`A,B` and `C` are three complexes of chromium(III) with the empirical formula `H_(12)O_(6)C_(13)Cr`. All the three complexes not react with concentrated `H_(2)SO_(4)` whereas complexes `B` and `C` lose `6.75%` and `13.5%` of their original mass respectively, on treatment on treatment with concentrated `H_(2)SO_(4)` Identify `A, B` and `C` .

Explanation

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, `3` `Cl` and `1` `Cr`.
As `Cl` and `H_(2)O` are present as ligands and complex 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`.