1.

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

Answer»


Solution :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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