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Step-by-Step Solution
Step 1: Identify the Coordination Sphere and Ionization Sphere
The given metal complex is MCl3·2L and is octahedral. When 1 mole of this complex is treated with excess AgNO3, 1 mole of AgCl is formed. This indicates that exactly one Cl− ion is free (or ionizable) and hence can be precipitated as AgCl. Thus, only one chlorine is outside the coordination sphere.
Step 2: Determine the Complex Formula
Because we have determined that only one chloride is outside the coordination sphere, the complex formula can be written as [MCl2L2]Cl. The chloride ion in square brackets is bound to the metal, while the chloride after the bracket is the ionizable chloride outside the coordination sphere.
Step 3: Use the Octahedral Coordination Number
An octahedral complex has a coordination number of 6. In [MCl2L2], the two chlorides account for 2 sites (each Cl is monodentate, providing one coordination site each). That leaves 4 more sites to be filled by 2 molecules of L.
Step 4: Deduce the Denticity of Ligand L
Because 2 molecules of L must occupy 4 coordination sites in total, each L molecule must bind through 2 sites (i.e., each L is bidentate). Therefore, the denticity of L is 2.
Answer
The denticity of the ligand L in the complex is 2.
