Problem 77 - Chemistry The Molecular Nature of Matter

Question

Predict the shapes of (a) \(\mathrm{IO}_{4}^{-},(\mathbf{b}) \mathrm{IF}_{4}^{-},(\mathbf{c}) \mathrm{TeF}_{6},(\mathbf{d}) \mathrm{SiO}_{4}\) and (e) \(\mathrm{ICl}_{2}\).

Step-by-Step Solution

Answer

\((\mathbf{a})\) Tetrahedral, \((\mathbf{b})\) Seesaw, \((\mathbf{c})\) Octahedral, \((\mathbf{d})\) Tetrahedral, \((\mathbf{e})\) Linear

Step 1: Understand the concept of VSEPR theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is used to predict the shape of molecules based on the notion that electron pairs repel each other and will therefore arrange themselves as far apart as possible in three dimensions to minimize this repulsion.

Step 2: Determine the Lewis structure for each molecule/ion

Draw the Lewis structure for each molecule/ion to identify the number of bonding pairs (BP) and lone pairs (LP) of electrons around the central atom.

Step 3: Predict the shape of \text{(a)} \(\mathrm{IO}_{4}^{-}\)

For \(\mathrm{IO}_{4}^{-}\), the central iodine atom has 7 valence electrons, plus 1 from the negative charge, minus 4 from the bonds with oxygen, resulting in 0 lone pairs. With 4 BP and 0 LP, the shape is tetrahedral.

Step 4: Predict the shape of \((\mathbf{b})\mathrm{IF}_{4}^{-}\)

In \(\mathrm{IF}_{4}^{-}\), the central iodine atom has 7 valence electrons, plus 1 from the negative charge, minus 4 from the bonds with fluorine, resulting in 1 LP. With 4 BP and 1 LP, the shape is seesaw.

Step 5: Predict the shape of \((\mathbf{c})\mathrm{TeF}_{6}\)

For \(\mathrm{TeF}_{6}\), the central tellurium atom has 6 valence electrons minus 6 from the bonds with fluorine, leading to 0 LP. With 6 BP and 0 LP, the shape is octahedral.

Step 6: Predict the shape of \((\mathbf{d})\mathrm{SiO}_{4}\)

In \(\mathrm{SiO}_{4}\), the central silicon atom has 4 valence electrons minus 4 from the bonds with oxygen, resulting in 0 LP. With 4 BP and 0 LP, the shape is tetrahedral.

Step 7: Predict the shape of \((\mathbf{e})\mathrm{ICl}_{2}\)

For \(\mathrm{ICl}_{2}\), the central iodine atom has 7 valence electrons minus 2 from the bonds with chlorine, resulting in 3 LP. With 2 BP and 3 LP, the shape is linear due to the LP-LP and LP-BP repulsions arranging the BP as far apart as possible.

Key Concepts

Lewis Structure

The foundation for predicting molecular geometry begins with drawing the Lewis structure. This visual representation displays the valence electrons around atoms and how they are shared or paired in a molecule or ion. It's essential to accurately depict bonding pairs (BP) of electrons that form covalent bonds between atoms and lone pairs (LP) that are not shared and remain on individual atoms. For example, in \textbf{IO}\(_4^-\), the central iodine (I) atom has seven valence electrons plus one from the extra negative charge. These electrons are shared with four oxygen atoms, resulting in a Lewis structure with four BP and no LP on iodine. Understanding the correct Lewis structure is pivotal because it lays the groundwork for determining molecular shapes using VSEPR theory.

Molecular Geometry

Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. The shape is primarily determined by the positions of nuclei and the electron pairs. VSEPR theory, which stands for Valence Shell Electron Pair Repulsion, guides the prediction of these shapes. After drawing the Lewis structure, one can visualize the molecular geometry by applying VSEPR logic, considering both BP and LP. For instance, with no LP on the central iodine atom in \textbf{IO}\(_4^-\), the atoms will adopt a tetrahedral shape to minimize repulsion. This concept is vital because the molecular shape affects many physical and chemical properties, including reactivity, polarity, and interaction with other molecules.

Electron Pair Repulsion

The principle of electron pair repulsion is integral to understanding molecular geometry. Electron pairs, whether they are in bonds or lone pairs, carry negative charges that repel each other. According to VSEPR theory, these pairs will orient themselves as far apart as possible to minimize repulsion. This concept explains why \textbf{IO}\(_4^-\) has a tetrahedral shape (as there are no LPs causing additional repulsion) while \textbf{IF}\(_4^-\) forms a seesaw shape due to the presence of one LP exerting repulsion. Recognizing how electron pair repulsion shapes molecular geometry is critical for comprehending the molecular form and, consequently, the behavior of substances.

Bonding Pairs and Lone Pairs

Distinguishing between bonding pairs and lone pairs of electrons is crucial for predicting molecular structures. Bonding pairs are those electrons that are shared in covalent bonds between atoms, contributing to the structural foundation of the molecule. In contrast, lone pairs are electrons that remain unshared on an individual atom. They influence molecular shapes, often taking up more space than bonding pairs due to their stronger repulsive force. For example, in \textbf{ICl}\(_2^-\), the central iodine atom has two bonding pairs and three lone pairs. The lone pairs will repel each other and the bonding pairs, leading to a linear shape as the bonding pairs are forced as far apart as possible. Recognizing the number and types of electron pairs is essential for applying VSEPR theory effectively to predict accurate molecular geometries.