Problem 100 - Chemistry and Chemical Reactivity

Question

Halogens form polyhalide ions. Sketch Lewis electron dot structures and molecular structures for the following ions: (a) \(\mathrm{I}_{3}^{-}\) (b) \(\mathrm{BrCl}_{2}^{-}\) (c) \(\mathrm{ClF}_{2}^{+}\) (d) An iodide ion and two iodine molecules form the \(\mathrm{I}_{5}^{-}\) ion. Here, the ion has five I atoms in a row, but the ion is not linear. Draw the Lewis dot structure for the ion, and propose a structure for the ion.

Step-by-Step Solution

Answer

The structures are: (a) Linear \(\mathrm{I}_3^-\), (b) T-shaped \(\mathrm{BrCl}_2^-\), (c) Linear \(\mathrm{ClF}_2^+\), (d) Trigonal bipyramidal \(\mathrm{I}_5^-\).

Step 1: Structure of \( \mathrm{I}_3^- \)

The \( \mathrm{I}_3^- \) ion consists of three iodine atoms. To draw its Lewis structure:- Place three iodine atoms in a line: I—I—I.- Add an extra electron for the negative charge, making a total of 22 valence electrons.- Distribute electrons to form complete octets around the outer iodine atoms, resulting in one lone pair on each terminal iodine and 3 lone pairs on the central iodine.- Hybridization of central iodine is \(\text{sp}^3d\), leading to a linear structure.

Step 2: Structure of \( \mathrm{BrCl}_2^- \)

The \( \mathrm{BrCl}_2^- \) ion has one bromine and two chlorine atoms.- Start with Br—Cl—Cl, placing bromine in the center.- The ion has 22 valence electrons, including one extra electron for the charge.- Each chlorine forms a single bond with bromine.- Distribute remaining electrons completing octets around chlorine atoms, and assign lone pairs to bromine.- The central bromine has a total of 3 lone pairs, making the structure T-shaped, with \(\text{sp}^3d\) hybridization.

Step 3: Structure of \( \mathrm{ClF}_2^+ \)

The \( \mathrm{ClF}_2^+ \) ion consists of one chlorine and two fluorine atoms.- Place Cl in the center with F—Cl—F.- Account for 20 valence electrons after removing one electron for the positive charge.- Allocate electrons to complete octets for fluorines; chlorine has 3 lone pairs.- The structure is linear due to \(\text{sp}^3\) hybridization of chlorine.

Step 4: Structure of \( \mathrm{I}_5^- \)

The \( \mathrm{I}_5^- \) ion includes five iodine atoms. To draw its structure:- Place five iodine atoms in a sequential arrangement: I—I—I—I—I.- Distribute 42 valence electrons (including the negative charge) to satisfy octets.- Central iodine has no lone pairs and is bonded to two others; terminal iodines each have 3 lone pairs.- Due to hybridization \(\text{sp}^3d\), the structure is trigonal bipyramidal with two iodine atoms on opposite axial positions and the rest forming a T-shaped arrangement in the equatorial plane.

Key Concepts

Lewis Structure

Understanding the Lewis structure is crucial for visualizing how atoms connect in a molecule. For polyhalide ions like \(\mathrm{I}_3^-\), \(\mathrm{BrCl}_2^-\), \(\mathrm{ClF}_2^+\), and \(\mathrm{I}_5^-\), the Lewis structure helps us see how valence electrons are distributed among the atoms.

Here's how to approach drawing Lewis structures for polyhalide ions:

• Identify the total number of valence electrons by adding up the electrons from each atom and adjusting for any charges.
• Arrange the atoms, placing the least electronegative atom in the center (usually). For \(\mathrm{I}_3^-\), align iodine atoms linearly: I—I—I.
• Bonds form between atoms with pairs of electrons, and lone pairs are distributed to complete the octets.

After you plot the basic structure with bonds and lone pairs, check electron count to ensure all atoms fulfill their octet needs, adjusting for any charges by adding or removing electrons. This method provides insight into molecular shape and reactivity.

Molecular Geometry

Molecular geometry describes the 3D shape of a molecule and is determined using the Lewis structure and hybridization concepts. For example, in the \(\mathrm{I}_3^-\) ion, although the atoms are arranged linearly in its Lewis structure, this is due to the central iodine's \(\text{sp}^3d\) hybridization, allowing a linear geometry.

In contrast, the \(\mathrm{BrCl}_2^-\) ion is T-shaped because of the central bromine's three lone pairs that repel the bonded pairs of electrons into a non-linear arrangement, also resulting from \(\text{sp}^3d\) hybridization.

• Types of geometries:
  • Linear: As seen in \(\mathrm{I}_3^-\) and \(\mathrm{ClF}_2^+\).
  • T-shaped: Illustrated by \(\mathrm{BrCl}_2^-\).
  • Trigonal bipyramidal: Found in \(\mathrm{I}_5^-\) due to the unique positioning of iodines around the central atom.

Understanding these geometries is vital for predicting how molecules interact with other chemical entities.

Valence Electrons

Valence electrons are the outermost electrons that play a critical role in bonding and chemical reactions. For polyhalide ions, counting and arranging valence electrons accurately is essential for crafting the Lewis structure.

• Each halogen generally brings 7 valence electrons.
• When forming ions, additional electrons are added for negative charges, while electrons are removed for positive charges.
• In \(\mathrm{I}_3^-\), the extra electron accounts for the negative charge, giving a sum of 22 valence electrons.

The distribution of these electrons among the atoms can influence properties like molecular stability and shape. By carefully assigning electrons to both bonds and lone pairs, one can predict the molecular geometry and potential interactions with other chemical structures. Remember, achieving a full octet for each atom (except hydrogen) often results in a more stable molecule.