Chemistry 795T. NC State University. Lecture 4. Vibrational and Rotational Spectroscopy

Transcription

1 Chemistry 795T Lecture 4 Vibrational and Rotational Spectroscopy NC State University

2 The Dipole Moment Expansion The permanent dipole moment of a molecule oscillates about an equilibrium value as the molecule vibrates. Thus, the dipole moment depends on the nuclear coordinate Q. µ Q = µ 0 + µ Q Q +... where µ is the dipole operator.

3 Rotational Transitions Rotational transitions arise from the rotation of the permanent dipole moment that can interact with an electromagnetic field in the microwave region of the spectrum. µ Q = µ 0 + µ Q Q +...

4 The total wave function The total wave function can be factored into an electronic, a vibrational and a rotational wave function. Ψ = ψ el χ v Y JM M rot = χ * v Y J M ψ * el µψ el dτ el χ v Y JM dτ nuc = χ * v χ v dq χ * v Y J M µ 0 χ v Y JM sinθdθdφ = χ * v Y J M µ 0 χ v Y JM sinθdθdφ

5 Interaction with radiation An oscillating electromagnetic field enters as E 0 cos(ωt) such that the angular frequency hω is equal to a vibrational energy level difference and the transition moment is M rot = µ 0 0 2π 0 π * Y J+1,M cos θ Y JM sin θ dθdφ

6 Interaction with radiation The choice of cos(θ) means that we consider z-polarized microwave light. In general we could consider x- or y-polarized as well. x sin(θ)cos(φ) µ 0 = µ X i + µ Y j + µ Z k y sin(θ)sin(φ) µ 0 = µ 0 sinθcosφi + sinθsinφ j +cosθk z cos(θ)

7 Vibrational transitions Vibrational transitions arise because of the oscillation of the molecule about its equilibrium bond configuration. As the molecule oscillates infrared radiation can interact to alter the quantum state. µ Q = µ 0 + µ Q Q +... M vib = µ Q * χ v +1 Qχ v dq

8 Classical Vibration of a Diatomic As was the case for rotation, we can consider a simple model of a mass on a spring attached to a wall of infinite mass and a diatomic molecule as two simple examples. Mass on a spring Diatomic k k m m 1 m 2 k is the force constant Reduced mass µ = m 1m 2 m 1 + m 2

9 Harmonic approximation V(Q) =V(Q 0 )+ V Q (Q Q 0) V Q 2 (Q Q 0) At equilibrium V Q =0 Assume terms higher than quadratic are zero By definition 2 V Q 2 = k, the force constant

10 Classical approach to vibration Solution is oscillatory Any energy is possible µ 2 Q t 2 + k 2 Q2 =0 V(Q) k force constant Q Q

11 Classical vibrational motion A particle undergoes harmonic motion if it experiences a restoring force that is proportional to its displacement, x. F = -kq (k is a force constant) F = -dv/dq and V = 1/2kQ 2. The classical harmonic oscillator can also be written as: ma = kq, µ 2 Q + kq =0 t 2 Solutions have the form of sin(ωt) or cos(ωt) depending on the initial conditions. These solutions imply that ω = k µ

12 Classical potential function The potential is V = 1/2kQ 2, which is a parabolic function. This potential is called a harmonic potential. The force constant k has units of Newtons/meter (N/m) or Joules/meter 2 (J/m 2 ). The angular frequency ω = 2πν, ν is the frequency in Hz.

13 Quantum approach to vibration h 2 2µ Solution is Gaussian Energy is quantized 2 Ψ Q 2 + k 2 Q2 Ψ = EΨ E = v hν Q v is the quantum number Allowed transitions v v + 1, v v - 1

14 Vibrational wavefunctions Energy levels are given by E v = (v +1/2)hω Wavefunctions are Ψ v = N v H v e -y2 /2 where H v is the Hermite polynomial Typical energies are of the order of cm -1

15 Solutions to harmonic oscillator The Hermite polynomials are derivatives of a Gaussian, y = α 1/2 Q H v (y) = ( 1) v e y 2 d dy The normalization constant is N v = 1 2 v v! α π 1/4 α = µk h v e y 2 v H v (y) N v 0 1 α/π 1/4 1 2y 1/ 2 α/π 1/4 2 4y 2 2 1/ 8 α/π 1/4 3 8y 3 12y 1/ 48 α/π 1/4

16 The zero point energy The lowest level is E 0 = 1/2hω The lowest vibrational level is not zero in energy. This is consistent with the uncertainty principle. If atoms were completely still at absolute zero then we would know both their position and moment to arbitrary accuracy. The width of the wavefunction is related to positional uncertainty of an atom. We call E 0 the zero point energy.

