VIBRATIONAL SPECTRA (INFRARED SPECTRA) - Nalanda Open University

Vibrational spectra (Infrared Spectra) ( 4 2 2 m1 2 k ) A1 KA2 0.(2.12 (a))Similarly from equation (2.11 (b)), we getm2 x 2 k x x 21x2 A2 C os 2 t x 2 x2 dx2 A2 2 S in 2 t dtd 2 x2dt2 A2 –4 2 2 C os 2 t m2 A2 4 2 2 C os 2 t k A2 A1 C os 2 t m2 A2 4 2 2 kA1 kA2 0 kA1 4 2 2 m2 k A2 0Thus equation (2.12(a)) (2.12(b)) are simultaneous equation of first degreeSolution : (i) A2 A2 0 0It is a trivial solution and so is meaningless(ii)For non-trivial solution we construct secular determinant of A1 & A2 4 m 212 K K K 4 2 m2 2 K 0From this secular equation, we can obtain expression for frequency 4 m k 4 m k k 02 212 22 16 4 m4 m2 4 2 2 m1k 4 2 2 m2 k k 2 k 2 031

Vibrational spectra (Infrared Spectra) 4 2 2 4 2 2 m1m2 m1k m2 k 0 4 2 2 m1m2 m1 m2 k 0 4 2 2 m1m2 m1 m2 k 2 2 1 k m1 m2 m1m24 21k. ;4 2 where m1m2m1 m21 k2 is reduced mass.(2.13)It gives the expression for frequency of vibration of two particles and itsrelation with force constant (k) and reduced mass ( ) of the system. The frequency of thevibration is written as osc in Hz and osc in cm–1 as :thenv 2 0 and hence v 0 osc 1 k2 . (2.13 a) osc 1 k2 c .(2.13 b)Also, if4 2 2 0then 0 and hence 0It corresponds to the motion in which both particles are displaced by the same amountin same direction i.e. x1 x2.Thus 0 corresponds to translational motionOn substituting the osc in equation (2.12 a) & (2.12 b), we getA1 m2 x1 A2 m1 x2.(2.14)32

Vibrational spectra (Infrared Spectra)If m2 is lighter than m 1, the vibrational amplitude of m 2 will be correspondinglygreater than that of m1.On substituting the value of force constant k 4 2 2osc in equation (2.10),we getU 1 2 1kx 4 2osc x 2222 U 2 2 oscx2.(2.15)It shows that in simple harmonic motion (S.H.M) the potential energy (U) isproportional to the square the displacement of the centre of gravity of themolecule. The potential energy is parabolic.The concept of reduced mass ( ) reduces the vibration of two atoms in amolecule to the vibration of a single mass point, whose amplitude equal theamplitude change (A2-A1) of the vibrating atoms in the molecule. An increase in energywill make the oscillations more vigorous, i.e. the degree of compression or extension will begreater but the vibrational frequency osc will be the same. Such a model gives a vibrationalfrequency independent of the amount of bond distortion. However classical mechanicsallows amplitudes and therefore the energy of vibration to attain any value contrary to thequantum nature of energy.(Fig. 2.4): Some of the vibrational energy levels & allowed transition of H.O.diatomic molecule (NO)2.5. SCHRÖDINGER EQUATION APPLIED TO HARMONIC OSCILLATIONThere are a few simple systems where the potential energy is not constant, yet theSchordinger equation can be exactly solved. For example, vibration of a diatomic moleculeand motion of an atom in a crystal lattice.33

Vibrational spectra (Infrared Spectra)Let us consider a particle of mass ‘m’ attached to a weightless spring and restricted inthe same, way so that it can move only in the x-direction. The force acting on this particle isgiven by Hooke’s law as :fdfd xmfrSince direction of driving force (fd) and restoring force (fr) are opposite to each other, fd fr fr x fr kx where k is proportionality constant known as force constant.If fr 1 dyne, x 1 cm k 1 dyne / cm.It is defined as equal to force per unit displacement. This type of force is called harmonic.Whenever the motion of a particle can be described by the simple law known as Hooke’slaw, then the system is said to be harmonic oscillator.The potential energy, U is given by dU fdx dU fdx kxdx dU kxdxOn integrating, we getUx dU k xdx0 U 01 2kx2The kinetic energy, T is given by11 dx T mv 2 m 22 dt 2211 px2 2 m v 2m2 mwhere px mv is linear momentum. For two particle system having mass m1 and m2,the kinetic energy, is given by34

