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Chapter 9: Problem 32

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{PF}_{6}^{-}, \mathrm{SbF}_{5}, \mathrm{SF}_{4}\) ?

Short Answer

Expert verified
Both \( \mathrm{SbF}_{5} \) and \( \mathrm{SF}_{4} \) have more than one \( \mathrm{F}-\mathrm{A}-\mathrm{F} \) bond angle.

Step by step solution

01

Determine Molecular Geometry of Each Species

First, identify the central atom in each molecule or ion and determine the electron-domain geometry according to the VSEPR (Valence Shell Electron Pair Repulsion) theory. \( \mathrm{PF}_{6}^{-} \) has a central \( \mathrm{P} \) atom with 6 bonds, \( \mathrm{SbF}_{5} \) has a central \( \mathrm{Sb} \) atom with 5 bonds, and \( \mathrm{SF}_{4} \) has a central \( \mathrm{S} \) atom with 4 bonds and a lone pair. The geometries are octahedral for \( \mathrm{PF}_{6}^{-} \), trigonal bipyramidal for \( \mathrm{SbF}_{5} \), and see-saw for \( \mathrm{SF}_{4} \).
02

Analyze Bond Angles in \( \mathrm{PF}_{6}^{-} \)

In the octahedral geometry of \( \mathrm{PF}_{6}^{-} \), all \( \mathrm{F}-\mathrm{P}-\mathrm{F} \) bond angles are identical and equal to \( 90^\circ \) or \( 180^\circ \). This structure does not have more than one angle type.
03

Analyze Bond Angles in \( \mathrm{SbF}_{5} \)

The trigonal bipyramidal geometry in \( \mathrm{SbF}_{5} \) has different \( \mathrm{F}-\mathrm{Sb}-\mathrm{F} \) bond angles: equatorial \( 120^\circ \) and axial \( 90^\circ \). Therefore, more than one type of bond angle exists.
04

Analyze Bond Angles in \( \mathrm{SF}_{4} \)

The see-saw shape of \( \mathrm{SF}_{4} \) results in varying bond angles due to the presence of a lone pair. The \( \mathrm{F}-\mathrm{S}-\mathrm{F} \) bond angles differ between equatorial positions (approx. \( 120^\circ \)), axial positions (\( 90^\circ \)), and others are distorted because of lone pair repulsion, resulting in multiple distinct bond angle types.
05

Conclusion

Both \( \mathrm{SbF}_{5} \) and \( \mathrm{SF}_{4} \) have more than one type of \( \mathrm{F}-\mathrm{A}-\mathrm{F} \) bond angle, while \( \mathrm{PF}_{6}^{-} \) does not.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molecular Geometry
Molecular geometry is like a three-dimensional map showing how atoms are arranged around a central atom in a molecule. This shape is crucial because it affects how molecules interact, their polarity, and even their chemical reactivity.
Based on the number of bonds and lone electron pairs around a central atom, we can predict molecular geometry using the VSEPR theory.
  • For example, in the case of \( \text{PF}_6^- \), the central phosphorus (P) atom is bonded to six fluorine atoms in an octahedral shape. This uniform geometric figure is known for its equal bond angles.
  • For \( \text{SbF}_5 \), the central antimony (Sb) atom attaches to five fluorine atoms, resulting in a trigonal bipyramidal shape.
  • Finally, \( \text{SF}_4 \) features a sulfur (S) atom with four bonded fluorine atoms and a lone pair of electrons, forming what's called a 'see-saw' shape.
In sum, molecular geometry gives us a clear picture of a molecule's possible shape and its underlying chemical behavior.
Bond Angles
Bond angles are the angles between adjacent lines representing bonds on a 2D projection of a molecule. They are a vital aspect of understanding molecular geometry because they can indicate how a molecule might behave chemically and physically. These angles are measured in degrees and give insights into the spatial arrangement of atoms.
In molecules like \( \text{PF}_6^- \), all F-P-F bond angles are consistent at either 90掳 or 180掳 because of its symmetric octahedral shape.
On the other hand:
  • \( \text{SbF}_5 \) exhibits bond angles of 120掳 between the equatorial F atoms and 90掳 between axial and equatorial fluorines due to its trigonal bipyramidal shape.
  • In \( \text{SF}_4 \), the lone pair causes deviation from standard angles, with bond angles around 90掳 and slightly less than 120掳 as the structure favors a 'see-saw' conformation.
Seeing these angle variations helps us predict interactions and properties of different molecules.
Electron-Domain Geometry
Electron-domain geometry provides a broader aspect on how the regions of electron density around the central atom are distributed. This not only includes bonds but also lone electron pairs, which significantly influences a molecule's overall shape.
The VSEPR theory uses the concept of electron pairs repelling each other to explain molecular structure. For instance:
  • In \( \text{PF}_6^- \), the electron-domain geometry is octahedral, with six bonding pairs and no lone pair, ensuring all domains are evenly spaced out.
  • For \( \text{SbF}_5 \), it possesses a trigonal bipyramidal electron-domain geometry with no lone pairs to disrupt angles.
  • However, in \( \text{SF}_4 \), the electron-domain geometry is also trigonal bipyramidal, but the presence of a lone pair at the equatorial position creates a see-saw shape as it pushes the bonded electrons, altering the expected angles.
This understanding helps in predicting the molecule's reactivity and stability by assessing how lone pairs can distort structures.

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Most popular questions from this chapter

How would you expect the extent of overlap of the bonding atomic orbitals to vary in the series IF, ICl, IBr, and \(\mathrm{I}_{2}\) ? Explain your answer.

In the sulphate ion, \(\mathrm{SO}_{4}{ }^{2-}\), the sulphur atom is the central atom with the other 4 oxygen atoms attached to it. (a) Draw a Lewis structure for the sulphate ion. (b) What hybridization is exhibited by the \(S\) atom? (c) Are there multiple equivalent resonance structures for the ion? (d) How many electrons are in the \(\pi\) system of the ion?

Ethyl propanoate, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{3},\) gives a fruity pineapple-like smell. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\sigma\) and how many \(\pi\) bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around the carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X}: \mathrm{PF}_{3}, 96.3^{\circ} ; \mathrm{PCl}_{3}, 100.3^{\circ}\); \(\mathrm{PBr}_{3}, 101.0^{\circ} ; \mathrm{PI}_{3}, 102.0^{\circ} .\) The trend is generally attributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the halide electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCl}_{4}\).

Consider the molecule \(\mathrm{PF}_{4}\) Cl. (a) Draw a Lewis structure for the molecule, and predict its electron-domain geometry. (b) Which would you expect to take up more space, a \(\mathrm{P}-\mathrm{F}\) bond or a \(\mathrm{P}-\mathrm{Cl}\) bond? Explain. (c) Predict the molecular geometry of \(\mathrm{PF}_{4}\) Cl. How did your answer for part (b) influence your answer here in part (c)? (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?
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