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When one mole CrCl3 .6H2O is treated wit...

When one mole `CrCl_3 .6H_2O` is treated with excess of `AgNo_3`, 3 mole of AgCl is obtained. The formula of complex is

A

`[CrCl_3 (H_2O)_3].3H_2O`

B

`[CrCl_2(H_2O)_4]Cl.2H_2O`

C

`[CrCl(H_2O)_5]Cl_2 . H_2O`

D

`[Cr(H_2O)_6]Cl_3`

Text Solution

Verified by Experts

The correct Answer is:
D
`[Cr(H_2O)_6]Cl_3`
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