Q.When 1 mol $CrCl_3 \cdot 6H_2O$ is treated with excess of $AgNO_3$, 3 mol of $AgCl$ are obtained. The formula of the complex is:

Concept: Coordination Compound Nomenclature — the number of $AgCl$ moles precipitated equals the number of chloride ions present outside the coordination sphere (i.e., free or ionizable $Cl^-$).

Reasoning:

1. Excess $AgNO_3$ precipitates only the chloride ions that are not coordinated to the metal centre.
2. 3 mol of $AgCl$ means 3 mol of free $Cl^-$ per mole of complex. …

This problem tests your understanding of coordination compound nomenclature and the difference between coordination sphere and counter ions. When you treat a complex with $AgNO_3$, the silver ions only precipitate chloride ions that are free — those outside the square brackets. Chloride ions inside the coordination sphere (bonded directly to the metal) do not dissociate and will not react with $Ag^+$.

The question gives you a critical experimental fact: 1 mol of the complex yields 3 mol of $AgCl$. That means all three chloride ions present in the formula unit are outside the coordination sphere. None are bonded to chromium.

Let’s check each option systematically.

1. Option (i): $[CrCl_3(H_2O)_3] \cdot 3H_2O$

   Here, all three chlorides are inside the coordination sphere. The three water molecules outside are just water of crystallization. So zero free $Cl^-$ ions — this would give 0 mol of $AgCl$. Eliminated.

2. Option (ii): $[CrCl_2(H_2O)_4]Cl \cdot 2H_2O$

   Two chlorides are inside the sphere, one is outside as a counter ion. So only 1 mol of $AgCl$ would precipitate. Not matching the given 3 mol. Eliminated.

3. Option (iii): $[CrCl(H_2O)_5]Cl_2 \cdot H_2O$

   One chloride inside, two outside. That gives 2 mol of $AgCl$. Still not enough. Eliminated.

4. Option (iv): $[Cr(H_2O)_6]Cl_3$ …

Method: Conductometric / Precipitation Analysis of Ionizable Chloride

This method uses the fact that only free (ionizable) chloride ions outside the coordination sphere react with $AgNO_3$ to give $AgCl$ precipitate. Chloride ions inside the coordination sphere (ligands) do not precipitate.

Steps

Step 1: Identify the given data

• 1 mol of complex $\rightarrow$ 3 mol of $AgCl$
• This means 3 mol of free $Cl^-$ are present per mol of complex.

Step 2: Count total chloride in each option

All options have total 3 Cl atoms per formula unit. But only those outside the square bracket are free.

Step 3: Check each option for number of free $Cl^-$

• (i) $[CrCl_3(H_2O)_3] \cdot 3H_2O$

  → All 3 Cl are inside coordination sphere → 0 free Cl → gives 0 mol $AgCl$ ✗

• (ii) $[CrCl_2(H_2O)_4]Cl \cdot 2H_2O$ …

Common Mistakes & How to Avoid Them

Mistake 1: Confusing total chloride with ionizable chloride

The error: Students see 3 mol of AgCl formed and assume the complex contains 3 chloride ions in the coordination sphere — picking option (i) or (iv) without thinking.

Why it's wrong: AgNO₃ only precipitates free chloride ions (outside the coordination sphere). Chloride ions inside the coordination sphere (bonded to Cr) do not react with Ag⁺.

How to avoid: Always ask: "Which Cl⁻ are free to precipitate?" Only the counter ions (outside square brackets) react with AgNO₃.

Mistake 2: Forgetting water can be inside or outside the coordination sphere

The error: Students count all 6 water molecules as hydrate water (outside brackets) or all as coordinated water (inside brackets), ignoring that water can be both.

Why it's wrong: The formula $CrCl_3 \cdot 6H_2O$ tells total composition — it doesn't show how water is distributed. Some water may be coordinated to Cr, some may be outside as water of crystallization.

How to avoid: Remember: water molecules can be inside the coordination sphere (ligands) or outside (lattice water). The dot ($\cdot$) separates the complex from crystallization water.

Mistake 3: Not balancing charge and coordination number

The error: Picking an option without checking if Cr's coordination number (usually 6) and oxidation state are consistent.

Why it's wrong: Cr³⁺ typically has coordination number 6. Each option must have exactly 6 ligands (Cl⁻ + H₂O) in the coordination sphere.

How to avoid: For each option:

• Count total ligands inside brackets = must be 6
• Check charge balance: Cr³⁺ + (ligand charges) + (counter ion charges) = 0

Mistake 4: Rushing to pick the first option that "looks right"

The error: Seeing 3 mol AgCl and immediately choosing option (iv) $[Cr(H_2O)_6]Cl_3$ because it has 3 Cl⁻ outside.

Why it's wrong: Option (iv) gives 3 mol AgCl — but so does option (iii)! Both have 3 ionizable Cl⁻. You must check all conditions.

How to avoid: Test every option systematically: …