Molecular Vibrations and Spectroscopy

Beer–Lambert law absorbance

Absorbance A measures how much light a sample removes from a beam. It is defined through the transmittance T = I/I_0 as A = -\log_{10} T. The Beer–Lambert law states that A = \varepsilon l c, where \varepsilon is the molar absorption coefficient in L mol^{-1} cm^{-1}, l is the path length in cm, and c is the concentration in mol L^{-1}. Path length must be converted to cm before substitution, so a 1 mm cell gives l = 0.1 cm. The fraction of light absorbed is 1 - T = 1 - 10^{-A}.

Because absorbance is linear in concentration, doubling c doubles A, but it does not double T. For a doubled concentration the transmittance becomes T^2. For example, if \varepsilon = 5000 L mol^{-1} cm^{-1}, l = 1 cm and c = 1\times10^{-4} M, then A = 0.50 and T = 0.316, so about 68 % of the light is absorbed. If a solution transmits 80 % of the light, then A = 0.097, and doubling its concentration gives A = 0.194 with T = 0.64.

For a mixture of non-interacting species, absorbances are additive at the same wavelength: A = \sum_i \varepsilon_i c_i l. The transmittance of the mixture is obtained from this total absorbance, not by adding the individual transmittances.

Infrared activity of normal modes

A vibration is infrared active only if it produces a change in the permanent dipole moment. In group-theory terms, an IR-active mode transforms as one of the components x, y or z of the dipole operator. A vibration is Raman active only if it changes the polarizability, which transforms as quadratic functions such as x^2, xy or z^2. The totally symmetric stretch of a homonuclear diatomic produces no dipole change, so it is IR inactive but Raman active.

In a linear molecule such as CO_2, the bending mode and the asymmetric stretch change the dipole and are IR active, whereas the symmetric stretch leaves the dipole unchanged and is IR inactive. In the trigonal planar SO_3 molecule of D_{3h} symmetry, the symmetric in-plane breathing mode of A_1' symmetry is IR inactive, while the out-of-plane umbrella mode of A_2'' symmetry and the doubly degenerate E' modes are IR active.

A high-spin d^4 complex such as Mn(III) is Jahn–Teller active. A tetragonal distortion lowers the symmetry of an octahedral \mathrm{ML_6} unit, and the set of six M–L stretches splits into groups of different symmetry. Only some of the resulting groups are IR active, and a doubly degenerate set carries about twice the intensity of a singly degenerate band, so the number of observed IR bands and their intensity ratio follow from the degeneracies.

Boltzmann population of vibrational levels

At thermal equilibrium the ratio of populations of two levels is N_{\text{upper}}/N_{\text{lower}} = (g_{\text{upper}}/g_{\text{lower}})\exp(-\Delta E/kT), where g is the degeneracy of each level. For non-degenerate levels the degeneracy ratio is 1. If the ground-state population is 10 times the upper-state population, then \exp(\Delta E/kT) = 10, which gives T = \Delta E/(k\ln 10). When \Delta E is given in cm^{-1}, the constant k must be given in cm^{-1} K^{-1}.

At ordinary temperatures most molecules occupy the vibrational ground state v = 0. Transitions that start from v = 0, namely the fundamental 0 \to 1 and the overtone 0 \to 2, have intensities that are nearly independent of temperature. A hot band starts from an excited level such as v = 1, and its intensity grows as temperature rises. Because of anharmonicity, the spacing between v = 1 and v = 2 is smaller than the spacing between v = 0 and v = 1, so a 1 \to 2 hot band lies at lower wavenumber than the fundamental. The overtone 0 \to 2 lies at the highest energy of these three transitions.

1H NMR spin-spin multiplicity

A proton coupled to n equivalent nuclei of spin 1/2 gives n + 1 lines, following the n+1 rule. More generally, n equivalent nuclei of spin I give 2nI + 1 lines. A proton with no spin-active neighbours gives a singlet. Nuclei that are not equivalent to one another, but that have different coupling constants, multiply their patterns, so the total number of lines is the product of the individual counts.

