Question

Iodine $\left({\mathrm{I}}_2\right)$ is only sparingly soluble in water (left photo). Yet upon the addition of iodide ions (e.g., from KI), iodine is converted to the triiodide ion, which readily dissolves (right photo): $${\mathrm{I}}_2(s)+{\mathrm{I}}^{-}(aq)\rightleftarrows {\mathrm{I}}_3^{-}(aq)$$ Describe the change in solubility of ${\mathrm{I}}_2$ in terms of the change in intermolecular forces.

Short answer

Iodine becomes more soluble in water as triiodide ions form, strengthening ion-dipole interactions.

Step-by-step solution

Understand Solubility Effects

Iodine (${\mathrm{I}}_2$) is solid and only sparingly soluble in water due to its non-polar nature, while water is polar. Non-polar solutes have weak solubility in polar solvents as the attractive intermolecular forces between the solute and solvent are minimal.

Recognize the Role of Iodide Ions

When iodide ions (${\mathrm{I}}^{-}$) are introduced into the solution (often from potassium iodide, KI), they interact with iodine (${\mathrm{I}}_2$). Iodide ions, being charged species, have stronger interactions with iodine due to the formation of temporary dipoles and ionic bonds.

Formation of Triiodide Ion

The introduction of iodide ions allows the formation of triiodide ions (${\mathrm{I}}_3^{-}$), which are more soluble in water. This is explained by the fact that the intermolecular forces become predominantly ion-dipole, which are stronger than the forces between non-polar ${\mathrm{I}}_2$ and polar water alone.

Describe Change in Intermolecular Forces

Initially, the intermolecular forces between ${\mathrm{I}}_2$ and water are weak van der Waals forces due to polarity differences. With ${\mathrm{I}}^{-}$ ions present, the iodine molecules form ${\mathrm{I}}_3^{-}$ ions that dissolve readily due to the formation of stronger ion-dipole interactions with water.

Key concepts

Intermolecular Forces

Intermolecular forces are the attractive forces that exist between molecules. These forces dictate how molecules interact with each other. There are several types of intermolecular forces, including van der Waals forces, dipole-dipole interactions, and hydrogen bonds.
Understanding intermolecular forces is key to explaining why certain substances dissolve in others. Stronger intermolecular forces lead to greater solubility, as molecules are more effectively attracted to and interact with the solvent.

• Van der Waals forces are weak forces that occur due to temporary dipoles in molecules.
• Dipole-dipole interactions occur between polar molecules with permanent dipoles.
• Hydrogen bonds are a special, strong type of dipole-dipole interaction involving hydrogen atoms.

When studying solubility, it's crucial to consider which of these forces are at play and how they influence the behavior of molecules in solution.

Non-polar and Polar Interactions

The difference between non-polar and polar interactions is fundamental to understanding solubility. Non-polar molecules, such as iodine $ext{I}_2$, do not have a permanent dipole moment because they share electrons equally. In contrast, polar molecules have a permanent dipole due to an uneven distribution of electrons, like water ($ext{H}_{2}extO$).
When a non-polar molecule is placed in a polar solvent like water, the intermolecular forces between them are weak. This results in limited solubility. Non-polar dissolves non-polar and polar dissolves polar; this is due to like dissolving like principle.

• Non-polar solutes generally dissolve better in non-polar solvents.
• Polar solutes tend to dissolve well in polar solvents.

Recognizing these interactions helps to predict solubility and understand why certain substances do not naturally mix.

Ion-Dipole Interactions

Ion-dipole interactions occur when an ion is sorrounded by molecules with a permanent dipole. These interactions play a significant role in the solubility of ionic compounds in polar solvents like water.
When you add iodide ions $ext{I}^{-}$ to iodine $ext{I}_2$, it facilitates the formation of triiodide ions $ext{I}_3^{-}$. The newly formed ions have stronger interactions with water. This happens because ion-dipole forces, which are interactions between an ionic charge and a dipole, are stronger than van der Waals forces.

• Ion-dipole forces are present when salts dissolve in water.
• They increase the solubility of the ionic compounds significantly.

In the context of iodine, the shift from weak van der Waals forces to stronger ion-dipole interactions accounts for the increased solubility observed in the presence of iodide ions.

Iodine Chemistry

Iodine chemistry is fascinating due to the unique properties of iodine, a non-polar diatomic molecule that behaves differently in various environments. Being non-polar, iodine $ext{I}_2$ is not easily soluble in water. However, its chemical behavior changes significantly in the presence of iodide ions.
The reaction $ext{I}_2(s)+ext{I}^{-}(aq)\rightleftarrows ext{I}_3^{-}(aq)$ demonstrates this transformation. In the reaction, iodide ions interact with iodine molecules to form triiodide ions $ext{I}_3^{-}$, which dramatically improves solubility.

• Triiodide ions are more stable in aqueous solutions than $ext{I}_2$.
• This reaction highlights the importance of ion-dipole interactions in increasing solubility.

Studying iodine chemistry offers insights into how simple molecular interactions can lead to profound changes in solubility and chemical behavior.