When 1 mol CrCl3.6H2O is treated with an excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is : (a) [CrCl3 (H2O)3].3H2O (b) [CrCl2(H2O)4]Cl.2H2O (c) [CrCl(H2O)5]Cl2.H2O (d) [Cr(H2O)6]Cl3

contexto.130

3 years ago

Solution:

(d) 3 mol of AgCl implies 3Cl are given in the arrangement henceforth, the equation of the complex will be [Cr(H2O)6]Cl3.