Rotational Spectroscopy- QM model of motion: rigid rotor- Moments of inertia (16.4)- Rotational IR, Millimetre Wave and Microwave

Spectra (16.5, 16.6)i. Diatomic & Linear Polyatomic Moleculesii. Symmetric Rotorsiii. Stark Effectiv. Asymmetric Rotors and Spherical Moleculesv. Applications in Astronomy

- Rotational Raman Spectroscopy (16.7, 16.8)- Structure Determination from Rotational Spectroscopy

Rotational Spectroscopy

Rigid Rotor: Diatomic RotationThe rigid rotor model is an excellent approximation to theproblem of freely rotating molecules. There can be (i) noadditional torques, (ii) no angular potential energydependence, (iii) no significant intermolecular interactions(makes it good for gas phase analysis), (iv) no majorenergetic contributions from molecular vibrations.

m1 m2

r1 r2

r = r1 + r2

Centre ofgravity

Here r = r1 + r2, and r1 and r2 are defined such thatm1r1 = m2r2

rotation axis

The moment of inertia about this rotation axis isI = m1r1

2 + m2r22

Substituting in the above equation, we write

I = m2r2r1 + m1r1r2 = r1r2(m1 + m2)

Since r2 = r - r1,

m1r1 = m2(r - r1)

m1r1 + m2r1 = m2r

Moment of Inertia & Reduced MassWe can now solve for ri:

r1 'm2

m1 % m2

r , r2 'm1

m1 % m2

r

and finally calculate the moment of inertia:

I 'm1m2

m1 % m2

r 2' µr 2

where the reduced mass is given by

µ 'm1m2

m1 % m2

m1 m2

r

µ

r

Thus, the two bodyrigid rotor can bereduced to a onebody problem witha single reducedmass, meaning thatwe can use one-bodied angularmomentum to treatthese problems!

L = r × p, angularmomentum vectorz to rotation

Schroedinger Equation for Rigid RotorsThe problem of the rotating diatomic molecule can now bereduced to treating the motion of the particle on a sphere(much like for the hydrogenic wavefunctions). Recall thatfor translational motion the Schroedinger equation is:

&£ 2

2mM2

Mx 2%

M2

My 2%

M2

Mz 2Q(x,y,z,t) % V(x,y,z,t)Q(x,y,z,t) ' i£ MQ(x,y,z,t)

Mt

and since the V = 0, things are greatly simplified. In onedimension, the kinetic energy operator for linear motion is:

T ' &£ 2

2mM2

Mx 2'

p 2x

2m, p x '

£iMMx

and recalling that linear momentum, p, and angularmomentum, L, are related by similar classical equations:

T 'L 2

2I' &

£ 2

2IM2

MN2

If N is the azimuthal angle in the polar coordinate system,then this operator is for calculating the energy associatedwith rotation about a ring. If three dimensional rotation isto be examined (i.e., kinetic energy anywhere on a sphere)

If the potential energy is assumed to be zero for the rigidrotor model, the above operators are the total hamiltonians.

T ' &£ 2

2I1

sin2MM2

sin2 MM2

%1

sin22

M2

MN2

Angular Momentum ReviewThe angular momentum of the molecule is directedperpendicular to the plane of rotation, pointing up or downdepending on whether rotation is clockwise or counter-clockwise (see the right hand rule).Consider angular momentum in the x, y and z directions. In turns out that the operators do not commute with oneanother, implying we cannot simultaneously specify allthree components of angular momentum:

[A, B] ' AB & BA[L x, L y] ' i£Lz [L y, L z] ' i£Lx [L z, L x] ' i£Ly

[L y, L x] ' &i£Lz [L z, L y] ' &i£Lx [L x, L z] ' &i£Ly

However, a special operator, , commutes with all threecomponents of angular momentum:

[L x, L2] ' [L y, L

2] ' [L z, L2] ' 0

L2

This means we may specify the magnitude (via ) andone component (e.g., ) of angular momentum. Whentwo operators commute, they share a common set ofeigenfunctions. is the simplest form:

L2

L z

L z ' &i£ MMN

It turns out that the eigenfunctions shared by and are the spherical harmonics, !

