Question

1 mol of an octahedral metal complex with formula $MCl_3 \cdot 2L$ on reaction with excess of $AgNO_3$ gives 1 mol of $AgCl$. The denticity of Ligand L is _________. (Integer answer)

Hint:

Always fix the number of ions outside the bracket first using the $AgCl$ data. Then, ensure the sum of (number of ligands $\times$ denticity) inside the bracket equals 6 for octahedral complexes.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
In coordination chemistry, the reaction with $AgNO_3$ precipitates chloride ions that are present outside the coordination sphere. An octahedral complex always has a coordination number (C.N.) of 6.
Step 2: Key Formula or Approach:
Number of ionizable $Cl^-$ ions = Moles of $AgCl$ precipitated per mole of complex.
Coordination Number (C.N.) = $\sum (\text{Number of ligands} \times \text{Denticity})$.
Step 3: Detailed Explanation:
Since 1 mole of $MCl_3 \cdot 2L$ produces 1 mole of $AgCl$, only one $Cl^-$ ion is outside the coordination sphere.
The formula of the complex can be written as $[MCl_2 L_2]Cl$.
For an octahedral complex, the C.N. is 6.
The ligands inside the sphere are two $Cl^-$ ions and two $L$ molecules.
$Cl^-$ is a monodentate ligand (denticity = 1).
Let the denticity of $L$ be $d$.
\[ C.N. = (2 \times 1) + (2 \times d) = 6 \]
\[ 2 + 2d = 6 \]
\[ 2d = 4 \]
\[ d = 2 \]
Step 4: Final Answer:
The denticity of Ligand L is 2.