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Classical Normal Mode Analysis: Harmonic Approximation

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Classical Normal Mode Analysis: Harmonic Approximation Classical Normal Mode Analysis: Harmonic Approximation The vibrations of a molecule are given by its normal modes. Each absorption in a vibrational spectrum corresponds to a normal mode. The four normal modes of carbon dioxide, Figure 1, are the symmetric stretch, the asymmetric stretch and two bending modes. The two bending modes have the same energy and differ only in the direction of the bending motion. Modes that have the same energy are called degenerate. In the classical treatment of molecular vibrations, each normal mode is treated as a simple harmonic oscillator. • + • Symmetric stretch Asymmetric stretch Bend Bend Figure 1. Normal Modes for a linear triatomic molecule. In the last bending vibration the motion of the atoms is in-and-out of the plane of the paper. In general linear molecules have 3N-5 normal modes, where N is the number of atoms. The five remaining degrees of freedom for a linear molecule are three coordinates for the motion of the center of mass (x, y, z) and two rotational angles. Non-linear molecules have three rotational angles, hence 3N-6 normal modes. The characteristics of normal modes are summarized below. Characteristics of Normal Modes 1. Each normal mode acts like a simple harmonic oscillator. 2. A normal mode is a concerted motion of many atoms. 3. The center of mass doesn’t move. 4. All atoms pass through their equilibrium positions at the same time. 5. Normal modes are independent; they don’t interact. In the asymmetric stretch and the two bending vibrations for CO 2, all the atoms move. The concerted motion of many of the atoms is a common characteristic of normal modes. However, in the symmetric stretch, to keep the center of mass constant, the center atom is stationary. In small molecules all or most all of the atoms move in a given normal mode; however, symmetry may require that a few atoms remain stationary for some normal modes. The last characteristic, that normal modes are independent, means that normal modes don’t exchange energy. For example, if the symmetric stretch is excited, the energy stays in the symmetric stretch. The background spectrum of air, Figure 2, shows the asymmetric and symmetric stretches and the bending vibration for water, and the asymmetric stretch and bending vibrations for CO 2. The symmetric stretch for CO 2 doesn’t appear in the Infrared; a Raman spectrum is needed to measure the frequency of the symmetric stretch. These absorptions are responsible for the vast majority of the greenhouse effect. We will also use CO 2 as an example, below. The normal modes are calculated using Newton’s equations of motion. 1-4 Molecular mechanics and molecular orbital programs use the same methods. Normal mode calculations are available on-line. 5 2 40 38 36 34 32 30 28 26 24 Bend: -1 22 H2O 1595 cm Single Beam Single 20 18 Asymmetric stretch: -1 16 H2O 3756 cm 14 12 10 Symmetric stretch: Asymmetric stretch: 8 Bend: -1 -1 -1 CO 2 2349 cm 6 H2O 3652 cm CO 2 667 cm 4000 3500 3000 2500 2000 1500 1000 500 Wavenumbers (cm-1) Figure 2. The Infrared spectrum of air. This spectrum is the background scan from an FT-IR spectrometer. Harmonic Oscillator Review Lets first review the simple harmonic oscillator. Consider a mass m, supported on a spring with force constant k. Hooke’s Law for the restoring force for an extension, x, is F = -kx. In other words, if the spring is stretched a distance x>0, the restoring force will be negative, which will act to pull the mass back to its equilibrium position. The potential energy for Hooke’s Law is obtained by integrating dV F = - = -kx (1) dx 1 to give V = k x 2 (2) 2 In molecular mechanics and molecular orbital calculations, the force constant is not known. However, the force constant can be calculated from the second derivative of the potential energy. d2 V k = (3) dx 2 The Hooke’s Law force is substituted into Newton’ Law: d2 x F = ma or m 2 = -kx (4) dt and solved to obtain the extension as a function of time: x(t) = A sin(2 πνt) (5) where ν is the fundamental vibration frequency and A is the amplitude of the vibration. Taking the second derivative of the extension gives 2 d