17 Polyatomic Molecules There are 3N total degrees of freedom in a molecule that contains N atoms. There are three translational degrees of freedom. These correspond to motion of the center of mass of the molecule. In a linear molecule there are two rotational degrees of freedom. In a non-linear molecule there are 3 rotational degrees of freedom. The remaining degrees of freedom are vibrational.

18 Normal modes - water Symmetric stretch ν cm -1 Symmetric Stretch ν cm -1 Asymmetric stretch Stretch ν cm cm -1 Bend ν cm -1 There are 3 normal modes (3N - 6). All of them are infrared active since all show a dipole moment change in their motion. The harmonic approximation can be applied to each normal mode.

19 Normal modes - CO 2 Symmetric stretch (infrared ν inactive) cm -1 Symmetric stretch (Raman active). +. Asymmetric stretch 3 ν -1 Bends cm -1 ν cm -1 (IR active) (infrared active) ν cm -1 (IR active) There are 4 normal modes (3N - 5). Three of them are infrared active since they show a dipole moment change in their motion.

20 Vibrational Transition

21 Vibrational Transition

22 Vibrational Selection Rule v = v + 1 v = v - 1 v = 2 v = 1 v = 0 Q

23 Overtones of water Even in water vapor ν 1 ν 3, but symmetries are different, Γ 1 Γ 3. However, the third overtone of mode 1 has the same symmetry as the combination band Γ 1 Γ 1 Γ 1 = Γ 1 Γ 3 Γ 3. Strong anharmonic coupling leads to strong overtones at 11,032 and 10,613 cm -1. These intense bands give water and ice their blue color. ν 1 symmetric stretch 3825 cm -1 ν 2 bend 1654 cm -1 ν 3 asymmetric stretch 3935 cm -1

24 Frequency shift due to molecular interactions Hydrogen bonding lowers O-H force constant and H-O-H bending force constant. vapor liquid ν ν ν

25 Question Which expression is correct? µ A. k = ω B. k = 1 µω C. k = µω D. k = µω 2

26 Question Which expression is correct? A. k = B. k = µ ω 1 µω ω = C. k = µω D. k = µω 2 k µ ω 2 = k µ k = µω 2

27 Question How many normal modes of vibration does methane have? A. 15 B. 12 C. 9 D. 6

28 Question How many normal modes of vibration does methane have? A. 15 B. 12 C. 9 D. 6

29 Vibrational transitions As an example we can calculate the transition moment between the state v = 0 and v = 1. χ 0 = α π 1/4 e αq2 /2, χ 1 = α π 1/4 2αQe αq2 /2 M vib = µ Q = µ Q α π 1/2 α π 1/2 2α 2α e αq2 /2 Q 2 e αq2 /2 dq π µ = 3/2 2α Q 1 2α

30 Vibrational transitions Note that this result is a statement of the vibrational selection rule. Within the harmonic approximation transitions can only occur between states separated by one quantum number ( v = 1 or v = -1). This general rule can be seen by considering integrals of the type shown in the previous slide.

31 Rotation in two dimensions The angular momentum is J z = pr. J z r m Using the debroglie relation p = h/λ we also have a condition for quantization of angular motion J z = hr/λ. p

32 Classical Rotation In a circular trajectory J z = pr and E = J z2 /2I. I is the moment of inertia. Mass in a circle I = mr 2. Diatomic I = µr 2 r m r m 1 m 2 Reduced mass µ = m 1m 2 m 1 + m 2

33 The 2-D rotational hamiltonian The wavelength must be a whole number fraction of the circumference for the ends to match after each circuit. The condition 2πr/m = λ combined with the debroglie relation leads to a quantized expression,j z = mh. The hamiltonian is: h2 2I 2 Φ φ 2 = EΦ

34 The 2-D rotational hamiltonian Solutions of the 2-D rotational hamiltonian are sine and cosine functions just like the particle in a box. Here the boundary condition is imposed by the circle and the fact that the wavefunction must not interfere with itself. The 2-D model is similar to condition in the Bohr model of the atom.

