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

Short Answer

Expert verified
The molecular geometry of each molecule/ion is as follows: 1. PF6-: Octahedral 2. SbFs: Trigonal bipyramidal 3. SF4: Seesaw or distorted tetrahedral All three molecules or ions, PF6-, SbFs, and SF4, have more than one F-A-F bond angle.

Step by step solution

01

1. PF6-

Phosphorus (P) has 5 valence electrons and Fluorine (F) has 7 valence electrons. The PF6- ion has 5+6(7)+1(extra electron) = 48 electrons. In this complex, P is the central atom, surrounded by 6 fluorine atoms. To minimize the repulsion between electron pairs, the molecular geometry of PF6- will be an octahedral, with 6 F-A (P-F) bonds. This structure will have several F-P-F angles of 90 and 180 degrees. Therefore, PF6- has more than one F-A-F bond angle.
02

â‚‚. SbFs

Antimony (Sb) has 5 valence electrons, and Fluorine (F) has 7 valence electrons. The overall molecule has 5+5(7) = 40 electrons. In this molecule, Sb is the central atom, surrounded by 5 fluorine atoms. To minimize the repulsion between electron pairs, the molecular geometry of SbFs is trigonal bipyramidal, with 5 F-A (Sb-F) bonds. SbFs has 3 equatorial (F-Sb-F) angles equal to 120 degrees (at the central plane), and 2 axial (F-Sb-F) angles equal to 180 degrees, perpendicular to the central plane. Therefore, there are multiple F-A-F bond angles in SbFs.
03

₃. SF4

Sulfur (S) has 6 valence electrons and Fluorine (F) has 7 valence electrons. In the SF4 molecule, there are 6+4(7) = 34 electrons. In this molecule, S is the central atom, surrounded by 4 fluorine atoms with one lone pair on S. The molecular geometry of SF4 is called seesaw or distorted tetrahedral. The SF4 molecule has two S-F bond angles in the equatorial plane, equal to 120 degrees; and on each of two axial positions perpendicular (F-S-F) to the central plane, the angle is 180 degrees. Thus, there are more than one F-A-F bond angles in SF4. All three molecules or ions, PF6-, SbFs, and SF4, have more than one F-A-F bond angle.

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Key Concepts

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

Molecular Geometry
Molecular geometry refers to the three-dimensional arrangement of atoms within a molecule. It plays a crucial role in determining the molecule's properties, such as reactivity, polarity, and phase of matter. Understanding molecular geometry allows us to predict the angles between atoms and how they influence molecular interactions.
To determine the molecular geometry, we use the Valence Shell Electron Pair Repulsion Theory (VSEPR). This theory suggests that electron pairs around a central atom arrange themselves to minimize repulsion. For instance, the molecule \( \text{PF}_6^- \) exhibits octahedral geometry, with the phosphorus atom at the center surrounded symmetrically by six fluorine atoms. This results in optimal spacing of electron pairs, reducing repulsion.
In contrast, \( \text{SbF}_5 \) has a trigonal bipyramidal structure, where three fluorine atoms are located in the equatorial plane with bond angles near 120 degrees, while two fluorines are axially positioned, creating 180-degree angles with the central atom. In \( \text{SF}_4 \), the presence of a lone pair forces the molecule into a "seesaw" or distorted tetrahedral shape, influencing the bond angles due to the extra repulsion from the lone pair.
Valence Electrons
Valence electrons are the outermost electrons of an atom and are crucial in forming chemical bonds. In VSEPR Theory, the number of valence electrons informs the arrangement of atoms in a molecule by determining how electron pairs (bonding and lone pairs) will orient themselves in space.
  • In \( \text{PF}_6^- \), phosphorus contributes 5 valence electrons, with each fluorine contributing 7, and an extra electron adds up to 48 electrons in total, influencing the molecular shape.
  • Antimony in \( \text{SbF}_5 \) brings in 5 valence electrons, with each fluorine atom having 7. Collectively, 40 electrons dictate the molecule's trigonal bipyramidal shape.
  • For \( \text{SF}_4 \), sulfur's 6 valence electrons, combined with those from four fluorine atoms, total 34 electrons, resulting in its unique "seesaw" geometry.
By understanding valence electrons, we can predict how atoms bond and the resulting molecular geometry, guiding how molecules interact with each other.
Bond Angles
Bond angles are the angles formed between adjacent bonds at an atom, and they're crucial for defining molecular shape and behavior. Different molecular geometries result in varying bond angles, which can influence physical and chemical properties.
In molecules like \( \text{PF}_6^- \), the octahedral geometry creates bond angles of 90 degrees between adjacent fluorine atoms and 180 degrees across the phosphorus atom. This symmetrical structure helps minimize the repulsion between electron pairs, achieving stability.
The trigonal bipyramidal arrangement in \( \text{SbF}_5 \) results in 120-degree angles between equatorial bonds and 180-degree angles between axial bonds. This asymmetry gives \( \text{SbF}_5 \) its distinct geometric properties.
In \( \text{SF}_4 \), with its seesaw shape, the lone pair slightly distorts bond angles from the idealized values. Bonds within the equatorial plane measure approximately 120 degrees, while those in the axial position are around 180 degrees, influenced by the lone electron pair's repulsion. Understanding these angles is crucial for predicting molecular behavior and interactions.

