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Chapter 9: Problem 40

(a) The \(\mathrm{PH}_{3}\) molecule is polar. Does this offer experimental proof that the molecule cannot be planar? Explain. (b) It turns out that ozone, \(\mathrm{O}_{3}\), has a small dipole moment. How is this possible, given that all the atoms are the same?

Short Answer

Expert verified
(a) The polarity of the \(\mathrm{PH}_{3}\) molecule alone does not offer experimental proof that the molecule cannot be planar. It is the presence of a lone pair on the central phosphorus atom that makes the molecule non-planar, which in turn contributes to its polarity. (b) Ozone, \(\mathrm{O}_{3}\), can have a small dipole moment even though all the atoms are the same because of the difference in formal charges resulting from varying bond types (single and double bonds) between the oxygen atoms. This difference in bond types leads to an uneven distribution of charges and, as a result, a molecular dipole moment.

Step by step solution

01

(a) Polarity of \(\mathrm{PH}_{3}\) molecule

In order to determine if the polarity of the \(\mathrm{PH}_{3}\) molecule can be considered as proof for it being non-planar, we first need to understand what causes a molecule to be polar. A molecule is considered polar when there is an uneven distribution of charges, leading to an overall molecular dipole moment.
02

(a) Molecular geometry of \(\mathrm{PH}_{3}\)

The central atom in \(\mathrm{PH}_{3}\) molecule is phosphorus (P) which forms bonds with three hydrogen (H) atoms. Phosphorus has five valence electrons, and it shares one electron with each hydrogen atom to form three single covalent bonds. This leaves a lone pair of electrons on the phosphorus atom, which will affect the shape of the molecule. The electron repulsion due to the lone pair will push the hydrogen atoms downward, resulting in a tetrahedral geometry. The bond angle between the hydrogen atoms is approximately 93.5° which deviates from the ideal angle of 109.5° in a perfect tetrahedron due to the presence of the lone pair.
03

(a) Polarity and planarity of \(\mathrm{PH}_{3}\)

Now that we know the molecular geometry of \(\mathrm{PH}_{3}\), we can analyze its polarity. Since the molecule has a lone pair on the central atom, there will be an uneven distribution of charge, and the molecule will be polar. However, the fact that the molecule is polar alone does not necessarily mean that it cannot be planar. There are polar molecules that are planar as well (e.g., water). In the case of \(\mathrm{PH}_{3}\), it is the tetrahedral geometry due to the presence of a lone pair on the central atom that makes it non-planar, not the fact that it is polar.
04

(a) Conclusion for \(\mathrm{PH}_{3}\)

The polarity of the \(\mathrm{PH}_{3}\) molecule itself cannot offer experimental proof that the molecule cannot be planar. It is the presence of the lone pair on the central atom that makes the molecule non-planar, which in turn contributes to its polarity.
05

(b) Molecular geometry of ozone, \(\mathrm{O}_{3}\)

Ozone, \(\mathrm{O}_{3}\), consists of three oxygen atoms. Each oxygen atom forms a bond with one other oxygen atom, and one of the oxygen atoms is doubly bonded to the central oxygen atom, forming a bent or V-shaped molecular geometry with an approximate bond angle of 116°.
06

(b) Polarity of ozone, \(\mathrm{O}_{3}\)

In ozone, there is a formal charge on both the terminal oxygen atoms: one being negative and the other positive due to the difference in electronegativity between the singly bonded and doubly bonded oxygen atoms. This difference creates a molecular dipole moment along the axis connecting the central oxygen atom to the singly bonded oxygen atom, making the molecule polar.
07

(b) Conclusion for ozone, \(\mathrm{O}_{3}\)

It is possible for ozone, \(\mathrm{O}_{3}\), to have a small dipole moment even though all the atoms are the same because of the difference in formal charges due to varying bond types (single and double bonds) between the oxygen atoms. This difference in bond types leads to an uneven distribution of charges and, as a result, a molecular dipole moment.

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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 reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$\mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6}\), you run into a problem. What is it? (c) What could you do to resolve the difficulty in part \((\mathrm{b}) ?(\mathbf{d})\) The molecule \(\mathrm{IF}_{7}\) has a pentagonalbipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\).

(a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, and following the model of Figure 9.46 , how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) It turns out that the difference in energies between the valence atomic orbitals of \(\mathrm{H}\) and \(\mathrm{F}\) are sufficiently different that we can neglect the interaction of the \(1 s\) orbital of hydrogen with the \(2 s\) orbital of fluorine. The \(1 s\) orbital of hydrogen will mix only with one \(2 p\) orbital of fluorine. Draw pictures showing the proper orientation of all three \(2 p\) orbitals on \(\mathrm{F}\) interacting with a \(1 s\) orbital on \(\mathrm{H}\). Which of the \(2 p\) orbitals can actually make a bond with a \(1 s\) orbital, assuming that the atoms lie on the \(z\) -axis? (d) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy-level diagram for HF. These are called "nonbonding orbitals." Sketch the energy-level diagram for HF using this information and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (e) Look at the Lewis structure for HF. Where are the nonbonding electrons?

(a) What does the term diamagnetism mean? (b) How does a diamagnetic substance respond to a magnetic field? (c) Which of the following ions would you expect to be diamagnetic: \(\mathrm{N}_{2}^{2-}, \mathrm{O}_{2}^{2-}, \mathrm{Be}_{2}^{2+}, \mathrm{C}_{2}^{-} ?\)

(a) Write a single Lewis structure for \(\mathrm{SO}_{3}\), and determine the hybridization at the \(\mathrm{S}\) atom. (b) Are there other equivalent Lewis structures for the molecule? (c) Would you expect \(\mathrm{SO}_{3}\) to exhibit delocalized \(\pi\) bonding? Explain.
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