Q.A coordination compound $CrCl_3 \cdot 4H_2O$ precipitates silver chloride when treated with silver nitrate. The molar conductance of its solution corresponds to a total of two ions. Write structural formula of the compound and name it.

Concept: Coordination Compound Nomenclature — determining the number of ionizable chlorides from precipitation and conductance data.

Reasoning:

1. Precipitation with AgNO₃: Only chloride ions outside the coordination sphere (free Cl⁻) precipitate as AgCl. Since the compound gives AgCl, at least one Cl⁻ is ionic.

2. Molar conductance: A solution with two ions total means the complex dissociates into exactly two particles. For a Cr(III) complex, the cation is always complex; the anion must be a single Cl⁻ (giving one cation + one anion = two ions). …

The key to solving this lies in understanding two separate experimental clues and letting them converge on a single structure.

Clue 1: Precipitation with silver nitrate.

When you add $\ce{AgNO3}$ to a solution of the compound, only the chloride ions that are outside the coordination sphere (i.e., free, ionisable chloride) will react to form $\ce{AgCl}$ precipitate. Chloride ions that are directly bonded to the metal as ligands do not dissociate and therefore do not precipitate. The fact that the compound does precipitate silver chloride tells you that at least one chloride is outside the coordination sphere.

Clue 2: Molar conductance corresponds to two ions total.

Conductance depends on the number of charged particles in solution. If the solution contains only two ions total, that means the compound dissociates into exactly one cation and one anion — a 1:1 electrolyte. For example, $\ce{NaCl}$ gives two ions; $\ce{CaCl2}$ gives three. So your complex must break into exactly two charged species.

Now, the compound is $\ce{CrCl3·4H2O}$. Chromium(III) has a coordination number of 6 (almost always). So the central $\ce{Cr^{3+}}$ ion must be surrounded by six ligands. The available ligands are water molecules and chloride ions. You have 4 water molecules and 3 chloride ions total.

Let’s work through the possibilities step by step.

1. Determine the number of ionisable chlorides. Let $x$ be the number of $\ce{Cl^-}$ ions outside the coordination sphere (these will precipitate with $\ce{AgNO3}$). Then the number of $\ce{Cl^-}$ ligands inside the sphere is $3 - x$. The total number of ligands around $\ce{Cr}$ must be 6. So:

You have 4 water molecules total. Some may be inside the sphere, some outside as water of crystallisation. Let $y$ be the number of water molecules inside the sphere. Then:

But $y$ cannot exceed 4 (you only have 4 water molecules). So $3 + x \le 4$, which gives $x \le 1$. Since $x$ must be a non-negative integer, $x$ is either 0 or 1.

2. Use the conductance clue.

   If $x = 0$, all three chlorides are inside the sphere, so from $y = 3 + x$ the sphere holds only $y = 3$ water molecules, with the fourth water sitting outside as water of crystallisation: $\ce{[Cr(H2O)3Cl3]·H2O}$. This complex has no chloride outside the coordination sphere at all, so it would give no free chloride ions on dissolving — it would not precipitate AgCl. But the problem states that it does precipitate silver chloride. So $x=0$ is ruled out.

   Therefore $x = 1$. That means exactly one chloride is outside the sphere (ionisable), and the other two chlorides are ligands inside the sphere.

3. Now find the water ligand count.

   With $x=1$, $y = 3 + 1 = 4$. So all four water molecules are inside the coordination sphere. There is no water of crystallisation. The complex cation is $\ce{[Cr(H2O)4Cl2]^+}$, and the anion is the single free $\ce{Cl^-}$.

   The structural formula is:

4. Check the conductance. In solution, this dissociates into:

That’s exactly two ions — matches the conductance clue.

5. Check the precipitation. Only the free $\ce{Cl^-}$ reacts with $\ce{AgNO3}$: $\ce{Cl^- + AgNO3 -> AgCl v + NO3^-}$ …

Method: Conductance & Precipitation Analysis for Coordination Compound Structure

This problem uses conductance and precipitation data to deduce the coordination sphere and counter ions.

Step 1: Interpret the conductance data

• Molar conductance corresponds to two ions total in solution.
• This means the complex dissociates into 1 cation + 1 anion (or possibly 2 ions of opposite charge).
• The complex must have only one ion outside the coordination sphere.

Step 2: Interpret the precipitation data

• $CrCl_3 \cdot 4H_2O$ treated with $AgNO_3$ gives silver chloride precipitate.
• This means chloride ions are present outside the coordination sphere (free $Cl^-$ ions).
• The precipitate confirms that at least one $Cl^-$ is ionic (not coordinated).

Step 3: Determine the coordination sphere

• Total composition: $CrCl_3 \cdot 4H_2O$ → 1 Cr, 3 Cl, 4 H₂O.
• Chromium(III) has coordination number 6 (common for $Cr^{3+}$).
• The coordination sphere must contain 6 ligands (water molecules and/or chloride ions).

