Explain why a green solution of potassium manganate turns purple and a brown solid is precipitated when `CO_(2)` gas is bubbled into the solution.

Explanation

(i) Iodide `(I^(-))` ion is a stronger reducing agrent 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 dissociates to 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 undergoes disproportionation to form purple coloured `MnO_(4)^(-)`ion and the brown solid to `MnO_(4)`

(iv) Yellow solution is due to formation of `Na_(2)CrO_(4)` as follows `:`
`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)`