Pure rotational spectra

Moment of inertia: i

2ii rmI

rotational axis through centre of mass.

key molecular parameter.

depends on molecular geometry.

information about bond lengths / angles.

In general, the rotational properties of anymolecule are expressed in terms of themoments of inertia about threeperpendicular axes set in the molecule Ia,Ib, and Ic (convention abc III ).

We initially suppose that molecules are rigid rotors (that is, nodistortion under stress of rotation).

Spherical rotors: 3 equal momentsof inertia (CH4, SiH4, and SF6).

Symmetric rotors: 2 equal momentsof inertia (NH3, CH3Cl, and CH3CN).

Linear rotors: 1 moment of inertia(molecular axis) equal to zero (CO2,HCl, OCS, CHHC ).

Asymmetric rotors: 3 differentmoments of inertia (H2O, H2CO,CH3OH).

The rotational energy levels

Energy of a body rotating around an axis a is:a

2a2

aaI2

JI

2

1E

a = angular velocity (rad/s), Ia = moment of inertia, and Ja =angular momentum. Hence, taking into account the 3 rotationalaxes:

c

2c

b

2b

a

2a

I2

J

I2

J

I2

JE This is the key equation!

Spherical rotors (3 equal moments of inertia)

I2

J

I2

JJJE

22c

2b

2a

J2 = square of the magnitude of the angularmomentum, and we established that:

22 )1(J JJ J = 0, 1, 2,…

Rotational energy levels of a spherical rotorare confined to:

,...2,1,0I2

)1(E2

JJJJ

We define the rotational constant B of the

molecule:I2

B2

, hence:

,...2,1,0)1(BE JJJJ

Separation of adjacent levels is: JJJ B2EE 1

Because B decreases as I increases, large molecules haveclosely spaced rotational energy levels.

Symmetrical rotors (2 equal moments of inertia)

Unique moment of inertia labelled: ||I , the other two: I .

The rotor is classified as: oblate (pancake) when ||I > I .

prolate (cigar) when I > ||I .

||

2a

2c

2b

I2

J

I2

JJE

; since 2c

2b

2a

2 JJJJ , we can write:

2a

||

2

||

2a

2a

2

JI2

1

I2

1

I2

J

I2

J

I2

JJE

and we established that:

JKK

JJJ

,,...1,0J

,...2,1,0)1(J

222a

22

Rotational energy levels of a symmetrical rotor are confined to:

22

||

2

I2

1

I2

1

I2

)1(E

K

JJKJ,

Using the rotation constants:||

2

I2A

,

I2

B2

, we have:

2)BA(1BE KJJKJ, J = 0, 1, …; K = 0,…, ±J

The equation for the rotational energy levels matches what weshould expect for the dependence of the energy levels on thetwo distinct moments of inertia of the molecule:

K = ±J, almost all angular momentumarises from rotation around the principalaxis, energy levels determined largely by

||I . Sign of K does not affect the energy

(since K2) because independent onsense of rotation.

K = 0: no component of angularmomentum about the principal axis,energy levels depend only on I .

Linear rotors (1 moment of inertia equal to zero)

Nuclei treated as mass points rotation occurs only about an axis to the line of atoms zero angular momentum around the line similar to the case of a symmetrical rotor BUT with K = 0.

Rotational energy levels of a linear rotor are confined to:

,...2,1,0I2

)1(E2

JJJJ

Same expression as for spherical rotor but K = 0 and A = B!

Degeneracies

For a symmetric rotor: K)(J,EE and each level (except K = 0)

is 2-fold degenerate (±K).

However, the angular momentum of the molecule has acomponent on an external, laboratory-fixed axis. Thiscomponent is quantized with permitted values JM , with

quantum number MJ = 0, ±1, ±2, …, ±J, giving 2J+1 values inall.

(a) (b) (c)

All drawings refer to a state with K = 0.(a) MJ close to its max value J. Molecular rotation is mostly around the

laboratory z-axis(b) Intermediate value of MJ.(c) MJ = 0, no angular momentum about the z-axis.

)(MJEE , but it is necessary for a complete specification of

the state of the rotor. Consequently, all 2J+1 orientations of therotating molecule have the same energy.

It follows that Erot is: 2(2J+1)-fold degenerate for 0K (2J+1)-fold degenerate for 0K

For a linear rotor: K = 0, but the angular momentum may stillhave 2J+1 components on the laboratory axis, so itsdegeneracy is 2J+1.

For a spherical rotor: seen as a symmetric rotor with A = B, thequantum number K still takes any of 2J+1 values, but (K)EE .

Hence the degeneracy is: 2J+1 due to orientation in space. 2J+1 due to orientation about arbitrary molecular axis.

The overall degeneracy is therefore: (2J+1)2. Example: J=10implies 441 states of same energy.

Centrifugal distortion

Atoms of rotating molecules are subject to centrifugal forces distortion of the molecular geometry stretching of the bond moments of inertia increase rotational constant decreases energy levels are slightly closer than rigid predictions

Treated empirically by subtracting a term from the energy:

22 )1(D1BE JJJJ JJ

DJ = centrifugal distortion constant = ameasure of a bond stiffness. Largewhen the bond is easily stretched.

For a diatomic molecule:22

3

h

B4D

J

Hence the observation of the convergence of the rotationallevels as J increases can be interpreted in terms of the rigidityof the bond.

Rotational transitions

Typical values of hc/BB for small molecules are in theregion of 0.1 to 10 cm-1 (for example, 0.356 cm-1 for NF3, and10.59 cm-1 for HCl), so rotational transitions lie in themicrowave region of the EMR spectrum.