Let the formula be: $[Cr(H_2O)_x Cl_y]Cl_z \cdot (4-x)H_2O$

• Total water: $x + (4-x) = 4$ ✓
• Total chloride: $y + z = 3$
• Coordination number: $x + y = 6$

Step 4: Solve for $x$, $y$, $z$

From $x + y = 6$ and $y + z = 3$:

• Possible integer solutions:
  • If $z = 1$ (one ionic $Cl^-$), then $y = 2$, $x = 4$ → $[Cr(H_2O)_4Cl_2]Cl$
  • If $z = 2$, then $y = 1$, $x = 5$ → but $x$ cannot exceed 4 (only 4 water molecules total)
  • If $z = 3$, then $y = 0$, $x = 6$ → impossible (only 4 water molecules) …

Here are the common mistakes students make on this Coordination Compound Nomenclature problem, along with how to avoid each.

1. Mistake: Misinterpreting the Conductivity Data

The error: Students see “molar conductance corresponds to a total of two ions” and think the compound has only two atoms or that the complex itself is a single ion.

Why it’s wrong: Conductance tells you the number of ions in solution, not the number of atoms. “Two ions” means the compound dissociates into one cation and one anion (like NaCl → 2 ions).

How to avoid:

• Remember: Conductance ∝ number of ions.
• “Two ions” = 1 cation + 1 anion.
• For $CrCl_3 \cdot 4H_2O$, the total ions must be 2, so the complex must be neutral overall (no extra counterions beyond the complex itself).

2. Mistake: Forgetting the Role of Water in Coordination Sphere

The error: Students treat all 4 water molecules as lattice water (outside the coordination sphere) or, conversely, put all 4 inside the sphere without checking charge balance.

Why it’s wrong:

• If all 4 water are outside, the complex would be $[CrCl_3(H_2O)_4]$ — but then the complex is neutral, and there are no free ions → conductance would be near zero (not 2 ions).
• If all 4 water are inside, the complex might be $[Cr(H_2O)_4Cl_2]^+$ with one free $Cl^-$ → that gives 2 ions (correct number), but then the precipitation test fails (see next mistake).

How to avoid:

• Water can be inside (coordinated) or outside (lattice).
• Use the precipitation test to decide:
  • $AgNO_3$ precipitates only free chloride ions (outside the coordination sphere).
  • Count how many $Cl^-$ are free → that tells you how many are outside.

3. Mistake: Ignoring the Silver Nitrate Test

The error: Students write a formula that gives 2 ions but doesn’t match the precipitation result.

Example: $[Cr(H_2O)_4Cl_2]Cl$ gives 2 ions ($[Cr(H_2O)_4Cl_2]^+$ and $Cl^-$), but it would precipitate 1 mole of AgCl per mole of compound. The problem says it precipitates silver chloride — but doesn’t say how much. However, the key is: if all chloride were free, you’d get 3 AgCl. The fact that it precipitates at all means at least one $Cl^-$ is free.

How to avoid:

• The precipitation test tells you how many chloride ions are outside the coordination sphere.
• Here, the compound precipitates AgCl → at least one $Cl^-$ is free.
• Combined with “2 ions total”, the only possibility is:
  • Complex cation: $[Cr(H_2O)_4Cl_2]^+$
  • Free anion: $Cl^-$
  • Total ions = 2 ✓
  • Free Cl⁻ = 1 → precipitates AgCl ✓

4. Mistake: Wrong Oxidation State of Chromium

The error: Students assign Cr an oxidation state that doesn’t match the formula or charge balance.

Why it’s wrong:

• In $CrCl_3 \cdot 4H_2O$, total charge = 0.
• If the complex is $[Cr(H_2O)_4Cl_2]^+ Cl^-$, the complex cation has charge +1.
• Let oxidation state of Cr be $x$: $x + 4(0) + 2(-1) = +1 \implies x - 2 = +1 \implies x = +3$
• Cr is in +3 oxidation state (common for Cr).

How to avoid:

• Always write the charge balance equation.
• Remember: $H_2O$ is neutral, $Cl^-$ is -1.
• Common Cr states: +2, +3, +6. Here it’s +3.

5. Mistake: Incorrect Naming (IUPAC)

The error: Students name the compound incorrectly — e.g., “Tetraaquadichlorochromium(III) chloride” but forget parentheses, oxidation state notation, or alphabetical order.

Common naming errors:

• Writing “tetraaquadichloro” without hyphen or wrong order.
• Forgetting the Roman numeral for oxidation state.
• Writing “chromium” before ligands (ligands come first).
• Not using “-ate” for anionic complexes (not needed here, it’s cationic).

Correct name:

Tetraaquadichloridochromium(III) chloride

(Note: “chlorido” is IUPAC preferred over “chloro”, but “chloro” is still accepted in many Indian exams — check your syllabus.)

How to avoid:

• Follow IUPAC order: ligands alphabetically → metal → oxidation state in Roman numerals.
• Use aquo for $H_2O$, chlorido or chloro for $Cl^-$. …