Question

The solubility of ${{I}_{2}}$ increases in water in the presence of
A. $KI$
B. ${{H}_{2}}S{{O}_{4}}$
C. $KMn{{O}_{4}}$
D. $N{{H}_{3}}$

Answer 1

Hint: The solubility of any chemical in water depends upon the ionisation of the specific molecule. If any molecule converts into ions easily then they have good solubility and if the molecule does not convert into ions when added to water then the molecule has less solubility in water.

Complete Step by Step Solution:
We know iodine is a non polar compound and has very low electron affinity whereas water is a polar solvent. So, the non- polar iodine does not dissolve in the polar water.

And $KI$ is also a polar molecule but it is soluble in water. The polar KI tens somewhat charge separation in the iodine molecule hence makes the iodine to be soluble in KI.
Secondly, iodine molecules have vacant antibonding molecular orbitals of lower energy.

So, the iodine molecule get take the electrons from $KI$ in antibonding orbitals and form an anion with $KI$ as shown in the reaction:-
$KI+{{I}_{2}}\to K{{I}_{3}}$
${{I}_{2}}+{{I}^{-}}\to {{I}^{-}}_{3}$ ( complex )
Thus, If $KI$is present in water, then the solubility of iodine increases.
Thus, Option (A) is correct.

Note: Remember that the solubility of some chemicals in water can be increased by making them as salts or complexes by the addition of suitable chemicals. Sometimes, the solubility goes on decreasing due to the common ion effect.