Question

Halogens combine with one another to produce interhalogens such as BrF \(_{3}\). Sketch a possible molecular structure for this molecule, and decide if the \(\mathrm{F}-\mathrm{Br}-\mathrm{F}\) bond angles will be less than or greater than ideal.

Short answer

The BrF3 molecular structure is T-shaped, and F-Br-F bond angles are less than 90°.

Step-by-step solution

Determine the Total Number of Valence Electrons

To sketch the molecular structure of \(\text{BrF}_3\), we first need to calculate the total number of valence electrons. Bromine (Br) has 7 valence electrons and each Fluorine (F) atom also has 7 valence electrons. Hence, the total number of valence electrons in \(\text{BrF}_3\) is \(7 + 3 \times 7 = 28\) electrons.

Sketch the Basic Structure

We determine the central atom and lay out the basic structure. Since Bromine is less electronegative than Fluorine, it becomes the central atom. We connect Bromine with three Fluorine atoms, forming three single bonds which utilise 6 of the 28 valence electrons available.

Place Remaining Electrons as Lone Pairs

After creating the bonds, we have used 6 electrons. Now, we have 28 - 6 = 22 electrons left. These electrons are placed as lone pairs around the central Bromine atom. Given that each bond takes the equivalent of 2 electrons, Bromine will prefer to have additional lone pairs placed around it to achieve an octet or expanded valence. Br will have two lone pairs, and each F will also take the remaining electrons to fill their octets.

Consider Electron Domain Geometry

With three bonding pairs and two lone pairs around the central Bromine, we examine the electron domain geometry. It exhibits a "trigonal bipyramidal" electron geometry. However, the presence of lone pairs affects the actual bond angles, leading to deviations from the ideal 90° or 120° bond angles.

Determine Actual Molecular Geometry and Bond Angles

Lone pairs occupy more space than bonds, pushing the \(\text{F}\) atoms closer together. Therefore, the \(\mathrm{F}-\mathrm{Br}-\mathrm{F}\) bond angles will be less than the ideal 120° for equatorial positions on a trigonal bipyramidal geometry. Indeed, the molecular shape will be "T-shaped."

Key concepts

Molecular Structure

Understanding the molecular structure of a compound involves identifying how atoms are connected and arranged spatially. In the case of interhalogens like \( ext{BrF}_3\), the molecular structure is determined by the number of atoms bonded and the arrangement of electron pairs. Here, bromine (Br) serves as the central atom because it is less electronegative than fluorine (F).
This configuration features bromine bonded to three fluorine atoms. The significant factor here is the electron pairs around bromine, which influence the overall shape. Knowing the molecular structure helps predict other chemical properties and reactions.

Valence Electrons

Valence electrons are the outermost electrons of an atom. They determine how an atom interacts and bonds with others. For interhalogens like \( ext{BrF}_3\), knowing the number of valence electrons is crucial to sketch its molecular structure. Bromine has 7 valence electrons, similar to each fluorine atom, which also holds 7.
Adding them, the total valence electron count in \( ext{BrF}_3\) is \(7 + 3 \times 7 = 28\) electrons. This electron count guides how the atoms are bonded: three bromine-fluorine single bonds use 6 electrons (3 pairs), with the remaining electrons allocating as lone pairs. Such arrangements define an atom's approach to achieving a stable configuration.

Electron Geometry

Electron geometry refers to the spatial configuration of electron domains around a central atom. For \( ext{BrF}_3\), bromine, being the central atom, showcases a trigonal bipyramidal electron geometry. This shape is determined by all regions of electron density, including bonds and lone pairs.
With three bonding pairs and two lone pairs positioned around bromine, the electron geometry sets the foundation for the molecular shape. The arrangement seeks to minimize repulsion, allowing lone pairs to occupy equatorial positions, leaving a more definitive shape with potentially deviated bond angles from those predicted by ideal models.

Bond Angles

In molecular structures, bond angles are crucial for understanding the shape and reactivity of a molecule. In \( ext{BrF}_3\), bond angles are affected by both bonding and lone electron pairs. Typically, one might expect 120° bond angles in a trigonal bipyramidal setup for equatorial positions.
However, lone pairs exert greater repulsive forces than bonding pairs, leading to smaller bond angles. Their presence makes the \( ext{F-Br-F}\) angles less than the ideal 120°, resulting in a "T-shaped" molecular geometry. Such deviations help predict the molecule’s behavior and interactions.