35 The 3-D rotational hamiltonian The Laplacian is the second derivative operator. For the 3-D problem in spherical polar coordinates it is: 2 = 1 r 2 r r 2 r + 1 r 2 sin θ θ sin θ θ + 1 r 2 sin 2 θ 2 φ 2 The Schrodinger equation is: h 2 2m e 2 Ψ = EΨ The angular part is obtained by separation of radial and angular variables. Ψ r,θ,φ = RrY θ,φ

36 Separation of variables After substitution of the product wavefunction we have: h2 Rr r r 2 R r + 2m er 2 h 2 RrE h 2 Y θ,φ The separated equations are: 1 sin θ θ sin θ Y θ Y =0 sin 2 θ φ 2 1 Rr r r 2 R r + 2m er 2 h 2 RrE = β 1 Y θ,φ 1 sin θ θ sin θ Y θ Y sin 2 θ φ 2 = β The angular part is: sin θ θ sin θ Y θ + 2 Y φ 2 + βsin2 θ Y θ,φ =0

37 Separation of angular variables The angular equation can be further separated into the θ and φ variables: sin θ Θθ θ sin θ Θ θ The separation variable is M 2 : + βsin 2 θ + 1 Φφ 2 Φ φ 2 =0 sin θ Θθ θ sin θ Θ θ + βsin 2 θ = M 2 1 Φφ 2 Φ φ 2 = M 2

38 Spherical harmonics The general solutions of the rotational hamiltonian are called spherical harmonics. They have the form: Y JM θ,φ = 1 M 2J +1 J M! 4π J + M! The LeGendre polynomial satisfies the equation: 1 x 2 d 2 P dx 2 Where x = cos(θ). P JM cosθ e imφ 2xdP dx + β M 2 P(x) =0 2 1 x

39 LeGendre polynomials The solution requires that β=j(j+1) with J=0,1,2.. Where J is the rotational quantum number. The azimuthal quantum number M is also called J z. The magnitude of J z <J. The solutions are Legendre polynomials P 0 (x)=1 P 2 (x)=1/2 (3x 2-1) P 1 (x)=x P 3 (x)=1/2 (5x 3-3x) The shapes of the spherical harmonics are those of the s, p, and d orbitals etc. The degeneracy of each level J is 2J+1. J = 0 is s, M = 0 J = 1 is p, M = -1, 0, 1 J = 2 is d, M = -2, -1, 0, 1, 2

40 Space quantization in 3D z J = 2 M=+2 M = +1 M = 0 M = -1 M = -2 Solutions of the rotational Schrödinger equation have energies E = h 2 J(J +1)/2I Specification of the azimuthal quantum number m z implies that the angular momentum about the z-axis is J z = hm z. This implies a fixed orientation between the total angular momentum and the z component. The x and y components cannot be known due to the Uncertainty principle.

41 Spherical harmonic for P 0 (cosθ) Plot in polar coordinates represents Y 00 2 where Y 00 =(1/4π) 1/2. Solution corresponds to rotational quantum numbers J = 0, M or J z = 0. Polynomial is valid for n 1 quantum numbers of hydrogen wavefunctions

42 Spherical harmonic for P 1 (cosθ) Plot in polar coordinates represents Y 11 2 where Y 11 =(1/2)(3/2π) 1/2 sinθe iφ. Solution corresponds to rotational quantum numbers J = 1, J z = ±1. Polynomial is valid for n 2 quantum numbers of hydrogen wavefunctions

43 Spherical harmonic for P 1 (cosθ) Plot in polar coordinates represents Y 10 2 where Y 10 =(1/2)(3/π) 1/2 cosθ with normalization. Solution corresponds to rotational quantum numbers J = 1, J z = 0. Polynomial is valid for n 2 quantum numbers of hydrogen wavefunctions.

44 Spherical harmonic for P 2 (cosθ) Plot in polar coordinates of Y 22 2 where Y 22 =1/4(15/2π) 1/2 cos 2 θe 2iφ Solution corresponds to rotational quantum numbers J = 2, J z = ± 2. Polynomial is valid for n 3 quantum numbers of hydrogen wavefunctions

45 Spherical harmonic for P 2 (cosθ) Plot in polar coordinates of Y 21 2 where Y 21 =(15/8π) 1/2 sinθcosθe iφ Solution corresponds to rotational quantum numbers J = 2, J z = ± 1. Polynomial is valid for n 3 quantum numbers of hydrogen wavefunctions

46 Spherical harmonic for P 2 (cosθ) Plot in polar coordinates of Y 20 2 where Y 20 =1/4(5/π) 1/2 (3cos 2 θ-1) Solution corresponds to rotational quantum numbers J = 1, J z = 0. Polynomial is valid for n 3 quantum numbers of hydrogen wavefunctions