Saved Bookmarks
| 1. |
An inorganic compound (A) gives the following reactions : (i) Produces a violet flame when burnt. (ii) When heated with manganese dioxide, it produces oxygen. (iii) When treated with iodine, it produces chlorine and a compound (B). (iv) When heated alone, it produces compounds (C) and (D). Identify compounds (A-D) and explain the reactions at steps (ii) to (iv). |
|
Answer» Solution :(i) Since compound (A), when burnt produces violet FLAME, therefore, compound (A) is a potassium salt. (ii) Since compound (A), when heated with `MnO_(2)` produces `O_(2)`, therefore, compound (A) must be potassium chlorate `(KCIO_(3))` `UNDERSET("Pot. chlorate")(2KCIO_(3))OVERSET(MnO_(2),Delta)rarr 2KCI + 3O_(2)`. (iii) When treated with `I_(2)`, it produces `CI_(2)` and potassium iodate `(KIO_(3))`. Therefore, compound (B) is `KIO_(3)` `underset("Pot. chlorate")(2KCIO_(3))+I_(2)overset(Delta)rarr underset("Pot. iodate (A)")(2KIO_(3))+CI_(2)` In this reaction, `I_(2)` acts as a reducing agent and hence decreases the oxidation state of CI from + 5 in `KCIO_(3)` to 0 in `CI_(2)` while that of I increase from 0 in `I_(2)` to + 5 in `KIO_(3)`. (iv) When heated alone, `KCIO_(3)` undergoes disproportionation reaction to give `KCIO_(4)`(C) and KCI (D) `underset("Pot. chlorate (A)")overset(+ 5)(4KCIO_(3)) overset(Delta)rarr underset("Pot. PERCHLORATE (B)")overset(+7)(3KCIO_(4))+underset("Pot. chloride (D)")overset(-1)(KCI)` The reason why `KCIO_(3)` undergoes disproportionation reaction is that it can be simultaneously oxidised (oxidation state of Cl increases from + 5 in `KCIO_(3)` to + 7 in `KCIO_(4)`) as well as reduced (oxidation state of Cl decreases from + 5 in `KCIO_(3)` to - 1 in KCI). |
|