Animated Solution for Chemistry - Coordination Compounds: 1 mol of an octahedral metal complex with formula $MC{l}_3\cdot 2L$ on reaction with excess of $AgN{O}_3$ gives 1 mol of $AgCl$. The denticity of ligand $L$ is .........

A Journey through Werner's Theory

Imagine you are a chemical detective, and you are handed a mysterious coordination complex with the empirical formula $MC{l}_3\cdot 2L$. Your mission is to uncover the true identity of the ligand $L$, specifically its denticity. How many 'teeth' does this ligand use to bite onto the central metal? Let's break down the clues step by step.

The Precipitation Clue

Our first major clue comes from a classic chemical interrogation technique: adding silver nitrate ($AgN{O}_3$). The problem states that 1 mole of our complex reacts to form exactly 1 mole of silver chloride ($AgCl$) precipitate.

According to Werner's Theory, a coordination compound has two types of valencies. The primary valency corresponds to the ionizable counter ions located outside the square brackets. The secondary valency corresponds to the non-ionizable ligands trapped inside the coordination sphere.

Because we get exactly 1 mole of $AgCl$, we know with absolute certainty that exactly one $C{l}^{-}$ ion is acting as a counter ion outside the coordination sphere.

Constructing the Coordination Sphere

Now, let's do some simple accounting. The empirical formula gave us a total of 3 chloride ions. If 1 of them is outside the bracket, the remaining 2 must be inside, directly bonded to the central metal $M$.

We also have two molecules of our mystery ligand $L$. Since they aren't precipitating, they must be inside the sphere too. This allows us to write the true structural formula of our complex:

The Octahedral Mandate

The final piece of the puzzle is the geometry. The problem explicitly states that the complex is octahedral. In the realm of coordination chemistry, an octahedral geometry is a strict mandate: the central metal must have a coordination number of exactly 6. It needs 6 coordinate bonds to be stable in this shape.

Let's sum up the bonds we have inside our square brackets: - We have 2 chloride ($C{l}^{-}$) ligands. Chloride is a well-known monodentate ligand, meaning each provides 1 bond. That's $2\times 1=2$ bonds. - We have 2 mystery $L$ ligands. Let's call the denticity of $L$ as $x$. They provide $2\times x$ bonds.

To satisfy the octahedral mandate, the total number of bonds must equal 6:

Solving this simple algebraic equation:

The math doesn't lie! Each ligand $L$ must form exactly two coordinate bonds with the central metal. Therefore, $L$ is a bidentate ligand, and its denticity is 2.