Question

The colourless $KBr$ or $KI$ Solutions turns red and deep brown respectively on passing chlorine through it. Give reasons.

Answer 1

Hint: It is a type of a halogen displacement reaction in which more reactive metal or compound of the halogen group will replace the other metal or the compound which is less reactive in nature.

Step by step answer: Chlorine is more reactive than any other metal of its group because of its high valence electrons and high electronegative nature. The order of the electronegative nature of the halogen compounds is chlorine $Chlorine \geqslant bro\min e \leqslant Iodine$. It means the chlorine is the most electronegative and reactive compound in the halogen group. Therefore, when potassium bromide and potassium iodide is passed through chlorine , the chlorine being highly reactive and highest electronegative displaces the less reactive metal which are bromine in the potassium bromide and iodine in the potassium iodide
The reaction takes place:
$2KBr + C{l_2} \to 2KCl + B{r_2}$
\[2KI + c{l_2} \to 2KCl + {I_2}\]
In both the reactions we can observe that the Single displacement reaction takes place in which the less reactive metal ion is replaced. When the Potassium bromide is passed through the chlorine solution, the chlorine reacts with potassium forming potassium chloride and free bromine ions of the red colour are left in the solution. Similarly, when the potassium iodide is passed through a chlorine solution the potassium chloride is formed leaving the free iodine ions of deep brown colour in the aqueous solution.

Note: In the halogen displacement reaction the reactivity of the halogens compounds decreases as we move down the series which means a more reactive halogen will replace the less reactive halogen compound from the solution of its salt. It means the most reactive halogen will replace all the other halogen while the least reactive halogen will get displaced by any other halogen, in the displacement type reaction.