Question

Predict the trend in the \(\mathrm{F}(\) axial \()-\mathrm{A}-\mathrm{F}\) (equatorial) bond angle in the following \(\mathrm{AF}_{n}\) molecules: \(\mathrm{PF}_{5}, \mathrm{SF}_{4}\) and \(\mathrm{ClF}_{3}\)

Short answer

In the given molecules, PF5, SF4, and ClF3, the F(axial)-A-F(equatorial) bond angles are predicted as 90° for all three molecules using the VSEPR theory, as they have trigonal bipyramidal, see-saw, and T-shaped molecular geometries, respectively.

Step-by-step solution

Calculate the number of valence electrons for each molecule

Before determining the electron domain geometry and the molecular geometry, let's find the number of valence electrons for each molecule. - For \(\mathrm{PF}_{5}\): P has 5 valence electrons and each F has 7 valence electrons, so the total is \(5 + (5 \times 7) = 40\) valence electrons. - For \(\mathrm{SF}_{4}\): S has 6 valence electrons and each F has 7 valence electrons, so the total is \(6 + (4 \times 7) = 34\) valence electrons. - For \(\mathrm{ClF}_{3}\): Cl has 7 valence electrons and each F has 7 valence electrons, so the total is \(7 + (3 \times 7) = 28\) valence electrons.

Determine the electron domain geometry and molecular geometry for each molecule

Now, we will use the VSEPR theory to find the electron domain geometry and molecular geometry for each molecule. - \(\mathrm{PF}_{5}\): With 5 bonding groups and no lone pairs, the electron domain geometry is trigonal bipyramid, and the molecular geometry is also trigonal bipyramid. - \(\mathrm{SF}_{4}\): With 4 bonding groups and 1 lone pair, the electron domain geometry is trigonal bipyramid, and the molecular geometry is "see-saw" or distorted tetrahedron. - \(\mathrm{ClF}_{3}\): With 3 bonding groups and 2 lone pairs, the electron domain geometry is trigonal bipyramid, and the molecular geometry is T-shaped.

Determine the F(axial)-A-F(equatorial) bond angles for each molecule

Now that we have the molecular geometry, we can predict the F(axial)-A-F(equatorial) bond angles in these molecules. - \(\mathrm{PF}_{5}\): In a trigonal bipyramid, the F(axial)-P-F(equatorial) bond angle is 90°. As there are no lone pairs causing distortion, this angle remains the same. - \(\mathrm{SF}_{4}\): In the see-saw geometry, the F(axial)-S-F(equatorial) bond angle is still 90° as the lone pair does not influence this angle. - \(\mathrm{ClF}_{3}\): In a T-shaped geometry, the F(axial)-Cl-F(equatorial) bond angle is 90° because the lone pairs force the bonds to remain perpendicular to each other. In conclusion, the F(axial)-A-F(equatorial) bond angles for all three molecules—PF5, SF4, and ClF3—are equal to 90°.

Key concepts

Electron Domain Geometry

Electron domain geometry plays a crucial role in understanding the shape and structure of molecules. It is determined by the arrangement of electron pairs around a central atom, including both bonding pairs that form bonds and lone pairs that do not participate in bonding.

This concept is best explained through the Valence Shell Electron Pair Repulsion (VSEPR) theory, which suggests that electron pairs arrange themselves as far apart as possible to minimize repulsion between them. This results in specific geometric arrangements known as electron domain geometries. Common examples include:

• Linear: With two electron domains, the angle is 180°.
• Trigonal Planar: With three electron domains, the angle is 120°.
• Tetrahedral: With four electron domains, the angle is 109.5°.
• Trigonal Bipyramidal: With five electron domains, as in the mentioned \( ext{PF}_5\), the angles are 120° in the equatorial plane and 90° axial to equatorial.
• Octahedral: With six electron domains, all angles are 90°.

Understanding the electron domain geometry is essential because it forms the foundation for determining the molecular geometry of a molecule.

Molecular Geometry

Once the electron domain geometry is established, we can delve into molecular geometry, which describes the actual shape of the molecule. Unlike electron domain geometry, molecular geometry considers only the positions of atoms, not the lone pairs.

The distinction becomes apparent in molecules like \( ext{SF}_4\) and \( ext{ClF}_3\). These molecules have lone pairs that distort the shape defined by their electron domain geometry:

• Trigonal Bipyramidal with no lone pairs: Results in the same molecular geometry, as seen in \( ext{PF}_5\).
• Trigonal Bipyramidal with one lone pair: The lone pair in \( ext{SF}_4\) forces a "see-saw" shape, differing from the original geometry.
• Trigonal Bipyramidal with two lone pairs: As in \( ext{ClF}_3\), where the molecular shape is T-shaped due to the influence of two lone pairs.

The molecular geometry is significant for predicting how molecules will interact, affecting properties like polarity and reactivity.

Valence Electrons

Valence electrons are the outermost electrons of an atom and are pivotal in determining how atoms bond and form molecules. These electrons are involved in forming bonds, as they can be shared with other atoms, given away, or accepted to achieve a stable electron configuration.

To predict molecular behavior and geometry, calculating the total number of valence electrons in a molecule is the first step. For example:

• Phosphorus in \( ext{PF}_5\) has 5 valence electrons, with each fluorine contributing 7, leading to 40 valence electrons for the molecule.
• Sulfur in \( ext{SF}_4\) has 6 valence electrons, combining with four fluorines' 28 for a total of 34 valence electrons.
• Chlorine in \( ext{ClF}_3\) brings 7 electrons, and three fluorines add 21, totaling 28 valence electrons.

The number and arrangement of valence electrons determine the electronic structure and ultimately the chemical properties and reactions of the molecule.