Werner's Theory

Werner examined a family of cobalt-ammine chlorides — $\text{CoCl}_3\cdot 6\text{NH}_3$, $\text{CoCl}_3\cdot 5\text{NH}_3$, $\text{CoCl}_3\cdot 4\text{NH}_3$ — that all look like simple addition compounds but behave completely differently in solution. Some precipitate all their chlorine with $\text{AgNO}_3$; some precipitate only part of it. Some conduct like a solution with four ions; others like three, or two. Classical valence theory, which allowed a metal only one fixed valence, could not explain this. Werner's resolution was to split the metal's bonding capacity into two kinds.

• Primary valence is ionisable and non-directional. It equals the metal's oxidation state and is satisfied by anions that sit outside the coordination sphere, free to dissociate in water and precipitate with $\text{AgNO}_3$.
• Secondary valence is non-ionisable and directional. It equals the coordination number (CN) — the number of ligand donor atoms bonded straight to the metal — and it fixes a definite geometry: CN 6 gives octahedral, CN 4 gives tetrahedral or square planar.

From this split, the working vocabulary of the chapter follows directly:

• A ligand is an ion or molecule that donates an electron pair to the metal through a coordinate bond, occupying one secondary-valence position.
• Denticity counts how many donor atoms one ligand uses: monodentate ($\text{NH}_3$, $\text{Cl}^-$), bidentate (ethylenediamine "en", oxalate $\text{ox}^{2-}$), polydentate (EDTA$^{4-}$, hexadentate). Denticity is a property of one ligand; coordination number is the total across all ligands bound to the metal.
• An ambidentate ligand has two possible donor atoms but uses only one at a time, e.g. $\text{NO}_2^-$ (through N or O) or $\text{SCN}^-$ (through S or N).
• A chelate forms when a bi- or polydentate ligand grips the same metal at two or more points, closing a ring; chelation generally raises stability (the chelate effect).

The figure below is the picture Werner's postulates draw: a coordination sphere, written in square brackets, holding the metal and its secondary-valence ligands in a fixed octahedral arrangement, with the primary-valence anions sitting outside it, free in solution.

The reasoning that lets a chemist read primary and secondary valence off two simple pieces of experimental data — molar conductance and $\text{AgNO}_3$ precipitation — is what actually justifies Werner's split, and tracing it through explains why the same raw formula $\text{CoCl}_3\cdot n\text{NH}_3$ can hide several genuinely different structures depending only on how its atoms are distributed between the coordination sphere and the free ionic space outside it.

Step 1. Start with precipitation. Silver nitrate precipitates $\text{AgCl}$ only from chloride ions that are free in solution, because only free $\text{Cl}^-$ can meet $\text{Ag}^+$ and form the insoluble salt. A chloride bonded directly to the metal, inside the coordination sphere, is not free — it does not ionise, so it does not precipitate. The moles of $\text{AgCl}$ formed per mole of compound therefore count exactly the ionisable chlorines, which is exactly the primary valence satisfied by chloride.

Step 2. Bring in conductance. A solution's molar conductance rises with the number of ions it produces per formula unit: a 1:1 electrolyte conducts less than a 1:2, which conducts less than a 1:3. Counting the total ions in solution — one complex ion plus however many free chloride ions — fixes both the charge on the complex cation and, by elimination, how many chlorides must be bound inside the sphere rather than free outside it.

Step 3. Combine the two readings. Whatever chlorine does not show up in the precipitation count but is still needed to balance the formula must be sitting inside the coordination sphere, occupying a secondary-valence position alongside the neutral ammonia ligands. This is the crux of Werner's method: precipitation counts primary valence directly, conductance counts total ions, and subtracting one from the other recovers the composition of the coordination sphere itself, without ever seeing a single molecule.

Step 4. Generalise. Because primary valence must equal the metal's oxidation state and secondary valence must equal a fixed coordination number (6 for cobalt(III)), any distribution of ligands and anions consistent with both totals is a valid Werner structure — which is exactly why the same $\text{Co} : \text{Cl} : \text{NH}_3$ ratio can produce a family of related but distinct compounds, distinguished only by how the chlorines partition between "inside" and "outside" the bracket.

Problem. $\text{CoCl}_3\cdot 6\text{NH}_3$ gives 3 moles of $\text{AgCl}$ precipitate per mole of compound with excess $\text{AgNO}_3$, and its solution conducts as if it contains 4 ions per formula unit. Deduce the Werner formula, the primary and secondary valence, the coordination number, and write the modern formula.

Step 1. All 3 chlorines precipitate, so all 3 $\text{Cl}^-$ are ionisable — none is bound to cobalt, all three sit outside the coordination sphere.

Step 2. 4 ions in solution = 1 complex cation + 3 free $\text{Cl}^-$ anions, so the complex cation must carry a $3+$ charge to balance those three anions.

Step 3. Primary valence = 3, matching the 3 ionisable $\text{Cl}^-$ and the $\text{Co}^{3+}$ oxidation state. Secondary valence = coordination number = 6, satisfied entirely by the six neutral $\text{NH}_3$ ligands, which occupy the coordination sphere without affecting the ion count.

Step 4. The modern formula is $[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$: an octahedral $[\text{Co}(\text{NH}_3)_6]^{3+}$ cation, coordination number 6, with three free chloride ions. Answer: $[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$, primary valence 3, secondary valence (CN) 6, octahedral.

Competitive stretch

Change only the ammonia count to $\text{CoCl}_3\cdot 5\text{NH}_3$, and the same raw ratio of atoms gives a different structure. Here only 2 moles of $\text{AgCl}$ precipitate and the solution conducts as a 1:3 electrolyte (3 ions). Precipitation says only 2 chlorines are free; the third must be bound inside the sphere, taking up one secondary-valence position alongside the 5 $\text{NH}_3$ ligands, so CN is still $5 + 1 = 6$. Conductance confirms it: 3 ions = 1 complex cation $[\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+}$ + 2 free $\text{Cl}^-$. The formula is $[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$ — same $\text{CoCl}_3\cdot n\text{NH}_3$ family, one chloride moved from "outside" to "inside" the bracket, dropping both the precipitated $\text{AgCl}$ and the ion count by one.

• Confusing primary valence (ionisable, equals oxidation state, satisfied by anions) with secondary valence (non-ionisable, equals coordination number, satisfied by ligands) — these are two separate counts, never added into one number.
• Assuming every chlorine written in a formula like $\text{CoCl}_3\cdot 5\text{NH}_3$ will precipitate with $\text{AgNO}_3$ — only the ones outside the coordination sphere actually do.
• Forgetting that a neutral ligand such as $\text{NH}_3$ or $\text{H}_2\text{O}$ can fill a secondary-valence position without changing the ionic charge or the ion count in solution.
• Mixing up denticity (how many donor atoms one ligand itself uses) with coordination number (the total donor-atom count around the metal, which may come from several ligands together).

Ask the tutor: if you want to practise pulling primary valence, secondary valence, and the modern formula out of conductivity and precipitation data, ask the AI tutor to walk through one or two more $\text{CoCl}_3\cdot n\text{NH}_3$-style problems with you.