L z L2

YRm R(2,N)

L z

Solution for Rigid RotorsSince the spherical harmonics are simultaneouslyeigenfunctions of and , the solutions of the S.E. areL z L2

L2YRm R' R(R % 1)£ 2 YRm R

, R ' 0, 1, 2,..., n&1

L zYRm R' m £YRm R

, mR ' 0, ±1, ±2,..., ±RR

These operators are of paramount importance in manyproblems dealing with angular momentum.The angular momentum vector in a molecule has aparticular magnitude:

* PL* ' R(R % 1)£

and for any value of R, there are 2R + 1 possible orientationscorresponding to the allowed values of Lz = m£. Since thex and y components of angular momentum are unspecified,this means the vector can lie anywhere on a conemaking and angle 2 w.r.t. the z axis, with angles

PL*

PL

cos2 'm

R(R % 1)

This also implies that cannot coincidewith the z direction,as this would impartdefinite values on Lxand Ly (i.e., they’d beequal to zero!)

PL

Energy Levels of the Rigid RotorSchroedinger’s equation is now deceptively simple:

J 2

2IR ' EJR '

J(J % 1)£ 2

2IR

Here we have replaced L with J, which is a special notationto denote angular momentum of rigid rotors (it is the sameas L in every way, with mJ = 0, ±1, ±2, ..., ±J).

Thus, the energy can be calculated as:

EJ '£ 2

2IJ(J % 1)

The rigid rotor has quantized angular momentum:

Remember, the energy comes from EJ = J2/2I. J and mJdefine the spherical harmonics, but only J defines theenergies (mJ levels are degenerate (same energy) unless inthe presence of a magnetic field).

*PJ* ' J(J % 1)£

This notation is usually simplified even further, by usingthe rotational constant, B:

B = £2/2I (in Joules)

since we know that I will be constant for a given rigidrotor. Then, energy is given by:

EJ = B J(J + 1)

The spectrum of a pure rigid rotor has transitions evenlyspaced by )E = 2B(JO+1), where JO is the lower rotationalenergy level. The selection rule for this model is: )J = ±1

Spectrum of a Rigid Rotor

20B

12B

6B

2B0

4

3

2

10

EJ J

2B 4B 6B 8B 10B 12B etc.

2BE

- There is no zero-point energy,i.e., when J = 0 the energy EJ =0 as well.

- The “rotational ladder” ofenergy levels is not evenlyspaced - spacing increases withincreasing J

- Spherical harmonic functionscan be used as solutions to thisSchroedinger equation, sincethis motion is described by aparticle on a sphere

- Other types of motion (e.g.,vibration) can shift these levelsand create anharmoncity

Moments of InertiaIn order to use rotational spectroscopy to study thestructure of molecules, they must be classified according totheir principal moments of inertia. The moment of inertia,I, of any molecule about any axis through the centre ofgravity of the molecule is given by:

I ' ji

mir2i

where mi and ri are the mass and distance of the atom ifrom the axis.There are three principal axes which define the moment ofinertia, conventionally labelled a, b and c, which are allmutually perpendicular. The principal moments ofinertia, Ia, Ib and Ic are defined such that:

Ic $ Ib $ Ia

There are a variety of molecular shapes for which relativeprincipal moments of inertia can be defined:Ic ' Ib > Ia ' 0

Ic ' Ib > Ia

Ic > Ib ' Ia

Ic ' Ib ' Ia

Ic … Ib … Ia

Ic • Ib > Ia

Ic > Ib • Ia

linear molecule

prolate symmetric rotor or top

oblate symmetric rotor or top

spherical rotor

asymmetric rotorprolate near-symmetric rotor

oblate near-symmetric rotor

Molecules & Principal Inertial Axes

CH N

C O

H

H

C

I

H HH

C

H

H HH

F

S

F

F F

FF

N

NH

H H

H

H

H

H

H

H

H

O

CC

C H

H

H

H

a

c

b

a

bc

ca

b a

cb

a

b

c

b

a

b

aa

b

Ic ' Ib > Ia ' 0 Ic ' Ib > Ia

Ic > Ib ' Ia Ic ' Ib ' Ia Ic ' Ib ' Ia

Ic … Ib … Ia Ic • Ib > Ia Ic > Ib • Ia

Moments of Inertia, 2

In the example on the right, thecentre of mass lies along the C3axis, and perpendicular distancesare measured from this axis

