A) \[\left[ Co{{\left( N{{H}_{3}} \right)}_{4}} \right]C{{l}_{3}}\]
B) \[\left[ Co\left( N{{H}_{3}} \right)4C{{l}_{2}} \right]Cl\]
C) \[\left[ Co{{\left( N{{H}_{3}} \right)}_{4}} \right]C{{l}_{3}}\]
D) \[\left[ CoC{{l}_{3}}\left( N{{H}_{3}} \right) \right]{{\left( N{{H}_{3}} \right)}_{3}}\]
Correct Answer: B
Solution :
[b] Mole of \[AgCl=\frac{1.435}{143.5}=0.01;50mL\,of\,0.2\] M complex \[\equiv \] 0.01 mole \[\left[ Co{{\left( N{{H}_{3}} \right)}_{4}}C{{l}_{2}} \right]Cl\to {{\left[ CoN{{H}_{3}} \right)}_{4}}C{{l}_{2}}{{]}^{+}}+C{{r}^{-}}\]You need to login to perform this action.
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