When 1 mol CrCl3 ⋅6H2O is treated with excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is :

1
[CrCl3 (H2O)3]⋅3H2O
2
[CrCl2 (H2O)4]Cl⋅2H2O
3
[CrCl(H2O)5]Cl2 ⋅H2O
4
[Cr(H2O)6]Cl3

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