Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Ammonia provides a particularly simple example of the dependence of vibrational frequencies on the atomic masses and of the use of vibrational frequencies to distinguish between a stable molecule and a transition state. First examine the vibrational spectrum of pyramidal ammonia ("ammonia" on the precalculated Spartan file). a. How many vibrational frequencies are there? How does this number relate to the number of atoms? Are all fre- quencies real numbers or are one or more imaginary numbers? Describe the motion associated with each frequency and characterize each as being primarily bond stretching, angle bending, or a combination of the two. Is bond stretching or angle bending easier? Do the stretching motions each involve a single \(\mathrm{NH}\) bond or do they involve combinations of two or three bonds? b. Next, consider changes to the vibrational frequencies of ammonia as a result of substituting deuteriums for hydrogens ("perdeuteroammonia" on the precalculated Spartan file \() .\) Are the frequencies in \(\mathrm{ND}_{3}\) larger, smaller, or unchanged from those in \(\mathrm{NH}_{3}\) ? Are any changes greater for motions that are primarily bond stretching or motions that are primarily angle bending? c. Finally, examine the vibrational spectrum of an ammonia molecule that has been constrained to a planar geometry ("planar ammonia"' on the Spartan download). Are all the frequencies real numbers? If not, describe the motions associated with any imaginary frequencies and relate them to the corresponding motion(s) in the pyramidal equilibrium form.

Short Answer

Expert verified
In summary, pyramidal ammonia has 6 vibrational frequencies, all of which are real numbers, with some involving bond stretching and angle bending. Angle bending is generally easier in this molecule. When comparing perdeuteroammonia (ND3) with ammonia (NH3), the vibrational frequencies of ND3 are smaller due to the larger atomic mass of deuterium, especially in bond stretching motions. For planar ammonia, there is at least one imaginary frequency associated with the unstable planar geometry, corresponding to the nitrogen atom moving between planes of hydrogen atoms, and eventually resulting in the restoration of the pyramidal structure.

Step by step solution

01

Determine Number of Vibrational Frequencies

According to vibrational analysis, the number of vibrational frequencies for a nonlinear molecule is given by \(3N-6\), where N is the number of atoms. In the case of NH3, N = 4 (1 nitrogen atom and 3 hydrogen atoms), so the number of vibrational frequencies = \(3 \cdot 4 - 6\) = 6.
02

Analyzing Real/Imaginary Frequencies

All expected vibrational frequencies of pyramidal ammonia should be real numbers since it is a stable molecular configuration.
03

Describing and Characterizing Motions

The motions of these vibrational frequencies can be described and characterized as follows (this might differ slightly): 1. Frequency 1 - Bond bending: Two N-H bonds bending in and out. 2. Frequency 2 - Bond bending: Three N-H bonds bending in and out simultaneously. 3. Frequency 3 - Bond stretching: All three N-H bonds stretching together symmetrically. 4. Frequency 4 - Bond stretching: Two N-H bonds stretching and the third one contracting. 5. Frequency 5 - Bond stretching: One N-H bond stretching and the other two contracting. 6. Frequency 6 - Mixture of both (involving both bond stretching and angular bending). Angle bending in pyramidal ammonia is generally easier than bond stretching, as bending vibrations typically have lower energy. b.
04

Comparing Vibrational Frequencies

The vibrational frequencies of ND3 are smaller compared to those of NH3 since the masses of deuterium atoms are larger. This has a stronger effect on the frequencies related to bond stretching motions since these motions are more sensitive to changes in the atomic mass than angular bending motions. c.
05

Real/Imaginary Frequencies for Planar Ammonia

Planar ammonia is a vibrationally unstable (transition-state) structure, so it is expected to have at least one imaginary frequency. This indicates that the molecule tends to undergo motion in the direction associated with the imaginary frequency, leading to the more stable pyramidal geometry.
06