Vibrational spectra (Infrared Spectra)2 dr dr 11T m1 1 m2 2 22 dt dt 211m r 2 m r 22 11 2 22 T The reduced mass, is given by111 m1 m2p 211 T x 2 2 x2 x22 2 Total energy associated with the system is given asH T U H px22 1 2kx2This is expression of energy through classical mechanicsClassically the equation of motion is expressed asm.d2 xdt 2 kx (Newton’s law)which has the general solution as k x x0 sin t m where x0 & are arbitrary constants, x0 being the amplitude of the oscillation. Accordingto classical mechanics, the particle oscillates from x x0 to x x0sinusoidally with the time at a frequency 12 kmThe energy then changes back and forth from the kinetic energy form to the potentialenergy form, being exclusively kinetic when x 0 and exclusively potential when x x0. Thetotal energy is constant given bykx02. It can have any positive value, there being no limit235

Vibrational spectra (Infrared Spectra)on the value of x0 . Now let us cansider the properties of such a system according to the lawof quantum mechanics.Let us first set up Hamiltonian operator for Harmonic oscillator (H.O.)We know thatp2 1H U T x kx 22m 2 2 2 1 2px2ˆ 1 2ˆ kxˆˆ H kx2 x 2 22m 2 ( pˆ x i x2 pˆ 2x22 i 2 2 2 i ) x x 2 x2 The schodinger equation for harmonic oscillator then may be written as : weknow that S.E. in operator form is as H E 2 2 1 2 kx E 2 x 2 2 2 x 2 2 x2 2 1 2 2 kx E 2 2 2 2 1 E kx 2 02 2 .(2.16)This is Schrödinger equation for harmonic oscillator. The problem is now to find thewell behaved functions which satisfy the equation (2.16) and the allowed energy levels.The solution of this equation vanishes at infinite and is single valued andfinite, and the energy is discontinuous but changes by integral value of thevibrational quantum number v given by1 h k Eosc v 2 2 .(2.17)where v 0, 1, 2, 3 . known as vibrational quantum number (can take onlypositive integer values, including zero).The quantum energy levels with the simple harmonic oscillator as a model areequidistant and have been represented in figure (2.4)36

Vibrational spectra (Infrared Spectra)It should be particularly noted that energy at v 0 is not zero but1h and is called2 osczero point energy. It has no counterpart in the classical approach. This is also in accordancewith the Hiesenberg’s uncertainty principle, i.e. at 0 K (–273ºC) when even translational,rotational motion have been frozen, uncertainty of the position of the molecules still existdue to zero point energy and is equal to1h per vibrational mode.2 oscFor example in case of NO, osc 1904 cm 1zero point energy 11h 6.625 10 –34 1904 3 101022 18923 10–24 JIt is conventional to express vibrational modes and energy levels in cm–1 as follows:Eosc1 G (v ) osc cm 1hc2 where G(v) is called term value, v osc the molecular vibration in wave numbers, and vis the vibrational quantum number.Zero point energy in cm–1 vosc2zero point energy of NO in cm–1 Again, E v 1 h 2 2π1904 952 cm 12k Thus the quantum mechanics requires that only certain discrete energies areassumed by the vibrator. The termh2 k appears in both classical as well as quantum mechanical treatment37

Vibrational spectra (Infrared Spectra)Also, from classical mechanics treatment we know that m 1 k1 k m1 m2 2 m 2 cm1m2.(2.18)From equation (2.17) and (2.18), we have1 E v h m2 .(2.19)where m is the vibrational frequency of mechanical model.If we now assume that transitions in vibrational energy levels can be brought about chesthedifference in energy levels E between the vibrational quantum states andprovided that the vibration causes a change in dipole. This difference is identical betweenany pair of adjacent levels, because v in equation (2.17) & (2.19) canassume only whole numbers. That is , E h m h k2 .(2.20)At room temperature majority of molecules use in the ground state ( 0). ThusE0 1h 2 mIn order to move to the first excited state with energy E1 3h requires radiation of2 m 3 1energy h m h m h m2 2 The frequency of radiation that will bring about this change is identical to the classicalvibrational frequency of the bond m . Thus,Eradiation h E h m m h k2 .(2.21)1 kHz2 .(2.22)38

