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(C) The complex $MCl_{3} \cdot 2 \ L$ reacts with excess $AgNO_{3}$ to yield $1 \ mol$ of $AgCl$,which indicates that only $1 \ mol$ of $Cl^{-}$ ion i

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$2$

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(C) The complex $MCl_{3} \cdot 2 \ L$ reacts with excess $AgNO_{3}$ to yield $1 \ mol$ of $AgCl$,which indicates that only $1 \ mol$ of $Cl^{-}$ ion is present in the ionization sphere.Thus,the coordination formula can be written as $[MCl_{2} \ L_{2}]Cl$.For an octahedral complex,the coordination number is $6$.Let the denticity of ligand $L$ be $x$.The coordination number is calculated as: $(2 	imes 	ext{denticity of } Cl^{-}) + (2 	imes x) = 6$.Since $Cl^{-}$ is a monodentate ligand (denticity $= 1$),we have: $(2 	imes 1) + 2x = 6$.$2 + 2x = 6 \implies 2x = 4 \implies x = 2$.Therefore,the denticity of ligand $L$ is $2$.

$1 \ mol$ of an octahedral metal complex with formula $MCl_{3} \cdot 2 \ L$ on reaction with excess of $AgNO_{3}$ gives $1 \ mol$ of $AgCl$. The denticity of ligand $L$ is ...... . (Integer answer)

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