Question

On boiling an aqueous solution of $KCl{{O}_{3}}$ with iodine, the following product is obtained:
A. $KI{{O}_{3}}$ ​
B. $KCl{{O}_{4}}$
C. $KI{{O}_{4}}$
D. $KCl$

Answer 1

Hint: Potassium chlorate or $KCl{{O}_{3}}$ is a white coloured crystal. It is a strong oxidising agent that means it helps to oxidise other substances and itself get reduced.

Complete Step by Step Answer:
On boiling an aqueous solution of $KCl{{O}_{3}}$or potassium chlorate with iodine gas potassium iodate forms as the major product along with a gaseous halogen compound termed as chlorine gas. The reaction can be given as follows:
${{I}_{2}}+2KCl{{O}_{3}}\to C{{l}_{2}}+2KI{{O}_{3}}$

Thus in this reaction one mole of iodine gas reacts with two moles of potassium chlorate, a crystalline white solid to produce two moles of potassium iodate as a major product along with one mole of chlorine gas. In this reaction the potassium chlorate acts as an oxidising agent which helps iodine to get oxidised and form potassium iodate and itself gets reduced from chlorine gas of lower oxidation state.

Thus we can write that on boiling an aqueous solution of $KCl{{O}_{3}}$ or potassium chlorate with iodine gas potassium iodate or $KI{{O}_{3}}$ is formed as a major product.
 Thus the correct option is A.

Note: Chemical reactions are carried out by different mechanisms, among which is the displacement of an atom in one molecule by another. When the elements have a chemical relationship like they belong to the same family as chlorine and iodine, it is easy to verify the substitution reaction.