Vibrational spectra (Infrared Spectra)This expression can also be written in terms of wave number of radiation.1k1 k m1 m2 1 cm.(2.23)2 c 2 cm1m2Thus, where, wave number of absorption peak in cm–1k force constant (in dynes/cm)c velocity of light ( 3 1010 in cm/s)m1 & m2 are masses of two atoms–1Energy (cm )The allowed vibrational energy levels and transition between them for a diatomicmolecule undergoing simple harmonic motion may be shown as :192 osc.172 osc.152 osc.132 osc.112 osc.92 osc.72 osc.52 osc.32 osc.12 osc. 0r eq.Internuclear distancecm–1 osc.Fig. (2.5). The allowed vibrational energy levels and transition between them for adiatomic molecule undergoing simple harmonic motion.39

Vibrational spectra (Infrared Spectra)2.6 ZERO POINT ENERGY (E0)According to the old quantum theory, the energy levels of a harmonicoscillation were given byEn n h If this were true, the lowest energy level would be that with n 0, and would thereforehave zero energy. This would be state of complete rest and represent the minimum inpotential energy curve. The uncertainty principle does not allow such a state of completedefined position and completely defined momentum (in this case zero). As a result wavemechanical treatment show that the energy levels of the oscillator are given by 1 Ev v h 02 .(2.24)where v is the vibrational quantum number which may take on the values,v 0, 1, 2, 3 . The vibratory motion of the nuclei of a diatomic molecule can be representedas vibration of a simple harmonic oscillator. In such an oscillator the vibrational energy Ev isrelated to the fundamental vibrational frequency 0 by the above wave mechanical1relationship. The above equation shows that such an oscillator retains the energy E0 h 02in the lowest vibrational level v 0. This residual energy, called zero point energy of theoscillator cannot be removed from the molecule even cooling it to 0 K (–273ºC). The enrgy1E0 h 0 must be added to the planck’s expression for the mean energy of an oscillator.2The implication is that the diatomic molecule (and indeed any molecule) can neverhave zero vibration energy; the atoms can never be completely at rest11h 0 Joules or 0 cm 1 , the zero point energy;22depends only on the classical vibration frequency and hence on the strength of the chemicalbond (k) and atomic masses ( ).relative to each other. The quantityThe prediction of zero point energy is the basic difference between the wave mechanicaland classical approaches to molecular vibrations. Classical mechanics could find no objectionto a molecule possessing no vibrational energy but wave mechanics insists that it mustalways vibrate to some extent, the latter conclusion has been amply borne out by experiment.2.7 SELECTION RULEFurther use of the Schrödinger equation leads to the simple selection rule for theharmonic oscillator undergoing vibrational changes : v 140

Vibrational spectra (Infrared Spectra)To this, we must of course add the condition that vibrational energy changes will onlygive rise to an observable spectrum if the vibration can interact with radiation, i.e. if thevibration involves a change in the dipole moment of the molecule. Thus the vibrationalspectra will be observable only in hetronuclear diatomic molecules since homonuclearmolecules have no dipole moment.Applying the selection rule we have immediately : 1 1 Ev v 1 v 1 osc v osc2 2 osc cm 1.(2.25)for absorption, whatever the initial value of v.Such a simple result is also obvious from the figure given above; since thevibrational levels are equally spaced, transitions between any two neighboring states willgive rise to the same energy change. Further since the difference between energy levels weexpressed in cm–1 gives directly the wave number of the spectral line absorbed or emitted. spectroscopic oscThis, again, is obvious if one consider the mechanism of a absorbing or emission inclassical terms. In absorption for instance, the vibrating molecule will absorb energy onlyfrom radiation with which it can coherently interact and this must be radiation of its ownoscillation frequency.For example, NO (nitric oxide) molecule.The expected vibrational energy levels for NO molecules are equally spaced. Transitionbetween any two neighbouring states will give rise to the same energy change and thusonly one line. The NO as harmonic oscillator should absorb at 1904 cm–1c10 110 1or osc c osc 3 10 1904 s 5712 10 s The energy of this quanta of radiation isE h osc 6.626 10 34 5712 1010 J 37847 10 24 JIt is the energy which must correspond to the energy difference, even for av 0 v 1 transition.2.8 BOLTZMANN DISTRIBUTIONWith the help of Boltzman distribution we can calculate the number of molecules in v 1 state relative to the ground vibrational state at room temperature , i.e. 27ºC or 300 K41

