Question 7 of 8

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