1.

B +NaOH rarrDirty green ppt. overset(NH_(3))underset(("excess"))rarr Insoluble Find the option where blut ppt is obtained.

Answer»

When A is treated with `K_(4)[FE(CN)_(6)]`
When B is treated with `K_(4)[Fe(CN)_(6)]`
When A is treated with `K_(3)[Fe(CN)_(6)]`
When B is treated with `K_(3)[Fe(CN)_(6)]`

Solution :`X rarr Fe_(2)O_(3), A rarr FeCI_(3), B rarr FeI_(2)`,
`FeI_(2)+2NaOH rarr underset(("Dirty green ppt"))(Fe(OH)_(2)darr+2NaI)`
`Fe^(3+) +[Fe(CN)_(6)]^(4-)` gives same blue colouration as
`Fe^(2+) +[Fe(CN)_(6)]6(3-)`


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