A) \[CrC{{l}_{3}}.6{{H}_{2}}O\]
B) \[[Cr{{\left( {{H}_{2}}O \right)}_{3}}C{{l}_{3}}].{{\left( {{H}_{2}}O \right)}_{3}}\]
C) \[\left[ CrC{{l}_{2}}{{\left( {{H}_{2}}O \right)}_{4}} \right]Cl.2{{H}_{2}}O\]
D) \[[CrCl{{({{H}_{2}}O)}_{5}}]C{{l}_{5}}.{{H}_{2}}O\]
Correct Answer: C
Solution :
[c] The ions present in the ionisation sphere are precipited Hence\[\left[ CrC{{l}_{2}}{{\left( {{H}_{2}}O \right)}_{4}} \right]C{{l}_{2}}.{{H}_{2}}O\] contains 1/3 Cl in ionisation sphere to be precipited by \[AgN{{O}_{3}}\] as \[AgCl\]
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