Molecular Vibrations

Chem 3240 · Lecture 4.3

Davit Potoyan

From the ladder to a spectrum

A photon is absorbed when h\nu matches a gap. Spectroscopists measure light in wavenumbers, \tilde{\nu} = 1/\lambda:

E = h\nu = \frac{hc}{\lambda} = hc\,\tilde{\nu}

G(v) = \frac{E_v}{hc} = \tilde{\nu}\left(v + \frac{1}{2}\right), \qquad \tilde{\nu} = \frac{1}{2\pi c}\sqrt{\frac{k}{\mu}}

Lifting v \to v + 1 takes a photon of wavenumber \tilde{\nu}: the bond’s own vibration

What one wavenumber is worth

quantity	value	what it tells you
1 cm^{-1}	1.24\times10^{-4} eV = 0.0120 kJ/mol	the conversion factor hc
1 eV	8066 cm^{-1}	chemical bonds: a few eV
k_BT at 300 K	208 cm^{-1}	below this, levels are warm
bond vibrations	200 to 4400 cm^{-1}	2.3 to 50 μm: the infrared

Italic v: the quantum number. \tilde{\nu} (“nu tilde”): the wavenumber

Example: the HCl line

HCl absorbs at 2886 cm^{-1}. Find the wavelength, frequency and photon energy, and compare with k_BT = 208 cm^{-1}.

\lambda = \frac{1}{\tilde{\nu}} = 3.47\ \mu\text{m}, \qquad \nu = c\tilde{\nu} = 8.65\times10^{13}\ \text{Hz}

E = hc\tilde{\nu} = 0.358\ \text{eV} = 34.5\ \text{kJ/mol} \approx 14\,k_BT

The line sits below the harmonic 2990 cm^{-1} of Lecture 4.2: anharmonicity, later today

Light needs a transition dipole

\text{absorbed only if}\quad \langle v'|\hat{M}|v\rangle = \int \psi_{v'}\,M(x)\,\psi_v\,dx \neq 0

M(x) is the dipole at stretch x. No area, no absorption: the dipole must change

Selection rules

\langle v'|\hat{M}|v\rangle \approx M'\,\langle v'|\hat{x}|v\rangle, \qquad M' = (\partial M/\partial x)_{R_e}

One rung at a time: \hat{x} connects only neighbors (Lecture 4.2), so \Delta v = \pm 1

Seeing it: HCl and N_2 vibrating

HCl’s dipole swings with the bond. N_2, O_2 (99% of air): no IR spectrum
At 300 K nearly all molecules start in v = 0: one strong line, the fundamental

Stiffness and mass set the wavenumber

\tilde{\nu} \propto \sqrt{k/\mu}. Halides: same \mu, weaker bonds down the group. Bond order: N\equivN > O=O > Cl-Cl

Example: a carbon-13 label

^{12}C^{16}O absorbs at 2143 cm^{-1}. Where does ^{13}C^{16}O absorb? (Same k: the electrons do not change.)

\mu(^{12}\text{C}^{16}\text{O}) = 6.856\ \text{u}, \qquad \mu(^{13}\text{C}^{16}\text{O}) = 7.172\ \text{u}

\tilde{\nu} = 2143\ \text{cm}^{-1}\times\sqrt{\frac{6.856}{7.172}} = 2095\ \text{cm}^{-1} \quad\text{(measured: 2096)}

A 48 cm^{-1} shift: isotope labels tag one bond in a big molecule

Real bonds are anharmonic

The real well flattens toward dissociation: the steps shrink, and the ladder ends

The anharmonic oscillator

G(v) = \tilde{\nu}_e\left(v + \frac{1}{2}\right) - \tilde{\nu}_ex_e\left(v + \frac{1}{2}\right)^2

\tilde{\nu}_{0\to v} = v\,\tilde{\nu}_e - v(v + 1)\,\tilde{\nu}_ex_e

\tilde{\nu}_e: the harmonic wavenumber. x_e: the anharmonicity, about 0.005 to 0.02
The gaps shrink by 2\tilde{\nu}_ex_e per rung
Breaking the bond from v = 0 costs D_0 = D_e - G(0)

Overtones

Anharmonicity makes \Delta v = 2, 3 weakly allowed, each tens of times fainter
Hot bands start from v = 1: faint at 300 K, stronger in a hot gas

Example: HCl constants from two lines

HCl: fundamental 2886 cm^{-1}, first overtone 5668 cm^{-1}. Find \tilde{\nu}_e and \tilde{\nu}_ex_e; predict the second overtone.

