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Give reasons for the following: (i) [CuI_(4)]^(2-) does not exist while [CuCl_(4)]^(2-) exists. (ii)HgCl_(2) and SnCl_(2) cannot exist together in aqueous solution . (iii) A green solutionof potassium manganateturns purpleand a brown solid is precipitated when CO_(2) is passes through the solution. (iv) Deep greenprecipitate ofCr(OH)_(3) gives yellow solution on addition of H_(2)O_(2). (v) FeCl_(3) solution turns brown on standing. |
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Answer» Solution :(i) Iodide `(I^(-))` ion is a stronger reducingagrent than chloride `(C l^(-))` ion. Consequently , iodie reduce`Cu^(2+)` ion( of `CuI_(2)` ) to `Cu^(+)` ion thereby converting `CuI_(2)` to CuI, i.e., `2CuI_(2) rarr 2CuI+I_(2)` Hence `[CuI_(4)]^(2-)` does not exist. (ii) `SnCl_(2)` is a reducing agent . It reduces agent. It reduces `HgCl_(2)` to `Hg_(2)Cl_(2)` and then to Hg according to the following reactions `:` `SnCl_(2)+ 2HgCl_(2) rarr SnCl_(2)+ Hg_(2)Cl_(2)` `Hg_(2)Cl_(2)+ SnCl_(2) rarr SnCl_(4) + 2Hg` (iii) `CO_(2)`on passing through aqueous solution forms `H_(2)CO_(3)` which dissociatesto give `H^(+)`and `HCO_(3)^(-)` ions. `CO_(2)+ H_(2)O hArr H_(2)CO_(3) hArr 2H^(+) + CO_(3)^(2-)` In presence of `H^(+)` ions , `MnO_(4)^(2-)` ion undergoesdisproportionation to form PURPLE COLOURED `MnO_(4)^(-)`ion and the brown solid to `MnO_(4)`  (iv) YELLOWSOLUTION is DUE to formation of `Na_(2)CrO_(4)` asfollows `:` `underset("Green ppt.")(2Cr(OH)_(3))+ 4NaOH + 3H_(2)O_(2) rarr underset("Sodium chromate (Yellow solution)")(2Na_(2)CrO_(4)) + 8 H_(2)O` (v)`FeCl_(3)` solution on standing undergoes hydrolysis to form brown `Fe_(2) O_(3). x H_(2)O` `FeCl_(3) + 3H_(2)O rarr Fe(OH)_(3) +3HCl` `2Fe(OH)_(3) rarr underset("Hydrated ferric oxide (Brown )")(Fe_(2)O_(3).3H_(2)O)` |
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