1.

Which of the following statement(s) is/are correct with reference to Fe^(2+) and Cu^(2+)

Answer»

Both react with NaOH
`Cu^(2+)` GIVES BLUE ppt of `Cu(OH)_(2)` where as `FE^(2+)` gives `Fe(OH)_(2)` [green ppt]
`Cu^(2+)` gives CHOCOLATE brown ppt whereas `Fe^(2+)` gives sky blue ppt with `K_(4)[Fe(CN)_(6)]`
Both have four unpaired electrons

Solution :`Fe^(2+)+2NaOHrarr underset("(green ppt.)")(Fe(OH)_(2))+2Na^(+)`
`Cu^(2+)+2NaOHrarrunderset("(blue ppt.)")(Cu(OH)_(2))+2Na^(+)`
(b) is correct as explained above.
(a) is also correct as both react with `NaOH`.
`2Cu^(2+)+K_(4)[Fe(CN)_(6)] rarr underset("(chocolate brown ppt.)")(Cu_(2)[Fe(CN)_(6)])+4K^(+)`
`2Fe^(2+)+K_(4)[Fe(CN)_(6)]rarr underset("(sky blue ppt.)")(Fe_(2)[Fe(CN)_(6)])+4K^(+)`
(d) is not correct `because Cu^(2+)` has 1 unpaired electron whereas `Fe^(2+)` has 4 unpaired electrons.


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