Atomic mass, wavenumber

The S-H absorption is typically weak and occurs at a lower wavenumber (frequency) than the corresponding 0-H vibration because of the higher atomic mass of sulfur (and so increased reduced mass of SH). The difference in reduced mass, however, does not completely account for the reduction in the stretching frequency of the S-H bond as this bond is also generally weaker (smaller k) than the corresponding O-H bond (typical bond dissociation energies OH, 460 kJ mol SH, 340 kJ mol ). 

The Vs (A/) mode (a) is Raman-active, whereas the II (A2") mode (b) is IR-active. The vj ( ) and 5 ) modes are both IR- and Raman-active. Thus, three peaks are expected in the IR spectrum of BCI3 and occur at 243,460, and 956 cm . Three peaks are also expected in the Raman spectrum of BCI3, having wavenumbers of 243, 471, and 956 cm . The Vj (A/) and Vj ( ) vibrational modes (a) and (c) show little or no isotope effect when the atomic mass of B is changed because the B atom does not move significantly in either of these two vibrations. On the other hand, the II (A2") and ( ") modes (b) and (d) show a significant isotope effect because the B atom is involved in both of these vibrational modes. 

The vibrational wavenumber, a = 2897 cm calculated above is for a H— C1 molecule. But protium, H, is not the only hydrogen isotope found in nature there is deuterium, (D), with a mass equal to two atomic units, and tritium, (T), with atomic mass of 3 amu. The hydrogen atoms found in nature are a mixture of all three isotopes and, like in any mixture, its properties - and mass - are the average value, m(H) = 1.0078 amu. Likewise, chlorine in nature is a mixture of about 75% of isotope C1 and about 25% of isotope C1, so the average mass of Cl atoms is 35.453 amu or that of CI2 molecules twice this. How does this affect molecular vibrations and vibrational spectra ... 

If the atomic masses in the G matrix elements are expressed in atomic mass units and the frequencies expressed in wavenumbers, then the equation may be written as... 

The validity of Equation 8.4 for predicting stretching frequencies is illustrated by its use to give v as 3040 cm (see Exercise 2) the experimental value for this vibration is in the range of 2960-2850 cm . Other consequences of Equation 8.4 are noteworthy. For instance, experimentally the wavenumber for stretching of a C-D bond is about 2150 cm a lower value than that for a C—H bond this difference reflects the effect of atomic mass on the absorption frequency as accounted... 

Atomic mass - Smaller atoms give bonds that undergo vibrational excitation at higher frequencies, thereby corresponding to a higher wavenumber of absorption. For example, compare the bonds below. The C—H bond involves the smallest atom (H) and therefore appears at the highest wavenumber. 

Figure B2.3.17. REMPI spectra of the HCl and DCl products from the reaction of Cl atoms with (CH3)3CD [63], The mass 36 and 2 ion signals are plotted as a fiinction of the 2-photon wavenumber. Assigmnents of the...

The positions of both absorptions in methyl and n-propyl cyclopropenone (1838/1605 cm-1,1835/1600 cm-1) are displaced to lower wavenumbers compared to the symmetrically substituted species (dimethyl cyclopropenone 1848, 1866/ 1657 cm-1 di-n-propyl cyclopropenone 1840/1630 cm-1). If the 1640 cm-1 band were a pure C=0 stretching vibration, the observed large effect of three-ring substitution could not be explained, since no change in the relative mass and geometry of participating atoms takes place. 

If we know the masses of the atoms involved in a vibration, and the wavenumber can be determined from a spectrum, then we can readily calculate a value for the force constant k. 

For such molecules, all of the vibrations are active in both the infrared and Raman spectra. Usually, certain of the vibrations give very weak bands or lines, others overlap, and some are difficult to measure, as they occur at very low wavenumber values.40 Because the vibrations cannot always be observed, a model of the molecule is needed, in order to describe the normal modes. In this model, the nuclei are considered to be point masses, and the forces between them, springs that obey Hooke s law. Furthermore, the harmonic approximation is applied, in which any motion of the molecule is resolved in a sum of displacements parallel to the Cartesian coordinates, and these are called fundamental, normal modes of vibration. If the bond between two atoms having masses M, and M2 obeys Hooke s law, with a stiffness / of the spring, the frequency of vibration v is given by... 

This corresponds to radiation in the microwave or far-intirared regions of the spectrum. For example, with H35C1 the wavenumber of the J = 0 to J = 1 transition is 21.186 cm-1. Measurement of such rotational transitions is an accurate way of determining the moments of inertia of gas-phase molecules. If the masses of atoms are known, the bond length can then be estimated. 

For a diatomic molecule A—B, the only vibration that can occur is a periodic stretching along the A—B bond. The masses of the two atoms and their connecting bond may be treated, to a first approximation, as two masses joined by a spring and Hooke s law may be applied. This leads to the expression for the frequency of vibration v in wavenumbers (cm-1) ... 

In the treatment of two atoms connected together, a simple harmonic oscillator model can be adopted involving the two masses connected with a spring having a force constant fk. Thus, the vibrational frequency in wavenumbers 2 depends from the reduced mass p, from fk with c being the velocity of light. 

A deuterium atom is heavier than a hydrogen atom, but the real point is not its weight, which involves gravity, but its mass, which does not. The vibrational frequency of a bond depends on its stiffness (the force constant) and on the masses of the atoms involved. For a diatomic molecule A-B the vibrational frequency (in wavenumbers) is governed by the simple formula... 

Bonds involving a light hydrogen atom have smaller reduced masses and therefore absorb at higher wavenumbers than bonds involving only heavier atoms. 

As the mass of the atom bonded to carbon increases (i.e. C-H to C-C), the reduced mass increases, less energy is absorbed and the wavenumber decreases. 

If the masses v4, B,C,D... are all similar and the force constants/,/2,/j... are of the same magnitude, the vibrations of individual atoms are strongly coupled with the result that no band can be assigned solely to any particular group of atoms. If, however, the mass of atom A is considerably smaller than those of the other atoms, one mode of vibration will involve the stretching of the bond between A and the rest of the molecule. The system can be considered as approximating to a diatomic molecule A—X, and the wavenumber of the vibration is then given by... 

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