An asymmetric rotor will havethree different moments ofinertia, and all three rotationaxes coincide at the centre ofthe molecule

Spherical rotors, which have three equal moments ofinertia, include CH4, SiF4 and SF6, and belong to a cubic oricosahedral point group.

Symmetric rotors have two equal moments of inertia,include NH3, CH3Cl and CH3CN, and have at least a 3-foldaxis of symmetry

Linear rotors have at least one moment of inertia equal tozero, include all diatomic and linear polyatomic molecules.

Asymmetric rotors are molecules without a threefold (orhigher) axis - other elements of symmetry may exist, butthese generally have very complicated sets of energy levels

Moments of Inertia for Different MoleculesPictured below are different types of molecules/rotors withtheir corresponding moments of inertia.

Diatomic & Linear Polyatomic MoleculesRecall that we developed the Schroedinger equation for therigid rotor in Lecture 6, and found that the rotationalenergy levels of a rigid rotor (or in this case, a linearmolecule) can be written as:

EJ '£ 2

2IJ(J % 1)

where I = µr2 and µ is the reduced mass, equal to

µ 'm1m2

m1 % m2

and J is the rotational quantum number, J = 0, 1, 2, ...

In rotational spectroscopy the frequency (µ-wave or mm-wave) or wavenumber (far-IR) is measured in practice. The energy levels above can be converted to term valuesF(J) which are expressed as frequency (divide above by h)or wavenumbers (divide above by hc):

F(J) 'Er

h'

h8B2I

J(J % 1) ' BJ(J % 1)

F(J) 'Er

hc'

h8B2cI

J(J % 1) ' BJ(J % 1)

where hcB = £2/2I, or B = £/4BcI = h/8B2cI, where B isknown as the rotational constant. The interchangeable useof the symbols F(J) and B with frequency and wavenumberis unforunate, but is a very common symbolism.

Pure Rotational Spectrum of COIn the picture below, we see the rotational term values,relative populations and transition wavenumbers for CO

Selection Rules & ConventionsTransition intensity is proportional to the square of thetransition moment which is given by

R r ' mR)

r µR))

r dJ

The selection rules for a pure rotational spectrum are:

Rule 1:Transitions are allowed in diatomic molecules such as CO,NO, HF, even 1H2H (µ = 5.9 × 10-4 D), but not H2, N2, Cl2. Transitions are allowed in polyatomics such as H-C/N, O=C=S, 1H-C/C-2H (µ = 0.012 D), not H-C/C-H, S=C=S.

Rule 2:)J refers to JN - JO, which are the rotational quantumnumbers of the upper and lower states, respectively.)J = -1 really has no meaning, as )J = +1 refers to eitherabsorption or emission because )J = JN - JO. Transitionfrequencies or wavenumbers are given by

L (or L) ' F(J % 1) & F(J) ' 2B(J % 1)¯

where unfortunately J is used instead of JO by convention. This means transitions are spaced by an interval of 2B.

1. Molecule must have a permanent dipole moment(i.e., µ … 0)

2. )J = ±13. )mJ = 0, ±1 (important if molecule is in an

electric or a magnetic field - covered later)

Rotational Transitions and SpectraA J = 1-0 transition (writtenconventionally as JN-JO) occurs at 2B. Whether the transition falls in the µ-wave, mm-wave or far-IR regiondepends upon the values of B and J.

20B

12B

6B

2B0

EJ

2B 4B 6B 8B 10B 12B etc.

2BE

Most linear polyatomic molecules havesmaller B values and more transitionstend to occur in the µ-wave and mm-wave regions.

