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Lithium provides a very simple example of the effect of oxidation state on overall size. Perform HF/6-31G* calculations on lithium cation, lithium atom, and lithium anion, and compare the three electron density surfaces corresponding to enclosure of \(99 \%\) of the total electron density. Which is smallest? Which is largest? How does the size of lithium relate to the number of electrons? Which surface most closely resembles a conventional space-filling model? What, if anything does this tell you about the kinds of molecules that were used to establish the space-filling radius for lithium?

Short Answer

Expert verified
The HF/6-31G* calculations on lithium cation (Li+), lithium atom (Li), and lithium anion (Li-) result in three electron density surfaces corresponding to the enclosure of 99% of the total electron density. It is observed that as the oxidation state of lithium increases, the size of the surface decreases, while the opposite occurs when the oxidation state decreases due to changes in electron-electron repulsions. Among the three surfaces, the lithium atom (Li) electron density surface most closely resembles a conventional space-filling model, which represents the van der Waals radius of the atom. This analysis indicates that the space-filling radius for lithium may have been established based on molecules containing lithium in its neutral oxidation state.

Step by step solution

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1. Perform HF/6-31G* calculations

Start by performing the Hartree-Fock calculations using the 6-31G* basis set for Li+, Li, and Li-. You can do this using software such as Gaussian, ORCA, or any other computational chemistry software.
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2. Compare electron density surfaces

Once you have the results from your calculations, generate the electron density surfaces for Li+, Li, and Li- at the 99% total electron density level. Compare the size of these surfaces to determine which one is smallest and which one is largest.
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3. Analyze the size of lithium and its relationship with the number of electrons

Analyze the results to see how the size of the electron density surfaces (hence the size of lithium species) is related to the number of electrons. Generally, as the oxidation state increases (loses electrons), the size decreases due to the reduction of electron-electron repulsions. On the other hand, as the oxidation state decreases (gains electrons), the size increases due to the increase in electron-electron repulsions.
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4. Determine which surface most closely resembles a conventional space-filling model

Identify which electron density surface best resembles a conventional space-filling model out of the three surfaces generated for Li+, Li, and Li-. The space-filling model generally represents the van der Waals radius of the atom or ion.
05

5. Draw conclusions regarding the molecules responsible for the space-filling radius of lithium

Based on your comparison of the electron density surfaces and identification of which surface resembles a conventional space-filling model, infer what kinds of molecules may have been used to establish the space-filling radius for lithium. Consider the oxidation states of lithium in those molecules and the role they play in determining the size of lithium species.

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Key Concepts

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

Lithium Cation
Lithium cation, often denoted as Li+, is a lithium atom that has lost an electron. This loss of an electron gives the cation a positive charge.

Due to the absence of an electron, the electron-electron repulsions that usually result in a larger atomic radius are reduced. This makes a cation smaller than its corresponding neutral atom.
  • The HF/6-31G* calculation for Li+ involves computing its electron density surface and understanding its impact on the size and charge distribution.
  • It is important to note that Li+, being smaller due to its positive charge, binds more effectively with negatively charged species compared to its neutral counterpart.
Understanding the nature of Li+ is crucial for interpreting how various oxidation states affect the physical properties of lithium, particularly its size.
Electron Density Surfaces
Electron density surfaces represent the distribution of electrons around an atom within a molecule. In chemistry, these surfaces help visualize the spatial regions with high probability of finding electrons.

For lithium species, electron density calculations are performed at 99% of the total electron density to best approximate atomic size.
  • These surfaces allow scientists to understand atomic interactions and reactivity, essential for molecular modeling and drug design.
  • In lithium cation (Li+), lithium atom (Li), and lithium anion (Li-), electron density surfaces show clear differences in size and shape.
The comparison of various electron density surfaces, therefore, offers insight into phenomena such as atomic and molecular stability as well as reactivity.
Oxidation State
The oxidation state in chemistry refers to the degree of oxidation of an atom within a substance.

It’s crucial in predicting the chemical behavior of an element, including lithium, as seen in various oxidation states:
  • As electrons are added or removed, the oxidation state changes, impacting the molecular geometry and electron density surfaces.
  • Higher oxidation states (e.g., Li+) result in fewer electrons, leading to reduced repulsive forces and, thus, smaller atomic size.
  • Conversely, a lower oxidation state (e.g., Li-) means more electrons, increasing repulsion and atomic size.
By analyzing these relationships, chemists can predict how changes in oxidation state will affect molecular interactions and properties.
Space-Filling Model
The space-filling model is a way to represent molecules, emphasizing the relative sizes of atoms and their spatial arrangement.

