Answer
64.8k+ views
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.
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)