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

(i) [CrCl3 (H2O)3]⋅3H2O

(ii) [CrCl2(H2O)4]Cl⋅2H2O

(iii) [CrCl(H2O)5]Cl2⋅H2O

(iv) [Cr(H2O)6]Cl3

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(iv) [Cr(H2O)6]Cl3

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