Vibrational And Rotational Analysis Of Hydrogen Halides

Vibrational and RotationalAnalysis of Hydrogen HalidesGoalsQuantitative assessments of HBr molecularcharacteristics such as bond length, bondenergy, etcCHEM 164

HumaneyesNear-InfraredInfrared4.0

What do things appear in differentcolors?highlydangeroussun screanMicrowaveoven E hn hc/l For visible lights, the photon energy is within 1 to 4 eV, verycomparable to transitions involving two electronic states

Vibrational and Rotational Energyvibrationcoupledrotation

Infrared Spectra Spectral absorption peaks correspond to atransition of two vibrational-rotationalstates residing in the same electronic state The energy (wavelength) of the peakreflects the energy difference betweenthese two vib-rot states So, what are the specific energies for thevibrational-rotational levels?

Basics of Quantum MechanicsThe hydrogen bromide (HBr) molecule has two nucleiand 36 electrons, and the coordinates of the nuclei aredenoted with R, and the coordinates of the electronsare r. The total Hamiltonian H consists of the kineticenergy of all the particles plus the electrostaticinteractions between the two nuclei, the electrons andthe nuclei, and the electrons with other electrons. Thistotal Hamiltonian has the form of H (R,r) E (R,r), and the eigenvalues, E, correspond to theelectronic, vibrational, and rotational energy levels.

Born-Oppenheimer Approximation Max Born (December 11, 1882 –January 5, 1970) was a Germanphysicist and mathematician. He wonthe 1954 Nobel Prize in Physics. J. Robert Oppenheimer (April 22, 1904– February 18, 1967), Americanphysicist, known as the “Father of theAtomic Bomb”. As a scientist,Oppenheimer is remembered most forbeing the chief founder of the Americanschool of theoretical physics while atthe University of California, Berkeley.

Born-Oppenheimer ApproximationBorn and Oppenheimer found that the electron motions can beseparated from the motion of the nuclei by writing thewavefunction, (R,r) (R,r) (R), where (R,r) is anelectronic wavefunction, and (R) is a nuclear wavefunction.With this fixed nuclei approximation we can introduce apotential energy, V(R), for nuclear motion, and we get theCentral Force Hamiltonian:[-(ħ2/2m) 2 V(R)] (R, , ) E (R, , )where spherical coordinates are used for the center-of-masscoordinates for the nuclei, and the reduced mass,m mHmBr/(mH mBr)The potential energy, V(R), is determined only by the atomiccharges and the number of electrons, so it is the same for allisotopic variants for the hydrogen or bromine atoms.

Central Force problems can be further simplifiedby separating the angular parts ( , ) of thewavefunction from the radial part (r). Hence wetake (R, , ) R(R)S( , ), and S( , ) YJM( , ),where the YJM( , ) are the Spherical Harmonicsfunctions which are the same angular functionsas are found for the hydrogen atom. In theSpherical Harmonics J assumes integer values 0,1, 2 and M is also an integer with –J M J.There are (2J 1) possible values for Mfor each value of J.

Now the radial functions satisfy thedifferential equation:Note that this equation does not depend on Mso the vibration-rotation energies do notdepend on the orientation of the angularmomentum. Now if we define (R) R(R)/R,we find

Rigid-Rotor Approximation We expect that the rotational energywill depend on the angular momentumJ, so we identify this form of energywith the second term of the previousm mm equation: E J (2Jm R1)hm m We note that the rotational energy isJ 3zero for J 0.2Hrot2HBrBrJ 2J 1J 0

Rotational EnergyWhen there is no vibrational motion we expectthe molecule to have the internuclear separation(bond length) R Re, and the rotational energyin cm-1 or wavenumbers becomeshF(J) BeJ(J 1) with Be 8 2 m R 2cewhere Be is the Rotational Constant and c isthe speed of light and h is the Planck’sconstant. This expression for the rotationalenergy corresponds to the Rigid Rotorapproximation.

Vibrational Energy We investigate the form of thevibrational energy by subtracting theJ ( J 1)hterm 2m R (R) from the vibrationrotational Hamiltonian given above.The result is22h d V ( R) Evib 22m dR22

Harmonic Oscillator Now if we set x R – Re and k V”(Re) , andwe neglect the terms with powers of x greaterthan 2, we have after a little manipulation theSWEh 2 d 2 ( x) k 2 x ( x) E ( x)22 m dx2Hook’s law

Harmonic Oscillator The harmonic oscillator energy levels are given1 kn as Evib (v ½)hn where 2 m and v is aninteger (0, 1, 2, ). Note that the harmonic oscillator has nonzeroenergy at v 0. The energy for the harmonicoscillator for v 0 is the zero point energy,and this zero point energy is different fordifferent isotopic variants of a molecule.

Harmonic Oscillator For vibrational energies in cm-1 or wavenumberswe have Gv we(v ½) where the VibrationalConstant is1kwe 2 c mv 3v 2J 3J 2J 1J 0v 1v 0

Energy levels in the Rigid RotorHarmonic Oscillator Approximation If the rigid rotor and the harmonic oscillatorapproximations are combined we have theenergy in wavenumbers (cm-1) for vibration rotation as G(v) F(J) we(v ½) BeJ(J 1) We expect that this will be a goodapproximation for the energy levels for small vand J, and this approximation is called theRigid Rotor-Harmonic Oscillator or RRHOapproximation.

