Animated Solution for Chemistry - Coordination Compounds: On treatment of $100\text{ mL}$ of $0.1\text{ M}$ solution of ${\text{CoCl}}_3\cdot 6{\text{H}}_2\text{O}$ with excess of ${\text{AgNO}}_3$; $1.2\times 10^{22}$ ions are precipitated. The complex is

A Stoichiometric Journey

Imagine a beaker containing $100\text{ mL}$ of a $0.1\text{ M}$ solution of cobalt chloride hexahydrate (${\text{CoCl}}_3\cdot 6{\text{H}}_2\text{O}$). When we introduce an excess of silver nitrate (${\text{AgNO}}_3$), a brilliant white precipitate of silver chloride ($\text{AgCl}$) instantly forms. But how much precipitates? The problem states that exactly $1.2\times 10^{22}$ ions are precipitated. Our mission is to decode the exact molecular structure of this coordination complex using this stoichiometric clue.

Finding the Moles of the Complex

First, we need to establish our baseline: exactly how many moles of the cobalt complex did we start with? We are given the molarity and the volume, so we can rely on the fundamental molarity equation:

Substituting the given values, we multiply $0.1{\text{ mol L}}^{-1}$ by $0.1\text{ L}$ (which is $100\text{ mL}$ converted to liters).

We have exactly $0.01\text{ moles}$ of the complex in our beaker. Keep this number locked in your mind.

Finding the Moles of the Precipitate

Now, let's shift our focus to the white precipitate at the bottom of the beaker. The problem tells us that $1.2\times 10^{22}$ ions were precipitated. These are the chloride ions that successfully reacted with the silver ions. To make this number useful, we must convert it into moles using Avogadro's number ($N_A=6.022\times 10^{23}$).

We have generated approximately $0.02\text{ moles}$ of silver chloride precipitate.

The Ratio and Werner's Theory

Here is where the magic of Werner's Coordination Theory comes into play. According to Werner, only the ions located outside the coordination sphere (the primary valency) are ionizable and free to react with ${\text{AgNO}}_3$. The ligands trapped inside the square brackets (the secondary valency) are non-ionizable.

The number of moles of $\text{AgCl}$ formed tells us exactly how many chloride ions were present outside the coordination sphere per molecule of the complex. Let's find this critical ratio:

This elegant result means there are exactly two ionizable chloride ions per complex molecule. They must reside outside the square brackets.

Formulating the Complex

Let's piece the puzzle together. The original empirical formula is ${\text{CoCl}}_3\cdot 6{\text{H}}_2\text{O}$, which contains a total of three chloride ions. Since we just proved that two chloride ions are outside the coordination sphere, exactly one chloride ion must be trapped inside.

Cobalt(III) almost exclusively forms octahedral complexes, meaning it requires a coordination number of six. With one chloride ion inside, we need five water molecules to fill the remaining coordination sites.

This leaves one water molecule out in the cold, acting as water of crystallization outside the sphere. Assembling these pieces, we get the final structure:

This perfectly matches option (d). The problem beautifully illustrates how macroscopic stoichiometric measurements can reveal the microscopic geometric reality of coordination compounds.