Question

A solution containing 2.675g of $CoC{{l}_{3}}.6N{{H}_{3}}$ (molar mass=267.5g/mol) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $AgN{{O}_{3}}$ to give 4.78g of AgCl (molar mass=143.5g/mol). The formula of the complex is: (Atomic mass of Ag=108u).
A. \[\left[ Co{{\left( N{{H}_{3}} \right)}_{6}} \right]C{{l}_{3}}\]
B. \[\left[ CoC{{l}_{2}}{{\left( N{{H}_{3}} \right)}_{4}} \right]Cl\]
C. \[\left[ CoC{{l}_{3}}{{\left( N{{H}_{3}} \right)}_{3}} \right]\]
D. \[\left[ CoCl{{\left( N{{H}_{3}} \right)}_{5}} \right]C{{l}_{2}}\]

Answer 1

Hint: The ligands present inside the square brackets of the coordination compounds are not ionizable. This information is vital for solving this question. Only the ligands outside the square brackets are ionisable and knowing the moles of these ions, we can find the chemical formula of the compound.

Complete step by step solution:
-The given compound is a coordination compound. Such compounds are different from other compounds. They retain their identity in the solution.
-For any compound, we can find the number of moles of the compound by the formula
\[moles=\dfrac{wt.\text{ in grams}}{\text{Molecular wt}\text{.}}\]
-In the question, we see that the weight and the molecular weight of the complex compound is given from which we can find the moles of the compound. The moles for the complex compound will be given as moles = $\dfrac{2.675}{267.5}=0.01$
-Similarly, we can find the moles of AgCl also by the same formula as we are given the value of both its molecular weight and the weight of the compound. The moles of AgCl can be given as
Moles = $\dfrac{4.78}{143.5}\approx 0.03\text{ }moles$
-In the question we can see that the solution is passed through the cationic exchanger and from the reaction in the exchanger, AgCl is being produced. Thus we can say that the chloride ion responsible for making AgCl comes from the complex solution itself.
-It means that the moles of the chloride ions of the complex solution should be equal to the moles of the AgCl compound as AgCl dissociates to give the chloride ions.
-Now from the values of moles, we see that the moles of the AgCl chloride ions is thrice that of the complex. It means that there were 3 chloride ions present in the complex compound that ionized to form the AgCl compound.
-Thus our compound should contain 3 ionizable chloride ions. Thus, 3 chloride ions should be present outside the square brackets of the coordination compound. From all the given options, we can see that only 1 option satisfies our desired result.

Therefore the correct option is A.

Note: Units must be taken into account before solving the question. Mole is a unitless quantity as it represents the ratio. So both the quantities required to find mole should have the same unit else the answer will be wrong. The SI unit of the weights for finding mole is taken as grams although the SI unit of weight is kilogram where 1kg=1000g