Centrifugal Distortion ConstantHowever a much more accurate representation of thevibration-rotation energy levels is obtained byconsidering the fact that the potential energy has termswith larger powers of x, and that the effectiveinternuclear separation depends on the vibrational state(v) of the molecule. Also the effects of centrifugaldistortion can be included. The resulting expressionsfor F(J) and G(v) are the following:F(J) BvJ(J 1) – DvJ2(J 1)2 where Dv is the centrifugal distortionconstant; and Bv Be – a(v ½).rotationalconstantVibrationalquantum number

Anharmonic CorrectionsG(v) we(v ½) – wexe(v ½)2 weye (v ½)3 In this course we will neglect the centrifugaldistortion and the higher terms ( 2) in thevibrational energy expression.For purely harmonic oscillator, theoreticallythe molecule will never dissociate, i.e., thebond never breaks.

Vibration-Rotation EnergyHence for a state with quantum numbers v andJ, we have the vibration-rotation energy21 1 G (v) F ( J ) we v we xe v Bv J ( J 1)2 2 21 1 1 we v we xe v Be J ( J 1) a J ( J 1) v 2 2 2 We see that the energy levels are determinedby four parameters: we, weye, Be, and a, andyou will determine them from the analysis ofspectra.

Vibration-Rotation Spectroscopy of HBrIn the final project you will analyze the transitionenergies from a state with quantum numbers v”, J” toanother state with quantum numbers v’, J’. Thetransition energy in wavenumbers is given by [F’(J’) G(v’)] – [F”(J”) G(v”)]where the rotational and vibrational energies are givenby the above expressions.Note that we use F’ and F” but not G’ and G”. This isbecause F’(J’) depends on v’ and F”(J”) depends on v”.

Selection Rulesfor Rotational Transitions It appears that there might be many transitionsfrom a state with (v”,J”). However there areactually few transitions because of SelectionRules. Selection Rules for Rotational Transitions:– There must be an oscillating dipole moment.– For a diatomic molecule like HBr, DJ 1.

Selection Rulesfor Vibrational Transitions There must be an oscillating dipole moment– For diatomic molecules, this means the two nuclei mustbe differentThe value of the integral i(x)m(x) f(x)dx must benonzero. In this equation we have the vibrationalwave functions for the initial (i) and final (f) statesand the dipole moment (m) function for themolecule. In other words the transition dipolematrix element must be nonzero.

Selection Rules We find two sets of rotational transitionsdepending whether DJ 1 or DJ –1. Theformer type of transitions are called R branchtransitions, and the latter are P branchtransitions. Hence from J” 0 only the Rbranch transition R(0) is found, while for J” 1R(1) and P(1) are found. The R and P branchtransitions are labeled with the value for J”.

Selection Rulesfor Vibrational Transitions For pure harmonic oscillators, we get the selectionrule that Dv 1. If the molecule has mechanical anharmonicity (V(x)has higher order terms) or electric anharmonicity (mhas quadratic and higher order terms), then themolecule will exhibit Dv 2, 3, 4, transitions. The Dv 1 transitions are usually the most intenseand are called fundamental transitions while theother transitions are weaker and are called overtonetransitions.

Vibrational-Rotational TransitionsDJ 1DJ 0DJ -1

Isotope EffectsThe isotopic substituted molecules H79Br andH81Br have different absorption peaks becausethe spectroscopic constants depend on thereduced mass, m, which is different for the twomolecules. In particular we m–½ since k is thesame for all isotopic variants. Similarly Be m–1since Re is the same for all isotopic variants. Youare expected to find the reduced mass scaling forthe spectroscopic constants a and wexe.

Data Analysis The expected transition energies are given bythe equation: [F’(J’) G(v’)] – [F”(J”) G(v”)]x And for v” 0xx43yx12 1 1 n we [v '] we xe [v '2 v '] Be J '( J ' 1) J "( J " 1) a v' J '( J ' 1) J "( J " 1) LINESTy a1x1 a2x2 a3x3 a4x42 2

Linest Hence we can obtain the four desiredspectroscopy constants by multiple regressionwith the independent variables given in thesquare brackets. LINEST will give all thespectroscopic constants provided data from allyour group members is used. LINEST willalso give the standard errors for eachspectroscopic constant

Bond Dissociation Energy If we define DG(v) G(V 1) – G(V), then D00v max DG (v)v 0

Bond Dissociation Energy If only the harmonic and the first anharmonicterms are used in G(v), we haveG(v) we(v ½) – wexe(v ½)2andDG(v) we – 2wexe(v 1) Now As vbh DG (0)v maxDG (0)Area where222we 1 , D 0 w e (w e 2w e xe )02w e xe4w xD00 Area maxe eoDe 2wwxeDe0 e e4we xe4 SinceDoo G(0), we find This is the Birge-Sponer expression for the DissociationEnergy.

Rotational TemperatureNow the number of v" 0 HBr molecules in agiven J" state is given by the Boltzmannexpression: F "( J ") N J " 2 J " 1 exp kT where k has units of cm-1K-1. The intensity of atransition from the J" state is proportional toS(J” J’)NJ’, where S is a Honl-London Factor.

Thus for R and P branch transitions in HBr, the intensity B J "( J " 1) isI ( J " J " 1) ( J " 1) expR branch v"kT B J "( J " 1) I ( J " J " 1) J "exp v" kT P branchThese two equations can be combined to give an expression that is valid forboth R and P branch transitions: B J "( J " 1) I ( J " J " 1) ( J " J ' 1) exp v" kT We can extract the rotational temperature, T, from a plot ofyxI ( J " J ')lnvs J "( J " 1)J " J ' 1This will be a straight line with slope –Bv”/kT, and Bv” Be – a(v” ½).Since the rotational constant is known, you can extract T. Error propagationmust be used to determine the error in T.

Summary Bond length (Re) Bond Energy (De)Be h8 2 m Re2 c2wwe xe0eDe 4we xe4 Experimental TemperatureIsotopeeffect?

FTIR Spectrometer