Rotational Spectroscopy

Home , Rigid rotor, Rotational spectroscopy

Recommended Reading: 1. Banwell and McCash: Chapter 2 2. Atkins: Chapter 16, sections 4 - 8

Aims In this section you will be introduced to 1) Rotational Energy Levels (term values) for diatomic molecules and linear polyatomic molecules 2) The rigid rotor approximation 3) The effects of centrifugal distortion on the energy levels 4) The Principle Moments of Inertia of a molecule. 5) Definitions of symmetric , spherical and asymmetric top molecules. 6) Experimental methods for measuring the pure rotational spectrum of a molecule Microwave Spectroscopy - Rotation of Molecules Microwave Spectroscopy is concerned with transitions between rotational energy levels in molecules.

Definition d Electric Dipole: p = q.d +q -q p

H Most heteronuclear molecules possess Cl a permanent dipole moment -q +q e.g HCl, NO, CO, H O... p 2

Molecules can interact with electromagnetic radiation, absorbing or emitting a photon of frequency ω, if they possess an electric dipole moment p, oscillating at the same frequency Gross Selection Rule: A molecule has a rotational spectrum only if it has a permanent dipole moment.

Rotating molecule

_ + _ +

_ + _ + dipole momentp dipole

Homonuclear molecules (e.g. O2, H2, Cl2, Br2…. do not have a permanent dipole moment and therefore do not have a microwave spectrum! General features of rotating systems m Linear velocity v angular velocity v = distance ω = radians O r time time

v = ω × r Moment of Inertia I = mr2. A molecule can have three different moments of inertia IA, IB and IC about orthogonal axes a, b and c.

2 I = ∑miri i R Note how ri is defined, it is the perpendicular distance from axis of rotation ri Rigid Diatomic Rotors

ro IB = Ic, and IA = 0. m1 m2 C = centre of gravity. C Express I in terms of m1, r1 r2 m2 and r0.

r1 + r2 = r0 (1) m r m r 2 m1⋅m2 2 2 from 1⋅ 1 = 2 ⋅ 2 ( ) I = r0 = μ⋅r0 2 2 m1 + m2 I = m1⋅r1 + m2 ⋅r2 (3) Derive this expression NOTE: Units of I = kg.m2. μ = reduced or effective mass of the molecule Energy and Angular Momentum

Angular momentum about axis a is JA = IAωA Energy of a body rotating about an axis ‘a’ with constant angular velocity 1 is E = Ia ω2 a 2 A A

A body free to rotate about all three axes has a total rotational energy 1 1 1 E = I ω2 + I ω2 + I ω2 2 A A 2 B B 2 C C And a total angular momentum given by

J2 J2 J2 J2 = A + B + C Recall that J is 2 2 2 a vector = (IAωA ) + (IBωB ) + (ICωC ) Energy Levels of a Rigid Diatomic Rotor

For diatomic molecule IB = Ic = I and IA = 0 so:

2 2 2 1 2 1 2 JB Jc J E = IBωB + ICωC = + = 2 2 2IB 2Ic 2I

Can find quantum energy levels using Correspondence Principle: 2 2 2 J(J+ 1)⋅h J → J(J+ 1)⋅ = 4π2 1. Rotational energy for a rigid diatomic rotor is quantised

J(J+ 1)⋅h2 Units: Joules EJ = 8π2I J = 0, 1, 2, 3, … = rotational quantum number Energy Levels of a Rigid Diatomic Rotor

2. Rotational energy is usually expressed in units of cm-1. E Recall E = h ν = h ⋅ c ⋅ ~ν therefore ~ν = ε = then h⋅c 2 EJ h J(J+ 1) h -1 εJ = = = J(J+ 1) cm h⋅c 4π2 ⋅2I⋅hc 8π2Ic

B - Rotational Constant, units cm-1

h  ε = B⋅J(J+ 1) where B = = J 8π2Ic 4πIc

εJ gives the allowed energy levels of the rigid rotor (in cm-1). B can be determined experimentally by spectroscopy Notation: Rotational Term F(J) The energy of a rotational state is normally reported as the Rotational Term, F(J), a wavenumber divided by hc "E F(J) ! !J = = B# J (J +1) hc

Ritz Combination Principle The wavenumber of any spectral line is the difference between two terms. Two terms T1 and T2 combine to produce a spectral line of wavenumber. ΔE ~ν = T − T = 1 2 hc Energy Levels of a Rigid Diatomic Rotor

εJ = B⋅J(J+1) J = 0, 1, 2, 3 ..... 6 42B

F(J) (= ε ) J J 5 30B

4 20B

3 12B

2 6B 1 2B 0 0 Can we predict nature of a microwave rotational spectrum? Rotational Transitions in Rigid Diatomic Molecules Selection Rules: J = 1. A molecule has a rotational spectrum 5 only if it has a permanent dipole moment. 2. ΔJ = ± 1 +1 = adsorption of photon, -1 = emission of photon. 4

