Question

Why is ${{\text{I}}_{\text{2}}}$ more soluble in KI than water?

Answer 1

Hint:We know that like dissolves like means polar solvent dissolves polar molecules and nonpolar molecules dissolve non-polar molecules. We will decide if the iodine, water, and KI are polar and non-polar. The solubility of a non-polar compound in polar compounds depends upon the polarization of a nonpolar molecule.

Complete answer:Iodine is a non-polar compound. The water is a polar solvent.According to the Like dissolves Like rule, the non-polar iodine molecule cannot be soluble in polar water.Even the KI is also polar but the iodine molecule is soluble in KI. The solubility of the iodine molecule in KI depends upon various reasons as follows:First, as we add iodine molecules in KI, the ion-dipole creates. The polar KI causes somewhat charged separation in the iodine molecule hence makes it somewhat polar which helps the iodine to be soluble in KI.Second, iodine molecule has vacant antibonding molecular orbitals of lower energy. So, iodine molecules get electrons from KI in antibonding orbitals and form an anion with KI shown as follows:
\[{\text{KI}}\,{\text{ + }}{{\text{I}}_{\text{2}}}\, \to \,{\text{KI}}_3^ - \]
So, the iodine molecule is soluble in KI rather than water.Therefore, due to the compound formation, ${{\text{I}}_{\text{2}}}$ more soluble in KI than water.

Note:Iodine cannot form such anion with water. The iodine reacts with KI and form \[{\text{KI}}_3^ - \]. Now, the \[{\text{KI}}_3^ - \] is a polar compound. So, it is soluble in water. Iodine can be solubilized in water by dissolving iodine in KI and then into the water. As two iodine atoms share two electrons to form a bond, so iodine is a covalent molecule. Iodine accepts the electrons from another molecule in its antibonding orbital. The compound formed in this way shown colour due to charge transfer.