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When 1 mol CrCl(3).6H(2)O is treated wit...

When 1 mol `CrCl_(3).6H_(2)O` is treated with excess of `AgNO_(3)`, 3 mol of `AgCl` are obtained. The formula of the coplex is

A

`[CrCl_(3)(H_(2)O)_(3)].3H_(2)O`

B

`[CrCl_(2)(H_(2)O)_(4)]Cl.2H_(2)O`

C

`[CrCl(H_(2)O)_(5)]Cl_(2).H_(2)O`

D

`[Cr(H_(2)O)]Cl_(3)`

Text Solution

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The correct Answer is:
D
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