On treatment of 100 mL of 0.1 M solution of `COCl_(3).6H_(2)O` with excess of `AgNO_(3), 1.2 xx 10^(22)` ions are precipitated. The complex is

Options

A. `[Co(H_(2)O)_(6)]Cl_(3)`
B. `[Co(H_(2)O)_(5)Cl]Cl_(2).H_(2)O`
C. `[Co(H_(2)O)_(4)Cl_(2)]Cl.2 H_(2)O`
D. `[Co(H_(2)O)_(3)Cl_(3)].3H_(2)O`

Answer

B

Explanation

100 ml of 0.1 M `CoCl_(3).6 H_(2)O` solution contains. Complex
`=100xx0.1` millimole = 10 millimole = 0.01 mole
As the ions produced are precipitated by `AgNO_(3)`, the ions produced must be `Cl^(-)` ions.
Thus, 0.01 mole of the complex give `Cl^(-)` ions
`=1.2xx10^(22)`
`=(1.2xx10^(22))/(6.0xx10^(23))` mole = 0.02 mole
`therefore` 1 mole of the complex gives `Cl^(-)` ions = 2 moles (which are precipitated as AgCl) Remembering that coordination number of Co is 6 and two `Cl^(-)` ions are present outside the coordination sphere, the formula of the complex will be `[Co(H_(2)O)_(5)Cl]Cl_(2).H_(2)O`.