Transitions observed in absorption spectrum. 3 J = 0 è J = 1 2 ~ν = Δε = ε − ε = 2B − 0 = 2B cm-1 J J=1 J=0 1 J = 1è J = 2 0 6B ~ν Δε ε ε 6B 2B 4B cm-1 = J = J=2 − J=1 = − = 4B 8B 2B 10 B ~ -1 In general νJ→J+1 = 2B(J+ 1) cm

Spacing between lines in spectrum = 2B cm-1 Rotational Spectra of Rigid Diatomic Molecules

h Line separation in the rotational spectrum of HCl is ≈ B 21.2 cm-1 è B = 10.6 cm-1; I can be found from = 2 HCl 8π Ic m m Once I is known, then r (the bond 1⋅ 2 2 2 HCl HCl I = r0 = µ ⋅r0 length) can be determined from m1 + m2

Need to know mH and mCl, but these are known and tabulated. Summary so far: 1. Microwave spectroscopy is concerned with transitions between rotational energy levels of molecules

2. General features of rotational systems: I, ω, μ

I m r2 m1⋅m2 2 2 = ∑ i i I = r0 = μ⋅r0 i m1 + m2 3. Energy Levels of a rigid diatomic rotor h F(J) ≡ ε = B⋅ J(J+ 1) B = J 8π2Ic J = 0, 1, 2, ….; B = rotational constant, units cm-1

4. Selection rules: (1) permanent dipole moment, (2) ΔJ = ± 1 only 5. Spacing between lines of in rotational spectra of rigid diatomic molecules is constant and equal to 2B cm-1. Why is Rotational Spectroscopy important?

1. bond lengths From pure rotational 2. atomic masses spectra of molecules we 3. isotopic abundances can obtain: 4. temperature

Important in Astrophysics: Temperature and composition of interstellar medium

Diatomic molecules found in interstellar gas:

H2, OH, SO, SiO, SiS, NO, NS, HCl, PN, NH, CH+, CH, CN, CO,

CS, C2. S.Taylor, D. Williams, Chemistry in Britain, 1993 pg 680 - 683 Population of Rotational Energy Levels

NJ+1 # "E & # ! J hc& Boltzmann distribution: = exp% ! ( = exp% ! ( NJ=0 $ kT ' $ kT '

Orientation of the angular momentum L is quantised. This results in a (2J+1) degeneracy of the energy level PopulationJ $ #E ' $ BJ (J +1)hc' ! (2J +1)exp& " ) = (2J +1)exp& " ) PopulationJ=0 % kT ( % kT (

Spectral Line Intensity ∝ Population Example: Rotational Spectrum of HCl

m1 m2 C

r1 r2

B = 10.6 cm-1, T = 300K Maximum Value of J At what value of J will the intensity be a maximum, at a given temperature?

We can find the value of Jmax by finding the value of J that maximises the population: $ #E ' $ BJ (J +1)hc' Population ! (2J +1)exp& " ) = (2J +1)exp " % kT ( %& kT ()

Differentiate w.r.t. J and set equal to zero for a maximum. This gives kT 1 J = − max 2hcB 2 must round this value to the nearest integer (J can only be an integer) Bond lengths and Rotational Constants of some Diatomic Molecules

Note effect of r0 and m on the rotational constant B.

I ∝ r0 and m and B ∝ 1 / I Effect of Isotope Substitution

From spectra we can obtain: bond length or atomic weights and Isotopic Abundances Effect of Isotope Substitution on Spectra For 12C16O B’ = 1.92118 cm-1. 13C16O B’’ = 1. 83669 cm-1. h 2 B B' h 8π I''c I'' µ'' = 2 = = = = 1.046001 8π Ic B'' 8π2I'c h I' µ' Atomic weight 12C = 12.0000, 16O = 15.9994 13C = 13.0007, 12C16O J = 5 13C16O

2 1 0 Non-Rigid Rotor For a non-rigid rotor the bond-length increases as the angular velocity increases Centrifugal Distortion h B 1 B decreases as J = 2 B ∝ 8π Ic r2 increases

For a non-rigid rotor

2 2 εJ = B⋅ J(J+ 1) − D⋅ J (J+ 1)

D - Centrifugal Distortion constant (stiffness constant) units: cm-1.

Like B, D depends on the molecule rigid rotor non-rigid rotor 2 2 εJ = B⋅J(J+1) εJ = B⋅ J(J+ 1) − D⋅ J (J+ 1) J = 5

2 1 0

cm-1 2B 4B 6B 8B 10 B 12 B Non-Rigid Rotor

2 2 εJ = B⋅ J(J+ 1) − D⋅ J (J+ 1)

~ 3 ν = εJ+1 − εJ = 2B⋅(J+ 1) − 4 D⋅(J+ 1) The rigid rotor selection rules still apply, i.e. ΔJ = ±1 Typical values: B ~ 1 - 10 cm -1 and D ~ 10-3 – 10-2 cm-1. 4B3 D can be related to the D = ~2 vibrational frequency of the molecule νvib Summary 1. Population of energy levels: Boltzmann Distribution degeneracy of rotational states

2. Efect of bond length and mass on rotational constant of diatomic molecule

3. Efect of isotopic substitution atomic weights and isotopic abundances

4. Non-rigid rotor D - centrifugal distortion constant