Problem 106 - Chemistry and Chemical Reactivity

Question

Iodine, \(I_{2}\), is much more soluble in a water solution of potassium iodide, KI, than it is in pure water. The anion found in solution is \(I_{3}^{-}\) (a) Draw an electron dot structure for \(\mathrm{I}_{3}^{-}\). (b) Write an equation for this reaction, indicating the Lewis acid and the Lewis base.

Step-by-Step Solution

Answer

\( \mathrm{I}_3^{-} \) is linear; \( \mathrm{I}_2 \) + \( \mathrm{I}^- \rightarrow \mathrm{I}_3^{-} \), \( \mathrm{I}_2 \) is the Lewis acid.

Step 1: Understand the Problem

We are asked to represent the iodine ion \( \mathrm{I}_3^{-} \) using Lewis dot structures and write an equation showing the reaction in which \( \mathrm{I}_2 \) becomes \( \mathrm{I}_3^{-} \).

Step 2: Draw Lewis Dot Structure for \( \mathrm{I}_3^{-} \)

First, note that \( \mathrm{I}_3^{-} \) involves three iodine atoms. Iodine has 7 valence electrons. Because the ion has a -1 charge, there is one extra electron. Distribute the electrons to make each iodine with a full octet, remembering that the overall charge of the ion is -1. The structure is linear with a central iodine atom bonded to the other two iodine atoms, each with a complete octet and the entire structure carrying a -1 charge.

Step 3: Identify Lewis Acid and Base

In the reaction where \( \mathrm{I}_2 \) reacts with \( \mathrm{I}^- \) to form \( \mathrm{I}_3^{-} \), \( \mathrm{I}_2 \) accepts an electron pair, thus acting as a Lewis acid. \( \mathrm{I}^- \) donates an electron pair, acting as a Lewis base.

Step 4: Write the Reaction Equation

The reaction can be represented by: \[ \mathrm{I}_2 + \mathrm{I}^- \rightarrow \mathrm{I}_3^- \]. This equation indicates that \( \mathrm{I}_2 \) and \( \mathrm{I}^- \) react to form the triiodide ion, \( \mathrm{I}_3^- \). \( \mathrm{I}_2 \) is the Lewis acid and \( \mathrm{I}^- \) is the Lewis base.

Key Concepts

Iodine Chemistry

Iodine is a fascinating element, known for its vital role in chemistry and biology. In its natural state, iodine is found as a diatomic molecule, represented as \( I_2 \). This form is only slightly soluble in water. However, when introduced to a solution containing potassium iodide (KI), the solubility of iodine increases. This is because iodine can react with iodide ions \( (I^-) \) from the potassium iodide to form the triiodide ion \( (I_3^-) \).

In this reaction, iodine becomes more soluble due to the formation of \( I_3^- \), which is more stable in solution. The triiodide ion formation showcases the versatile chemistry of iodine, where it can easily form larger structures by bonding to other iodine atoms.

Lewis Acid-Base Theory

Understanding Lewis acid-base theory is crucial when analyzing chemical interactions, like the one involving iodine and iodide ions. According to this theory, a Lewis acid is a substance that can accept an electron pair, while a Lewis base is a substance that donates an electron pair.

In the formation of \( I_3^- \), the \( I_2 \) molecule functions as a Lewis acid. It accepts an electron pair from the \( I^- \) ion. Here, \( I^- \) acts as a Lewis base since it donates an electron pair to \( I_2 \).

• \( I_2 \): Lewis Acid - accepts electron pair.
• \( I^- \): Lewis Base - donates electron pair.

Understanding these roles makes it easier to identify how specific reactions occur at the molecular level.

Chemical Reactions

Chemical reactions are processes where substances are transformed into new products. The reaction of iodine with potassium iodide is a clear example of a redox reaction, although it focuses primarily on the electronic shifts through the Lewis Acid-Base Theory.

The reaction equation \( I_2 + I^- \rightarrow I_3^- \) specifically illustrates the interaction between iodine and the iodide ion. This process shows the transformation of iodine’s chemical structure through the acceptance and donation of electron pairs, resulting in the formation of the more stable \( I_3^- \) ion.

• This type of chemical reaction highlights how ionic species can enhance solubility.
• It also serves as a model to comprehend other similar transformations in chemistry.

Mastering these reactions is key in understanding the broader impact on solutions and compound formation.