Problem 33 - Chemistry The Central Science In Si Units

Question

(a) Explain why \(\mathrm{BrF}_{4}^{-}\) is square planar, whereas \(\mathrm{BF}_{4}^{-}\) is tetrahedral. (b) How would you expect the \(\mathrm{H}-\mathrm{X}-\mathrm{H}\) bond angle to vary in the series \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2}\) Se? Explain. (Hint: The size of an electron pair domain depends in part on the electronegativity of the central atom.)

Step-by-Step Solution

Answer

\(\mathrm{BrF}_{4}^{-}\) is square planar due to lone pairs, \(\mathrm{BF}_{4}^{-}\) is tetrahedral with no lone pairs. \(\mathrm{H}_2\mathrm{O}\) has a smaller bond angle than \(\mathrm{H}_2\mathrm{S}\) and \(\mathrm{H}_2\mathrm{Se}\) due to higher electronegativity of oxygen.

Step 1: Determine the Electron Geometry of \(\mathrm{BrF}_4^{-}\)

\(\mathrm{BrF}_4^{-}\) consists of 4 fluorine atoms surrounding a bromine atom with a negative charge. Bromine has 7 valence electrons and gains one more due to the negative charge, for a total of 8 electrons. Using VSEPR theory, distribute these electrons: 4 bonding pairs and 2 lone pairs on the bromine atom. According to VSEPR, this results in an octahedral electron geometry, but square planar molecular geometry due to the lone pairs occupying positions opposite each other.

Step 2: Determine the Electron Geometry of \(\mathrm{BF}_{4}^{-}\)

\(\mathrm{BF}_{4}^{-}\) consists of 4 fluorine atoms surrounding a boron atom with a negative charge. Boron has 3 valence electrons and gains one more due to the negative charge, for a total of 4 electrons. In VSEPR theory, these electrons form 4 bonding pairs. With no lone pairs, the geometry is tetrahedral because all positions around boron are occupied by fluorine atoms equally, and no lone pairs alter this geometry.

Step 3: Analyze \(\mathrm{H}-\mathrm{X}-\mathrm{H}\) Bond Angles in \(\mathrm{H}_2\mathrm{O}, \mathrm{H}_2\mathrm{S}, \mathrm{H}_2\mathrm{Se}\)

For molecules \(\mathrm{H}_2\mathrm{O}\), \(\mathrm{H}_2\mathrm{S}\), and \(\mathrm{H}_2\mathrm{Se}\), we need to consider the central atom (O, S, Se respectively) and their electronegativity. Oxygen is the most electronegative, creating the largest repulsion between bonding pairs and lone pairs, leading to a smaller actual bond angle than the ideal 109.5° of tetrahedral geometry. As electronegativity decreases from O to Se, the repulsion decreases, and the actual bond angle approaches the ideal, slightly increasing as we move from \(\mathrm{H}_2\mathrm{O}\) to \(\mathrm{H}_2\mathrm{Se}\).

Step 4: Interpret the Influence of Electronegativity on Bond Angle

The size of electron pair domains is influenced by the electronegativity of the central atom. More electronegative atoms hold onto electron pairs more tightly, compressing bond angles. Thus, in \(\mathrm{H}_2\mathrm{O}\), the bond angle is less than in \(\mathrm{H}_2\mathrm{S}\) and \(\mathrm{H}_2\mathrm{Se}\), which have less tightly held electron pairs, allowing for increased bond angles.

Key Concepts

VSEPR theory

The Valence Shell Electron Pair Repulsion (VSEPR) theory is a fundamental concept in chemistry that allows us to predict the shape of a molecule. This theory is based on the idea that electron pairs, whether they are bonding or non-bonding (lone pairs), will repel one another and arrange themselves as far apart as possible around a central atom.

For example, in the molecule \( \text{BrF}_4^- \), VSEPR theory tells us that the six electron pairs (four bonding with fluorine atoms and two lone pairs) adopt an octahedral arrangement. However, the two lone pairs tend to occupy positions opposite each other, leading to a square planar molecular geometry. This is because lone pairs exert a stronger repulsive force compared to bonding pairs, affecting the final shape of the molecule.

Conversely, in \( \text{BF}_4^- \), there are no lone pairs and only four bonding pairs, all equally spread out around the boron atom, leading to a tetrahedral shape. This demonstrates how the presence or absence of lone pairs significantly influences molecular geometry according to the VSEPR theory.

Electron pairs

Electron pairs are crucial in determining the shape and geometry of molecules. They can be classified into bonding pairs, which are shared between atoms forming a bond, and lone pairs, which are non-bonding and belong to a single atom.

In molecules like \( \text{BrF}_4^- \), bromine is surrounded by four bonding pairs and two lone pairs. These electron pairs affect the molecular shape due to electron pair repulsion. The lone pairs try to stay as far away from other electron pairs as possible, which can lead to a different arrangement than might be expected if only considering bonding pairs.

This influence can be seen in how the presence of lone pairs in \( \text{BrF}_4^- \) creates a square planar shape, as opposed to \( \text{BF}_4^- \) which is tetrahedral, having no lone pairs to change the basic geometry dictated by bonding pairs alone. Understanding electron pairs is essential for applying VSEPR theory effectively to predict molecular shapes.

Electronegativity

Electronegativity refers to the ability of an atom to attract and hold onto electrons. It plays a significant role in molecular geometry because different electronegativities result in varying electron pair distributions and interactions.

In a molecule like \( \text{H}_2\text{O} \), oxygen exhibits higher electronegativity than sulfur and selenium in \( \text{H}_2\text{S} \) and \( \text{H}_2\text{Se} \). This high electronegativity means oxygen holds electron pairs more tightly, leading to greater repulsions that decrease the \( \text{H}-\text{O}-\text{H} \) bond angle compared to \( \text{H}_2\text{S} \) and \( \text{H}_2\text{Se} \).

As you move down a group in the periodic table, electronegativity decreases. Thus, sulfur and selenium hold onto their electrons less tightly, resulting in decreased repulsion between electron pairs and slightly larger bond angles as observed in \( \text{H}_2\text{S} \) and \( \text{H}_2\text{Se} \). The concept of electronegativity is vital in understanding how chemical behavior and molecular structure are often closely related.

Bond angles

Bond angles are the angles formed between the lines that represent the bonds emanating from a central atom. They are determined by the number of electron pairs around the central atom and the presence of lone pairs.

In an idealized situation, a tetrahedral molecule such as \( \text{CH}_4 \) has bond angles of \(109.5^\circ\). However, when lone pairs are present, like in \( \text{H}_2\text{O} \), the bond angles are reduced. This is due to lone pairs creating stronger repulsions than bonding pairs, compressing the bond angles to \(104.5^\circ\) in the case of water.

As we consider molecules like \( \text{H}_2\text{S} \) and \( \text{H}_2\text{Se} \), the bond angles increase slightly due to the reduced repulsive force of electron pairs, thanks to the lower electronegativity of sulfur and selenium. Understanding how to predict bond angles is crucial for visualizing the three-dimensional arrangement of atoms in molecules and their subsequent chemical properties.