How does iodine react with iodide ions to form triiodide?

Iodine molecules ($I_2$) react with iodide ions ($I^{-}$) to form the triiodide ion ($I_3^{-}$) through a process where the iodide ion acts as a Lewis base, donating an electron pair to the iodine molecule, which functions as a Lewis acid. This reaction is represented by the equation:1

$$I_2+I^{-}\to I_3^{-}$$

Because the iodide ion can act as a Lewis base, the presence of soluble iodide significantly enhances the solubility of iodine in water. This interaction is also observed in the formation of potassium triiodide ($K{I}_3$) when iodine is dissolved in a solution of potassium iodide, a process often used to prepare aqueous iodine solutions. These polyiodide complexes are easily detected in solution and are classified as charge-transfer complexes.1