This model shows atoms as spheres whose radii are proportional to their van der Waals radii, offering an intuitive visualization of their real 3D form.
  • In lithium species, space-filling models can closely resemble the most accurate electron density surface. This is crucial for correctly representing molecular shape and electronic environment.
  • The model provides clarity on how atoms within a molecule pack and interact with one another, significantly influencing their chemical properties and stability.
Understanding which electron density surface of lithium resembles this configuration aids in identifying the most suitable representation based on electron configurations and atomic size.

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Most popular questions from this chapter

Hydrazine would be expected to adopt a conformation in which the NH bonds stagger. There are two likely candidates, one with the lone pairs on nitrogen anti to each other and the other with the lone pairs gauche: On the basis of the same arguments made in VSEPR theory (electron pairs take up more space than bonds) you might expect that anti hydrazine would be the preferred structure. a. Obtain energies for the anti and gauche conformers of hydrazine using the HF/6-31G* model. Which is the more stable conformer? Is your result in line with what you expect from VSEPR theory? You can rationalize your result by recognizing that when electron pairs interact they form combinations, one of which is stabilized (relative to the original electron pairs) and one of which is destabilized. The extent of destabilization is greater than that of stabilization, meaning that overall interaction of two electron pairs is unfavorable energetically: b. Measure the energy of the highest occupied molecular orbital (the HOMO) for each of the two hydrazine conformers. This corresponds to the higher energy (destabilized) combination of electron pairs. Which hydrazine conformer (anti or gauche) has the higher HOMO energy? Is this also the higher energy conformer? If so, is the difference in HOMO energies comparable to the difference in total energies between the conformers?

VSEPR (valence state electron pair repulsion) theory was formulated to anticipate the local geometry about an atom in a molecule (see discussion in Section 25.1). All that is required is the number of electron pairs surrounding the atom, broken down into bonded pairs and nonbonded (lone) pairs. For example, the carbon in carbon tetrafluoride is surrounded by four electron pairs, all of them tied up in \(\mathrm{CF}\) bonds, whereas the sulfur in sulfur tetrafluoride is surrounded by five electron pairs, four of which are tied up in SF bonds with the fifth being a lone pair. VSEPR theory is based on two simple rules. The first is that electron pairs (either lone pairs or bonds) will seek to avoid each other as much as possible. Thus, two electron pairs will lead to a linear geometry, three pairs to a trigonal planar geometry, four pairs to a tetrahedral geometry, five pairs to a trigonal bipyramidal geometry, and six pairs to an octahedral geometry. Although this knowledge is sufficient to assign a geometry for a molecule such as carbon tetrafluoride (tetrahedral), it is not sufficient to specify the geometry of a molecule such as sulfur tetrafluoride. Does the lone pair assume an equatorial position on the trigonal bipyramid leading to a seesaw geometry, or an axial position leading to a trigonal pyramidal geometry? The second rule, that lone pairs take up more space than bonds, clarifies the situation. The seesaw geometry in which the lone pair is \(90^{\circ}\) to two of the SF bonds and \(120^{\circ}\) to the other two bonds is preferable to the trigonal pyramidal geometry in which three bonds are \(90^{\circ}\) to the lone pair. Although VSEPR theory is easy to apply, its results are strictly qualitative and often of limited value. For example, although the model tells us that sulfur tetrafluoride adopts a seesaw geometry, it does not reveal whether the trigonal pyramidal structure (or any other structure) is an energy minimum, and if it is, what its energy is relative to the seesaw form. Also it has little to say when more than six electron pairs are present. For example, VSEPR theory tells us that xenon hexafluoride is not octahedral, but it does not tell us what geometry the molecule actually assumes. Hartree-Fock molecular orbital calculations provide an alternative. a. Optimize the structure of \(\mathrm{SF}_{4}\) in a seesaw geometry \(\left(C_{2 v} \text { symmetry }\right)\) using the HF/3-21G model and calculate vibrational frequencies (the infrared spectrum). This calculation is necessary to verify that the energy is at a minimum. Next, optimize the geometry of \(\mathrm{SF}_{4}\) in a trigonal pyramidal geometry and calculate its vibrational frequencies. Is the seesaw structure an energy minimum? What leads you to your conclusion? Is it lower in energy than the corresponding trigonal pyramidal structure in accordance with VSEPR theory? What is the energy difference between the two forms? Is it small enough that both might actually be observed at room temperature? Is the trigonal pyramidal structure an energy minimum? b. Optimize the geometry of \(\mathrm{XeF}_{6}\) in an octahedral geometry \(\left(\mathrm{O}_{\mathrm{h}} \text { symmetry }\right)\) using the HF/3-21G model and calculate vibrational frequencies. Next, optimize \(\mathrm{XeF}_{6}\) in a geometry that is distorted from octahedral (preferably a geometry with \(\left.C_{1} \text { symmetry }\right)\) and calculate its vibrational frequencies. Is the octahedral form of \(\mathrm{XeF}_{6}\) an energy minimum? What leads you to your conclusion? Does distortion lead to a stable structure of lower energy?