The far-IR spectrum from 15 to 40 cm-1 with JO = 3 to 9 isshown below

Frequencies & wavenumbers ofrotational transitions of CO observin the mm-wave region

Uses of rotational spectroscopyVery powerful tool for determining:1. accurate molecular mass2. relative isotopic abundances3. extremely accurate bond length determinations4. identity of molecules (astronomy)

20B

12B

6B

2B0

EJ

12C16OB = 1.92118 cm-1

2B 4B 6B 8B 10B 12B etc.

2BE

20B

12B

6B

2B0

EJ

13C16OB = 1.83669 cm-1

Superimposed rotational spectra of 12C16O and 13C16O:

Spectral IntensitiesTwo important factors determine the intensities of thetransitions: (1) the numerical square of the transitionmoment equation (varies little with J) and (2) the relativepopulation of the energy levels involved in the transition.Once again, the Boltzmann distribution can be used todetermine the relative populations of the rotational energystates, NJ, and the ground rotational state, N0:

NJ

N0

' (2J % 1)e &EJ /kT

where gJ = 2J + 1, the degeneracy of the Jth level (againnote that in the absence of electric or magnetic fields, thelevels with mJ = 0, ±1 are all degenerate).The 2J + 1 term of course increases as J increases, but theexponential term decreases as J increases - eventually theexponential term dominates at high J values and NJ/N0approaches zero. As a result, the population is at amaximum value at J = Jmax:

d(NJ /N0)

dJ' 0, Jmax '

kT2hcB

1/2&

12

For CO, Jmax = 7 (highest population), but the actualspectrum has a maximum intensity at JO = 8, because thereare some other factors we have omitted from this analysis.

for B in cm-1.

Spectral Intensities, 2The number of molecules in the energy levelscorresponding to the quantum number J relative to thenumber of molecules in the ground state is shown as afunction of J, for three temperatures of 100, 300 and 700 K.

As long as the ratio NJ/N0 increases, the intensity of thelines originating from quantum states with the quantumnumber J will increase.

The varying intensities can be used to monitor the excitedrotational states of molecules of molecules say in flames,other “gaseous objects”, interstellar spectroscopy, etc.which have temperature gradients.

Centrifugal DistortionThere is actually a small decrease in the transition spacingswith increasing J, so the approximation of treating adiatomic molecule as a rigid rotor is not perfectly valid. Avibrational component must be included which when Jincreases, the nuclei tend to be thrown outwards bycentrifugal forces (i.e., spring stretches, r increases, andtherefore B decreases).This was originally accounted for by changing the termvalues of the equations to

F(J) ' B[1 & uJ(J % 1)]J(J % 1)but is now more commonly written as

F(J) ' BJ(J % 1) & DJ 2(J % 1)2

where D is the centrifugal distortion constant, and isalways positive for diatomic molecules. Transitionfrequencies are modified toL (or L) ' F(J % 1) & F(J) ' 2B(J % 1) & 4D(J % 1)3¯

D depends upon the stiffness of the bond, and can berelated to the vibrational frequency, T = (k/µ)1/2, from theharmonic approximation:

D '4B 3

T2

Interestingly, this implies that rotational spectroscopy canbe used to estimate vibrational frequencies!

Centrifugal Distortion, 2Where does the equation below come from?

F(J) ' BJ(J % 1) & DJ 2(J % 1)2

m1 m2

r

Recall that in the rigid rotor approximation the internucleardistance r does not change.

But in reality it will increase for higher values of J due tothe centrifugal force:

m1 m2

rJO

m1 m2

rJN

Lower J Higher JSo in the “rigid rotor” hamiltonian, the kinetic energy termis affected by the moment of inertia, and there is a now apotential energy term of the form V = ½k(r - re)2

We will not attempt the formal proof of this here, but if Te= £(k/µ)1/2 and B = £2/2I, then D comes from:

D '£ 4

2I 3(k/µ)'

£ 4

2I 3(T2e/£ 2)

'£ 6

2I 3(T2e)

'4B 3

T2e

Centrifugal Distortion, 3Thus the centrifugal forces in the molecule result innarrowing of the spacings between peaks corresponding tohigher values of J:

20B

12B

6B

2B0

EJ

2B 4B 6B 8B 10B 12B2B

E

Eetc.