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Most popular questions from this chapter

(a) What does the term paramagnetism mean? (b) How can one determine experimentally whether a substance is paramagnetic? (c) Which of the following ions would you expect to be paramagnetic: \(\mathrm{O}_{2}^{+}, \mathrm{N}_{2}{ }^{2-}, \mathrm{Li}_{2}^{+}, \mathrm{O}_{2}^{2-} ?\) For those ions that are paramagnetic, determine the number of unpaired electrons.

The following is part of a molecular orbital energy-level diagram for MOs constructed from 1 s atomic orbitals. (a) What labels do we use for the two MOs shown? (b) For which of the following molecules or ions could this be the energy-level diagram: $$ \mathrm{H}_{2} \mathrm{He}_{2}, \mathrm{H}_{2}^{+}, \mathrm{He}_{2}^{+}, \mathrm{or} \mathrm{H}_{2}^{-} ? $$ (c) What is the bond order of the molecule or ion? (d) If an electron is added to the system, into which of the MOs will it be added? [Section 9.7\(]\)

(a) Does \(C S_{2}\) have a dipole moment? If so, in which direction does the net dipole point? (b) Does \(\mathrm{SO}_{2}\) have a dipole moment? If so, in which direction does the net dipole point?

Butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6},\) is a planar molecule that has the following carbon-carbon bond lengths: $$ \mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{134 \mathrm{pm}} \mathrm{CH}=\mathrm{CH}_{2} $$ (a) Predict the bond angles around each of the carbon atoms and sketch the molecule. (b) From left to right, what is the hybridization of each carbon atom in butadiene? (c) The middle \(\mathrm{C}-\mathrm{C}\) bond length in butadiene \((148 \mathrm{pm})\) is a little shorter than the average \(\mathrm{C}-\mathrm{C}\) single bond length (154 pm). Does this imply that the middle \(\mathrm{C}-\mathrm{C}\) bond in butadiene is weaker or stronger than the average \(\mathrm{C}-\mathrm{C}\) single bond? (d) Based on your answer for part (c), discuss what additional aspects of bonding in butadiene might support the shorter middle \(\mathrm{C}-\mathrm{C}\) bond.

Consider the following \(\mathrm{XF}_{4}\) ions: \(\mathrm{PF}_{4}^{-}, \mathrm{BrF}_{4}^{-}, \mathrm{ClF}_{4}^{+},\) and \(\mathrm{AlF}_{4}^{-}\) (a) Which of the ions have more than an octet of electrons around the central atom? (b) For which of the ions will the electron-domain and molecular geometries be the same? (c) Which of the ions will have an octahedral electron-domain geometry? (d) Which of the ions will exhibit a see-saw molecular geometry?

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