Motions Associated with the Imaginary Frequencies

The imaginary frequency in planar ammonia often corresponds to motion of the nitrogen atom switching between one plane of hydrogen atoms to another, which relates to the angular bending and restoring of the molecular structure to the pyramidal equilibrium form.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Vibrational Frequencies
Vibrational frequencies of a molecule help us understand its dynamic movements. For a nonlinear molecule like ammonia (NH_3), the number of vibrational frequencies is calculated using the formula \(3N-6\), where \(N\) is the number of atoms. In NH_3, \(N = 4\) (1 nitrogen and 3 hydrogen atoms), resulting in six vibrational frequencies. These frequencies are incredibly valuable as they demonstrate how different atoms in the molecule move, whether through stretching or bending.
  • Frequencies involving bond bending often feature atoms moving in and out in a synchronized way, affecting the bond angles.
  • Bond stretching frequencies involve changes in distance between bonded atoms, reflecting the bond length's slight expansion and contraction.
The NH_3 molecule is stable in its pyramidal form, thus all vibrational frequencies are real numbers. Any presence of imaginary frequencies would indicate a transition state, implying an unstable configuration.
Molecular Geometry
The molecular geometry of ammonia is essential when discussing its vibrational characteristics. In its equilibrium state, ammonia ( NH_3 ) has a pyramidal shape with a nitrogen atom sitting above the base formed by three hydrogen atoms. This structure leads to specific vibrational movements:
  • Bond bending occurs as the nitrogen atom shifts slightly, altering the angles between the nitrogen and hydrogen atoms.
  • Bond stretching happens when the bonds between nitrogen and each hydrogen atom stretch or compress.
When ammonia is forced into a planar geometry, it results in a less stable structure, evident by the occurrence of imaginary vibrational frequencies. This geometry has the nitrogen atom in the same plane as the hydrogen atoms, which is a transition state rather than the stable pyramidal form. Imaginary frequencies in this state denote instability, showing how the molecule naturally reverts to the pyramidal configuration.
Isotopic Substitution
Isotopic substitution involves replacing an atom in a molecule with another isotope, such as substituting hydrogen with deuterium in ammonia to form perdeuteroammonia ( ND_3 ). This substitution has a notable effect on vibrational frequencies:
  • The larger mass of deuterium compared to hydrogen reduces vibrational frequencies, especially those related to bond stretching.
  • Frequencies associated with angle bending are slightly less affected, as bending motions are generally less mass-dependent.
This change is crucial because it shows how vibrational spectra can be fine-tuned by the molecular mass, providing insights into molecular dynamics and stability. In essence, the heavier isotope adjusts the vibrational motion, typically slowing it down due to the increased mass.
Bond Stretching and Angle Bending
Vibrational motions in molecules involve both bond stretching and angle bending, each playing a key role in defining the vibrational spectrum. In ammonia, these motions represent:
  • *Bond Stretching*: This occurs when the distance between nitrogen and hydrogen atoms in NH_3 alters. Stretching can involve one bond changing while the others contract or multiple bonds adjusting simultaneously.
  • *Angle Bending*: This motion represents changes in the NH bond angles. The nitrogen atom may move closer or farther from the plane of hydrogen atoms, altering the bond angles and leading to angle bending vibrations.
Generally, angle bending requires less energy than bond stretching, making bending vibrations occur at lower frequencies. Understanding these motions helps to analyze the stability and dynamic behavior of the molecule, offering insights into molecular vibrations and their impact on molecular properties.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Pyramidal inversion in the cyclic amine aziridine is significantly more difficult than inversion in an acyclic amine, for example, requiring \(80 \mathrm{kJ} / \mathrm{mol}\) versus \(23 \mathrm{kJ} / \mathrm{mol}\) in dimethylamine according to HF/6-31G* calculations. One plausible explanation is that the transition state for inversion needs to incorporate a planar trigonal nitrogen center, which is obviously more difficult to achieve in aziridine, where one bond angle is constrained to a value of around \(60^{\circ},\) than it is in dimethylamine. Such an interpretation suggests that the barriers to inversion in the corresponding four- and fivemembered ring amines (azetidine and pyrrolidine) should also be larger than normal and that the inversion barrier in the six-membered ring amine (piperidine) should be quite close to that for the acyclic. Optimize the geometries of aziridine, azetidine, pyrrolidine, and piperidine using the HF/6-31G* model. Starting from these optimized structures, provide guesses at the respective inversion transition states by replacing the tetrahedral nitrogen center with a trigonal center. Obtain transition states using the same Hartree-Fock model and calculate inversion barriers. Calculate vibrational frequencies to verify that you have actually located the appropriate inversion transition states. Do the calculated inversion barriers follow the order suggested in the preceding figure? If not, which molecule(s) appear to be anomalous? Rationalize your observations by considering other changes in geometry from the amine to the transition state.