F(J) ' BJ(J % 1)& DJ 2(J % 1)2

F(J) ' BJ(J % 1)

rigid rotor spectrum

non-rigid rotor spectrum

Excited Vibrational States: VibrotorsEvery vibrational level, including the ZP level, have a stackof rotational levels associated with them

The ratio of the population NL of the Lthvibrational level to N0 (ZP level) is

NL

N0

' e &£TL/kT

where L is the vibrational quantum no.Since this ratio is only 0.10 for L = 1and v = 470 cm-1, rotational transitionsin excited state are generally very weakunless the molecule is quite heavy or thetemperature is quite high.

The rotational constants B and D are slightly vibrationallydependent so that the term values and frequencies are:

FL (J) ' BLJ(J % 1) & DLJ 2(J % 1)2

v (or v) ' 2BL(J % 1) & 4DL(J % 1)3

and the vibrational depedence of B is (to a good approx.):BL ' Be & "(v%1/2)

where Be is the equilibrium state rotational constant and " isthe vibration-rotation interaction constant. To obtain Be,(and therefore re), BL must be obtained in two vib. states. Ifthe L = 1 state is insufficiently populated, B0 is onlyobtained. The vibrational dependence of DL is generallyvery small (forget it for now) - but we will look atvibrational-rotational spectroscopy later in the course.

012345678

012345678

J

J

vO = 0

vN = 1

Symmetric RotorsIn linear molecules, the rotational angular momentumvector P is directed along the axis of rotation, and isquantized in terms of J. In a prolate symmetric rotor, Pcan be oriented in any direction in space (depending uponthe molecule), and the molecule rotates about P.

PCI

H

HH

P

aPa

The component of P along the a axis, Pa, is quantized,taking on values only of K£, where K is a secondrotational quantum number, giving the projection ofangular momentum along the principal axis: K = 0, 1, ..., JRotational terms, neglecting centrifugal distortions and thevibrational dependence of rotational constants are given by

F(J,K) ' BJ(J % 1) % (A & B)K 2

and these constants are related to Ia and Ib by (L units)

A 'h

8B2Ia

; B 'h

8B2Ib

;

In an oblate symmetric rotor (e.g., NH3):

F(J,K) ' BJ(J % 1) % (C & B)K 2

C 'h

8B2Ic

The energy level spacings for these symmetric rotors aresomewhat more complicated for than a linear molecule

Symmetric Rotors, 2The selection rules for transitions for these molecules are:

)J = ±1 ; )K = 0and the expression for transition frequencies is:

L (or L) ' F(J%1, K) & F(J, K) ' 2B(J%1)

Rotational energy levels for (a) a prolate and (b) an oblatesymmetric rotor are shown below:

Transitions again show an equal spacing of 2B, and therequirement that a molecule must have a permanent dipolemoment still applied for the symmetric rotors.

Symmetric Rotors, 3Including the effects of centrifugal distortion for a prolatesymmetric rotor, the term value is:

F(J,K) ' BJ(J % 1) % (A & B)K 2

& DJJ2(J % 1) & DJKJ(J % 1)K 2

& DKK 4

where there are now three possible centrifugal distortionconstants, DJ, DJK and DK. There is an analogous equationfor an oblate symmetric rotor. The transition frequenciesor wavenumbers for both cases are given by:

L (or L) ' 2(Bv & DJKK 2)(J%1) & 4DJ(J % 1)3

The -2DJKK2(J+1) term separates the J + 1 components ofeach (J + 1) 7 J transition with different values of K.For example, the eight components of the J = 8-7 transitionof isothiocynate (H3Si-N=C=S) are shown below, whichhas a linear SiNCS chain and is a prolate symmetric rotor.

Here, K = 0 to 7, and peaks are separated by centrifugaldistortion.

