## Vibration and Vibrational Spectra of Diatomic Molecules

We have got knowledge of energy levels and wavefunctions for harmonic vibrations as an approximation of diatomic molecule potential curve. Since electric dipole moment of molecule should change during vibration homonuclear molecules don't have vibrational spectra. We obtain the selection rule in harmonic approximation for heteronuclear molecule:

Δv = ± 1
E = h w0(v + ½) | v = 0, 1, ...

Derivation of selection rules for harmonic oscillator:

The dipole moment µ can be expanded into distance r and correspondingly z:

µ = µ0 + dµ/dz½zo· z , i.e. µ ≈ µ0 + const. · z.

The harmonic oscillator eigenfunction is ym~e-z²/2 · Hm(z). We obtain:

µ = -∞∫+∞ ym* · µ · yndz = -∞∫+¥ e-z²/2 Hm(z) [µ0 + const · z] e-z²/2 Hn(z) dz

= µ0 -∞∫+∞ e-z²· Hm(z) · Hn(z) dz + const -∞∫+¥ e-z² Hm(z) · z · Hn(z) dz

The first term gives its contribution for m = n where Hindex are orthogonal Hermitian polynomials. For transition m ¹ n, however. For the second term estimation one can use the following recursion formula

Hn(z) = 2z Hn-1(z) - 2(n -1)Hn-2(z)

Substituting n → n+1 and transforming into: z · Hn(z) = ½ Hn+1(z) + n · Hn-1(z) one can finally obtain:

const -∞∫+∞ e-z²Hm(z)½Hn+1(z) dz | + | const · -∞∫+∞ e-z²Hm(z) · n · Hn-1(z)dz
½¾¾¾¾¾¾¾¾¾¾¾¾½ |  | ½¾¾¾¾¾¾¾¾¾¾¾¾¾½
¯ |  | ¯
it's not zero only when m = n + 1 |  | it's not zero only when m = n - 1

I.e. the quantum number is changed for transition between two levels only on 1.

The best potential description is given by Morse potential

V(r) = De[1 - e-α(r-re)]2

where De is the potential minimum depth in the point xe and

α = (µ/2De)½ · w0

Having this potential one has the following energy levels:

E = h w0(v + ½) - h w0xe(v + ½)2 | xe = α2h/(2µw0) = h w0/(4De)

Unfortunately, the Morse potential is seldom used; it's rather to solve the Schroedinger equation for each new problem again. We will write down the general expression for the data analysis:

Ev = h we(v + ½) - h wexe(v + ½)2 + h weye(v + ½)3 + ...

where quantities wexe, weye, ... are corresponding constants (and they are not products). The separation two levels is then as follows (v + 1 ← v):

ΔEvib = h[we(v + 3/2) - wexe(v + 3/2)2 + ... - we(v + ½) + wexe(v + ½)2 - ...]

ΔEvib = h[we− 2wexe(v + 1) + ... ] (i.e. v-dependent !)

The ground vibration v = 1 ← v = 0:

w0 = we - 2wexe + ...

The zero point energy Ev=0 is as follows:

Ev=0 = 1/2h we−1/4h wexe + 1/8h weye + ...

Example: (w's in cm-1)

| we | wexe | w0 | re/pm |
H2 | 4401 | 121 | 4159 | 74,1 | ← IR-inactive !
HCl | 2991 | 52,8 | 2885 | 127,5 |
N2 | 2359 | 14,3 | 2330 | 109,8 | ← IR-inactive !
I2 | 214 | 0,61 | 212,8 | 266,6 | ← IR-inactive !
OH | 3738 | 84,9 | 3568 | 96,97 |

Since vibrational energy is bigger comparing with kT for room temperature there are only transition from v = 0 till v = 1 in absorption. Surely, product can be made up during chemical reactions into some very high vibrational levels and one can observe emission from these levels.
For instance, reaction H + O3 → OH(v = 9) + O2 occurs in the upper stratosphere and one can observe weak vibrational transitions.

Two more useful expressions:

The molar binding dissociation enthalpy ΔHm and dissociation energy D0 are connected by the following relation:

ΔHm = NAD0 + RT | NA: Avogadro number

The difference between dissociation energy D0 and depth De of the potential well is as follows:

De = D0 + ½ h we - ¼ h wexe

Overtones

According to anharmonicity one can observe lines that are forbidden by the selection rule: Δv = ± 1. These overtones have weaker intensities. However we can use lasers with which one can easily induce overtones by using their high powers.

Example: Transition probabilities for OH
| v' ← v'' | n / cm-1 | λ / nm | σ0/norm. by 1 ¬ 0 º 100
| 1 ¬ 0 | 3568 | 2803 | 100
| 2 ¬ 0 | 6966 | 1436 | 16
| 3 ¬ 0 | 10195 | 980 | 0,5
visual range | 4 ¬ 0 | 13253 | 755 | 0,03
spectral range! | 5 ¬ 0 | 16142 | 619,5 | 0,01

Polyatomic molecule vibrations

Nonlinear N-atomic molecule has 3N-6 normal vibrations; linear molecule - 3N-5. CO2 as linear molecule has 4 vibrational degrees of freedom (the bending vibrations are met twice in it); whereas water has 3 vibrational degrees of freedom. When having normal vibrations the excited vibrational mode doesn't influence on other normal vibrations.

When we have a deal with more huge molecules it's worthwhile to divide normal vibrations according to its symmetry into few series (see theory of groups, molecular symmetry). The same is also true when we want to know which vibrations are active in IR.

In this additional figure one can see 3 normal vibrations for H2O (all IR-active).

Vibrational levels for gas-like water given in cm-1. The values in brackets are the vibrational modes (n1, n2, n3).