The three vibrational frequencies in \(\mathrm{H}_{2} \mathrm{O}\left(1595,3657, \text { and } 3756 \mathrm{cm}^{-1}\right)\) are all much larger than the corresponding frequencies in \(\mathrm{D}_{2} \mathrm{O}(1178,1571,\) and \(2788 \mathrm{cm}^{-1}\) ). This follows from the fact that vibrational frequency is given by the square root of a (mass-independent) quantity, which relates to the curvature of the energy surface at the minima, divided by a quantity that depends on the masses of the atoms involved in the motion. As discussed in Section \(26.8 .4,\) vibrational frequencies enter into both terms required to relate the energy obtained from a quantum chemical calculation (stationary nuclei at 0 K) to the enthalpy obtained experimentally (vibrating nuclei at finite temperature), as well as the entropy required to relate enthalpies to free energies. For the present purpose, focus is entirely on the so-called zero point energy term, that is, the energy required to account for the latent vibrational energy of a molecule at \(0 \mathrm{K}\) The zero point energy is given simply as the sum over individual vibrational energies (frequencies). Thus, the zero point energy for a molecule in which isotopic substitution has resulted in an increase in mass will be reduced from that in the unsubstituted molecule: A direct consequence of this is that enthalpies of bond dissociation for isotopically substituted molecules (light to heavy) are smaller than those for unsubstituted molecules. a. Perform B3LYP/6-31G* calculations on HCl and on its dissociation products, chlorine atom and hydrogen atom. Following geometry optimization on HCl, calculate the vibrational frequency for both HCl and DCl and evaluate the zero point energy for each. In terms of a percentage of the total bond dissociation energy, what is the change noted in going from HCl to DCl? \(\mathrm{d}_{1}\) -Methylene chloride can react with chlorine atoms in either of two ways: by hydrogen abstraction (producing HCl) or by deuterium abstraction (producing DCl): Which pathway is favored on the basis of thermodynamics and which is favored on the basis of kinetics? b. Obtain the equilibrium geometry for dichloromethyl radical using the B3LYP/6-31G* model. Also obtain vibrational frequencies for both the unsubstituted and the deuterium-substituted radical and calculate zero point energies for the two abstraction pathways (you already have zero point energies for HCl and DCl). Which pathway is favored on the basis of thermodynamics? What would you expect the (thermodynamic) product ratio to be at room temperature? c. Obtain the transition state for hydrogen abstraction from methylene chloride using the B3LYP/6-31G* model. A reasonable guess is shown here: Calculate vibrational frequencies for the two possible structures with one deuterium and evaluate the zero point energies for these two structures. (For the purpose of zero point energy calculation, ignore the imaginary frequency corresponding to the reaction coordinate.) Which pathway is favored on the basis of kinetics? Is it the same or different from the thermodynamic pathway? What would you expect the (kinetic) product ratio to be at room temperature?

A surface for which the electrostatic potential is negative delineates regions in a molecule that are subject to electrophilic attack. It can help you to rationalize the widely different chemistry of molecules that are structurally similar. Optimize the geometries of benzene and pyridine using the HF/3-21G model and examine electrostatic potential surfaces corresponding to \(-100 \mathrm{kJ} / \mathrm{mol}\). Describe the potential surface for each molecule. Use it to rationalize the following experimental observations: (1) Benzene and its derivatives undergo electrophilic aromatic substitution far more readily than do pyridine and its derivatives; (2) protonation of perdeuterobenzene \(\left(\mathrm{C}_{6} \mathrm{D}_{6}\right)\) leads to loss of deuterium, whereas protonation of perdeuteropyridine \(\left(\mathrm{C}_{5} \mathrm{D}_{5} \mathrm{N}\right)\) does not lead to loss of deuterium; and (3) benzene typically forms \(\pi\) -type complexes with transition models, whereas pyridine typically forms \(\sigma\) -type complexes.

For many years, a controversy raged concerning the structures of so-called "electron-deficient" molecules, that is, molecules with insufficient electrons to make normal two-atom, two- electron bonds. Typical is ethyl cation, \(\mathrm{C}_{2} \mathrm{H}_{5}^{+}\) formed from protonation of ethene. Is it best represented as an open Lewis structure with a full positive charge on one of the carbons, or as a hydrogenbridged structure in which the charge is dispersed onto several atoms? Build both open and hydrogen-bridged structures for ethyl cation. Optimize the geometry of each using the B3LYP/6-31G* model and calculate vibrational frequencies. Which structure is lower in energy, the open or hydrogenbridged structure? Is the higher energy structure an energy minimum? Explain your answer.

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?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free