Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empi

Q: Why do only the ions outside the square bracket precipitate?

The square bracket represents the coordination sphere, where ligands are tightly bound to the central metal atom via coordinate covalent bonds. These bonds don't break in aqueous solution. Ions outside the bracket are held by weaker ionic bonds and dissociate freely in water, allowing them to react and precipitate.

Q: How did we know the coordination number is 6?

The problem states it's an 'octahedral' complex. In coordination chemistry, an octahedral geometry strictly requires a coordination number of $6$, meaning exactly six ligand donor atoms must be attached to the central metal.

Q: Could water or ammonia act as counter ions outside the bracket?

Ammonia and water are neutral molecules. Counter ions must be charged to balance the charge of the complex ion. While water can exist outside as 'water of crystallization' (hydrate), it doesn't act as a counter ion to balance charge. Ammonia is almost always a ligand inside the sphere.

Q: What if the question said 2 moles of AgCl precipitated?

If only $2$ moles of $\text{AgCl}$ precipitated, it would mean only $2$ $\text{Cl}^-$ ions are outside the sphere. The third $\text{Cl}^-$ would have to be inside the bracket, acting as a ligand to satisfy the secondary valency.

Q: Is it possible for a ligand to satisfy both primary and secondary valencies?

Yes! If an anion like $\text{Cl}^-$ is inside the coordination sphere, it acts as a ligand (satisfying secondary valency) AND contributes to balancing the oxidation state of the metal (satisfying primary valency). However, it will NOT precipitate.

Question

Three moles of AgCl get precipitated when one mole of an octahedral co-ordination compound with empirical formula $	ext{CrCl}_3 \cdot 3	ext{NH}_3 \cdot 3	ext{H}_2	ext{O}$ reacts with excess of silver nitrate. The number of chloride ions satisfying the secondary valency of the metal ion is ……… .

Sigma Percentile · Accepted Answer

### The Mystery of the Coordination Sphere

Imagine you are a chemical detective, and you are handed a mysterious crystalline powder. You run some basic elemental analysis and determine its empirical formula:

$$ 	ext{CrCl}_3 \cdot 3	ext{NH}_3 \cdot 3	ext{H}_2	ext{O} $$

But an empirical formula is just a list of ingredients; it tells you nothing about how the molecule is actually built! Is it a simple salt? Is it a complex network?

The problem gives us a massive clue: it is an **octahedral coordination compound**. This means the central chromium atom ($	ext{Cr}^{3+}$) is acting like a king, surrounded by exactly six loyal subjects (ligands) that form its inner court, known as the coordination sphere.

### Werner's Masterpiece: Decoding Valency

To solve this puzzle, we must channel the genius of Alfred Werner, the father of coordination chemistry. Werner proposed a revolutionary idea: transition metals possess two distinct types of valencies.

The **primary valency** is the traditional oxidation state. It is ionizable, meaning these ions sit outside the coordination sphere and can freely swim away when the compound is dissolved in water.

The **secondary valency** is the coordination number. It is strictly non-ionizable. These ligands are trapped inside the square brackets, bound tightly to the metal via coordinate covalent bonds. They dictate the 3D geometry of the complex and will never break free in a simple aqueous solution.

### The Chemical Interrogation

So, how do we find out which atoms are trapped inside the inner court and which are freely swimming outside? We bring in a chemical interrogator: Silver Nitrate ($	ext{AgNO}_3$).

Silver ions ($	ext{Ag}^+$) have an intense affinity for free chloride ions ($	ext{Cl}^-$). When they meet in solution, they instantly lock together to form a heavy, white precipitate of Silver Chloride ($	ext{AgCl}$).

The problem states a beautiful, definitive fact: when one mole of our mysterious complex reacts with excess $	ext{AgNO}_3$, it yields exactly **three moles of $	ext{AgCl}$ precipitate**.

$$ 1 	ext{ mole complex} \xrightarrow{	ext{excess AgNO}_3} 3 	ext{ moles AgCl} \downarrow $$

What does this mean? It means all three chloride ions from our empirical formula were free to react! Therefore, all three $	ext{Cl}^-$ ions must be acting as primary valencies, sitting completely outside the coordination sphere.

### Assembling the Puzzle

Now, the pieces of the puzzle fall perfectly into place. If all three chloride ions are outside the bracket, who is inside?

Remember, the geometry is octahedral, which strictly demands a coordination number of $6$. We look at our remaining ingredients: we have exactly three ammonia ($	ext{NH}_3$) molecules and three water ($	ext{H}_2	ext{O}$) molecules.

To satisfy the king's demand for six ligands, all of these neutral molecules must enter the inner court. We can now write the true, structural formula of our complex:

$$ [	ext{Cr}(	ext{H}_2	ext{O})_3(	ext{NH}_3)_3]	ext{Cl}_3 $$

### The Final Verdict

Let's return to the core question asked by the examiners: *What is the number of chloride ions satisfying the secondary valency?*

We know that secondary valency refers exclusively to the ligands trapped **inside** the square brackets.

Looking at our beautifully deduced formula, $[	ext{Cr}(	ext{H}_2	ext{O})_3(	ext{NH}_3)_3]	ext{Cl}_3$, we can clearly see that the inside of the bracket contains only water and ammonia. There are absolutely no chloride ions inside.

Therefore, the number of chloride ions satisfying the secondary valency is a resounding **zero**.

### The Way Forward: A Thought Experiment

Before we close this case, let's stretch our minds. What if the reaction had only produced **two moles** of $	ext{AgCl}$ precipitate?

That would mean only two chlorides were free outside. The third chloride would have been forced inside the bracket to help satisfy the octahedral geometry, pushing one water molecule out as a hydrate. The formula would have been:

$$ [	ext{Cr}(	ext{H}_2	ext{O})_3(	ext{NH}_3)_2	ext{Cl}]	ext{Cl}_2 \cdot 	ext{H}_2	ext{O} $$

In that scenario, exactly one chloride ion would be satisfying the secondary valency! Always use the moles of precipitate as your ultimate key to unlock the hidden architecture of coordination compounds.

Enter Numerical Value:

Visualized Solution

Empirical Formula \& Geometry

Werner's Theory

Precipitation Reaction

Formulating the Complex

Secondary Valency of Chloride

What If?

The Sigma Insight: Nomenclature, Isomerism, Importance and Werner's Theory

The Mystery of the Coordination Sphere

Werner's Masterpiece

The Chemical Interrogation

Assembling the Puzzle

The Final Verdict

The Way Forward

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