\tilde{\nu}_e - 2\tilde{\nu}_ex_e = 2886, \qquad 2\tilde{\nu}_e - 6\tilde{\nu}_ex_e = 5668

\tilde{\nu}_ex_e = 52\ \text{cm}^{-1}, \qquad \tilde{\nu}_e = 2990\ \text{cm}^{-1}

\tilde{\nu}_{0\to3} = 3(2990) - 12(52) = 8346\ \text{cm}^{-1}

Measured: 8347 cm^{-1}. And 2990 is the \hbar\omega of Lecture 4.2.

One oscillator per normal mode

G(v_1, v_2, \dots) = \sum_i \tilde{\nu}_i\left(v_i + \frac{1}{2}\right)

3N - 6 modes (3N - 5 linear), each an independent oscillator: the energies add
A mode absorbs only if its own motion changes the dipole
Silent modes show up in Raman spectra, which sense the polarizability instead

CO_2: which modes absorb?

Symmetric stretch: the two halves cancel at every instant. Antisymmetric stretch and bend: the dipole swings

Example: why CO_2 is a greenhouse gas

The Earth (288 K) radiates most strongly near 10 μm. N_2, O_2 and CO_2: which can absorb it?

N_2 and O_2: no dipole at any length, transparent.

CO_2 bend: 1/(667\ \text{cm}^{-1}) = 15.0\ \mu\text{m}, right inside the Earth’s emission. Antisymmetric stretch: 4.26 μm. Symmetric stretch: silent.

0.04% of the air, absorbing where the other 99% cannot

Reading an infrared spectrum

Bonds to H absorb highest: the light H atom, not a stiffer bond. Then triple, double, single

Proteins: the amide I band

The backbone C=O stretch near 1650 cm^{-1} shifts with helix or sheet: infrared reads protein structure

Where the force-constant matrix comes from

One bond (Lecture 4.1): V \approx \frac{1}{2}kx^2 with k = V''. Many atoms: Taylor in every displacement; at the minimum all slopes vanish:

V \approx \frac{1}{2}\sum_{i,j} K_{ij}\,x_ix_j, \qquad K_{ij} = \frac{\partial^2 V}{\partial x_i\,\partial x_j}\ \ \text{(the Hessian)}

CO_2 with one spring per bond:

V = \tfrac{1}{2}k(x_2 - x_1)^2 + \tfrac{1}{2}k(x_3 - x_2)^2 \quad\Longrightarrow\quad K = k\begin{pmatrix} 1 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 1 \end{pmatrix}

Newton: \;m_i\ddot{x}_i = -\sum_j K_{ij}x_j. Off-diagonal entries couple the atoms

Look for a mode: an eigenvalue problem

Weight by mass, q_i = \sqrt{m_i}\,x_i: \quad\ddot{\mathbf{q}} = -F\mathbf{q}, \quad F_{ij} = K_{ij}/\sqrt{m_im_j} is symmetric

Guess a normal mode: every atom in step, at one frequency, in fixed proportions \mathbf{u}

\mathbf{q}(t) = \mathbf{u}\cos\omega t \quad\Longrightarrow\quad -\omega^2\,\mathbf{u}\cos\omega t = -F\mathbf{u}\cos\omega t

F\,\mathbf{u}_k = \omega_k^2\,\mathbf{u}_k

Eigenvectors: the mode shapes. Eigenvalues: \omega_k^2. Six (five) zeros: translations, rotations
F symmetric: the eigenvectors are perpendicular, \mathbf{u}_k\cdot\mathbf{u}_l = \delta_{kl}

