Q.On the basis of the following observations made with aqueous solutions, assign secondary valences to metals in the following compounds: Formula — Moles of AgCl precipitated per mole of the compounds with excess $AgNO_3$

Concept: Werner Coordination Theory — secondary valences (coordination number) are satisfied by neutral molecules or negative ions that do not precipitate with $AgNO_3$; only free chloride ions (outside the coordination sphere) give AgCl.

Reasoning steps:

1. Each mole of AgCl precipitated corresponds to one mole of free $Cl^-$ ions in solution.
2. The total number of $Cl$ atoms in the formula minus the free $Cl^-$ gives the number of $Cl$ inside the coordination sphere (satisfying secondary valence).
3. The secondary valence (coordination number) is the total number of ligands (neutral molecules + coordinated $Cl$) attached to the metal.

| Compound | Free $Cl^-$ (from AgCl) | Total Cl | Coordinated Cl | Neutral ligands | Secondary valence | …

Werner’s theory is the key here. He proposed that metals have two types of valency: primary valency (ionisable, satisfied by anions, shown as oxidation state) and secondary valency (non-ionisable, satisfied by ligands or water, fixed for a given metal). In solution, only chloride ions that are outside the coordination sphere (i.e., not directly bonded to the metal) will precipitate as AgCl with AgNO₃. Chloride ions inside the coordination sphere are covalently bonded and do not precipitate.

So, the number of moles of AgCl precipitated tells us exactly how many Cl⁻ ions are ionic (outside the sphere). The total chloride in the formula minus that number gives the chloride inside the sphere. The secondary valence (coordination number) is the total number of ligands (NH₃, H₂O, or Cl⁻) directly attached to the metal.

Let’s work through each compound step by step.

1. Compound (i): $PdCl_2 \cdot 4NH_3$ — 2 moles AgCl

   • Total Cl atoms in formula = 2.
   • AgCl precipitated = 2 → both Cl⁻ are ionic (outside sphere).
   • So, inside the coordination sphere: 0 Cl⁻, but 4 NH₃ molecules.
   • Secondary valence of Pd = number of ligands attached = 4 (all NH₃).
   • The complex is $[Pd(NH_3)_4]Cl_2$.

2. Compound (ii): $NiCl_2 \cdot 6H_2O$ — 2 moles AgCl

   • Total Cl = 2.
   • AgCl = 2 → both Cl⁻ are ionic.
   • Inside sphere: 0 Cl⁻, but 6 H₂O molecules.
   • Secondary valence of Ni = 6 (all H₂O).
   • The complex is $[Ni(H_2O)_6]Cl_2$.

3. Compound (iii): $PtCl_4 \cdot 2HCl$ — 0 moles AgCl

   • Total Cl = 4 (from PtCl₄) + 2 (from 2HCl) = 6.
   • AgCl = 0 → no chloride is ionic; all Cl⁻ are inside the coordination sphere.
   • So, inside sphere: all 6 Cl⁻ are bonded to Pt.
   • Secondary valence of Pt = 6.
   • The complex is $[PtCl_6]^{2-}$ (the 2H⁺ are counterions, but the question asks for the metal’s secondary valence). …

Method: Werner's Coordination Theory – Secondary Valence Assignment via Conductivity / Precipitation Data

Concept First (Why this works)

Werner proposed that metal ions have two kinds of valences:

• Primary valence (ionizable, satisfied by anions, corresponds to oxidation state)
• Secondary valence (non-ionizable, satisfied by ligands, corresponds to coordination number)

When a coordination compound dissolves in water, only the ions outside the coordination sphere (satisfying primary valence) are free to react. AgCl precipitation with excess $AgNO_3$ tells us how many $Cl^-$ ions are free (outside the coordination sphere) — each free $Cl^-$ gives 1 mole of AgCl.

