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The number of chloride ions which would ...

The number of chloride ions which would be precipitated, when `CrCl_(3).4NH_(3)` is treated with silver nitrate solution

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The correct Answer is:
1
Co-ordination no. of Cr is 6.
`[Cr(NH_(3))_(4)Cl_(2)]Cl""to[Cr(NH_(3))_(4)Cl_(2)]^(+)+Cl`
`Cl^(-)+AgNO_(3)toAgCldarr+NO_(3)^(-)`
Thus only one `Cl^(-)` ion is precipitated.
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