Three symbols

symbol	what it is	in time	CO_2 example
\mathbf{q}(t)	the configuration: every mass-weighted displacement	changes	(\sqrt{m_O}\,x_1,\ \sqrt{m_C}\,x_2,\ \sqrt{m_O}\,x_3)
\mathbf{u}_k	mode k: a fixed shape, an eigenvector of F	fixed	\mathbf{u}_{\text{sym}} = (-1, 0, 1)/\sqrt{2}
Q_k(t)	how much of mode k is present, Q_k = \mathbf{u}_k\cdot\mathbf{q}	changes	Q_{\text{sym}} = A\cos\omega_{\text{sym}}t

\mathbf{q}(t) = \sum_k Q_k(t)\,\mathbf{u}_k \qquad\text{like}\qquad \psi = \sum_n c_n\,\psi_n

Configuration = (how much) \times (shape). The guess of the last slide: one Q_k = \cos\omega t, the rest zero

Project onto the modes: they come apart

Start from \mathbf{q} = \sum_k Q_k\,\mathbf{u}_k and project Newton’s equation onto one axis \mathbf{u}_k:

\begin{aligned} \ddot{Q}_k &= \mathbf{u}_k\cdot\ddot{\mathbf{q}} = -\mathbf{u}_k\cdot(F\mathbf{q}) && \text{Newton} \\ &= -(F\mathbf{u}_k)\cdot\mathbf{q} && F \text{ symmetric} \\ &= -\omega_k^2\,\mathbf{u}_k\cdot\mathbf{q} && \text{eigenvector} \end{aligned}

\ddot{Q}_k = -\omega_k^2\,Q_k

Only Q_k on the right: each mode is its own spring, whatever the others do

The Hamiltonian becomes a sum

Mass-weighted: kinetic \frac{1}{2}\dot{\mathbf{q}}\cdot\dot{\mathbf{q}}, potential \frac{1}{2}\mathbf{q}\cdot F\mathbf{q}. Insert \mathbf{q} = \sum_k Q_k\mathbf{u}_k:

\dot{\mathbf{q}}\cdot\dot{\mathbf{q}} = \sum_{k,l}\dot{Q}_k\dot{Q}_l\,\mathbf{u}_k\cdot\mathbf{u}_l = \sum_k\dot{Q}_k^2, \qquad \mathbf{q}\cdot F\mathbf{q} = \sum_{k,l}Q_kQ_l\,\omega_l^2\,\mathbf{u}_k\cdot\mathbf{u}_l = \sum_k\omega_k^2Q_k^2

H = \sum_k\left(\frac{1}{2}\dot{Q}_k^2 + \frac{1}{2}\omega_k^2Q_k^2\right) = \sum_k H_k

\mathbf{u}_k\cdot\mathbf{u}_l = \delta_{kl} kills every cross term, like the -kx_1x_2 that passes energy between atoms
Quantum: \hat{H} = \sum_k\hat{H}_k, so energies add, G = \sum_k\tilde{\nu}_k(v_k + \frac{1}{2}) (Lecture 4.1)

CO_2: coupled atoms, independent modes

\omega_{\text{sym}}^2 = \frac{k}{m_O}, \qquad \omega_{\text{anti}}^2 = \frac{k}{m_O}\left(1 + \frac{2m_O}{m_C}\right)

Modes (-1, 0, 1) and (1, -2.7, 1); frequency ratio 1.91 (measured 1.69)

Live: energy in atoms and in modes

Takeaway

An infrared spectrum reads the vibrational ladder in wavenumbers, G(v) = \tilde{\nu}(v + \frac{1}{2}) with \tilde{\nu} = \frac{1}{2\pi c}\sqrt{k/\mu}. Light is absorbed only if the transition dipole \langle v'|\hat{M}|v\rangle is not zero: the dipole must change as the bond stretches, and a harmonic bond climbs one rung at a time. Real bonds are anharmonic: the steps shrink, the bond can break, and weak overtones appear. A polyatomic molecule is a set of independent oscillators, one per normal mode, found as the eigenvectors of F\mathbf{u} = \omega^2\mathbf{u}; CO_2’s bend absorbs the Earth’s infrared, N_2 and O_2 cannot.