Question:
When 1 mol CrCl3.6H2O is treated with an 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
Solution:
Option (iv) [Cr(H2O)6]Cl3 is the answer.