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A coordination compound $CrC{{l}_{3}}.4{{H}_{2}}O$ gives white precipitate of $AgCl$ with $AgN{{O}_{3}}$. The molar conductance of the compound corresponds to two ions. The structural formula of the compound is:
(A) $\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{3}} \right]$
(B) $\left[ Cr{{({{H}_{2}}O)}_{3}}C{{l}_{3}} \right]{{H}_{2}}O$
(C) $\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]Cl$
(D) $\left[ Cr{{({{H}_{2}}O)}_{4}}Cl \right]C{{l}_{2}}$

Last updated date: 13th Jun 2024
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Hint: In a coordination compound, the species inside the square brackets are non-ionizable and are collectively called a coordination sphere. The species present outside the coordination sphere are ionizable and are called counter ions.

Complete step by step solution:
We have been given the following facts about the coordination compound $CrC{{l}_{3}}.4{{H}_{2}}O$.
It gives white precipitates of silver chloride (AgCl) on reaction with silver nitrate ($AgN{{O}_{3}}$)
Its molar conductance corresponds to two ions.
Let us try to deduce the structural formula of the compound with the above-given information.
We know that the compound gives precipitates of AgCl with $AgN{{O}_{3}}$, this infers that the compound must contain chloride ions as counter ions.
So based on the above fact, we can rule out $\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{3}} \right]$ and $\left[ Cr{{({{H}_{2}}O)}_{3}}C{{l}_{3}} \right]{{H}_{2}}O$ since both the structural formulas contain no ionizabe $C{{l}^{-}}$ ions as the counter ions.
Now, the structural formula of the compound could be $\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]Cl$ or $\left[ Cr{{({{H}_{2}}O)}_{4}}Cl \right]C{{l}_{2}}$.
We know that the central atom and the ligands inside the square do not dissociate into ions and form a single entity. Only the counter ions present outside the coordination are ionizable.
But the molar conductance of the compound $\left[ Cr{{({{H}_{2}}O)}_{4}}Cl \right]C{{l}_{2}}$ is due to three ions as it dissociates as
\[\left[ Cr{{({{H}_{2}}O)}_{4}}Cl \right]C{{l}_{2}}\rightleftarrows {{\left[ Cr{{({{H}_{2}}O)}_{4}}Cl \right]}^{+}}+2C{{l}^{-}}\]
So it cannot be the correct structural formula of the given coordination compound.
Consider the dissociation of the compound $\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]Cl$ in solution.
\[\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]Cl\rightleftarrows {{\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]}^{+}}+C{{l}^{-}}\]
The molar conductance of $\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]Cl$ corresponds to two ions, therefore, the structural formula of the coordination compound $CrC{{l}_{3}}.4{{H}_{2}}O$ will be $\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]Cl$.
As $\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]Cl$ contains one $C{{l}^{-}}$ ion as the counter ion and therefore, on addition of $AgN{{O}_{3}}$ it gives one mole of AgCl.
\[\left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]Cl+AgN{{O}_{3}}\to \left[ Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}} \right]N{{O}_{3}}+AgCl\]

Hence, the correct option is (C).

Note: Note that the species inside the square brackets do not dissociate in solution into simple ions hence they do not contribute to the conductance of the solution. The total number of ions produced on the dissociation of a coordination compound is equal to the number of counter ions plus one coordination sphere.