A solution containing 2.675 g of `CoCl_(3).6NH_(3)` (molar mass = 267.5 g `"mol"^(-1)`) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of `AgNO_(3)` to give 4.78 g of AgCl (molar mass = 143.5 g `"mol"^(-1)`). The formula of the complex is :
(At. mass of Ag = 108u)

Options

A. `[CoCl_(3)(NH_(3))_(3)]`
B. `[CoCl(NH_(3))_(5)]Cl_(2)`
C. `[Co(NH_(3))_(6)]Cl_(3)`
D. `[CoCl_(2)(NH_(3))_(4)]Cl`.

Answer

C

Explanation

Moles of complex `=(2.675)/(267.5)=0.01`
Moles of AgCl precipitated `=(4.78)/(143.5)=0.03`
`therefore Cl^(-)` ions produced = 0.03 moles.
It shows that 3 moles of `Cl^(-)` ions are produced per mole of complex.
`therefore` Possible structure of complex is `[Co(NH_(3))_(6)]Cl_(3)`.
`underset(0.01"mol")underset(1"mol")([Co(NH_(3))_(6)]Cl_(3)) to [Co(NH_(3))_(6)]^(3+) +underset(0.03"mol")underset(3"mol")(3Cl^(-))`