Steps to Solve

1. Identify the metal's primary valence (oxidation state) from the formula.
2. Count total $Cl$ atoms in the formula.
3. From moles of AgCl precipitated, find how many $Cl^-$ are free (ionizable).
4. Subtract free $Cl^-$ from total $Cl$ → these $Cl$ atoms satisfy secondary valence (inside coordination sphere).
5. Assign secondary valence = number of ligands (including coordinated $Cl$ and $NH_3$ or $H_2O$) directly attached to the metal.

Application to Given Compounds

(i) $PdCl_2 \cdot 4NH_3$ — AgCl = 2

• Total $Cl$ = 2
• Free $Cl^-$ = 2 (all $Cl$ are ionizable)
• Coordinated $Cl$ = 0
• Secondary valence = number of ligands = $4NH_3$ = 4

(ii) $NiCl_2 \cdot 6H_2O$ — AgCl = 2

• Total $Cl$ = 2
• Free $Cl^-$ = 2
• Coordinated $Cl$ = 0
• Secondary valence = $6H_2O$ = 6

(iii) $PtCl_4 \cdot 2HCl$ — AgCl = 0

• Total $Cl$ = 6
• Free $Cl^-$ = 0 (no AgCl)
• Coordinated $Cl$ = 6 …

Common Mistakes & How to Avoid Them (Werner Coordination Theory)

Mistake 1: Confusing Primary Valence with Secondary Valence

The error: Students often think the number of $AgCl$ precipitated equals the total chlorine atoms in the formula. For example, in $PdCl_2 \cdot 4NH_3$, they see 2 Cl atoms and assume secondary valence = 2.

Why it's wrong:

• Primary valence = oxidation state (ionizable Cl⁻ ions)
• Secondary valence = coordination number (ligands directly attached to metal)
• $AgNO_3$ only precipitates free Cl⁻ ions — those outside the coordination sphere.

How to avoid:

• Remember: Precipitated Cl⁻ = Primary valence (ionizable)
• Total Cl – Precipitated Cl = Coordinated Cl
• Secondary valence = number of all ligands (NH₃, H₂O, Cl⁻ inside sphere)

Mistake 2: Forgetting that Neutral Ligands Also Count

The error: Students count only Cl atoms when assigning secondary valence, ignoring NH₃ and H₂O.

Example: For $NiCl_2 \cdot 6H_2O$, they see 2 Cl precipitated → secondary valence = 2. Wrong.

Correct reasoning:

• 2 Cl⁻ precipitated → 2 Cl⁻ are outside sphere (primary valence = 2)
• But the complex has 6 H₂O molecules as ligands
• Secondary valence = 6 (from 6 H₂O)

How to avoid:

• Secondary valence = total number of donor atoms (neutral + anionic) directly bonded to metal
• Count NH₃, H₂O, and any Cl⁻ that is not precipitated

Mistake 3: Misinterpreting Zero Precipitation

The error: When 0 moles of $AgCl$ precipitate (e.g., $PtCl_4 \cdot 2HCl$), students think there are no Cl atoms at all.

Why it's wrong:

• Zero precipitation means all Cl atoms are inside the coordination sphere
• They are covalently bonded to the metal, not free ions

Example: $PtCl_4 \cdot 2HCl$

• Total Cl = 6
• Precipitated Cl = 0 → All 6 Cl are coordinated
• Secondary valence = 6 (all Cl ligands)

How to avoid:

• Zero precipitation ≠ zero chlorine
• It means 100% of Cl is in coordination sphere

Mistake 4: Forgetting to Account for All Ligands in Secondary Valence

The error: In $CoCl_3 \cdot 4NH_3$, students see 1 Cl precipitated → they assign secondary valence = 1.

Correct approach:

• 1 Cl⁻ precipitated → 1 Cl⁻ outside sphere
• Total Cl = 3 → Coordinated Cl = 3 – 1 = 2
• Also 4 NH₃ ligands
• Secondary valence = 2 (Cl) + 4 (NH₃) = 6

How to avoid:

• Always: Secondary valence = (Coordinated Cl) + (Neutral ligands)
• Never skip counting neutral ligands

Mistake 5: Mixing Up Oxidation State and Coordination Number …