Asymmetric RotorsThese molecules form the largest class of molecules -however, there are no closed formulae to generally predictterm values and spectroscopic transition frequencies due totheir asymmetry.Term diagonal values can be determined accurately by fullmatrix diagonalization for each value of J. Still, theselection rules )J = ±1, )mJ = 0, ±1 still apply, and themolecules must have a permanent electric dipole moment.

Pure rotational spectra of near-symmetric prolate oroblate rotors can be roughly interpreted with

F(J,K) • BJ(J % 1) % (A & B)K 2

F(J,K) • BJ(J % 1) % (C & B)K 2

where and for prolate and oblate rotors, respectively. Since themolecules are only approximately symmetric, K is notstrictly a good quantum number (i.e., not always integervalues).

B ' (1/2)(B % C) B ' (1/2)(A % B)

Dipole moments of asymmetric rotors (components alongvarious inertial axes) can be determined using the Starkeffect).

Spherical Rotor MoleculesSpherical rotor molecules, which have no permanent dipolemoment, should in theory have no IR, mm-wave or µ-waverotational spectra.In CH4 it is possible that rotation about any of the C3 axesin the molecule will result in centrifugal distortion of thethree H atoms not along the rotation axis, therebytemporarily converting the molecule to a symmetric rotorwith a very small dipole moment.The very small transient dipole moment permits thedetection of a very weak rotational spectrum. The far-IRspectrum of silane (SiH4) is shown below, from which thedipole moment is estimated as 8.3 × 10-6 D.

The term values for a spherical rotor are

L (or L) ' F(J % 1) & F(J) ' 2B(J % 1)¯

F(J) ' BJ(J % 1)and the transition frequencies are:

which is the same for linear rigid rotors (spacings: 2B). Tdmolecules give these weak spectra, but Oh molecules donot, since rotation about a C4 axis does not produce adipole moment (e.g., no rotational spectrum for SF6).

The Stark EffectRecall that space quantization of rotational angularmomentum (linear molecule) can be expressed as

(PJ)z ' mJ£

where mJ = J, J - 1, ..., -J. Under normal conditions the (2J+ 1) components of each J level remain degenerate. In thepresence of an electric field õ, the degeneracy becomespartially removed: each is split into J + 1 componentsaccording to *mJ* = 0, 1, 2, ..., J.

This splitting in an electric field isknown as the Stark effect. For a linearmolecule, the energy levels EJ aremodified to EJ + Eõ, where

Eõ 'µ2õ2[J(J % 1) & 3m 2

j ]

2hBJ(J % 1)(2J & 1)(2J % 3)

The above expression has mJ2, so this

term is independent of the sign of mJ,and it involves the molecular dipolemoment.

Increasingly complicated equations areavailable for more complicated rotors.

This makes µ-wave and mm-wave spectroscopy verypowerful techniques for determining the magnitude (notdirection) of molecular dipole moments. Studies arenormally on volatile species (sometimes at very high temp!)

Applications in AstronomyRadiotelescopes are used to scan the universe for radiationin the rf EM region, and are used to detect a variety ofmolecules.

parabolicreflectingdish

radiofrequencydetector

Examples: atomic H, 8 = 21 cm (varying quantities anddensities throughout universe), and OH, 8 = 18 cmMolecules are detected in regions of space where nebulaeare found (Milky Way and other galaxies), and in luminousclouds composed of interstellar dust and gasStarlight passing through dust clouds is reddened due topreferential scattering, % to 8-4, of the blue light of the dustparticles (estimated to be 0.2 µm in diameter on average)

New stars are formed by gravitational collapse in the regionof nebulae, which contain the material from which the newstars are formed - this is why they are of interest (e.g.,Sagitarius B2, near the centre of our galaxy, is muchstudied). NH3 was detected in 1963 in Hat Creek, CA,using a radiotelescope designed to detect in the mm-waveregion. Emission lines in the 8 = 1.25 cm region weredetected indicating the v2 = 0 and v2 = 1 levels of theinversion vibration of NH3!

Rotational Spectroscopy: Instellar MoleculesFor the molecules below, the transitions detected usingradiotelescopes are rotational in nature (except OH andNH3). C2H, HCO+ and N2H+ were detected in space beforethey we ever detected in the laboratory!!

