When 1 mole of ${\mathrm{CrCl}}_{3}6{\mathrm{H}}_{2}O$ is treated with excess of ${\mathrm{AgNO}}_{3},$ 3 moles of ${\mathrm{AgCl}}_{}$ are obtained. The formula of the complex is:

1. [CrCl3(H2O)3].3H2O
2. [CrCl2(H2O)4]Cl.2H2O
3. [CrCl3(H2O)5]Cl2.H2O
4. [Cr(H2O)6]Cl3

Hint: Complex dissociate and generate 4 ions in the solution

1 mole of AgNO3, precipitates one free chloride ion (${Cl}^{-}$).

Here, 3 moles of AgCI are precipitated by an excess of AgNO3. Hence, there must be three free Cl${}^{-}$ ions

So, the formula of the complex can be$\left[Cr{\left({H}_{2}O\right)}_{6}\right]C{l}_{3}$ and the correct choice is option 4th