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(NH_3)_5Cl_2`
B. `[Co(NH_3)_6]Cl_3`
C. `[CoCl_2(NH_3)_4]Cl`
D. `[CoCl_3(NH_3)_3]`

Answer

B

Explanation

No. of moles of `CoCl_3 . 6NH_3 = (2.675)/(267.5) = 0.01`
No. of moles of` AgCl = (4.78)/(143.5) = 0.03`
Since, 0.01 moles of the complex `CoCl_3. 6NH_3` gives 0.03 moles of AgCl on treatment with `AgNO_3` , it implies that 3 chloride ions are ionisable, in the complex. Thus, the formula of the complex is `[Co(NH_3)_6]Cl_3`