When 1 mol `CrCl_(3).6H_(2)O` is treated with excess of `AgNO_(3)`, 3 mol of `AgCl` are obtained. The formula of the coplex is

Options

A. `[CrCl_(3) (H_(2)O)_(3)].2H_(2)O`
B. `[CrCl_(2)(H_(2)O)]Cl_(2).2H_(2)O`
C. `[CrCl(H_(2)O)_(5)]Cl_(2).H_(2)O`
D. `[Cr(H_(2)O)_(6)]CL_(3)`

Answer

D

Explanation

when 1 mole of ` CrCl_(2) .6H_(2)O` is treated with excess of `AgNO_(3) ` , 3 moles of Ag Cl are obtained . The formula of the complex is `[Cr (H_(2)O)_(6)]Cl_(3)`