x 2 2 = -4π ν x (6) dt 2 Substituting Eq 6 back into Eq 4 gives: -4π2ν2 m x = -kx (7) which is the basis for the classical calculation of the normal modes of a molecule. Colby College 3 Normal Mode Analysis For molecules the x, y, z coordinates of each atom must be specified. The coordinates are: Atom 1: X 1, Y 1, Z 1, Atom 2: X 2, Y 2, Z 2, etc. …… The extensions are the differences in the positions and the equilibrium positions for that atom: Atom 1: x 1 = X 1 – X1,eq y1 = Y 1 – Y1,eq z1 = Z 1 – Z1,eq (8) Atom 2: x 2 = X 2 – X2,eq y2 = Y 2 – Y2,eq z2 = Z 2 – Z2,eq Atom i: x i = X i – Xi,eq yi = Y i – Yi,eq zi = Z i – Zi,eq Where X i,eq , Y i,eq , and Z i,eq are the equilibrium (energy minimized) positions for atom i. For example, if x 1, y 1, and z 1 are all zero, then atom 1 is at its equilibrium position. Molecular mechanics or molecular orbital calculations are used to find the potential energy of the molecule as a function of the position of each atom, V(x 1, y 1, z 1, x 2, y 2, z 2, x 3, y 3, z 3,...,x N,y N,z N). The second derivative of the potential energy can then be used to calculate the force constants, Eq 3. However, there are now 3Nx3N possible second derivatives and their corresponding force constants. For example, ∂2V 11 2 = k (9) ∂ x1 xx is the change of the force on atom 1 in the x-direction when you move atom 1 in the x-direction. Similarly, ∂2V 12 = k (10) ∂x1∂y2 xy is the change of the force on atom 1 in the x-direction when you move atom 2 in the y-direction. The various types of force constants are shown in Figure 3. ∂2V 11 = k same atom same direction 2 ∂ x 2 xx 1 1 ∂2 11 V 1 2 ∂ 2 = kyy same atom same direction y1 ∂2V 11 ∂ ∂ = kxy same atom different directions 1 2 x1 y1 ∂2V 12 = k different atom same direction 2 ∂x1∂x2 xx 1 ∂2V 12 = k different atom and direction 2 ∂x1∂y2 xy 1 Figure 3. Types of second derivatives and force constants These force constants are not the force constants for individual bonds, they are force constants for the motion of a single atom subject to all its neighbors, whether directly bonded or not. The Colby College 4 complete list of these force constants is called the Hessian, which is a 3Nx3N matrix. Eq 7 is then applied for each force constant. 1,2 π2ν2 11 11 11 12 12 1N -4 m1x1= -kxx x1 - kxy y1 - kxz z1 - kxx x2 - kxy y2 -…- kxz zN (11) π2ν2 11 11 11 12 12 1N -4 m1y1= -kyx x1 - kyy y1 - kyz z1 - kyx x2 - kyy y2 -…- kyz zN : π2ν2 21 21 21 22 22 2N -4 m2x2= -kxx x1 - kxy y1 - kxz z1 - kxx x2 - kxy y2 -…- kxz zN : π2ν2 N1 N1 N1 N2 N2 NN -4 mNzN= -kzx x1 - kzy y1 - kzx z1 - kzx x2 - kzy y2 -…- kzz zN In words, the right-hand sides of the above equations simply state that the total force on atom i is the sum of the forces of all the atoms on atom i. In addition, we need to keep track of the x, y, and z directions for each atom. There are a total of 3Nx3N terms on the right. All these terms are confusing. A simple example will help at this point. For our example consider a symmetrical linear triatomic molecule that can only vibrate along the x-axis, Figure 4. CO 2 is a good example of a symmetrical linear triatomic. O1 C2 O3 x Figure 4. A symmetrical triatomic molecule with vibrations limited along the internuclear axis. Because we have limited the vibrations to the x-axis, which is the internuclear axis, this model will provide the symmetric and asymmetric stretching modes, only. Eqs 11 then reduce to π2ν2 11 12 13 -4 m1x1= -kxx x1 - kxx x2 - kxx x3 (12) π2ν2 21 22 23 -4 m2x2= -kxx x1 - kxx x2 -kxx x3 (13) π2ν2 31 32 33 -4 m3x3= -kxx x1 - kxx x2 - kxx x3 (14) since we only need to keep the x-terms. Several numerical techniques are available to solve linear sets of simultaneous equations such as this. Conventionally, however, the problem is simplified by converting to mass weighted coordinates, for example: ~ ~ x1 = m1 x1 x2 = m2 x2 , etc. (15) and mass weighted force constants: 12 ~12 kxx kxx = (16) m1 m2 In the new mass weighted coordinates, Eqs 12-14 become: Colby College 5 ~ ~ ~ π2ν2 ~ 11 ~ 12 ~ 13 ~ -4 x1 = - kxx x1 - kxx x2 - kxx x3 (17) ~ ~ ~ π2ν2 ~ 21 ~ 22 ~ 23 ~ -4 x2 = - kxx x1 - kxx x2 - kxx x3 (18) ~ ~ ~ π2ν2 ~ 31 ~ 32 ~ 33 ~ -4 x3 = - kxx x1 - kxx x2 - kxx x3 (19) For example, we can show that Eq 17 is equivalent to Eq 11, by substituting Eqs 15 and 16 into Eq 17.
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