Diatomics: OH, CO, CN, CS, SiO, SO, SiS, NO, NS, CH,CH+, SiC, NH, CP, HCl, CO+, SO+

Triatomics:H2O, HCN, HNC, OCS, H2S, N2H+, SO2, HNO,C2H, HCO, HCO+, HCS+, H2D+ (cyclic)

4 atoms: NH3, H2CO, HNCO, H2CS, HNCS, N/C-C/C,H3O+, C3H (linear), C3H (cyclic)

5 atoms: N/C-C/C-H, HCOOH, CH2=NH, NH2CN,C3H2 (linear), C3H2 (cyclic), H-C/C-C/C

6 atoms: CH3OH, CH3CN, NH2CHO, CH3SH, CH3NC,HC2CHO, HC3NH+, C5H

7 atoms: CH3-C/C-H, CH3CHO, CH3NH2, N/C-C/C-C/C- H, CH2=CHCN, C6H

8 atoms: HCOOCH3, CH3-C/C-C/N9 atoms: CH3OCH3, CH3CH2OH, N/C-C/C-C/C-

C/C-H, CH3CH2CN, CH3C4H11 atoms: N/C-C/C-C/C-C/C-C/C-H13 atoms: N/C-C/C-C/C-C/C-C/C-C/C-H,

Frequencies of such molecules detected in the laboratoryare completely unambiguous and accurate - but interstellarspectra must be corrected for the Doppler effectC2, N2, O2 and H-C/C-H cannot be detected with thesemethods - but the presence of the remarkable cyanopoly-acetylenes was a complete surprise!!

Researcher Profile: Lucy M. ZiurysLucy M. Ziurys, Professor of Chemistry & AstronomyDirector, Arizona Radio Observatory

Chemical Physics/Spectroscopy/Astrochemistryhttp://www.chem.arizona.edu/ziurys/ziur-group.html

At present, more than 120 different chemical species have beendetected in interstellar space, primarily in giant gas cloudsscattered throughout our Galaxy. One of the primary objectives ofour research is to study these molecules and their chemistry viaan interdisciplinary approach that involves high resolutionmolecular spectroscopy in the laboratory, radio astronomicalobservations, and chemical modeling. Of particular interest aresmall molecules composed of a metal atom and a simple ligand.These species are also of significance for organometallicchemistry. Small organic molecules related to sugars and nucleicacids are another area of investigation, as part of the NASAAstrobiology Institute.

The Arizona Radio Observatory’s 12m telescope at Kitt Peak, Arizona

They have recorded the spectra of a wide rangeof metal-bearing hydride, hydroxide, cyanide,acetylide, methyl, nitride and carbide radicals,including AlNC, CuCH3, MgCN, CaC, AlH,CoCN, and MnCl

Almost all of these species haveunpaired electrons, and thus theirspectra exhibit complex fine andhyperfine splittings. Analysis ofsuch data requires a detailedknowledge of quantummechanics and the coupling ofangular momentum.

Key Concepts1. Moments of inertia can be calculated accurately for

molecules of a variety of shapes. The “rotationalshapes” are classified as linear, symmetric, sphericaland asymmetric rotors. Symmetric rotors may beprolate or oblate.

2. For all rotors but the asymmetric rotors, exactexpressions for energy levels (term values) androtational transition frequencies can be calculated.

3. The rotational selection rules are: (1) Molecule musthave a permanent dipole moment (i.e., µ … 0), (2) )J= ±1 and (3) )mJ = 0, ±1 (important if molecule in anelectric or a magnetic field)

4. Centrifugal distortion of molecules results in unevenspacing in the pure rotational spectra. The centrifugaldistortion constant D is used to modify the rigid rotormodel to accurately predict energy levels andtransition frequencies.

5. Rotational spectra of symmetric rotors are furthercomplicated by the second rotational quantumnumber, K, which quantizes the molecular rotationalangular momentum.

6. Rotational transitions are split into multiple peaks inthe presence of an electric field: the Stark effect