• Spin-1/2 neighbours — ^{31}P (I = 1/2, 100 % abundant) and ^{1}H couple normally, so a proton bonded to a single phosphorus appears as a doublet, and a proton with four equivalent phosphorus neighbours appears as a pentet. Example: a ^{1}H signal split by four equivalent spin-\frac{1}{2} nuclei is a pentet.
• Spin-1 neighbours — deuterium (I = 1) gives 2nI + 1 lines. Example: a CHD_2 proton with two equivalent deuterium neighbours appears as a quintet.

Deuterium nuclei do not appear in the ^{1}H spectrum, so only the proton signal is observed.

Halogen isotope mass patterns

Chlorine has the isotopes ^{35}Cl and ^{37}Cl in an abundance ratio of about 3:1, and bromine has ^{79}Br and ^{81}Br in a ratio of about 1:1. For a molecule containing n halogen atoms, the isotope pattern of the molecular ion follows the binomial expansion of (a + b)^n, with a and b the relative abundances of the two isotopes. Two chlorine atoms therefore give the pattern (3+1)^2 = 9:6:1 for [M]^+, [M+2]^+ and [M+4]^+. Two bromine atoms give (1+1)^2 = 1:2:1, and one chlorine plus one bromine give 3:4:1.

The pattern is used together with NMR. A para-disubstituted dibromobenzene has four chemically equivalent aromatic protons, so its ^{1}H spectrum shows a singlet in the aromatic region, whereas an ortho isomer gives two multiplets.

Normal modes of triatomic molecules

A nonlinear molecule with N atoms has 3N - 6 vibrational modes, and a linear molecule has 3N - 5. A nonlinear triatomic molecule therefore has three normal modes. In each normal mode all atoms oscillate at the same frequency, and the centre of mass does not move and the molecule does not rotate. For an angular XY_2 molecule such as H_2O (point group C_{2v}), the three modes are a symmetric stretch of A_1 symmetry, a bend of A_1 symmetry and an asymmetric stretch of B_2 symmetry. In the symmetric stretch both bonds lengthen and shorten in phase. In the bend the bond angle opens and closes. In the asymmetric stretch one bond lengthens while the other shortens.

A linear triatomic molecule such as CO_2 has four modes, and its bending mode is doubly degenerate.

Force constant from vibrational frequency

In the harmonic oscillator model, a diatomic molecule has the fundamental frequency \tilde{\nu} = \dfrac{1}{2\pi c}\sqrt{k/\mu}, where \tilde{\nu} is in m^{-1}, c is the speed of light in m s^{-1}, k is the force constant in N m^{-1}, and \mu is the reduced mass in kg. Rearranging gives k = \mu(2\pi c\tilde{\nu})^2. The reduced mass is \mu = m_1 m_2/(m_1 + m_2). Molar masses in g mol^{-1} must be divided by N_A and multiplied by 10^{-3} to give the mass of one molecule in kg.

Unit conversion is the most common error. A wavenumber in cm^{-1} must be multiplied by 100 to give m^{-1}. For a light diatomic such as H–Cl, where the reduced mass is close to the mass of a hydrogen atom, a stretching wavenumber near 3000 cm^{-1} corresponds to a force constant of roughly 500 N m^{-1}.

Raman Stokes shift

In Raman scattering, a photon of exciting wavenumber \tilde{\nu}_0 is scattered with a shift equal to a vibrational wavenumber \tilde{\nu}_{\text{vib}}. The exciting wavenumber in cm^{-1} is \tilde{\nu}_0 = 10^{7}/\lambda, where \lambda is the wavelength in nm. In the Stokes process the molecule ends in an excited vibrational level, so the scattered line appears at \tilde{\nu}_S = \tilde{\nu}_0 - \tilde{\nu}_{\text{vib}}, at lower wavenumber than the laser line. Anti-Stokes lines appear at higher wavenumber.

For example, a 633 nm laser has \tilde{\nu}_0 \approx 15\,798 cm^{-1}. A vibration of 1000 cm^{-1} therefore gives a Stokes line at about 14 798 cm^{-1}. Stokes lines are usually more intense than anti-Stokes lines, because most molecules occupy the ground vibrational state.