Benzyne has long been implicated as an intermediate in nucleophilic aromatic substitution, for example, Although the geometry of benzyne has yet to be conclusively established, the results of a \(^{13} \mathrm{C}\) labeling experiment leave little doubt that two (adjacent) positions on the ring are equivalent: There is a report, albeit controversial, that benzyne has been trapped in a low-temperature matrix and its infrared spectrum recorded. Furthermore, a line in the spectrum at \(2085 \mathrm{cm}^{-1}\) has been assigned to the stretching mode of the incorporated triple bond. Optimize the geometry of benzyne using the HF/6-31G* model and calculate vibrational frequencies. For reference, perform the same calculations on 2 -butyne. Locate the \(\mathrm{C} \equiv \mathrm{C}\) stretching frequency in 2 -butyne and determine an appropriate scaling factor to bring it into agreement with the corresponding experimental frequency \(\left(2240 \mathrm{cm}^{-1}\right) .\) Then identify the vibration corresponding to the triple-bond stretch in benzyne and apply the same scaling factor to this frequency. Finally, plot the calculated infrared spectra of both benzyne and 2-butyne. Does your calculated geometry for benzyne incorporate a fully formed triple bond? Compare with the bond in 2 -butyne as a standard. Locate the vibrational motion in benzyne corresponding to the triple bond stretch. Is the corresponding (scaled) frequency significantly different \(\left(>100 \mathrm{cm}^{-1}\right)\) from the frequency assigned in the experimental investigation? If it is, are you able to locate any frequencies from your calculation that would fit with the assignment of a benzyne mode at \(2085 \mathrm{cm}^{-1} ?\) Elaborate. Does the calculated infrared spectrum provide further evidence for or against the experimental observation? (Hint: Look at the intensity of the triple-bond stretch in 2-butyne.)

It is well known that cyanide acts as a "carbon" and not a "nitrogen" nucleophile in \(\mathrm{S}_{\mathrm{N}} 2\) reactions, for example, How can this behavior be rationalized with the notion that nitrogen is in fact more electronegative than carbon and, therefore, would be expected to hold any excess electrons? a. Optimize the geometry of cyanide using the HF/3-21G model and examine the HOMO. Describe the shape of the HOMO of cyanide. Is it more concentrated on carbon or nitrogen? Does it support the picture of cyanide acting as a carbon nucleophile? If so, explain why your result is not at odds with the relative electronegativities of carbon and nitrogen. Why does iodide leave following nucleophilic attack by cyanide on methyl iodide? b. Optimize the geometry of methyl iodide using the HF/3-21G model and examine the LUMO. Describe the shape of the LUMO of methyl iodide. Does it anticipate the loss of iodide following attack by cyanide? Explain.

One of the most powerful attractions of quantum chemical calculations over experiments is their ability to deal with any molecular system, stable or unstable, real or imaginary. Take as an example the legendary (but imaginary) kryptonite molecule. Its very name gives us a formula, \(\mathrm{KrO}_{2}^{2-}\), and the fact that this species is isoelectronic with the known linear molecule, \(\mathrm{KrF}_{2}\), suggests that it too should be linear. a. Build \(\mathrm{KrF}_{2}\) as a linear molecule \((\mathrm{F}-\mathrm{Kr}-\mathrm{F}),\) optimize its geometry using the HF/6-31G* model, and calculate vibrational frequencies. Is the calculated KrF bond distance close to the experimental value \((1.89 \AA) ?\) Does the molecule prefer to be linear or does it want to bend? Explain how you reached this conclusion. b. Build \(\mathrm{KrO}_{2}^{2-}\) as a linear molecule (or as a bent molecule if the preceding analysis has shown that \(\mathrm{KrF}_{2}\) is not linear), optimize its structure using the HF/6-31G* model, and calculate vibrational frequencies. What is the structure of \(\mathrm{KrO}_{2}^{2-} ?\)

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