Vibrational spectra (Infrared Spectra)N v 1N v 0 exp EKT 37847 10 24 exp 1.38 10 23 300 e 9.1428 0.000107It shows that less than 1% of molecule are in the v 1 state and still smallnumber in higher levels. It follows that in experiment at room temperature only transitionfrom the ground vibrational state v 0 are of major importance. Also the force constant canbe calculated as : k 4 2 2oscwith 12.495 10 27 kg for NO molecule. osc 5712 1010 Hz or s 12 k 4 3.14 12.495 10 27 5712 1010 2k 1609.4 Nm 1 SI units k 16.094 10 5 dyne cm 1 (CGS units)Force constant (k) in a quantitative way is a measure of how strong are force of attractionbetween the two atoms of a molecule.The horizonal lines in figure in the previous section represent some of the vibrationalstates. The points of intersection of the potential energy curve and horizontal lines give thevalue req q, where Evib U, the kinetic energy is zero so thatAt req q, Evib U 1 1 v h osc kq 22 2 where q is amplitude of the vibrationfor v 0, the maximum value of amplitude of vibration (qmax) can be11 2hvosc kqmax22 hv qmax osc k 12for NO molecule,1qmax 37847 24 2 0.00485 nm 1609.4 42

Vibrational spectra (Infrared Spectra)Stiffness of a chemical bond, which is a measured by the force constants (k) is reflectedin the amplitude of vibration. The greater rigidity and lower amplitude of the HF than thoseof HI molecules.2.9 FORCE CONSTANT (K) AND BOND STRENGTHSIn a harmonic oscillator, the restoring force per unit displacement is called force constantk by (fr –kx) and is given byK 4 2 2 4 2 m1m2m1 m2.(2.26)where is vibrational frequency (in Hz) and m1 and m2 are masses of the oscillatingatoms.The force constant for diatomic molecule can be obtained by using above equationprovided the vibrational frequency ( ) is known. Force constant for polyatomic moleculecannot be determined directly. A method has been suggested by assuming that each valencybond has a certain definite value for the force constant, which is characteristic of the bondand independents of the molecule in which it occurs. Force constants of some bonds aregiven in the table below.Table 2.1. Force constants (k) of some bonds in dynes/cmBondForce cons tan t( k ) 105Bondk 10 5Bondk 10 5C O4.9C—N4.8C —C4.6C OC O12.318.6C NC N12.117.5C CC C9.515.8It is an interesting fact that, the force constant (k) increases approximately in proportionto the multiplicity of the bond, and so the former can be used to give an indication of thelatter. For example, the force constant for the carbon oxygen bond in carbon dioxide hasbeen found to be 15.7 105 dynes/cm. This value lies between C O and C O as shown inthe table. This result is in good agreement with the resonance structure of carbon dioxide,which is– O —C O O C O O C OSimilarly, the force constant given for C O is that for carbon monoxide (CO). Thusthe value 18.6 10 5 dyne/cm provides support to the triple bond structure of carbonmonoxide.The bond length varies inversely with the bond order and bond energies vary directlywith bond order. Force constant (k) of the bond is directly related to bond order. Hence, asbond order increases, bond energy as well as force constant increase. As bond energyincreases, bond length decreases and force constant increases. As bond length increases43

Vibrational spectra (Infrared Spectra)bond energy decreases and force constant also decreases. As force constant increases, bondlength decreases and bond energy increases.2.10. THE ANHARMONIC OSCILLATORThough a simple harmonic oscillator gives a good picture, it does not explain thefollowing points :(i) The potential energy and therefore the restoring force increases infinitely withincreasing distance from the equilibrium position. Therefore it places no limit on how far abond can be stretched, while in an actual molecule, when atoms are at a great distance fromone another, the attractive force is zero, the bond will break and correspondingly the potentialenergy has a constant value.(ii)

infrared radiation with molecular vibration gives infrared spectrum. If the average position and orientation of a molecule remains constant but the distance between the atoms in a molecule change, molecular vibrations are said to take place. A vibrational spectrum is observed experimentally as Infrared as well as Raman Spectra.