Selection rules for observing a pure rotational spectrum

Gross selection rule: polar molecule(permanent electric dipole moment, hence, itappears to possess a fluctuating dipole whenrotating).

Rotationally inactive molecules: non-polar, homonucleardiatomic, symmetrical linear, spherical rotors (unless distortedby rotation)

Specific selection rules: evaluation of the transition dipolemoment between rotational states. We can show (tedious!)that, for a linear molecule, fi = 0 unless:

1,01 JMJ

with

emission1Δ

absorption1Δ

J

J

When the transition moment is evaluated for all possiblerelative orientations of the molecule to the line of flight of thephoton, it is found that the total JJ 1 transition intensity isproportional to:

20

20

2,1

2

1

12

1

J

JJJ for J >> 1

where 0 = permanent electric dipole moment of the molecule.

Hence, strongly polar molecules yield much more intenserotational lines than less polar molecules.

The appearance of rotational spectra

For convenience< we use: hc/BB

Energy levels of a rigid symmetric or linear rotor expressed aswavenumbers of the allowed JJ 1 absorptions are:

)1(B21 JJJ

for J = 0, 1, 2, … Spectra consists of:

lines with ,...B6,B4,B2

line separation = B2

line spacing gives B or B.

B gives moment of inertia.

Intensities of spectral lines increase with J and pass through amaximum because of the existence of a maximum in thepopulation of rotational levels. From Boltzmann (for a level J):

Tk/E BJegNN JJ

N = total number of molecules, gJ = degeneracy. Hence:

2

1

Bhc2

Tk0

d

dN2/1

Bmax

J

JJ

For a typical molecule (e.g. OCS, with B = 0.2 cm-1) at roomtemperature, hcB1000TkB , so 30max J .

Rotational Raman spectra

Gross selection rule: anisotropicallypolarisable: = E with = molecule’spolarizability.

Rotationally Raman inactive molecules: spherical rotors.

Specific selection rules:Linear rotors: 2,0 J

Symmetric rotors: 02,1,0 KJ

The J = 0 transitions do not lead to a shift of the scatteredphoton’s frequency and contribute to the unshifted Rayleighradiation.

Prediction of the form of the Raman spectrum of a linear rotorby applying the selection rule 2J to the rotational levels:

2 J : the molecule is left in a higher rotational state,hence a decrease in i :

)32(B22 i JJJ

Stokes lines < i and at displacements of 6B ,

10B , 14B ,... from i for J = 0, 1, 2,…

2 J : the molecule is left in a lower rotational state,hence a decrease in i :

)2(B22 i 1JJJ

Anti-Stokes lines > i and at

displacements of 6B , 10B , 14B ,... from

i for J = 2, 3, 4,… (notice that 2J is

the lowest state that can contribute underthis selection rule).

Separation of both Stokes and anti-Stokes lines is 4B , hence I can be

extracted and used to find the bondlengths as in microwave spectroscopy.

An example of application:

Let’s predict the form of the rotational absorption spectrumand the rotational Raman spectrum (when exposed tomonochromatic 336.732 nm laser radiation) of 14NH3.

First, we calculate the moment of inertia for 14NH3

Symmetric rotor with bond length 101.2 pm and HNH bondangle of 106.7°. Using trigonometry, with the bond angle andbond length r, one shows that for a symmetric molecule:

2H|| r)cos1(m2I

2

NH

NH2H r)cos21(

mm

mmr)cos1(mI

Numerically, with:mH = 1.0078 u and mN = 14.0031 u,

247|| mkg10413.4I & 247 mkg10806.2I

.

Second, we calculate the rotational energy levels of 14NH3

Rotational constants: A = 6.344 cm-1 and B = 9.977 cm-1.

Energy levels: 2K633.31977.9E JJ cm-1.

Upon multiplication by c (speed of light) E acquires units of

frequency: 2K9.10811.299E JJ GHz.

For J = 1, the energy needed for the molecule to rotate mainlyabout its figure axis (K = ±J) is equivalent to 16.32 cm-1 (489.3GHz). But end-over-end rotation (K = 0) corresponds to 19.95cm-1 (598.1 GHz). Not a surprise, it is microwaves!

Finally, we can predict the form of the rotational spectrum:

Since NH3 is a polar symmetric rotor, the following selectionrules apply:

1J and 0K

For absorption, 1J and we can use )1(B21 JJJ . Because B = 9.977 cm-1, we can draw

up the following table for the JJ 1 transitions:

J 0 1 2 3 … (cm-1) 19.95 39.91 59.86 79.82 … (GHz) 589.1 1197 1795 2393 …

The line spacing is 19.95 cm-1 (589.1 GHz).

The molecule is Raman active since non-spherical. i =336.732 nm corresponds to i = 29697.2 cm-1. We use the

equations defined to obtain the following line positions:

J 0 1 2 3Stokes lines:

(cm-1) 29637.3 29597.4 29557.5 29517.6Anti-Stokes lines:

(cm-1) 29757.1 29797.0

EXAM

Of the molecules N2, CO2, OCS, H2O, CH2=CH2, C6H6, only OCSand H2O are polar, so only these two molecules have microwavespectra. Which of the molecules H2, NO, N2O, CH4 can have a purerotational spectrum?

For symmetric rotors, an additional selection rule states that K =0. To understand this rule, consider the symmetric rotor NH3, wherethe electric dipole moment lies parallel to the figure axis. Such apolar molecule cannot be accelerated into different states ofrotation around the figure axis by the absorption of radiation, so K= 0.

From a rot spectrum, because the masses of the atoms are known,it is a simple matter to deduce the bond length of a diatomicmolecule.