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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.

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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.

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Most popular questions from this chapter

Chemists know that nitric and sulfuric acids are strong acids and that acetic acid is a weak acid. They would also agree that ethanol is at best a very weak acid. Acid strength is given directly by the energetics of deprotonation (heterolytic bond dissociation); for example, for acetic acid: \\[ \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H} \longrightarrow \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}+\mathrm{H}^{+} \\] As written, this is a highly endothermic process, because not only is a bond broken but two charged molecules are created from the neutral acid. It occurs readily in solution only because the solvent acts to disperse charge. Acid strength can be calculated simply as the difference in energy between the acid and its conjugate base (the energy of the proton is 0 ). In fact, acid strength comparisons among closely related systems, for example, carboxylic acids, are quite well described with practical quantum chemical models. This is consistent with the ability of the same models to correctly account for relative base strengths (see discussion in Section \(26.8 .3)\) Another possible measure of acid strength is the degree of positive charge on the acidic hydrogen as measured by the electrostatic potential. It is reasonable to expect that the more positive the potential in the vicinity of the hydrogen, the more easily it will dissociate and the stronger the acid. This kind of measure, were it to prove successful, offers an advantage over the calculation of reaction energy, in that only the acid (and not the conjugate base ) needs to be considered. a. Obtain equilibrium geometries for nitric acid, sulfuric acid, acetic acid, and ethanol using the HF/3-21G model, and compare electrostatic potential maps. Be certain to choose the same (color) scale for the four acids. For which acid is the electrostatic potential in the vicinity of (the acidic) hydrogen most positive? For which is it least positive? Do electrostatic potential maps provide a qualitatively correct account of the relative acid strength of these four compounds? b. Obtain equilibrium geometries for several of the carboxylic acids found in the following table using the HF/3-21G model and display an electrostatic potential map for each. "Measure" the most positive value of the electrostatic potential associated with the acidic hydrogen in each of these compounds and plot this against experimental \(\mathrm{p} K_{\mathrm{a}}\) (given in the preceding table). Is there a reasonable correlation between acid strengths and electrostatic potential at hydrogen in this closely related series of acids?

Aromatic molecules such as benzene typically undergo substitution when reacted with an electrophile such as \(\mathrm{Br}_{2},\) whereas alkenes such as cyclohexene most commonly undergo addition: What is the reason for the change in preferred reaction in moving from the alkene to the arene? Use the Hartree- Fock \(6-31 G^{*}\) model to obtain equilibrium geometries and energies for reactants and products of both addition and substitution reactions of both cyclohexene and benzene (four reactions in total). Assume trans addition products (1,2-dibromocyclohexane and 5,6-dibromo-1,3-cyclohexadiene). Is your result consistent with what is actually observed? Are all four reactions exothermic? If one or more are not exothermic, provide a rationale as to why.

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.

Diels-Alder reactions commonly involve electronrich dienes and electron- deficient dienophiles: The rate of these reactions generally increases with the \(\pi\) -donor ability of the diene substituent, \(Y\), and with the \(\pi\) -acceptor ability of the dienophile substituent, \(\mathrm{X}\) The usual interpretation is that electron donors will push up the energy of the HOMO on the diene and that electron acceptors will push down the energy of the LUMO on the dienophile: The resulting decrease in the HOMO-LUMO gap leads to a stronger interaction between diene and dienophile and to a decrease in the activation barrier. a. Obtain equilibrium geometries for acrylonitrile, 1,1 dicyanoethylene, \(c i s-\) and \(t r a n s-1,2-\) dicyanoethylene tricyanoethylene, and tetracyanoethylene using the HF/3-21G model. Plot the LUMO energy for each dienophile versus the \(\log\) of the observed relative rate for its addition to cyclopentadiene (listed below the structures in the preceding figure). Is there a reasonable correlation between LUMO energy and relative rate? b. Obtain transition-state geometries for Diels-Alder cycloadditions of acrylonitrile and cyclopentadiene and tetracyanoethylene and cyclopentadiene using the HF/3-21G model. Also obtain a geometry for cyclopentadiene. Calculate activation energies for the two reactions. How does the calculated difference in activation energies compare with the experimental difference (based on a value of 7.61 for the difference in the log of the rates and assuming \(298 \mathrm{K}) ?\)

Discussion of the VSEPR model in Section 25.1 suggested a number of failures, in particular, in \(\mathrm{CaF}_{2}\) and \(\mathrm{SrCl}_{2},\) which (according to the VSEPR) should be linear but which are apparently bent, and in \(\operatorname{Se} \mathrm{F}_{6}^{2-}\) and \(\mathrm{TeCl}_{6}^{2-},\) which should not be octahedral but apparently are. Are these really failures or does the discrepancy lie with the fact that the experimental structures correspond to the solid rather than the gas phase (isolated molecules)? a. Obtain equilibrium geometries for linear \(\mathrm{CaF}_{2}\) and \(\mathrm{SrCl}_{2}\) and also calculate vibrational frequencies (infrared spectra). Use the HF/3-21G model, which has actually proven to be quite successful in describing the structures of main-group inorganic molecules. Are the linear structures for \(\mathrm{CaF}_{2}\) and \(\mathrm{SrCl}_{2}\) actually energy minima? Elaborate. If one or both are not, repeat your optimization starting with a bent geometry. b. Obtain equilibrium geometries for octahedral \(\mathrm{SeF}_{6}^{2-}\) and \(\mathrm{TeCl}_{6}^{2-}\) and also calculate vibrational frequencies. Use the HF/3-21G model. Are the octahedral structures for \(\operatorname{Se} \mathrm{F}_{6}^{2-}\) and \(\mathrm{TeCl}_{6}^{2-}\) actually energy minima? Elaborate. If one or both are not, repeat your optimization starting with distorted structures (preferably with \(C_{1}\) symmetry).

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