A, B and C are three complexes of chromium with the empirical formula H_(12)O_(6)Cl_(3)Cr. All the three complexes have Cl and H_(2)O as ligands. Complex A does not react with conc. H_(2)SO_(4). Complexes B and C lose 6.75% and 13.5% of their original weight respectively on heating with conc. H_(2)SO_(4). Identify A, B and C.

A, B and C are three complexes of chromium with the empirical formula

1.	A, B and C are three complexes of chromium with the empirical formula H_(12)O_(6)Cl_(3)Cr. All the three complexes have Cl and H_(2)O as ligands. Complex A does not react with conc. H_(2)SO_(4). Complexes B and C lose 6.75% and 13.5% of their original weight respectively on heating with conc. H_(2)SO_(4). Identify A, B and C.
Answer» Solution :Remembering that coordination NUMBER of Cr is 6, different conditions may be discussed as under : The formula `H_(12)O_(6)Cl_(3)Cr` suggests that there are 6 `H_(2)O` molecules, 3Cland 1 Cr. As Cl and `H_(2)O` are present as ligands andcomplex A does not LOSE `H_(2)O` on adding conc. `H_(2)SO_(4)`, this means, no `H_(2)O` is present outside the coordination sphere. Hence, the formula of A is `[Cr(H_(2)O)_(6)]Cl_(3)` Molecular mass of the complex `H_(12)O_(6)Cl_(3)Cr=12+96+106.5+52=266.5` amu As B loses 6.75% of `H_(2)O, H_(2)O` lost from 1 molecule `=(6.75)/(100)xx266.5=18` amu= `1 H_(2)O` molecule Thus, in B, one `H_(2)O` molecule is present outside the coordination sphere. Hence, complex B is `[Cr(H_(2)O)_(5)Cl]Cl_(2).H_(2)O`. As C loses 13.5% of `H_(2)O, H_(2)O` lost from 1 molecue `=(13.5)/(100)xx266.5=36` amu = `2 H_(2)O` molecules Thus, in C. two `H_(2)O` molecules are present outside the coordination sphere. Hence, complex C is `[Cr(H_(2)O)_(4)Cl_(2)]Cl.2 H_(2)O`.