• Why it’s wrong: In coordination compounds, the dot ($\cdot$) means the ammonia molecules are ligands attached to the central metal ion, not separate molecules.
• How to avoid: Always interpret $M X_n \cdot y L$ as a complex where $L$ are ligands coordinated to the metal $M$, and the anions ($X$) may be inside or outside the coordination sphere.

2. Confusing which ion is inside vs outside the coordination sphere

• Mistake: Assuming both $SO_4^{2-}$ and $Cl^-$ are outside the sphere.
• Why it’s wrong: The two isomers differ precisely because one anion is inside the coordination sphere and the other is outside.
• How to avoid: Use the precipitation tests as clues:
  • $AgNO_3$ gives white precipitate ($AgCl$) → free $Cl^-$ ions present.
  • $BaCl_2$ gives white precipitate ($BaSO_4$) → free $SO_4^{2-}$ ions present.
  • If a test gives no precipitate, that anion is inside the coordination sphere (not free).

3. Writing the wrong structural formula

• Mistake: Writing $[Co(NH_3)_5Cl]SO_4$ for isomer A when it should be $[Co(NH_3)_5SO_4]Cl$, or vice versa.
• How to avoid: Match the precipitation result:
  • Isomer A reacts with $AgNO_3$ → free $Cl^-$ → $Cl$ is outside → formula: $[Co(NH_3)_5SO_4]Cl$
  • Isomer B reacts with $BaCl_2$ → free $SO_4^{2-}$ → $SO_4$ is outside → formula: $[Co(NH_3)_5Cl]SO_4$

4. Naming the isomerism incorrectly

• Mistake: Calling it linkage isomerism or geometrical isomerism.
• Why it’s wrong: Linkage isomerism involves ambidentate ligands (like $NO_2^-$ vs $ONO^-$). Geometrical isomerism requires different spatial arrangements (cis/trans). Here, the difference is which anion is a ligand.
• How to avoid: Recognize that the two isomers have the same molecular formula but give different ions in solution — this is ionization isomerism.

5. Errors in IUPAC naming

• Mistake: Forgetting to name the anion outside the sphere correctly, or misordering ligands alphabetically.
• How to avoid: Follow IUPAC rules:
  • Name ligands alphabetically (ignoring prefixes like penta-).
  • Name the central metal with its oxidation state in Roman numerals.
  • Name the counter ion (outside sphere) last.
  • For sulfate as a ligand, use sulfato; for chloride as a ligand, use chlorido.

✓ Correct Solution

(i) Identify A and B with structural formulas

Isomer A (gives $AgCl$ precipitate, no $BaSO_4$):

• Free $Cl^-$ outside → $SO_4^{2-}$ is a ligand.
• Formula: $[Co(NH_3)_5SO_4]Cl$

Isomer B (gives $BaSO_4$ precipitate, no $AgCl$): …