A) \[[Cr({{H}_{2}}{{O}_{6}})]C{{l}_{3}}\]
B) \[[Cr({{H}_{2}}{{O}_{5}})Cl]{{H}_{2}}O.C{{l}_{2}}\]
C) \[[Cr{{({{H}_{2}}O)}_{4}}C{{l}_{2}}]Cl.2{{H}_{2}}O\]
D) \[[Cr{{({{H}_{2}}O)}_{3}}C{{l}_{3}}]3{{H}_{2}}O\]
Correct Answer: A
Solution :
When the solution of the complex is passed through cation exchanger, \[nC{{l}^{-}}\] ions will combine with H4' (of cation exchanger) to form HCI \[nC{{l}^{-}}+n{{H}^{+}}\to nHCl\] \[\because \]1 mol of the complex will form n moles of \[HCl\]. \[\therefore \]1 mole of complex n mol \[HCl\] = n moles of\[\text{NaOH}\] Mole of the complex \[=\frac{0.319}{266.5}\] \[=0.0012\]mol mole of \[\text{NaOH}\]used \[=\frac{28.5\times 0.125}{1000}\] \[=\text{ }0.00\text{36 mol}\] 0.0012 mol of complex = 0.0036 mol \[\text{NaOH}\] \[=0.00\text{36 mol HC}l\] mol complex \[=\frac{0.0036}{0.0012}=\] 3 mol HCI Thus, all the \[C{{l}^{-}}\]ions are outside of the coordination sphere. Hence, complex is \[[Cr{{({{H}_{2}}O)}_{6}}C{{l}_{3}}]\].
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